thermo-structural analysis manual

WADD-TR-60-517 OFFICIAL F1 Z COPY

Volume I •

THERMO-STRUCTURAL ANALYSIS MANUAL

TECHNICAL REPORT NO. WADD-TR-60-5 17, Vol. Ii N,

August 1962

Flight Dynamics LaboratoryDirectorate of Materials and Processes

Aeronautical Systems DivisionAir Force Systems Command

Wright-Patterson Air Force Base, Ohio

Project No. 1367, Task No. 136710

(Prepared under Contract AF 33(616)-6066 byI • the Republic Aviation Corporation, Farmingdale, L. I., New York;H. Switzky, M. J. Forray, and M. Newman, authors.)

NOTICES

When Government drawings, specifications, or other data are used for anypurpose other than in connection with a definitely related Government procure-ment operation, the United States Government thereby incurs no responsibihty

nor any obligation whatsoever; and the fact that the Government may haveformulated, furnished, or in any way supplied the said drawings, specifications,or other data, is not to be regarded by implication or otherwise as in anymanner licensing the holder or any other person or corporation, or conveyingany rights or permission to manufacture, use, or sell any patented inventionthat may in any way be related thereto.

Qualified requesters may obtain copies of this report from the ArmedServices Technical Information Agency, (ASTIA), Arlington Hall Station,Arlington 12, Virginia&

This report has been released to the Office of Technical Services, U.S.Department of Commerce, Washington 25, D.C., in stock quantities for saleto the general public.

Copies of this report should not be returned to the Aeronautical SystemsDivision unless return is required by security considerations, contractualobligations, or notice on a specific document.

B

0 RIALE T m•• CAL LIBRARY

WADD-TR-60-517 W Fe-, CLIA

Volume I

THERMO- STRUCTURAL ANALYSIS MANUAL

'-.51A1

TECHNICAL REPORT NO. WADD-TR-60-517, Vol. I

"August 1962

Flight Dynamics Laboratory

Directorate of Materials and Processes

Aeronautical Systems Division

Air Force Systems Command

Wright-Patterson Air Force Base, Ohio

k Project No. 1367, Task No. 136710

(Prepared under Contract AF 33(616)-6066 by

the Republic Aviation Corporation, Farmingdale, L. I., New York;H. Switzky, M. J. Forray, and M. Newman, authors.)

- OR

"J RC WK ti

NOTICES

When Government drawings, specifications, or other data are used for any 3I purpose other than in connection with a definitely related Government procure-

ment operation, the United States Government thereby incurs no responsibilitynor any obligation whatsoever; and the fact that the Government may have

formulated, furnished, or in any way supplied the said drawings, specifications,or other data, is not to be regarded by implication or otherwise as in anymanner licensing the holder or any other person or corporation, or conveyingany rights or permission to manufacture, use, or sell any patented invention

L that may in any way be related thereto.

Qualified requesters may obtain copies of this report from the ArmedServices Technical Information Agency, (ASTIA), Arlington Hall Station,Arlington 12, Virginia&

This report has been released to the Office of Technical Services, U.S.

Department of Commerce, Washington 25, D.C., in stock quantities for sale

to the general public.

Copies of this report should not be returned to the Aeronautical SystemsDivision unless return is required by security considerations, contractualobligations, or notice on a specific document.

* B

S- .. . ..

This manual was prepared by the Structural Rese•rch and AosvelopuontGroupe the Structures oection Roesearch and Development Ldvision of theRepublio Aviation Corporation. The vork vas accomplished Under Contract

J 33(616)-.6066 and the 750A A oplied hesoarch rogram. entitled MOechanicsof Plight.o Project Noe 1367. Structural Design Criteria in Task Noe13002. Structural Analysis Methods. This vork was initiated under the Vdirection of the btructural Analysis Unit. Structures Branch, AircraftLaboratory* Lirectorate of lb oratories, Wright Air Wovelokemnt Genter*.1w1. is Winnegad acted initially as project engineer and was succeededby Mr. Go Richard. The Manual vaa completed under the direction of the

Structural Analysis Unit. Configuration liosearch Section, Structuresbranch, Flight Pynmics laboratory, wdth Mr. Go S. Haddux as Projectagineer.

The work was coordinated and supervised by D. Re. So I•evy Head ofStructural Research and Development roupo., His valuable suggestions andcritici•is are gratefully acknowledged as are names of the followdigpersonnel of the Applied oeseorch and DeveloAmnt vision of RepublicAviation Corporation. Nt. A. Alberi, Acting Mtmbger of Technical

tgiveering; e. C. Roaeakraz•• o cti¢ g Chief Structures Engineer; andMr. C.o toasner, Principal Structures E ngioeer1

S(Now designated Aeronautical Systems Division)

WM260517 4-"M.4 500. 9-Z4-62

I77-

ABSTRACT

This anual includes lcompilation of methods of solution for thermal stress pro-blems of t• types freqgntly encountered by aircraft designers. Soqe/f thf' methodsrepresent original work done at Republic Aviation Corporation and others woee obtainedfrom t~ general literature. Wl~re fea•1e, graphs and formulas awe presented from

which tl user my obtain answers directly. *Tlese are presented in non-dimensionalform to extend tbk,4*& applicability. P( otho cass, tables a.%4 furnished which describemethods of solutin, Liberal use is made of illustrative problems and examples.

[Within t~e limitations of linear elastic theory, ge following problems a~e treatedin deM:

(1) ticauy determinate beams•(2) edundant beams and fr si(3) *vete or bolted joints,(4)•lts,•.

(5) ially symmetric shells/

For more complex linear problems, a general method of attack is presented whichreduces the thermal stress problem to an equivalent mechanical loading problem. Thisapproach permits utilization of the great variety of analytical methods which have beendeveloped for stress analysis of structures imder purely mechanical loads. A brief re-view of some of these methods is included with pertinent remarks on their applicability tothermal stress problems.

In many cases of practical interest, thermal effects introduce non-linearity by causinglarge deflections, by affecting the mechanical properties of the material, or by introducingcreep. Solutions for these problems are quite limited. However, they are discussed in somedetail and a generalized stress-strain-time-temperature relationship is postulated which isapplied to buckling of columns and plates.

PUBLICATION REVIEW

This report has been reviewed and is approved.

FOR THE COMMANDER:

W1l1.1"i C. lýColonel , USAFChief, Flight bynnmics Laboratar

WADD TR 60-517 iii

TABLE OF CONTENTS*

Page

Abstracti i,List of Illustrations x

Notes on Using the Mannal xviii

Introduction xix

SParagraph Page

SECTION 1 - THEORETICAL CONSIDERATIONS 1.0

Table of Contents 1.1

1 Theoretical Considerations 1.2

1.1 Strain 1.2

1. 1.1 Compatibility 1.6

1.2 Traction and Stress 1.7

1.2.1 Stress Equations of Equilibrium in Cartesian Coordinates 1.9

1.3 Boundary Conditions 1.11

1.4 Stress-Strain Relationship 1.12

1.5 Uniqueness 1.15

1.6 Special Orthogonal Coordinate Systems 1.15

1.7 Energy Principles 1.18

1. 7. 1 Principle of Virtual Work 1.18

1.7.2 Variational Principles 1.19

1.8 References 1.27 A

* This Table of Contents lists the prefatory material, sections, and sub-sections of theManual. Paragraphs with numbers of more than three tiers are omitted. However,each section has its own complete table of contents.

WADD TR 60-517 iv

-- - ~ - - - - -

TABLE OF CONTENTS (Cont'd)

Paragraph Page

SECTION 2 - STRUCTURAL TECHNIQUES 2.0


2 Structural Techniques 2.3

2.1 Linear Structures 2.5

2.1.1 Virtual Work Techniques 2.5

2.1.2 Strain Energy in Terms of Internal Loads and Deformations 2.9

2.1.3 Application of Virtual Force Technique - Flexibility Coefficient 2.11

2.1.4 Applications of Virtual Displacement Techniques 2.16

2.1.5 Properties of Influence Coefficients Matrix 2.19

2.1.6 Equivalent Thermal Load 2.21

2.1.7 Plane Stress and Plane Strain Problem 2.23

2.1.8 St. Venant' s Principle 2.25

2.2 Non-Linear Structures 2.26

2.2.1 Incremental Linear Solutions 2.27

2.2.2 Inverse Solution 2.27

2.2.3 Approximate Solutions 2.28

2.3 References 2.29

SECTION 3 - MATERIAL CONCEPTS 3.0


3 Material Concepts 3.2

3.1 Deformation Mechanism 3.4

3.2 Stress- Temperature- Time-Strain Relationship 3.5

3.2.1 Uniaxial Stress-Strain Relationship 3.6

3.2.2 Experimental Determination of the Material Constants 3.9

3.2.3 Complex Loading 3.14

3.3 References 3.19

WADD TR 60-517 v


Paragr PphSECTION 4 - THERMO-ELASTIC ANALYSIS OF BEAMS 4.0


4 Thermo-Elastic Analysis of Beams 4.4

4.1 Thermo-Elastic Analysis of Statically Daterminate 4.6(Unrestrained) Beams

4.1.1 General Solution 4.7

4.1.2 Evaluation of Integrals 4.11

4.2 Statically Indeterminate (Externally Restrained) Beams 4.58

4.2.1 Beam Deflections (with illustrative problem) 4.58

4.2.2 Fixed End Reactions 4.63

4.2.3 General Solution of Statically Indeterminate Beams 4.86

4.2.4 Mechanics of Solution by the Flexibility Method 4.88

4.2.5 Mechanics of Solution by the Stiffness (Slope-Deflection) 4.92Method

4.2.6 Selection of Method of Analysis 4.95

4.2.7 Curved Beams 4.105

4.3 References 4.117

SECTION 5 - THERMO-ELASTIC ANALYSIS OF JOINTS 5.0


5 Thermo-Elastic Analysis of Joints 5.2 45.1 The One-Dimensional Problem 5.4

5.1.1 The Joint Equations 5.5

5.1.2 The Special Case of Constant Bay Properties 5.8

5.1.3 Constant Bay Properties - Rigid Sheets 5.24

5.1.4 Constant Bay Properties - Rigid Attachments 5.28

5.1.5 The Influence of "Slop" on the Load Distribution 5.30

WADD TR 60-517 vi I

TABLE OF CONTENTS (Cont' d)

Paragraph Page

5.2 The Two-Dimensional (Bowing) Problem 5.36


5.3 Determination of the Attachment-Hole Flexibility Factor 5.39

5.4 References 5.40

SECTION 6 - THERMO-ELASTIC ANALYSIS OF PLATES 6.0


6 Thermo-Elastic Analysis of Plates 6.3

6.1 Theory of Deformation of Plates 6.4

6.1.1 Definition of a Plate 6.4

6.1.2 Assumptions for Linear Plate Theory 6.4

6.1.3 Three Kinds of Plate Problems 6.6

6.1.4 Fundamental Equations of Thermo-Elastic Plate Theory 6.7

6.2 Bending of Plates 6.13

6.2.1 Bending of Rectangular Plates with Linear Gradient Through 6.13the Thickness

6.2.2 Bending of Circular Plates with Linear Temperature Gradient 6.17Through the Thickness

6.2.3 Circular Plates - Temperature Difference as a Function of the 6.18Radial Coordinate

6.2.4 Approximate Solution of Free Plate with Arbitrary Temperature 6.26Variation Through the Thickness Only

6.3 Slab Problems - Plates 6.29

6.3.1 Thermal Stresses in Rings- Asymmetrical Temperature 6.31Distribution

6.3.2 Thermal Stresses in Solid Circular Plates Due to Asymmetrical 6.34Temperature Distribution

6.3.3 Circular Disk with Concentric Hole Subject to a Power Law 6.42Temperature Distribution

WADD TR 60-517 vii


Pararh Page

6.3.4 Circular Plate - Central Hot Spot 6.48

6.4 References 6.57

SECTION 7 - THERMO-ELASTIC ANALYSIS OF SHELLS 7.0


7 Thermo-Elastic Analysis of Shells 7.2

7.1 Truncated Conical Shells 7.3

7.1.1 Radial Loading Perpendicular to the Cone Axis 7.4

7.1.2 Meridional Loading - Membrane Analysis 7.20

7.1.3 Meridional Temperature Variation 7.26

7.2 Approximate Solution for Non-Conical Axisymmetric Shells 7.28

7.3 Cylindrical Shells 7.30

7.3.1 Cylindrical Shells with Axisymmetric Loading and Meridional 7.30Temperature Variation

7.3.2 Cylindrical Shells with Temperature Gradient Through the 7.37Thickness

7.4 References 7.51

SECTION 8 - STRUCTURAL ASSEMBLIES 8.0


8 Structural Assemblies 8.2

8.1 Shells of Revolution with Bulkheads 8.3

8.2 Determination of Interaction Forces and Moments 8.6

8.2.1 Solution for the Interior Bulkhead 8.7

8.2.2 Solution for the External Bulkhead 8.8

WADD TR 60-517 viii


Paragraph Page

SECTION 9 - STABILITY OF STRUCTURES 9.0


9 Stability of Structures 9.2

9.1 Stability Criteria 9.4

9.1.1 Stability by Eigenvectors 9.7

9.1.2 Stability by Virtual Work 9.14

9.1.3 Properties of Eigenvectors(Deflection Modes) and Eigenvalues 9.15(Critical Loads)

9.1.4 Approximate Methods of Determining Stability 9.18

9.1.5 Shear Energies 9.26

9.1.6 Plasticity and Eccentricity 9.29

9.1.7 Temperature 9.31

9.2 Non-Dimensional Buckling Curves 9.32

9.2.1 Non-Dimensional Stability 9.32

9.3 Curved Plates and Shells 9.48

9.4 References 9.50

WADD TR 60-517 ix

LIST OF ILLUSTRATIONS

Figure Page

SECTION 1 - THEORETICAL CONSIDERATIONS

1. 1-1 Displacement of a Structure Under Mechanical and/or Thermal 1. 4Loads

1.1-2 Types of Strain 1.6

1.2-1 Tractions at the Point Q Acting on the Plane with Normal n" 1.8

1.2-2 Stress Components 1.9

1.2.1-1 Stress Components on Rectangular Parallelepiped 1. 10

1.3-1 Specification of Surface Tractions 1.12

1.7.1-1 Elastic Body with Applied Forces 1.181.7.2.2-1 Variation of Extremizi.g Function 1.221.7.2.3-1 Elastically Propped Cantilever Beam 1.24

1.7.2.4-1 Beam with 'hree Supports 1.27

SECTION 2 - STRUCTURAL TECHNIQUES

2.1.1.1-1 Body Under Lcad 2.6

2.1.1.2-1 Vertically Loaded Beam in Equilibrium 2.8

2.1.2-1 Beam Under Load 2,9

2.1.3.2-1 Superposition of Flexibilities 2. 12

2.1.3.2-2 Equivalent Structures in Series 2. 13

2.1.4.3-1 Stiffness Coefficients for a Double Beam 2.19

SECTION 3 - MATERIAL CONCEPTS

3. 1-1 Deforn stion and Energy Potential Under Shearing Action 3.5

3.2.1-1 Non-Dimensional Stress-Strain Curves 3.7

3.2. 1-2 Non-Dimensional Secant Modulus Curves 3.7

3.2. 1-3 Non-Dimensional Tangent Modulus Curves 3.7

3.2. 1-4 Non-Dimensional Tangent-Secant Modulus RaTWo (.:,rves 3 7

3.2.2.1-1 Short Time Stress-Strain Curve 3.9 12

3.2.2.2-1 Determination of Material Cznstants from Short Time Stress- 3.11Strain Data

3. 2.2.2-2 Determination of Material Constants from Constant Creep 3.12Rate Data

3.2.2.3-1 Creep Curve 3.13

3.2.2.4-1 Variation of Material Parameters with Temperature 3.14

WADD T. 60-517 x

9-I

UIST OF ILLUSTRATIONS (Cont' d)

Figure Page

3.2. 3.1-1 Principal Directions and Octahedral Planes 3.16

3.2.3.2-1 Uniaxlal Stress-Strain Curve 3.18

3.2.3.2-2 Hydrostatic Stress-Strain Curve 3.18

3. 2. 3. 2-3 Octahedral Stress-Strain Curve 3.18

3.2.3.2-4 BIaxial Stress-Strain Curve 3.18

SECTION 4 - THERMO-ELASTIC ANALYSIS OF BEAMS

4.1.1-1 Elastic Area Cross Section 4.8

4. 1.2.1-1 Finite Sum Method - Element Breakdown of Cross Section 4.12

4.1.2.1-2 Finite Sum Method 4.15

4. 1.2. 1-3 Stress Distribution for Problem of Figure 4. 1. 2. 1-2 4.18

4.1.2.2-1 Sandwich Beam Cross Section 4.20

4. 1.2.3-1 Typical Configurations of Elastic Cross Section Areas 4.22

4.1.2.3-2 Determination of Average ciT Profile for Bending About UU 4.22Principal Axis

4. 1.2.3.1-1 aT Profile 4.23

4. 1.2.3.1-2 aT Values at Five Equally Spaced Points 4.24

4.1.2.3.1-3 Polynomial aT Piflles of Illustrative Problem 4.27

4. 1. 2.3.2-1 Continuous Cross Section with Monotonic Elastic Width 4. 28Variation

4.1.2.3.2-2 Monotonic Geometric Shapes Defined by Parameters p and K 4.294.1.2.3.3-1 Elongation, yL for Continuous Geometry of the Form 4.32

L4.32S= Eobo (1 +ft s), with K = 1 (Straight Sides)00

4.1.2.3.3-2 Curvature, 6 L., for Continuous Geometry of the Form 4.33

SEb = E b^ (1- + 8 sK), with K = 1 (Straight Sides)

4.1.2.3.3-3 Elastic Cross Section 4.34

4. 1.2.3.3-4 Thermal Stresses for Polynomial aT Distributions 4.39

4. 1.2.3.3-5 Decomposition of Total aT Profile into Symmetrical ond 4.41Antisymmetrical Components

4. 1.2.3.3-6 Symmetric Elastic Cross Section with Unsymmetric aT 4.43Profile i

4.1.2.3.3-7 Thermal Stresses for the Beam of Figure 4. 1. 2. 3. 3-6 4.45

WADD TR 60-517 xi

LIST OF ILUSTRATIONS (Cont' d)

Figure Page

4.1.2.3.4-1 Gei~eral Multi-Rectangular Section 4.46

4. 1.2.3.4-2 Unrestrained Beam Cross Section and Temperature 4.48Distribution

4. 1.2.3.4-3 Multi-Rectangular Section with a Bending Axis of Symmetry 4.51

4.1.2.3.4-4 Approximate Representation of a Zee by a Two Rectangle 4.53Symmetrical Geometry

4. 1. 2. 3.4-5(a) Elongation s Versus Width Parameter (e) 4.54

4. 1. 2. 3. 4-5(b) Curvature (6 L)a Versus Width Parameter (e) 4.55

4. 1.2.3.4-6 Bi-Metallic Box Beam with Symmetrical Geometry 4.56

4.2.1-1 Curvature wt(x) and Elongation Zt(x) Due to 4.59

Combined Mechanical and Thermal Loading as a Function ofDistance Along the Span

4.2.1-2 Indeterminate Beam Deflections 4.61

4.2.2. 1- 1 Sign Convention for Combining Fixed End Reactions of Beams to 4.64Determine Resultant Fixed End Reactions at a Joint

,. 2.2.2-1 Mechanical and Thermal Loads Acting on Cantilever Beam 4.66

4.2.2.3-1 Fixed End Reaction for Constant EI; Transverse End Load (Po) 4.700

4.2.2.3-2 Fixed End Reaction for Constant EI; End Moment (Mo) 4.71

4.2.2.3-3 Fixed End Reactions for Linear 1/EI; Transverse End Load 4.72(P ), (c=1, Y=1)

4.2.2.3-4 Fixed End Reactions for Linear 1/EI; End Moment (Mo), 4.73(c=1, V=1)

4.2.2.3-5 Fixed End Reaction for Parabolic 1/EI; Transvere End Load 4.74(Po0) (ce 2, Y= 14.2.2.3-6 Fixed End Reaction for Parabolic 1/EI; End Moment (Mo), 4.75

(c=2, v=1)

4.2. 4-1 Flexibility Method 4.90

4.2.5-1 Stiffness Method 4.93

4.2.6.1-1 Beam with Degrees of Freedom 4.97

4.2.6.1-2 Temperature and Load on Beam 4.98

4.2.6.1-3 Solution 4.101

4.2.7-1 Sign Convention 4.106

4.2. 7-2 Determination of Virtual Forces m, f, p 4. 107

4.2. 7.2-1 Elaatic Center Method 4.111

WADD TR 60-517 xii

lo

LIST OF ILLUSTRATIONS (Cont'd)

Figure Page

4.2.7.3-1 Resolution of Force into Symmetrical and Antisymmetrical 4. 113Components

4.2.7.3-2 Frame Problem 4.114

SECTION 5 - THERMO-ELASTIC ANALYSIS OF JOINTS

5-1 Mechanical Joint 5.2

5.1-1 Splice Strap for a Rigid Beam Flange 5.4

5.1. 1-1 One-Dimensional Joint Equilibrium 5.5

5.1.1-2 One-Dimensional Compatibility 5.6

5.1.1-3 Deformation of the Joint Due to Axial Stretching of the Sheets 5.7

5.1.1-4 Deformation of the Joint Due to Local Distortions of the Holes 5.8and Attachments

5. 1. 2-1(a) Joint Coefficients for Constant Bay Properties (AIN vs. Z) 5.10

5.1.2-1(b) Joint Coefficients for Constant Bay Properties (BIN vs. Z) 5.11

5. 1. 2-1(c) Joint Coefficients for Constant Bay Properties (A2 N vs. Z) 5.125. 1. 2-1(d) Joint Coefficients for Constant Bay Properties (B2 N vs. Z) 5.13

5. 1. 2-1(e) Joint Coefficients for Constant Bay Properties (A3 N vs. Z) 5.14

5.1.2-1(f) Joint Coefficients for Constant Bay Properties (B3 N vs. Z) 5.15

5. 1. 2-1(g) Joint Coefficients for Constant Bay Properties (A4 vs. Z) 5.16

5. 1. 2-1(h) Joint Coefficients for Constant Bay Properties (B4 N vs. Z) 5.17

5. 1. 2-1(i) Joint Coefficients for Constant EBy Properties (A5N vs. Z) 5.18

5. 1. 2-1(j) Joint Coefficients for Constant Bay Properties (B5N vs. Z) 5.19

5.1. 2-2 Joint with Constant Bay Properties 5.21

5. 1.2-3 Alternate Top and Bottom Sheet Designation 5. 225.1. 3-1 Rigid Sheet Solutiron 5.26

5.1.23-2 Attachment Loads Due to Thermal Loading when the Joint 5.27

Has Constant Bay Properties

5.1.5-1 Attachment in an Oversize Hole 5.30

5.1.5-2 Attacbment Displacements Due to Slop - Positive Loads 5.31

5.1.5-3 Scarfed Splice 5.34

5. 2. 1-1(a) Joint in Unbowed Configuration 5.37

5. 2. 1-1(b) Bowing of Joint Due to Combined Effects of Non-Uniform 6.37Temperature Distribution and Mechanical Loading

5.3-1 Attachment-Hole vs. Deflection Curve 5.39

WADD TR 60-517 xiii


Figure Page

SECTION 6 - THERMO-ELASTIC ANALYSIS OF PLATES

6.1.2-1 Cross Section of Plate Before and After Deformation 6.5

6.1.4-1 Positive Directions of Forces and Moments 6.8

6.1.4-2 Clamped Circular Plate with Linear Gradient Through 6. 11Thickness

6.2. 1-1 Plate GeomeLry Showing a Positive Linear Thermal Gradient 6.13Through the Thickness

6.2.1.3-1 Deflection (w) vs. Position (x) for y Parameters 6.15

6.2. 1.3-2 Bending Moment (M ) vs. Position (x) for y Parameters 6.16y

6.2.1.3-3 Bending Moment (M ) vs. Position (x) for y Parameters 6.16x

S6.2. 2-1 Plate Geometry 6.17

6.2. 3.1-1 Non-Dimensional Deflection 6.20

6.2.3.1-2 Non-Dimensional Radial Moment 6.21

6.2.3.1-3 Non-Dimensional Tangential Moment 6.226.2.4-1 Plate with Arbitrary Planform and Constant Thickness 6.26

6.3.2-1 Radial Stress Distribution in a Disk when a Temperature 6.36

Distribution T = T (r/b)K cos ne (n = 0, 1,2, 3) is Maintained

6.3. 2-2 Hoop-Stress Distribution in Disk when Temperature Distribution 6.37KT=T 0 (r/b) cos ne (n = 0,1, 2,3) is Maintained

6.3. 2-3 Shear-Stress Distribution in Disk when Temperature Distribution 6.38KST = T (r/b) cos no (n = 1, 2,3) is Maintained

0(

"6.3.2-4 Radial and Hoop-Stress Distribution for the Axisymetric 6.39Temperature Case, T = T (r/b)K

o0

6.3.3-1 Temperature Profiles for n = 1/3, 1/2, 1, 2 and 3 6.45

6.3.3-2 Radial Variation of Radial Stress for a Disk with a Hole (a/b = 0, 6.450.4, 0.8;n = 1/3, land3

6.3.3-3 Variation of Maximum Radial Stress with n and a/b for a Disk 6.45with a Hole

6.3.3-4 Variation of Tangential Stress at Inner Boundary with n and a/b 6.45for a Disk with a Hole

6.3. 3-5 Variation of Tangential Stress at Outer Boundary with n and 6.46a/b for Disk with a Hole

6.3.3-6 Variation of Radial and Tangential Stress at Center and Boundary 6.46with n for a Solid Disk

WADD TR 60-517 xiv

LIST OF ILLUSTRATIONS (Cont' d)

Figure Page

6.3.3-7 Variation of Radial and Tangential Stress at Inner Boundary 6.46of Disk with n and a/b for a Disk on a Shaft

6.3.3-8 Variation of Tangential Stress at Outer Boundary of Disk with 6.46n and a/b for a Disk on a Shaft

6.3.4-1 Plate and Hot Spot 6.49

6.3.4-2(a) Radial Displacement at Boundary of Hot Spot (Y = 0. 3) 6.51

6.3.4-2(b) Radial Displacement at Outer Boundary of Plate (Y = 0.3) 6.52

6.3.4-3 Stresses in Plate (Y = 0.3) 6.53

6.3. 4-4 Variation of Stress Difference with b/a (Y = 0.3) 6.54

6.3.4-5 Variation of Maximum Stress Difference with t for b/a 6.55Large. Maximum Occurs at r = a.

SECTION 7 - THERMO-ELASTIC OF SHELLS

7.1-1 Truncated Conical Shell Geometry 7.3

7. 1. 1-1(a) Conical Shell with Axisymmetric but Otherwise Arbitrary 7.4Loading

7. 1. 1-1(b) Radial and Meridional Load Components Ph and P x 7.4Respectively

7.1. 1.1-1 Conical Shell - Hooprings and Meridional Beam Strips 7.5

7. 1. 1. 1-2(a) Non-Dimensional Particular Solution of Eq. (2) for the 7.10Radial Deflection of a Conical Shell, K = 0, 1,2

7. 1. 1. 1-2(b) Non-Dimensional Particular Solution of Eq. (2) for the 7.11Radial Deflection of a Conical Shell, K = 3,4

7.1.1.2-1 Self-Equilibrating Loads and Moments • and • at a Conical 7.13Shell Edge

7.1.1.2-2 Edge Loading Functions 7.15

7.1.1.2-3 Conical Shell Under Uniform Radial Loading 7.16

7. 1. 1. 2-4(a) Edge Forces and Moments Corresponding to the Particular 7.19 ISolution

7. 1. 1. 2-4(b) Edge Loading Equal in Magnitude but Opposite in Direction to 7.19the Particular Solution Edge Loading

7.1.1.2-5 Me•idional Moment and Radial Load Measured from Base of 7.21the Shell Shown in Figure 7.1.1.2-4

7.1.2-1 Meridional Loading on a Conical Shell 7.22

7.1.2-2 Conical Shell Under Vertical Loading 7.23

7.1.3-1 Reduction of Thermal Problem to Equivalent Mechanical 7.27Loading Problem

7.2-1 Approximation of a Non-Conl. al Shell by a Series of Conical 7.29Sections

WADD TR 60-517 xv


Figure Page

7.3.1.2-I Self-Equilibrating Loads and MomentsW and . at at 7.23Cylinder Edge

7.3.1.2-2 Infinitely Long Cylinder with Discontinuous Axisymmetric 7.34Temperature Distribution

7.3.1.2-3 Horizontal Loads and Moments at Cut Edges 7.36

7.3.1.2-4 Non-Dimensional Stresses and Radial Deflections 7.38

7.3.2.1-1(a) Shell Geometry 7.39

7.3.2. 1-1(b) Variation of Stresses Through the Cylinder Thickness 7.39

7.3.2.1-2 Maximum Tangential and Longitudinal Stress in Cylinder per 7.41OF Temperature Difference Between the Outer and Inner Radii

7.3.2.1-3 Tangential Stress at Outside Radius per °F Temperature 7.42Difference Between the Outer and Inner Radii

7.3.2.1-4 Maximum Radial Stress per OF Temperature Difference 7.43Between the Outer and Inner Radii

7.3.2.1-5 Location of the Position of Maximum Radial Stress 7.447.3. 2. 2-1 Non-Dimensional Radial Stress as Function of •for the 7.49

2case ah = 5 aKt

7.3.2.2-2 Non-Dimensional Tangential Stress as Function of L for the 7.49case ah = 5 a

7.3.2.2-3 Non-Dimensional Surfaces Tangential Stress as Function of 7.50Kt2 with ah as a Parameter2a

7.3.2. 2-4 Non-Dimensional Surfaces Tangential Stress for a Hollow 7.50Cylinder with Outer Surfaces at Temperature T1 and InnerSurface Insulated

SECTION 8 - STRUCTURAL ASSEMBLIES8.1-1 Deflections, Loads and Moments at a Shell and Bulkhead 8.4

Junction

8.1.2 Edge Rotation of Solid Circular Bulkhead Due to Uniform 8.5Pressure Loading

8.2-1 Long Cylinder with Interior and End Bulkheads 8.7

8.2. 1-1 Interior Bulkhead 8.8

8.2.2-1 External Bulkhead 8.9

SECTION 9 - STABILITY OF STRUCTURES

9.1-1 Stability of Columns 9.6

9.1.1.1-1 Loads on Plate 9.8

9.1.1.2-1 Pin-Ended Column 9.11

WADD TR 60-517 xvi

- -r •" '••• •- • - . . . `.. .. .`.`..`>••• • ; .T .`• • •-. o - :. . . .. ..

LIST OF ILLUSTRATIONS (Cont' d)

Fgr Page

9.1.1.2-2 Eigenvectors and Eigenvalues of Column 9.13

9.1.4. 1.2-1 Virtual Force on Pin-Ended Column 9.19

9.1.4. 2-1 Pin-Ended Column with Variable Stiffness 9.23

9.1.4.2-2 Clamped Column 9.24

9.1.5-1 Shear and Torsional Load on a Pin-Ended Column 9.27

9.1.7-1 Deflection and Load in .. Restrained Column 9.32

9. 2. 1-1(a) Stability of Simple Free Flange, Shear Panel, or Shells 9.35

9.2. 1-1(b) Stability of Simple Free Flange, Shear Panel, or Shells 9.36

9. 2. 1-2(a) Stability of Column or Plate Column 9.37

9. 2. 1-2(b) Stability of Column or Plate Column 9.38

9.2. 1-3(a) Stability of Simple-Simple Plate 9.39

9. 2. 1-3(b) Stability of Simple-Simple Plate 9.40

9.2.1-4 Stability of Clamped-Free Plate 9.41

9.2. 1-5 Stability of Clamped-Clamped Plate 9.42

9.2. 1-6 Sheet Stringer Cross Section 9.45

9.2.1-7 Sandwich Panel 9.47

9; 3-1 Cylinder In Compression 9.50

WADD TR 60-517 xvii

NOTES ON USING THE MANUAL

This Manual consists of nine basic sections, divided into numbered sub-sectionsand paragraphs. For simplicity in cross-referencing material in the text, all portionsof the Manual designated with a two-tier number (i.e., 1. 1) are considered sub-sections,and all portions designated by numbers of three or more tiers (i. e., 1 1. 1 or 1. 1. 1. 1)are considered paragraphs.

Throughout the . lanual, the numbered paragraphs (or sub-sections) have beenused as the basis for numbering figures, tables, and equations, with new sequencesbeginning with each numbered paragraph. Figure and table numbers consist of an appro-priate paragraph number, followed by a sequence number for the particular figure ottable. For convenience the paragraph designations have been omitted from the equationnumbers. When an equation from another paragraph is cited in the text, the number ofthe paragraph in which that equation occurs is also cited. When a paragraph number isnot given in conjunction with the citation of an equation, it is to be assumed that theequation is included in the paragraph in which the citation occurs.

References are listed at the end of those sections which have more than onereference. In addition,each section contains its own complete table of contents and listof symbols.

W3

A.WADD Th 60-5 17 xviii

INTRODUCTION

Aerodynamic heating of bodies moving at supersonic and hypersonic speeds resultsin non-uniform temperature rises. High velocity airplanes and missiles are thus sub-jected to transient and steady state non-uniform temperature distributions which producethermal stresses. In its broadest sense, the thermal problem encompasses both thefields of mechanics and thermodynamics of non-rigid bodies. This Manual is not concernedwith the solution of the thermodynamic problem. The purpose of the Manual is to provideanalytical techniques for the determination of deformations and stresses in structures sub-jected to mechanical loads and prescribed temperature distributions.

In order to render problems amenable to engineering application, simplifying assump-tions and idealizations are made. These are enumerated in the appropriate sections of theManual. The subject matter is oriented toward the engineer rather than the mathematician

E in that detailed derivations of techniques and formulas are not emphasized. Instead, morespace is devoted to application and qualitative discussions. Mathematical derivations canbe found in the references listed at the end of each section.

The choice of subject material was primarily dictated by those problems which aretractable analytically and which occur most frequently in practice. However, in manyinstances, suggestions are made for the extension of techniques to more complex problems.Most of the problems treated are linear elastic, in which the stiffness may vary pointwiseover the structure but does not vary with load. Furthermore, in the more difficult problems(e. ., plates and shells), it is assumed that temperature variations do not affect the stiff-nesses significantly. Non-linear problems are, for the most part, discussed qualitatively.

Brief summaries of the nine sections of the Manual follow.

Sections 1 and 2 of the Manual briefly discuss the theoretical considerations and funda-mental techniques which underlie the structural applications. Section 1 presents the essen-tial ideas of strain and stress, the concepts of equilibrium and compatibility, and discussesthe important energy theorems. General techniques (e.g., virtual work, virtual displace-ments, flexibility and stiffness methods) are covered in Section 2. It is recommended thatstandard texts and the references listed at the ends of the sections be consulted for a morecomplete development of these subjects.

Elevated temperature environments generally cause the non-linear behavior ofengineering materials to become more pronounced, and the designer is forced to reassessthe interacting effects of stress, temperature, and time upon structural materials. Section 3discusses the deformation mechanism and postulates a stress-strain-temperature-time re-lationship which takes these non-linear effects into account. Methods of determining thenecessary material parameters from simple test data are indicated.

Section 4 presents the thermo-elastic analysis of beams. Since beam analysis is afrequently occuring structural problem, much space is devoted to the presentation of approxi-mate time-saving techniques. The deformations and stresses in unrestrained beam cross

WADD TR 60-517 xix

INTRODUCTION (Cont'd)

sections are discussed in detail. Temperature distributions are represented by polynomials,cross sectional geometries are expressed parametrically, and non-dimensional solutions aredeveloped. The common form of tabular solutions is also included. General techniques arethen presented for the solution of indeterminate beam systems. The concept of equivalentfixed end reactions is employed in reducing the thermal problem to a mechanical loadingproblem.

Section 5 deals with the thermo-elastic analysis of joints subjected to mechanicalloads and temperature. The loads in the attachments are presented in non-dimensionalgraphical form. The flexibility of the plate material and of the hole-pin combination are con-sidered in the solutions. The effects of rigid pins, rigid sheets and "slop" are evaluated.The problem is initially presented for a joint which does not bend and then is modified toindicate the change in the non-dimensional parameters with bending.

Section 6 is concerned with the determination of thermal stresses and deflections inplates due to temperatures which vary through both the plate plane and-the plate thickness.The major problem areas considered are

(1) The bending of ciruclar and rectangular plates.(2) The axisymmetrical and asymmetrical slab problem (plane stress and plane strain)

for circular plates and rings.

Sections 7 and 8 present the thermo-elastic analysis for the stresses and deformationsof axisymmetric shells subjected to axisymmetric loads and temperatures. Solutionsare given for the conical shells which can approximate the solutions ýfor other structuralshapes. Solutions for the compatibility forces which are generated at boundaries of conicalsegments, edges, or bulkheads are included to complete the analysis.

Section 9 deals with the instability of structures. Approximate methods are emphasizedin order to include the effects of temperature on stiffness and plastic behavior. Use is madeof non-dimensional buckling curves to reduce the amount of data required by instability pro-blem solutions.

%3

Manuscript released by the authors September 1960 for publication as a WADD TechnicalReport.

WADD TR 60-517 x

SECTION 1

THEORETICAL CONSIDERATIONS

MY

",6

•tWADD Ta 60-517 1.0 !

XcY 4Z1 r.. - 4%.l l.w. -'',,'' -

SECTION 1

-THEORETICAL CONSIDERATIONS

TABLE OF CONTENTS

Paragraph Title ?age

1 Theoretical Considerations 1.2

1.1 Strain 1.3

1.1.1 Compatibility 1.6

1.2 Traction and Stress 1.7

1.2.1 Stress Equations of Equilibrium in Cartesian Coordinates 1.9

1.3 Boundary Conditions 1.11

1.4 Stress-Strain Relationship 1.12

1.5 Uniqueness 1.15

1.6 Special Orthogonal Coordinate Systems 1.15

1.7 Energy Principles 1.18

1.7.1 Principle of Virtual Work 1.18

1.7.2 Variational Principles 1.19

1.7.2.1 Definitions of Terms 1.20

1.7.2.2 Variational Techniques 1.21

1.7.2.3 Minimum Potential Energy 1.23

1.7.2.4 Minimum Complementary Energy 1.25

1.8 References 1.27

WADD TR 60-517 .i

SECTION 1 - THEORETICAL CONSIDERATIONS

The general problem considered is the determination of the stresses and deformationsin a structure when subjected to a given load-temperature distribution. The problem can beexceedingly complex and it is essential that the engineer have some concept of the structure' sbehavior. Many difficult problems can be approximated by neglecting parameters which areof small significance. The significance of a parameter can only be estimated if the behaviorof the structure is known. The introduction of temperature into the probleta may complicatethe solution but it does not alter the structural techniques. It is, therefore, mandatory thatthe engineer become cognizant of the structural principles before he attempts to solve thethermal problen.

When a structure is subjected to mechanical =d thermal stimuli, it responds by de-formingand storing energy. The deformation is characterized by a "strain distribution" withan accompanying "stress and energy density distribution". The determination of these distri-butions requires an understanding of the concepts of strain and compatibility, stress and equil-ibrium, boundary conditious, uniqueness of solutions, and the stationary characteristics ofthe energy forms. These concepts are briefly discussed in this section.

The following symbols are used throughout this section:Length of beam

n Unit vector normal to surfacen, Cosine of angle between normal to the surface and i axisr Cylindrical or spherical coordinatet Traction vectoru Displacement vector with components u1 , u2 , u3x, y, z Distances along the coordinate axes

B BodyE Young' s modulusE Secant modulusFs Body force per unit of volume acting in i directionG Shear modulusK Spring constantL Potential energyL* Complementary potential energyP Point in a body; Concentrated loadQ Point in a bodyR Region of body; ReactionS SurfaceT TemperatureU Strain energyU* Complementary strain energyUV Strain energy per unit volume

UV Complementary strain energy per unit volume

V VolumeW Loss in potential energy of the surface tractions and body forcesWt Loss in potential energy where the tractions are prescribed

WADD TR 60-517 1.2

5N

Loss in potential energy where the displacements are precribedX, Y, Z Traction in x, y, z directions

a Coefficient of linear expansion6 Operator indicating a small changeA Deflection

lj 1=j: Extensional strain in the i directioniij: %earing strain. Half the angle change between two initially perpendicularlines in the i and j directions.

e Cylindrical or spherical coordinate9o" +0a +0o

xx yy zzp Poisson' s ratioa Stress

Stress components acting on plane perpendicular to i direction and in the j011j direction4'Spherical coordinatew Rotation

Subscripts

i, j Dummy subscriptsn Acting on plane perpendicular to n directionx, y, z Referring to x, y, z directionsr, 0, 4 Referring to r, G, 4 directions. Wu4

Symbol for differentiation; e.g., ui, j =-- i I

Superscripts

Displaced position

1.1 STRAIN

A body is said to be strained when the relative positions of points in the body arealtered. The changes in the relative positions of points are called deformations, and thestidy of deformations is the province of the analysis of strain.

Although all material bodies are to some extent deformable, it is useful to introducetae ideal case of a rigid body, i.e., one which does not deform. A rigid body is one for whichthe distance between every pair of points remains the same throughout its history.

ta Let the non-rigid body B, in the undeformed state, occuP3 some region R referred

to an orthogonal set of Cartesian axes O-x1 x2 x3 (Figure 1.1-1) fixed in space. Let

P (x1 , x2 , x3 ) represent a typical point P. In the strained state, the points of B will occupy

some region RI and the point P (x, x2 , x3 ) will displace to the point P' (xi, xj, x1). Thedisplacement u is a vector and is given by

x xu (u1, u2 , u3 ) = (x -x 1 ' x 2' 2 -x 3 ).

WADD TR 60-517 1.3

1. 1 (Cont' d)

x•3

x2

1

FIGURE 1.1-1 DISPLACEMENT OF A STRUCTURE UNDERMECHANICAL AND/OR THERMAL LOADS

Now consider two neighboring points 0 (0, 0, 0) and 0' (x 1 ,X2 ,X3 ) before deformation. Deter-

minination of their relative displacements due to some external stimulus is of interest. From theTaylor' s Expansion Theorem of the calculus,

u. (XlX 2 ,X3 ) =u (0,0,0) x1 + x( +000VV Ot e (0invovO) (0 )t 2 +O )

-- 1 ) ( ,0 ,0

+ terms involving the higher derivatives, (Ia)

or, more compactly,3 a u

uW a xi , X+.. (1b)J=1 0

aluLet the symbol u W be defined by u, = -j

Then the expression (la) becomes3

(x)=u, (0)+ E (Uj)O xj+. (0c)=11

WADD TR 60-517 1.4

24

1.1 (Cont'd)

Furthermore,

(UjJ) = (ui, + ujj) + . (Uil - Uj,)

is an algebraic identity.

Introduce the symbols1

CiJ = 2 (u ij + u,) d

and (2)

SiJ = - Uj,i)

so that the relative displacement of the neighboring points is3

o, W 0 - -i (0 C ÷+..J=1 i+ i 0

The quantities c are symmetrical in the indices, i.e., 6Ej C Ji and are known as the

components of strain (Reference 1-1).

pUlExample 1: C.1 is the longitudinal component of strain in the xI direc-

tion, (Figure 1. 1-2(a)) ui 1 = - is the longitudinal or extensional component of strain in

the x, direction.

Example 2: E12 = = 2 is half the change in the angle between two

line elements which were originally at right angles to each other and is referred to as theshearing component of strain in the xl-x 2 plane (Figure 1. 1-2(b)).

The quantities wo i are skew symmetrical, i.e. = - wji and are known as the

components of rotation because they can be shown to represent components of a rigid body rota-tion Reference 1-1).

At each point of the body, there exists a set of three mutually orthogonal directions forwhich the shearing strains are zero. These directions are called the principal axes of strainand the corresponding extensional strains are the principal strains. The principal axes ofstrain remain perpendicular to each other after the deformation, and an elemental rectangularparallelepiped with edges parallel to the principal axes remains a rectangular parallelepiped qafter deformation. In general, it will have also undergone a small rotation (Reference 1-2).

WADD TR 60-517 1.5

1.1 (Cont' d)

ddXLx 2, u 2 x 2, u2 •2

I dxI + dx1

d. xX ax1u

SXlU1•.•_.Xl,U 1

(a) Extensional Strain (b) Shearing Strain

FIGURE 1.1-2 TYPES OF STRAIN

1.1.1 Compatibility

From a physical point of view, the displacement ui in a simply-connected continuous

body must be single valued and continuous. Certain restrictions must be placed on the strainsS. in order that this be so. These restrictions constitute the so-called strain-compatibilityequations.

The compatibility equations are obtained by considering the defining formulas for thestrain components in rectangular coordinates:

(1)•-1[ ui, j + uj,i = ij '(1

as a system of partial differential equations from which the displacements ui are to be deter-

mined when the strain components c E are prescribed functions of the coordinates. If the bodyIs simply-connected, it can be'shownJ (Reference 1-1) that the conditions on the strains neces-sary and sufficient to ensure single-valued, continuous solutions for.the displacements consti-tute six independent partial differential equations. Expressed in Cartesian coordinates, they are:

WADD TR 60-517 1.6

111 (Cent' d)

a2 a f a C a 4Exx a yz+ z x xy)

-a z = 5ax - x + y + az

ae a ( xz a ac )azax =- ay aY+ yz + I X

a2c a2 a24X U•a

Sz= y + zzaaz az2 ay2

D2 a2 z 2t

azxop ax2 ax 2

T.................................................................................

2 +E D

Y Oaz 8x 2x 2 y O

222

areneesar to enur snl-aedy d zslcmnsTeecniiossxiythttelmt

2 -z (- +~- Zxx

2. 2 2

The abovera corcestinbiliy eqation usuallyddescribedhnhermal and mehanca psressstaems)alsoughoute the bdetemiatonse sofmte relaive dislyangmon the suromc Eq (1). oyhwn

Figre necessar Lt teui omvco o enesurefage-auddipaeents. thepsin e odtossecifyntatted limit

thrugou the bory. Ccigonswider snompasste point, by lyn on" thThraenSo the bodyction acing

across the surface S at the point Q is defined as

tn = Lira 6Pn ()I

2 s-+

•]•WADD TR 60-517 1.7

-xy

1.2 (Cont' d)

-a nnnn• 6S

-S aS

QQ

(a) Uncut Body (b) Free Body

FIGURE 1.2-1 TRACTIONS AT THE POINT Q ACTING ON THE PLANE WITH NORMAL n

The traction Is thus a vector which specifies the force acting per unit of area at apoint. The components of the traction vector in a given set of directions are defined as thestress components. The stress components at a given point are dependent upon the orienta-tion of the plane through the point with respect to the chosen coordinate system. Thus, forexample, in a rectangular coordinate _system (Figure 1. 2-2), the stress components in thex,y,and z directions of the traction t , acting on the face normal to the x axis, are ax' or

and o- , respectively. Here, the first subscript indicates the direction normal to the areaon wh~h the stress component acts and the second subscript refers to the direction of thecomponent.

All the remarks pertinent to principal strains apply to the stresses. In addition, whenthe material is isotropic, then the principal directions of stress and strain coincide.

,6 1

WADD TR 60-517 1.8

,".--,,

1.2 (Cont' d)

tyz x

1.2.~ ~~~~~~~~0 1 StesEutosoCqi arimi atsa oriae

0~ xxzx

dy -~x

4 zz

dx

z

FIGURE 1. 2-2 STRESS COMPONENTS

1. 2. 1 Stress Equations of Equilibrium in Cartesian Coordinates

The equilibrium equations state that the sum of the forces acting on a differential ele-ment of material are in static or dynamic equilibrium.

Figure 1. 2. 1-1 illustrates an elemental rectangular parallelepiped with edges parallelto the x, y and z axes. In addition to the stresses acting on the faces, there may exist bodyforces r dxdydz, where Y' is the body force per unit of volume. In most practical ap-plications, the body forces are due either to the weight of the body in a static problem orD' Alembert inertia forces in a dynamic problem.

Equilibrium of forces in the x direction requires that:xx + + +F =0

Ox ay fz x

Similarly, for the y and z directions,

WADD TR 60-517 1.9

1. 2. 1 (Cont' d)

W + yy+ aoy+ F =0

(1a)

+ + -+ F =0.

'14

Gy + dayy

yxy + 83 'd aaux

8z +Id

dy Oýy dz

a- x dx

u +-zzdzzz az

zd

FIGUREa 1..-1SRSSCMPNNS NRETNULRPAeLLLEPPE orI

WADD TR 60-517 1.10

1.2. 1 (Cont' d)

Conservation of angular momentum (equilibrium of moments) leads to equality of cross-shears,or

~ (1b)ox =~

Consequently, the significant number of stress components is reduced from nine to six. Theequations of equilibrium in other than Cartesian coordinates take different forms (Section 1.6).

1.3 BOUNDARY CONDITIONS

In order to have a complete solution, something must be known about each point on thebounding surface of a structure. Either the surface tractions (applied stresses) must be knownor the displacements must be specified (e. g., zero displacements at a support) or a knownrelationship between the tractions and displacements must be prescribed ((,. g., flexible sup-port of known stiffness).

The prescribed traction components in the directions of the chosen coordinate axes maybe expressed In terms of the surface stresses by means of the equilibrium equations

X =o n +or n +a nxxx yx y zxz

Y =axy nx +y n +* n (1)

Z =axz nx+ yz ny zz nz

where X = component of traction in the x direction, etc. (Figure 1.3-1), nx = cos (n,x) =cosineof the angle between the normal to the surface and the x axis.

WADD Th 60-517 11

A1.3 (Cont' d)

y

Y

0121

a X

FIGURE 1. 3-1 SPECIFICATION OF SURFACE TRACTIONS

1.4 STRESS-STRAIN RELATIONSHIP

The complete solution of a problem in elasticity requires the determination of sixstress components, three displacement components and six strain components at each point

S~of a body. Fifteen indepoudent equations are required for the evaluation of these quantities.S~Nine of these equations have already been presented, namely the six strain displacement re-

• ~lations (Eqs. (2) of Sub-section 1. 1, which ensure that the comp teatibilit equations are sat- :Sisfied when the displacements are single-valued) and the three equilibrium equations (Eqs.S~(1) of Paragraph 1. 2. 1). Both of these sets of equations are independent of the material [S~properties of the body. t

WADD T 60-57 1.1

1.4 (Cont' d)

The remaining six equations are obtained from a stress-strain relationship betweenthe six stresses and six strains. It is through this relationship, which is a property of thematerial, that the effects of plasticity, temperature, creep, etc., come into play.

If the material is isotropic and linearly elastic, then the stress-strain relationshipcan be expressed as

I ii [TI *i-{ ('j + 'kk) + aT

l~v ( )(1a)

E - .. i,j,k = 1,2,34 ij E Ij

where E = elastic modulus,z/ = Poisson' s ratioa = linear coefficient of thermal expansionT = Temperature above room temperature as a datum.

The shear modulus is defined byE

G = 2(l- -) (1b)

Because of the isotropy of the material, the expansions due to temperature cause noshearing strains. The linear elastic-stress-strain relations for an isotropic material are ofthe form of Eq. (la) for any orthogonal coordinate system (rectangular or curvilinear). Forexample, in a rectangular coordinate system,

i,j,k = 1,2,3 x,y,z

while for a cylindrical coordinate system,

i,j,k = 1,2,3 = r, 0, z

The linear stress-strain relationships of Eq. (la) can be modified to provide stress-strain relationships for materials which are not linear elastic by employing an equivalent

secant modulus E= a in place of E. Unfortunately, E5 will not be constant andseatmouu S c - ciT

direct solution of problems may be extremely difficult. A progressive solution may be pos-sible by solving the problem for small increments of load-time histories where it is assumedthat the modulus is constant in each interval, evaluating new constant moduli for the nextincrement, and continuing the process until the final load-time history is applied. This tech-nique must assume the sense of the incremental stress since a material may have differentmoduli dependent on the direction of loading.

The strain-compatibility equations can now be converted to stress-compatibility equa-

tions through the stress-strain relationships. Thus for a linear-elastic stress-strain relation-ship with temperature, Eqs. (2) of Paragrapd 1.1. 1 become

2____• __ lO 2T+ T 1(1+V1) V2 + .2@V+

xx 8 x2 E -V a0

(21+ 1 ) TV2

(1+yy 2 V- - + atE [291 V2 " + 2 = 0 (2)

WADD TR 60-517 1.13

1*!

1.4 (Cont' d)

2+~r 1+v/ V2 T, 2T]

( v _ + aE -0(ly 2zz 1+ Yz 2-

Oxaz

(1+1) V2 a2@ a 2 T (2)aXy+ )x--•- + a E Oxft 0

P2S

aa =xp cont ' d

(lu 2 ay + z 2@ + aE•--- 2T=0,a1VV + +2 21Eyz Opy Oz

where

aj-0 + a + axx yy zz

or, in shorthand notation,

(l+Y) V CTij + E L +% 17.'__ V2T+Ti = 0,2 + @ 'ij + a E qj 1 v 2 T ,i0

where

1 =0,l¢j•lj =• =j

The stress compatibility equations can be employed to show that the only time thethermal stresses are identically zero in an unrestrained linear-elastic body is when thetemperature is a linear function of the rectangular (x,y,z) coordinates. Assume all stresscomponents oxx ...... ayz are zero. Then certainly the equilibrium equations are satisfied

throughout the interior of the body and on the surface. Equations (2) reduce to

82T _ T1+" V2T + = 0 = 01 - 2/ a2 a z

1 V+ T+ - 0 = 0 (3)

V2T LT 0 T 0

WADD TR 60-517 1.14

1.4 (Cont' d)

(I+v) V2 a+ a 2E v 2T 0[1-& ZJ

22a + 026 + aE -= 0•xz axaz C•xft

(2r2 a2+ 2 (2)S--•-- VE 0 co-ttdax Oy axOpycont' d

(I+v) V2 + a20 + a E =0,yz -Yf Oy &Z

where=a + a + a"

xx yy zz

or, in shorthand notation,

(1+Y) V2 +[ ®+ V2a+ T, = 0ij i J -Y Ij

where

iJ =1, i=j

The stress compatibility equations can be employed to show that the only time thethermal stresses are identically zero in an unrestrained linear-elastic body is when thetemperature Is a linear function of the rectangular (x,y,z) coordinates. Assume all stresscomponents oxx ...... a are zero. Then certainly the equilibrium equations are satisfied

throughout the Interior of the body and on the surface. Equations (2) reduce to

1 V2 T 82T OT = 0V +" T + 0._ ax ft 0.T• =

1 +v 8y2 32T81+V 2T +0T= 0 (3)

1 +'v 2 a2 T _03 T

WADD TR 60-517 1.14

1.4 (Cont' d)

which yields the condition

2T a2T 82T 2T 2 2 T -22 2 2(4)

Equations (4) have the unique solution that T must be of the form

T = Ax + By + Cz + D , (5)

where A, B, C, D are constants; i.e., T is at most a linear function of the rectangularcoordinates x, y, and z.

NOTE: If T = Ar for an unrestrained circular plate where r is the radial coordinate measuredfrom the center, there will be thermal stresses. This follows because T is linear in r but notin x and y where

r=Fx2 + y

1.5 UNIQUENESS

The strain displacement relations, equilibrium equations, and linear stress-strain re-lations provide a set of equations from which the stresses, strains, and displacements are to bedetermined at each point of a linear elastic body. From this set of equations, together withappropriate boundary conditions, it can be proven not only that there exists a solution to thelinear elasticity problem but also that the solution is unique.

In a simple form (see Reference 1-1 for a more complete discussion), Kirchhoff IsUniqueness Theorem can be stated as follows: If the initial displacements, velocities, bodyforces and temperatures are specified throughout the volume, and if the compatibility con-ditions discussed in Paragraph 1.1.1 are satisfied and the appropriate boundary conditions(Sub-section 1.3) are specified over the entire surface, there exists only one form of equili-brium in the sense that the distribution of stresses and strains is determined uniquely.

The above theorem applies to elasticity problems with infinitesimal strains and dis-placements. If the strains and displacements are finite, the solution may not be unique as inproblems concerning elastic stability (Section 9), where different equilibrium configurationsare possible. The uniqueness of the solution for infinitesimal strains suggests the utilizationof the incremental method for closer approximations of the non-linear problem. Considerationshould be given to changes in stiffness, geometry, etc., with load-temperature history.

1.6 SPECIAL ORTHOGONAL COORDINATE SYSTEMS

The representation of the stress-strain relationships, strain-displacement equationsand the equilibrium equations depends upon the syt.tem of reference employed. It is oftenconvenient to select a coordinate system which will simplify the boundary conditions and re-quired solution. The three most useful orthogonal coordinate systems are the rectangular,cylindrical and the spherical. For linear elastic, isotropic bodies, the stress-strain relation-ships (Eq. (la) of Sub-section 1.4) are of the same form for all orthogonal coordinate systems.The strain-displacement equations and equilibrium equations have already been presented inrectangular coordinates. The cylindrical and spherical counterparts are shown in Tables 1.6-1and -2. These expressions are simplified if the three dimensional problem reduces to a two orone dimensional problem.

WADD TR 60-517 1.15

1.6 (Cont d)

CD

$4 +

+ :3 (

;4 1) k ~P4 - co

1-. (D (

H 5.

c-J$ W

P4.

+ ~ +

H-lo " I ,-1

N4 (D N N

wk U) t $4 $4 (D

C- +

N N

w M %U

WADD TR 60-517 1.16

LA

1.6 (Cont'd)

TABLE 1.6-2

EQUATIONS OF EQUILIBRIUM

ij,+F =0

Rectangular

•xx, x yx, y zx, +z • F=

S +o +or +F =0xy,x yy,y zy,z y

+y +o r +F 0xz,x yz,y zz,z z

Cylindrical

+ 1 rr ee 1rrF 0arr, r r re,e rz, z r r

1, 2 + F2r - E,e +zO,z + 're r Fe= 0

or +1e + C + a + Fz0rz,r r Oz, e zz, z r rz

Spherical

O=rr, r +rler,e + 1 + 2 c - t

3a rreee + ae - q0e cot E) F0 0= rO), r + 1r (r E) e•r + (sEE '0 E) 0• r o

a1 + __ 1 +3 +2cote )Oe +F00 0r4,r r - 60 ,e 0rsine 4i0- r

WADD TR 60-517 1.17

_____...____,__,,__.*•• • • . .. -... ... .. .

IJ

1.7 ENERGY PRINCIPLES

The behaviour of a structure can be described by means of energy principleswhich yield relationships between the stresses, strains, displacements and forceswhen adequate conditions (boundary, temperature-load-time history, material proper-ties, etc.) are defined. These principles can be derived from the "Principle of VirtualWork" which is a restatement of conservation of energy.

1.7.1 Principle of Virtual Work

"If a body is in equilibrium under the action of prescribed body and surfaceforces, the work done by these forces in a small additional displacement, the virtualdisplacement 6u, is equal to the change in the internal strain energy, second orderterms in the increments of strain being neglected." The strain energy is defined inParagraph 1.7.2.1.

A qualitative argument supporting this theorem is as follows:

"The body "B" with surface S (Figure 1.7.1-1) is in equilibrium under the actionof the applied external force, body forces, and internal stress systems, represented

t ~6 t

B

at O tt

FIGURE 1.7. 1-1 ELASTIC BODY WITH APPLIED FORCES

symbolically by 'I" "1F" and "o*", respectively, If an incremental, self-equilibratingexternal force system "6 t" is imposed, there result corresponding internal stresses6 ar, strains 6 e and continuous displacements 6 u. From conservation of energy,

ft6udS + F6udV+ (terms of order 5u6t)

S=-f• e5c dV + (terms of order 65 6c)

B

WADD 1'R 60-517 1.18

1.7.1 (Cont' d)

Neglecting second order terms,

which is the mathematical statement of the theorem.

A consequence of the theorem of virtual work is the Reciprocal Theorem of Betti

and Rayleigh: If an elastic body is subjected to two systems of body and surface forces,then the work that would be done by the first system ti, F (surface tractions and body

forces, respectivelyl in acting through the displacements u! due to the second system of1

forces, is equal to the work that would be done by the second system t , F, in acting through

the displacements u. due to the first system of forces. In symbols,the theorem becomes:

ItiuidS+ F udV Vtu.dS+ 1 F!u.dV (2a).Sl ~B 1 VS B

where the repeated subscript I implies summation.

An alternate form of the re :iprocal theorem is

SdS + d = F udVwhere ftfuij di1S Bj B

where the unprimed quantities correspond to surface tractions, body forces, and stressesof the first system and the primed quantities to the displacements and strains of the secondsystem. It is important to note that the forces and stresses of the first system can be un-related to the displacements and strains of the second system, and may include thermaleffects. The only requirements are that the force-stress system be in equilibrium and thestrain displacement system be compatible. This feature of the reciprocal theorem makesit extremely useful in the solution of statically indeterminate problems (Paragraph 2.1.1).

REMARK: The virtual work and reciprocal theorems are quite general and are not restrictedto elastic problems.

1.7.2 Variational Principles

The solution of three dimensional problems in the theory of elasticity re.presentsin many cases an almost impossible task even for the isothermal or mechanical problem.For isotropic and homogeneous bodies, the literature contains only approximate solutionswhich are based upon simplifying assumptions. One very powerful method of attack whichwill yield approximate solutions is based upon that branch of mathematics which is knownas the calculus of variations. The application of the calculus of variations to the field ofelasticity and thermoelasticity depends upon the following variational principles:

(1) The principle of minimum potential energy, and

(2) The principle of minimum complementary energy (sometimes called comple-mentary potential energy).

These in turn follow from the principle of virtual work which was discussed in Paragraph1.7.1.

4 WADD TR 60-517 1.19

1.7.2 (Cont' d)

Before discussing the variational principles, in detail, it will be desirable to firstdefine the energy terms employed and the mathematical operations which are essential totheir proper application.

1.7.2.1 Definitions of Terms

Potential Energy of Surface Tractions and Body Forces: The loss in potentialenergy of the surface tractions t, and body forces Fi is defined by

W f tudS + FudV (1)#SB

where the tractions ti have the dimensions of force per unit of area and the body forces F1

have the dimensions of force per unit of volume. The displacements ui, and hence the poten-

tial energy, are measured from a datum defined by the undeformed position of the body.

Strain Energy: The strain energy density in a deformable body is defined as thework done per unit of volume by the internal stresses when the stresses and corresponding"strains vary from zero to their terminal values. The strain energy density may be expressedin terms of the stresses and strains as

ud= , (2)

where, in Cartesian coordinates,

ode= a de +0a de +o de +2a dexx xx yy yy zz xy xy

(3)+ 2cr de + 2o de a de

yz yz zx zx = ij

Similarly

= ~0

The strain energy of the body is then

U= ýUVdV (4)

In most problems it is convenient to express the strain energy in terms of internal forcesand displacements rather than stresses and strains. Appropriate expressions are given inSection 2.

Potential Energy: The potential energy L of a body is defined as

L = U-Wt (5a)L -- t ,

where U is given by Eq. (4) and Wt is the loss in potential energy of the body forces over the

WADD TR 60-517 1.20 '-

1.7.2.1 (Cont' d)

total volume (assuming the body forces to be prescribed throughout the volume) and of the

surface tractions over that portion of the surface where the tractions are prescribed.

That is,

Wt B FiuidV t Stt.u.dS , (5b)

where St denotes that portion of the surface where the traction and not the displacements are

prescribed. This statement is amplified in Paragraph 1.7.2.3 where the principle of minimumpotential energy is discussed. Note that an elastic support may be considered to be part of thestructure so that displacements or tractions may still be prescribed on the boundary.

Complementary Potential Energy: The complementary potential energy L* isdefined as

L* =U* - Wu4u

where U* is the complementary strain energy of the body and (assuming the body forces tobe prescribed throughout the body) W is the loss in potential energy of the surface traction

uover that portion of the surface, S where the displacement and not the traction is pre-scribed. u

That is,

Wu=f t.u.dS• (5c)

u

1.7.2.2 Variational Techniques

Stationary Values: Application of minimum energy principles involves the defini-tion of stationary values for functions or, more generally, functionals (integrals of functions).A stationary condition is necessary in order that an extremum (maximum or minimum)exist. In the case of a function f (x1,, x ... , xn), the conditions for a stationary value

at a point P are

1 2 nP P

In the case of functionals, the problem is more complex. For example, given the functional

I (y) F (x,y,y') dx, where F is a given explicit function of x,y,y', the question may

be asked, what is the function y (x) (if it exists) which makes I&y) stationary where y(x ) andy(xl) are prescribed? The aim of the calculus of variation is to solve problems such Rs this.

A typical procedure is shown as follows:

WADD TR 60-517 1.21

1.7.2.2 (Cont'd)

Consider the integral

ýx0i-fF (x, y, y', y) dx

and seek the function y = y (x) which makes I stationary where y and y' are prescribed atx and x The existence of such a function is assumed and the effect on I is examined of a

aatioy in y by a small amount 6y where 6y =y' =0 at x=x 0 and x =x (Figure 1. 7.2.2-1).0)

y

6y(x)

(x19yExtremizing

Function

FIGURE 1.7.2.2-1 VARIATION OF EXTREMIZLNG FUNCTION

If I is an extremizing function, then

j Xl [ F y F 6 F Y, +f

68= .X [ - 6y B y' + ,6y" ] dx =0 (2)xO

The left hand side is called the first variation of the integral I.

In order to eliminate the variations 5y' and 6y" from Eq. (2), the second andthird terms are integrated by parts:

r F 1F X1 1 FX6ydx= __a-,y - d (L, I ydx (3a)

0 Xo

WADD TR 60-517 1.22

1.7.2.2 (Contd)

x•x10 0 0(3b)

Xf1 d2 OF

Because of the assumed boundary conditions, the bracketed terms in (3a) and (3b) vanish sothat

f, [F d d2 8F6=d F +_ d+ -" 6ydx =0 (4)

0 d2

The above Integral must vanish for all admissible 6y, which requires that the expression inthe bracket of Eq. (4) be zero, i.e.,

2OF d O2F d OF(,• d •)+ x , 0 .(5)

This differential equation is known as the Euler 0ifferential equation from which the function

y can be determined.

1.7.2.3 Minimum Potential Energy

The functional called potential energy was defined in Paragraph 1.7.2.1. We nowstate the following important theorem for elastic bodies:

Theorem of Minimum Potential Energy: Of all compatible displacements for astable structure satisfying given displacement boundary conditions, those which satisfy theequilibrium equations make the potential energy, L, a minimum. The converse theorem, whichis the one used most often, is as follows:

"t"Of all the displacements satisfying the boundary conditions, those which make the

potential energy a minimum satisfy the equilibrium equations." The proofs of these theoremsand the other theorems discussed may be found in Reference 1-1.

Since the potential energy ( L = U - Wt) is a minimum for the true equilibrium

state, it must also be stationary, i.e.,

6L =(U-Wt)=o , (la)

where the quantities U and Wt are defined in Paragraph 1. 7. 2. 1. Equation (1) is stated quite

simply; however, a full understanding of its meaning requires further explanation. According toKirchhoff's Uniqueness Theorem, either the displacements or the tractions (but not both) must beprescribed at each point of the boundary of the body. Variations of the potential energy, Eq. (la),

WADD TR 60-517 1.23

1.7.2.3 (Cont' d)

are effected by varying the displacements only over those portions of the body where thetractions are prescribed. The displacement variation must vanish over those portions of theboundary where the displacements are prescribed. As a consequence, If the displacementsare given over the entire surface and there are no body forces (Wt = 0), then Eq. (la)becomes

6U =0. (1b)

This is, in effect, a stationary (minimum) strain energy principle:

If there are no body forces present then, of all the arbitrary sets of continuousdisplacements which satisfy the compatibility equations and the specified displacement con-ditions over the entire boundary, those displacements which satisfy equilibrium make thestrain energy a local minimum with respect to neighboring displacements.

(Remark: Since variations of the potential energy are expressed in terms of the displace -ments, the strain energy must be given in terms of displacements.)

Example: The following elementary problem illustrates the application of theprinciple of minimum potential energy.

A cantilever beam of flexural rigidity EI is propped by an elastic spring withspring constant K (force/unit of deflection). Find the end deflection A due to an appliedload P.

PElI

42K

FIGURE 1.7.2.3-1 ELASTICALLY PROPPED CANTILEVER BEAM

The internal (strain) energy for the beam-spring system, expressed in terms ofthe end displacement is

U = + -- (see any elementary text on strength of materials).

213 2

According to the th.eorem, the loss In potential energy of the external loads Wt is takenonly over those pertions of the body where the forces are prescribed. In this problem theonly prescribed force is the end load P. Thus,

WADD TR 60-517 1.24

1.7.2.3 (Cont' d)

Wt =PA,

so that

L=U-Wt•L _ U- W t

=(2-E---+K --2- PA.3 2

It follows from Eq. (la) that

6L=dL 6A M-1+K A 6A 06LjdA 3~~KA~]AO

The factor 6 A is arbitrary and, therefore, the bracketed term is zero, i. e.,

A P

M + K

Note that

d2LL 3E + K) (A) >0,dA2 (5A2 52 / .

d,&2 3

which establishes a minimum.

1. 7.2.4 Minimum Complementary Energy

Another important variational principle relates to the complementary energyfunctional (L* = U* - Wu) defined in Paragraph 1. 7. 2. 1.

The Theorem of Minimum Complementary Energy: "Of all states of stress in astable structure satisfying equilibrium and given traction boundary conditions, that statewhich satisfies compatibility requirements makes the complementary energy L* aminimum."

Since the complementary energy Is a minimum for the compatible state, it mustalso be stationary, i.e., ,

5L* = 6 (U*-)W ) 0 (la)

where the quantities U* and W are as defined In Paragraph 1.7.2. 1.

The strain energy, in this case, Is expressed in terms of the stresses or forces,and variations of the complementary energy are effected by varying the stresses such thattractions vary only over those portions of the surface where the displacements are prescribed.

WADD TR 60-517 1.25

ZM

A1.7.2.4 (Cont' d)

The stress variation must vanish over those portions of the boundary where the tractions areprescribed. Consequently, if the tractions are given over the entire surface (Wu = 0 ) thenEq. (la) becomes

6 U* = 0 (Ib)

Equation (1b) is known as The Minimum Strain Energy Theorem, which states:

"Of all arbitrary stress states which satisfy equilibrium and the specified tractionsover the entire boundary, that state which satisfies compatibility makes the complementarystrain energy a minimum." When a problem is linearly elastic U* = U.

Example: The theorem of minimum complementary energy will now be demon-strated for the case of a beam on three supports.

A beam (constant EI) is Icaded and supported as shown in Figure 1. 7. 2. 4-1. Itis assumed that there is no displacement at the supports. The support reactions are to bedetermined.Solution: This is a statically indeterminate beam of 1 degree of indeterminancy. Let 2Rrepresent the center reaction. Because of symmetry, the other two reactions are equal toW-R. External equilibrium of the structure is satisfied for all values of R. However, thereis only one value of R which will also satisfy compatibility conditions, i.e., that the slope iscontinuous, and the displacement is zero at the middle support.

Since the loads are prescribed everywhere except at the supports, force variations,according to the theorem, are permitted only at the supports where displacements are prescribedas zero (W = 0 ). Thus the complemeptary strain energy, expressed in terms of the forcesmust be a iinimum with respect to variations in R, i.e., 6 U* = 0, 6 2 U* > 0.

Considering only bending energy, the complementary strain energy

21 2 M2U* = E-" dx=2 f - dix ,

0 u*= 0 2

where

i•M =(w-R)x ; Ox:x<L12

=i--Rx; _._W1

or

U * = 2 2W- R)2 xRd) +

2EI 2EI

2

WADD TR 60-517 1.26

1.7.2.4 (Conts d)

•'* =• ( - -R w3 • =El 12 W 1

W-R 2RW-R

FIGURE 1.7.2.4-1 BEAM WITH THREE SUPPORTS

Since SR is arbitrary, the bracketed expression is zero, which yields

11

The two end reactions are, therefore,

5W -R = T6 W

and the center reaction is

112R= -W .8

Note also that

4 3 [8U* = 13 (8 R)2 > 0El 12 >02

1.8 REFERENCES

1-1 Sokoln:koff, I. S., Mathematical Theory of Elasticity, McGraw-Hill BookCompany, Inc., 1956.

1-2 Timoshenko, S. and Goodier, J. N., Theory of Elasticity, McGraw--illBook Company, Inc., 1951

WADD TR 60-517 1.27

SECTION 2

STRUCTURAL TECHNIQUES

4 6

WADDRRU60-1RA2 0EHI•E

A

SECTION 2

STRUCTURAL TECHNIQUES

TABLE OF CONTENTS

Paragraph Title Page

2 Structural Techniques 2. 3

2.1 Linear Structures 2.5

2.1.1 Virtual Work Techniques 2.5

2.1.1.1 Virtual Forces 2.5

2.1.1.2 Virtual Displacements 2.7

2.1.2 Strain Energy in Terms of Internal Loads and Deformations 2.9

2.1.3 Application of Virtual Force Technique - Flexibility Coefficient 2. 11

2.1.3.1 Flexibility Coefficient for a Beam-Like Structure 2.11

2.1.3.2 Addition of Flexibility Coefficients 2.12

2.1.4 Applications of Virtual Displacement Techniques 2.16

2.1.4.1 Load on a Structure 2.16

2.1.4.2 Stiffness Coefficients 2. 16

2.1.4.3 Addition of Stiffness Coefficients 2. 17

2.1.5 Properties of Influence Coefficients Matrix 2.19

2.1.5.1 Symmetry 2.20

2.1.5.2 Positive Definiteness 2.20

2.1.6 Equivalent Thermal Load 2.21

2.1.6.1 Three Dimensional Hydrostatic Loading 2.21

2.1.6.2 Fixed End Beam Reactions 2.23

2.1.7 Plane Stress and Plane Strain Problem 2.23

2.1.7.1 Plane Strain 2.23

2.1.7.2 Plane Stress 2.25 i

WADD TR 60-517 2.1


Paragraph T.tle Page

2.1.8 St. Venant s Principle 2.25

2.2 Non-Linear Structures 2.26

2.2.1 Incremental Linear Solutions 2.27

2.2.2 Inverse Solution

2.282.2.3 Approximate Solutions

2.3 References 2.29

WADD TR 60-517 2.2 F

SECTION 2 - STRUCTURAL TECHNIQUES

The solution of the structural equations can be performed in many ways. The tech-nique selected depends upon the type of problem, the type of boundary conditions, the com-plexity and the required accuracy, etc. The problem is simplified if the structure is linearand permits the addition of simple solutions to obtain the final solution.

The following symbols are used throughout this section:

a,b,i,j,k,r Degrees of freedom of a structuref. Internal axial load in beam due to a unit load at "i"1

r rf. Deflection at "i" due to a unit load at "j" with respect to a datum at "r"

k Moduius of rupture factor = Mult. /Melasticm. Internal moment in beam due to a unit load at "i"

n Unit normal to bcunding surfacen x, n ynx Direction cosines of normal with x, y, z axes

qi Internal torque in beam due to a unit load at "i"

r Distance from origin in polar coordinates; distance from shear centeru i Generalized displacement

6u Arbitrary displacement satisfying compatibilityu, v, w Displacement in x, y, z directionv iInternal transverse shear due to unit load at 'fi"

x,y,z Coordinates in x,y, z directionsy Distance in beam cross section from neutral axis

A AreaA Effective transverse shear areaE Linear modulus of materialF. Surface or body force acting in x i direction

5 •Arbitrary f.arce system satisfying equilibriumG Linear sbear modulusSI Moment of inertia of cross section!K3i Load at "j" due to a unit (virtual) displacement at "i"

L Length of beamM Internal moment due to applied loadsP Internal axial load due to applied loadsQ Internal torque load due to applied loadT Temperature

Internal strain energy = f .. d0.. dV

4 U* Complementary strain energy = Cvf t e i doai dV

V Transverse shear load; volume; potential for a conservative force systemW Loss in potential energy of surface tractions and body forces6 W E 6 5W Change in potential due to a virtual displacement5 W• 6 W* Change in complementary potential due to a virtual force system

WADD TR 60-517 2.3

4I

AE Axial stiffness of cross section f E5 dA

AvG Transverse shear stiffness of cross section = GS dAV

El Bending stiffness of cross section = fES y2 dA

JG Torsional stiffness of cross section = f Gs r 2 dA

0 Linear coefficient of thermal expansionY Average shear strain - transverse shear displacement per unit length6 Incremental operator6 Krcnecker delta = 1 if i = J

0iflDeflection at "i"Strain

C xC ;C Extensional strainsxx yy zz,;fyz;e Shear strains

SM %.a d -T Axial strain of cross section due to applied loads and tem perature, respectively

E Polar coordinate - angle between r and x axis11i Rotation at "i"

KM and KT Curvature of cross section due to applied loads and temperature, respectively

Transformation matrix converting applied loads to loads on the substructuresV Poisson' s ratioa Stressa Stress on plane perpendicular to x axis and in y directionxyT Torsion - change in twist of cross section per unit lengthVp Airy functionV Gradient operator

Subscripts

a,b,i,j Pertaining to degrees of freedom a, b, i, Ji,j Pertaining to xi, x directionsJr Pertaining to datum structureA Pertaining to axial energy, stresses or strains; allowableB BodyM Pertaining to bending energy; deformation due to mechanical loadsQ Pertaining to torsional energy, stresses or strainsS Surface; secant modulusT Deformation due to temperatureV Pertaining to transverse shear energy, stresses or strains

Superscripts

1, 2 SubstructurePertaining to a rotational degree of freedom (e.g., b')

WADD TR 60-517 2.4

2.1 LINEAR STRUCTURES

If the structural response to mechanical and thermal loads is directly proportion-al to these loads and independent of past history, the structure is said to be linear elastic.For a linear elastic structure, it is possible to solve the general structural problem by thesolution of a few unit-type problems, and to add these solutions to obtain the final completesolution. The technique is that each unit solution shall satisfy equilibrium and compatibilityand, when superposed upon each other, satisfy the given boundary conditions. Thus, it ispossible to remove boundary conditions and solve for the new structure and superpose self-equilibrating force systems which will have the net effect of satisfying the boundary conditions(flexibility method). It is also possible to divide the structure into smaller sub-structures,solve each sub-structure and impose compatibility conditions to satisfy the boundary condi-tions of the over-all structure and the sub-structures (stiffness method). These techniquesare illustrated in the flexibility and stiffness coefficient methods of Sub-section 4. 3 and arederivable from the "Principle of Virtual Work" discussed in P,. agraph 1.7. 1.

No structure is truly linear elastic with respect to thermal stimulation since thestiffness of the structure will change with temperature. The structure can be treated aslinear elastic, for approximate solutions, where the changes in temperature or stressesare not severe enough to significantly change the stiffness of the material and where thestrains are small and geometry changes can be ignored.

2.1.1 Virtual Work Techniques

As a consequence of the reciprocal theorem (Paragraph 1.7. 1), the loss in poten-tial energy of the body and surface forces acting on a structure in its true equilibrium state,due to a compatible set of arbitrary virtual displacements (not necessarily satisfying anydisplacement boundary conditions), is equal to the gain in strain energy. In equation form,

6U= 6W W, (1)

where 6 U= 0 edV

6W= fF6u dV+f½6uidS

In addition, the loss of complementary potential energy of the body and surface forces acting,on a structure in its true equilibrium state, due to a self-equilibrating set of arbitrary virtualforces or stresses (not necessarily satisfying traction boundary conditions) is equal to the gainin complementary strain energy, i.e.,

6U* =W* =•6Wu (2)

where6U* = •oV

fBE d6W*= fBui6FidV+ sui6tidS

An inequality results if the structure is not in its true equilibrium state.

2. 1.1.1 Virtual Forces

Figure 2.1.1. 1-1(a) shows a body in equilibrium under the force systems F and Fk

WADD TR 60-517 2.5

2. 1. 1. 1 (Cont' d)

F4• // Lr. Original Boundary

I kDeformed Boundary

4k

F kr

(a) Deformed Structure

•2

WAD F 1 6

/ F

F- k

FIGURE 2. 1.1.1-1 BODJY UNDER LOAD

WADD TR 60-517 2.6 2

2.1.1.1 (Cont'd)

"Determine the displacement in some chosen direction of point i. To do this, apply a unitforce 6F. = 1, acting at the point in the chosen direction (Figure 2.1.1.1-1(b), and react

this force by 6F r This force ystem is in equilibrium and thus satisfies the requirements

of a virtual force system. From Eq. (2) of Paragraph 2.1. 1,

6U* W*= 1 - u. 6F u (Ia)I1 r r

If the reaction points of 65F are chosen as the datum with respect to which the deflection isr

to be measured, then u = 0 andr

6 U* =u =J 6udV (1b)B

The deflection can always be measured with respect to a datum defined by thereaction points for the applied virtual load. Statically determinate virtual reactions atpoints which are reactions for the actual statically indeterminate structure will alwaysyield deflections with respect to the structural datum with a minimum amount of calcula-tion. A rigid body correction must be made to determine the deflection of a point withrespect to a structural datum if the chosen datum is not the same as the structural datum.The rigid body correction is the displacement of the chosen datum with respect to thestructural datum.

2. 1,1.2 Virtual Displacements

To determine a load F. which is in equilibrium with the force systems F and Fk

F Figure 2. 1. 1. 1-1(c)]1, a displacement 6u. = 1 is imposed in the direction of F. with corres-

"ponding arbitrary but compatible deformations throughout the body. This displacement sys-tem satisfies the requirements of a virtual displacement system. From Eq. (1) of Paragraph

S~2.1.1,

6U= 6W= Fi 1 F. 6u. +Fk 6uk (la)

If the displacement system is chosen in such a manner that

6u 1 = cuk =0,

then

6 U= F. = 6 dV . (lb)¢B

The virtual displacement principle can also yield an alternate statement of the equili-brium equations. To illustrate this, consider the beam of Figure 2. 1. 1. 2-1 to be subjectedto a rigid body virtual displacement in the vertical direction,

6u=l1

Since no straining occurs in a rigid body displacement,

6 U= f acdV=0,B n

and n6W= XFIU= 6u F •

WADD TR 60-517 2.7

2.1.1.2 (Contt d)

However

6U= 6W,

which results in n

F. =0.i=l 1

The above is a statement of equilibrium of vertical forces.

Consider next a rigid body virtual rotation 6 0 1 about point 0.

Again

6U O

But n n

6W= Z F1 6x i 60 F xii=l i=1

Therefore, n

. Fixi =0,i=1

which is a statement of equilibrium of moments.

- ---- --- --

--- ----- - - - -

6u

FIGURE 2.1.1.2-1 VERTICALLY WOADED BEAM IN EQUI-IBRIU11

WADD TR 60-517 2.8

2.1.2 Strain Energy in Terms of Internal Loads and Deformations

It is most often convenient to express the strain energy of the body in terms of in-ternal forces and the deformations of these forces rather than as stresses and strains. Con-sider a body in equilibrium and compatibility under a given set of loads and a temperaturedistribution. Cut the body into elemental volumes and apply tractions at the cuts which areequal to those stresses which existed in the body before it was cut. These tractions areexactly those necessary to insure a perfect fit between the elemental volumes. It can beshown by the Divergence Theorem that the strain energy expressed b- these internal forcesacting through the displacement of the cuts is equal to the work done by the body and surfaceforces acting through their displacement, and that the complementary strain energy is equalto the complementary work of the body and surface forces.

Figure 2.1.2-1 illustrates a beam subjected to temperature and mechanical loads.The equivalent internal loads and the deformation of the cross section are defined by an

T YY ~ F

Sx Principal Axisr Mhear Center

z - - .\ Elastic AreaP Center (C. G.)

x Section B-Bx

FIGURE 2.1.2-1 BEAM UNDER LOAD

axial load P and an axial deformation per unit length of c + E- a bending moment A andM T' digmmnMada change in slope per unit length (curvature) of KM + KT; a twisting moment Q and a change

in twist per unit length (torsion) of T ; a transverse shear load V and a relative shear dis-placement per unit length (shear strain) of 7. The subscript M implies the deformation as-sociated with the mechanical load and the subscript T implies deformations associated withtemperature.

Assuming linear strains over the beam cross section, let

e = Axial strainA

6 = Torsional shearing strainQ

andEV = Transverze shearing strain (may be nonlinear)

WADD TR 60-517 2.9

2.1.2 (Cont'd)

These strains may be expressed in terms of the internal deformations by the relations

C M + T + KM + KT) yA= M T

C = Tr (1)

fGSeV dA= (AvG) ).

Similarly let

•A = xial stress

qQ = Torsional shearing stress

= Transverse shear stress.

These internal stresses can be expressed in terms of the internal loads by the relations0A = ESEAA-aT) = E+ eT + (K + K y- Ca

ESLEM +T M T)

GS Qr(2a),7Q =GS C Q (2a)-

f TV dA= fGS eV dA= V

where

- pEM EA JG

fES aTdA___ VT = V (2b)

EA V

KM= M

fEs EyaTdAKT=KT El

and = JE Sy 2 dA

SE-A fEsdA

° = Gsr 2 dA (for circular sections only) (2c)

AVG= f GS dAv

= fE y dA

Evaluating the strain energy f f0o dedV and the complementary strain energy f EdodV

results in jU = UA + UM + UQ + UV

S=_ + , + , + ,UV3a)

WADD TR 60-517 2. 10

2.1.2 (Cont'd)

where 00 M+C L PUA I=P('MITd UX =f (M+ET) dPdxTPd(9-M+...)dx LMPd

LIKM+ T LOMUM =f _Md(KM+KT)dX U = -( +KT) dMdx

00 MT 0f-KL L Q b)

UQ = j Qdrdx U = rdQ dxUQf f Q -f fVdyd V fJLJf dV dx

In addition,

6U =f0 [{o P E6(M+ CT)} M{16 (K.+KTJ+ Q{6 T}11V{6Y I]dx (4a)

6 UL* J[IO1PI(EM +CT) -tOMI(KM+K T) +16QIT+t6VJsvjdx . (4b)

Thus, the change in strain energy is equivalent to the internal loads acting through thechange in the internal deformations. The change in complementary strain energy is equi-valent to the internal deformations acting through the change in the internal loads.

Using expression (4b) and the virtual force technique, the displacement at somepoint on the beam (when shear energy is neglected), is given by the expression

L L

u. 0 + KT) mi dx + f0( M+ CT)f. dx (5)

where 6M = m.6p = f.•

Beam deflections are further discussed in Paragraph 4.2.1.

2.1.3 Application of Virtual Force Technique - Flexibility Coefficient

The flexibility coefficient rf.. is defined as the deflection (or rotation) at point i

with respect to some datum r, when the structure is subjected to a unit load at point j andreacted at points defining the datum r.

2.1.3.1 Flexibility Coefficient for a Beam-Like Structure

Using the virtual force technique and the equivalent loads and displacements for a

WADD TR 60-517 2. 11

2.1.3.1 (Cont t d)

6 beam-like structure, results in

mjm qjq1f.. = i dx. 1

E--I -J- AVG

The deflection of the structure can be obtained by employing Eq. (5) of Paragraph2.1.2 or by adding the effects of each load times the corresponding flexibility coefficient,

A, = Z f ij P. f where repeated indices indicate summation. (2)jijf j Pi

When the structure is linear elastic, the values of AE, EI, JG, and AvG do notchange with load or temperature stimuli and linear simultaneous equations can be set upto solve for the desired quantities (see Paragraph 4.2.4).

2.1.3.2 Addition of Flexibility Coefficients

Equation (1) of Paragraph 2.1.3.1 can be visualized as a group of springs in series.The energy content of a spring is directly proportional to its flexibility and the approximation1 1 1of ignoring all energies but the bending energy is equivalent to assuming 0 = - JG GThe flexibility of a structure can be analyzed by superposing the flexibilities of structures Vwhere all flexibilities except one are assumed zero as shown in Figure 2.1.3.2-1.

AXIAL BENDING TWIST SHEARLf.f.IfijP -J E j i dx + 0 + 0 + 0 Axial Energy

Lm.mf.ijM 0 + -] dx + 0 + 0 Bending EnergyfijM~ ~ = T+ 1

f. = 0 + 0 +f 1 dx + 0 Twisting Energy0 JG

0 + 0 + 0 +0 Lv dx %ear Energy

v

fij f iijP + ijM + fjQ + fi Total Energy

FIGURE 2.1.3.2-1 SUPERPOSITION OF FLEXIBILITIES

A structure (e.g., a cantilever) can be analyzed by the solutions of individual portions(springs) which are then placed together by satisfying compatibility and equilibrium at eacb of thecommon contacts. This is also equivalent to analyzing the over-all structure where all flexibi-lities but one portion are identically zero and adding the solutions as shown in Figure 2.1.3.2-2.

WADD TR 60-517 2. 12

2.1.3.2 (Cont' d)

OR b,M a b

(a) Actual Structure

1 - Ua booo ~ y a aU ,Fa +Fb 1 •F

'(M+MbF L+ -Fb L) Mb

S• (FbL+M)(F L Mb)

(b) Series Components of Actual Structure

LI

b *b

fb

+ Ma MbS+b

"Rigid a b

FM

(c) Supcrposable Equivalents

FIGURE 2.1.3.2-2 EQUIVALENT STRUCTURES IN SERIESWADD TR 60-517 2. 13

2.1.3.2 (Cont' d)

In Figure 2.1.3.2-2(a) and -2 (b) the actual structure is shown to be equivalentto two substructures 1 and 2 connected in series. Substructure 1 has the same displacementdatum as the actual structure, with end loads equal and opposite to the reactions of sub-structure 2. Substructure 2 has the same loads as the actual structure but the displacementsare corrected for a rigid body motion corresponding to the translation and rotation (ua) of

the end of substructure 1. The same results and calculations would be obtained by adding thesolution of the two rigid flexible beam problems shown in Figure 2.1.3.2-2(c).

The elemental substructures are much easier to analyze than the over-all structure.The loads on the substructures (F) are expressed in terms of the applied loads on the actualstructure (F) by means of the equilibrium equations.

{F=B ]fFl= 1010 F5

1 L 1 Ma (1)

010 Fb

The displacements of the actual structures can be similarly expressed in terms of the loadsand flexibility of substructures.

1 1A = f (F +F) + fa, (M +Mb +FbL)a aa a b aa a+M b+F

a 1fat a (Fa+ Fb) + lfat a' (Ma * Mb + FbL) S(2)

a a+L + 2 fbb b bb, Mb

)b Oa fb'b b + fb'b' Mb

In matrix form:

a if if 1 1 + L F aaaa aa ~ aa +Lfaa' IaaII i I l"i

~~- - --- - -r - -- _- . . . . - _ _ -laa fata a I faa +Lf +a 2a If +a2a M

f~ If fa +L 1f 2 Ii 2f1i= aa aat aa aat + fbb fat 'A bb?

+L I f.L +Lf +L I+L Fa' a a' a't a' a fa, aa a '

1 1 1A 1 2 1 2 Mfata I fat at fat a +L 1 ata + fbtb' fatat fbtb? bJ

[f I JFJJ (3)

WADD TR 60-5 17 2.14

2.1.3. 2 (Cont' d)

The above influence coefficients of the actual structure can be expressed in terms of theinfluence matrices of the substructures and the matrix expressing the loads on the sub-structures in terms of the applied loads, as follows:

faa faaf 0 0

f a fat a' 0 0

[fi] = [A'] 0 0 2f [x] (4a)

2 20 0 2fbb 2fbIb'

where [Al'] is the transpose of [A].

The above transformation of the influence coefficients matrices of the substructuresto the influence matrix of the actual structure can be readily verified by utilizing the fact thatthe total work done on the substructures is equal to the total work done on the actual structure.

Thus

But from (3)

{[f ]{F

therefore:

{Fij} [ ij~ {Fj} I {=fk} F (4b)

From Eq. (1),

{-fr} = {rj} {FJ}

so that

Substituting the above in Eq. (4b) yields

{ F'ý I [fij] { Fj} f {F'j[ý [ý] [An]j {Fj}

or

[fij] = [ki] [fKr] [rj]

F3

WADD TR 60-517 2.15

2.1.4 Applications of Virtual Displacement Techniques

The loadR on the structure and the stiffness coefficients can be obtained by applying

the principle of virtual displacement. The stiffness coefficient K,1 is defined as the force (ormoment) at a point j when the structure is subjected to a unit displ9cement Rt point i.

2.1.4.1 Load on a Structure

Apply a virtual displacement pettern which does not -Aolate compatibility and isunity at Fi and zero at all other points where other forces act (see Figure 2.1.1.1-1(c)).

From Eq. (lb) of Paragrapr, 2.1.1.2,

6U F,= a 6cdV.

For beam-like structures using the equivalent internal loads and deformation of

LEq. (4) of Paragraph 2.1.2,

6U=F =rf [P 5 - M 6K + Q6 T + V(5y)lIdx, 1

where the changes in the deformation are due to the virtual displacement and the loads actingon the cross section constitute the self-equilibrating thermal stress system.

For beam-like structures the bending energy predominates and the following is obtainedas an approximation

LF i M6 Kdx. (2)

Thus, to obtain the force at a point in the structure, evaluation of the virtual work due to thevirtual displacement is required. 2his is equal to the internal forces in the structure actingthrough the virtual internal deformation. This procedure is employed in Section 9, "Stability ofStructures."

2.1.4.2 Stiffness Coefficients

Applying a displacement at point j causes loads to be generated at j and points wheredisplacements are kept zero. These forces are the counterpart of the flexibility coefficientswhich give the deflection at points of interest due to a unit load at a given point and zero loadeverywhere else. The stiffness coefficients are the loads at points of interest due to the unit

displacement at the given point and zero displacement a! the other points of interest. Theelements of the stiffness coefficients matrix are the loads which result in the unit deflectionmatrix and can be obtained by inverting the flexibility coefficient matrix. The forces at thedatum can be obtained from the equilibrium equatior. Similarly, the influence coefficient Imatrix can be obtained by inverting the non-singular stiffness matrix. The non-singular matrixis obtained by removing the elements associated with the datum of the structure.

(1a)fij 1KJI

Sf -1 (1b)

f -11

The stiffness coefficients for an elemental beam can be obtaJned by inverting the flex-ibility coefficients as shown in the following example.

WADD T.R 60-517 2.16

2.1.4.2 (Cont'd)

Example - Cantilever Beam of Constant El (Bending Energy Only):

a f L= I a Ina dx x2 L3•'• -•• __•saEl = E0 l 3EI-'d

La f amaa L (2)

a f~ a EI

where prime (a') refers to rotational loads or displacements at point a.

L3 L 12 6E _Kaa Kaa2 3 2 3 2

E L L (L3)

6E El L E 6 4K K' L L 2 L

L 2

From equilibrium and symmetry the entire stiffness matrix can be obtained for the loads whichare required when a joint is moved a unit displacment and all other joints are fixed. Ibis isshown in Table 2.1.4.2-1

TABLE 2.1.4.2-1 STIFFNESS COEFFICIENTS OF BEAM-LIKE STRUCTURE

Displacement at Joint

12EI 6EI 12EI 6EIL3 -- 2- 3 L2

t •) 6EI 4EI 6 El 2 El

2 L 2 L

LA L

]() 12EI 6EI 12EI 6EI

L 3 L 2 L 3 L 2

(), 6EI 2EI 6EI 4EIL 2 L L 2 L

2.1.4.3 Addition of Stiffness Coefficients

If two or more elements of a structure have a common point (joint) of interest,then the load necessary to obtain the displacement at that point is the sum of the forcesnecessary to apply the displacement to each element. This is analogous to a set of springsin parallel.

WADD TR 60-517 2.17

2.1.4.3 (Cont' d)

It is important to maintain a sign convention wHich is invariant in space for thestiffness coefficients since the forces of elements with a common point may be so orientedthat the load in the adjacent elements may be of opposite sign (References 2-1 and 2-2).

Coupled stiffness coefficients can be obtained by repeating basic stiffness coef-ficients and adding those terms which are in common. Thus the stiffness at a joint is addi-tive (the springs -are in parallel) and the total stiffness at a joint is the sum of the stiffnessesof all members coming into the joint. This is shown in Table 2.1.4.3-1 and Figure 2.1.4.3-1.

TABLE 2.1.4-3-1. COUPLED STIFFNESS COEFFICIENTS

Displacement at Joint

00 I12E11 6Ek 12Ei ik I

S3 2 3 2L• L LI L1!

6EliI 4E 1 1 6ElII 2EliI127 2 2 L1 I 0 0

L L L1

12L 6L 111 L1__ _____ __1 ___ __L3L26EL1h 26E21 1E 1 6 12E 212 4E1 4E2 6E22 2 12E2 2 6E2123 2 L 2 2 2 3 2

L1 L1 L 1 L2 1 2 j L2

6E_1h 2E,1 1 6E I2 6E212 4E61 1 4E2I 2 E2E2 2 2E2! 23L 2 2 L 2L L2 1 2 L 2 22

I12Ef 6E I 12EI GE I00- 22 22 2222003 2 3 2

2 ~ 2E219 0 E22 22 6E 2 I2 4E 2 I20 2 2 LL2 L2 LL

22 2 2

WADD TR 60-517 2. 18

2.1.4.3 (Cost'd)

12E 1

3L I

7 ~6E 1'

+J6E 1 2Ei1

1__1 1 12 L

L1

12E 1 12E I 6E 11 6E2I1h 22 E__ 2'2

3 3 2 21 L2,

-E211 4E 212A A 11

6E1 6E12 L1 L2

L12 2L2

1 2

-- I2

12E 2 I2 6E212

L2223 2

6E ~ _L_ 2E 1 236E 212 2 2

LL2

FIGURE 2.1.4.3-1 STIFFNESS COEFFICIENTS FOR A DOUBLE BEAM

2.1.5 Properties of Influence Coefficients Matrix

The influence (flexibility and stiffness) coefficients are utilized in the solution ofindeterminate structures (e.g., Sub-section 4.3). The coefficients can be obtained by calcula-tion (References 2-1 and 2-3) or by experimental techniques. It is possible to employ availablestructural relationships to simplify or check the calculation or experimental determination ofthese coefficients. The resalts are presented for flexibility coefficients but apply equally wellto stiffness coefficients.

WADD TR 60-517 2.19

2.1.5.1 Symmetry

The deformation of a point I due to a unit load •tt a second point j is equal to the de-formation of the second point j due to a unit load at the original point I. The deformation(linear or rotational) at a point corresponds to the type of load (force or moment) at the otherpoint.

fiJ =fji (la)

"Aj =Ai ; i'=i (ib)

e. = A.. ; O A j , (ic)ijt 13

where Aij = linear displacemement at i due to unit linear load at J

oi = rotational displacement at i due to unit load at j

A.., = linear displacement at I due to a unit moment at j

= rotational displacement at i due to unit moment at jThe symmetry of the matrix is evident by noting the equivalence of the energy terms

similar to f xs aL (mi)(mj) dxL (m )(mi)

0dx = oJ dx .

This symmetry can be utilized to reduce the number of calculations or to average the

experimental results.

2.1.5.2 Positive Definiteness

The fact that each structure absorbs energy in deforming requires that the displace-ment of a point on a structure cannot have a component which is opposite to fie direction of theapplied load. This is demonstrated by the positiveness of the energy terms similar totL(mil2 dx

EI) . Thus the influence coefficient defining the deflection of a point due to a load at

that point must never be negative (always positive except for the reaction for which it is zero).Thus the main diagonal of the influence coefficient matrix must be non-negative,

f.. > 0. (1)11

Another relationship derivable from the positive definiteness of the structural matrixis that the product of two main diagonal elements must never be less than the square of the cor-responding off-diagonal element.

"'" fii 2t (f iJ)2 (2)

This is evident from the Schwartz inequality,

Ef I )(m ) d (dx) (3)

WADD TR 60-517 2.20

.4:

2.1.5.2 (Cont'd)

The positive definiteness property can be employed to check on the accuracy of theexperimentally determined values.

2.1.6 Equivalent Thermal Load

The linear elastic structure permits a solution by the superposition of variousloading conditions as long as the total effects add up to the original problem. This approachis particularly adaptable to structures subjected to temperature. The effect of a temperatureis to cause an axial expansion of each elemental volume of the structure. The stresses whicharise due to temperature are the self-equilibrating stresses necessary to make each elementalvolume compatible with the adjacent elemental volumes and the boundary conditions.

There are basically two methods of solving the equivalent mechanical loading pro-blem for the linear-elastic structure subjected to both temperature and mechanical loading.In each method, the structure is first decomposed into elemental units. The behavior ofeach elemental unit for temperature and mechanical loading is readily determinable. In thefirst method, the elemental units are permitted to deform independently of each other and thedisplacement of the boundaries of adjacent elemental units are observed. Attempt is then madeto put the decomposed structure together again. This requires that the elemental boundariesbe compatible with each other. Self-equilibrating force systems (internal forces) are applied,wherever necessary, at adjacent boundaries to guarantee compatibility. For beam-like struc-tures, this is called the flexibility method. In the second method, the elements are restrainedfrom deforming when subjected to the temperature and load by infinitely rigid restraints. Themechanical loads generated at the restraints are supplied by the structure if this restraintactually exists. If the restraint does not exist then the boundaries of the elemental units arepermitted to deform (always maintaining compatability) until the load in the artificial restraintvanishes. For beam-like structures, this is called the stiffness method. These techniques aredescribed for beam-like structures in Sub-section 4-3.

The solution for beam-like structural units is usually expressed as influence coef-ficients, that is, in terms of the deformation due to unit type loading, or the loading due to aunit type deformation. It is advantageous, therefore, to convert a thermal loading to an equiva-lent mechanical loading so as to treat a temperature problem as an equivalent mechanical pro-blem and simplify the amount and type of calculations. All the structural techniques availablefor mechanical loads could then be employed to solve the structure subjected to mechanicalloads and temperature.

2.1.6.1 Three Dimensional Hydrostatic Loading

Assume that the body is cut into elemental volumes which are then supported withinfinitely rigid restraints so that compatibility between the elemental volumes is maintained.Application of temperature distribution to the structure will now try to expand these elementalvolumes which will be counteracted by hydrostatic stresses equal to - - 2 at the artificialI - 2z/

-EcaVTrestraints. The artificial restraints must supply the difference 1 - 21/ of the restraining

hydrostatic stress loads applied to adjacent elemental volumes. Since the restraints are arti-ficial, the structure must be allowed to deform so as to balance out these "equivalent bodyforces." By this means the original problem has been changed into two problems. The stressc.

in the first problem of the artificially restrained body are simply:

WADD TR 60-517 2.21

- ~ --.-. "I- =

IIt2.1.6.1 (Cont' d)

a (1 a~ EaTxx yy zz 1 -21.V

(1)1o 1 1a = -or a 0xy yz zx

The second problem is the classical linear elastic problem with body forceswhich are proportional to the thermal gradient.

F Ea 3x 1-2z2 ax

F Ea 8Ty 1-2v Dy (2)

FEct aTz 1-2Y/ az

and the boundary conditions,

+ y + 2 o EaT,ox yy xz z 1 -2y x

2a. + 2 o n + on = n T3) (3yxx yyy yz z 1-2v y

2az + 2o + 2 .n = EaT

zy zzz 1-2v

The equilibrium equations,

a a)(2 ) (2

)x a 2 ._ _ _

+ vy + - E2 y =0 (4)

2 +-,-(2x + 1 - E2 OTY aaO y &z 1- 2v ay

and the strain displacement (or compatibility) equations (Paragraph 1.1), together with thestress-strain relations (Paragraph 1.4), provide the necessary and sufficient equations fora solution.

The final solution is the sum of the individual solutions,

1 2 sS+20 (5)

Thus the temperature problem is the mechanical equivalent of applying a hydro-static loading, proportional to the temperature, to all the elements and the boundary of thestructure and then apply body forces to the structure, equivalent t, th. gradient of the hydro-

static loading, along with boundary forces which negate the 1 boundar 'y forces.

WADD TR 60-517 2.22

2.1.6.2 Fixed End Beam Reactions

If the behavior of the elemental structures when subjected to mechanical andthermal loads is known (e. g., determinate beams of Section 4) within acceptable engineering

accuracy, the artificial restraints (Jacks) -.eed not be constructed at each elemental volumebut only at the common boundaries (joints) of these larger elemental structures. First thethermal stimulation is applied to the structure. Each joint of the elemental beams will tendto deform but will be restrained by the artificial Jacks. The loads introduced in the artificialjacks are commonly termed fixed end reactions. The artificial jacks must supply the unbalanceof the fixed end reactions coming into the common joint from the various structural elements.The original structural problem is thereby reduced to two simpler problems. The first ',ro-blem is the solution of the elemental (fixed end) beams which are restrained from any motionat the ends. The solution of this type of problem is indicated in Sub-section 4.2. The secondproblem is the deformation necessary to eliminate the artificial restraints and internal loadsin a structure with mechanical loads equal to the negative of the unbalanced fixed end reactionsacting at each joint. The solution of this type problem is indicated in Sub-section 4.3. Thesolution of the original problem is thus the sum of the solutions of the two simpler problems.Reference 2-2 indicates that the deformation at the joints of the structure, due to the appliedtemperatures and loads, is identical to those calculated by the equivalent fixed end reactions.The relative deformations of the elemental beams between the joints must be added to the jointdeformations to obtain the total deformations.

2.1.7 Plane Stress and Plane Strain Problem

The three dimensional problem, in general, requires the determination of 15 quanti-ties (6 stress, 6 strain, and 3 displacements), given the body forces and the boundaryconditions. This problem can be greatly simplified if the given geometry or loading makes someof the unknown quantities zero (or insignificant). This is the case in the plane strain and planestress problems.

The state of plane strain is defined as one in which the displacement component in agiven direction is zero (e.g., z di, ection, w = 0) and the other displacement components areindependent of this direction (u z = v z = 0). This condition arises in a long (axial dimen-sion large with respect to cross sectional dimensions) prismatic body under loads and tem-peratures which are independent of the axial coordinates.

The state of plane stress is defined as one in which the stresses acting on faces

parallel to a given plane are zero. This condition arises in a short (axial dimension small

with respect to cross sectional dimensions) prismatic body, e.g., a flat plate, whose faceare unrestrained and unloaded. The axial dimension is sufficiently small so that the axialstress cannot change significantly.

Solutions can be obtained by an "Airy" function which satisfies the equation ofequilibrium identically and is made to satisfy the compatibility equations. A generalsolution can be obtained by the sum of such functions (which also satisfies compatibilityand equilibrium) and whose total boundary condition duplicates the actual boundary con-dition. The St. Venant' s principle (Paragraph 2.1.8) can be employed to obtain approxi-mate solutions which are satisifactory at distances away from the area where the boundaryconditions are not satisfied identically.

2.1.7.1 Plane Strain

The technique is demonstrated for a rectangular coordinate system out is equallyapplicable to other orthogonal systems.

WADD TR 60-517 2.23

2.1.7.1 (Cont d)

Plane strain is defined by the condition w = 0; u = u(x,y) ; v = v(x,y).

From Table 1.6-1 the following strain conditions are obtained:

EC = x xxy) ; E = Xy x,y); Ix = (x'y)'cc ocYY yy xy xy

42 I E =0.(1zz =xz yz

Assume that the body forces are derivable from a potential function (-V), (e.g., gravity,temperature, etc.). Thus -V F,x x

From Table 1.6-2, since a = = 0, the equilibrium equations are obtained,yz xz

(xx -V) x +yx'y 0 (2a)

, + (a - V), = 0. (2b)xy, x yy

From Eq. (1) of Paragraph 1.1.1 the compatibility equations reduce to

xx,yy + 4yy,x -x 2 xy,y = 0, (3)

and from a linear stress-strain relationship this becomes

E 1V0 -V ac Jo ] 2 c l=0 (4)E (1• xx •yy ,yy yy xx ,'xx - xy,xyI

The Airy function (pis defined as

Ux -v = ,yy, U -V 9',xx -x = ',x' (5)

which satisifies the equilibrium equations identically and substituting (5) in (4) results in0+ 2+ V, +I- +I +V , (6a)0 D = ,xx 2OxxyT Y yyyy V ,xx ,yy

or

V4 ( 1 --2- V2 = 0 (6b)

which satisfies the elastic compatibility condition.

An Airy function satisfies the equations of equilibrium and linear compatibilityand each one will determine characteristic boundary conditions.

The potential (-V) can be viewed as the restraint forces generated by the artificialrestraints to maintain compatibility by restraining the elemental volumes when stimulated byload or temperature. In a gravity field it is simply the density and in a temperature ield it

is the hydrostatic stress ( -

The temperature problem reduces to

v 4 9 + E V 2 T=O N

WADD TR 60-517 2.24 14

1j4

2.1.7.1 (Cont' d)

and for zero mechanical boundary tractions,

S= = o. (8)

211In polar coordinates, V T =T + r T +- T (9a),rr r ,r r 2 GO

and

24 2 + 12+ r - 2+ 2 41. (9b)

ar r a2

The total solution is the sum of the stresses obtained from the above formulation with

the stresses from the restrained conditions 01 =y = z - EaT

2. 1.7.2 Plane Stress

o•Ya ned In the case of plane stress, the technique is similar. Utilizing a_ = a iand the equilibrium, compatibility, and stress-strain relationship, tf foilwO g isS~offalned:

S(4 12 )

For thermal stimulation it is no longer necessary to restrain the thermal ex-pansion in the axial (z) direction since the structure is free to expand in this direction.The necessary artificial restraints are just the xz and yz planes without the xy plane.

The potential or equivalent hydrostatic pressure becomes (-V) = -c 1 Thethermal plane stress problem then becomes

V 4 + cE V2 T = 0 , (2)

¶ and for zero mechanical boundary tractions,

0•= , 0. (3)

The total solution is the sum of the stresses obtained from the above formulation with

the stresses from the restrained conditions a = - EaTyy 1-v

2.1.8 St. Venant' s Principle

In many instances the solution to the problem can be simplified by obtaining anapproximate solution In which the surface tractions (which may be impossible to prescribe)are not satisfied identically. In general, the engineer only knows the resultant force andnot the traction distribution. "If the local applied tractions are replaced by a staticallyequivalent load system, then the solution will be satisfactory at points sufficiently removedfrom the region of application of the local tractions." This is known as the "St. Venant' sPrinciple" and is extremely useful when the exact solution is too complex (or may not besolvable with available mathematical tools). This principle is employed in Section 4 % herethe zero tractions on the free end of the thermoelastic beam were replaced by a self-equili-brating thermal stress system. The solutions are satisfactory at a cross section away(approximately 2 to 3 depths) from the free end and a local approximation is employed inthe vicinity of the free end rather than attempting an exact solution.

WADD TR 60-517 2.25

2.2 NON-LINEAR STRUCTURES

The necessary conditions that a structure will be linear-elastic in going from itsoriginal state to its final state is seldom fully realized in an actual airframe design whereminimum weight is required.

The advent of high speed venicles has forced the designer of airframe structuresto re-assess the non-linear effects of stress, temperature and time upon the structuralmaterial. Formerly the design was based on a linear elastic analysis of the structure be-cause of the simplicity and wealth of analyses in the literature employing classical"elasticity" (linear) theories.

An examination of experimental data, however, revealed some discrepancies betweenthe linear analyses and the actual results. In some cases the analysis underestimated themaximum load while in other instances it overestimated the load.

The problem of a beam in bending is a well known example of underestimating themaximum load. The classical linear theory results in a stress distribution which is propor-

tional to the distance from the neutral axis (or - ) and the maximum allowable momentI (A

(MA) would occur when the maximum allowable stress (arA) is attained M = 1

This value, obtained by assuming a stress distribution, is significantly lower than t!& experi-Or I

mental value of MA = k A (modulus of rupture) by a factor of k > 1 whica de-SYmiax

pends upon the geometry and the stress-strain curve of the material. The discrepancy is pri-marily due to the fact that the linear analysis results in a significant stress gradient which willbe reduced as the stresses approach the maximum allowable stress due to the actual stress-strain curve of the matt rial. Additional examples are stress-concentrations, thermal stresses,interaction stresses of shells, etc. The problem of the buckling of a structure is a well knownexample of overestimating the maximum load. The classical linear theory assumes a stressdistribution with a constant upper bound on the modulus. The calculated value can be signif-icantly above the actual value which depends upon the stress level and the stress gradient. Thisis qualitatively discussed in Section 9 on the effect of plasticity and eccentricity upon stability.Additional examples of overestimating maximum 16ad is the ignoring of stress concentrations,fatigue damage, etc.

The designer was forced, therefore, to modify the linear analysis by introducing aplasticity factor to account for the non-linearity of the stress-strain relationship. This pro-cedure is fairly successful when the effects of time and thermal gradients are negligible andthe non-linear behavior of the structure can be estimated from the experimentally determineduniaxial (short-time) stress-strain relationship at the design temperature.

The problem of the effects of a general stress-temperature-time history upon a structureis quite oomplex and available analyses are usually lacking in reliability and ease of application.Attempts to employ modified linear procedures are fraughtwith danger and may lead to excessiveerrors in estimating the life of a structure. The errors can occur in either direction; either inunderestimating the structural life and resulting in an excessively heavy structure, or in over-estimating the structural life which would cause a premature failure. Any attempt to experi-mentally determine adequate empirical analysis procedures would be defeated by the tremendousamount of test time and the high costs necessary to determine the effect of all possible stress-temperattwe-time histories upon various geometric configurations. The time available to dothis experimental work before a design must be selected would be far from adequate in the rapidlyidvancing technology of materials, structures, etc. It is imperative, therefore, that all analysisnethods be based on fundamental concepts and simple procedures which best approximate thetvailable experimental data. The empirical design factors should be obtainable with as little

VADD TR 60-517 2. 26

a-

.- I.

2.2 (Cont' d)

experimental data as possible and the analysis procedures should be presented in as simplea manner as possible to minimize the amount of labor required to analyze the structure. Anon-dimensional approach should be investigated in order to reduce the amount of experi-mental data and analysis curves required.

An incremental time technique as described in Section 1 can be employed in specialcases. In general, however, the amount of necessary computations will be prohibitive.Approximations based on deformation theory could be attempted to make the problem solvableand the approximations would give a clue as to whether the solution overestimates or under-estimates the true behavior of the structure. The effect of approximations (neglectingenergies, approximating variables by constants, additional assumptions, etc.) can be morereadily evaluated by simplifying the non-linear general equations than by trying to modifythe solutions of over-simplified linear equations.

The equations of equilibrium and of compatibility, it was shown, do not depend upona stress-strain relationship. They can be obtained from energy principles alone (virtualwork). The equations of equilibrium are basically a relationship of the stresses, and theequations of compatibility are a relationship of strains. The stress-strain relationshipsare the additional conditions necessary to combine the equations cf equiibrum and com-patibility in terms of the stresses or strains alone. The material behavior is discussedin Section 3 to obtain reasonable and non-dimensional stress-strain relationships.

Methods of analyzing non-linear structures are briefly discussed in the followingparagraphs. Most non-linear problems are considered beyond the present scope of thisManual. The non-linear problem of stability is presented in more detail in Section 9 be-cause of the great need, the utility, and availability of approximate methods.

2.2.1 Incremental Linear Solutions

Incremental linear solutions were briefly discussed in Section 1. The techniquewill give the correct results.provided the incremental history is sufficiently small and thestructure is elastic. An elastic structure may be linear or non-linear and is defined bya stress-strain relationship which is uniquely defined so that a reversal of stress remainson the initial loading curves. The incremental technique is equivalent to approximating thestress-strain relationship by a series of straight lines and solving a set of linear elasticstructures with different geometries and stiffnesses. The necessity for recalculating thegeometry or including higher derivatives depends upon the effect of these parameters andthe required accuracy. The amount of calculation and accuracy are inversely related andgood judgement is necessary to obtain a sufficiently accurate solution without prohibitivelylarge numbers of calculations. This technique can be readily adapted to a digital or analogtype of calculation machine.

2.2.2 Inverse Solution

The inverse technique can be employed to solve for the loading conditions whichresult from an assumed displacement and corresponding compatible strain distribution.This technique can be applied to linear or non-linear problems. The stress distributionand the loading can be obtained byutilizing the known stress-strain (linear or non-linear)relationship with the assumed strain distribution. The St. Venant principle (Paragraph 2.1.8)can be employed to extend the solution to structures in which the boundary loadings are notidentical but statically equivalent.

WADD TR 60-517 2.27

2. 2. 2 (Cont' d)In many cases it is possible to determine the correct displacement or strain

distribition pattern by employing the principle of minimum complementary potential(Paragraph 1.7.2.4). This procedure was utilized in Reference 2-4 to prove the existenceof a linear axial strain distribution in a prismatic beam subjected to terminal loads.General solutions can be obtained for beams subjected to axial loads, transverse loadsand moments by assuming linear axial strain distributions and employing non-dimensionalstress-strain relationships (Section 3). The applied load can then be plotted versus a suit-able strain parameter for various material parameters.

Assuming Airy functions (Paragraphs 2.1.7.1 and 2.1.7.2) results in inverse solu-tions of the linear plane strain and plane stress problems for the boundary conditions satis-fied by the Airy functions. A combination of several Airy functions can sometimes beemployed to obtain boundary conditions sufficiently close to the actual boundary conditions.

The inverse technique is the upper bound approximation technique described inParagraph 2.2.3. When the correct strain distribution pattern is assumed, the approxi-mation solution is the correct solution.

2.2.3 Approximate Solutions

The principles of minimum potential and complementary potential energy as de-scribed in Paragraphs 1.7.2.3 and 1.7.2.4 do not depend upon the stress-strain relation-ship. They can be employed to obtain upper and lower bounds to the solution of linear ornon-linear problems. For example, assuming an incorrect displacement or strain distri-bution which satisfies compatibility and the boundary displacement conditions will under-estimate the stiffness of the structure, i.e., will overestimate the strain energy (expressedin terms of the disDlacements) as weil as the work done by the surface and body forces. Ifa virtual displacement is now applied to the incorrectly assumed displacement pattern, thecalculated change in strain energy Is greater than the work done by the true forces in actingthrough this virtual displacement. If only one force is acting (.e.g., axial load on column),the calculations will result in an upper bound of the applied force if the change of strainenergy is expressed in terms of the displacements. If the change of strain energy can beexpressed in terms of the forces, a lower bound of the applied force can be calculated.

Assuming an incorrect force or stress distribution, which satisfies equilibrium andthe boundary force conditions, will overestimate the stiffness of the structure, i.e., willoverer.timate the complementary strain energy (expressed in terms of the stresses) as wellas the complementary work done by the surface and body forces. Applying a virtual forcesystem to the incorrectly assumed force system can result in a lower or upper bound on thedisplacement by expressing the change of complementary strain energy in terms of the forcesor displacements, respectively.

The above techniques are employed in obtaining the stability of structures (Section 9).They can also be employed to analyze plastic beams subjected to axial and bending loads (inwhich linear strain distributions are assumed across the beam cross section). Anotherapplication is the determination of an upper bound to the ultimate load of an axial tensionmember subjected to a stress concentration such as a hole. The strain distribution whichcorresponds to the linear elastic solution to the problem satisfies compatibility even whenmultiplied by an arbitrary constant. Employing this strain distribution with a non-linearstress-strain relationship, such as.presented in Section 3, will result in stresses and anintegrated axial load. Assuming a maximum strain criterion of failure will result in anupper bound on the axial load when the maximum strain is attained at the "stress" concen-tration.

4WADD TR 60-517 2.28

44 4

'4

2.3 REFERENCES

2-1 Meissner, C., Stress Analysis of Aerodynamic Surfaces Using Method ofElastic Coefficients, Republic Aviation Corporation Report ESAM-21,July 1958.

2-2 Switzky, H., Solution of Indeterminate Beam-Like Structures, RepublicAviation Corporation, RDSR-2, to be issued.

- 3 Switzky, H., Influence Coefficients, Republic Aviation Corporation ReportESAM-22, August 1956.

2-4 Switzky, H., Determination of Mechanical Loads and Thermal Stressesfrom Experimental Data, Republic Aviation Corporation Report RDSR-5,to be issued.

.6

SWADD TR 60-5 17 2.29

-4

.4

A

1.

-.4SECTION 3

.4T.

MATERIAL CONCEPTS4-A

4:.4,

-4

4

-4

44

-4.4-

.4,I,

I.

.4.4

.4 ii

-.4""-.4

WADD TR 60-517 3.0

- � I&�A�--� � ____

SECTION 3

MATERIAL CONCEPTS

TABLE OF CONTENTS


3 Material Concepts 3.2

3.1 Deformation Mechanism 3.4

3.2 Stress-Temperature-Time-Strain Relationship 3.5

3.2.1 Uniaxial Stress-Strain Reiationship 3.6

3.2.2 Experimental Determination of the Material Constants 3.9

3.2.2.1 Apparent Modulus (EA) 3.9

3.2.2.2 Slip Stress (ao) 3.10

3.2.2.3 Work Hardenability Factor (K= Kt Kr) 3.13

3.2.2.4 Interpolation of Material Parameters 3.13

3.2.3 Complex Loading 3.14

3.2.3.1 Invariants 3.15

3.2.3.2 Transformation of Stress-Strain Relationship 3.17

3.3 References 3.19

OF

WADD TR 60-517 3.1 .

F SECTION 3 - MATERIAL CONCEPTS

The ability of a structure to withstand the applied loads is dependent upon the materialproperties as well as the structure geometry. The material properties are sensitive to thetype and magnitude of loading, the temperature, and the chronological history (time) of thesefactors. The problem would not be so complex if the principle of superposition could be ap-plied. Unfortunately, this Is only approximately true and then for only small stresses, tem-peratures and times.

In general, the structural stiffness will be different from point to point in the structureand will change with load (stress). temperature, and time. The structure becomes an ex-ceedingly difficult non-linear problem. Recourse is made to deformation theory and approxi-mate methods to solve the problem as if a series of incremental changes occur, each of whichis a line:." problem with a unique solution until the total loading history is applied.

The exceedingly large number of materials and load-temperature-time histories thatare possible makes it absolutely imperative to try to represent the behaviour of structuralmaterials as functions of a iew material parameters. From a practical design analysis pointof view, it is desirable to limit the number of parameters while still retaining an acbeptableengineering approximation to the strain relationship. This would permit a relatively simplemathematical expression to interrelate the equilibrium and compatibility equations of Section 1(which are independent of the stress-strain relationship) to obtain structural solutions. It isalso desirable to present the strain relationship in a non-dimensional form to minimize thecomputational work and to permit non-dimensional graphical solutions of the structural problem.This technique can be utilized in the design of a structure with minimum guaranteed properties.

The stress-strain relationship suggested in this manual is predicated upon empirical re-lationships which approximate experimental data. A deformation mechanism is assumed whichdoes not contradict experimental evidence and which leads to three (temperature dependen')parameters for the short time stress-strain relationship and in addition, two additional para-meters to include the effects of time. Methods of determining these parameters from simpletest data are indicated.


t Time, secondsC Short time (instantaneous) measure of slip (CIKT sinh ar/6) in which no primary

1 T 0or secondary -'reep occurs, non-dimensional

02 Measure of primary creep (C2(1-e) KT slnh a/a) due to "relaxation," etc.non dimensional

C3 Measure of secondary creep (C3 t KT sinh a/%r0 ) as the rate of "strain hardening"is balanced by the rate of "softening by recrystallization," 1/sec

C4 Material constant representing the number of particles per unit time exceedingthe thermal potential barrier

E Elastic modulus (material-temperature parameter) associated with hydrostaticstress and dilation, psi

EA Apparent elastic modulus of material (obtained from short time stress-straincurve), psi

E Secant modulus = ale (psi)

WADD TR 60-517 3.2

1

r

ET Tangent modulus = do/d e (psi)

K Time-temperature work hardenability material parameter directly proportionalto slip = (Kt) (KT). non-dimensionalTime component of K, non-dimensional

KT Temperature component of K, non-dim ensional

KT C4e Q/T (Suggested by Maxwell-Boltzmann Energy Distribution, Reference 3-2)

L. M. Larson-Miller parameter = T (C + log t)Q Material constant - activation energy per gram atom divided by the universal gas

constant, °K

KE =temperature, time, material parameter (f = 0 elastic material;A/fo= I viscous material)

6 Deviation =e-O/EA

AU Stress incrementC Strain, In/int Strain rate, in/in/sec (dot over symbol indicates derivative with respect to

time, e.g., kt, &)

E Instantaneous strain due to stress= qiE + CIKT sinh ir/ , In/in

ar Applied stress, psiStress associated with slip. This is a material -temperature parameter which isa function of the past history of the material, i.e. , stress-temperature-time, etc.,psi

Subscripts

1,2,3, Principal directionsI, II, I1 Invariant values of stress and strainI Dummy index

Superscripts

Prime Deviations, e.g.; a' and C'1

WADD TR 60-517 3.3

3.1 DEFORMATION MECHANISM

Deformation is viewed as the movement of atoms from one equilibrium position toanother. The energy potential necessary to overcome the potential barrier, i. e., to ini-tiate and continue the motion, is absorbed from the environment; stress potential, tem-perature, etc. (See Figure 3. 1-1). This movement, or slip, occurs initially along planesin crystals where the stress and temperature potential first attains the value of the potential"barrier. The external potential (temperature and external loading) necessary to initiatethis slip is reduced by dislocations along the shear plane which locally increases the stresspotential. The dislocation then migrates to the boundary of the crystal at which time thedeformation along the "favored" planes ceases. No more deformation can occur until theenergy potential is increased. This phenomenon is known as "strain hardening". If thethermal and residual stress potential is quite small, a reapplication of the applied load wouldindicate no significant slip would occur until the applied load exceeds the previously appliedload.

Continuing to increase the applied load causes planes in other crystals to attain thenecessary potential to slip. Simultaneously, the movement of dislocations to the crystalboundaries increases the internal stress potential at the boundaries which. when coupledwith the external potential (thermal and stress), causes the deformed crystal to grow at theexpense of its neighbors. This phenomenon is known as "softening" or recrystallization".The growth is in the orientation of the "favored" crystal so that the initial slip can now bepermitted to continue by moving the dislocation to the new boundary of the crystal (whichcontinues to expand). Thus the total slip is a function of the previous slip and implies anexponential type of stress function (Reference 3--1). The amount of slip is extremely sen-sitive to temperature (thermal energy). Statistical thermodynamic considerations indicatethat an exponential expression which expresses the probability that the atoms will have therequired thermal energy would define the effect of temperature on slip (Reference 3-2).

Anothe" factor should be considered. As the deformation proceeds. it induces-esidual stresses in the surrounding crystals which then seek a state of equilibrium corres-ponding to a lower energy state. This phenomenon is known as "relaxation" if it occurs underload and "retarded memory" if It occurs under no load after being stressed beyond the"elastic limit" of the material. Both are evidenced by an increase in strain which is ex-ponential in form since its value depends on the instantaneous value of the residual stress.This residual stress, in turn, depends upon the amount of alleviating strains. The residualstress and strain hardening process can account for the "Bauschinger effect" wherein theyield stress is inoreased in the direction of original loading to a greater extent than in thereverse direction, as well as for the heat treating and'annealing phenomena.

The phenomenon of creep can be viewed as a continuation of slip under constant

load. The first stage of creep is a relaxation which proceeds exponentially with time

[c 2 )] while the second stage occurs when the strain hardening rate equals the re-

crystallization rate resulting in a constant slip rate [ C3 t ]. The third stage o1 'creep is

considered to be a manifestation of material failure and is not included in the strain rela-tionship.

WADD TR 60-517 3.4

A.

3.1 (Cont'd)

SHEAR

(a) INITIAL (b) UNSTABLE (C) FINALEQUILIBRIUM EQUILIBRIUM EQUILIBRIUM

b

ENERGYPOTENTIAL POTENTIAL BARRIER

Oa- C

(d) SLIP- SHEAR DIRECTION

FIGURE 3.1-1 DEFORMATION AND ENERGY POTENTIAL UNDER SHEARING ACTION

3.2 STRESS-TEM PERA TURE-TIME-STRAIN RELATIONSHIP

This relationship, presented in Paragraph 3. 2. 1,is predicated on experimental data ob-tained from short and long time uniaxial load tests. The mathematical formulation is in termsof material constants which are selected so that the calculated strain curve closely approxi-mates the experimental curve in the regions of interest. Methods of obtaining the materialcons'ants from simple uniaxial tests are indicated in Paragraph 3. 2. 2. Methods of extrap-olating the relationship for complex loading conditions (biaxial, etc.) are shown in Paragraph3.2.3.

WADD TR 60-517 3.5

.---

3.2.1 Unlaxial Stress-Strain Relationship

Strain can be viewed as composed of two components. One component is elastic(assumed linear) and is associated with the volumetric change (dilation) without distortiondue to hydrostatic load. The constant ratio of stress to strain (E) is a material temperatureparameter. The other component is inelastic an.l is associated with distortion (slip) withoutvolumetric change due to shearing load. The slip is assumed proportional to the hyperbolicsine of the stress ratio (as suggested by Reference 3-1) and proportional to a time-temperatureconstant which varies exponentially with temperature and time. The initial slip for smallstresses is almost linear, thus the apparent (initial) modulus (EA) of the stress-strain curve

AIis slightly less than the true modulus E. This would suggest that the (EA) modulus determined

by dynamic methods would be more consistent and higher than that determined by stress-straincurves at high temperatures (which will contain some inelastic effects, e.g., creep, relaxation,etc. ). The following stress-strain curve is assumed:

e= a•E + K sinh a/le (la)

where

o = strain (in/in)E = elastic modulus (material-temperature parameter) associated with hydrostatic

stress and dilation, psia = applied stress, psi

= stress associated with slip. This is a material-temperature parameter which isa function of the past history of the material, ". e., stress-temperature-time, psi

K time-temperature work hardenability material parameter directly proportionalto slip = (Kt) (KT), non-dimensional

Kt = time component of K = C1 + C2 (1 - e-t) + C t, non-dimensional (Ib)

C1 = short time (instantaneous) measure of slip (ClK sinh a/la) in which no primaryor secondary creep occurs, non-dimepsional d t " tc

C2 = measure of primary creep (C2 (1 - e- ) KT sinh a/a 0 ) due to "relaxation", etc.non-dimensional

C3 = measure of secondary creep (C3 t KT sinh a/ao ) as the rate of "strain hardening"

is balanced by the rate of "softening by recrystallization", 1/secKT - temperature component of K, non-dimensional

C~-Q/T=Ce (Suggested by Maxwell-Boltzmann Energy Distribution, Reference 3-2)

Q = material constant - activation energy per gram atom divided by the universal gasconstant, *K

C4 = material constant representing the number of particles per unit time exceeding thethermal potential barrier .

The generalized strain relationship, which includes the effects of temperature and timeas well as stress, can be employed to obtain short time or isochronous (constant time) stress-strain and stress modulus curves (See Figures 3. 2. 1-1 through 4). The temperature effectis contained inodepandannyof the material parameters upon the temperature and the time

WADD TR 60-517 3.6 .i

3.2.1 (Cont'd)

I3 NoOKEA

KEA

EAE" 0- ETos

E0 EA

NON- DIMENSIONAL NON-DIMENSIONAL

STRNESS-TRAINLU CURVES TNN-SECANT MODULUS CRE

EKEA

FIGURE 3.2.1-3 FIGURE 3.2.1-2NON - DIMENSIONAL NON -DIMENSIONAL

STANENS-TRMOULU CURVES TNE-SECANT MODULUS CRE

• ~RATIO CURVES .

• WADD 'FR 60-517 3.7

0 05

,I3.2.1 (Cont' d)

effect is incorporated in the I Item described below. The variation of the material para-meters with temperature must be determined from experimental data. Initial investigationin this direction has Indicated that the functional relationship is smooth and that interpola-tion should give satisfactory accuracy.

Another Justification of the assumed strain relationship is the similarity in formulationto the various diffusion rate processes found in the literature (e. g. , Reference 3-3 and 3-4).These formulations agree satisfactorily with the experimental evidence and are employed withavailable strength and creep data to obtain interpolated or extrapolated strength data. As anexample, the Larson-Miller parameter, employed as a correlation parameter (Reference 3-3and 3-4), is equivalent to the energy of activation Q for an unstressed material, and shouldbe modified by a stress factor if it soaks under load,e. g., L.M. = T (In C4 + in t + 1%).

The generalized relationship, Eq. (la) can be manipulated to obtain equivalent stress-strain, secant modulus, tangent modulus, etc., curves and to devise methods to determinematerial parameters used in design from simple unlaxial short time and long time data.The generalized relationship can be utilized in the same way that a room-temperature stress-strain curve is employed to get allowables. The results are presented in a non-dimensionalform in terms of the material design parameters.

LetEA = apparent elastic modulus of material (obtained from short time stress-strain

curve)ET = tangent modulus = do/de (psi)ES = secant modulus = q/e (psi)

= K E /o = temperature, time, material parameter (8 0 elastic material;Io p 1 viscous material).

ThenE /I-P •

EAe

= - (1-f) + Psinh a/lo (2.a a, 0

0 0

E 0 (2b)

EAA = (1-) + P cosh a•a (2c)

ET 0

Figures 3.2. 1-1 through 4 illustrate the non-dimensional form of the stress-strain, secantmodulus, and tangent modulus curves resulting from Eqs. (la), (2a), (2b) and (2c). This datais utilized in Section 9 on the stability of structures-in which the effective modulus is somecombination of EA, ET and ES, and also for use in obtaining approximate solutions of non-

linear Droblems (Reference 3-5).

"WADD TR 66-517 3.8

3.2.2 Experimental Determination of the Material Constants

The material constants are selected so that the mathematical formulation of

the strain relationship satisfactorily approximates the experimental data.

3.2.2.1 Apparent Modulus (EA)

The apparent modulus E is obtained as the initial slope of the short timestress-strain curve as shown in Fiure 3.2.2.1-1. It may be obtained from the naturalfrequency of an axial bar or torsional rod if inelastic effects cannot be eliminated fromthe short time test. The time at stress and the magnitude of the stresses should be smafiso that the non-linear components are insignificant.

EA

/e=A4+K SINHO/cre*/ / E

8=/ //"(

EE

FIGURE 3.2.2. 1-1 SHORT TIME STRESS-STRAIN CURVE

WADD TR 60-517 3.9

3.2. 2. 2 Slip Stress (ao)

The stress (o- ) which is associated with slip can be determined from a short

time stress-strain curve if the creep effect is negligible. The technique is shown inFigure 3.2.2. 2-1 in which the deviation a = r-u/E is plotted logarithmically versus a.

AIf the creep effect cannot be ignored, then a can be determined from constant creep

rate data shown in Figure 3. 2. 2. 2-2 in which the constant creep is plotted logarithmicallyversus a. Justification of this technique follows and is based on the property that thesinh x is approximately eX/2 for large x. From Eq. (la) of Paragraph 3. 2. 1

e= alE + Kt KT sinh ,l%

C K6 f -aE KK sinh "1 ,-- 1KT e°/ o (1a)tT

where t-.- 0 and oa/a >10

ClKTIn =In 2 + 7/•a (b)

CIKT 1 (.logS = log 2 + . 3 (la 0 ) (Ic)

On a semi-log plot, (1c) is an equation of a straight line of the form

y =b+mxClKT

wherey=log 5, b= log 2 x=a

I'and m=-. (1/a" 1 =tan a Ax A '(10 6

2.3 AX0A

=I a 1 2- 61 (1)0 2.3 A(log 6) 2.3 log (612/611) (See Figure 3.2.2.2-1)

Similarly,

&= a/E + KtKT sinh Iaol + KtKT ok/a (3a)

=KtKT sinh a/r°0 (for constant stress creep test) (3b)

C 3 KT2 ea/v-° (for t-- o , al/a >1) (3c)

In In 2 + (70 (3d)

WADD TR 60-517 3.10

3. 2. 2. 2 (Cont'd)

E A =

LCG SCALE

T

-' 2

T

I I!C. KT

2I -I

2 2

21 I

0 2 .3 Io ( 812 / 8 1,.

A LINEAR SCALE

02FIGURE 3.2.2. 2-1 DETERMINATION OF MATER-I•. CONSTANTS FitOM SIhORT

WADD 60-517 STESS-SRAIN DATA

WAD r 0-17 . M

3. 2. 2. 2 (Cont' d)

LOG SCALE

biT2

i1

,Ay

0-A

log e C3KT I

- 2 +1 'ii/0 22- (71"0 2 .3 log ( , 2 /* )

C31KT ,e

2

LINEAR SCALE

FIGURE .. '..- ETERMINATION OF M.VrI \. CONST. .NTrS FIRoM ('ONST'.I'. Tl

W A D I) I1R 6 0 -5 1 7 C R E E P R A 'r E l) . '3. 1 2"• " ~ 3.12 I

MTAM

3.2.2.2 (Cont' d)

CSKT 1log; log C3K 1 Ova.~)2 (

1 (4)Uo = 2.3 gog t) (Reference Figure 3.2.2.2-2)

1 02 - (O1

0*% =-

lg( 12/13. 2.2.3 Work Hardenability Factor (K = KtKT)

The value of K can be obtained from data shown in Figures 3.2.2.2-1

and -2 and Figure 3.2.2.3-1. CjKT Is obtained from the intercept of the straight

line in the deviation versus stress curve (Figure 3. 2. 2. 2-1). C2 KT Is obtained from

the Intercept of the straight line In the strain versus time curve (Figure 3.2. 2.3-1).C3 KT is obtained from the Intercept of the straight line7 In the creep rate versus strees

curve (Figure 3.2.2.2-2).

• ~( TI," a-

C KT SINH q•/ T'o

C1 KT SINH 01Oo

OlE EO-(ZO)

FIGURE 3.2.2.3-1 CREEP CURVE

3. 2.2.4 Interpolation of Material Parameters

An increase in temperature or a soaking at temperature will decrease themagnitudes of the material parameters. The temperature potential reduces the mag-nitude of the stress potential necessary to induce slip. The temperature also increasesthe rate of recrystallization and relaxation which reduces the strain hardening. Pre-liminary plots of the material parameters versus either a temperature or an equivalentsoai:ing parameter (e. g., Larson-Miller Parameter) have indicated that the resultsshould be relatively smooth curves as exemplified in Figure 3. 2. 2. 4-1. In order toemploy the non-dimensional techniques proposed in this section, it would be expedient tocompile data (EA, ao, ClKT, C2 KT, C3 KT) on the various structural materials in the

manner shown in Figure 3.2.2.4-1.

WADD TR 60-517 3.13

___ __ ___ __ ___ __ ___ __,__ __ ___ .__ __ ___ __ ___ - -.. • • - • " v • ,

3.2.2.4 (Cont'd)

EA

112 KT

T OR T (C+logt

FIGURE 3.2.2.4-1 VARIATION OF MATERIAL PARAMETERS WITH TEMPERATURE

3.2.3 Complex Loading

It is possible to describe the stresses and strains in a material by functions(tensors) which are capable of expressing the state of the stresses and strains at a point.They can always be expressed by the direction and magnitude of their principal components(stresses or strains). For an isotropic material, the principal directions of stress andstrain coincide. Associated with the stress or strain are three invariant magnitudes whichare independent of the direction of reference. These three invariant magnitudes. or anythree combinations of all three quantities, define the tensor. It is assumed that a relation-

ship exists between the invariants of the stress and strain tensors. The unlaxial stress-strain curve would then be an expression of this relationship which is quite simple to cbtainexperimentally and could be.!mployedto obtain the stress-strain relationship for more com-plex loadings.

WADD TR 60-517 3.14

3.2.3.1 Invariants

The three quantities which are characteristic of the stress and straintensor can be expressed as follows:

Let

0l,02,03 = principal stresses

C1 ,E 2 ,E 3 = principal strains

Then

first stress invariant - hydrostatic pressure (la)

=1/3 (aI + a2 + a3)

Cr = second stress invariant - octahedral shear stress

= 1/3 -/(aI - a2) + (a -)2 + (a -)2 (ab)

ai = third stress invariant

= 1/3 (o") (1"2) (03) (lc)

a. = principal deviatoric stress = a. -or (1d)

= first strain invariant - dilation

= 1/3 (E1 + E2 + e3 ) (2a)

EII = second strain invariant - octahedral shear strain

=1/3 (E 1 + (E 2 - E + (E 3 - 1 )2 (2b)

= third strain invariant

= 1/3 (C 1) (C 2 ) (C3 ) (2c)

= principal deviatoric strain = -I (2d)

Associated with the principal directions (1, 2, 3) are 8 (octahedral)planes which make equal angles (arc cos 1/J/-) with the principal axes. On theseplanes, the normal stress and strain are 0I and c, respectively, and the tangential

(shear) stress and strain are a and c11 (See Figure 3. 2. 3. 1-1). If it is assumed that

the relationship between the normal stress and strains and the tangential stresses andstrains are independent (orthogonal) of each other and are uniquely defined for anymaterial, then the invariant stresses and strains for a complex loading can be related tothose obtained for a simple one, such as uniaxial tension. Thus, the physical modelcorresponding to Figure 3.2.3. 1-1 suggests that those crystal planes which are orientedso that the applied loads produce higher octahedral shear stresses will slip first (assum-ing dislocations are uniformly distributed) and that these crystals "strain-harden" re-quiring higher stresses to cause further slip in these and less favored crystal planes.

WADD TR 60-517 3.15

MUM

3.2.3.1 (Cont'd)

03

~. LINE OM NORMAL TOPAEABC

...........

..........

. . . . . . ..... B

FIGURE 3.2.3.1-1 PRINCIPAL DIRECTIONS AND OCTAHEDRAL PLANES

WADD TR 60-517 3.16

3.2.3.2 Transformation of Stress-Strain Relationship

The hydrostatic components, stress (ai) and strain (EI) are assumed to vary

linearly with volumetric deformation but without any permanent deformation. Experi-mental evidence indicates that the assumption is quite satisfactory for the ranges ofstress-temperature experienced. The relationship between the octahedral componentsof stress Of!) and strain ( e is assumed to be non-linear, with no volumetric defor-

mation, and to constitute the inelastic portion of the stress-strain relationship (Eq. (la)a (I +Z/ ),/2CI1K T 3aiIof Paragraph 3.2.1). The relationship assumes c (I+)E (1 +/2c°

11 +1/)EA + 12- aFigures 3. 2. 3. 2-1 through -4 illustrate how a uniaxial stress-strain curve

can be employed to derive a complex stress-strain curve. The uniaxial stress-strainlaw is obtained experimentally as shown in Figure 3.2.3.2-1. The hydrostatic stressand strain and the octahedral stress and strain are computed in Figures 3.2. 3.2-2 and-3. The complex loading is employed to obtain strains from stresses (as shown inFigure 3.2.3.2-4) or stresses from strains, by assuming that the deviatoric to octahedralstress and strain ratios are equal. This is implied by the von Mises yield criteria andthe Mises-Levy stress-strain rate relationship (see References 3-6 and 3-7), i.e.,

a.'I a.Y' - aY a1I1- - (3)

Ei 6i -6I 11

The invariant stress-strain relationships can be employed to postulateallowables of combined loading conditions from allowable invariants of simple (uniaxial)loading conditions wherever applicable, and to obtain approximate solutions to complexproblems (assuming a compatible strain distribution results in an upper limit of theapplied load, whereas assuming an equilibrium stress distribution results in an upperlimit on the displacement of the structure).

WADD TR 60-517 3.17

-i- - - ,--- --. a-ý----~ ;

3.2.3.2 (Cont'd)

"T zT T=Tl

ASSUMED LINEAR

I VEAE

3FIGURE 3.2.3.2-1 FIGURE 3.2.3.2-2

UNIAXIAL STRESS-STRAIN HYDROSTATIC STRESSt-STRAINCURVE CURVE

T zTj T:T,

ti: __5:CONSTANT

A SSUMED 114/ a- (I+Z,)tiC;KT 3-

E, (JI-'v A+ C SINH 3 ""

( +1.)%r2z:•(I ~2~~/' E•4 : L(O 4 -o'z) + E

3 1

SFIGURE 3.2.3.2-3 FIGURE 3.2.3.2-4OCTOHEDRAL STRESS-STRAIN BIAXIAL STRESS-STRAIN

CURVE CURVE

WADD TR 60- 17 3.18

3.3 REFERENCES

3-1 Nadai, A., The Influence of Time Upon Creep; The Hyperbolic Sine CreepLaw, S. Timoshenko, Anniversary Volume, MacMillan Co., p 155, New York,1938.

3-2 Tolman, R. C., Principles tf Statistical Mechanics, Oxford University Press,p 71, 1938.

3-3 Conrad, H., Correlation of High Temperature Creep and Rupture Data, ASMEPreprint Paper No. 58-A 96.

3-4 Heimerl, G. J., Time-Temperature Paramet,,rs for Rupture and Creep ofAluminum Alloys, NACA TN-3135.

3-5 Mendelson, A., Hirschberg, M. H.1, Manson, S.S., A General Approach tdthe Practical Solution of Creep Problems, ASME Preprint 58-A 98.

3-6 Von Mises, R., G6ttinger Nachr., Math. -phys. K1. p 582, (1913).

3-7 Le'vy, M., C.R. Acad. Sci., Paris, 70,-1323 (1870).

WADD I'R 60-51- 3.19

r

L.

I

I

-'.

SECTION 4

THERMO-ELASTIC ANALYSIS OF BEAMS

"WADD TR 60-517 4 0

N5

SECTION 4

THERMO-ELASTIC ANALYSIS OF BEAMS

TABLE OF CONTENTS

Paragraph Title P

4 Thermo-Elastic Analysis of Beams 4.4

4.1 Thermo-Elastic Analysis of Statically Determinate (Un- 4.6restrained) Beams

4.1.1 General Solution 4.7

4.1.1.1 Elastic Section Properties of Cross Section 4.9

4.1.1.2 Stresses and Deformations of the Cross Section Due to 4.9Temperature

4.1.1.3 Stresses and Deformations of the Cross Section Due to 4.10Combined Mechanical and Thermal Loading

4.1.2 Evaluation of Integrals 4.11

4.1.2.1 Solution by Finite Sum (with illustrative problem) 4.11

4.1.2.2 Geometry of Concentrated Elastic Areas - Sandwich Beam 4.20

4.1.2.3 Power Series Solution - Bending About One Principal Axis 4.21

4.1.2.3.1 Polynomial Approximation of a Given aT Distribution (with 4.23illustrative problem)

4.1.2.3.2 Binomial Representation of Continuous Elastic Cross Sections 4.28

4.1.2.3.3 Solution for Continuous Elastic Cross Sections (with 4.30illustrative problem)

4.1.2.3.4 Solutions for Discontinuous (Multi-Rectangular) Elastic Cross 4.46Sections (with illustrative problem)

4.1.2.4 Power Series Solution - Corrections for Bending About Both 4.58Axes

WADD TR 60-517 4.1

F1


Paragra TePageT

4.2 Statically Indeterminate (Externally Restrained) Beams 4.58

4.2.1 Beam Deflections (with illustrative problem) 4.58

4.2.2 Fixed End Reactions 4.63

4.2.2.1 Sign Convention 4.63

4.2.2.2 Applied Beam Loads 4.65

4.2.2.3 Determination of Fixed End Reactions 4.67

4.2.2.4 Determination of Fixed End Reaction for Special Cases 4.76

4.2.2.5 Use of Equations and Graphs 4.81

4.2.3 General Solution of Statically Indeterminate Beams 4.86

4.2.3.1 Degrees of Freedom 4.86

4.2.3.2 Equivalent Loading 4.87

4.2.4 Mechanics of Solution by the Flexibility Method 4.88

4.2.5 Mechanics of Solution by the Stiffness (Slope-Deflection) 4.92Method

4.2.6 Selection of Method of Analysis 4.95

4.2.6.1 Problem IA - Statically Indeterminate Beam by the Flexibility 4.96Method

4.2.6.2 Problem IB - Statically Indeterminate Beam by the Stiffness 4.101Method

4.2.6.3 Problem II - Sandwich Beam by the Flexibility Method 4.103

4.2.7 Curved Beams 4.105

4.2.7.1 Values of i', w? and M 4.108

WADD TR 60-517 4.2



4.2.7.2 Special Cases 4.109

4.2.7.3 Sample Problem - Frame Subjected to Thermal and 4.112Mechanical Loads

4.3 References 4.117

WADD TR 60-517 4.3

tISECTION 4- THERMO-ELASTIC ANALYSIS OF BEAMS

The thermal loading on a beam reduces the stiffness (increasing deformations dueto mechanical load), and in general induces thermal stresses and deflections. If the beamis unrestrained externally (statically determinate, then thermal stresses can only occurdue to the internal requirements that plane cross sections before bending remain plane afterbending, provided that the undeformed beam does not exhibit sharp curvature (R/d S 5).Experimental evidence does not violate this assumption. If the beam is restrained external-ly (statically indeterminate), additional thermal stresses will be caused by the redundantrestraining forces. A

The thermo-elastic analysis of beams, with all of the above factors contributing tothe problem, is presented. General solutions are given and methods are presented for thenumerical solution of specific problems.


a Non-dimensional thermal strain ratio; Distance from fixed end to initiation ofload A

a, a' Coefficients of polynomialsb Width of cross sectionc Exponent defining variation of 1/(-I)d Distance from reference axis to an extreme fibere Non-dimensional width parameterf Virtual axial force caused by unit virtual loadsf. Influence coefficient for fixed end axial reaction (see text, Paragraph 4.2.2.3)

h Heightj Exponent defining variation of curvaturek Exponent defining variation of axial load; Integer indicating rectangle locationm Virtual moments caused by unit virtual loadsmc Influence coefficient for fixed end moment reaction (See text, Paragraph 4.2.2.3)

m, n Integersp Virtual shear loads caused by unit virtual loads

Influence coefficient for fixed end transverse reactionpi cq Variable transverse loadr Exponent defining variation of transverse loadings Non-dimensional distance u/du, v Distance from reference axes which are oriented parallel to elastic principle

axesu Distance from v reference axis to elastic centroidw Change in rotationof cross section per unit distance (curvature) due to

mechanical loadingwl Change in rotation of cross section per unit distance (curvature) due to

thermal loadingw Total change in rotation of cross section per unit distance, wt = w + w'

x Distance along length of beam(x) Function of xy, z Distances from arbitrary but perpendicular reference axes

WADD Th 60-517 4.4

41

4

-- fEydA f EzdAy, z Distances f EdA f EdA from Y, Z reference axes to elastic centroid

y•," istncs Eyci TdA fE'zaTdADistances fETdA from Y, Z reference axes to location of7i~d ' EciTdAresultant force on the cross section

A Area =fdAAA, dA Small element of areaB. Level of aT at point "J" above the reference axis value of aT, B =a T-ct T

j J j o00E Modulus of elasticity = E(T)Eb Elastic width = E baj

(avg)EA Effective axial stiffness or force required for unit axial deformation, fEdAE'l Effective bending stiffness rr moment required for unit rotation,

Eli - fE (y-y) 2 dA

F Axial force due to mechanical loadingF' Restoring axial force due to temperature, F' = f EaTdA2L Total cantilever axial load at L due to mechanical loads

I Moment of inertia of cross section, Ij= (y-_2 dA

I0 Moment of inertia of an element of area about its centroidal axis

K Non-dimensional shape parameterL Length of beam; Power of general term of the aT polynomialM Bending moment due to mechanical loading

Restoring bending moment due to temperature, MN = -•'-y) f EaTdA

7n Total cantilever moment ,nt L due to applied mechanical loadL

P Transverse-load due to mechanical loadingTotal transverse cantilever load at L due to applied mechanical loads

T Temperature change from a datum valueUU, VV Elastic principle axesY, Z Arbitrary reference axes

Q Mean coefficient of expansionaT Strain due to free thermal expansionp Non-dimensional taper parameter'Y Axial strain parameter6 Non-dimensional rotation parameterC Strain?I Axial strain at elastic centroid due to thermal loading

Axial strain at elastic centroid due to mechanical loadingTotal axial strain at elastic centroid, t= + ?I

Non-dimensional length parameter defining distance from initiation of load on beam71 Non-dimensional step parameterX Non-dimensional axial stiffness, X = EA/dEobo

S"r Variable axial load (shear flow)p Non-dimensional elastic centroid location, 1 = u/d Iv Non-dimensional length parameter - a/L; Non-dimensional bending stiffness

T' =EI/d3 Eob

a StressSA Deflection

WADD TR 60-517 4.5

"•.:•-_ •,.. -•-.• •. -••= '.••-• • •e-•.• • • ."• •••-:-'.•.-• .•'; • •-, . • •..•'•• ••'-,•- •:- x~>• =:-:•;• •• : • • '.•". .• .'--:.•- •:• • •'• •"•:• •• 1

Subscripts

a Anti-symmetricalc Combined thk At location of k rectanglem Due to mechanical loadingn At extreme fiber locationo At reference axis locations Symmetri .alt Due to temperatureu, v, uu, vv About principal axesy,z,yy,zz About arbitrary but perpendicular reference axes

About y, z centroidal axes0 Free end of cantilever beamL Fixed end of cantilever beam, left side; Due to L term of cT polynomialR Fixed end of cantilever beam, right side

4.1 THERMO-ELASTIC ANALYSIS OF STATICALLY DETERMINATE (UNRESTRAINED)BEAMS

This Sub-section 4.1 presents methods for determining the deformation andstresses of an unrestrained beam subjected to temperature variations through the beamcross section. The section properties for the general case of variable modulus (due totemperature and construction) are given in integral form for the purpose of determiningthe response of the cross section to both temperature and load. The general solution isthen presented in integral form and is evaluated by the methods of finite sum and powerseries.

The finite sum solution is the most general of those discussed and requires aminimum amount of engineering ingenuity to set up and evaluate tables. It should be usedin cases where the directions of the principal axes are not obvious. The power seriesmethod, though not as accurate as the finite sum solution, gives a more rapid means ofobtaining analytical solutions through the use of approximating analytical functions. Theuse of approximating functions is further justified by the fact that the functions are to beintegrated, thus improving the accuracy of the approximation.

If the beam is unrestrained externally (statically determinate), then the solutionpresented in this sub-section defines the deformations (axial strain and ch ,nge of curvature)and the cross sectional stresses. If, however, the beam is restrained ext .rnally (statical-ly indeterminate), additional thermal stresseswill arise due to compatibility forces (re-dundants'., -t the boundary. The deformations of the cross section, obtained in this Sub-section, allow the determination of beam deflections and are thus essential to the thermalsolution of indeterminate beams as outlined in Sub-section 4.2.

WADD TR 60-517 4.6

U- ----- ',-,>'< - 4

4.1.1 General Solution

The following paragraphs present the general thermo-elastic solution for an un-restrained beam in integral form, as derived in Reference 4-1. Evaluation of these integrals,which causes most of the difficulty in obtaining numerical solutions to specific problems, isdiscussed in Paragraph 4. 1. 2.

The following assumptions and limitations apply to the thermo-elastic solution of theunrestrained beam:

(a) Plane cross sections before bending remain plane after bending. This is thebasicassumption in all bending problems. Experimental data indicate that this requirement for thedeformations of the cross section is sufficiently accurate for engineering analysis.

(b) Material is linear elastic at any temperature. Thus a single relationship ofstress to strain (a = E E ) can be utilized to connect the equations of deformation and equili-brium and the principle of superposition can be employed. Any plasticity, buckling, or creepwould modify the results, usually by reducing the peak stress but increasing the deflections.

(c) The variation of the cross section and temperature along the length of the beamis both continuous and smooth, and does not produce any significant shear forces. Significantshear forces occur in the vicinity of these abrupt changes (heat sinks, free and clamped ends,and abrupt changes in cross section) and the solution must be modified to account for these ef-fects. These effects should be insignificant at distances greater than the order of the dimen-sions of the cross section (St. Venant Principle). For example, Reference 4-2 shows that atdistances of approximately three beam dopths away from free ends, solutions obtained underthe assumptions of plane cross sections remaining plane are valid.

The unrestrained beam subjected to temperature (see Figure 4. 1.1-1) is analyzedby subjecting the beam to a set of force systems which satisfy equilibrium and produce defor-mations which are compatible with the requirement of plane cross sections remaining planeafter bending.

Consider a unit length of beam. Initially, each fiber of the beam is liberated fromthe influence of its neighbors. The temperature distribution is then applied to the beam whichcauses each fiber to expand by an amount aT. In general, the thermal expansion of the fiberswill cause the cross sectional plane formed by the ends of the fibers to warp. To satisfy therequirement of plane cross sections remaining plane, a pressure loading of -EcaT is appliedto eliminate the thermal expansion and return the cross section to its original position and con-dition (plane). This pressure loading upsets the equilibrium of the cross section. An axialload (F' =f EaTda, equal in magnitude but opposite in direction to the force on the cross sec-tion due to the pressure) is applied at the elastic centroid of the cross section. This balancingaxial force causes pure translation without rotation of the cross section plane so that the re-quirement of plane cross sections remaining plane is not violated. Rotational equilibrium stillremains to be satisfied since the location of the resultant restoring force -Y, in general, willnot coincide with the centroid of elastic area y. Equil!ibritun is achieved by applying a bal-ancing moment to the cross section of sufficient magnitude (M' = [ y7]F' ) to cause pure ro-tation of the cross sectional plane.

WADD TR 60-517 4.7

4.1.1 (Cont' d)

The superposed force systems now satisfy equilibrium and produce deforma-tions which are compatible with the requirement of plane cross sections remaining plane.Thus, the procedure outlined above must result in a stress-deformation distribution foran unrestrained beam which is consistent and unique under the a3sumptions.

d z

dyy•

Location of Resultant d~~~Restoring Force !••

x

FIGURE 4.1.1-1 ELASTIC AREA CROSS SECTIONs

' i

4. 1. 1. 1 Elastic Section Properties of Cross Section

The structural response to both mechanical loads and thermal stimulae is governedby both the effective bending (EI) and axial (EA) stiffness of the cross section. The equationsfor these properties, which are stated below, are Identical to those for cross sections havingconstant E except that E is retained within the Integral sign since it is permitted to varyover the cross section.

E"A = fEdA (Z)

y =f E y dA/f EdA (2a)

z = fEz dA /fEdA (2b)

ETI-- = fE z2 dA - 2 fEdA (3a)yy

El-- fEy 2 dA - 2 fEdA (3b)zz

Ef-- E yzdA -y ZfEdA (3c)yz

.E- I-- + El-- EI-- - 2---(Eyyuu' ElIv) YY 2 zz yy 2 zz) + (E'-Iy) 2 (4)

The distances ý and 1 (Figure 4. 1. 1-1) to the elastic centroid are given bEqs. (2).Thus, the elastic centroid, Eq. (1), is the centroid of the effective elastic area EA, not ofthe geometric area. Similarly, the geometric moments of inertia are of no significance whenE varies over the cross section. The effective bending stiffnesses, Eq. (3), must be employed.

Equation (4), expresses the bending stiffnesses about the elastic centroid principalaxes in terms of the bending stiffnesses about arbitrary centroidal axes. The elastic princi-pal axes are defined as those orthogonal axes for which

El = fEuvdA = 0.uv

4. 1. 1.2 Stresses and Deformations of the Cross Section Due to Temperature

No thermal stresses are caused in an unrestrained isotropic, homogeneous, linearlyClearly, in the particular case of an unrestrained beam, a linear aT distribution over the cross

section produces free thermal expansions which cause plane cross sections to remain planeafter deformation. No stresses are required to maintain the plane cross section

In general, a non-linear temperature distribution over the cross section would causefree thermal expansions of the beam fibers which would warp a plane cross section out of plane.Thermal stresses are produced which restore the plane cross section. In this respect, a tem-perature distribution can be considered as a "thermal load" which, when applied in the absenceof mechanical load, still produces stresses and deformations.

Given below are the equations, derived in Reference 4-1, for the stresses and defor-mations in an unrestrained beam due to thermal load.

WADD TR 60-517 4.9

4. 1. 1.2 (Cont d)

E A

( -Ti-M,'XT ' M-

W1 y-,'z..y y3J (2)

El-y

:• ~where .

SF ' = 'E •T d A (4) •- 2 = •EaTy dA/f•E•T dA (5a)z = fEaTz dA/ETdA (Sb)Mz (y-)F' (6a)

W = (Y - 'F) F' (6b)

yy

4. 1.1.3 Stresses and Deformations of the Cross Section Due to Combined Mechanical and :Thermal Loading

For a linearly elastic beam, mechanical loads can be combined with thermal loadssimpl y f superposition. Thus

(E'+ýF) (1a)

(Wzt L1j(M'v + M--)]+j E (IV'•, +M- M(2a)

(M z :

M+-t (zi z-) F (6b)

4. Stand, e sse s ausefompressive Cross asStnecParatioraphe 4.1.1.2tant and

t Moments about the yy and zz centroidal axes are positive when their sense is suchy by and z - z have positive values). F is positive when tensile, and T is positive when

i ~above a datum value. "

WADD TR 60-51741

zt (FI F

~ ~ .s - - .E

4. 1.2 Evaluation of Integrals

The solution of the thermo-elastic beam problem requires the evaluation of the integralsgiven in Paragraph 4. 1. 1. Two methods of sufficient engineering accuracy are presented. In thefirst, integrals are approximated by finite sums; in the second, the goemetry and temperature dis-tributions are approximated by simple integrable functions (polynomials or power series). Approxi-mate methods are justifiable since the functions must be integrated thereby improving the accuracyof approximation.

4. 1. 2. 1 Solution By Finite Sum

The cross section is broken up into a finite number of elemental areas, selected so thatthe variation of otT and E in each element Is small. The procedure is adaptable to all cross sec-tions and the degree of accuracy increases with the decrease in the size of the elements. Digitalcomputing machines can be employed (see Reference 4-3) to solve the problem.

The general solution is outlined in tabular form in Table 4. 1.2.1-1 and illustrated bythe example which follows, with numerical results given in Table 4. 1. 2. 1-2. The number of tabularcolumns and labor involved in solving the problem are considerably less when the following sim-plifying conditions are realized:

(a) y and z axes coincide with principal centroidal axes u and v (example: axis ofsymmetry)

(b) ctT is symmetrical about a principal axis of symmetry of the elastic geometry(M t -- or M'--= 0)

yy zz

Additional tabular columns are eliminated if E and ay are constant over the crosssection. Table 4. 1. 2. 1-3 outlines the solution for the case in which all of the above simplificationspertain.

The finite sum solution for the deformations of the cross section is based on an approxi-mating geometry consisting of a finite number of points of concentrated elastic area located at thecentroids of elements. Once the deformations ( Z, w , Wz) have been calculated from the tabular

solution, stresses can be obtained at any points on the cross section as, for example, extremefiber points (see illustrative problem).

?4

WADD TR 60-517 4.11

ra 0 x0 it...)

IT H

w ®a

- 0 .4

+ ! 1,4 4 *., (4

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[NCCCI II 1 1 i 111 1

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4.1.2.1 (Cont d)

The solution of an unrestrained beam by the finite sum method is illustratedbelow for an unsymmetrical section with an unsymmetrical temperature distribution,T = T (y, z), on the cross section of the Halcomb 218 "'zee" spar shown in Figure4.1.2.1-2. Determination of the following is required:

(1) Thermal stresses and deformations

(2) Stresses and deformations when subjected to a moment about the verticalaxis of 10,000 inch-pounds and an axial force of 10,000 pounds, i.e.:

M-- = 10,000 inch-pounds

M =_ 0

"F 10,000 pounds

The solution is worked out in Table 4.1.2.1-2 by the finite sum method. Stressesare plotted in Figure 4.1.2.1-3.

X

'.4

3

A

WADD TR 60-517 4.14

4.1.2.1 (Cont'd)

MECHANICAL LOADS

M-- = 10,000 inch poundsYy

M-- = 02.0

z

" F = 10,000 pounds,• d' 1.0' ]•c

.30 d 0 @ c THERMAL LOADSd"l 6 6 50

Element Mean Temp. E/10 o(106).20

7 1 1000°F 20.0 6.60

2 950 21.1 6.53

3 850 23.3 6.35.80 4 770 24.7 6.23

4.0 5 700 25.8 6.03

6 650 26.5 5.88.80 :5 7 700 25.8 6.03

8 800 24.2 6.259 900 22.2 6.44

.80 4

y

.80 3)

FIGURE 4.1.2.1-2 FINITE SUM METHOD

WADD TR 60-517 4.15

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- �- 0

-N N 40CC. 00 14 *,. *"

4 ICI 4 �CI C 0.0 � � -

__________________ = - - -� - sJ.qli 0m�z1E3 -

4~ 1 1 4

0 'r

C~C

"ow l-c c c . .

a- m V

'1A -,

Ik C - 14I.

o �

'5 C

I

-a SO

* - 2

I 0 SO C

.1 '5 0-I

5

2:

i

CC -t

o 04 C

'5

'C C

S.- 4

I.;

- C

-C

2

caJ

P-a

JCD

2 Qn

C5.

oftC

4.1.2.2 Geometry of Concentrated Elastic Areas - Sandwich Beam

The finite sum method (described in paragraph 4. 1. 2. 1) reduces the continuous )cross section into a number of discrete elements, such that each element is considered tobe a concentrated point of elastic area (EA), and free thermal strain (aT). This approxi-mation is especially applicable to beam cross sections made up of localized, high areaflanges (or caps) and thin shear webs. In beams of this type, the deformations of the crosssection and thermal stresses in the caps can usually be determined with sufficient accuracy(especially if peak temperatures occur in the caps) by neglecting the web material, thusconsiderably reducing the amount of calculation.

As an example, consider the sandwich beam cross section showr in Fig. 4.1.2.2-1.It is assumed that E, a, and T are constant in each face, the face thickness is small com-pared to the depth of the cross section, and the core carries only shear. Thus the crosssection is symmetrical about the y axis and the faces can be considered as concentrationsof elastic area (EA) and free thermal strain (aT).

Y E1IA 1, fT1I

V>CORE FACES

hZ

E A aT2 2' 2 2

FIGURE 4. 1.2.2-1 SANDWICH BEAM CROSS SECTION

The formulas at the bottom of Table 4.1.2.1-3 are directly applicable. They yield

lEA +1 A h (1)

_ E AIofT 1 + E2 A2 2T2EA +EA(2)

w = 1 T1 - a 2 T2'z (3)

h

o1 = 02 = (4)

WADD TR 60-517 4.20

A.N

4. 1. 2.2 (Cont' d)

Note that no thermal stresses occur in the faces, Eq. (4). In fact, no thermalstresses can occur in an unrestrained beam when the cross section consists of less than fourconcentrated areas because, for two or three such areas, a plane can always be passedthrough the thermally deformed section.

4.1.2 3 Power Series Solution - Bending About One Principal Axis

The power series method gives a non-dimensional solution to the unrestrainedthermo-elastic beam problem in terms of non- dimensional parameters which define the thermalloads and section properties.

This paragraph discusses the application of the power series method to problemsin which the directions of the principal axes are known and bending occurs about only one ofthese axes (as, for example, when one of the principal axes is an axis of symmetry for boththe elastic area of the cross section and the aT distribution). Under such conditions, thepower series method usually has a distinct advantage over the finite sum method.

The method can be extended easily to the general case of bending about both prin-cipal axes (Paragraph 4.1 2.4). However, because of the additional numerical computationrequired, expecially when the directions of the principal axes must first be determined, themethod of finite sum should be used in problems involving bending about both axes so as toobtain solutions with the least amount of labor.

Paragraph 4 1. 1 1 shows that, in general, the modulus of elastic)ty, E, variesover the cross section and thus appears under the integral sign in the formula.R for the sectionproperties. Accordingly, the structural behavior is goverred by the elastic properties of thecross section (EA, EI), rather than by the geometric properties of the cross section (A, I).

The power series method will be applied to beams having the following types ofelastic cross section configurations:

(a) Cross sections where the elastic width (Eb) is monotonically continuousthrough the depth or on each side of a bending axis of symmetry. (Figure 4. 1. 2.33-1 (a) and-1(b)).

(b) Cross sections with discontinuous elastic width variation of the multi-rect-angular type as exemplified by tees, channels, I-beams, etc. (Figure 4.1.3-1(c)). Crosssections that do not have the above configurations can usually be approximated by them witha reasonable degree of accuracy.

The following steps are taken in arriving at the solution:

(a) The average aT profile (Figure 4.1.2.3-2 and Paragraph 4.1.2.4) in a direc-tion normal to the principal bending axis is approximated by a power series (polynomial) of theform

L n n-IaiT= a Ls ans +a s- +. . .

as ann + nI + . . o (I)

L=o

(b) Continuous cross sectional configurations of the types discussed above arerepresented by binomials of the form

Eb = E0b° (1 + 3s) (2)

where and K are non-dimensional parameters defining the elastic width variation.

WADD TR 60-517 4.21

IL WADT

4.1.2.3 (Cont' d)

Eb _nbn_-

71 d

0E0 Bending Axisd Id

Ref. dof Symmetry

y 5 4A~i L.-4Ref.00Axis E-- Eb -]

(i) Monotonically Continuous (b) Elastic Widths Monotonically (c) Multi-rectangular Dis-Elastic Width Variation Continuous and Symmetrical continuous Elastic Width

About the Bending Axis Variation

FIGURE 4.1.2.3-1 TYPICAL CONFIGURATIONS OF ELASTIC CROSS SECTION AREAS

Discontinuous multf-rectangular configurations are expressed by means of stepparameters 77 m defined in Eq. (1) of Paragraph 4.1.2.3.4.

(c) The parameters which define the aT distribution and elastic cross section inthe above manner are substituted into non-dimensional expressions for the stresses anddeformations.

u b+ u

f E aT(uiv) dv

u.1) aT

//d 'and T arFntin

f b E(ui 'v) dvaF b--b

UU - bt :[ a

(a) Cross Section With Two Dimensional (b) One Dimensional Average a T ProfileDistribution of E, a , T in u Principal Direction.

FIGURE 4.1.2.3-2 DETERMINATION OF AVERAGE aT PROFILE FOR BENDINGABOUT UU PRINCIPAL AXIS

4A

'WD R605742

4.1.2.3.1 Polynomial Approximation of a Given aT Distribution

Consider the temperature (ctT) profile shown in Figure 4.1.2.3.1-1, whichrepresents the average variation in the direction of one of the principal axes.

S,U a T

u T R r

FIGURE 4.1.2.3.1-1 aT PROFILE

It is desired to determine a polynomial expression of the form

j /n

•T= ns +a~ln- . .. .+ s+ o n L agL (1)

0 L=o

where s =u/d, which will match the profile values of aT at discrete points equally spacedthrough the depth. The solution consists of solving n linear algebraic equations for the nunknown coefficients an, an. . . . a1 where n is numerically equal to the number of

equal subdivisions of the profile. In matrix form, the coefficients are determined from theequation:

2 1 n -1F1 2 1/1*-a• Ni) N-O•(÷ ... .+ . -2 2 2 nI

Ia2 nj- '2 .. B2aj (n•) n•) . . .. Nn-) .. ... n) B i (2

an j (1) (1) . . . . ... (1) Bn .

where Bj (ojTj - %TO). (3)

WADD TR 60-517 4.23

4.1.2.3.1 (Cont'd)

Obviously, as more and more subdivisions are taken, higher order polynomials areobtained which more accurately match the correct temperature profile. However, the addi-tional work and time expended in obtaining high order, extremely accurate, polynomial repre-sentations is seldom justified. Since the temperature distribution is to be integrated in theprocess of obtaining the solution, the accuracy of the approximation using a low order poly-nomial is improved. Also,an exact representation of the temperature profile may becomemeaningless if the available data from which the profile is constructed is scant and inaccurate(as is often the case). In most cases, solutions of engineerin3 accuracy can be obtainedwith polynomials no higher than the 4th order (which matches aT at the extreme fiber, at thereference axis and 3 intermediate equally spaced points). The matrix solutions for the co-efficients, presented in Eq. (2) in general form, are given in Table 4.1.2.3.1-1 for polynomialsup to the 4th order.

The order of polynomial to be used in any given problem should be determined onthe basis of factors such as the completeness of available data, the shape of the temperatureprofile, and the desired degree of accuracy. Tbre coefficients of polynomials higher than the4th order can be determined from Eq. (2).

The method of polynominal approximation of the aiT profile is illustrated belowfor the case where the values of aT are given at five equally spaced points, as shown inFigure 4.1.2.3. 1-2. The following will be determined:

(1) The polynomial which fits the 3 points®, ® and®

(2) The polynomial which fits all 5 pointss

3.50S:7--® , : 2.25 _

1.50 -3/41. 00 L=/T s=1/ 2

S 1. ooL Ref. 3 x-- Axis -to aTx 103N

FIGURE 4.1.2.3.1-2 aT VALUES AT 5 EQUALLY SPACED POINTS

For a 3 point fit, Table 4.1.2.3.1-1 gives2

2L 2CLT = L = as2 + a0S + 0oTo

wheree oTo 1. 0x10 -3

and

a2 -4 2 3.5 - 1.0 -4 2] 2.5{:}I

WADD TR 60-517 4.24

.~ ..

.

5..

ZA4.

1:: *o: -•

.I- -I I %

All

0- d- e

o.

o. .4 (.4 +4

0 14 04444 1111

14144 " ~ .4 .

aaat ty a C a a a aa6a

OR S

4.1.2.3.1 (Cont'd)

The matrix multiplication indicated above yields the desired column matrix

(la}= HIs B}In general the value of each element, a. offal is obtained by selecting the i th row of

[s] and summing the products of its elements with the corresponding elements of thecolumn B}.

Thus:

ax103 - (4) (0.5) + (-1) (2.5) = -0.5

a 2 x103 = (-4) (0.5) + (2) (2. 5) = 3.0

and therefore

aTxl03= 3.08-0.5s+1.0

For a 5point fit, Table 4.1.2.3.1-1 gives

T = as4 +a 83 + a s2 + als + T

where aoTo = 1.0x10 3

and

a( 16 )(-12) ( 5:y)(-1) .01 1

a2 (-6V~ (76 ) (-37j) (7j. .50x 10 3=

a 96 ) (-128) ( 74) (-16) 1.252 22 2S(-4 (64) (-42j) (14 250La4 -

or

3 1ax10 = (16 )(0) + (-12) (.50) + ( 5.j )(1.25) + (-1) (2.50) = -1. 8333a 1 11

a2 x103 = (-69) (0) + ( 76 ) (.50) + (-37-- (1. 25) + 7-j) (2.50) = +9.667

3 2a3 x10 = ( 96 )(0) + (-128) (.50) + ( 74V (1.25) + (-16) (2.50) = -10.66?

322a4 x (-4 (0) + (64)(.50) + (-42) (1.25) + (10) (2.50) = +5.333

Thus123 4 3 2aT x 10 = +5.333s - 10.667s + 9.667s - 1.833s + 1.000

Plots of the aT profiles resulting from the above polynomial representationsare shown in Figure 4.1.2.3.1-3.

WADD TR 60-517 4.26

4.1.2.3.1 (Cont'd)

1.0

.8

.. 6

.6c~x1 3 =5333T x 103 5. 33382 -10. 667s+9.667s -1.833s+1.000

.4 3 2aTx10 = 3.Os -0.5s+ 1.0

.2

0 Fitted Points

aT x 1030 1 2 3 4 5

FIGURE 4.1.2.3.1-3 POLYNOMIAL aT PROFILES OF ILLUSTRATIVE PROBLEM

WADD TR 60-517 4.27

4.1.2.3.2 Binomial Representation of Continuous Elastic Cross Sections

Approximate representations of continuous cross sectional geometries havingmonotonic elastic width variation (Figure 4. 1. 2.3.2-1) are given by the following basicexpression:

r K 3 6 2!1" (la)

El) =Eobo I + K 2! 0uwhere s d

and Enbn

Eob I (Taper Parameter) (ib)00

The value of K (shape parameter) is determined from the equation[Eub - EbK .2 o' Eh. - Ebo (2)

U,SE b

nn7u ds I

E b.5.5 -

u = .50ds = .50

Reference Axis

FIGURE 4.1.2.3.2-1 CON.TINUOUS CROSS SECTION WITH MONOTONIC ELASTIC WIDTHVARIATION

The taper parameter, 1 , defines the ratio of the elastic width at the extremefiber to the elastic width at the reference axis; the shape parameter, K , defines the shapeof the sides (elastic width variation). Figure 4.1.2.3.2-2 presents a summary of the geo-metric shapes represented through the complete range of the parameters 13 and K.

* < - 1 gives negative width and is thus meaningless. K < 0 occurs for non-monotonicwidth variation and results in infinite width at the reference axis.

WADD TR 60-517 4.28

4.1.2.3.2 (Cont'd)

ft 13=-I -1< 1<0 J9=0 13>0

0< K< 1 _

Kl

(straightSides)

K > 1

FIGURE 4.1.2.3.2-2 MONOTONIC GEOMETRIC SHAPES DEFINEDBYPARAMETERS 13AND K

In many cases, the temperature range is such that the variation of modulusof elasticity is insignificant. The actual cross sectional geometry then has the sameshape as the elastic geometry, and geometric shapes such as circles, ellipses, etc.,can be closely approximated by Eq.(la). Note that the actual shape of the elasticcross section need not be symmetrical (see Paragraph 4.1.2.4). Figure 4.1.2.3.2-2is presented in this manner only to emphasize the shape of the sides.

WADD TR 60-517 4.29

4.1.2.3.3 Solution for Continuous Elastic Cross Sections

For continuous cross sections having monotonic elastic width variation(Figure 4.1.2.3.2-1), Lhe elastic section properties for bending about a principal axisare obtained by substituting Eq. (la) of Paragraph 4.1.2.3.2 into the general integralequations for elastic section properties which are presented in Paragraph 4. 1. 1.1. Theresults, derived in Reference 4-1, are given below in non-dimensional form.

AEA

d dE 0b0 K + I

u .. (+fi (2)d Eob K + 1 2

E___ =(1 __Pf--____(3T + AP 2 (3)

d3 Eb K +3

where 13 and K are determined from Eqs. (1b) and (2) of Paragraph 4.1.2.3.2. These"elastic" section properties should be employed in the solution of the deformation and stressesof beams subjected to mechanical and thermal loads.

The total thermal deformation of the cross section can be considered as thesuperposition of the deformations due to the thermal loads represented by each term ofthe caT polynomial (Eq. (1), Paragraph 4.1.2.3.1). The contribution of the Lth term ofthe aT polynomial to the total deformation is obtained by substituting aT = a LsL, andthe expressions for the section properties given above, into the general integral equationsfor the deformations (Paragraph 4.1.1.2).

Thus, in non-dimensional form, the deformations due to an aT profile of the

form aT aLsL are

L/ a+l) (4)6L L L'+I/a L -+-( + K +KL + 1

5L dw I /aL + + K L 2++I (5)L L +k + ---

Equations (4) and (5), taken from Reference 4.1, are shown graphically in Figures 4. 1. 2.3.3-1

and -2 for the case K=1 (elastic cross section with straight sides). Referring to these graphs,note that for the case L=0, corresponding to aT constant through the depth, the non-dimensionalelongation y 0 and rotation 60 have the numerical values one and zero, respectively, for allvalues of taper parameter fl. The case L=I corresponds to an aT distribution which varieslinearly through the depth and thus 6 =1 for all Pt. Note also that the deformations are relativelyinsensitive to changes in the taper ratio for large values of 13.

The total thermal deformations for an aT profile represented by a polynomial oforder "n" are, by superposition:Sn n

L D 01aL YL (6)L=o L__o

WADD Th 60-517 4.30

4. 1.2.3.3 (Cont d)

n n

wt WL= aLJL" (7)

SL=o L=o

The stresses due to thermal loading are determined from the total thermaldeformations by the relationship

" = E -•T + cI + w' (-u-)] (8a)

or alternatively byn

a= E X aJ [ -SL + Y/L+6 L(s (8b

The stresses due to mechanical loading are, from Eqs. (1) and (3) for the sectionproperties,

A E F M~(-)

-dEbo (u-u )] ' 8)

where in the above equations, s, u, E and aT are prescribed for the point at which the stressis to be determined.

WADD TR 60-517 4.31

-.-

4.1.2.3.3 (Cont'd)

L1.0aT aLs

L=0

.8 .6 L--o

L=2

.- - -.4

L=4

.2.

0

-1.0 -. 5 0 .5• 1.0 1.5 2.0Enb

Eb

n n

5 ~ ~ ~ ~E b> IL~0 0 Typical Cross Sections

FIGURE 4.1.2.3.3-1 ELONGATION, -y' FOR CONTINUOUS GEOMETRY OF THE FORMEb = E b (I +s sK), WITH K I (STRAIGHT SIDES).

0 0

WADD TR 60-517 4.32

4.1.2.3.3 (Cont'd)

LatT= a s

L1.2

L=2

1.0 L I

L //000

_.0 11115

0 I," I,0=O-1.0 -. 5 0 .5 1.0 1.5 2.0

Eb00 Eb

APCross --o

SSections a. Eb° 3.310 00

FIGURE 4.1.2.3.3-2 CURVATURE, 6 L' FOR CONTINUOUS GEOMETRY OF THE :FORM Eb =Eb° (1I+(3 K), WITH K = I1(STRAIGHT SIDES).iI

• WADD) TR 60-517 4.33

!C

4.1 .2.3.3 (Cont'd)

The solution of the stress and deformation for an unrestrained beam, with thecross section * shown in Figure 4. 1. 2.3.3-3, is illustrated below.

It is desired to calculate and compare the deformations and stresses produced byeach of the following polynomial approximations to the aT profile:

(a) aTx103 =3.0s 2 -0.5s +1.

3 4 3 2(b) aT x i•0 = 5.333s -10.667s +9.667s - 1. 833s -1. 000 (Refer to

Figure 4.1.2.3-5)

E b =3x106

n nl

E b 8x 10s 1.05 5 d = 3.0"

s= .5u = 1.5"

Ref.Axis I

E b = loxl06

FIGURE 4.1.2.3.3-3 ELASTIC CROSS SECTION

• The elastic cross section shown in Figure 4.1.2.3.3-3 makes physical sense in that across section which has a physical geometry of constant width b has an elastic width Ebwhich decreases with increasing temperature.

WADD TR 60-517 4.34

4.1.2.3.3 (Cont'd)

Solution:

(1) The parameters P and K are determined from Eqs. (lb) and (2) ofParagraph 4.1.2.3.2. Thus

Enbn3Ebo - 1 -. 7000

E~b 10 -

E 5nb. - Eobo

3.32 Log (3.5) = 1.8063

(2) Substituting the above values of • and K in Eqs. (1), (2) and (3) gives thenon-dimensional section properties:

A.=l ~.= -.7000=1+ =1 + 2.80 = .7506K+1 2.8063

1+1 . - - + -. 7 0 0 . 4 2 1 1A= (2 K+2 .7506 2 3.8063

1 + _ ;L = u2 + -. 7000 .7506 (.4211)2 .054b3 K+3 3 4.8063

(3) The solutions of Eqs. (4) through (7) for the deformations caused by each of

the polynomial ciT distributions, are given below in Tables 4.1.2.3.3-1 (a) and -1 (b).

(4) The stresses are obtained by substituting the deformations into Eq. (8).

Thus:

For 2nd order cT polynomial

or = E [-aT + V' + w' (u-u)]

= E -aT x 10 3 + 1.5398 +..7563 (u-u)] x 10-3

For 4th order aT polynomial

c= E T x 10 3 + 1.4974 + .7580 (u-U)] x 10-3

where u = pcd = (.4211) (3.0) = 1.2633

The above stresses are plotted in Figure 4.1.2.3-.3-4.

WADD TR 60-517 4.35

4.1.2.3.3 (Cont'd)

If it is desired to determine the points of maximum tensile and compressivestrains without resorting to a plot through the depth, use of the alternate expression forthe stress, Eq. 8(b) is advantageous. Thus, by substituting into Eq. 8(b), the 2nd orderpolynomial approximation

22 L 2 -3

aT = . a~s = (3.000s -0.500s 4 1.000) x 10L=o

p = .4211

and the values of 6 and -y obtained from Table 4.1.2.3.3-1(a), the following analyticexpression is obtained:

a X10 3 = _3.000s2 +2.769s- .4157E

To obtain the extremum points of strain, set the derivative of q/E equal to zero,

i.e.,

d- E~-~ = (-6.000s + 2. 769) x0-3 = 0

which gives an extremum point at

2.769S 6.000 - .4615 2dThis corresponds to a point of maximum strain since the second derivative is

ds2

negative. Since only one extreme point exists, the minimum (maximum compressive) strainfor the cross section must occur at one of the extreme fibers, as verified by Figure 4.1.2.3.3-4.

Evidently, if E is constant over the cross section, then the points of maximum andminimum strain correspond to points of maximum and minimum stress.

A comparision of the deformations obtained from the two polynomial approximationsof the aT distri:bution (Tables 4.1.2.3.3-1(a) and -2(b) shows good agreement. This was to beexpected since the aT distributions are integrated in the process of obtaining the deformations,thus improving the accuracy of lower order polynomial approximations. The stresses, on theother hand (Figure 4.1.2.3.3-4), do not show as good an agreement. This is due mainly to thefact that the unrestrained beam thermal stresses are obtained from Eq. (8) as small differencesbetween large numbers, and are therefore sensitive to small changes in the deformations. Thusa very accurate determination of the thermal stresses requires an even more accurate calcula-tion of the deformations, which in turn requires the use of higher order, more accurate, poly-nomial approximations of the aT profile, or the use of the finite sum method. It becomesapparent that the temperature distribution must be known to a high degree of accuracy (which isusually unobtainable in actual structures) to obtain accurate thermal stresses. The high degreeof accuracy of temperature distributions is not required for the determination of thermal de-formations.

WADD TR 60-517 4.36

-- 7: 2

S÷ -

--,

a °. .

÷ • ° . . I

S• ÷i~

- ai

________________ � � � �

A

- 0N 0

0 C N0 � 0 Q0�No a � - eec..

ICC, 01.00 .. * e CIa.

* eec- c-a

a CI*0 ��-' -C CC

- -, ICa 0

N -

I- *��C N

0 a

-F* C

a

I' 'C -

4a� �a 'I* 0

o -

CC 0 0 N

0 j 0 0.C�0 e

N � C- S * - -

- a '-, 1-

'CS

-kmZ

< �- I

I

- 1( CN a

N

o 0

� t**O 01-1.400 0 1-

* S N o.- � aa '� ICC-C -

IN

-I, 04 U

zo - - 0

o *

1.1 � 0a �- .� U.4 + 0a 001-c-na 5 0� a NO - 00000 a

t.CC.0 00000 o0.0 .� .,.< U0 N-eec

2 0 I

'a U

Ia NI-C 0-I, i-o

* 0 'C-CA 0� 77 ___ _____ � - - ______

____ _________ __________ p

c4

4.1.2.3.3 (Cont d)

3.0

3 22.d 17x 10 = 3.Os -0.5s+ 1.0

aTx 10 3 =5.333s4 -10.667s3 +9.667s2

-1. 833s 4 1.000

u

1.11

0 Ref. Axis

-. 60 -. 40 -. 20 0 +. 20 +.40 +.60

x 10

FIGURE 4.1.2.3.3-4 THERMAL STRESSES FOR POLYNOMIAL UT DISTRIBUTIONS

WADD TR 60-517 4.39

4. 1.2.3.3 (Cont d)

The relationships for the section properties and deformations given in Eqs. (1)through (5) can be simplified if the elastic geometry is symmetric about the bending axis,Figure 4.1.2.3.3-5(a). For this case, the non-dimensional section properties reduce to

= EA 2_ (9)s dE b K + I0 0

-S0 (10)d

andZI 2 EI1 +

8 d3 K+30 0

where d is the distance from the reference axis to an extreme fiber and 1 and K are deter-mined from Eqs. (ib) and (2) of Paragraph 4.1.2.3.2 using the elastic width variation oneither side of the reference axis (bending axis of symmetry).

The solution for the deformations is obtained by first decomposing the totala T distribution, Figure 4.1.2.3.3-5(b), into components which are symmetrical and anti-symmetrical about the bending axis (Figure 4.1.2.3.3-5(c) and -5 (d)).

Thus if, as shown in Figure 4.1.2.3.3-5(b), the aT profile above the bending

axis is represented by the polynomial

n,v = •.+....................... +ansn (122)

L=o

and the caT profile below the bending axis by

noT a (-s) = a' + a' (-s) + . a' (-S)n (12b)

L o 1 nL=o

then the symmetri( al component of the total aT profile is

nnS a L a + a - aa

____ rolOaa + ~ ~ at

n 0

S(13a)

+, 2 (±on

and the anti-symmetric component is

WAD- at L 0 -5 o7 4a 2

( -~ a' nis•. +(n nlsn (13b)

2?

i ~ the reference axis and the lower (negative) signs for points below the reference axis. •

SWADD TR 60-517 4.40 144

I ea

.3 *4.

- a

£ - 0

a-I" aN.� �

0fl[ aL�%F.j 0

C.)

a dlaj U

�F.

a0

aH!

a

a

a F. 2

aNl * 0

I- aNS F.� A" I. � 0F. ab aV.

* 3-. AF. -� * A±

-iF.

a 0- a a 1.3

-. 0 a1

3* mm n a a C

a: a: aIC�- 9 a

U a OIAJA �' V.!V. *an 3 01-j - aL.

C TF.0 3-an p

A aq� AC

a 3.3L.a �0 A z........ 0

1.3 -

ai �1*0 * I

-. - a'.a . - 1.2a

t. a C

- 00 I

4.1.2.3.3 (Cont'd)

For the symmetrical component of the aT profile, Eqs. (4) and (5) reduce to

'L~ a +ats _5 L L+I + + 1T) (14)( aLaL s

2

dwt L, s 0dwLs = 0 (15)

(L~s - aL +a' L

and for the anti-symmetrical component of the aT profile

L, a 0 (14a)(L)a= aL + a' L

2

(6 dw L, a 2 1 + (15a)L)a (a. a L' ) L +2 K +L +2

L zs

The total deformations are obtained by superposition of the symmetrical and anti-

symme•-•ical components as

n n

Lo

n nI F-= 1 aL - a? LW1 W1 ,' a - 2, '2 )d( ) (17)

L=o L=o -_

Note that the elongation, ? I is caused solely by the symmetrical component of the aTprofile, while the rotation, w' is caused solely by the anti-symmetric component ot theprofile.

4Once the deformations have been determined as above, the stresses can be obtained

from the relationship

a = E (-aeT + ' + w ' u) (18)

where u and aT are prescribed for the point at which the stress is to determined.

The method of solution is illustrated for an unrestrained beam having the geometryand aT profile shown in Figure 4. 1. 2.3.3-6. Stresses and deformations are determinedas follows:

WADD TR 60-517 4.42 i

4.1.2.3.3 (Cont'd)

51 2

= 50S +5150ll0 10-3

___Ref.

0 Axis 2 L_

sI L

S• •~L=o as

-s, -u [5.T00 1.50(-s) +.50] 1O-

u-2

S(Straight Sides) (bP aT Profile

FIGURE 4.1.2.3.3-6 SYMMETRIC ELASTIC CROSS SECTION WITH

UNSYMMETRIC aT PROFILE

(1) From Eqs. (lb) and (2) of Paragraph 4.1.2.3.2 or Figure 4.1.2.3.2-2,

K= +1

(2) Substituting the above values of P and K in Eqs. (9) and (11) gives the non-dimensional section properties

As •=2 (1+ P+1 2 () =2 - =1

3 K+3 3 4 6

(3) The solutions of Eqs. (14), (15a), (16) and (17) for the deformations are workedout in Table 4.1.2.3.3-2.

(4) The stresses are obtained by substituting the deformations in Eq. (18), thus

a = E (-aT + Z' + w' u)

= E (-aT x 103 + .9583 + .0750 u) x10

These stresses are plotted in Figure 4.1.2.3.3-7.

WADD TR 60-517 4.43

r~pm ý-i

4. 1.2. 3.3 (Cont d)

++ '

+

CD cv) toa '.~J-.+ + -.4 .

e CdE).

£0 to

s-4 (i Lco

00

0 t

N 0A0S - - 0

£00

£0 Cd

0V co~0£

z. x,.

-c-S

-+ 1+I1 . 4Q

+ 0 1

£0I +~ +.

x C4 0 0 0to' Q £0 £0 0+D 00

r4 93- 0

£00 0-C4+ +q 4 0

0m

If 01;? AP4' :3 £0 £0 £0

U 'u-. $

WADD ~ a Th6051 44

4.1.2.3.3 (Cont'd)

2.0

1.5

1.0

.5

Ref. 0

-3.0 -2.0 -1.01.2030

-1.0

-2.0

FIGURE 4.1.2.3.3-7 THERMAL STRESSES FOR THE BEAM OF FIGURE 4.1.2.3.3-6

WADD TR 60-517 4.45

4.1.2.3.4 Solutions For Discontinuous (Multi-Rectangular) Elastic Cross Sections

A wide variety of beams are of the composite or built-up type, fabricatedfrom extruded shapes, bent-up sheet, flat plates, etc. In most such cases, the elasticcross section can be considered to consist of a finite number of rectangles as shown inFigure 4.1.2.3-1(c) where the elastic width Eb varies discontinuously through the depth.

The solution for the deformations and thermal stresses in an unrestrainedbeam of this type, for bending about a principal axis, as derived in Reference 4-1, is

presented below. In order to systematize the solution, a step parameter 7) is introduced,defined

as

In= + (kek-1ek) (1 Skm (1)

where, from Figure 4.1.2.3.4-1,

Ekbkek E b (width parameter)

k 1, 2,. .

"uk'k=-d

n = number of rectangles less one

m = an integer.

Enbn.

nu=d

5 kd Ekbk u sd

1 dRef.

Eobo

FIGURE 4.1.2.3.4-1 GENERAL MULTI-RECTANGULAR SECTION

WADD TR 60-517 4.46

S4.1.2.3.4 (Cont'd)

4. . Non-dimensional elastic section properties are expressible In terms of the,• step parameters as

EA(2S = dEb 71 (2)

"U 772 (3)

EV 3 773 -)/2(4) "/=d3 Eobo

Note that the above expressions correspond to Eqs. (1), (2), and (3) of Para-graph 4.1.2.3.3 for continuous section properties, where

+ +___ , _ ) I ___( +(12 K - + +2 3 K-- + 3

have been replaced by 711, 772' and 773, respectively.

The deformation modes, for an aT profile expressed as a power seriesn L

aT a ,are- 0

V L 7- L +I (1)

(5)i

L= aL A (6)6L = aL V1 7-- L+2 -'71L +1)()L 1

The total stresses and deformations are obtained, as in the case of continuous geometry,from Eqs. (6), (7) and (8) of Paragraph 4.1.2.3.3. If the aT polynomial is higher thanthe first order, then to obtain the stresses and defarmations, one must evaluate, in addi-tion to '?1' 712 and 773, values of '7 up to '7n+2 where n is the order of the aT polynomial.

WADD TR 60-517 4.47

4.1.2.3.4 (Cont'd)

The solution for the deformations and stresses of the unrestrained beam shownin Figure 4.1.2.3.4-2 is illustrated below.

7. b E =10 x 10 (Constant)7,=b3960 x 10

10.0 =d

10.=d OT= (1360s 2 +1300s +1300)x10 6

1.0

___T /_ V _ __

10.0b -61300 x 10

(a) Cross Section (b) aT Profile

FIGURE 1.1.2.3.4-2 UNRESTRAINED BEAM CROSS SECTION ANDTEMPERATURE DISTRUBITION

(1) The numerical calculaticu for the necessary step parameters, Eq. (1),is carried out in Table 4.1.2.3.4-1(a)

(2) Non-dimensional section properties are calculated by substituting theabove step parameters in Eqs. (2), (3) and (4). Thus

= =7.2320

172 .0950X - = .2320 = .4095

]= 17 - x ,2 = .0728 - .2320 (.4095)2 = .0339

(3) The solution of Eqs. (6) and (7) of Paragraph 4.1.2.3.3 and of (5) and (6)above for the deformations is carried out in Table 4.1.2.3.4-1(b).

(4) The stresses are determined' Eq. (8) of Paragraph 4.1.2.3.3 by

a= E [-aT + C' + w' (u-

10E-6 1-6)= -aT + (2259x10-) + (26. O (u-4.095)]

wherei = ii = (.4095) (10.0) = 4.095

WADD TR 60-517 4.48

14

- -~ ýZ

4. 1. 2. 3.4 (Cont'd)

to r

1-4 00-

-V 0o V1 L

. .

40 CO0- -

0I

r-4C 0CoC

1- c. C Cco -10t.- 0 0 0o

bO C4. 0 0M

P4 .ý M-

.C' 'P- *4 r m

C44

V-4 0-+ + + +V- 0 -To-4 -4

0 o?0 Mo Go 0r 0 tC o / -% 0 -4 o o 0J L. 1

.0®C .0C

.~~~ ~ I q I 4 ,e00,IC ~

0~~ v-4 UU

rz~-r-4

00

-4 -4 4 -4

V-4 c-)jc'

WADD TR 60-517 4.49

4. 1. 2. 3.4 (Cont d)

xf

Cd

0

80 C-3

'44-

00

o o-4c

LO to 0co I ~I 0

ot0.P4- CD

0o 0 m

oH

-..44 -

Ito-

m. M

~~C110)OD 00 0

_m t- - _ _

-. r4

+a IV n CDC

Q 0ol to II

+ 4o C) II

E)r~ CD 31

WA04 0R 6514.0 0

4.1.2.3.4 (Cont'd)

As in the case of continuous elastic geometry, the equations for the section pro-perties and deformations are simplified if the elastic geometry is symmetric about thebending axis (see Figure 4.1.2.3.4-3).

Ek bkE Eb

________n nf

s --kkuk

Sl utd 0o0 -Ref. Axis

S I (Bending Axisof Symmetry)

I I 8I-

----------------------------L--------I--I I S = -

LT_ __ ... u i-

r- L• .I d

FIGURE 4.1.2.3.4-3 MULTI-RECTANGULAR SECTION WITH ABENDING AXIS OF SYMMETRY

in this case,the non-dimensional section properties of Eqs. (2), (3), and (4) reduce to

= EAX s - dEob' 271 (7)

U _

Is d (8)

and

El -=2 73 (9)d Eb

where the step parameters are determined from Eq. (1) using the geometry on either sideof the reference axis and d is the distance from the reference axis to an extreme fiber.

The solution for the deformations is obtained, as in the case of continuous elasticgeometry, by decomposing the total aT distribution into components which are symmetricaland anti-symmetrical about the bending axis (see Eqs. (12) and (13) of Paragraph 4.1.2.3.3and Figure 4.1.2.3.3-5).

WADD TR 60-517 4.51 r

4.1.2.3.4 (Cont'd)

For the symmetrical component of the aT profile, Eqs. (5) and (6) reduce to

ELs 2(YL)s (a + a 17L+1 (10)

LwsLa s = 0 (11)(5Ls=(aL 0 'L/2

and for the anti-symmetrical component of the aT profile

( = L,a -0 (12)(YL)a (aL - a'L)/ •

dwt

6 La(a= aL a' L)/2 s7 (13)

Once the deformation modes have been determined,the total deformations andstresses are obtained by superposition from Eqs. (16), (17) and (18) of Paragraph4.1.2.3.3.

Solutions of Eas. (10) and (13) are plotted in Figures 4.1.2.3.4-5(a) for tiespecial case of two rectangles* (sections such as channels, "I" beams, tees and cruti-forms conform to this configuration) with s = .97.

nI

* Practically all aircraft structural sections can be approximated by multi-rectangularconfigurations with respect to the principal axes. See, for example Figure 4.1.2.3.4-4in which a zee section is approximated by a symmetrical two rectangle geometry.

WADD TR 60-517 4.52

4.1.2.3.4 (Cont'd)

tt

ago wt' 2

ICOOO

m ;2 _ _ _ _ _ _ ............. .......

FIGURE 4.1.2.3.4-4 APPROXIMATE REPRESENTATION OF A ZEE BYA TWO RECTANGLE SYMMETRICAL GEOMETRY

WADD TR 60-517 4.53

4.1.2.3.4 (Cont'd)

Symmetric Temperature Distribution(NOTE: (6 L)s = 0)

-En n•

E= d Ref.

Axis

u L

n = .97 TSn dT

1.0

.9

.8

.7

.6"

.4

00 2 4 6 8 10 12 14 16 18 20 22

n ne E o b

FIGURE 4.1.2.3.4-5(a) ELONGATION (yL)s VERSUS WIDTH PARAMETER (e)

WADD Th 60-517 4.54

4.1.2.3.4 (Cont'd)

Anti-Symmetric Temperature Distribution(NOTE: (yL)a 0)

Eb

E- n bn-_

E 9b =17d d

Ref.

Axle

uan d • .97

1.5 _

1.4

1.3 L

1.2

1.04:4

33

.3 .9

.8 4

.7

000-

1' .4

.33

0 2 4 6 8 10 12 14 16 18 20 22

E be=

0 b0

FIGURE 4.1.2.3.4-5(b) CURVATURE (6 L~ VERSUS WIDTH PARAMETER (e)

WADD TR 60-517 4.55 L

4.1.2.3.4 (Cont'd)

The solution for a symmetrical elastic geometry Is illustrated below.

The box beam shown in Figure 4.1.2.3.4-6 is subjected to an aT distribu-tion which varies through the depth as shown on the right. It is assumed that the aTdistribution is constant across the width. To find the deformations Z 1, w' and the thermalstresses, it is necessary that the step parameters be calculated from Eq. (1) in Table4.1.2.3.4-2(a); the deformation modes as given by Eqs. (10) and (131 are calculated inTable 4.1.2.3.4-2(b).

Thus, from Eqs. (16) and (17) of Paragraph4.1.2.3.3,

nz X ()L . x 10- 6 1518 x10- 6

L;=0

n

t (W')L, = • Col. x 10-6 73X 10-6

L1-0

and from Eq. (18) of Paragraph 4.1.2.3.3.a E [ -cT + (1518 x10-6) + (73x10-6)u

Covers: Titanium, E = 15 x 106; Webs: Aluminum, E 10 x 106

4iT = a~s = (600s + 600s + 600) 106

50:/ " •-

2

S4.011 ,+ Ref .10 -Axis

.10 (ryp)-TP

3 -

t) L [__(3 _

L =L a'= L(-s)L _[900-s)3 +600] o0-6

FIGURE 4.1.2.3.4-6 BI-METALLIC BOX BEAM WITH SYMMETRICAL GEOMETRY

WADD TR 60-517 4.56

!..

0) -.

in a - --

C S Q

CC a- - -ma --

S 3 a 0

iflx~dlf:-1 S Ca

It -- 0

* Ký

-Q 0 C - . q

4.1.2.4 Power Series Solution - Corrections for Bending About Both Axes

The power series solution discussed in Paragraph 4.1.2.3 assumes that bendingtakes place about only one principal axis UU . This is the case when the VV principal axisis an axis of symmetry for both the elastic geometry and the aT distribution. Bending aboutthe VV axis (w. J 0) is caused by unsymmetrical geometry and aT distribution about theVV principal ais. The correction requires the addition to the solution of the bending aboutthis principal axis VV. The final deformations are the axial elongation and curvature w' U

about the UU axis solved previously plus the curvature w' about the VV axis.v

The final thermal stress is

a = E [-IT + V' + wv (u-u) + W' (v-V)1 (1)

Note that any distribution of aT can be solved by superposing the solution into two parts,each being the solution for the average distribution (Figure 4.1.2.3-2) about that principalaxis. The solution about any axis can be accomplished by the power series or finite skmmethod, whichever is most readily solvable.

4.2 STATICALLY INDETERMINATE (EXTERNALLY RESTRAINED) BEAMS

In Sub-section 4.1 it is shown that in externally unrestrained (statically deter-minate) beams, thermal loads produce deformations and self-equilibrating internal stresseswhich are compatible with the internal requirement of the plane cross sections remainingplane. In externally restrained (statically indeterminate) beams, compatibility forces aregenerated at the restraints in order to make the beam deflections satisfy the external (bound-ary) conditions. These compatibility forces (redundants) and applied mechanical loadsproduce stresses and deformations within the beam which are superimposed on the unre-strained thermal stresses and deformations.

Paragraph 4.2. 1 discusses beam deflections which must be known in order tosatisfy compatibility conditions. A knowledge of the deflected shape of a structure is alsoimportant from a design point of view.

Fixed end reactions due to distributed thermal and mechanical loadings are pre-sented in Paragraph 4.2.2. It is subsequently shown in Paragraph 4.2.3 how distributedthermal and mechanical loads can be replaced by their equivalent fixed-end reactions atdiscrete structuaral load points, thereby facilitating the application of the methods of influenceand stiffness coefficients to the solution of indeterminate beam problems.

4.2.1 Beam Deflections

Under the assumption of plane cross sections remaining plane after defor-

mations have occurred (internal shear strain energy neglected), the deflections ofa beam subjected to combined thermal and mechanical loading can be determinedfrom a spanwise integration of the total unit deformations of the cross sections.

The total deformations, which are a function of the distance x along the span(see Figure 4.2. 1-1), are obtained by superimposing the deformations due tothermal and mechanical loading, i.e.; i

WADD TR 60-517 4.58

A.

4.2.1 (Cont'd)

wt) W 'W, W + w(X) ( , + MEl El

jt(x) --- •t() + 7(x)= + (2)AE AE

where

wt(x) = Total rotation (curvature) of cross section per unit of length.

w' (x) = Rotation of cross section of unrestrained beam, per unit of length .M1due to thermal loading = :

w(x) Rotation of cross section per unit of length due to applied and

redundant (mechanical) loading = -

EI

W t(x) = Total axial strain at elastic centrold

' (x) = Axial strain of unrestrained beam at elastic centrold due to thermal

loading = F'TE_

S(x) = Axial strain at elastic centroid due to applied and redundant (mechan-

ical) loading - FAE

The total curvature wt(x) and axial strain it(x) have the same significance in theM Fcalculation of thermo-mechanical deflections that the familiar El and F for purely

M_ F' Mmechanical loading problems. Thus, once the quantities , __ , _ and

FE1 AE E1

F have been determined by the methods of Sub-section 4.1, the beam deflections can

be calculated, as usual, by energy methods.

y

(x)= ' +

%Z M' M

-- Wt (X) Elý ;

FIGURE 4.2.1-1 CURVATUREwx) AND ELONGATION Zt(x) DUE TO COMBINEDMECHANICAL AND THERMAL LOADING AS A FUNCTION OFDISTANCE ALONG THE SPAN

WADD Th 60-517 4.59

4.2.1 (Cont'd)

By the principle of virtual work, derivable from energy considerations (Paragraph2.1.2), the deflection and slope at any point on a line structure subjected to combined me-chanical and thermal loading (neglecting shear strain energy) are

A = - mdx +fit-fdx (3a)

e = -fwt ml dx +fit fI dx (3b)

where m and f are the virtual moments and axial forces, respectively, caused by applyinga unit virtual load at the point in the direction of the desired deflection component; m' andfI are the virtual moments and axial forces caused by the application of a unit virtualmoment at the point at which the slope is to be determined.

The principle of virtual work equates the external work done by the virtual forcesystem acting through the actual external displacements to the corresponding internal energydue to the internal virtual stresses acting through the actual internal strains. Because ofthis, the deflection or slope at the point of applied virtual force is determined relative to thedatum defined by the reactions to the virtual force. The inherent advantage in this conceptlies in the fact that once the actual stresses are determined for a statically indeterminatestructure, the deflection or slope of any point on the structure can be determined relativ2to a datum located in the vicinity of the point by dealing with the portion of the structurelying in that vicinity, since the virtual stresses outside the region affected by the virtualforce system are identically zero. The virtual force system can be taken as a simpledeterminate one, thereby avoiding the solution of an additional indeterminate problem in-volving the virtual forces and considerably minimizing the work involved.

The preceding concepts are illustrated by the following problem in which Figure4.2.1-2(a) shows the distribution of curvature and axial elongation in an indeterminatebeam; this information has been obtained from the solution of the indeterminate beam pro-blem given in Paragraph 4.2.3.

Under the action of thermal and mechanical loads, a point initially located at 0moves to the deflected position 0 . In order to determine the absolute deflection and ro-tation of this point (Figure 4.2. 1-2(b)), the virtual loads must be reacted at points whichdefine a fixed datum. Point A, located at the fixed wall, provides one such datum sincethis point has zero vertical and horizontal deflection and zero slope. TIus, as shown inFigure 4.2.1-2(c), the virtual force systems are obtained by reacting unit virtual loadsand moments applied at point 0 by determinate shears, axial loads, and moments at point A.

WADD TR 60-517 4.60

Iw t Ox2 - 14.l b, 261.6):x0 0 w(7.$ x -219.5x.1590.8)1 P04

>Wt~ (4.0 *1.X-33. tlo (161.1) 1 10

At 0

(a) titri-lrtibUo of Curvaitur and Axial E0lo0aptin In ao ltsdetrmlna". BeM

B C

)b) Deflectino and Rotation at Point "0" M'Iasured From a Fixed Dat-m

Ilb

Sin-lb .,LljA0 BC

b/V7 1 n -lb S1

S(C) tM~t VlrmuI/ Lod Al at Point '0" am igce at Point "A" (3 cases)

(d) Unit Vertial Virta load Applied at Point "0" and Reacte at Point "B".

CI -A0 "C"••

(c) D ct i OS Aon at Point BE" a m teafcted at Point DEFLE CTmeNS

0IU E..- IDEEM AT l b EA ELCI

* 'A*-'--

4.2.1 (Cont' d)

Substituting in Eqs. (3a) and (3b), the absolute values of the deflections andslope obtained are

5 5A -f wt- mvo +fot - fdx

-f(4.0 x2 _114. 1x + 2,31. 6)(x -5) dx +] xl10 6

0

= 1100 x 10-6 inch.5 5 _

A oh f wtmhdx +f

= [0 + (2.0 x +0. 0 x-138.9) (-1) dx 0-6 = 486 x 10 inch.

5 5E). f wt- ml dx+f: it V

"5-o (4.0x 2_-114.1x+261.6)(1)dx+0] xl0-6 = -49x10-6 radians.

Figure 4.2.1-2(d) shows the vertical unit virtual load reacted at point B. Point Bdoes not deflect vertically. However, a rotation doe3 occur at this point. Thus, the verticaldeflection of point 0 obtained by reacting the virtual load in this manner is equal to the ver-tical distance from the tangent drawn at B tc the curve at 0' (Figure 4.2. 1-2(e)), or

10 2 -SAt : (4.0 x + 10.9 x -363.4) (x-5) X 10-6 138 x 10-6inch.

OV f

The deflections of a beam-like structure can always be obtained by integrating thecurvatuxes and elongations as shown above. However, as shown In Paragraph 4.2.3, a moreefficient procedure for obtaining deflections, which lends itself conveniently to digital com-puting techniques, consists of using flexibility coefficients (Paragraph 2.1.1.1). Ignoringshear and axial energy due to mechanical loads, the flexibility coefficients are

mi mi dx5 ij El (4)

through which it is possible to determine the deflection at a degree of freedom by super-position, i.e.,

E 6i] p', (5)

where the Pj's are the negative of the fixed end reactions due to mechanical and thermalloads, at thWe locations of the predetermined j degrees of freedom. One of tle advantagesof using this procedure lies in the fact that the flexibility coefficients are approximatelyconstant structural properties wh 4ch do not depend on the mechanical and only slightly on the"thermal loading, Once calculated, they can be used to obtain the deflections for any set ofapplied loads.

WADD TR 60-M17 4.62

! fK

• i

4.2.2 Fixed End Reactions

The fixed end reactions are those mechanical lcads applied at the ends of abeam-like structure which do not allow the ends to move when the beam is subjected tomechanical and thermal stimuli. The negative of the fixed end reactions to the appliedloads and temperatures can be shown to be equivalent mechanical loads acting at thejoints which will produce the same deformations at the joint. This is discussed in

Paragraph 2.1.6 and Reference 4-4. The transformation of mechanic-al and thermalloads acting all along the structure to equivalent loads at the joints assist in permit-ting the reduction of the structural problem to a finite number of unknowns. This as-sists in a systematic procedure for solving statically indeterminate structures (alsoreferred to as indeterminate structures) subjected to mechanical and thermal loads asshown in Paragrapl. 4.2.3.

4.2.2.1 Sign Convention

The sign convention for individual elemental beams is in accordance withstanclard engineering practices which state that loads which cause tension in a member orcompression in upper (positive) fibers are positive. However, care must be taken indetermining the resultant fixed end reactions at a joint. This is shown by Eqs. (1) through(3) and Figure 4.2.2. 1-1 which indicate that the resultant fixed end reactions at a joint inthe structure are obtained by first reversing the signs of the axial and moment reactions ofthe beam to the right of the joint (the sign of the shear reaction of the beam to the right ofthe joint is not changed) and then adding algebraically with the corresponding axial, momentand shear reactions of the beam to the left of the joint. The fixed end reaction on each beamis determined from Eqs. (1b), (2b) and (3b) of Paragraph 4.2.2.2 and (1), (2) and (3) cfParagraph 4.2.2.3. The analyst need not bother to combine fixed end reactions on the leftand right hand side of a joint until the final loads at a joint are required. This last step issimple to do with Eqs. (1), (2) and (3) or Figure 4. 2. 2. 1-1, and is demonstrated inIllustrative Problem IA of Paragraph 4. 2. 6.1 and in Figure 4. 2. 4-1.

TI

WADD TR 60-517 4.63

4.2.2.1 (Cont'd)

x-()Joint

For Joint:

Positive P Loads are up (PL PLL + PLR) (1)

Positive F Loads are to the right (FL = FLL - FLR) (2)

Positive M Loads are counterclockwise (ML = MLL - MLR) (3)

Positive direction aad displacements and rotations are in the same direction aspositive loads and moments.

F_ F

r 1 svi M LL LR LROMP, M 31 POR

FOL (b) Adjacent Beams OR FOR

For Adjaceit Beams:

Positive curvature w when curve is concave downward/ Inside of curve to right when going to right

Tensiie strains for upper fibers

Positive moment causes compression in upper fibersOpposite in sense to positive curvaturePositive moment on right side of beam is counterclockwisePositive moment on left side of beam is clockwise /

Positive trausverse loads are in positive y direction

Positive axial load causes tension in memberPositive axial load on right side of beam is to thc rightPositive axial load on left side of beam is to the left

FIGURE 4.2.2.1-1 SIGN CONVENTION FOR COMBINING FIXED END REACTIONS OFBEAMS TO DETERMINE RESULTANT FIXED END REACTIOS'S ATA JOINT

WADD TR 60-517 4.64

.g

4.2.2.2 Applied Beam Loads

Various types of loading conditions are consider, I to act on the beam. Thelocation of the load is arbitrary (Figure 4.2.2.2-1) but its .Stribution is assumed to beexpressible as a power function. Any loading can be apprc.imated by the superpositionof such loads at different initiation points (see illustrative problems in Paragraph 4.2.2.5)The type of loading (Figure 4.2.2.2-1) is as follows:

(1) Concentrated Moment, M(2) Concentrated Transverse Load, P(3) Varying Transverse Load, q = q LI(4) Concentrated Axial Force, F, with eccentricity, y k(5) Varying Eccentric Axial Force (Shear Flow), =T

(6) Varying Thermal Curvature, w' = w'L

(7) Varying Thermal Axial Elongation, L'= E

With reference to Figure 4.2.2.2-1(e), letTotal transverse cantilever load at L due to applied mechanical loads

ML = Total cantilever moment at L due to applied mechanical load¢.1

= Total cantilever axial load at L due to applied mechanical load

ThenP f1d aqL

L a dt + P + +P (a)1 0

S7L = a q (1-9)dt + Pa + M -a ryd• -0 0a a2q a rLyS•LL

(IL (rftl)(r+2) + M k+1 Fy (2a)

L a f df + F aL + F (3a)"ok+L - ÷ =f + k + 1

L o L 0 rL1L2Sarq+PL+L+= +PL (1by

ML Mo + PoL o+ PoL +rl)(r+2) + Pa M k+1 Fy (2b)

S~a TL

"F = F + •r= + a- L +F (3b)L o L -o k +1where P' Mo, F are to be determined in order to find the fixed end reactions (P

Mo, ML Fo and FL).

WADD TR 60-517 4.65

___ __ S 77 = ý_ I -

4.2.2.2 (Cont d)

SL

I~roqL' q -- qLr rr tk =

(a) Transverse Mechanical Loads (T) Axial Mechanical Loads

(M, P and q) (F and r')

Ka

(e) Unrestrained Thermal Curvature (dll Unrestrained Thermal Axial Elongation

-4---- F X---T

4-X 0

Mý- p .a a .-

(e) Internal Cantilever Load

S~FIGURE 4.2.2.2-1 MECHANICAL, AND THERMAL LOADS ACTING ONCANTILEVER BEAM

WADD TB 60-517 4.66

"zt L

- * ..z.---L~.:. ~ " -..L

4.2.2.3 Determination of Fixed End Reactions

The mechanical loads (fixed end reactions) that are required at the end of thebeam (P , Mo, Fo) to negate the deflections due to the applied loads and thermal stimuli

are presented below in a general form (as derived in Reference 4-4) and are then reducedfor particular types of problems. The reactions at the other end (PL' ML' FL) are

obtained by employing the equilibrium Eqs. (1b), (2b), and (3b) of Paragraph 4.2.2.2.

The general solution for the fixed end reactions at the end of the beam (seeFigure 4.2.2.2-1) is as follows:

V ~c1Po .-0L E0 w'L Pj,c +(Fy-M) po, c +e E (cC n)' (1 z)c-npLj~c o~c \ nI n, c

0Ln=o

YLyr L C~+ k+ 1 Pk+1,c n( cCnnP k+l +n, c

- Ln-nc-n(a

O E 'L) 2 c

+r-f-)(r-2) k r+2,c Ie (cCJ) n (1- ,)c Pmr+2+n,c

-Io~ IPm~~•cn•nl- )e-n mlcnnc]

([ L)2qL)v e)Cfl nnn ]

(rl E r+2 m j +2,c + IF-M (CC e (1-nc

n=o

r+1 rL e r c (1 _,)c-n m

Fo=EIo•f + + e•_cCn r+r,c]

k 1,c fok , c +v n11-k•~n

v0 L fP c m +FC + (cC'n ()zl ( 1 _))"n f,4n,c'] a

nn-+c1 ic (c Cnn( 1) cIn

where Lnc'k4lnc

= r Jc = 0+1)D [ + e ~l n+2 j+2(1 +-.)- (eb)

SMo /M

F 0 V F f , +e : 1 IC V __ (I-V

•72• =m fl~ir 0 I -- -" e ( c± - '$ )' + ( + c+c'2-/ (2b) I.

Eo Lj,cL 6 + c Iý[j°c I

% V

WADD TR 60-517 4.67

- - ."Ali

4.2.2.3 (Cont' d)

F- 1 (3b)

whereD[=-4. +e[ e 3 - (c +1)(c +2) (c+3) + e2 2 (4)

(c +1) (c +2) (c +3)

and

j= exponent describing curvature variation due to temperature or themoment variation due to mechanical loading, e.g.,

j = o, pure moment or eccentric axial loadwhen J = 1, concentrated transverse loadJ = r+2, for a varying transverse load, q = q Lr

kj = k+1, for a varying eccentric axial load, T = rL T

and alsoEI

e ELI 1 O bending stiffness ratio parameter (5a)ElIL L

EAel -A 1 = axial stiffness ratio parameter (5b)

LAL

i = a/L = ratio of loaded length of beam to length of beam (5c)

c = power variation of 1/EI = Eo 1 e ()c (5d)

c'= power variation of 1/EA = EAo e x (5e)

E0 LIL = Bending stiffness at x = 0 and L, respectively

L'LEoAo ELAL Axi-•l stiffness at x = 0 and L, respectively

c Cn = (n+1) term of binomial expansion, i.e., the number of combinations ofc items taken n at a time = c!/n! (c-n)!

cCo = First term of binomial expansion = 1

The above equations are obtained by applying the mechanical and thermal loads tothe statically determinate cantilever fixed at L and determining the deflections at the freeend 0. The loads necessary to negate these deflections are P Mo ,and FO. The solutionis given in terms of each type of load and includes the effect of varying bending and axialstiffness. Solutions are presented in terms of influence parameter (p. , m. , f. ) which

J, c J'c J'care the fixed end reactions (P 0 M0 , Fo) to a unit type of mechanical or thermal load. For

example, m MoJc - is the negative of the ratio of the fixed end moment, M, to the

WADD TR 60-517 4.68

4.2.2.3 (Cont'd)

cantilever moment, L which would cause a Jth power variation of curvature on a beamwhose 1/EI varies as a power of c. Similiarly, p j c = ;'L and fJ, c = -L The

formulations and graphical aids can be employed to obtain the fixed end reactions due tomost types of mechanical or thermal loadings. The mechanical and thermal loads arefirst allowed to act ,, Lon the cantilever beam. Then the loadings and thermal deformationare decomposed into a sum of loadings as described in Paragraph 4.2.2.2 and illustratedin Figure 4.2.2.2-1. Each component is then analyzed for the fixed end reactions (PM , and F ) it causes. This analysis is done in accordance with Eqs. (1) through (5S orw % the gr'aphical aids, Figures 4.2.2.3-1 through -6. The total fixed end reactions arethe sum of all such effects.

Illustrative examples follow to illustrate the computation technique. Graphicalsolutions of the equations are shown for special conditions which simplify the generalequations by eliminating terms inside the summation sign (i.e., c = 0 or v = 1).These cases are of practical utility and are shown in Figures 4.2.2.3-1 through -6. Thespecial conditions are:

(1) El constant and any distribution of applied load ( 4(2) i/E1 varying linearly and moment expressible as a sum of power functions

initiating at point 0 ( v= 1)(3) 1/El varying parabolically and moment expressible as a sum of power functions

initiating at point 0 (v = 1).

Equations (ib), (2b) and (3b) of Paragraph 4.2.2.3 are simpler to apply then (la),(2a) and (3a) thereof, but care must be taken that the correct value of j is employed foreach type of load.

WADD TR 60-517 4.69

4.2.2.3 (Cont'd)

1.6 ---

---- P L' P LT-0

0

1.4------

1.00

00

'.00

00,04

WAD T 6051 4.000

7--

.4 w ji

4.2.2.3 (Cont' d)

0

Eror " _

-.05 L

- - - - -- --

.10--- or/

-,0 I W,o

.15- - - I I.N

.10 3

inan1,00

'N Z

-. 20 .-

- -- I- I"

0 1.

00o0

WADD T-.R 0-5 4 I.30i .4

• : :.2

:••.4 .6 . 8 1.0

-•:• FIGURE 4.2.2.3-2 FEXED END REACTION FOR CONSTANT EI; END MOMENT (Mo

SWADD TR 60-517 4.71 •

& a`

IN~

l ! It I "A I~F fF t -1 i - I ,+,I!

co

W. c4

OfEiI -P cYi

+ ,1 L V 4

+ i ~i~ l l I I.---- ! I I m+•ol l ! f l t i l l i ! i l ! I t I I I i l t i t I I Io=° , . o. , +1..< x

+ e •-.÷ m

0 1 00 .- 0

+ = 0

S•. =, ..+ .,. ,+ •,. o . Y

¼++ r ½-

o I

1 -1

""i

IV I Ila

4.2.2.4 Determination of Fixed End Reaction For Special Cases

CASE A: Constant EI (c=0, e=0, .. D=l/12)

The beam of constant bending stiffness (El) simplifies the general solution byc

making the term j in Eqs. (la), (2a) and (3a) of Paragraph 4.2.2.3 identically zero.

Solutions for P and M are obtained by elemental types of loads and the solutions arethe curves plotted in FAgures 4.2.2.3-1 and -2. The values of P and M are obtainedfrom the equilibrium Eqs. (1b), (2b) and (3b) of Paragraph 4.2.2. L

(1) Pure Moment M (J=O)

M

-a= v L

L 0

From Eq. (2a) of Paragraph 4. 2. 2. 2 and (ib) of Paragraph 4.2. 2. 3

L-POL

+ [ 0 + Y +/]PO, 0 = L = (9+1)0/12) 2• +0 +2'•

P = 6V (1 -Y) (Plotted in Figure 4.2.2.3-1) (la)0,0

-L -MPo =-L---P, o =

From Eq. (2b) of Paragraph 4.2.2.3.

-Mo- - v - +0SM0,0 o -'2Z = 0+1)(1/12) 62 2

M = 11 (3Y/ -2) (Plotted in Figure 4.2.2.3-2) (1b)0,0

S -MY (3v -2).

a.1

SWADD Tht 60-517 4.76 A

SI

/I AA

- J I

z41

~.0 - -

" Limt I

_FT -~'i

t

1 07-v 1 ° 1 ,

I ~

I.J

1 -.

i i b iii

I CI0

°".....~

0

I4 I

,

•

4.2.2.4 (Cont'd)

(2) Concentrated Transverse Load P, (U i)

p0

.a V 1= a/L

Employing the same equations as (1) above gives

L = Pa. PLL *

P1,0= (1+1)(1/12) [2 1-1+2

-PoL, -PoL-PL 0PL V (3 -2 VI) (Plotted in Figure 4.2.2.3-1) (2a)

L YPL=

P2P= (3-2 Y) (2b)

*0 +I - VL - o-Mo + 1 0)]

= m 0 =2(1/12) - 6 + 3 12 +

-M0o m = - (1 - V) (Plotted in Figure 4.2.2.3-2) (2c)

p---L = m1,0

Mo 2p-- = (1- V) (2d)

r(3) Distributed Transverse Load L = --=r+2)

LL o

a 0

Employing the same equations givesI22z/ /L2 q L

L = (r+1)(r +2)

V 6- ] (Plotted in Figure 4.2.2.3-1) (3a)Pr+2,0 - 77L r T+3 r +4

WADD TR 60-517 4.77- !4

4.2.2.4 (Cont'd)

I qL (r+l) (r+2) (r+3) 6- r +- 43b)

M r~In -2 + 6 (Plotted in Figure 4.2.2.3-2) (3c)

_ = - + 6___

r+2, 0 L r +3 r+4

(3dqLL2 =-(r +1) (r +2) (r +3) -2 +4r

If r = 0 (uniform load),

0- (3e)

2 "3( 1 v (3f)

If r =1 (linearly varying load),

= 3 ( 1 1q 4 -O (3g)qLL

M _L 11 v=qLL2 312 20 (3h)

(4) Thermal Loading w ' w JLL

L 4L

From Eqs. (la) and (ib) of Paragraph 4.2.2.3,

PL

oo L ; 12V 1•Pj, 0-6 10 0 w 0 j2 v ORa)

pi, WL = +

! oL L6- 12+Y (4b)

PWADD T 6+1 j 4.2

WADTR 60-517 4.78

4.2.2.4 (Cont'd)

From Eqo. (2a) and (2b) of Paragraph 4.2.2.3,

m°= EoI~WL - • [-2+ 6t (4c)

=o (EJL w'L TF 2 +i 6 (4d)

(5) Distributed Axial Load T =T L •:k0=k +1)

LL

From Eqs. (2it) and (3a) of Paragraph 4.2.2.2 and (lb), (2b) and (3b) of Paragraph 4.2.2.3,

-aT Ly - YLTLY •

"'L k+l k+ 1

a T L YL T L':L =k"l = k+ 1

POL

121,

Pk +l,0 (-IV (-LyTL)/k+l= k +2)1 k+3)

2IlL

Fro .12 an 1 oP a 4

YTL (k +l) (k+2) 2 a)+

LP

M 12, +(ink+l, 0 (-1YLyTL)/+1 ( k +2 6 2(k + 3)

+10 6L v[YF°L (k+1) 2(k +2) 3 k +3]

Fo(k+1

_k +1, 0 L +

(5c)

W LL D T+1) k 4+2)

WADD Th 60-517 4.79

4.2.2.4 (Cont'd)

Equations (1) through (5) define the fixed end moments for all types of mechanicaland thermal loads on a beam of constant cross section. The solution can be obtained by directsolution in the equations or from the graphical solutions presented in Figures 4.2.2.3-1 and-2 for unit solutions which must be multiplied by 4-4L' -ýOL and -:L to obtain Mc, P andF.0

CASE B: Variable El (cj 3) and Load Distributed Over Complete Span (v = 1)

Another condition of structural interest and simplicity is a beam of variable Elwhere the loads are continuously distributed over the entire span (V = 1).

From substitution of V = 1 in Eqs. (1), (2) and (3) of Paragraph 4.2.2.3 thegeneral solution becomes:

P ' Lyrf L L qf2 eP oL= EoW LPjc +-'iPk + c + 1 +c, )(r+1(r-+2) Pr+2,c+ePr+2+c,c (6a)

M EoIw' m + LyL +emk+L2 +1 (6b)0 00 L j,c k+1 k+l,c 1c (r+1)(r+2) r+2,c+L r+2+c,c

F EAEFf + k+1c' le'f +c' (6c)0F o EooLfj,c +k• +1 + 1, +1 klC,

where the values of p- c and mic are given in Figures 4.2.2.3-3 through -6. Terms suchas F, M and P are ignored since they are reacted directly at the support. (Note that recoursemust be taken to Eqs. (1), (2) and (3) of Paragraph 4.2.2.3 if cj0 or v #1).

1Example: The equations are illustrated for a uniform load on a beam of linear - and

FElFIEE =10.

LEL

q lb/in p j

K M~j0

From the data,EI

c =1, 00 = 0 0, r=0, q =qEl. L E ~LO L~q=

and from Figures 4.2.2.3-3 and -4,

Pr+2, c P2 , 1 110 r+2+c, c P3,1 = 115

mn Im -. 010 In In -028r+2, c 2,1 r+21c,c = 3,1

Substituting in Eqs. (6a) and (6b) results in

WADD TR 60-517 4.80

4.2.2.4 (Cont'd)

P L 110 + 9 (.115)]P Lo (0+1)(0+2)

P .573 q L

Lq [-.010 - 9 (.028)] 2=- (0+1)(0+2) =.131q

Substituting in Eqs. (ib) and (2b) of Paragraph 4.2.2.2,

-PL=-.573qL+ Lq= " 0+1

I ~PL . 427 q LLq LI2 2QML .131 qL2 +(-.573 qL)L +To+,)+) =.058q L

L (+

The above solution to the problem is illustrated by the following sketch:

q lb/in.

.058 q jj ).131qL2

.42L q4 573 qL

LL

Note how the fixed end reactions increase on the stiffer end.

4.2.2.5 Use of Equations and Graphs

The technique of using the equations or graphs of the previous paragraph isillustrated in a few simple problems which are purposely made simple to compare withknown solutions and to show the effect of variable stiffness.

PROBLEM A: Concentrated Load at Center of Span L

If PM

ML 0PT _•t .• • ° .•

4 L/2L

(1) Constant EI c = e =o (Note: Eq. (2b) and (2d) of Paragraph 4.2.2.4 could beapplied directly)

Using the graphical solutions,

PL PL0 0=1

L, 1.0 (Figure 4.2.2.3-1, j 1)

WADD TR 60-517 4.81

4.2.2.5 (Cont'd)

1.0.PO= -P,0 = -1.0 .5 PL )PO = -. 5P

From Eq. (lb) of Paragraph 4.2.2.2,

PL - (P+P)= (-.5P + P) -. 5 P-M -MP v mL r 1,0 = -. 25 (Figure 4.2.2.3-2, J 1)

MO = - ml,o ML = .25 (.5PL)

= .125 PL

From Eq. (2b) of Paragraph 4.2.2.2,

ML Mo + PL + = PL (.125-.5 +.5) = .125 PL

The solution (using sign convention) follows:.125PL +P .125 PL

.5P L/2 5P

Checking the solution by Eq. (2b) and (2d) of Paragraph 4.2.2.4,

P2 ~~ 3-2 (.5J -.5

M. = (.5) (1- .5)= .125

(2) Linear EI and EooELI 10 (c =1, e =9, Z/=.5)

From Eq. (4) of Paragraph 4.2.2.3,

1 12 12D +e+(e 2/6) 23.5

From Eq. (la) of Paragraph 4.2.2.3 for Transverse Load Po'SP I 0 1 = CLp i p ,, c + e _. [ ( C C n if ( I c -n &' .o L4l+ n'o Pl' +n, cj

%PL = -Y LP {pl1 1 +e [(1e 1)Pl 1 1 + ,p2 j}%L -Y -'LP {P [ 1 + e (1 -,v)] v e

WADD TR 60-517 4.82

4.2.2.5 (Cont' d)

The influence parameters P 1, 1 and P2 , 1 are obtained from Eq. (ib) of Paragraph 4.2.2.3.

(12 )+51_ {2 2 .5 [1+.5( } ( 1383pit,1 23.5 +1 2 6 1 +2 )

.(12) .(2 ) {+ 2+2 [1+.5(9)]} .1115P29 1 (2.5 2 ++ 1 6 2 +2

O -. 5p 5 5 (.1383) +4.5 (.1115) 631 P

PL (PO + PL) = P(.6312 - 1.00) = -. 369P

Similarly, from Eqs. (2a) and (2b) of Paragraph 4.2.2.3 for Moment

t12 N 9 +3-5- -02mI =(23- )(--) 7 --0426

2 1 . 1 9 3I=-

M =-LP m 1 ,1 [1 +e (1 -Yl) +m2,1 ve

Mo -0- 1-.5LP [5.5 (-.0426) + 4. .209PL

M =M +PL+ 2PL + (-.6312) + .5 078PLL o L + 5 ]

ELIL P EOol 10 E I

.369 P ].631 P

+.078 PL +.209 PLL/2

L I

PROBLEM B: Uniform Load in Center Third of Span and Constant Elq lb./in q lb./in q lb./in

v =.67 v.33

S L L/31 T -- 2 L/3- L/3- L/3 2 L/3 -.From Eqs. (3e) and (3f) of Paragraph 4.2.2.2 (r=O),

P0 -(.67)3 - 12(.67) (3) .

qLL (1) (2) (3) 6 4 ]

8 -1 -3245o *- (4) •- (5) 27o 27272 -1/6

qLL 6 6 6

WADD TR 60-517 4.83-4•

4.2.2.5 (Cont' d)

L 1 13 1/6__-- = -(--i- +-- = -1/

M ~818 (-2+1) (-2+ 1/2)0o 27 2-7 8 -3/2 _-13.012 6 6 (27)(6) 12(27)

ML= .0401 - .1667 + .1667 = .0401q LL

q lb./in

.167 qL L/3 L/3 L/3 .167 qLL

+.0401 qL2 +.0401 qL2

Using Figures 4.2.2.3-1 and -2,

v = .67, P 2 0 = .89, In 2 0 -. 225

V= .33, P20 = .56, In 2 0 =-.167

PoL = -(.67 L)2 q 89)- J.33 L)2 -. 167 q 2

2 ((856)g.'56 Graphical solution

_(.67 _ L)332 jchecks analytical

MO -(.67L) 2 (1.225)- 2 (-.167) = .0407 q L2 equations.

PROBLEM C: Thermal Loading (v= 1)

A description of the thermal deformation is approximated by a power series.

1 2 ( n i

W LOD + Li (ML ~ L2 (WLLJ - wLi L.J=

j nLOE + C L-

1 = xLo +Li (LI " " " J

where x is the distance from the 0 end.

(1) Constant El and EA

From Eq. (4b) of Paragraph 4.2.2.4, and v = 1,

WADD TiR 60-517 4.84

4.2.2.5 (Cont' d)

Fo WL=f 6- + --. 6 -1'-122 + Jl 12PO L + 0 +2) W+l 6-+1+2/+

00 L 12 1+121i+

-PLWL1 + WL2 " " ( - 12)] =6 0 -0 +2 L +1)(J+2) (7)

From Eq. (4d) of Paragraph 4.2.2.4 and v =1,

= E1 0 (-2 +o ) + (-2 + 6 (-)++Mo 1 671 0 )/ 66

M. -- 6- wL2 +" + +1 ++2

(l-J) wL

=2E EIo (J+1)(J+2)

From Eq. (2b) of Paragraph 4.2.2.2,

ML 1Mo +P oL

W~ 6Eolo 2 J 19)

ML= Eo0 o wIo + WL1 + 6- WL2 +" J+1 2 (j+1 )(J 2 )

From Eqs. (3a) and (3b) of Paragraph 4.2.2.3 and v1,

F L Fo -EoAo 00 L + 2 "+ -J =-EoAo1j+ (10)

E11; =10; e 9)

(2) Variable (Linear) El and EA (c=1; EL 10 e9

From Eqs. (6a) and (6b) of Paragraph 4.2.2.4,EI

00+. )-PL P o = L (wLoPo, 1 +WLIPI,I+WL 2 P2, 1 +"

ML = Mo = EVIo (wL m 0 , 1 +WL ml,1 +wL2 2, 2 +... )

and mj 1 are obtained from Figures 4.2.2.3-3 and -4.

E I-PE 10 o L LWI.o (-.41)+ wLl (.05) +wL2 (.115)+...

ML =M= 00o [ WLo (+.46) +WL (+.067) +wL2 (.007) +.E°Ao

Nf c'=1 and EA - 10 and el=9, then from Eqs. (3b) of Paragraph 4.2.2.3

and (6c)of Paragraph 4.2.2.4,

WADD TR 60-517 A.85

2. t

4.2.2.5 (Cont'd)

FL FE 00A AL fJc 0 c f clL_= Fo -1 -- EjA, L

00 i 2 ___

E 0 EAo 0 2;LJ = -- EoAo 2: j +I 1

4.2.3 General Solution of Statically Indeterminate Beams

The solution of a structural problem requires the simultaneous solution of theequations of compatibility and equilibrium. For linear stable structures it is poss-bie toarrive at the final solution by a series of intermediate steps, cuts, (releases) and ,.estralnts,each of which is an artifice to simplify the solution and whose total effect is zero. This isanalogous to the addition of a chemical catalyst which aids in the chemical combination ofvarious elements but does not enter into the final form of the chemical solution. Compat-ibility and equilibrium are maintained (for each degree of freedom) at each Intermediatestep in the solution so that the final solution simultaneously satisfies equilibrium and com-patibility but removes any cuts or restraints that do not exist on the actual structure.

The statically indeterminate problem is simplified with the introduction ofgeneralized coordinates (degrees of freedom). The relationships between these coordinates,and the corresponding loads are sufficient to determine the stresses, and deformations inthe structure. The relationships are either the equilibrium and flexibility matrices or thestiffness matrix (which incorporates equilibrium in its derivation). The loads are thenegative of the fixed end reactions determined in Paragraph 4.2.2. TIe solution is accom- 3plished by solving the linear simultaneous equations obtained from the above relationships.

4.2.3.1 Degrees of Freedom

The solution to a stable structural problem is complete when all the loads anddisplacements (stresses and strains) at each point of the structure are known to an accept-able degree of accuracy. A beam-like structure can be visualized as a network of smallbeams between points of structural interest, i.e., joints, sudden changes in geometry,changes in direction, etc. The linear solution is also considered complete if the loads anddeformations of all such points are known. Thoý relative deformation between the ends ofthe small beams, together with the loads acting on these beams, could then be employed todetermine the internal loads (stresses) acting on the beams. Thus the solution of a structureresolves itseli into the determination of loads and deformations at the ends of the elementalbeams (all the degrees of freedom).

The possible deformations and loads which each end (point) of the elemental beamscould undergo are defined as degrees of freedom of the structure and are the "generalizedcoordinates" of the structural problem. In a generalized spatial problem of three dimensionsthere are a maximum of six possible degrees of freedom for a point. These are the displace-ments in the directions of the three orthogonal axes and the rotations about these axes. In aplanar problem of two dimensions there is a maximum of three possible degrees of freedomfor a point. These are the displacements in the direction of the two orthogonal axes in theplane of the structure and the rotation about the third axis which is perpendicular to the planeof the structure. In many problems the structure will be such that some of the degrees offreedom can be ignored. In some cases, the loads and deformations coresponding to a degreeof freedom tre independent of other degrees of freedom (uncoupled) and the structural pro-blem can be solved as separate problems. In any case, all possible degrees of freedom should .be investigated and simplified (omitted, uncoupled, etc.) before attempting to continue thesolution.

WADD TR 60-517 4.86

4.2.3.1 (Cont'd)

In a given linear problem of n degrees of freedom, either the applied load ordeformation corresponding to each degree of freedom is known, or a known linear relation-ship exists between the load and displacement. Thus there are essentially n quantities to bedetermined before the structural problem is solved. The necessary n simultaneous equationsare supplied by the equations of equilibrium and/or compatibility at each generalized coordinate.

The solutions will be presented in matrix form to simplify the manipulation andmeaning of the equations.

4.2.3.2 Equivalent Loading

In order to minimize the amount of computations necessary to solve the structuralproblem it is convenient to convert any arbitrary loading on a linear structure to a series ofequivalent loadings acting at the points of interest of the structure, which will cause the samedeformations as the actual loads and temperatures, at the points of interest. The interrelation-ship between the generalized displacements and loads can be expressed by a stiffness (K) orflexibility (f) inatrix for a given geometry and material prior to applying the loads. Thereare various methods of obtaining K and f, some of which are demonstrated in Paragraph 2. 1.4. 2and 2. 1. 3 and in References 4-5 and 4-6. The solution would then require obtaining equiva-lent loads from any arbitrary loading and employing the stiffness and/or flexibility matrix Andthe requirements of compatibility and equilibrium at each point of interest to obtain n (n de-grees of freedom) simultaneous equations for the n unknowns. It is possible to solve the pro-blem for unit loadings. This considerably reduces the work involved in analyzing a structurefor more than one set of loads, which can be obtained from linear combinations of the unitloads.

The equivalent loadings at the points of interest are the negatives of the fixed endreactions (see Paragraph 4.2.2). The fixed end reactions are exactly those mechanical loadswhich, when applied to the elemental beams, will negate the end displacements caused by theapplied mechanical and thermal stimulation. Applying the negative of these fixed end reactionsto the elemental beams would cause the same defoi. mations as the applied mechanical andthermal loads. If more than one beam has the same degree of freedom (adjacent beams, etc.)then the equivalent mechanical load associated with that degree of freedom would be the nega-tive of the sum of the fixed end reactions of all such beams.

The beam-like structure can be analyzed as follows:

(1) Impose rigid restraints (imaginary jacks) at all the n degrees of freedom ofthe structure. The jacks are applied at all the joints. This includes thereactions.

(2) Apply the mechanical and thermal loads on the structure and determine theforces in the jacks. These are the fixed end reactions whose solution isshown in Paragraph 4. 2.2.

(3) Remove the jacks and applied loads from the structure and apply the equiva-lent loading (negative of fixed end reactions) to the actual structure.

(4) Employing the flexibility or stiffness method to solve the generalized forcesand displacements at the n degrees of freedom.

(5) Employ the forces and displacements at each end of the elemental beams,together with the applied loads and temperatures, to determine the stressesand strains anywhere in the structure.

WADD TR 60-517 4.87

4.2.4 Mechanics of Solution by the Flexibility Method (Figure 4.2.4-1)

(1) Make sufficient cuts to the indeterminate structure to make it staticallydeterminate. That is, the internal forces acting at any cross section dueto an arbitrary mechanical load can be computed from the equations ofequilibrium alone.

(2) Every degree of freedom can be characterized by the following subscripts(Figure 4.2.4-1 and Table 4.2.4-1):

(a, b) = subscripts for cases in which the applied (Pa = - Ra) is knowna )abut the deformatior Ab is unknown.

(x,y) = subscripte . )r cases in which the applied load is the unknowncut (redundant = PxI load but the deformation (Ay = 0) is known and Aequal to zero.

(r, s) = subscripts for cases in which the load is unknown (P = ?) butthe deformation (A = 0) is known and equal to zero. The s sibscriptsrefer to degrees of'freedom which establish the datum of the cut (sta-tically determinate) structure. The datum defines the position in spacefrom which the displacements are measured. The number of subscriptsof the cut structures are limited to the number of equilibriumequations of the cut structure (e.g., two subscripts for a straight beamwith no axial load, 2ZV = EM = 0). These coordinates are a sub-groupof group (x,y). The selection of this group is arbitrary. However, awise selection will reduce the computations in determining the flexibilitymatrix and/or the accuracy and ease of the solution.

(c,d) = subscripts for cases in which the cut load (Pc) is unknown aswell as the deformation (Ad). There exists, however, a linear relation-ship between the deformation and loads. The forces of the c type can bevisualized as redundants which are on elastic supports rather than onfixed supports such as x type fdrces. The deformation Ad can be com-puted, once all the loads acting on the structure are know, by the linearrelationship which exists between the loads and deformations.

All degrees of freedom described above can exist at a point in a beam, e.g.,a rigid transverse support (y or s) with a flexible axial support (d) and with norotational support (b). Each degree of freedom is associated with two consecu-tive symbols. One symbol denotes a load and the other symbol denotes a de-formation. When a matrix is described by two consecutive subscript symbolsit expresses a relationship between degrees of freedom of the same type. Whenthe subscripts are not consecutive the matrix expresses relationships betweentwo different types of degrees of freedom.

(3) The properties of the sub-structure are computed. The equilibrium matrix gijrelates the loads to the reactions. The flexibility matrix fij ,the elements ofwhich express the deflection at i due to a unit load at J, is computed by variousmethod available in the literature, e.g., Paragraph 2. 1. 1. 1 and 2.1.1. 2 andReferences 4-5 and 4-6.

WADD TR 60-517 4.88

owl,

'.4

4.2.4 (Cont'd)

(4) The equivalent loads (loads in imaginary jacks) due to mechanical andthermal stimuli are applied at all the degrees of freedom. Each point ina planar structure may have three degrees of freedom and each degreeof freedom can be a different type (see (2) above).

(5) Equal and opposite unknown internal loads (Px+ - P and P = - P)

and applied at each side of the cuts. Pc and Pc act on sub-structure and

P and P act on supporting structure (Figure 4.2.4-1). These unknownX- c-N

(redundant) loads are exactly those required to maintain equilibrium andcompatibility at each cut.

(6) In essence there are n unimown quantities to find in order to be able to com-pletely solve the problem. They are the b deformations (Ab), the x loads(P ), the r loads (Pr), and the c loads (Pc). Thus n = b + x + r+ c.

There are r equations of equilibrium, b equations of deformation and x + cequations of compatibility. These are sufficient, in total, to solve for then unkmowns. The following equations are in matrix notation:

"r" Equations of Equilibrium (E.M = 0, MV = 0, 2.F = 0, "r" < 3) for eachsub-structure)

-Pr =gra Pa + grc (Pc + Pea) + gr (P + Pa)

Pr (gra Pa + 9rc Pca +grx Pxa) + grc c grx Px -a)

-Pr - ra aP + grPc + gx (•b)

"b" Equations of Deformation

" b = a P + fbc (PC + PPa) + fb) + Pa)

" b = (fba Pa + f'bc Pca + fbx Pxa) + fbc PC + fbx Px (2a)

bA (2b)AD = fba Pa +fbce Pc + fbx Px 2b

"Ix" Equations of Compatibility of Cut Supports

0 = A =f P f ( p)f ( P )(3a)y ya Pa ycf c ( c +Pa) +x x % xa

=ya a +fb Pc + fbx Px (3b)

"tc"l Equations of Compatibility of Cut Flexible Supports

-1 -1 (P +P&d = K) PC- = K ) -PC+) fda a d c cPa)(4a)

-K ~c~ ~x(4b)-'Q1 Pc f •da T-a + fdc PC + fdx Px 4b

WADD TR 60-517 4.89

4.2.4 (Cont'd) pa

1131s 8 () 9 (b) 10(b) 11(b) 14(y

7 (s) 2 (b) 3 (y) 4 (d) 5 () 12 (b]1 (s) 6 (yl

(a) Structure Under Load

~r7a ~a8 ~a9 ~al0 'all ~a12

(lPria p a2 p x3a Pc4a %a5 (tPx6a

x1Prl3 p! Pxl4aPr7 ýtprl l~3 lc4 px • -x14

(b) Load on Substructures

Psitive Sense ShownR =Fixed End Reaction

+ -'R R x3L Rx3R p .. )x3a (Rx3R x3L) RaR a9 (Ra9R Ra9L)SRa9L Ra9R

(c) Addition of Equivalent Load

FIGURE 4.2.4-1 FLEXIBILITY METHOD

TABLE 4.2.4-1 DEGREES OF FREEDOM

Type Total External Load On the -Beam Deformation Degrees of Freedom

(a,b) known unknown 2,5,8,9,10,11,12(pd) unknown unknown 4(xy) unknown (compatibility) known 3,6,14(r, s) unknown (equilibrium) known 1,7,13

WADD TR 60-517 4.90

4k

4.2.4 (Cont'd)

0 =fda Pa + (fdc + 1cd) Pc Px (4c)

where

o geometry of structure employed in equilibrium equations (either 1Sor moment arm ratios li)

f = flexibility matrix of statically determinate sub-structure = deflectionat i due to unit load atj.(First subscript denotes the row and seconddenotes the column of matrix. Second subscript of pre-multipliermust be same as first subscript of post-multiplier.)

= flexibility of elastic supports of "c-d" coordinates = deflection of ddue to a unit load at c (note inversion reverses subscripts)

f = grouping of flexibility sub-matrices (f f fx)ja ja Jc jx

P= grouping of load vectors = Pa which represents the a

equivalent applied loads.

AI = deflection at "T"

P =load at "all (due to applied loads = negative of fixed end reactions)a

P. = equivalent load at coordinate "i" due to applied loads (neg. of fixed endia reactions)

P c Px = redundant (cut) loads.

The solution of the simultaneous equation results in unit type solutionsin terms of (g. , f. K 1d) and the applied loads ( =r P. (Reference 4-5).

in ij, ii cd~ ple a ra-" rncSPca.

The loads and displacements at each point of interest are as follows:

PC = H P(5a)Pc = ca PaP =La Pa (Sb) "

xP x ~a a

A b = mbaa (6a)

Ad c Hca Pa cd c (6b)

WADD TR 60-517 4.91

4.2.4 (Cont'd)SA=A =0 (6c)

y s

where

H -- d c) -ueo foj (7)

L,~~ H-f f influence coefficients of the (8)Lxa (fyx)-1 ya + fy ca uncut structure (similiar to

influence line)

r g.I gr. I grx) I-I- influence coefficients of the (9)I H uncut structure

ca

tuba = (fba 'c ,x= flexibility of the uncut (10)

Equation (10) represents the flexibility of the statically indeterzinate

structure whereas fi represents the flexibility of the cut structure.

The problem is simplified to a great extent if no flexible supports (c, ddegrees of freedom) exist. In that case,

P = f-I f P (1a)yx ya a

= ~ ''~C-f~a}(l1b)A (ba fb.) ! } a ~ (12a)

A = A =0 . -(12b)y S

4.2.5 Mechanics of SoUton by the Stiffness (Slope - Deflection) Method (Figure 4.2.5-1)

(1) Place imaginary jacks (restraints) at each degree of freedom which fixes alldegrees of freedom in space and will not allow them to move no matter whatloads are applied.

(2) The stiffness matrix Ki is obtained by forcing one of the imaginary Jacks tomove a unit deformatiol while all the others remain fixed in space. Theeloads Imposed on the imaginary jacks are noted as K1t (load at I due to aunit movement at J). Note that element KiJ = 0 for alA points i which are not

WADD TR 60-517 4.92

4.2.5 (Cont'd)

aK =k a + kh+

Imgnr Jacks ab~f~~ tab .

Ab-1

(Note: I[ '- kj

Ill.iab rb)

A1 2 3

1111111-7 7k rbt b r

A) Imaginary Structure Krb(kr-b+ kr,+ kr•

b) Determination of KII

Pa (Applied Load)-Pa (Fixed end reacloD.n n I Kb= Pa

*mgnr f "b LrP

c) Applied Stimulation To Imaginary Structure d) Structure With Imaginary Jacks Removed

aPa tpatp. where PJ a aa ab N a KabIl -1

SAb P / =krbj Ab =kr~ alP

r -rb rbj'-I Kabl raS Pl = kab plAb' r bAp bP; =k2

~ rb 'Ab

p2 -kb2AAD. P2 =k 2~a Ab' r rbAb it~ a Lb' r rb A

e) Loads In Individual Members

FIGURE 4.2.5-1 STIFFNESS METHOD

WADD TR 60-517 4.93

4.2.5 (Cont' d)

in the Immediate vicinity of J since there is relative motion of the Jack onlyat J. It should also be noted that all elemental beams meeting at a commondegree of freedom are springs in parallel since they all have the same

motion and the K = 0 kJ where 0 is a force at i due to a unit deforma-

tion at J of elemental beam bounded at J. The values of the elemental 0 1 sare given in Paragraph 2.1.4.

(3) The mechanical and thermal stimuli are now applied to the restrainedstructure and the equivalent fixed end loads are resisted by the imaginaryjacks.

(4) Since the displacements at the degrees of freedom resisted by the imaginaryjacks, it is assured that compatibility at each point of interest is satisfied.Equilibrium at each joint is maintained by loads in the imaginary jacks whichexactly balance the equivalent applied loads resulting from loads acting on theelemental beams between the joints.

(5) The degrees of freedom can be characterized by the following subscripts:

(a, b) = subscripts for cases in which the deformation is unknown (Ab) Ibut the applied loads are known (Pa).

(e, f) = subscripts for cases in which the deformation (Af is unknown

but the applied load is zero (P = 0). This coordinate does not requirean imaginary jack except for tte determination of Kej

(r, s) = subscripts for cases in which the deformation is known to bezero (As = 0, an actual support) but the redundant load (Pr) is unknown.This coordinate does not require an imaginary Jack since the structuresupplies an actual Jack (restraint).

Note that elastic supports (c,d) need not be specified by generalized co-

ordinates but are separated into two components. The elastic memberis incorporated into the overall structure and coordinates (c,d) whichdescribed where the elastic member supported the sub-structure be-comes (a, b); the other end of the elastic member which represented theactual datum becomes (r, s). The (x,y) coordinates are in the general(r, s) group.

(6) Remove all imaginary jacks. The structure will now deform. The load inthe imaginary Jack produces deformations. The final deformation will beone in which the equivalent applied forces are exactly balanced by the forcesproduced by the relative deformation of the elemental beams whose commonends all deform the same amount (KA = P). The jacks could have been re-leased one at a time and then reapplied; such a procedure would be a trailand error method (e.g., Hardy-Cross Moment Distribution Method).

(7) The requirement of equilibrium at each degree of freedom results in thesimultaneous equations

k ab AAb +KafAf +Kas A- Pa=0 (1)

WADD TR 60-517 4.94

4.2.5 (Cont' d)

Keb "b + ef Af + Kea As Pe O (2)

Krb Ab +KrfAf +Kr As -Pr =0 (3)

The boundary conditions are:

P =0 (4)

A 8 0 (5)

whose solution is

Ab =+ ab Pa 6a

Af =-Ke KebK P (6b)

P=K K P(7a)Pr Krb -- ab Pa

Pr + Pra =(K K ) p total reaction (including (7b)r r rb ab loads directly applied to

SraP reactions)

where

K = (Kab - Kaf Kef Keb) = flexibility of actual structure

= tba of flexibility method, (8a)

-1Krb = (Krb - Krf Kef b = reactions due to a unit de- (8b)deformation at b

Pý = equivalent loads acting at points other than the supports

Ira = equivalent loads acting at the supports

ab = load in Jth member at "a" due to unit displacement at 'b"

The end loads on each elemental beam are

SP. (9a)a ab aba

i -- 1r K Pa (9b)

which are combined with the loads and temperatures applied to the beam toobtain the stresses and strains. R

4.2.6 Selection of Method of Analysis

SThe best method to employ cannot be decided arbitrarily but must be evaluatedafter due consideration is given to the structural problem as regards the computing tools

and time available, supplementary structural information (e.g., available solution of sub-

WADD TR 60-517 4.95

ML4

It I

4.2.6 (Cont' d)

structures larger than the elemental beam), past experience (ease of visualizing pro-

blem), etc. Generally, if the structure is primarily a set of elemental beams (e.g.,cantilevers) in series, i.e., the number of d +y degrees of freedom is less than thenumber of b degrees of freedom (d + y < b). then the flexibility method is better*. Ifhowever, the structure is primarily a set of elemental beams in parallel (bents, etc.)in which d + y > b, then the stiffness (slope deflection) method is better. When thenumber of redundant degrees of freedom is relatively small it is usually simpler tovisualize the flexibility method. As the number of redundant degrees of freedom in-creases, howevir, the selection of the simplest datum becomes more and more complexand the stiffness method which does not require selecting a datum (the datum of theactual structure is employed) becomes less Involved. Because of the lack of engineeringintuition required in the stiffness method it is more easily adaptable to computingmachine techniques. The flexibility method becomes more and more involved as thegeneralized coordinates (c, d) increase in number, especially when loads can be appliedon these elastic supports.

The flexibility and stiffness methods of solving beam-like structures areillustrated below. The problems are treated in great detail to familiarize the analystwith the methods of analysis. The following problems are considered:

PROBLEM IA - Statically indeterminate beam by the flexibility method

PROBLEM IB - Statically indeterminate beam by the stiffness method

PROBLEM II - Sandwich beam by the flexibility method.

4.2.6.1 Problem IA - Statically Indeterminate Beam by the Flexibility Method

The beam shown in Figure 4.2.6.1-1 is loaded with mechanical loads andtemperatures as described in Figure 4.2.6.1-2. It is required to determine the re-dundant loads on the structure. The problem is solved in the following steps:

(1) Determine and classify the degrees of freedom of the beam from thegeometry (Figure 4.2.6.1-1).

(2) Establish the static equilibrium equations.

(3) Establish the flexibility matrix by methods described in Section 2 of thisManual.

(4) Invert the flexbillty.matrix associated with the cut redundant loads andexpress the cut redundants in terms of the applied loads.

S(5) Obtain unit solutions of the redundant loads in terms of the applied loadsby combining (2) and (4).

(6) Apply the mechanical loads and temperature and determine the fixed endreaction by employing Paragraph 4.2.2.

(7) Substitute the negative of the fixed end reactions into the relationshipestablished in step (5) to obtain the solution shown in Figure 4.2.6.1-3.

* The number of s degrees of freedom is not too significant since the geometry matrix(g,,) need not be inverted.

WADD TR 60-517 4.96

4.2.6.1 (Cont'd)

S10• 10 -

P 7 --- , ; -- P8

4 p5 p6

FIGURE 4.2.6.1-1 BEAM WITH DEGREES OF FREEDOM

(1) Classificaton of Degrees of Freedom

(Uncoupled Since Beam ,

Susrp edn oStra*gh) ,Axial

,s) 1 (2.V=0) 4(.M=0) ,7(ZH=0)(O,y) 2, 3(ab) 5, 6(c,d) -- 8

(2) Equilibrium EquaUons (Ref. Eq. (1b) of Paragraph 4.2.4)

-r= gra Pa + gr PC + gx 1ýr

_ [l r 1 1 0 0 0 1 1 Pl1

P4JL0 10 -0 1 1 1 10 20 P2a

p3a

L4a aý rP (1)lP8

P6a

Ip2

(3) Flexibility Matrices (Constant El and EA) 3

2 32 333 8 33

El =83 (Reference 4-6) AE f8 8 =20

3 83 26672 3 5 6

Elf = 2 333 833 50 50

3 833 2667 150 200]

WADD TR 60-517 4.97

4.2.6.1 (Cont'd)

(4) Inverse (Stiffness)

1 1 .01373 -. 00429 1 -1El yx -. 00429 +.00171 -f =.05

From Eq. (4a) of Paragraph 4.2.2.4, the cut red-udant loads can be solvedin terms of the applied loads.

FD_- = + l -1 -8=x ya K8 P8 f + P8 f88

( rw o 1 0 0 04 -. 172) 1' 8a8P3 0 0 1 0 .043 .129 P21 -8 f 58 +1/K 8

P P8 P 1 + AE/PsL

4a LaP4a 1AE fK 8,dI

P5a. _ _ _ _ _ __td

6 - P8 I+AE/K8 L

(5) Unit Solutions

-P2 -2+ .043 •5-. 1 71 66

-P3 P3a + .043 P5a +.129 P6 a

-P1-- 1= - .086 P5a +.043 P6a

-P4 = P4a- .29 P1 a +.14 p6a

b = fbaPa + fbxx P

EJ5 = 1.40 P5a - ..7a5

EI&6 = -. 75 a + 2.80 P6a A8=

(6) Applied Loads

aT = 400(10)-6K 8 = 106 lb/in.

-5 _ E =10 llb/in.2ST 100 b 1lb A = l sqin.

T 1 .08 in.4a T=_100(_6 d = in.

1ar--10(1-aT Linear ThroughaT-=200(10) a200(l0)4 Depth a.g. at (L of

-10" - I--- 10" • Cross Section2 3

FIGURE 4.2.6.1-2 TEMPERATURE AND LOAD ON BEAM

WADD TR 60-517 4 98

I4.2.6.1 (Cont' d)

Thermal Distortion of Cross Section:

aT - ToT T +Twt (Linear Distrl- g -I =d 2dbution Through

Depth) (c.g. at % of CrossISection)

The aT values are given at discrete points o! the structure and are assumedto vary continuously. Since there are valkt s at three station (end stationsand mid-station), the best parabolic curve passing through these three valuesIs selected. This results in

0-< x<-0 1 06wl = 4x - 20x 106 •. = 2 2 +10x

105x <20 10 6 w' - 200 106 1 =300

Fixed End Reactions:

Mechanical Loads (we illustrative problems given in Paragraph 4.2, 2. 5)

0!5< x _< 10 •-100 -12 lb. /in. 10<-x!520

T+50=R1 /R2L -+,50] + 60"R 2R +60=R

125R4RL = -125 1 -100=R -100=R 6

P= -100,'y=.5 Iq=-12, Y/=1.0

Thermal Loads (see Eqs. (7) of Paragraph 4.2.2.5)

-0W 400 1w

400 I 0200

=L2 I r7-7-w=LOkTR1 R2LT R5 R. RRR

R4 14 4/U///////////I

200- 200 5L

10 6 W ' = -200 - - " ' = W l

L WLi

00 w? +wI +... Li +--L 6L L, L2 -+7 -6 7 -6 I

SRL =1tI= 1 (.08)10) (-200 +400) (.0 (08)(10) -(0)=0 '

S0! = 2R :_R3

•i ~WADD TR 60-517 49

4.2.6.1 (Cont'd)

-R 2 L = R1 = -16 + 32= + 16

w:2 +6-+

Ro Eoo W 0 -"6 L+ .... 53.3J

4 (10)7 (.08)(10)6 (o -- 400 =-53.3 (

I +160=RSRR

ML= MO + PoL

RSL=-53.3 +10 (16)=106.7 160 +10 (0)=160=R 6

116 -161k1 ý60 +160

-55.3 +106.71

The total equivalent loads are obtained with the aid of Eqs. (1) and (3) of

Paragraph 4.2.2.1

R = RLL+R

S=50 + 16 =66= -a

+R =50 + 60 -16 =94= 2a

=R3.60 = p3a.

ML MLL M LR

+Ra 0-[ (-125)] [--,3.) = 178.2 ~4

+15.= -M -5(-100) + [ +106.7 - 160] =78.3 = -P55

46a = (-100 - 0) + (160 - 0) = 60 = -P 6a

(7) The Reaction Loads are Obtained From the Unit Solutions.

-P2 = -94 + .043 (+78.3) - .172 (-60) = -80.31 lb-P 3 = -60 + .043 (+78.3) + .129 (=60) = -64.37 lb

-P1 = -66 - .086 (-78.3) + .043 (-60) = -75.31 lb

-P 4 = -178.2 - .29 (+78.3) + .14 (-60) = -209.31 in-lb

WADD TR 60-517 4.100

4.2.6.1 (ContG d)

Since the beam is straight, it is easier to determine P8 by direct solu-tion of the compatibility equation.

Sdx °10 (2x +lox) dx +fo 300 dxt]

8a = L = 201Jo -x dx (1)10 7

1 [2 (0)3 + 10 (10)2 + 300(10)] = 2083 lb= -• -2 3 2 ( 0

- P8 a 2083 1389lb

1 + AE 1 (10)7

~KL i +106(20)

1389 1389

75.31 80.31 64.37209.31

FIGURE 4.2.6.1-3 SOLUTION

4.2.6.2 Problem IB - Statically Indeterminate Beam by the Stiffness Method

The bending portion of the above problem is solved in the following steps:

(1) The degrees of freedom are classified.

(2) The stiffness matrix is determined by methods described in Section 2of this Manual.

(3) The displacements are determined from the equilibrium equationsat each degree of freedom. This requires inverting the stiffnessmatrix.

(4) The redundant loads are expressed in terms of the displacements, toobtain a solution in terms of the applied loads.

(5) Repeat steps (6) and (7) of Problem IA.

Classification of Degrees of Freedom

(r,s) = (1, 2, 3, 4)

(a,b) = (5, 6)

WADD TR 60-517 4.101

4.2.6.2 (Cont'd)

Stiffness Matrix - The stiffness matrix is determined as shown inFigure 2.1.4.3-1.

Unit Displacement atK 3 6 2 5 1 4El

3 .012 -. 06 -. 012 -. 06 0 0

6 :-.06 .4 i.06 1 .2 0 0

-. 012 .06 +.024 0 -. 012 -. 06d = (.012 + .012) =(-.60 + .06)0o .8

5 -. 06 2 =(.06 -. 06) = (.4 +.4) +.06 .2

1 0 0 -. 012 .06 +.012 .06

4 0 0 -. 06 .2 +.0 6 .-4

Equilibrium Equations and Displacements

P5a + K5 5 A5 + A56 N 0 = P5a + .8 EI15 + .2EIA6

P6a + K6 5 5 + K6 6 % = 0 P6a + .2 EIA5 + .4 EIA 6

EI A5 = 1.43 P5a - 7 2 P6a (Check flexibility equations ofProblem 1A)

El A6 = -. 72 P5a + 2.85 P6a

Reactions

1 K15 •16 A 5 .06 0 1.43 72 [5a

P2 _ K2 5 K2 6 % _ 0 .06 -. 72 2.85 LP6JP3 5• K6 -. 06 -. 06

.2 0P4 K45 K6._ .2 0

IRIP1 .086 -. 043 P5a (Checks Flexibility Equations of

Problem 1A)P2 .043 .172 PI--a23 043 -. 129 136

P +.286 -.144]

WADD TR 60-517 4.102

}

M

4.2.6.3 Problem II - Sandwich Beaw by the Flexibility Method

"The sandwich beam of this problem is subjected to temperatures of Ti and T2

above the datum on the top and bottom faces, respectively. A general solution is presentedfor the stresses and redundant forces in the sandwich beam in terms of geometry and endfixity. The problem is solved by utilizing the geometric relations for a sandwich beamestablished in Paragraph 4.1.2.2, solving for the unrestrained deformations, determiningthe fixed end reactions, and solving for the reactions and stresses.

.Geometry (see Figure 4.1.2.2-1)

E1A1 h h (See Eq. (1) of ParagraphS1 AI+E 2 A+ E2A2 1 + e 4.1.2.2)IAI+ E A 1 A E 1---

Unrestrained Deformations

a1 TI- LTL a1TIT (1 2T2 ) ,ITIw1 = -h h ( 2 h (1-a) (SeeEq. (3) of

1 Paragraph 4.1.2.2)

- EIAlaT1 + E2A2•T a1 1 A1 + 2A T 2 a1T(1 • ae (See Eq. (2) of Paragraph

E AI + EA2 A911 e 4.1.2.2)

01 =2 1 0 2 T(See Eq. (4) of Paragraph4.1.2.2)

where

E2A2 •2T2e = EIA1 and a alT-

Fixed End Reactions

Po = 0 Mo TE w' Fo = - EA (See Eqs. (1) through (3), Paragraph4.2.2.3;Po, =0 m =1 f =1)where0 00 00

EA = EIA + E2A = A1(1+e)1 1 2 2 E1 1

El =EA h2 _ 2 (E A + E2 A2 )1 11 11 22

"= EAI h2 (1 I +e )= EAh2 e1i 1 e2 1 +e(1 + e)

E (IE-a) E EIA)a T1 h et1-a)MO= 1l1+eh+e

Fo = -EIA 1alT1 + ae)

j WADD TR 60-517 4.103

4.2.6.3 (Cont'd)

General Solution

fdc Pca +fdc Pce-K• Pc

=- fd. + KY fdc P~ca

Let KF Axial stiffness of support and KM = Rotational stiffness of support.

1 -1(-Fl•_ • • ~ ~EIA1 ( + e) +EI1 1 )e

EAllT (1 + ae) eAF = - EIAI( +e) +

FKFF[ 0I

SEIllElT1 h(e(1-a

=1 + e(2a)

K F

EAh 2 e

SKM+

I- -'&-e (3a)

F A _2 _ _e_1r1

___.... 1 (+e Axial Shortening Cau..- .by Flexible Re- (lb}S1 +1 stralnt (K F)

KM

[E (+e)+ ____

E0"1 1 +(1 +ae) + ( (+e)KF KM

WADD To 60-517 4.104

-~~~~~ ~ ~ ~ - 1e xa kreigCu b lxbeR- (1b)

4.2.6.3 (Cont' d)

E 2T el +a'2 1 +e EIA1 (14e) E1Ah (3c)KF +1 1 +(1

KM

If K K 0 (unrestraine2), then a, a2 = 0. (3d)

It KF = KM = o (completely restrained), then

-Elti Ta,. 1 +e (1 +e) = - EIaIT1 (3e)

-E 2E2 alT1S02 = 1 +e p(1 +e)=- EaT (3f)

4.2.7 Curved Beams

If the beam is straight, the axial problem can be separated from the bending problem.•his cannot be done in a curved beam, and the problem must be solved with all the degrees offreedom at a point coupled. Most curved beams in aircraft structures are frames of relativelyfew load redundants, and the flexibility method is utilized since the frame is essentially a re-strained cantilever beam. The utilization of equivalent loads requires determination of fixedend reactions which are quite complex (except for special geometries and loads) and which"do not reduce the computation work except in problems involving many redundants.

The flexibility of the frame which results from mechanical loads is primarily de-termined by the bending.flexibility. This is because of the bending energy of the frame, dueto mechanical load&. is usually much greater than the axial aid shear energies. This can be

seen by comparing the values of ds, r-jE ds, and fA--G ds. The axialenergy due to temperature Vfids), however, may be a significant portion of the thermal bend-ing energy (f -w' Ids) and should be included in the calculations. The change in axial andbending stiffn-esses due to temperature can be quite significant and can be calculated with theaid of equations in Paragraph 4. 1. 1. 1.

The frame problem is solved by the flexibility procedure. The structure is madestatically determinate by cutting the frame, fixing one end, and applying redundant loads(XM, XF and X - Figure 4.2.7-1) at the cut to enforce compatibility. The deflection at thecut Is computed by virtual work in terms of the mechanical and thermal deformations times

r the virtual forces and by the redundant loads times the flexibility coefficients.

Let fFp (etc4 = Element of flexibility matrix; deflection at cut in k' (axial) direc-tion due to a unit load in P (radial) direction

XM (etc) = Redundant load at cut in M (rotational) direction

ImF (etc.) = Internal virtual force of m (rotational) type due to a unit load atcut in F (axial) direction

A (eta) = Deflection at cut in P (transverse) direction due to applied loads,• "temperature, and redundants.

WADD TR 60-517 4..105

4.2.7 (Cont'd)

u a

: "\e Fiber •

AA

v

Sy Outer FiberSqXF

X Section A-A

1. Tension is positive (increasing axial strain T)2. Compression on inner (upper) fiber is positive (increasing curvature w)3. Inward shear is positive4. e = Angle from negative Y axis to normal through section measured counter-

clockwise5. Applied clock.Ase moments "are positive6. Applied horizontal loads are positive to the left7. Applied vertical loads are positive in the up direction.

FIGURE 4.2.7-1 SIGN CONVENTION

The flexibility coefficients, neglecting axial and shear energy, are evaluated from

the virtual moments shown in Figure 4.2.7-2. The results in matrix form are shown asfollows:

m am~m ds mM Fd- InmMmInp ds. f!ý fyds f xdsfMM fMF fM E EIJ E'I JEI EI

_____ r pm=mP Ey fy Ei

ff mF E ds E -ds P s o EI

FM fFF fFP JE! EI El El El El

o m Pxds fxyds rx 2dsfjm?•TM d fm~ ds J mEI _j EI JEI El

f M fP F -P P E ld_ l d E l j - l1

The following compatibility equations express the requirement that the relative

motions at the cuts are zero. The virtual forc3 system is the unit load at the counterclock-

wise side of the cut with reactions on the other side of the cut.

A AM =f[(-w -w') mM + ( +•')fM. ds + fMMXM + fMFXF +fMPXP= 0

.AF =f [L(-w mF + ( +' +•)fF] ds + fFMXM + fFFXF +fFPXP= 0 (2a)

lp=f -w -wf) mP + (" +W')fP] do + fPMXM + fPFXF +fPXP = 0

WADD TR 60-517 4.106

4.2.7 (Cont' d)

or

!{d~ftdsX~ fA.+XfV +xp/xfo=ods f w' ds +_XM f + X F yd, E--EI El rEI ÷. •_

-.X - Jwt yds +Jc coseds + XMf EI + XFf + PfE 0 (2b)

Exds JwI xds - c sinEdx + XMJ ?EL + Fd E + Xpf 0f l ffEI FET P.EI

where fM=0 and fI" cos e ds and.f• sin 0 dx are usually small and assumed equal tozero.

The redundants XM, XF# X p are obtained by solving the simultaneous Eqs. (2b).

x

(441M=3.

ii = m~=Fm=1riM IIn= mp=X

f =0 fF cos fp =-sine

P 0 PF + sin 0 pp= +cosO

FIGURE 4.2.7-2 DETERMINATION OF VIRTUAL FORCES m,f,p

"WADD TR 60-517 4.107

4.2.7.1 Values of e wt and M

The values of W' and w' can be obtained by an analysis of the cross sections asshown in Sub-section 4.1.

A simple approximation of the aT distribution in a cross section is to assumea linear variation through the depth. In the case of a linear distribution, the values of Twould be aT at the centroid of the cross section and the value of w' woule be the difference

of the (aT)Is divided by the depth (w = h ) . The linear varia-

tion of aT usually results in a good approximation of the distortions of a cross section ( 'and w' ) due to temperature. These are the parameters that determine the redundant loads.

The moments on the cantilever (cut frame) beam can be determined by statics ofthe applied loads,

Mn

i=l H

or can be obtained in a table based on incremental moments, as follows:

" "+ = Mn + AoMn + Mvn + AMHn (1)

where

VnI 1 Vn + AV ; Vn Vn (X,, - xn)

Hn+1 + AMn = Hn (n+i-n)

AM MR = Moment about (n-1, yn+i) of all loads between (xn yn) and (xn+1,Yndl)

AVn = Vertiepl load (positive up) batween (x , yn) and (xn+1 , yn+l)

AH = Horizontal load (positive to left) between (xn, yn) and Xn+1 n

(yY.) = Coordinates of centroid of nth element

AS n = Length of the nth element.Notes: (a) Frame length between xn,, y.) and (,n+1 , yn+,) is equal to .5 (Asn + Ash+)

(b) The loads on the frame must be self-equilibratiý.g. This -should be checkedbefore the deformations due to the applied loads are calculated. (See lastrow of columns 32, 34, and 42 of Table 4.2.7.3-1).

It is advantageous for the purpose of analysis to:

(1) Keep the lengths of As small so that section properties are fairly constant ineach AS

(2) Start new segments when sections change radically(3) Locate a station (centroid of segment of length As) where load changes slightly(4) Make As zero where maximum stresses are expected (concentrated loads, etc.)

WADD TR 60-517 4.108

-I f~J

4.2.7.1 (Cont' d)

The total loads acting on any section are

979Zn =Mn + XM + Yn XF + x Xp

n= (Hn + XF) (cos E) + (Vn +Xp) (-sinXe)

=-- (n + XF)(4sine) + (V + Xp) (+ cos e)

4.2.7.2 Special Cases

The solution of the curved beam problem can be modified in special cases toreduce the amount of computations. The following cases are considered:

Case I - Symmetrical StructureCase II - Elastic Center MethodCase III - Relaxation of Boundary Conditions

Case I: Symmetrical Structure

in many structural problems the geometry s, AE, El on one side of anaxis is a mirror image of the geometry on the other side, e.g., El (y , +x) = El (y, -

X). This is usually the case in aircraft structures (except for some slight variation due totemperature effects) because of the probability of the load being applied from either side ofthe frame. When the structure is symmetrical, some of the influence coefficients(f- ==f fP xyds =F ) are identially zero and the othersJ = fIMP = PM = El =FP = PF

need only be evaluated for half the frame and then doubled, e.g.,

4 s(27r) s(r)f yds y2 y2ds2. 2o EI o EI

where s(ir) = length of frame between E = 0 and 7r)

The solution is simplified as follows:

X= w( fxds -f d f 'sin 0 ds)/fx;ds

x• [f •'V +f Y ( .i-fwds kj f w'~sEl[fW l .Mds rfw{(fs +2sf 1cos ds)j ~ (a

WADD TR 60-517 4.109

Al)

4;•

4.2.7.2 (Cont d)

If the applied loads (thermal and mechanical) are also symmetrical, e.g., M (Yn' xn) =

M (Yn' -x ), then the radial deflection of the cantilever (in P direction) due to applied loads

is zero and the other deflections need only be evaluated for half the frame and then doubled.If the applied loads are antisymmetrical, e.g., M (Yn' Xn) = - M 'Yn' -Xn), then the hori-

zontal deflection and rotation of the cantilever is zero and the radial deflection is evaluatedfor half the frame and then doubled. The following equations summarize the solution of

symmetrical frames since any loading can be separated into symmetrical (S) or antisym-metrical (A) components.

For symmetric loads:

XPS 0 (1b)

oS ()T) 2 •E (fos(T) fos(T)

y El yL i- Mds f l wds)

= f0 00 0ýMs EIMoyds _ d

EE (lb)

For antisymmetric loads:

XMA = ; XFA 0 (ic)

s(T) s (r) s(r)_ Nxds w' (x)ds -f 9 sin 0 ds

lElcf

XPA 0 0 o1c

x dsEl

Case II: Elastic Center Method

By transforming the reference axes to the "Principal Elastic Axes", itis possible to diagonalize the flexibility matrix. This reduces the three simultaneous equationsto three single equations which can be solved quite readily. The solutiors can then be expres-sed as

X -(f dsfw'ds)

=-(fŽ•~.-fw'yds +f•,eos o' ds)x'F =: (,,2 d s I

WADD TR 60-517 4.110

--- .

4.2.7.2 (Cont' d)

X = (M-2IdS fwtxlds2-f 'sinetds) (2) cont'dPf(x)2 dsJ El

where all the above terms are described in Figure 4.2.7.2-1.

It is seldom advantageous to employ the elastic center method unless the location(6, y) and direction (V)I is readily discernible. In the case of symmetry about the "Y" axisthe value of p = 0 and x = 0 so that y' = y - y and x1 - x. If symmetry exists about both axesit would be advantageous to establish the origin at the center of symmetry.

yt YA

. x,

F -yA

00.- X -.

x

FIGURE 4.2.7.2-1 ELASTIC CENTER METHOD

The transformation equations are:y, = (i-Y) cos V - (X-x) sin P

S= (y-j)sin 4p + (x- )cosq9 (3)

01= e -2

WADD TR 60-517 4.111

-,•

4.2.7.2 (Cont' d) 2xyds

tan 2 47 2 EI

fv~ds .2 fxSEl El

= ffdsly %r' A I --

E ElX:xs JEI /JEI-

Case MI: Relaxation of Boundary Conditions

It is often convenient to determine the effect of changing the frame by insertinghinges, etc. It is also convenient for computational techniques (digital, etc.) to maintaina standard form to solve all frame problems. Employing the same virtual force systempermits to use of standard tabular forms or digital procedures without modification foreach special structural problem.

Equations (2a) of Paragraph 4.2.7 can be viewed as the relative deformationsat each side of the cut. if a hinge existed, then AM would not equal zero but the redundant

load XM would equal zero.

For the case with a hinge, the compatibility equations becomeI E-•I-If y ••F" fy2 dsE X r xyds (a

dE wt yds +.f_'cosOds + X + X= 0 (4a)

Mx -s Iw x ds -s 'sin 0 ds+ xFf ds + xpJIx = 0 (4b)

Solution of these equations results in the redundant loads X and X . Note that

the tabular form shown in Table 4.2.7.3-1 can be employed to obtain Wife desirA1 quantities.

4.2.7.3 Sample Problem - Frame Subjected to Thermal and Mechanical Loads

Table 4.2.7.3-1 is designed to solve a general frame problem with a maximumof three redundants. The problem selected was a special one wherein the structure and loadswere symmetrical. It was chosen in order to indicate the simplification of work and calcula-tions which result when symmetry of loads and structure exists. Only half the rows in thetable need be calculated and many-of the columns can be eliminated as indicated below.

(1) Column 8 (denoted by A ) need not be computed if the structure is symmet-rical since •2 (8) would-equal zero.

(2) Columns 15, 19, 22, 24 and 27 (denoted by 4B ) need not be computed if theloads are symmetrical since Z of these columnsu would equal zero

(3) Columns 13, 14, 17, 18, 23 and 26 (denoted by @©) need not be computedif the loads are assymmetrical since the Z would be zero.

The summation (value of integral) of the column is therefore eitl;er twice thefirst half (2 = 2 t ) or zero since the value of the symmetrically located station in thesecond half is either the same or opposite to that of the first half. The first half of columns21 and 22 is usually computed for the purpose of employing Eqs. (2) of Paragraph 4.2.7.1 inobtaining the total loads acting on a cross section.

WADD TR 60-517 4.112

4.2.7.3 (Cont'd)

Any load distribution can be decomposed into symmetrical and antisymmetricalcomponents by dividing the load vector into two equal parts and applying one of thehalves to the structure with its symmetrical equivalent and applying the other half with ItsI antisymmetrical equivalent as shown in Figure 4.2.7.3-1. Using the sign conventionfor applied loads, a symmetrical load would be defined as

Mi=-Mi,

Hi = Hi,

where iI is the station symmetrical to I and an antisymmetrical load would be definedwith opposite sign. II

i~I P/21 I IP2 P/21 IP2IPI I P/ 1IP 2

S- I t

(a) Load 0() Symmetrical Load (c) Antisymmetrical Load

FIGURE 4.2.7.3-1 RESOLUTION OF FORCE INTO SYMMETRICAL ANDANTISYMMETRICAL COMPONENTS

The mechanical and thermal load, together with the geometry of the frame,is shown in Figure 4.2. 7.3-2. Te solution of the frame problem is obtained byemploying Eqs. (lb) and (1c) of Paragraph 4.2.7. 2 with the values obtained in Table4.2.7.3-1.

6-

••WADD TR 60-517 4.113 •

4.2."7. 3 (Cont' d)

2000 (Neg)

9 8h (Constant)

21000 (Neg)

'4¥

1ý4

x 2

,aT

+H

Sign Convention forApplied Ioads

FIGURE 4. 2.7.3-2 FRAME PROBLEM

WADD TB 60-517 4.114 K

- £4

0 -c

C.4-

011

A -1-

-C4

c c

£4

Ic..

ta ..

DI J-

~'~,(j 1 4 *

I..Z- -- z~

.5,e

I.I

_Q I

__m 11111 r

:5 Im -

o Q A z z-- .

41 0

CD~IA

H. 14

I-HC'.4

CM :EF.

C-1 11

14

I J *1 1 W

to o ;

-~~~~~~~~~~C I I I 1 88 o -- fi4-t24-~s'S,.Lfi fi> u..c

4.2. 7.3 (Contf d)

For symmetrical geometry and load, Eqs. (1b) of Paragraph 4.2.7.2 areutilized.

= [v' ®ý (Z' Z© @ (Et V + ® E' fl ]M ~ -(~ ® )2

-218,006 (10) (-34.0755

- 376.995 (10)- -11,118,000 (10) 7 - (-427.56 + 9.761 aT-7 7 h lo-2

(12.5665) (10) (15,740) (10) - (ý76,995 11) 7) 2

lXM = 13652 in lb - 67. 388 (10)6 a•T + .361 (10)6 atTh

XF = 2

X = 10333 lb + 1.3424 (10)3 .T 022 (10)6 aTAtXF -. 2h(0

4.3 REFERENCES

4-1 Switzky, H., Thermal Stresses and Deformations in Unrestrained Beams,Report No. E-SAM 24, Republic Aviation Corporetion, December 1957.Appendix, January 1960.

4-2 Klosner, J. M., Thermal Stresses at the End of a Spar, Report NoE-SAM-14, Republic Aviation Corporation, October 1955.

4-3 Engineering Structures Data Sheets Manual - IBM Procedures, ESM- IV,Republic Aviation Corporation, January 1958.

4-4 Switzky, H., Solution of Statically Indeterminate Beam-like Structures,RDSR-2, Republic Aviation Corporation (to be issued).

4-5 Meissner, C., Stress Analysis of Aerodynamic Surfaces Using theMethod of Elastic Coefficients, Report No. ESAM-21, RepublicAviation Corporation, July 1958.

4-6 Switzky, H., Influence Coefficients, Report No. ESAM-22, RepublicAviation Corporation, August 1956.

WADD TR 60-517 4.117

ii

SECTION 5

THERMO-ELASTIC ANALYSIS OF JOINTS

S60I

•,WADD TU 60-517 5.0

SECTION 5

THERMO-ELASTIC ANALYSIS OF JOINTS

TABLE OF CONTENTS

Paragraph Titlse Pag

5 Thermo-Elastic Analysis of Joints 5.2

5.1 The One-Dimensional Problem 5.4


5.1.2 The Special Case of Constant Bay Properties (with 5.8illustrative problem)

5.1.3 Constant Bay Properties - Rigid Sheets (with 5.24illustrative problem)

5.1.4 Constant Bay Properties - Rigid Attachments (with 5.28illustrative problem)

5.1.5 The Influence of "Slop" on the Load Distribution (with 5.30illustrative problem)

5.2 The Two-Dimensional (Bowing) Problem 5.36


5.3 Determination of the Attachment-Hole Flexibility Factor 5.39

5.4 References 5.40

1-

WADD TR 60-517 5..4

SECTION 5- THERMO-ELASTIC ANALYSIS OF JOINTS

This section considers the thermo-elastic analysis of mechanical joints. Theanalysis presented is directly applicable to problems where the stress levels lie in theelastic range (low ductility materials such as the ceramics, refractory materials, beryl-lium, etc., generally remain in the elastic range almost to failure). In addition, certainaspects of the analysis are shown to be of value in the solution of problems with stresslevels in the inelastic and plastic ranges.

As shown in Figure 5-1, a mechanical joint is one that is composed cf plate-likematerials joined by fasteners (such as rivets, pins or bolts) for the purpose of load transfer.

I

FIGURE 5-1 MECHANICAL JOINT

It is assumed that the heat conduction problem has been solved for the temperaturedistribution in the joint. To find the loads in the attachments and the stresses in the crosszsections for the known temperpture distribution and a given applied mechanical load, twoclasses of problems will be considered. The first is the one-dimensional case, whereineaca plate is restrained from deflecting out of its own plane. The second case to be con-sidered is the two-dimensional problem wherein the plates bow or deflect out of plane.

General solutions are presented and it is shown how these degenerate to the morefamiliar elementary joint equations when the applicable simplifying assumptions are made.


e Difference between attachment and attachment hole diameter (slop), Inchesf Attachment-hole flexibility factor, in/lbh Plate thickness, inces

Runnin subscriptj t The jth attachment, or bay

w Rotation of a plate per unit distance due to thermal loading (thermal curvature,as e Paragraph 4.1.1.2)

WADD TR 60-517 5.2

--------------------------------------------------------------- -- d---------------------------eah lteisrstaie fro delcin-u--- t- w pae Tescndcs o e n

s 'dered ---------- -d-e--- l- rblmwhrinte-lte-o-o-d-ec-ut- pae

x Iongitudinal distance along splice, inchesDistances from elastic axes of top and bottom plates, respectively, to thecontact surface, inches

y Distance in the thickness direction, inches

A Cross section area, square inches

AE Effective axial stiffness of a plate; f EdA (see Paragraph 4.1.1. 1), lbs.AJNj BiN Nol'-dimensional coefficients for the determination of attachment loads

) J (Figure 5.1.2-1).D Diameter, inchesE Young's Modulus, psi

EI Effective bending stiffness of a cross section, (see Paragraph 4. 1. 1. 1), lbs-in2

I Cross sectional moment of inertia, in4

K Stiffness, lbs/InL Longitudinal spacing between adjacent attachments, inchesM Externally applied bending moment, in-lbsN Total number of joint attachmentsP Attachment shear oad, lbsT Temperature, degrees FX Mechanical load, lbsZ Non-dimensional ratio of axial flexibility to attachment-hole flexibility

ca Coefficient of thermal expansion, in/in-OFA Unear displacement, inches; an increment0 Slop displacement between adjacent loaded attachments, inches? Axial strain of a plate due to thermal loading; fEczTdA/ f EdA

Free axial thermal expansion in the plane of the plate; f Z dx, inches

Subscripts

B Refers to bottom plateIN Refers to the jth attachment, or bayJN Refers to the jth attachment of a splice having N total attachmentsT Refers to top plate

A

WADD TR 60-517 5.3

5.1 THE ONE-DIMENSIONAL PROBLEM

The general problem is simplified if the overall geometry of the joint does not allowthe joint to bend out of plane. Such conditions are realized when, for example, the jointconsists of a splice strap in a flexurally rigid beam flange as shown in Figure 5. 1-1. Insuch a case, the problem is one-dimensional in that the joint displacements and attachmentloads are essentially dependent on the axial flexibility of the joint components in the directionof the applied external loading.

/A

FIGURE 5.1-1 SPLICE STRAP FOR A RIGID BEAM FLANGE

The solution for the load distribution to the joint attachments is obtained by satisfyingcompatibility conditions for the joint displacements and the equilibrium equation.

The following analysis is applicable to a mechanical joint composed of two arbitrarilydissimilar elastic materials. It is assumed that the attachments initially fill the holes andthat each attachment-hole combination deforms elastically under load.

The presence -f "slop" (which results from a combination of manufacturing tolerancesand differential thermal expansions between the plate holes and attachments) and its influenceupon the load distribution, is then considered. A detailed derivation of all solutions is pre- 8sented in Reference 5-1.

WADD TR 60-517 5.4

5. 1. 1 The Joint Equations

Equilibrium

Figure 5. 1. 1-1 shows a longltudal section through a typical joint under consideration.Denoting the shear load acting at the j attachment by Pj, equilibrium of forces requires that

N

j=1

where a tensile applied load X and attachment loads acting to the right on the upper sheet areconsidered positive.

2L.

A AXX

A I

i ,

FIGURE 5.1. 1-1 ONE-DIMENSIONAL JOINT EQUILIBRIUM

S~Compatibility Conditions

As shown in Figure 5.1. 1-2, compatibility of displacements requires that the axial con-traction or expansion of the plate material at the common surface, measured from a datum(defined by the unloaded, unheated spacing between the centerlines of adjacent pins) must beidentical for the upper and lower plates.

A-.A

P 2

X4

kwamFIUR 5.DR6-1 5.5 N-IENINLJIT QIIR~

CopaiiltyCndtin

i.1.1 (Cont'd)

L Initial Pin Position

ai Final Pin Position

Conmumon LI+ASurface

24

FIGURE 5.1.1-2 ONE-DIMENSIONAL COMPATIBILITY

For the jth general bay the compatibility equation is

A. =IT = AjB" (2)

where the subscripts T and B refer to the top and bottom plates, respectively. For the

one-dimensional case with no "slop" present, there are basically two types of deformationwhich contribute tr. the Aj, I of the joint.

The first type of deformation is the uniaxial stretching or contraction of the sheets due

to the combined effects of temperature and mechanical loading. Referring to Figure 5.1.1-3,the uniaxial stretching for the jth bay is

Sj L.:-:" (3a)

SA (jT= X P L + -jT dx (

S~T

j Lj (3b) "

Pi L + f0 dx •:

where a positive A :n2 rcases the spacing between adjacent attachments and 23b

- EaTdy

f hEdy

If the thermal gradient is linear through the thi:-kn•ss, then i" is approximately equal to thevalue of caT at the plate midplane. -

WADD TR 60-517 5.6

4 am

5.. s Cont'diIN •, I- x-:x P,

ZA

AIJJ-1 Pi J•J

i-i~J £ J++1

FIGURE 5. 1.1-3 DEFORMATION OF THE JOINT DUE TO AXIAL STRETCHINGOF THE SHEETS

The second basic typn of joint defornuation occurs because the internal joint loadscreate local distort~ons of the holes and atttchraeats as shown in Figure 5. 1.1-4. Thedeform~tion is expru ssed in terms of an experimentally determined attachment hole flexi-bility factor f for the given attachment-sheet rombivation (Sub-section 5. 3). Thus, forthe top sheet,

jT (j= I j JT

and for the bo~ttom sheet,

AlljB P -ij+I f (J+l)B + P JfJA (4b).

Substituting A.T Alj + A~jT and AB=AB jBfromn Eqs. (3)1and (4) into

Eq. (2) yields, After rearranging terj-.-s,

-- . (T E I i Pj+I 151 AE j

SI

J

W AIAJ =• (JT-•.B d5.7

'4and II

F\.•IGURE•' 5.1.1-3 DEOMTO 5O.H7 ON DET XA SRTHN

5.1.1 (Contd)

S• = P.+ I f(j+l)T

Pj -- 1, • fjB Pj~ • j+If (j+I)B

j+11

FIGURE 5.1.1-4 DEFORMATION OF THE JOINT DUE TO LOCAL DISTORTIONSOF THE HOLES AND ATTACHMENTS

The compatibility Eq. (5), which constitues-N-1 equations, together with the equili-brium Eq. (1) provides N linear algebraic simultaneous equations for N unknown attach-ment loads, P. Since Eq. (5) is in the form of a recurrence equation, it can be used to

express all the attachment loads in terms of I'the load in the first attachment, that can

then be evaluated from the equilibrium equation. Solutions can also be obtained by solvingthe simultaneous equations directly or by iteration and relaxation techniques. In general,no simple relationships exist between successive Pj' I . This difficulty arises from the factthat the recurrence equation coefficients are variable since they are functions of flexibilitiesand temperature distributions which may vary from bay to bay.

5.1.2 The Special Case of Constant Bay Properties

Consider the case where the flexibilities of the sheets and attachments, as well asthe temperature distribution, are the same in each bay (constant bay properties). From apractical point of view this situation is attained when the sheet thicknesses are constant.the attachments are all of the same type, size and spacing, and when the temperature varia-tion through the splice thickness does not vary appreciably in the direction of mechanicalloading. Tae coefficients of the P Is in the compatibility-recurrence Eq. (5) then become

constant and a general solution can be obtained with simple relationships between successivePi8- ThusL Eq. (5)becomes

WADD TR 60-517 5.8

5.1.2 (Cont'd)

-AE B] PIC= i ifX ( L )E (1)L )T+ L )BI I I• T1)

(j=1, 2 .... N- 1)

whereL0¢ (C-T cB dx

0

and

f f'T+fB

Eq. (1), together with the equilibrium Eq. (1) of Paragraph 5. 1. 1 yields the following solu-tion for the joint loads:

AE T jN B. A (2)PJN = AN + BiN (L f1N

IT h

where the subscript JN refers to the jth attachment in a joint of N attachments. Values ofthe coefficients AjN and BiN vs. Z, where

are plotted in Figures 5. 1. 2-1(a) through -1 (j) for J = 1--5 and N= 2 - 20. Equations for

the numerical calculation of these coefficients are derived in Reference 5-1.

By interchanging the designation of the top and bottom covers, the curves presented canbe used to obtain all of the loads in joints having as many as ten attachments. When the totalnumber of attachments exceeds ten, the curves give the loads in the first five attachments fromeither end of the splice.

The first term on the right hand side of Eq. (2) represents the contribution of mechanicalloading to the attachment load. The second term represents the contribution of thermal loading.Thus, Eq. (2) conveniently separates the thermo-mechanical problem into its superimposablecomponents. Note that for constant bay properties the thermal component of load at the centerof the joint must be zero from a symmetry argument. Therefore, the center bolt of an oddnumber of bolts has no load due to thermal effects (Figure 5.1.3-2). Because of this, B239B3 5 = B 4 7 .... = 0, as indicated in Figures 5. 1. 2-1. In addition, it is significant that when the

joint has constant bay properties, the attachment loads due to thermal loading alone are sym-metrical about the center of the splice which yields BN = -BNN, B2 N -B etc.

N WN T(N-7)N5

WADD TR 60-517 5.9

=EQ,

5. 1. 2 (Cont'd)

4 ________

ZIA

=We

C~j-Z4

AD. TR605175.1

5. 1. 2 (Contl d)__0

ico.

wy a.0_il -

4i L7L-- _ju~

w wI

/ 'II jr izll.ý, - __

fAD T-- 605751

5. 1. 2 (Cont d)

U)

CLW

3*1 LULm o3

LL.1

-< loo I-

0 0

WADD TR 6-51? 5.1

Co'- ~ ~ ~ .*~'

-

-- - . - - -'- - -7 ,- - -~

IP FGR 5 51.2 -1 Cd -T f

20 -JOINT COEFMICENTS FOR-

-. 4 [CONJSTANT BAY PROPTER

12- -~

am7i79

I8 1 V I

a. __

rl1

~~~:~~ - , - . -'t~

5. 1. 2(Cont d)

0w - - / 4 -

(0 >

-rw J

Di -

-++ioA zH

M 0

4 00

-~~~ AD TR 60-17 001

0.000Tj 00 100F ol '00 1,,I-e -1I- -,

or 00 00-0- -

000 -1-1 1--00

ii - - JOINT COEFFICIENTS FOR - -

0-- CONSTANT BAY PROPERTIES-

0-3)

4-

-n il A

:1 INNo

5. 1. 2 (Cont' d)

0

00

ClD

00

$2 -

-ooo

4 -40

-00',

oo'1

H+-4---

JOINT COEFFICIENTS FORo - CONSTANT BAY PROPERTIES

B4NFRLL

z

S. 1. 2 (Cont' d)

C 0

> 0 :I 1 - -

_ Z=

w W

uz z

I J'I

y. V

0 0

5.18

WADD TR 60-517 51

JOINT COEFFICIENTS FOR j

-o- CONSTANT BAY PROPERTIES-~

12

M I

- [1z

140M4B99- FORALL

M.

rr

5.1.2 (Cont'd)

The solution for the attachment loads in a joint is illustrated by the followitg problem.A titanium plate and an aluminum plate are bolted together with six attachments as shown inFigure 5. 1. 2-2(a). Find the attachment loads when the titanium and aluminum plates aresubjected to uniform temperature rises of 300*F and 70°F, respectively, and a mechanicalload X of 20,000 pounds is applied to the joint.

It is assumed that the bolt hole flexibility has been determined experimentally to be

f = .900 x 106 in/lb (See Sub-section 5.3).

Considering the titanium plate to be the top plate, the plate flexibilities are:

T .1 1 = .356 x 0-6 in/lb,

S = =0-6 A

AEB (1.5)(.250)(10)x lob 6

Since the temperature rise in each sheet is uniform, the incompatibility due to unrestrainedthermal expansion in each bay is given by

, = [(AT)T - (c"AT)B] L

= - 6.5(300)- 12(70)] x x10

= 1110 x 10 inch.

The above quantities are now substitute. into Eq. (2) resulting in

P. A. 356 (20 000)0+

)jN AjN + BjN (.90) (20000)+BjN (i0)

or

P. 20,000 A + 9135 B.JN jN JN~ W)

The coefficients AjN and BJN -re now determined from Figures 6. 1. 2-1 for

) = (.356 + .267) = .692

and N= 6.

WADD TR 60-517 5.20

4i

5.1.2 (Cont'd) /-TITANIUMAT a 300OF

SI IN. / Eu15XIO LB/IN2

S Oa " 6.5x10"t IN/IN-OF

20.000 LBS. ~ 4n fri1 .125 IN.

.250 IN./,

I ZfALUMINUM

20,000 LOAD N TOP) PLAT 70F3

EzIOXlOS LB/IN2

(b)

759003 400D IN800 T390 P

3,84

200 0 ,0 I , I

7590 3400 1800 1390 1980 4030

W2BOLT LOADS

3400 440301 I00O 1390, 19801

•:FIGURE 5.1. 2-2 JOINT WITH CONSTANT BYPOETE

'• WADD TR 60-517 5.21 BYPOETE

5.1.2 (Cont'd)

A1 6 = . 0140 Figure 5. 1. 2-1(a)

A26= .0239 (c)

A 3 6 = .0500 -(e)

A4 .1090 -l(g)

A 5 6 = .2450 - (4)

B = .8000 Figure 5.1.2-1(b)16

B 2 6 = .3200 -1(d)

B3 6 = .0880 -

B4 6 =-.0860 -1(h)

B5 6 =-. 3206 -(J)

The curves give values of AjN and BiN up to j = 5 and the splice under consideration

has 6 attachments. In order to obtain the coefficients for the last attachment, the designation

of the top and bottom plates as shown in Figure 5. 1.2-3 must be Interchanged such that thelast attachment (J = 6) In the original designation becomes the first attachment (j 1) in theinterchanged designation.

Denoting quantities in the interchanged designation by primes, results in

Bottom SheetSsTop Sheet

xx

(3) 6

(a) Original Designation

Top Sheet

ottom Sheet

(b) Interchanged Designation

FIGURE 5.1.2-3 ALTERNATE TOP AND BOTTOM SHEET DESIGNATION

WADD TR 60-517 5.22

5.1.2 (Cont'i)

-6f' =f=.900x10 in/lb

L 267 x 10-6 in/lbT B

~~ (4 ) 356 x10 in11b

A0, = -Ad = -1110x 106 in

Z' = Z =.692

and from Eq. (2),

A' jN .'267)BIjN] (2 0,300) - B'tN 00

or

PN = 20,000A' + 4693 B' 5)j;N jN JN

As in expressions (4) .A'16 A16 .0140

B6 = B =.800016 16From (3) an i (4) ,

16 20,GOO .0140) + 9135 (.8000) = 7590 lbs

P = 20.,000 (.0239) + 9135 (.3200) 3400 lbs

P36 20,000 (.0500. + 9135 (.0880) 1800 lbs

P46 20,000 (.1090) + 9135 (-.0860)= 1390 lbs

P 5 6 20,000 (.2450) + 9135 (-.3200)= 1980 lbs,

and from (5 ) anJ (6).

P tP =20,000(. 0140) + 4693 (.8000) =4030 lbs.

Equilibrium Check

N

, P = 20, 190 lbst-20,000 lbs

j=l

% iifference = 20,000 x(10) 5%

WADD TR 60-517 5.23

5.1.2 (Cont'd)

The load in the top plate and the attachment loads are plotted in Figures 5. 1. 2-2(b)and (c). The results show that the maximum load occurs in the first attachment and thatthe two end attachments carry more than half of the total applied mechanical load. Whenthe bay properties are constant the maximum load always occurs in an end attachment.However, as discussed in Paragraph 5.1.3, when plastic deformations occur in the vicinityof the bolt holes, the bolts tend to carry equal loads.

5.1.3 Constant Bay Properties - Rigil Sheets

Now the case is considered in which the sheets have negligible axial deformationas ccmpared to the deformations caused by local distortions of the holes and attachments(rigid sheets).

From a practical point of view, this condition is realized when the plates are thickand the attachments have small diameters -- or, when local yielding causes the effectiveattachment-hole flexibility factor to become large as compared to the axial flexibility ofthe sheets.

In the limiting case,

L 0 and(L)'3T B

and the compatibility Eq. (1) Gf Paragraph 5. 1.2 reduces to

j +1 j f

•,r, in terms of the first attachment load .

P = P1- (-i) . (1)

P 1 is obtained by summing Eq. (1 over the total number of attachments:

N Nt. =X=NP1 -i 0-1)J=l f j I

NPI N (N-1) f

or

P -+ (2) -

1 N 2 f

Substituting ( 2 ) in (1 ) results in

P '1=.+(+ i (3)

Equation (3 ) gives the attachment loads for a given thermal and mechanica! loading and aknown value of attachment-hole flexibility, f

WADD TR 60-517 5.24

5.1.3 (Cont' d)

The solution ehows that for the cpqe of constant bay properties and infinitely rigidsheets, the mechanical load distributes equally to the attachments while attachment loadsdue to thermal effects vary symmetrically about the transverse cenierline of the joint withmagnitudes inversely proportional to the attachment-hole flexibility (Figure 5.1.3.2).

For high loads which cause extensive plastic deformation in .he vicinity of the attach-ment holes, the effective attachment-hole flexibility may become large as compared to thesheet flexibility, in which case the solution of Eq. (3) is approached. if these plastic effectsbecome large enough, the increase in f tends to wipe out the effects of thermal loading withthe result that:

P~ -j NThis indicites that near the failure of ductile mater'ials, the mechanical load '.ends to

distribute equally to the attachments regardless of temperature distribution. Theeffect of an infinite sheet rigidity for the splice of Figure 5.1.2-2(a) upon attachmentloads may be calculated where, as before in Paragraph 5.1.2,

X = 20,000 lbs

f = .900 x 10-6 in/lb

AO2 = 1110 x 10-6 in.

N= 6

However,

A0T.

Substituting the above in E4. (3),20,000 (6+1 )1110

Pj 6 2 = .90

or

P. = 3,333+ (3.5- j)1233

Thus:

11 = 3,333 + (3.5-1) 1,233 = 6,420 lbs

P2 = 3,333 + (3.5-2) 1,233 = 5,180lbsP3 = 3,333 + (3.5-3) 1,233 = 3,950lbs

P4 = 3,333 + (3.5-4) 1,233 = 2,720 lbsP5 = 3,333 + (3.5-5) 1,233 = 1,480lbs

P6 = 3,333 + (3.5-6) 1,233 = 250 lbs

and 6P. = 20,000 lbs

WADD TR 60-517 5.25S.•i•U

5.1.3 (Cont'd)

The results are plotted in Figure 5. 1. 3-1. It is seen that the attachment loads varylinearly with the distance along the splice. A comparison with the flexible sheet results ofFigure 5. 1. 2-2 shows that the attachment loads in the rigid sheet solution drop off constartlyas one proceeds from the first to the last attachment, while in the flexible sheet solution theattachment loads are minimum at the center of the splice and build up toward the ends.

Load In Top Plate

20,000

8,400

4,450

6,420 5,180 3,950 2,720 1,,t,. 25.)

Plot of Bolt Loads

6,420 5,180 3,950 2,720 1,480

250

FIGURE 5.1.3-1 RIGID SHEET SOLUTION

WADD TR 60-517 5.26

5.1.3 (Cont'd)

Qualitatively, the difference in the results is due primarily to the effect of thermalloading (Figure 5. 1.3-2) which becomes more severe as the sheet rigidities increase.

SThe Attachment Loads

Are Symmetrical Aboutvrrt rTn nn rnto• _ The Line Of Symmetry

S-S and P >P 2 .

Pp P1 2 P3 =0 P4 =P 2 P51

4-u '4

S

FIGURE 5.1.3-2 ATTACHMENT LOADS DUE TO THERMAL LOADING WHENTHE JOINT HAS CONSTANT BAY PROPERITIES

39

WADD TR 60-517 5.27

5.1.4 Constant Bay Properties - Rigid Attachments

The attachments may be considered rigid if the attachment-hole flexibility isnegligible when compared to the axial flexibility of the sheets as, for example, when thesheets consist of flexible (soft) materials and the attachments are housed in large hia-meter rigid bushings. Although this situation seldom occurs in practice, the problem isof interest from a qualitative point of view, since it represents another limiting case ofthe general one-dimensional problem. In the limiting case, f -- 0 and the compati-bilit, Eq. (1) of Paragraph 5.1.2 yields

AO+ X()P1 L L"b'l)t1 Ecj.B

P2 = 3 =P4 PN-i 0

Thus, from the equilibrium equation,

X APN = X-P (2)LI' + L (2)

The above equations show that for infinitely rigid pins and constant bay properties, thetwo end attachments carry all the thermal and zyiecnanical loads.

When the upper and lower sheets are each heated to lifferent uniform temperatures,then

,&0 [ &T)T - (aAT)B] L

so that

"P X + (AE)T r (cAT)T - (PAT)B (3)

I (AE•,

and the intermc:iiate attachment loads are all /er.,.

WADD TR 60-517 5.28 U

5.1.4 (Cont d)

7he effect of infinitely rigid attachments for the splice of Figure 5 .1. 2-2(a) upon theIi attachment loads may be calculated where, as before in Paragraph 5.1.2,

X =20,000 lbs.

L -6(AE)T .356 x 10 In/lb

*19

L -6(AE)B .267 x 10 in/lb

I. [(a-T)T- (aAT)Bj = 1110 x 10- inch

but f=0 .

Substituting the above quantities in Eq. (3) givesn4

20,000 + (-3-6-)(1)10)1 =3.267 =13.200 lbs.

.356

For rigid attachments as atated above,

P2 = P3 = P4 = P5 = 0

and for ecuilibrium

P6 =20, 00'-1"3,200 =6,800 lbs.

WADD TR 60-517 5.29

5.1.5 The Influence of "Slop" on the Load Distribution

The presence of "slop", due to manufacturing tolerance and differential thermalexpansions between the plate holes and attachments, affects load distribution through thebasic joint compatibility equation.

The "slop" at each attachment is indicated by the difference in diameters of theplate hole snd attachment (Figure 5.1.5-1), as expressed by

e = emfg + etemp ; e >0 (i.e., e must always be a clearance),

wheree Initial room temperature manufacturing tolerance (clearance or interfer-

ence. Interference has negative sign.)

andetemp thermal slop (clearance or interference due to differential thermal

expansion between plate holes and attachments)

[aT)sheet -(Tattach D Dhl

S~e e• -

FIGURE 5.1.5-1

.,

FIGURE 5.1.5-1 ATTACHMENT IN AN OVERSIZE HOLE

As discussed in Paragraph 5. 1. l, the joint displacements (for compatibility purposes)have been measured from a datum defined by the initial spacing between the centerlinesof adjacent attachments. When slop is present, and thermal and mechanical loads areapplied to the joint, the attachments are displaced from the centers of the holes untilthey bear up against sheet material, as shown in Figure 5. 1.5-2.

WADD TR 60-517 5.30

5.1.5 (Cont' d)

LJ+6 J

r/ -- /f"- ° + ' 2 '

fJ PJ+ICenters of

/Holes77]w

Lj +5B

FIGURE 5.1.5-2 ATTACHMENT DISPLACEMENTS DUE TO SLOP - POSITIVE LOADS

The algebraic sign of the slop displacements depends on the direction of the joint

loads. For the top sheet, the displacement between adjacent attachments is given by-e P.+ e P

jT 2 jP_ 2 fj 1

and for the bottom sheet, by

ejB P e(+l)B6.:_I_. - Pi+ I_jB 2 pJi - 2 P +1

where a positive 6 Increases the spacing between adjacent attachments and

• + 1 for a positive attachment load,

P = - I for a negative attachment load.

IPi

The Incompatibility due to slop is therefore given by

!(e + e 1T (eTeAj jT - B= 2 T Be)+1 p+ T (e~ Bj J (1)

WADD TR 60-517 5.31

5.1.5 (Cont'd)

Equation (1) is significant in that it brings out the very nature of the slop problem.Consider, for example. that the slop is the same at all attachments. In this case. Eq. (1)reduces to

A6j (e T ( e (2)

As the mechanical loads applied to the joint increase, all the attachment loads tend to act inthe same direction (opposite to the externally applied load) or,

p pIP l1'j AI I Pi I

The bracketed quantity in Eq. (2) thus becomes zero and therefore

A6 = 0 (3)j

Since the A 6 's determine the influence of slop on the load distribution, Eq. (3) in-dicates that for high joint loadings the effects of uniform slop are eliminated. TR solve forthe load distribution with slop, the basic one-dimensional compatibility expression. Eq. (5)of Paragraph 5.1.1, must be modified by the addition of A 6 The compatibility equationsthen become

L- - jT i Pi =(A~ 5 ÷ j)- PjfA + f(jOl) + A x(-+Tf (4)1 i=1 J JJ0+) A

The above equation, together with the equilibrium Prlantion

N

x=X ,.j=1 J

provides N equations for the N required attachment loads, P. .The solution of these equa-

tions, however, involves more than simply solving a set of simultaneous algebraic equations.The values of A6. on the right side of 'Eq. (4) are given by (1) from which, in order to determine

Jthe A6 3 s, the sign of the attachment loads (positive or negative) must be determined. But this

is not known in advance. This presents one of the major difficulties of the slop problem. Thesuggested method of attack is as follows:

(1) Assume a set of directions for the attachment loads.

(2) Determine the A 6 Is from Eq. (1) and solve the simultaneous compatibility and

equilibrium equations (Eq. (1) of Paragraph 5. 1. 1 and Eq. (4) above).

WADD TR 60-517 5.32

5.1.5 (Cont'd)

(3) If the directions of the attachment loads as obtained from the solution agreewith the initially assumed directions, the solution is correct.

(4) If the directions of the attachment loads as obtained from the solution do notagree with the initially assumed directions, the solution is incorrect. The proceduremust be repeated with a new set of attachment load directions, preferably the onesobtained from the solution.

The solution is the correct one when the assumed set of attachment load directions

yields a solution with the same set of directions.

Obviously, a digital computer program is desirable in solving problems of this sort.

The following example illustrates the method of solution.

Scarfed steek and aluminum plates are bolted together with three attachments as shownin Figure 5. 1.5-3. Find the attachment loads when the steel and aluminum plates are subjectedto uniform temperature rises of 640°F and 80°F, respectively, and a mechanical load ofX = 5000 lbs. is applied to the joint. The manufacturing tolerance is to be taken as em .0003

inch for all holes. It is assumed that the bolt-hole flexibilities have been determined experi-mentally to be:

-6fl = 1. 300 x 10 in/lb1 -6

f = 1. 200 x 10 in/lb (Refer to Sub-section 5.3 for a discussion of bolt-hole2 -

f = 1. 300 x 10-6 in/lb. flexibility)3

The appropriate quantities for substitution in Eq. (4) are first determined.Using average thicknesses for each bay,

= 1.25 = .119 X 10- inlbIT (.175)(2)(30) x 10

(1• = 1.25\AE/ 2 T 1.125)(2)(30) x 106 = 167 x 10.6 in/lb

= 1.25= .278 x 10 6 in/lbIB (.225)(2)(10) x 10

1.25 -6

2B (.275)(2)(10) x 106 x

V

WADD TR 60-517 5.33

5.1.5 (Cont'd)

Steel

AT =640°F511 6 2

-Dia. Steel Bolts E 30x:106 lbin

a =6. OxO in a =6.5x10 in/in/eF

i n. -OF

n44

S~5000 lb• I -0 " I

F R . -X I5000 Ibs

•; Aluminum

S1. 25 in. 1. 25 in --. AT = 80°F

Sot =12x:10- in/in/°F

Since the temperature rise iii each oay is unizorini the incompatibilities due .unrestrair-ed:• thermal expansion are giver, by

S•1 = [(6tAT)T -(1.T)B3 ] L

-6 Inch

- 4000 x 10-6 inch .

A ssuming the temperatnre of each bolt tc be the .aric as the surrotnding sheet

material, the si-ps due to ter..perature are given Ly

temp = (T)top sheet - (cT)top o; uatach. D hele= .tempi top (6.01640) x.31w Io6

= 100 x inch.

WADD TR 60-517 5.34

5.1.5 (Cont'd)

[etempj bottom = [(T)bottom sheet (aT)bottom of attach.] Dhole

= (12.0X80) - (6.0) (8o) x .3125 x 10-6

= 150 x 10- inch.

-6Since eng =300 x 10 in. the total slops are given by

"elT =e 2 T = e3T = {emfg +etemp top[ 3 00 o X

400 x 10-inch.

"elB =e 2 B =e3B m[eMfg +etemp] bottom

[300 + 150] x 10-6

= 450 x 10 6 Inch.

The incompatibilities due to slop, Eq. (1), are thusA6 1(0040 P2 (40 + 50

1'21 1pi

2 2 pI45- 4 25 2 p.9 P- 1 -_ 1 04

ýi% IN _ P

2 1 = 21)

Substituting the known quantities in the incompatibility Eq. (4 and collecting terms yieldsthe following expressions:

r2 to1.697 P -1. 200 P2 = 4595

1 2 \P 1 P 31) (a)1

.394 P + 1.594 P - 1.300 P3 =4835+425 - - _

and for equilibrium,A

P+ +1 + P = 5000.

Expressions (a), (b) and (c provide 3 simultaneous equations which can be solved for theattachment loads P1 .I 2 and P3 ., provided their Jirections are assumed correctly.

For a first trial, assume that all the attachment loads are positive. In this case theslop terms (2nd terms on the right in (a )and (b)) vanish and the expressions reduce to

1. 697 P I- 1. 200 P 2 =4595.3

.394 P I+1. 594 P 2 -1. 300 P34835

P1 + P2 + P3 =5000.

WADD TR 60-517 5.35

5.1.5 (Cont' d)

for which the solution is

P1 =+ 3880 lbs "•

P =+ 1650 lbsr (d)

P3 =-5301Ibis

The above solution would be correct if no slop were present. However, since jointslop is present, the solution contradicts the initial assumption that the bolt loads are allpositive and it is therefore incorrect. As a second trial, assume that the directions of

the attachment loads are as given by the first solution, namely, that the loads in the firsttwo bolts are positive while the load in the last bolt is negative. Expressions (a) and (b)then become

1.P697 P- 1.200 P2 =4595

394 P + 1. 54 P 2 1.300IP3 =3985.

and as before,

P 1 + P 2 + P 3 = 5000.

for which the solution is

P 1 = 3730 lbs

P2 = 1440 lbs (e)

P 3 = -170 lbs.

This is the correct solution, since the directions bf the bolt loads are in agreement with theinitial assumption.

Expressions (d) and (e) show that there is not much disagreement between the solutionswith and without slop. This was to be expected. since the slop for the joint under considerationis small.

5.2 THE TWO-DIMENSIONAL (BOWING) PROBLEM

When the bouniary conaitions are such that the joint is allowed to bow out .•f its own

plane, the solution is much more cumplicate.i than in the one-dimensional case. Additional

factors such as rotati.nal and out-of-plane ,iisphcements, beam-coluni effects, moments

at the attachments, etc., enter intOz the problem. The present state of the art makes an exact

analytical solution t- the problem impracticable.

The purpose of the analysis presented here is to -btain t first approxir, ation to the

solution of the two-dimensional (bnwing) problem by modifying tWe equations of the one-

dimensional (uniaxial) silution.

A letailed 3erivation rf the solution is presented in Reference 5-1.

WADD FR 60-517 5.36

I|

5.2.1 The Joint Equations

P,.wing 'f the joint (Figure 5. 2. 1-1) occurs .ue to the combined effects ofnon-uniform temperature -listribution and externally applied mechanical loading.

The scluti,-n presenteJ gives the shear loads in the attachmvents for a known-.et f applied mechanical and thermal loads where the following simplifying assumptionsare made:

(1) The bay properties are constant (sheet thicknesses, attachment size and* spacing, stiffnessess, etc., are the same for each bay). The thermal loading is assumed

n-'t to vary in the longitudinal direction, but may vary through the thickness.(2) Vertical .out-of-plane deflections and clamping loads are assumed to have a

negligible effect on the load distribution (negligible beam column effects).(3) Monen s at the attachments have a negligible effect on, or are included in,

A ithe attachment-hole flexibility.(4) The contact faces of the top and bottom plates of the joint are initially plane;

the external axial loading is applied parallel to this plane in the direction of the line of attach-ments.

(5) As in the one-dimensional case, the joint materials are assumed to deformelastically under load.

lastic Axis ,.f Plate

/"14, A- A- Aý. 1+1B X +Y y )-MV Il J! I III)

j-i j j+l N

(a)

FIGURE 5.2. 1-1: (a) JOINr IN UNBOWED CONFIGURATION(b) BOWING OF JOINT DUE TO COMBINED EFFECTS

OF NON-UNIFORM TEMPERATURE DIs'rRIBU'FIONAND MECIHANICAIL WADING

WADD TR 60-51; 5.37

4;-i

5.2.1 (Cont'd)

Under the above assumptions the requirement of compatibility at the attachmentsyields the following compatibility equations:

j5' ~ -P. P f Xf(j 1, 2,. N-1) ()

A i Pf + j'1 f ATfA =

whereA =(•)T~ ('E) B (EI)T +(E) (2)

A(- L [('OTwT+(Ef)BwB]= Z10" L (YT + YB) (3)

ATEjT + (EK) B

+ y

IAT(Y B•) ML r4)+, -fA +i L + (-)B 1

and w is the curvature due to temperature (Paragraph 4. 1.1). If the thermal gradient Islinear through the thickness, then w approiamately equals aAT where ATis the linear

h hthermal gradient through the plate thickness (positive for higher temperatures on the upperface of the plate).

Equation I ) together with the equilibrium equation

provides N simultaneous equations for the determination of the N unknown attachmentshears.

A comparison of Eq. (1) with the one-dimensional compatibility equation, Eq. (1) ofParagraph 5.1.2, shows that the two forms are identic.l. Thus, when the bay propertiesare constant, the procedure for the two-dimensional solution is exactly the same as for theone-dimensional case if the one-dimensional coefficients

L + L L

AE T AEB AE T

are replaced by the expressions on the right side of Eqs. (2), (3) and (4). respectively.Coefficients A. sa B can then be obtained, as before, from Figures 5.1.2-1 for deter-

I jNmination of the attachment loads (Eq. (2) of Paragraph 5. 1. 2).

WADD TR 60-518 5.38

5.3 DETERMINATION OF THE ATTACHMENT-HOLE FLEXIBILITY FACTOR

The flexibility of a given attachment-hole combination must be determined quantitativelyin order to make use of the solutions presented. If load deformation curves for the specifiedsplice materials and temperatures are available (Figure 5.3-1), an Initially determined slope

Ko = gives the stiffness (or flexibility) in the elastic, low-load range.

101

Initial Slope [_-Line

KjOKs = Secant Stiffness

V A- Secant Slope Line

/

A (Deflection)

FIGURE 5.3-1 ATTACHMENT-HOLE LOAD VS. DEFLECTION CURVE

This initial slope is larger than the secant slopes encountered at higher load levels:using this initial value would give a conservatively greater stiffness (lower flexibility) thanactual for succeedingly higher load ranges. This would result in an overestimate of themaximum attachment load. When load deformation data is available, and the joint analysisshows loads corresponding to secant stiffnesses that are appreciably different from thoseinitially assumed, then the stiffnesses should be corrected and the analysis repeated. Sucha procedure should converge rapidly to a valid solution.

If load deformation data is not available, the limit bearing load criteria of Reference 5-2may be used to obtain an estimate of the attachment-hole flexibility. These criteria result in anoverestimate of the attachment-hole flexibility and an underestimate of the maximum attachmentload at load levels below yield.

As an example of the way in which the criteria of MIL-IDBK-5 (Reference 5-2) may beused to estimate the attachment-hole flexibility factor, consider the joint(in the illustrativeproblem)of Figure 5. 1.2-2 in which the bolt diameter is 1/4 in.,the upper sheet is . 125 in.thick titanium and the lower sheet is. 250 in. thick aluminum.

Assuming the aluminum plate to be of clad 2024-T6 material, Table 3. 2. 3. 0(f) ofReference 5-2 givvs a bearing yield stress of 78, 000 psi. The load at this yield stress is

WADD TR 60-517 5.39

5.3 (Coat' d)

Palum. = 78,000 (.250) x 1/4 = 4875 lbs.

If the titanium sheet is taken to be 6AL-4V material, Table 5. 2.4. 0(b) of Referenca 5-2gives a bearing yield stress of 198, 000 psi. The load at this yield stress is

P = 198,000 (. 125) x 1/4 = 6200 lbsPtitan.

The average yield stress load is thus

6200 + 48?5P - = 5590 lbs.avg.- 2

To find the bolt-hole flexibility, the deformation for which the average yield load occursis taken as being equal to 2% of the nole diameter (Reference 5-3, Paragraph 3.6111). Forthis deformation

f 0 9(.o2x1/4)avg 5590avg

5.4 REFERENCES

5-1 Report RDSR-3, Analysis of Joints, Republic Aviation Corporation (to be issued)

5-2 MIL-HDBK-5, Strength of Metal Aircraft Elements, March 1959

5-3 ANC-5 Bulletin, Strength of Metal Aircraft Elements, March 1955

WADD TR 60-51: 5.40

SECTION 6

THERMO-ELASTIC ANALYSIS OF PLATES

WADD TR 60-517 6.0

SECTION 6

THERMO-ELASTIC ANALYSIS OF PLATES

TABLE OF CONTENTL


6 Thermo-Elastic Analysis of Plates 6.3

6.1 Theory of Deformation of Plates 6.4

6.1.1 Definition of a Plate 6.4

6 1.2 Assumptions for Linear Plate Theory 6.4

6.1.3 Three Kinds of Plate Problems 6.6

6.1.4 Fundamental Equations of Thermo-Elastic Plate Theory (with 6.7illustrative problem)

6.2 Bending of Plates 6.13

6.2.1 Bending of Rectangular Plates with Linear Gradient Through the 6, 13Thickness

6.2.1.1 Unrestrained Rectangular Plates 6.146.2.1.2 Clamped Rectangular Plates 6,14

6.2.1.3 Simply Supported Rectangular Plates 6.14

6.2.1.4 Square Plate - One Edge Free 6.17

6.2.2 B3ending of Circular Plates with Linear Temperature Gradient 6.17;.Through the Thickness

6.2.2.1 Unrestrained Solid Circular Plates 6.17

6.2.2.2 Clamped Solid Circular Plates 6.17

6.2.2.3 Simply Supported Solid Circular Plates 6.17

6.2.3 Circular Plates - Temperature Difference gs a Function of the 6.18

Radial Coordinate

6.2.3.1 Clamped Plates (with illustrative problem) 6.19

6.2.3.2 Simply Supported or Free Plates. aTa.rK 6.25

WADD TR 60-517 6.1



6.2.4 Approximate Solution of Free Plate with Arbitrary Temperature 6.26

Variation Through the Thickness Only

6.3 Slab- Problems - Plates 6.29

6.3.1 Thermal Stresses in Rings - Asymmetrical Temperature 6. 31Distribution

6.3.2 Thermal Stresses in Solid Circular Plates Due to Asymmetrical 6.34Temperature Distribution

6.3.3 Circular Disk with Concentric Hole Subject to a Power Law 6.42

Temperature Distribution

6.3.4 Circular Plate - Central Hot Spot 6.48

6.4 References 6.57

W6 1

. WADD Th 60-517 6.2

SECTION 6 - THERMO-ELASTIC ANALYSIS OF PLATES

This section is concerned with the determination of thermal stresses in plates. Forexample, solid and hollow bulkheads are plate-like major components in the semi-monocoquetype of construction used in air and space vehicles.

The thermal-mechanical problems related to plates may be divided into bending, slab,instability types. The first two are discussed in this Section 6 while the instability type ofproblem is treated in Section 9.


a, b Planform dimensions of rectangular plates; radiih Thickness of plater, e Polar coordinatesu Radial component of displacementu, v, w Middle plane displacement in the x, y, and z directionsx, y, z Rectangular coordinatesu*, v* Displacements in the x and y directions

Eh3

D 12 (1-zYZ)

E Young' s modulus

M T ctEf Tzdz-h/2

Mx, My, Mxy Moments per unit of length

h/2

NT ltE f- TdzT -h/2Nx, Ny, N Forces per unit of length

y xyP Distributed lateral loadT TemperatureTD,AT Temperature difference between the upper and lower faces

a Coefficient of linear expansionx •yy z

SZty ~ fComponents of strainE Xy : Ey z ' zXZ

V Poisson's Ratio

orr' 'rO' e00 Components of stress in polar coordinates

1xx yy zz 5a a Components of stress in rectangular coordinates

0 Stress function

WADD TR 60-517 6.3

2 2

6.1 THEORY OF DEFORMATION OF PLATES

6.1.1 Definition of a Plate

A plate is defined as a non-rigid three-dimensional structure in which one dimension,the thickness, is much smaller than the remainine two dimensions. In its simplest form, thetheory of plates can be regarded as an extension of beam theory.

6. 1. 2 Assumptions For Linear Plate Theory

The classical theory of plates is based upon the following assumptions:

(1) The material is assumed to be homogeneous, isotropic, and toobey Hooke's Law.

(2) The constant thickness of the plate is small compared to its otherdimensions.

(3) The deflections of the plate are limited in magnitude and are of theorder of the plate thickness. It can be shown that this restriction means that the effectsof the normal deflections upon stretching of the middle plane of the plate are negligible.Consequently, only negligible stresses are induced in the median plane of the plate whenthe plate is loaded normally.

(4) Plane sections which before bending are normal to the median p'aneof the plate remain, under the above conditions, plane and normal to the median planeafter bending.

The median plane of the plate is assumed to lie in the xy-plane and thethickness of the plate is h. If the plate is loaded by forces normal to the xy-planethen the element will distort as shown in Figure 6. 1. 2-1, where u* and w are thedisplacements in the x and z directions, respectively.

r'rom the assumption that the deflections are small and that no medianplane displacements or stresses are induced under deformation, the point 0 will bedisplaced normally to the strainless median surface (the upper one) and the planesection AOB will rotate into position A1 OB 1. The angle between the horizontal x-axis

and the tangent to the deformed plane at point 0 is 3w.ax

In accordance with the assumption that plane sections remain plane andnormal after bending, the angle of rotation of the cross section A OB will also beaw. Then for the small assumed deflections

-x

U* ;)wax

In a similar manner, the displacement in the y direction for any point adistance z from the median plane is given by

vw

WADD TR 60-517 6.4

-%- - -

.1.2 (Cont'd)

X, u

"F CC

CD FO MATONI•

Cll

FIGUDD R E60 6.1. - RS SECIO OF PLT EOE N FE`` . . . ..

DE ORATO

CI A,

6.1.2 (Cont'd)

(5) The normal stresses through the thickness of the plate are negligiblebecause the surface loads are small compared to the bending stresses induced in theplate.

6.1.3 Three Kinds Of Plate Problems

The thermal-mechanical problems to be considered may be divided into

(1) Bending Problems - occur when the temperature varies in the thick-ness direction and the mechanical loads are normal pressures.

(2) Slab Problems - occur when the plate is loaded by mechanical forcesparallel to the middle plane of the plate which are uniform through the thickness and thetemperature varies in the planform direction only. This is a plane-stress problem(see Section 2 for definition of plane-stress) and is considered in Sub-section 6.3.

(3) Instability Problems - take place when there Is edge restraint toexpansio- in the direction parallel to the middle plane of the plate (see Section 9).

i60

WAdDD TRll60-517 6.6

6.1.4 Fundamental Equations Of Therwo-Elastic Plate Theory

The dlrec' approach to plate problems is to derive the equal-ions ofequilibrium of forces and compatiblity of displacementm. These equations are Inthe form of partial differential equations which must be golved subject to prescribedboundary conditions.

The differential equations or the thermal-mechanical problom of platesare discussed next, followed by an illustratlive example. The xy-plaae Is ansamed tobe the middle plane of the plate of contant thickness h and the shape of the bound~ryIs arbitrary. The components of displacement In tie x, y, and z directions are de-

noted by u*, v*, w (the symbols u, v. in this section are reserved for middle planedisplacements In the x- and y- directions, respectlvely). Under the assumptions thatplane sections remain plane, the displacements u*, v*, w are of the form:

u* (x,y,z) u (x,y) - z 2 (x,y)Ow

v* v (YZ)v(x,y) - z (x,y) (1)

w - w (xy).

The strain components in planes parallel to the xy-plane are:

2Exx= =o - z

4 V 2 w(2)

4 Ou*÷ WO au 2zv f 2 w

au 8Vau WvNote: 5i 8 + y- are the strain components in the middle plane of theBx' ay' xy

plate and arise as a result of in-plane deformations. The additional terms are dueto the bending of the plate.

The corresponding stress components are (T - temperature above roomtemperature datum): :4

E 5

xx E f *xW -(1 +V'

- •-E (6•3 • yy (1 +') -T) (3)

D- 2(1' V) *,cY

WD R60-517 6.7

6.1.4 (Cont'd)

which introduce the following forces and moments per unit of length (Figure 6. 1. 4-1):

h/2 h/2

N - f dz Mx f axZdzIx -h/2

-h/2 -h/2h12 h/2 -

N- dz Ma--My of zdz (4)yd yy yyy

-h/2 -h/2

h/2 h/2N JX a, Ydz M -X-MxY j a XYz&

MI "y X yx- M-b/2 -h/2

I

YY

S • Nyx

FIGURE 6.1.4-1 POSITIVE DIRECTIONS OF FORCES AND MOMENTS

Substitution of Eqs. (2) and (3) into Eqs. (4) yields

N X E h - ' + - -"

2 T1-13 15)

N - Eh W•, .4.

xy 2(1T 0 -) 6. f

WADD TR 60-517 6.8

6.1.4 (Cont'd)

fA2W a2W\ MT

My ~ - -- •+¥ -a2 I-P(5 otdax ay/2 22

xy (5 - a-x-],

Swhere the bending rigidity per unit of length of the plate is

3D Eh21_2Eh (6)

and h/2

NT fE_ TdzT -h/2

h/2 (7)

MT aE f Tzdz,-h/2

where a and E are assumed to be constant.

The equations of equilibrium in planes parallel to the xy-plane are

ON ONxay

(8)ON ON

Sxy+ y 0ax ay

and these equations imply the existence of a stress function 0 (x.y) such that

32N2 x

'10- Nx•y xy

WADD TR 60-517 6.9

6.1.4 (ContId)

Integration through the thickness of the compatibility equation for the planestress problem (Reference 6-1) gives

T4'

where2 2 a2

ax2 ay2 (10)

4-2 2 2V4 3- + -ay

The equation of equilibrium in the z direction (Reference 6-2) is

aM2M 22w 2w w282Mx a2M•' ---- Y --- +(2N - N)

!S = - 2 _( -+ -ax2 axay )y2x 2x2 xy +xNy y )y2 1)

where P (x,y) is the distributed lateral load in the positive z direction; andMx, My, M xy are the conventional bending moments per unit of length. If

the quantities Mx, M , Mxy are expressed in terms of the displacements from

Eq. (5), then the equilibrium equation in the z direction becomes:

4 a2 22 2D V w= P+N ++ N L + 2N 1 2MSIx 2 ; 2y y Dx ay M - T (12)

The solution of a problem requires in general that Eqs. (10) and (12) besolved simultaneously subject to appropriate boundary conditions.

If the plate is supported so that Nx, N y, N are negligible, as would occur

if there was no restraint in the median plane of the plate and no applied in-planeforces, then the simpler equation may be used:

4 1 2D V w=P- V M1-' T (1Z

It Is clear that the equation expressed by (13) is equivalent to a "mechanicalproblem" with normal loading = P - 1 V2 M , so that all the known

techniques (Reference 6-1) for non-thermal problems may be used.

WADD TR 60-517 6. I0

6.1.4 (Cont'd)

A solution of the differential equations in the case of a simple plate problemfollows.

A clamped circular plate of radius R :s subject to a linear gradient through

its thickness h where the temperature is T1 and T10 at the top and bottom faces,

respectively (T 1 > To). Find the deflection and stress assuming that the middle

plane of the plate is free to expand.

z

FIGURE 6.1.4-2 CLAMPED CIRCULAR PLATE WITH LINEAR GRADIENTTHROUGH THICKNESS

The temperature variation is denoted by

T= T- (T T )

From (13), the differential equation4 to be solved is

4 1 2DV w = P - V MTD~w=P-V M

where 2

P=0, andV M = 0as M does not depend on x and y.T T

Therefore

7 w =0 ,(14)

or in polar coordinates, with r as the radial coordinate:

r4 d4w!+ 2 d~w _2 d2 w -'"- 0. (15) i43 2 dr 4dw+2r 3 d w r 2-d + r =ý0(5

dr4 dr3 dr

WADD TR 60-517 6.11

- . --w=.* - -- .--

6.1.4 (Cont'd)

This is an equi-dimensional differential equation and the solution is readilyfound in an elementary differential equations text,

22w=Alogr+Br logr+Cr + D (16)

However, the deflection and bending moment must remain finite as r.-0.

This implies that A = B = 0. Therefore, w reduces to

w =Cr2 +D . (17)

dwThe clamped boundary conditions are w = d---= 0 when r = R. There-

fore, sunstitation of Eq. (17) into the boundary conditions yields w-0,

i.e., the normal deflections are identically zero.

From Eqs. (5) or analogous equations in polar coordinates, theradial and transverse moments per unit length are given by

-MTMr Me =

(

where from Eq. (7)

h r(TO + T1) h

ST E f 2 + (T0 -T 1 ) z dz,

-h/2

or aE(T -T) h

M 01MT = 12 (19)

The components of maximum stress in the radial and transv,..rsedirections are

= 6 6 cE(Ti- To)

SrrI, Olmax h001r = r M== 2(1 -Y) (20)

a = or EATrrj a 0 a 2 (1 -Y) (21)

Smax max

It where A T = T 1 - To = difference in temperature between the upper and

lower faces, respectively.

Note that in the simply supported case with boundary conditionsw = M = 0, there result zero stress and paraboloidal deflec'iorrs (see Paragraph A

r6. 2. 2. 3). More precise analysis yields spherical deriections.

WADD TR 60-5 17 6.12

6.2 BENDING OF PLATES

Bending problems in plates occur when the temperature varies in thethickness direction (and possibly other directions) and the mechanical loads(if they exist) are normal pressures. In the succeeding paragraphs emphasiswill be placed upon the linear gradient through the thickness.

6.2. 1 Bending of Rectangular Plates With Linear Gradient Through TheThickness

•he rectangular plate is shown in Figure 6.2.1-1 and the temper-ature variation through the thiciuess is assumed to be linear. Ex-.Pressionsfor stresses and deflections corresponding to various support conditions aregiven in the following paragraphs.

h

b/2

-, K -

b/2

4kz z

yAD= ro-T1

FIGURE 6.2. 1-1 PLATE GEOMETRY SHOWING A POSITIVE LINEAR THERMALGRADIENT THROUGH THE THICKNEISS

WADD TR 60-517 6.13

6.2.1.1 Unrestrained Rectangular Plates

A free rectangular plate subjected to a linear temperature distri-bution through the thickness and constant temperature over the planform is unstressed.The plate becomes curved and fits a sphere of radius inversely proportional to thedifference in surface temperatures and inversely proportional to the thickness.

6. 2.1.2 Clamped Rectangular Plates

Maximum Stress:

EaATh• x ~y=• 2(1Et- ~ (+ sign corresponds to face z =+ --- ) (1)

where the temperature difference is defined by

AT = 2 2T

Deflection: The transverse deflection w is identically zero.

6.2. 1.3 Simply Supported Rectangular Plates

Deflection(see Figure 6.2.1-1):-AT(1+v.)2a C sin m--a cosh -

W (1 + )43 3 cosh aw h m mS~~m=l, 3,5S... ,,

where mxb

m 2a (lb)

Bending Moments Per Unit of Length:2D sin (a- cosh (J!Y)

_x = x h m cosh a (2a)

m=1,3,5... m

2 ifrocs(ýý"M -aAT(1-

2 )D 4DaAT (1 -Y Z "c a

y h A3h m cosh a .

(2b)

The series for w converges very rapidly while the series for the momentsconverge more slowly. Judgment must be used in choosing a suffiC -it number Iof terms for the required accuracy. However; the formulas for the maximumbending moments and stress are expressible in closed form as given below.

WADD TR 60-517 6.14

6.2. 1. 3 (Cont d)

Maximum Bending Moments:(.ieference 6-2)

(M( +Eh2•tAT• (x) =(My) 12=a

Sy= b/2 x=o, x=a 1(

Maximum Bending Stress:

SEaAT(ax) (c) 2 (3b)y =*b/2 Yx=o, x=a

Equations (la), (2a), and (2b) have been evaluated in non-dimensional formfor the square plate as shown in Figures 6. 2. 1. 3-1 through -3.

}K

0.0 0.2 04 " 060.0 0.8

w h

FU 6 -ION ( R A

SaAT• 008 y, -S% •• • •"" •, , EDGEJ"

S0.12 *46

S~0.16

DFIGURE 6.2.1-3- DEFLECTION (w) VS. POSITION (x) FOR y PARAMETERS

WADD TR 60-517 6.15

6.2.1.3 (Cont'd)

0.32)

MO: 4D0AT (IV2)h

024 y u

0.16

S1 "43 SU PPORT

ma0.08 %- qw-.

- -,., .,.EDGEA

0.00

0.0 0.2 0.4 0.6 0.8 1.04

FIGURE-6.2.1.3-2 BENDING MOMENT (My) VS. POSITION (x) FOR y PARAMETERSy

0.32

MOI 4D (AT I- Y2)h

Y =0,1 FREE EDGE O0.24 0 ____SIMPLE SUPPORT -

xX :O FREE EDGE

Y' W, w SIMPLE SUPPORT000a y I I ,0.16 4 ,

SIMPILE SUPPORT \

0.0 0.2 0.4 0.6 0.8 1.0

FIGURE 6.2. 1. 3-3 BENDING MOMENT (Ms) VS. POSITION (x) FOR .y PARAMETERS

WADD TR 60-517 6.16

6.2. 1.4 Square Plate -- One Edge Free

Figures 6.2. 1.3-1 through -3 demonstrate the deflections andbending moments per unit of length vs. position in non-dimensional form.This shows the effect that freeing an edge has on the deflections and moments(Reference 6-3). The quantity Ma in the denominators of the moment expressionsis defined by 2

4DAT ( 1 -V ) (4)

6.2.2 Bending of Circular Plates with Linear Temperature Gradient Through

the Thickness

Formulas and curves for the deflection and stresses of circular plates(see Figure 6.2.2-1) subjected to a linear temperature distribution through thethickness are now presented.

Upper Face

z

FIGURE 6.2.2. -1 PLATE GEOMETRY

6.2.2. 1 Unrestrained Solid Circular Plates

A free circular plate subjected to a linear temperature gradient throughthe thickness is unstressed. The plate becomes curved and fits a sphere of radiusdirectly proportional to the difference in surface temperatures and inversely pro-portional to the thickness.

6.2.2.2 Clamped Solid Circular Plates

EaATThe maximum stress oe 2(1-? ) occurs at the faces where

AT is the temperature difference between the upper and lower surfaces. The transversedeflection is identically zero.

6.2.2.3 Simply Supported Solid Circular Plates

The plate is unstressed. The deflection w is given by

a AT 2 2

* where a is the radius of the plate.

WADD Th 60-517 6. 17

6.2.3 Circular Plates - Temperature Difference as a Function of the RadialCoordinate

The differential equation for this radially symmetric case is

V4w .(I+)c 2 (1V h TD(

where TD is the temperature difference between the upper and lower faces. Thevariation of temperature is assumed to be linear through the thickness, however,the variation with r is arbitrary.

It is assumed that the temperature T is expressible by a convergent powerseries expansion

z a) KT _ -h Z aK r + C (aK9 C are constants)K=O

so that the temperature difference TD is given byK

T D E- aK r (2)K-0

(Note: Positive z is downward)

The solution of Eq. (1), with the temperature differenca representationas shown in Eq. (2)• is

2 r1+v)a CD aK rK+2

1 a 2 2 h (K+2)2K=-G

where C , CC are arbitrary constants so far and must be determined by theboundarQ conditions.

WADD TR 60-517 6.18

6.2.3.1 Clamped Plates

For power series temperature distribution, the solution of Eq. (3) of Paragraph

6.2.3 for a clamped plate of radius a Is

2 a [2r+ 2 + 2aK+2 r 2 aK(K+2) (1)K-R0 (K+2)2

The bending moments in the radial and transverse directions are

M E 2 aK rK1Eh2 ! e (K+l+)r V) TD 2)r 12(1-/) K=0 K+2 a-) (2)

Eh 2 (Y O a~a 1e I [ rKI ( K

M = 12{[1-(K+1) ( +-(1+V)J -T D].(3)

For monomial distribution, the deflections and moments in non-dimensional formwhen TD is expressed by the monomial TD= aK rK are

D K1

wh 1 Ir K+2 K r 2 K

2 IT -a+a (4)(1+v,)a2 aTD(a) 'K+2)2 L --

M rrK (1 +ri

r rK_(

Eh2 K+2 [K + (5)

at To(a)

Me 1 'K+Il r K 1 l__h2 __ = K+ I - -) K (1- ,) (6)•i

12 DT(a)

where TD(a) is the temperature difference at r = a. Curves of non-dimensional deflections

and moments are presented in Figures 6.2.3.1-1 through -3 for K= 0, 1, 2, ... 5. Super-position may then be used for TD given by polynomials in r. The determination of the poly-

nomial describing the radial variation of TD can be obtained in the same manner as shown in

Paragraph 4.1.2.3.1.

S6.

-k WADD TR'l 60-517 6.19 I

6.2.3.1 (Cont'd)

3.2 .

2.8

24

2.0

Wh(K +2) 2

a OTD(a)

1.6

'1.2__ _ _

.4\

0 .2 .4 .6 .8 1.0

FIGURE 6.2.3.1-1 NON-DIMENSIONAL DEFLECTION

WADD TR 60-517 6.20

1.2

1.0

.8 •Mr

44

6E2.3.1 (Co) d I

ii

1.2

.0

Mrr

0 22A. . .

4/ a

FIG;tilt E(;.2.3.1I-2 NON- DIM ENSIONA 1. IRADIA 1. MOM ENT

WADD TR 60-517 6.21-iý ýrto

6.2.3.1 (Cont'd)

S.. ... . K =O1.4

34

ID

Me .2. .88 .

1.

.2

0 .2 .4 .6 .8 1.0rio

FIGURE 6.2.3. 1-3 NON-DIMENSIONAL TANGENTIAL MOMENT

WADD TR 60-517 6.22

6.2.3.1 (Cont'd)

The following illustrative problem demonstrates the solution for a clamped aluminumplate for which a = 10 inches h = 0.20 inch E = 107 psi

S= 0.30 C1 12 x 10-6 in./An./OF

and where the radial variation of temperature difference through the thickness is 7iven by

2TD = 200 + r /2.

Solution of the clamped plate problem involves the determination of the maximum deflec-tions and stress. Thus, if the coefficients of the polynomial expressing the temperature differ-ence distribution alre all of the same sign, as above, then the maximum magnitude of deflectionoccurs at the center of the plate while maximum magnitudes of radial and tangential momentsoccur at the boundary. Thus, considering the first term of the polynomial,

Tr (r) = Tb (a) =200,

From Figure 6.2.3.1-1

w, a0

and from Figures 6.2.3.1-2 and 6.2.3.1-3

r = ef 1.43

Eh2 Eh 2F25- cr rb (a) rT-Y a rb (a)

a a

Eh2 10 x4x10 2 x12x10-6 x2xl02T-- a r' (a) = 12ff 8012 arD ()12 s

Therefore,

M1J = M•] - 1.43 (80)= 114.4 in.lb./in.

a a

For the second term of the polynomial,2 . 21_

r " (a) (10Y 50TB (r)=- ; 2 D 2

K= 2

From Figure 6.2.3.1-1

w•h (K+2) 2 1

a2 a) = 1.30r (0

WADD TR 60-5176 6.23

I6.2.3.1 (Ccnt'd)

21a2 sr(a) 10 2 x 12x 1G x5x 5x 1012 x 10-1 x 1C A1875

h(Kf-2)

Therefore,

w= .01875 (1.30) . 0 24in.

From Figures 6.2.3.1-2 and -3

Mil ]

r .71Eh2

-- ar,a

7 1Mh e = 1.21

Eh-- aT (a) 112 D r

a

Eh 2 io 7 x4x10- 2 x 12x10-6 x5x101 26

2-- alr• (a) = 12

Therefore,

M= 20(.71) = 14.2 in.lb./in. ,

M" = 20 (1.21) = 24.2 in.lb./in.

By superposition

W i r 0 = W0 U,24 = .;24 in.

W]r W W

Mr = M'+M"jj = 114.4 * 14.2 = 12b.6 in.lb./in.

MV, M,, M'6 114.4 + 24.2 =138.'J in. Ib./ifl.

r r

---r = 1 = 1 ..

WADD TR 60-517 6.24

6.2.3.1 (Cont'd)

The maximum stress is given by

eer h2 r (0.20)-ar 1 a2i

Note: If the polynomial representing the radial variation of the temperature difference consistsof mixed signs, the points of maximum moments and deflection occur at the boundary or in theinterior where the radial derivatives of the quantities are zero.

K6.2.3.2 Simply Supported or Free Plates, TD = aKr

Once the clamped plate problem due to TD = aK r Khas been solved then the

simply supported or free plates may be solved by using the principle of superposition. Itis only necessary to add the solution of the mechanical problem of a circular plate subjectto a uniform radial bending moment equal and opposite to

(to relieve the moments on the boundary) determined from Eq. (5) of Paragraph 6.2.3.1.A straight forward calculation yields the result that for a simply supported or free plateadd the following non-dimensional expression to the solution for a clamped plate:

2 Mr Me E 2 (1)

Eh cTD(a)/12 Eh - aTD (a)/12 (K+2) (v-1)

r 1~

•wh ____2 ___K a+2 (2)

a a2TD(a) K+2

WADD TR 60-517 6.25

l l •• . .,-.- . • **--.-->•--

C. 2.4 Approximate Solution of Free Plate with Arbitrary Temperature Varia-tion Through the Thickness Only

The problem of the determination of the stresses, strains, and displace-ments in an unrestrained plate of arbitrary planform (Figure 6.2.4-1) which issubjected to an arbitrary temperature variation in the thickness direction has beensolved approximately (Reference 6-1). The procedure used is to solve the problemas a three dimensional elasticity problem using the semi-inverse method of St. Venant.This simply means that the form of the stress distribution is assumed and then theequations of equilibrium, compatibility and boundary conditions are examined in thelight of this assumption.

&A

FIGURE 6.2.4-1 PLATE WITH ARBITRARY PLANFORM AND CONSTANT THICKNESS

The temperature of the plate varies in the thickness direction only, thatis, T r(z). With this assumption, it is further assumed that

(z =a =a =, =0 (1)zz zx zy xy()

S= U = f(z).xx yy

It will now be shown that the function f(z) may be determined so that theequations of equilibrium and compatibility are satisfied identically. The boundaryconditions of free traction (no restraints) are satisfied on the average.

Substitution of Eqs. (1) into Eqs. (1) of Paragraph 1. 2. 1 (no body forces)shows that the equilibrium equations are satisfied identically. Furthermore, substitu-tion into the compatibility equations (Section 1) reveals that three of the compatibilityequations are satisfied identically, while the other three equations will be satisfiedonly if

WADD TR 60-517 6.26

- zn...Ž ___

6.2.4 (Cont'd)

d2 =o (2)dz 2 1-

which has the solution

fC 1 = C 2 + C = a a (3)f= 1 1-V xx yy

where the constants C1 and C, are to be determined from the boundary conditionsof zero traction. It is not polsible to satisfy the boundary conditions pointwise.However, the constants C and C may be determined so that the resultant forceand momept per unit of lelgth pr&duced by axx and a are zero over the edge of theplate.

f axx dz = dz = 0

h/2 h/2 (4)

~zdz; f ah2c zdz= 0f-h/2 aJ-h/2 Yyyd=0,

Substitution of Eq. (3) into Eqs. (4) yieldsEh/2 fh/2a a. = -7 T rdz v-3zUxx yy h -f h zh/2 h"

so that finally

1 -aET +2.1 + 12z'c yy t h3 MT (5)

* where NT and MT are defined in Eq. (7) of Paragraph 6. 1. 4. The parallel to the

beam equation is apparent.

pAccording to St. Venant's principle. the solution (5) is a very good ap-

proximation for traction-free edges at distances from these edges larger than twoor three plate thicknesses.

Once the stresses have been determined, the strains are found from thestress-strain relations (Hooke's Law). The displacements are then determined fromthe strain-displacement relations. The complete solution then appears as follows:

Stress Components: I[ aEr + NT 127 m

az z = zx =zy j xy

WADD TR 60-517 6.27

S6.2.4 (Cont d)

Strain Components: yx = h _ 1 + 12z MT

-2?' [ NT 12z +1_z = M 1l- aT (7)

C ~=e* = C,4Exy = yz 'zx 5

Displacement Components: (Exclusive of Rigid Body Motion)

1 rNT 12zu=- ++- MT

E L~hh

v = 12z • T Y (8)6 (x2 2) 1 "{ (Z 2 .Z

...MT +y) f(I-)E Tdz - 2 NT T

--3 TT (l-)

2

WADD TR 60-518 6.28

UI

6.3 SLAB PROBLEMS - PLATES

Slab problems occur when the plate is loaded by mechanical forces which areparallel to the middle plane of the plate and uniform through the thickness, and thetemperature varies in the planform direction only. This is a plane stress problem,i.e., zx•rzy =ar = 0. The effects of the mechanical loads and temperature

may be superposed.

The differential equation for the slab problem is now derived. In the planestress problem only one equation of compatibility is satisfied, namely

2 2 22

XX + yy= xyy2 x2 ax ay (1)aay

(See Reference 6-4)

If the stress-strain law expressions

xx E (xx - P ayy) + otT(x,y)1

a= -( -a ) + T(x,y) (2)E yy xx

2(1 +•)"Xy E 0xy

Nzz =x =Zy = 0)

are substituted into Eq. (1) and the stress equilibrium equations

3a+ - + 0

(3)

+ =0.

there results the equation

V2 (axx + ayy) + EV 2 T =0 (4)

where 2 2

V 2 + - in rectangular coordinates.

WADD TR 60-517 6.29

l I

6.3 (Cont'd)

This is the compatibility equation In terms of stresses and temperature only.If a "stress function" 0 (x,y) is introduced such that the components of stressare derivable from* by

2 a2' 2x 9iXX 2' yy ~'~(5)

then the equilibrium Equations (3) are automatically satisfied, and the com-patibility equation on # becomes

V4-0 -EaV 2 T, (6) ý

whereA2 2 2 24 + av 82 9Y 2 in rectangular coordinates.2 2

Since the effects of the mechanical loads and temperature loads may besuperposed, It is sufficient in many areas to consider only the thermal problemwith the boundaries unrestrained so that the components of stress normal andtangential to the boundary are zero at all points of the boundary (condition ofzero traction).

The solution of Eq. (6) for circular plates or rings subject to generalasymmetrical temperature distributions has been solved in References 6-5 through6-8. The results of these investigations are shown in Paragraphs 6.3. 1 and 6.3.2.

The particular case of radial symmetry is treated in References 6-9through 6-11, and the results of these investigations are given in Paragraph 6.3.3.

tn Theoretical and experimental work has been done for the plane stress solu-tion o! rectangular plates subject to planform variation of temperature, for example,References 6-12 and 6-13. However, detailed parametr' la studies have not beenmade. The above problem indicates an area which should be explored. No furtherdiscussion will be given to the slab problem of rectangular plates in this Manual.

WADD TR 60-517 6.30

t~

/I

6.3. 1 Thermal Stresses In Rings - Asymmetrical Temperature Distribution

The plane stress linear elastic solution of a traction-free circular ringof inner radius a and outer radius b, subjected to a general two-dimensionaltemperature distribution, was considered in References 6-7 and 6-8. Theoryand formulae were developed for the determination of the polar coordinate stresscomponents corresponding to temperatures of the form

K /r :T=T0

1 J cosneorT=T 0 sin no,0r- b•. To.•r -o ( ,,....where K and n are restricted to non-negative Integers. The purpose of thissection is to give a tabular set of quantities for the direct calculations o( thestress components corresponding to values of K and n from 0 through 3 fora - 0. 2, 0.5, 0.8, respectively. This range for the parameters K and n should

adequate for most practical application. A numerical exumple is given for aparticular temperature distribution to demonstrate the application of Table6.3. 1-1.

The stress components corresponding to a temperature distribution of theform

KTb -O(E Cos no A 1.T°

may be expressed in the form

E-Orr BK cos n 0

SE--EaT° =C aK' m°nO'i

-• ,,,le orr,•o• = Do cosO

S~the stress components are

SE-•'To BK sinnO

EaT K,n

-re =aT-y -DCK cos ne

SWADD TRt60-517 6.31

. ~ E '- .4''I',-

6.3.1 (Cont'd)

S'Ooe = DK, sinnO. (4)Er ,Tnndmybe'n

The quantitiesB CK, and D are functions of K.n, m,K. n - K,n' K,n b b admyb

found from Table 6.3. 1-1. Use of this table is demonstrated by the following example.4-

2 22T=T (E) (1-cosO)+TI=T,(• -T( • cosO+T 1

b oa• 0.5,

thenSrr

.j B 0 -B 2 1 Cos 0

0

E dro 2,0 2,1E'G° = D, D C o,1snt 05

EaT - D2cos0

The numerically greatest stress component, for example, is given by

-. 375--.1733 0.55IEaTj IEaTII-.*-1301max =0MT I rr I

"Thiis corresponds to a compressive stress in the transverse direction at the outer boundary

WADD TR 60-517 6.32

6. 3.1 (Coat' d)TABLE 6.3. 1-1

THERMAL STRESSES IN RINGS

0.3 0 0. 0 0 0 0.2 0 3.6 -. 23w -.Us0 -.516 0.5038 ..3 -am -. 038" -am001 0 .2 0& 0 .6i -m -...9% :am87 ., -A"?06 -. W .A99

.3 .180 0 .3im. -am -W&* -.03m 1.0 0 0 ."a1A6 Jul1 0 .3672 .9-A6306 -.130 23am.5 .333 0 Am95 LO 0 0 .59 13. 0 A 33.6 "23 0 -0070 .6 -AM37 A:"1 _*an&6

.1 A" 0 .. 099s. 1 3.2 .0 0 .m$53-A.8 Q 0 -.1725 .3 -.0708 Z13 A062 -01 -AM:

j .03"t2 0 A.61 .6 .m5 .066 -. 3 .v -. 10" -AIM92 401361L0 0 0 -. 71 . .1m* .01 im.23 1A0 0 0 "W3

* 0 j 0 0 .632 7 .. 36 -A"9 Auo -. 6 83.5 0 0 .1lop.3 4m o .3036 A3 Q77f .05% -.3616 .6 .815 .18 -.000" :40A6 am51 .2w5. -02 -.071=1 .0 .7 -. 6"m3 .0025 -. 0336.5 am51 0 .122 LU 0 0 A~. -. mm1 jam03 .. 03W0.7 .211 0 -.0m102 *3.2 0 0 129LO 0 0 Alm2.s .0"62 0 -.366 .3 -1&$5 Am60 01.0 .0616 0 .. 3m A -.8399 J26UT -.006 3 3.5410 0 0

1.0 a 0 -Am3 .5 -. 001 ="67 -. 03W9.6 .01 -.05606 0.8 0 &4 0 0 0

3 0.2 0 0 JAW0 .? ow 2~:~, 0 .10.1UD 0 am6 .9 -.031P is 4, A.9

.5 1pt 0 .11 LU0 0 0 AM30 .95 .00336 -.0392

am.12 0 3.02 33J-21 0 0 0AP295m 0 ...13 20. 0 0t o 1

.8 .0* -.36k? AS0.. 0 0. 71 0 .66LU. 0 003 *104 0 0 27 .. 092.6 .065 A 023 .03m I 0 -.2066

* ~~.3 -.20 X.677. .039A -AM70A6 -.16N -. 016 Am.38 -.240 3 0. 0 0 .233.5 ..236 0 2.27 .0 0 0 .20.0 .2 0 .1ftA6 -. M5 0 .o3 .9 am.3 0 .00650

.7 AM052 0 AM36 20 a 0 -IS5 Owm071 0 -. =3

.$ -A" 400 A ~. .4. A 3 LU a 0 ..jmLA 6.3 0 .318 . .09 0W -.097m0.8 0 0 ..

24 0 0.9 Am30 0 Oo 1. .2 a.0 0. -.0"8I11.-14 0 0 0 LA 0 0 -.375 .9 -.00m0 -. 0909-.001*12.2 0 .2 95 -.3623 -.0063 .0511

.3 .026 .030 n 3 0.5 0 0 .161.0 0 0 am00A6 Am636 "W3 .6 .05220 0 .17.5 Am"3 .0" J7 A" .039 11.8-1.6 .1636 NAW9 074 .6 As"9 -am22 1. 0 A.7 .167. AMA16 Alm1 p AAR 0 -im i .009600AR 7A8 .0AD .08 -.033 1.A 0 0 -.262 .9.0886.U06JW.9 .01708 a11 -.101 is5 .03W6 AD%" -.0609Z

LA 0 0 -.199 01.5 0 a -.37 3.0 a 0 -. 36.6 -.0637 -Awn21 -.395;

31.2 0 0 .235 .7 -A"97 -Am92 -. @=6 3 1.A 0 0 .13013 .0336 .031 X"32 8 -A"00 -A .. 632 Awn 92 .06ArmA6 .05 .0"0 .110 .9 -AM8 -am03 .1 A~05 A20066 On*3.5 "a00 Most0 .6 LU0 0 a is AR" 00" . -vA' .06m7 Awl6 .1m6 LA a 0 -j.3M.7 .0583 .033 .. 04% 11.54 0 0 0.8 ago77 .181 -am9 2. 0 0 .13 2. A a a Am5

LU 0" 0o Z11253 .7 Arm3 AM72 .0so .9 -.28283 -Acomm -A286"A .8 363 .0"3 -.805 .95 -. 000"6 -.3260 -*0657

0 2.8 0 0 16?.9 am66 *666 -. 3 L 0 0 mm5.3 A*"31 AM1 .0333 1Lo 0 0 -an)3A6 .0920 -Am -- m 1. 0 0 iI5 _am3 -.0127 -.3205 3 1 .5 0 36.0 Awn600 Amn1 -.3278*

.6-.23 -06001 -am99 .6 .0w8 .039) 4736x .9 -.00673 -. WD" -. 00".8 -.06as "*373 amp2 A8 A0" .0"6 A303 IA 0 0 .9103?.t -Ar36 -.05m0 .2m3 .9 A"03 ~006 -.22

LU 0 Am6 LU 0 0 -ima 884 0 0382.60 _J"52

.3 A.01 .035 .03* .A .0Ls .006 -- As2 .9 AVWW .000973 AMU62A6 -.03032 .0360 -- ow .7 -.285336 4013P -.032 915 .000331 *0000 Oulu16.5 -036 -.00315 -.0L . -.*=1 -am23 -8.0 L 0 0 -. 0"a3.6 -A* -.006 -.0376 .9 -.0o"9 -.02676 7=68.7 -A"91 -A"21 -A"80 Le s 0 .4w3 0 3.8 0 0 *L%90A8 -am03 :.039" -.0S3 .0 03651 .322 -.0032.v -.G0SA OKW0 03 . .220 v9 -. 20033 .*0305 m*8a

LO 0 0 .1m0 A6 *33 .623 .0 .9% -.32953 -.A6.207 -AD321.7 -.2803 .7706 -.0368 LU0 0 0 .03313

2 2 .11-1 0 0A . -. 0032 .001025 0W3132. 0 .62 .9 -.0311 -.0066 .40068 1 3. 0a 02

.3 -.00832 -.0623 -A"63 LA 0 0 .030 J.; .032308 Aab38-.3lA6 .013 -.00M3 Awn68 .9 -.28O23 - OM6 :.021.'.5 AM -DOAN8 -am*2 12 .5-1 0q 05 -*0236 -.0mm7 -. 06

A"" Jun2 .1 .0506 32. 0 0 .5 2A aU 0 -arm6-Cm AM am,0 A43 08 -ADA"6 -.0638 -.000911

.8AI .091 U 286 .7 Am95 -.0018 .03165 23.8 0 0 .1*0 =T" Al.-02.8GVS .0" OM S 6? ALM,3 -. WM396

1. 0 1 .76 9 -.0018229 OW -.0289 *mm

.5 NA62 -.038W -. ml .7 -Amw O0010 -am360

WADD TR 60-517 6.33

6.3.2 Thermal Stresses in Solid Circular Plates Due to AsymmetricalTemperature Distribution

Formulas and curves are presented for the stresses and displace-ments in an unrestrained solid plate (plane stress solution) corresponding to thetemperatureKr

T-T° fl cos n 0(K =0, 1, 2.,. n =0, 1, 2....

Superposition may then be applied if there are several terms in the temperatureexpression.

The formulas for the stress components are:

Case L K 1 n-2

K - n-2 n .0 rr = (ap +b p + cp E dr cos no

"- (ApX + Bp + Cp ) EaT 0 c(sne (1)

K n-2 n) a inare = (pK + 0p + y) EaTo sin no

Case i: K = n-2

r (apn 2 +pb,Pn 2 1np +0cn) EaT coon 0'rrn-p2 - I + CpP) EaT% cos n (%9=(Alp- B B'pn-In p12

= 6•pn-2 + ,,p•-,-n p + ip") EaT 0 sin ,n

WADD TR 60-517 6.34

6.3.2 (Cont'd)

"*. where

P r -(K+1)(K+2)b' A= (K+2)2 2(K2-n

22'-n (K+1) -n(n-1)(n-K)_ B -)(K+2) -n 2 [(K+2)_n2

C -(n+)(n-2) (n+l)(n+2) (n+l)(n)2(K+2+n) 2(K+2-,'n) 2 2(K+2 +n)

a' (n+l)(n-2) A' =n2 3n +2 6, = (n+1)

4n 4n '4 •

2 _b' n-1 -B' a -!C-22 (K+2) 2 n2 2

The stresses arr' aee' qre are piotted in Figures 6.3.2-1 through -3 for n = 1. 2. 3with K as a parameter. The axisymmetric case, n = 0 is shown in Figure 6.3.2-4. Fromsymmetry, the shear component are equals 0.

The following should be noted:

(1) When K = n, the temperature is a harmonic function (i. e.. V2 T = 0) sincerncosnoistherealpartofzn (z = x iy). Thus the corresponding stress com-ponents

%r = re = •ee = or ro ... ,(2) When T = To sin no =( .. 2. then in the formulas for the

stress components, Eqs. (1) and (2). cos no and sin nO must be replaced bysin'ne and -cos nO. respectively.

In the following example in which a circular bulkhead of radius b is subjected toa temperature distribution

2T = T -b (1-cosO) + T

2 2To (To cost+ T1,

where To, TI are constants, the expressions are desired for the stress components.

WADD TR 60-517 6.35

6.3.2 (Cont'd)

II21I - -

(a)

.2G 4 IN )

4 -- -- • -

0 .4 -,i IJO

-.4

(C)

FIGURE 6.3.2-1 RADIAL STRESS DISTRIBUTION IN A DISK WHEN A TEMPER-ATURE DISTRIBUTION T= To(r/bK cn 0 (n =0, 1, 2,3) ISMAINTAINED

WADD TR 60-517 6.36

6.3.2 (Cont'd)

4

04-

EaTacoaO -

00

I L.

. .A4 -- 6.37

-2 -

44

WADD ~ r. 2Rto5763Ia~o3

6.3.2 (Cont'd)

EaTisiO 0 ww-EaTs~u ~ .2 4 6 S 1

.• --K:0

(a)

EaTosim :9

S'a _I

FIGURE 6.3.2-3 SHEAR-STRESS DISTRIBUTION IN DISK WHEN TEMPERATUREDISTRIBUTION T = To (r/b)K cos n e (n= 1,2,3) IS MAINTAINED

WADD rR 60-517 6.38

0

OD

*10

0

E.3

0:

gz H

C4

CiO

COOD

Vlh

!_ Zv -

6.3.2 (Cont'd) i

2(1) Stress components corresponding to T (K 2, n 0);

Eq. (1) yield the coefficients

-4 1 _3 6 0(4)2 4

b=0 -b 6 0,=0

C 2(4) 1 c = '1 2) 1 y C4 2(4) 4'

which in turn yield the desired stress components

"0 12 1

E 0rr = (_i. + )(ctTo +

0

0or

= 0EaT

2(2) Stress components corresponding to T o(ib cos 0: (K 2, n -1);

Eqs. (1) yield the coefficients

-3 1 -12 4 -3 1a =-• =-•, A=--=- , 6 =--

6=0, B-0, 0 =(

cC= -. = 6 32(5)5 2(5) 5 5-,

which in turn yield the desired stress components

Err - g !p + I ) cos

EaT0 5) 5

EaT ()

WADD TR 60-517 6.40

6.3.2 (Cont' d)

're I 2 1 s

FaT 0 5 5,0

(3) Stress components corresponding to T are: a = r = 0.rr rOE)oe

(4) Total Stress components are:

Err 1 2 + C(p-i) cos eEctT = 4 ( p (P5

a

KCr,.i :iooo ,

E--T P (p- 1) sin e .

EaT 5 (p-5 ) 6

WADD TR 60-517 6.41

6.3.3 Circular Disk with Concentric Hole Subject to a Power Law remperatureDistribution

The thermal stresses in thin circular disks of constant thickness with a

concentric hole subject to a power law distribution in the radial direction aredetermined. The following three cases will be considered:

I The disk with a central holeII The solid diskITM The disk mounted on a shaft

Formulas and curves will be given for the radial displacement and stress componentsassuming a condition of plane stress (Paragraph 2.1.7). The following symbols are

used in this paragraph:

a Inner radius of plateb Outer radius of platen Exponent in temperature distributionr Radial coordinateu Radial displacement of points on diskE, Y', a Modulus of elasticity, Poisson' s ratio, and coefficient of linear

expansion, respectivelyT TemperatuieT* Mean temperature when referred to the square of the radius =

2 _ r2 a2-1f rTrdr = (r 2 _ 2)'Ja Td(r 2 )

o Trr Radial component of stress0ee Tangential component of siress

"Suffix a Refers to the inner radius of the diskSuffix b Refers to the outer radius of the disk.

A circular disk of radius b and constant thickness and with inner concentrichole of radius a is subject to the radial temperature variation

nAT = Tb ( , (1)

where Tb is the temperature at r = b.

The variation of the temperature distribution versus r with n as a parameterb-a ý

is shown in Figure 6.3.3-1.

The expressions for the radial displacement, and the radial and transverse com-ponents of stress are:

alP)r2 a2 TU Ar + . + +Z)(r2-a)T]2(2)

WADD TR 60-517 6.42

6.3.3 (Cont'd)

E [A(I + V + B(V -1) 2 (1 ~ 2 a 2)T.] (3)rr 1 -V 2 r2 2r2

2 2Oo - E 2 [A(l+)BV - (V -2 2 )aT+(l- 2r 2 (4

where A and B are constants which must be determined from the boundary conditions

and I9r 2)_1 far"T* = 4 - a Trdr . (5)

Case I - Unrestrained Disk With Concentric Circular Hole

The boundary conditions arr = 0 at r = a, r = b are imposed.

The expression for the displacement u in non-dimensional form is

2ru (1-P)r r_* (6)

aa a a

while the expressions for a and a00 are:rr eorrr a a2 T -T*rr (i )( b )()aETb r 2 Tb(7

1 (I+ -2T + (1 a. (8)aETb 2b L1b r2)

aFigure 6.3.3-2 shows the radial variation of - for a disk with a hole

a b.O,0.4, 0.8) correspondng to n - 1/3, 1, 3 (-L 0 means a very small

hole, niot a solid disk).

Figure 6.3.3-3 shows the maximum radial components of stress versus n.

From Eqs. (I) and (8).Oaooea Tb* 2 t" {+l) b+a

a=- (8))( 42 a+aETb Tb n+l)'(D ÷2) a+b (

2 b -=2 ( )n+2)(s) 2 (10)ETb TbL al1+)ab 2j

WADD TR 60-517 6.43

I 4 l i l--- -4 i,,

6.3.3 (Cont'd)

'Te beCurves showing the variation of and b with a/b and n are given

ETb £XTb wit1a2in Figures 6.3.3-4 and 6.3.3-5.

Case II - Solid Disk

The displacement and the stress components for a solid disk (a = 0) aredetermined by imposing the conditions UJr=0 = 0 and arrj r-- 0.

The radial displacement is

2 (1-) 0 +V)

In particular, the radial displacement on the boundary is

U]r~b = aTb b .(12)

The equations for a rr and a in non-dimensional form are

a rr 1 TbT*(

a~Eb 2* T * - 2

aETb 2 - Tb Tb Tb (14)

When T varies according to the power law (Eq. (1)) with n positive, a fallsfrom its maximum (tensile) value at tOe center of the disk to zero at &re periphery.At the center, a rr = aO and a 0 falls montonically with increasing radius (in an

algebraic sense) having its maximum compressive stress at the boundary r = b.From Eqs. (13) and (14),

rra a 1 b Aa e 1 (15)

aETb otETb 2 -b rb n+2

and

)eeb= (1t,)

aETb n+2

Figure 6.3.3-6 shows the variation of &'b aE and -b withvaiaio Erb' oETb' an ETb wt

C, n for a solid disk.

WADD TR 60-517 o.44

6.3.3 (Cont'd)

aEll aET6 0.15 r0I b-o/

0 _ V _ -0.5 ,ojbb/00i00 0 6/6-.08

04 06 s0 1.00210 0/b

02!

0 20 04 06 04 10 0 02 04 06 08 10r-_o 0/b

b-aTEMPERATURE PROFILES FOR n I/3,1/2.I.203 RADIAL VARIATION OF RADIAL STRESS FOR A DISK

FIGIURE 6 33-I WITH A HOLE (o/b i:0O,04.0S~fh: 1/3, 1 G )FIGURE 6.3 3-2

ti

090

0. C ___o,

0 I G0. .

04 *3

h n

lMItTIOM OF MAX. RAOIAL TRESS WITH n a VARIATION OF TANGENTIAL STIESS AT INK"R OUMNART1

0/b FOR A DISK WITH A HOLE IrH nm a a/b FOR A DISK WITH A HOLEFIGURE 6.533- FIGURE 3•.3-4

WADD TR 60-517 6.45

I

6.63.3 (Cont'd)

0•0

.0, o, 0O• -05B

0.G , Ox

0.•T _GE 6*o_._ o, 0 z-7

4 4a ET a .03

0., I 3

VARIATION OF TANOEIrAL STRESS AT OUTER VARIATION OF RA0IAL B TANgENTIAL sYrESS

oUSNARY wilT na a o/b FOR DISK WIT AHA"OLE AT CENTER D ISou KART WiTH nFORA 5O1ID

FIGURE -3- 3 FIGURE 4 3.-4

- zobl

ON -0 8

03 O0

0. 2

I'

•UmATIWOPl 1 RADIAL • TANROENTIL ST21ESS AT WSIA1'O• OF TAIuUIA •TRESS ATiWYna 3O1R O DISK wiTH n

06 b

FrOR A DISK OR A DIAFTISK ON A SHAFT-

aIUR ET 35

FI.0R E . 3 3-7

SWADD TR 60 5 1 ? 6.46 -,Il

6.3.3 (Cont'.d)

Case MI - Disk With Central Shaft

This case is analyzed as a plane-stress problem, i.e., no account is takenof the stress concentration at the cylindrical interface between the shaft and disk.It is assumed that at the Interface no radial movement occurs, i.e., U3___ = 0.

Mwe other boundary condition is that arr W 0, i.e., the outer boundary is unre-stained.2 r•

The following expressions result for crr and '00:

a2a + + ab 2

-- 1 I -,, (17)Tb Tb b2'• 2

I+3 201~- 2 2

00OO 1 b a[ r2 + a2 2T 1

aETb -2 [ b b i ti 2 Tb (12) F-b (8

In particular, the expressions for o , ee ad Are:

rrar'0a

nETT*(I_ a (19)

b b2

2

'Gee TLb 2 Ia E aTVi(0

Tb 2 -b-) 2 P ,) P -V)bb 2+

Curve* showing the variation of err., "eeo' a d aeo, with n and a/ areshown in Figures 6.3.3-7 and 6.3.3-8.

WADD TR 60-517 6.47

; ,y,

6.3.4 Circular Plate - Central Hot Wot

The stresses in a circular plate due to a concentric hot spot are determined (Refer-ences 6-10, 6-11). The central hot spot may be another material at constsmt temperature ora hot fluid paesing through the plate. The method of solution is to solve the steady-state heatconduction equation subject to prescribed boundary temperatures. Once this temperaturedistribution Is determined, the equation of equilibrium in the radial direction and the enforce-ment of aingle-valuedness of displacements yields the solution.

The following symbols are used in this paragraph:

a Radius of hot spot (Figure 6.3.4-1)b Outer radius of plater Radial polar coordinatet Thicimess of plate/thicimess of hot spotu Radial displacementT Excess of temperature of hot spot over that relative to the outer edge

of plate0rr Radial component of stress(76 Transverse or tangential component of stress

The following assumptions are made:

(1) The hot spot is at constant temperature T relative to the outer edge of the plate(2) The plate is assumed to be insulated so that all heat enters at radius a and

leaves at radius b which is at temperature 0.(3) Plate is assumed to be thin so that plane stress analysis (Section 2) may be used.

The outer boundary Is assumed to be unrestrained.(4) Ea is constant.

"The following design equations pertain to this problem:

(1) Temperature Distribution in Plate:

Temperature - T Log (b/r) (Log (b/a) •

where a -r =b.

(2) Radial Displacement at Junction Line r = a:

u] = Ta + B-)aE.)T a)ru Era + B- 2'

where

B -EaT (I -(1-3/) (1 - t) log tb/a)] (2a)2log b/a b2/,2 +1

A =~2

WADD TR 60-517 6.48

6.3.4 (Cont'd)

TEMPERATURE 0

FIGURE 6.3.4-1 PLATE AND HOT SPOT

SWADD Th 60-517 6.49 •

-N%

6.3.4 (Cont'd)

u 31A graph of a ' versus---, with t as a parameter, is shown inaa?' baT a

Figures 6.3.4-2a and 6.3.4-2b.

The components of stress in the radial and transverse directions are

'rr 2 Log ba [(I1 (l+V)+( -~tb 2/laz + (I -P) (} (t)

andEaT ]" I -Y)(1-}( - t) LJog i)/&)] t(hZ/r2 + 1} )r}¢

e0 = 2 Log b/a (( +y) + (I -v)t]bZ/a2 + (I -W) (I - t) - ]-o

•'he variation of a and - a with r/a is shown in Flgre 6.3.4-3 forrr rr GoV= 0.3, b/a = 2, 3, aD andfor t=0, 0.2, 0.5, 1, 2, 5 andwD.Values ofa (, may

be found by combining the given curves.

Figure 6.3.4-4 shows the variation of a - a at r a and r=b with b/a,wltht as a parameter.

"The following special cases are considered:

If the extent of the plate is large compared with that of the hot spot (b/a large)then

-EaT + (-Y) (0-t) a2

=1) + (1 -,)t] r1

and

arr ('e E.T r q -) 0 t) a)

Note that the majority of the plate is subjected to bi-axial compression of magnitudeE&T The stress difference has a maximum value obtained by substituting r=a into

Eq. (7) which varies with t as shown in Figure 6.3.4-5.

If b/a is near unity, the plate becomes a thin ring, resulting in

ar -*(, and

rrrArr 00E.0 "

WADD TR 60-517 65

6.3.4 (Cont'd)

[ .U r ~ o _ •( , ...a ) t , ( b. . .) 2 t ( .b -f

1.2 If

1.0 tzO

.8

.6 \ % %%N

%2.64.4 .4

% .4,

.2I100

0*0 ....__ _ ____

0 I 2 3 0Ob/a

FIGURE 6.3.4-2(a) RADIAL DISPLACEMENT AT BOUNDARY OF HOT SPOT (v 0. 3)

WADD TR60-517 6.51

6.3.4 (Cont'd)

[U] ,'zb too

[.4 -- \ f~ [ -1nzI., t]

gg

bla

.6

o/

FIGURE 6.3.4-2(b) RADIAL DISPLACEMENT AT OUTER BOUNDARY OF PLATE (v =0. 3)

.WADD TR 60-517 6.52

.\li

I N1

6.3.4 (Cont'd)

0i

b a-: 2 g.2aA

- U0rrr -O

t~o EOrTA .2. E0CTA0.5

I EI[ • , E ýe4 ( T ,. • o

.2 1.0v ::5 s2

1 1.0

0 1 . 0 1 r 2

1.0 b00

FIGUR EA E6. CC T-55 5

S6.53

0 52

.2.5

1.0

20

WD Tt 60-517 .5

• ,• . , , ..~~ ... 2- • •. -: •, ..:,.,•:... .:.: .... ••.• •::•••,o,,,:...,o..-- •.-,,••,:,• • ,•, ;

6.3.4 (Cont'd)

1.0

.8 I

C~OD.6

E aT /.4-

i5

.220 rot,=l

-. 4 't=0-

-. 6--tzo

.5

5.0

FIGURE 6.3.4-4 VARIATION OF STRESS DIFFERENCE WITH b/ (v =0. 3)

WADD TR 60-517 6.54

P'WI

6.3.4 (Cont'd)

1.0

.8

.6 0.3(O'rr- o'ee)&,9 c b LARGE

E T

.4-A

.2 .4-- 618'1.0 2 34510o

Ct

-. 4

-. 6

FIGURE 6.3.4-5 VARIATION OF MAXIMUM STRESS DIFFERENCE WITHt FOR b/a LARGE. MAXIMUM OCCURS AT r a.

WADD TR 60-517 6.55

... . ..... ~f~

6.3.4 ICont' d}

If t is large, the stresses become

2 2EaT [Log (b/a) b2/r2-I) [ L Ir

U rr 2 Log b/a 2a2 Lo1ir)b/a -

and EaT + Log (b/a)(b 2 /r 2 + 1) -

'00 =2 Log b/a 2 Log (b/r) .(9(b2/a -1)

With the usual conversions fron, plane stress to plane strain (Section 2), these are thestresses in a pipe containing fluid at temperature r.

For t-1, the plate is of uniform thickness throughout and the stresses aresimplified to

= Ear [ (2

arr 2 Log b/a 2b 2 r -(

and

EaT a 2)b

'oO=2bogb/a 2b r - I Log(b/r) (01)

A

WADD TR 60-517, 6.56

6.4 REFERENCES

6-1 Boley, B. and Weiner, J., Thermal Stresses for Aircraft Structures,WADC Technical Note 56-102, August 1955.

6-2 Timoshenko, S., Theory of Plates and Shells, First Edition, McGraw-Hill Book Company, Inc., New York, 1940.

6-3 Zaid, M. and Forray M., Bending of Elastically Supported RectangularPlates, Proceedings of the Society of Civil Engineers, Vol. 84, No. EM 3,July 1958, Proc. Paper 1719.

6-4 Timoshenko, S and Goodier, J. N., Theory of Elasticity, Second Edition,McGraw-Hill Book Company, Inc., New York, 1951.

6-5 Horvay, G., Transient Thermal Stresses in Circular Disks and Cylinders,Transactions of the ASME, January 1954, pp 127-135.

6-6 Forray, M., Thermal Stresses in Plates, Journal of the Aero/SpaceSciences, Readers! Forum, Vol. 25, No. 11, pp 716-717, November 1958

6-7 Forray, M., Thermal Stresses in Rings, Journal of the Aero/SpaceSciences, Readers Forum, Vol. 26, No. 5, pp 310-311, May 1959.

6-8 Forray, M., Formulas forthe Determination of Thermal Stresses in Rings, Journalof the Aero/Space Sciences, Readers' Forum, Vol. 27, No. 3, 1960, pp 238-240.

6-9 Norburg, J. F., Thermal Stresses in Disks of Constant Thickness, Air-craft Engineering, pp 132-137, May 1957.

6-10 Parkes, E. W., The Stresses in a Plate Due to a Local Hot Spot, Air-craft Engineering, pp 67-69, March 1957

6-11 Forray, M. and Zaid, M., Thermal Stresses in a Solid Circular Bulkhead,Journal of the Aeronautical Sciences, pp 63-64, January 1958.

6-12 Heldenfels, R. R. and Roberts, W. M., Experimental and Theoretical De-termination of Thermal Stresses in Flat Plates, NACA TN 2769, August 1952.

6-13 Przemieniecki, J. S., Thermal Stresses in Rectangular Plates, TheAeronam.',"Il Quarterly, Volume X, Part 1, pp 65-78, February 1959.

WADD TR 60-517 6.57

~Ii

SECTION 7

THERMO-ELASTIC ANALYSIS OF SHELLS

WADD TR 60-517 70

-- W

OF"I"Mi

SECTION 7

THERMO-ELASTIC ANALYSIS OF SHELLS

TABLE OF CONTENTS

Par ~Title Pg

7 Thermo-Elastic Analysis of Shells 7.2

7.1 Truncated Conical Shells 7.3

7.1.1 Radial Loading Perpendicular to the Cone Axis 7.4

7.1.1.1 The Hoopring Approximation for Radial Loading 7.4

7.1.1.2 Solution for Edge Loading (with illustrative problem) 7.13

7.1.2 Meridional Loading - Membrane Analysis (with illustrative 7.20

problem)

7.1.3 Meridional Temperature Variatkbn 7.26

7.2 Approximate Solution for Non-Conical Axisymmetric Shells 7.28

7.3 Cylindrical Shells 7.30

7.3.1 Cylindrical Shells with Axisymmetric Loading and Meridion. 1 7.30Temperature Variation

7.3.1.1 Hoopring Analysis 7.30

7.3.1.2 Solution for Edge Loading (with illustrative problem) 7.32

7.3.2 Cylindrical Shells with Temperature Gradient Through the 7.37Thickness

7.3.2.1 Cylindrical Tube - Steady Heat Flow in Radial Direction 7.37

7.3.2.2 Transient Thermal Stresses in Circular Cylinders 7.45

7.4 References 7.51

WADD TR 60-517 7,.1

SECTION 7 - THERMO-ELASTIC ANALYSIS OF SHELLS

The deformations and stresses in shells which have axisymmetric geometry, loads,and temperature distributions are presented. The case of thermal loading is shown to beequivalent to a pressure loading which is proportional to the local meridional temperature.

Solutions are also given for the case of edge loads and moments on the shell. Thedeformations at the shell edges are then employed to calculate the forces required to enforcethe edge boundary conditions (compatibility). For the most part, the analysis cosiderstemperature distributions which are constant through the thickness of thin shells, althoughspecial consideration is given to right-circular cylindrical shells having temperature varia-tions through the thickness.


a,b Inner and outer radii, respectivelyh Thickness; Thermal conductivity

Attenuation lengthr Radiusr Reference radiusto Timew Normal deflectionx Distance along meridion

Eh 3

D Flexural rigidity 2)

"E Young' s modulusF Edge loading functionsSkHorizontal force per unit of circumferential length

Value of H at a shell edgeK Hoopring stiffness; Incremental stiffness due to restrained ends; Thermal

diffusivityL Meridional length of a shell sectionM Bending moment per unit of length

7YL Value of M at a shell edgeN Normal force per unit of lengthP Surface traction (force per unit of area)Q Meridional shear force, per unit of circumferential length, perpendicular to

shell wallR Radius of thin-walled cylinderT TemperatureAT Temperature rise above reference levelW Weight

a Linear coefficient of thermal expansion6 Radial (horizontal) deflection

WADD TR 60-517 7.2

2 A Vertical extensionE• Strain77 Value of N at a shell edgeEO Azimuthal angle

SV Poisson' s ratioa •Stress

* 4) Angle between axis of revolution and the normal to the shell surface

Superscripts

Quantities corresponding to a particular solution

o Quantities corresponding to the meridional loading membrane solution

Subscripts

e Resulting from edge loadingsh Horizontal componenti, o Inner and outer

Se Hoop componentx Meridional componentrr, OE, zz Components in cylindrical coordinates

7.1 TRUNCATED CONICAL SHELLS

An approximate analysis is presented which may be regarded as adequately describingthe behavior of truncated conical shells (Figure 7. 1-1) subjected to axisymmetric load andtemperature distributions (Reference 7-1) when the following conditions apply:

(1) The shell is thin walled (h_5. 2r).(2) The inclination angle 0 is between 45* and 1350 (preferably between 60* and

1200) and the shell is rather short.(3) Deformations due to thermal and mechanical loading are linear-elastic and I

the principle of superposition may be 9mployed.

For very flat conical shells ((A between 30° and 45*) the analysis gives reliable re-sults when the cone is analyzed in sections that are short (L< 31), i.e., where the meridionallength L of each section analyzed is less than three times the so-called "attenuation length" Iof the shell (Paragraph 7. 1. 1).

:,• O Ih

FIGURE 7.1-1 TRUNCATED CONICAL SHELL GEOMETRY

WADD TR 60-517 7.3

7. 1. 1 Radial Loading Perpendicular to the Cone Axis

Any general axisymmetric loading can be resolved into radial and meridional com-ponents as shown in Figures 7. 1. 1-1 (a) and (b). A shell analysis for the radial (horizontal)component of the loading, based on the so-called hoopring approximation of shells, is pre-sented. The analysis is valid when the conditions listed in Sub-section 7. 1 apply.

A membrane analysis for the meridional loading component Is presented in Paragraph7. 1.2. The solution of the conical shell problem for arbitrary axisymmetric loading is ob-tained by superposing the hoopring solution for the radial component of the loading and the mem-brane solution for meridional loading component.

P

X 0h

I (a) (b)

FIGURE 7. 1. 1-1: (a) CONICAL SHELL WITH AXISYMMETRIC BUTOTHERWISE ARBITRARY LOADING

(b) RADIAL AND MERIDIONAL LOAD COMPONENTSPh AND Px' RESPECTIVELY

7.1.1.1 The Hoopring Approximation for Radial Loading

The axisymmetric shell under axisymmetric radial loading is visualized as anassembly of beams in the meridional direction (Figure 7. 1. 1. 1-1(a)), supported by an elasticfoundation (hooprings) of stiffness

2K Eh sin 4 lb/inK- 2 ' 2(1)

r in2

WADD TR 60-517 7.4

4"

Meiioa Sc(Conti d) A

N1 •

x (oo)

( ~ ~~~ MeMdoi4Scto

Beam

FIGURE 7.1.1I.1I-1 CONICAL SHELL - HOOPRINGS AND MERIDIONAL BEAM STRIPS

WADD TR 60-517 7: 5

7..1.1 ot' d)

For this structural model, the differential equation for the radial deflection 5tbecomes

rr 4 d2 r dr(2)0 0 dx2(d 2So E-- I- . \•

where Ph is the distributed radial (horizontal) loading(AT and\in

SJroh/Sin4, 'A.

4/3(1I - V2 )

.778 rhfsin 0 for P =0.3 , (3)

where i " is the so-called "attenuation length" of the shell; it is a measure of the degreeto which the effects of self-equilibrating edge loads die out away from the edges. Al

The internal shell loads, moments and deflections are expressed in terms of 5' as

01 - -- sin 4,"

-M? D -d26, -sin 4, D= Eh 3

dx• 12(1-v2)

SM' = v/M 2- D d d 2HI 4r (4)

= H'sin dx

N' = H' cos4,

SNdth= Eh6'/r jand the vertical deflection relative to the base is ZR

,• ~=fX• I(N' -iN 'e) e-Ix Eh Sin 4 + E' cos

o 01where the positive directions for the above quantities are as shown in Figure 7. 1. 1. 1-1 (b)and (c).

WADD TR 60-517 7T6

k.~

7.1.1.1 (Cont'd)

It is convenient to find a particular solution for Eq. (2) where the loading is ex-pressed as a polynomial.

In general, this particular solution will not satisfy the boundary conditions at theedges of the shell; i.e., the horizontal edge loads, moments or edge deflections and rotationsderived from the particular solution do not satisfy prescribed conditions. Additional horizontaledge loads and moments must then be applied to the shell which, when superposed on the edgeloads and moments resulting from the particular solution, satisfy the prescribed boundary con-ditions.

Following this approach, the loading term on the right side of Eq. (2) is expressedby a polynomial, such that

NrOr (til

A~ Ih k=O = 'o'r .. ~ oEh C k (CO0 + C "(ro )C2 •r + "+ CN r(5)

k=0 0)0o

A given meriaional load distribution can be expressed in this manner by the method of Paragraph

4.1.2.3.1.

A particular solution of Eq. (2) can then be obtained in the form of a polynomial

k+2

6' = a J if k is evenj=2,4,6 .......

or'or k+2

6, = arJ, if k is odd, (6)j=1,3,5 .......

where

ak+2 1

k-j .k])2 k+1 k+2-jCL k 2 [(k+1)(k)...j+1)J2 j/(r0) (A 2 (7)

and j = k, k-2, k-4,... 4, 2if k is evenj = k,k-2, k-4,... 3, lif k isodd

The quantity "A" is defined by

4r 24A 0A 4 4L cos • (8)

WADD TR 60-517 7.7

7.1.1.1 (Cont' d)

The particular solutions for 5' are given in non-dimensional form in the following table

for monomials (Ck(r/ro) ) up to k = 4.

TABLE 7. 1. 1. 1-1

Particular Solutions for 6', Corresponding to Monomial Radial Pressure Distributions

roPh 5 k

Eh roCk

.20 0C

C -r 3 o )4

2 3 Icos ~ r'

2C -- - ) 1 8 C soo) 2

Cr 2 r

o o CO00

3 () - 0 o(r + 180 _)_r)o ro rrro

To illustrate the application of Eq. (7), consider the particular case k = 2. Then

0 0 0,004k+ 6 a4 4 18 (_

4, - (-L - 5(cs (L)+20 -o

and a 2 - (1)04 (32) 12) -72

2 roA rA

0 A0

so that innon-dimensional form

WADD TR 60-517 7.8

7.1.1.1 (Cont' d)

2 -~18 22 0 o ~ ro r

The solutions of Table 7. 1. 1. 1-1 are shown graphically in Figures 7. 1. 1. 1-2(a) and (b).

Once 6 ' has been determined as above, the internal shell loads, moments, anddeflections corresponding to this particular solution can be obtained from Eqs. (4). The re-suits in non-dimensional form are summarized in Table 7. 1. 1.1-2.

i6 1

WADD TR 60-517 7.-9

'I

0 W

IZ8

:R o

S440

I ow-

- 1 I I (0

-. 0

020

2(j 2

H I I

14 0

-- ~ - '- -~ - - 0. - -

00

So 0 C',~ +4

o Lo

04

.44

1.1. 4j

4

o too

040

34 0

t- c0

3403

S. I 0 S.0

00

1- U4 l 0 0 0

;I.- " -(D 10to 0 p 04

to

7.1.1.2 Solution for Edge Loading

When the conical shell Is subject only to an end moment and end force as inFigure 7. 1. 1. 2-1, then for a shell of lateral length greater than three times the attenuationlength, it is permissible to consider the shell as infinite. Assuming this condition to hold,it is only necessary to satisfy the homogeneous form of Eq. (2) of Paragraph 7. 1. 1. 1(Ph =0) subject to the boundary conditions

M(0) =? H(0) =

M(0)= 0 H(oD)= 0

He-fdH e.

'M M 4d

SMe Me

dxa,6 • He

e M e

FIGURE 7.1.1.2-1 SELF-EQUILIBRATING LOADS AND MOMENTS $ AND2F&AT ACONICAL SHELL EDGE

Recognizing that these are short range edge effects, the variation of r/r in the

region of significant stresses (i.e., r/r 0 1) are ignored. Under these conditions the solu-

tion of the edge loading problem may be written as

WADD TR 60-517 7.13

ftr

7.1.1.2 (Cont' d)6e • 2 sin 05 'hL F3 + N I• (sin p) F4

"e 2D34

e 2D 1 /LF+YXJ(~nt)~e = F [+ N,'t F(+s (sinp)F ]

e D(1)

Me = l)LF 1+ •/(sm 4) F2

2 2hzF 2 +He : ,sin

M = Z Me

Q = H sin

e e

N = He Cosxe e (2)

Eh6eN• = r

e r

Ae = o r hN"N sin + ee Cos dx

where the functions Fk are given by

-x/9F = ,2 (e) cos - )

F = (e) sin2 2e

-x/s (3)F3 = -42(e) sin\3 41r)

F4 = (e) cos

The above functions are damped slng soidals which cause the boundary loads to attenuatewith distance from the edge as e-x/1. They are plotted in Figure 7. 1. 1.2-2 for the range

0o< < 4.0.

WADD TR 60-517 7.14

-11,- -.

7.1.1.2 (Cont'd)

.8

.4---

.2

0

2

FI

-. 2 4_ 3_

-. 4

0 1x2 3 4

FIGTuRE 7.1.1.2-2 EDGE LOADING FUNCTIONS

WADD TR 60-517 7.15

7.1.1.2 (Cont' d)

To illustrate the application of the preceding equations, consider the radial (hori-zontal) loading problem for the conical shell of Figure 7. 1. 1. 2-3.

r1 =50"

30"

Ph= 500 lb

FIGURE 7.1.1.2-3 CONICAL SHELL UNDER UNIFORM RADIAL LOADING

The top and boL.^.n edges of the shell are unrestrained (free), the radial pressure

is constant.at Ph =500 l-and

= 20"lb0 Ph = !'oo 20-h =20" ~in

E = 30 (10)6 psi

4,~~ =40v 0. 30

D- Eh3 _ 30(10)6(1)3 = 2.75(10)6 in - lb.

12(1-v,) 12(1-. 32)

Netit is necessary to obtain the solution for the stresses and deflections in the

Svicinity of the bottom of the shell. The first step is to obtain the particular solution.

"From Eq. (5), since k = 0,

.~4•

roPhEh Co

WADD TR 60-517 7.;16

7.1.1.2 (Cont' d)

so that

SCO = (20) (500) = 3. 33 (10)-430(10) (1)

Then from Table 7. 1. 1.1-1,

oC0 0

orCor 2 3.33(10)-4r2 2 -6 (a)

= 0 _ r -16.67 r (10) .(a

20

Also, from Table 7.1.1.1-2, for k = 0,

-0' tan = 2Co ror

0 0r-r M' sin, (-r° sin 2

VC0 2ý(sn'C D cos2 -C0D Cos2

-H' r 2 sin2 _ -Nx'ro 2 sin 2 • -Q'ro2 sin r 2

C D cos 3 4 C D cos 4 4 CoD cos 3 4, r/ro0 0

N'e r

; C0 Eh r

Substituting the known quantities into the above expressions,

W = -33.33 r (10)-6 radiansSin-l._b

M' = -64.8 -i-•: in

in-lbM4 = - 19.4 in (b)

H- - 64. 7 lbr in

S-45.7 lbNx r in

NO' = 500 r lbe ~ inf

WADD TR 60-517 7.17

..... -fir € :,• • •°•-7w• •• • •,} • - , • '•• • • , ., . . • • •

7.1.1.2 (Cont' d)

From the second and fourth expressions of (b), the particular solution yields moments and

horizontal loads at the shell edges (r° = 20"I r =50") as shown in Figure 7. 1. 1. 2-4(a).

To obtain the solution for the shell problem with free edges ('t= A= 0), superposeon the particular solation given by expressions (a) and (b), the solution of the edge loadingproblem for equal and opposite edge loads and moments (Figure 7.1.1.2-4(b)).

In the edge loading problem,the attenuation length (Eq. (3) of Paragraph 7.1.1.1)for the bottom edge of the shell is

-- .778 /2 -= 4.15 inV.707and

h = 64.8 in(Positive directions are as shown in

4lb Figure 7. 1. 1. 2-1)=3..24 .-

in

Thus, from Eqs. (1) ,

2= -(4.2 .152)(1707) [(64. 8) F3 + (4. 15) (.707) (3. 24) F4 ]

e ~2 (2. 75)(10) 634

-41 (4-15)(.E) e 2.75(01 L (64. 8) F 4 + 2 707) (3. 24) F1

Me = 64.8 F 1 + 3.24 (4.15) (.707) F 2

-2(64. 8)He - 2648 F + 3.24 F3e 4.15 (.707) 2 3

or

6 -(143.5"x 10-6 F3 - (21 x 10-6 F4e34

-6 -6'ee = - (97.8 x 10-) F 4 -(7.15 x 0) F1 (c)

M e = 64.8F I + 9.5 F'2

He = -44.2 F2 3. 24 F ,

WADD TR 60-517 7.18

4

7.1.1.2 (Contl d)

64.8 in.-_lb

6 1. 8 • •__,;in

64.8 •-1_..InI

in64.8 In-l

in

in-l

3. 24kinIb)

FIGURE 7.1.1.2-4 (a) EDGE FORCES AND MOMENTS CORRESPONDING TO THEPARTICULAR SOLUTION

(b) EDGE LOADING EQUAL IN MAGNITUDE BUT OPPOSITEIN DIRECTION TO THE PARTICULAR SOLUTION EDGELOADING

WADD TR 60-517 7.19

Ll.1. 12 (Cent' d)

and from Eqs. (2)

M = 19.5 F1 +2.85F

= Nxe =31.2F 2 +2.29F 3 (d)

eN (4300F 3 +630F 4 )e

where F1 (1 = 1,2,3,4) are functions of x/A.

The final solution for the deflections and stresses In the vicinity of the bottom of theshell is then obtained by adding expressions (a) and (b) to expressions (c) and (d).

The results for the meridional moment M = MI + Me and the radial load H H' + He

(in the vicinity of the base) are plotted in Figure 7. 1. 1. 2-5.

The stresses can be obtained from the equations

Nx :L_6Mox h h2

Ne 6MO

7.1.2 Meridional Loading - Membrane Analysis

An analysis for radial (horizontal) axisyin'etric loading components was presentedin Paragraph 7. 1. 1. The present paragraph presents a membrane analysis for the meridionalcomponents of the general axisynmmetric loading (Figure 7. 1.2-1). Having done this, the pro-blem of a general axisynmmetric distributed loading can be treated by resolving the loadinginto radial (horizontal) and meridional components; the effects of radial loading are then evalu-ated by the hoopring approximation of Paragraph 7. 1. 1 and the effects of meridional loading areevaluated by the following membrane analysis.

WADD TR 60-517 7.20

01 2h.

-70 •_+H +H "

NI

-60

.4

-40 ___-

M,

-30

-20 /

-10

0 -Z I a - - -

0 2 4 6 8 10 12

x -Inches

FIGURE 7.1.1.2-5 MERIDIONAL MOMENT AND RADIAL LOAD MEASURED"FROM BASE OF THE SHELL SHOWN IN FIGURE 7.1.1.2-4

WADD TR 60-517 7.21

7.1.2 (Cont' d)

No N0x r

r0

FIGURE 7.1. 2-1 MEREDIONAL LO~ADING ON A CONICAL SHELLFor the membrane analysis, the moments Mo and M°0 are considered negligible.

The meridional and radial equilibrium and the stress strain relatons then give the requiredinternal loads and deflections az

[r +r\L0o (x% P - ro 7

N° 0q r P dxrjr xo

-'1 rP rd

N0o__ , [rP+~ 2 x) -€J 1)= sin¢ - Eh sin@ x 2

f [cN0 0AO 0 sin + ecos e dx~J 0 Ocos-

N E Q MoM o =H =o 0A

wherc r r°0 + x cos 0 and superscript "oil refers to membrane effects.

If the membrane deflections and rotations as given by the above equations do not satisfyprescribed zonditions at the edges, then horizontal edge loads and restraining moments mustbe applied such that the edge boundary conditions are satisfied. The effects of these edge load-ings, discussed in Paragraph 7. 1. 1, are then superposed on the me-ibrane effects to give therequired solution.

WADD TR 60-517 7.22

7.1.2 (Cont' d)

The following illustrative problem demonstrates the computational techniques ;orsolution of a conical shell (Figure 7.1.2-2) supporting a weight W. This solution requiresthe determination of the stresses at the fixedbottom of the shell (a similar procedure canbe employed at the top).

W 1,000,000 lb.

600

E E=30 x10 6 psjA r= 20 , V= .30

r0 =0

x

FIGURE 7.1.2-2 CONICAL SHELL UNDER VERTICAL LOADING

Assume first that the bottom of the shell is unrestrained. Then for verticalequilibrium, the statically determinate meridlonal membrane reaction at the base is

W 1,000,000 lb

" 29oSin - 27r(20)(. 866) n9200 i

and the significant parameters are

Eh3 = 30(10)6(1)3 62 2 12(.91) 2.74(10) inlb12(1-3 )

a 7.7 78 A7 ixi 3. 74 in

, _ 3. 74 1.37(10)-6 InD 2.74(10) in-lb

WADD TR 60-517 7.23

7. 1.2 (Cont' d)From the first and third of Eqs. (1), with P= 0 and NX , the membrane deflection and

slope at the base of the shell are

0 V .30 (20)9200)= h (-rol)) = = .00184 in30(10) 11)

0o = 7) cos 4, -9200 (.560) = -. 000177 radians.Eh 30(10)6(1)(. 866)

Since the base of the shell is fixed, an edge radial load, 2) and moment ;'z mustbe appVied such that

0

5=5°+5e=0

e=e0°+Oe=0,

i.e.,

0 (a)

ee=-e°.

The values of W and 2ptare then obtained by inverting the first two equationsfrom (1) of Paragraph 7. 1.1. 2. where ae and ee are given above by (a) and x = 0. Thisleads to

2e° 26_ o(Z7f) i__

and

Ssin - 4P+112 2/D) sin •

or

S= -2(.000177)~ 2(. 00184)

1. 37 (10)-6 1. 37(10)-t(3. 74)(. 8t&) = -1090 in

and

-sin ' 2 (.000177) 4(.0,018411= 1920 lb 1"'~'~~,--6 4 6(.0184) A 90l1.37(10) 1.37 (10)6(3. 74)(. 866)

1920 lb .

(3. 74)(. 866) in

WADD TR 60517 7.-24

7.1.2 (Cont' d)

The final results are now obtained by summing up the effects of membrane actiondue to •77 plus edge load action due to and Wf, or,

N N N+N Sx x xe

MN = 0 +N e

eN 4

M = 0 + M M

But from Eq. (1),

0 r0Nx = 17

and from Eqs. (1) and (2) of Paragraph 7. 1.1. 2,

N = -2•F2 cos + FNe x sin +

-Eh .t2Ne = r 2 sin• ( ?fF3+ V'A.sinn F4 )

me A t F 1 + W sin F 2 ,

where F 1,... F4 are the known edge loading functions ofx (Eqs. (3) of Paragraph 7. 1. 1.2).

Substituting the known values of • ,7n,,W, 4 , ro, D, E,/ and h into the above, and

making use of the relation r = r + x cos 4, the following is obtained:0

-9200Nx l+.5(x/ro) + 336 F2 +296 F 3

No= 1+ xro (3630 F3 - 6400 F4 )00

M =-1090F 1 +1920F 2

M -330 F +580 F2

The stresses at the extreme fibers are given by the expressionsN 6M Ne 6MO

0x h ± h2 and ae h h2

WADD TR 60-517 7.25

-Iw4

7. 1. 2 (Cont' d)

In particular, for the edge x = 0

F = F= F =11 3 4

and

F 2 = 0,

so that the stresses at the edge x = 0 become- 15500 psi outer fiber

XI = 0 -9200 + 296 -k (-6550) = 2400 psi inner fiber

-= 2770 (-1980) - f-800 psi inner fiber

eIx =0 - J-4750psi outer fiber

In the prnsent problem, the maximum stresses in the vicinity of the base occur at the edge

x = 0as given above. This is usually the case when the functions F 1 and F 2 are of the opposite

sign. However, in problems where the F, and F 2 terms appear with the same signs in themoment expression, the location of maximum stress will often be at a small distance from theedge, usually within 2.

2

7. 1.3 Meridional Temperature Variation

The purpose of the following is to reduce the problem of a truncated conical shellwith a meridional temperature variation (Figure 7. 1.3-1(a)) to an equivalent mechanical loadingproblem, which can then be treated by the methods presented in Paragraph 7. 1. 1 (Radial Load-ing).

To this end, ccnsider the shell to be sliced into elemental rings in which free thermalexpansions take place (Figure 7. 1.3-1(b)). TIhe free thermal strains in the meridional and radialstrains are

x = EO = (tAT . (1)C AT

Under these conditions, since AT varies with the element, the edges of adjacentrings do not match. To reestabiish radial continuity of the rings, a radi3l distributed pressure

' -Eh -Elh aAT (2)Ph = r CO r

is applied (Figure 7. 1.3-1(c)) which negates the radial thermal expansion of each ring, butcausee an axial extension of

dA* = (1+Y) c dx.sin= (1 4 Y) aAT dx sin 45

A* = (1 s) spin X ofAATd. (3)

WADD TR 60-517 7;26

7. 1.3 (Cont d)

r dx r(I + AT)

dx(/l• AT , (I x

II(a) (b)

PhPh

(c) (d)

FIGURE 7.1.3-1 1EDUCTION OF THERMAL PROBLEM TO EQUIVALENTMECHANICAL LOADING PROBLEM

WADD TR 60-517 7.27

7.1.3 (Cont'd)

Since the radial pressures Ph are really absent, a radial pressure Ph is*

applied to the rejoined, continuous shell (Figure 7. 1.3-1(d)) which negates Ph' or*

Ph+ Ph 0

so that *

Ph -Ph* (4)

The total stress in the shell is then obtained by superposition of the restoringhoop forces

N = P r= -Eh aAT (5), h

produced by Ph in rejoining the rings, and the stress distribution which results from the

radial, distributed pressure Ph in the continuous shell, where

h Eh OAT. (6)P r

The solution of this latter problem was presented in Paragraph 7. 1. 1.

Thus, in summary, to obtain the solution for the truncated opnical shell with a

meridional temperature variation AT(x):

(1) Find the hoop force No from Eq. (5) and the axial extension A' from Eq. (3).If the shell is free to elongate axially, then this latter quantity is of no interest.

(2) The total solution is obtained by superposing on the above effects the solutionof the radial (horizontal) loading problem (Paragraph 7. 1. 1) for Ph given byEq. (6).

7.2 APPROXIMATE SOLUTION FOR NON-CONICAL AXISYMMETRIC SHELLS

When an axisymmetric shell with axisymmetric loading is non-conical, the approxi-mate solutions previously staied may still be used. The shell structure can most often be ap-proximated by a series of separate conical shells (Figure 7. 2-1(a)) and a spherical cap "0"(the theory of shallow spherical caps is given in Reference 7-2). The individual shell sectionscan first be analyzed by the hoopring and membrane approximations of Sub-section 7. 1. Thecommon edges of adjacent sections are then joined (Figure 7. 2-1(b)) by satisfying the equilib-rium requirement that

A B

WADD Tf1 60-517 7.28

.I

7.2 (Cont'd)

A

B

(a)Ci

01

bb)

FIGURE 7.2-1 APPROXIMATION OF A NON-CONICAL SHELL BY A SERIES OFCONICAL SECTIONS

WADD TR 60-517 7.29

7.2 (Contf d)

and the compatibility conditions

M -1

6 A 6 B

(2)eA eB.

The solution is thus obtained by superposition.

To prevent interaction of the effects of the edge loads and moments 7? and 70 whichpropagate from the upper and lower edges toward the center, the meridional length I. of theconical sections should be made larger than three times the "mean" attenuation length I ofthe section. However, for reliable results in a very flat conical shell section (inclinationbetween 30 and 45"), the meridional length should be made very short ( L< 3,). In thiscase the interaction between upper and lower edges of the section cannot be ignored in analyz-ing each element and the more complex analysis of Reference 7-3 must be employed.

7.3 CYLINDRICAL SHELLS

The following paragraphs are concerned with the analysis of cylindrical shells subjectto axisymmetric thermal and mechanical loads. A thermal analysis is presented in Paragrapn7.3. 1 for the combined problems of normal pressure loading and temperatures which vary inthe axial direction only. The analysis is rigorous within the customary approximations of thinshell theory.

Paragraph 7.3. 2 discusses the cases of thermal stresses in cylindrical shells due toa variation of temperature through the thickness for both steady state and transient problems.

7.3. 1 Cylindrical Shells With Axisymmetric Loading and Meridional Temperature Variation

The preceding analysis becomes much simpler for the special case of a cylindrical shell(4) = 900; r = R = constant). Within the customary shell theory approximations, the hoopringand edge loading analysis of Paragraph 7. 1. 1 is rigorous for the thin-walled cylinder.

7.3.1.1 Hoopring Analysis

For the case of the cylinder, Eqs. (2) through (4) of Paragraph 7. 1. 1. 1 become

4 -x4-+ 6 R [ P(x)+ a&T(x) ,

where

-/vRh

_Vo

-778 /R- (forw v .30).

Both normal axisymmetric pressure loading and axisymmetr.c tempeht-ture variation are in-cluded on the right side Z Fq. (1). 7he oqjations for t'-e deflections, internal loads, andmoments become

WADD TiR 60-517 7.30

rn

7.3.1.1 (Cont' d)

d36' (2)_•

dx3

N'= 0,x

and

N' =E +V ATldx

RA

JAI I +Aflxf L Eh

where positive directions for the above quantities are as shown in Figures 7. 1. 1. 1-1 (b) and(c).

As in the case of conical shells, it is convenient to find a particular solution forEq. (1) where the pressure and thermal loading are expressed by a polynomial.

In general, this particular solution will not satisfy the boundary conditions at theedges of the cylinder. As in the case of the conical shell, additional edge forces and edgemoments must be applied to the cylinder whichwhen superposed on the edge loads and momentsresulting from the particular solution, satisfy the prescribed boundary conditions.

Following this approach, the pressure and temperature terms on the right sideof Eq. (1) are expressed by a polynomial such that

RP N (RkRP + aAT= N CEh I k(R)A ~~k=0 k 2 NQ (3)31

2 Nx

= 0 + R 2 i) +... + CN -

A given meridional distribution can be expressed in this manner by the method ofPsragraph 4. 1.2.3. 1

The varticular solutions for a ' can then be obtained by substituting Eq. (3) intoEq. (1) and solving the differmntial equation. The particular solutions for 6' are given Innondimensionai form in Table 7.3. 1. 1-1 for monomials Ck(x/R)k up to k = 4.

WADD Th 60-517 7.31

.45y. ..z~--- -

7.3.1.1 (Cont' d) TABLE 7.3.1.1-1

Particular Solutions for 8' Corresponding to Monomial Pressure Distributions

k RP + aAT 61/CkREhk

0 CO 1

1 C1 x/R x/R

2 C2 (x/R)2 (x/R)

C3 (x/R) (x/R)4 44 C4 (/) (x/R)4 6 (1-)

Once the 61 has been determined as above, the internal loads, moments, slopesand axial deflections corresponding to this particular solution, can be obtained from Eqs. (2).

7.3. 1.2 Solution for Edge Loading

When the cylindrical shell is subjected only to an end moment and end load asshown in Figure 7.3. 1. 2-1 and the longitudinal length of the shell is greater than three timesthe attenuation length, the edge loading equations (Eqs. (1) and (2) of Paragraph 7. 1. 1. 2)become

62

3

ee D 2t F43 + 4

(1)Me F + -2

4 J +-2 N F3S~~~H = Qe=""•'2e e e2

and

M M N 0= e eXe e

Eh6eANe e N( 'ZNe R eo)e d

Eh

WADD TR 60-517 7.32

7.3.1.2 (Cont' d)

where the Fk' s are those functions of x/1 given by Eqs. (3) of Paragraph 7. 1. 1. 2.

Me d

2.I~Qe dQ e

dxJ I.

•'i Me

FIGURE 7.3.1.2-1 SELF-EQUILIBRATING LOADS AND MOMENTS . AND 7T, ATA CYLINDER EDGE

The following illustrative problem demonstrates the solution for a cylinder subjectedto an axisymmetric temperature distribution. The infinitely long cylinder of Figure7.3.1.2-2 has a radius of R = a and is subjected to a temperature distribution given by

4AT = T' when x<0= 0 when x>0.

Find the deflections and stresses due to this thermal loading.

The boundary conditions are

e =0at x=+a)

WADD T17 60-517 7.33

7.3.1.2 (Cont' d)

and

a a oaT'at X =-O .

e =0 J

FIGURE 7.3.1.2-2 INFINITELY LONG CYLINDER WITH DISCONTINUOUS AXISYM-METRIC TEMPERATURE DISTRIBUTION

To begin with, the continuous shell is cut into upper and lower portions at x = 0.All the stresses and deflections in the upper half are identically zero (no loading or temper-ature exists in the upper half).

The loading for the lower half is defined by

P=0

aAT = aT'.

WADD TR 60-517 7.34

7.3.1.2 (Cont' d)

Therefore, from Table 7. .1. 1-1, aT' = C and the particular solution ofEq. (1) of Paragraph 7.3.1.1 becomes

aaT- (this is easily seen to be simply6 0 the thermal strain times the

cylinder radius)

where the upper and lower halves are now denoted by + and -, respectively.

Since 61 is a constant, Eqs. (2) of Paragraph 7.3. 1. 1 give

0' =-M' =~ 3 M-0= N' C)=Nx=0

or the slope and stresses are identically zero everywhere in the unrestrained lower half ofthe cylinder. Thus in this case, the particular solution satisfies the boundary conditionsidentically.

Crsi In general, it may be stated that a AT variations which vary linearly with

Cartesian coordinates in the form

ciAT = A+ Bx+Cy+Dz

U• cause no stresses if the displacements are unrestrained externally.

Compatibility at the cut edges requires that+

6 (0) 6 (0)+

Se(o)=e(0).

In other words,+ +a' (0)+6 (O)= ' (0)+6 (0)e e+ +0' (0) + t()e (0) =' (0) + e (0)

but+6' (0)=0

6•' (0) =adr'

' (0)=e' 1(0) =,

therefore+

a (01=aT' + 6 (0)+ e e

ee (0)= e (0)

WADD TR 60-517 7.35

_ '

7.3. 1.2 (Cont' d)

Edge loads and moments must be applied at the cut (Figure 7.3. 1. 2-3) to satisfythe above conditions.

xI,II

FIGURE 7.3.1.2-3 HORIZONTAL LOADS AND MOMENTS AT CUT EDGES

Expressing the edge deflections and slopes by Eqs. (1) with

F1 (0) = F3 (0) = F4 (0) = 1

gives

_12 +2

2D• a tl' 2

57 2 J TD + 2

WADD TR 60-517 7.36

ff,

7.3.1.2 (Cont' d)

Solution of the above yields

0I

-aaT' D

The deflection for x >0 then becomes

5 =' +6e)0+ o3 DF 4 = a2T'

and for x<0, 1

a = a +6e=aadr' -ad"' F4 (X acrT' 1-2 F4 -

Th2 slopes, moments and hoop forces can be obtained in a similar manner fromEqs. (1) and (2). The stresses are then obtained from

6M

x h2

) O 6VM (Since MO =M).

h h2

The resulting radial deflection 6 and stresses ax, aO at extreme fibers areplotted in Figure 7. 3.1. 2-4.

7.3.2 _ylindrical Shells with Temperature Gradient Throuwh the Thickness

Design nomographs and analysis are presented for the solution of the steady statethermal stress problem. Cylinders with free and completely fixed ends are both considered.

The transient problem for long unrestrained solid and hollow cylinders follows.

"Tables and curves are given for the determination of the stresses in nondimensional form.

7.3.2.1 Cylindrical Tube - Steady State Heat Flow in Radiai Direction

The temperature distribution in the instance of a steady state radial temperaturedistribution in a cylindrical tube (Figure 7.3. 2. 1-1(a)) is

T (T. - T0) (b) +To(1

WADD TR 60-517 7.37

tt t

0

r44

'.44

0s 0

rzl

z

t- 2

0 0

-~ z

'-4

6 ca

WAD4 4- 6057I.3

* 7.3.2.1 (Cont'd)

• (a)

zz

C +

-1 7

FIGURE 7.3.2.1-1: (a) SHELL GEOMETRY0) VARIATION OF STRESSES THROUGH THE CYLINDER THICKNESS

WADD TR 60-517 7.39

7.3.2.1 (Cont' d)

where Ti andTO are the temperatures In the Inside (r =a) and outside (r b), respectively.

The stress components in the radial, tangential, and longitudinal directionsare as follows for the case when the tube is completely unrestrained.

E 1(Ti-T°) 0 b a 2 2 l 1 (2)

rr 2(1-v)log(b) - r - b2a a

E(Ti-To) [ -g 222 a]b (3

ee1 0 _ a b1~lg~S= 2(i-z'Yog(b/a) b -a 2r2

Ea(Ti-To) [ b 2 2 l (zz 211-Z,)log(b/a) I - 2 log r -2 a og a (4)

NOTE: azz = 0rr + a a zz max. and '00 max. occurs at r = b.

The results given in Eqs. (2) through (4) are accurate away from the ends of thetube since the boundary conditions of zero traction at the ends are not satisfied pointwise.Instead, the average value of the normal stress is zero.

The character of the stress distribution in a thick-walled tube in steady heat con-duction with temperature varying only radially (corresponding to Eqs. (2) through (4)) is shownin Figure 7.3. 2. 1-1(b).

The following tabulation gives the signs of the stresses for internal heating

(T > T):

Stress Radius Kind of Stress

Tangential Inside CompressionTangential Outside TensionLongitudinal Inside CompressionLongitudinal Outside TensionRadial All Radii Compression

For external heating, T 0 >Ti, the signs of all the stressesare changed.

Figures 7.3.2. 1-2 through 5 are alignment charts giving values of the above stresscomponents in terms of the geometry and the physical constants for one degree fahrenheit dif-ferential between the outside and inside surfaces. For a temperature difference of T degreesFahrenheit, results found by the nomograph must be multiplied by T.

WADD TR 60-517 7.40

7. 3.2.1 (Cont' d)

240- 400

180- 300

6Z

-z z5 J 210--350150- 250

4 4

4 w

3 -- _I

z120- 2 0 0

-I ) .3000 CL

cr z

Cl)

90-150

1.2 ?50

A: FIGURE 7.3.2.1-2 MAXIMUM TANGENTIAL AND L( TGITUDINAL STRESS INCYLINDER PER F TEMPERATU .E DIFFERENCE BETWEEN S'THE OUTER AND INNER RADP

WADD TR 60-517 7.41

7.3.2.1 (Cont'd)

560- 600210 350

T-

150- 250 z 500 500I. I w

S~270--450

120 200 2-I.5 Id:0

"o, 240- 40090 150 1

0

•44

Zm- 210- 350

60- 100

180- 00-4

z-5 V)

U)

16 150--250

120--L200

FIGURE 7.3.2.1-3 TANGENTIAL STRESS AT OUTSIDE RADIUS PER 'F TEMPER-ATURE DIFFERENCE BETWEEN THE OUTER AND INNER RADII

WADD TR 60-517 7.42

7.3.2.1 (Contt d)

6.0

5.0

4.0

39--653.0 600 1000

2.5

w 450-750

x A.z

2.0 30--50 1w"" 24 40

.0 24P I

1.7< 18-30 300-500cr 0

270 450 w

1.5 I-"20 5 240-400

a W210-350-1.4 9-5 c

Ix

6L -80 3006-10

150-250

-123--5 120-200

FIGURE 7.3.2.1-4 MAXIMUM RADIAL STRESS PER OF TEMPERATURE DIFFERENCEBETWEEN THE OUTER AND INNER RADII

WADD TR 60-517 74

7.3.2.1 (Coxnid)

10-9

-7

6

20:4

..9.-7 -

"-5.-- O-2

,4

"1.-5

-4 U.

I0~30 2

1.2 -- -

FIGURE 7.3.2. 1-5 LOCATION OF THE POSITION OF MAXIMUM RADIAL STRESS

SWADD TR 60-517 7.44

[IN

7.3.2.. (Cont t d)

Stress formulas for unrestrained thin-walled cylindrical tubes are now presented.If the tlickress of the wall 'b-a" is small in comparison with the inner radius "a" of thecylinder, tbtn b = b-a where b-a Therefore, log b = log (1 + )4

a a a a

so that for a thin-walled unrestrained cylindrical tube the stress at the boundaries are givenapproximately by

-Ea(T. - T

zz~r-a 2(1 -z/ a)

(5)aTIE( -To) 1 b-a)

(cee)r--b = '7zzr=b= 2 ( 1-v) 3)

where a Is negligible.rr

It is now assumed that end displacement is completely prevented. This conditionis exemplified by a cylinder with completely restrained ends. The formulas for the radialand tangential stress components are the same as for the unrestrained case. The axial stressis modified by the stress necessary to prevent axial displacements. The additional stress "K"necessary to prevent axial displacements is

-Ect(Ti - To) 2a2 lbo-1 = & -c (6)•

2log b[

and the total stress is azzT K + zzwhere azz is given by Eq. (4).

7.3.2.2 Transient Thermal Stresses in Circular Cylinders

This paragraph presents the determination of transient thermal stresses in longunrestrained cylinders s-jbject to temperatures which vary only with the radial coordinate andthe time t , and where inertia effects are ignored.

For solid cylinders, the following results are presented:

The radial and tangential stresses for a solid cylinder (Reference 7-4) are

arr L~f~ rTdr - rTdr1 (1)[a o r

E i a 1 1r.A, srTdr + rTdr-T] (2)'EO0 = • a- r2d + -2• i

where a is the radius of the cylinder, E is Young' s Modulus, v Poisson' s Ratio, a thecoefficient of linear expansion of the solid, and T its temperature at radius r.

WADD TR 60-517 7.45

7.3.2.2 (Cont t d)

SLice the resultant of the normal forces over the cross section of the cylinder iszero, then

0 zz Orr +oe

The direct application of Eqs. (1) and (2) would require a knowledge of the temperature dis-tribution. The heat transfer problem from which the temperature distribution is determinedwill not be considered here.

The temperature distributions and the non-dimensional stresses for the followingtransient problem are found in Reference 7-5.

Case E Thermal stress in the cylinder o:__r<_a initially at constant temperature Twhose surface is kept at temperature T1 for t>o. 0

Tables 7.3.2. 2-1 and 7.3.2. 2-2 give values of -(1- )rr and (1-•' NeE a(T° - TI) Ea(To-TI)

as functions of the dimensionless parameters r/a and - where K is the ther aldiffusivity of the material and is defined by a

(thermal conductivity)(mass density) (specific heat)

Case H: Thermal s.resses in the cylinder o<_ r S a, initially at uniform temperature Twith heat transfer between the cylinder and a medium at temperature T for t > o. O•

It is assumed that the normal gradient of temperature at the surface is directlyproportional to the difference between the surface temperature and the medium. This relationis expressed by

[ T + h(T - T1)] =0 (3)Or 8r r=a

NOTES: (1) h = 0 corresponds to no heat transfer (complete insulation), 2T1-SOrr=o

(2) h = co corresponds to infinite heat transfer, TI r=a= T (Case I)

Therefore, the r_Isults will involve three dimensionless parameters which are

most conveniently taken as -K! ar and ah. In Figures 7.3. 2. 2-1 and 7.3. 2. 2-2, respectively,

a al-_____ (' g)oee Kt

the values of and as functions of 2 are shown forE '(T - T' E (T -T' aSrragig 0o 0 to a.0 '

ah = 5 and values of 1 ranging from 0 to 1.0.

To illustrate the way in which the stresses vary with ah, the variation with 2

of the surface tangential stress acr for r = 1, is shown in Figure 7.3.2. 2-3 for ah avalues of 0. 2, 0. 5, 1, 2, 5, 10, 2u,- and 0oa0ý

WADD TR 60-517 7.46

[I

7.3.2.2 (Cont'd) N

00 00000000 000000886§§08§

0 0

0 *0 000000 0 0008080H00

9 000i 0(l6 0(i090000090 00

o% moo co- CE - f §§Y -0 P44 10

'v 0 000.CW I *@@@r% 0 0 000 00

r4 M~ (K MuN- 0 Wt- "%00 0 t--W -4 I 8 rR:,

0 40 04,I-14 4 r-4 -4 r4-4 t-4 r4 .- 400 0 888

0 0 i 0* 0 0 50 V 0 0 900

Z 4v t- 4 rI -1^ C% m O C-m c cm - m10 r4-O

0 ý U\ t -.- 0Cir-t-X\a 0 t2~~~~~j~~~C ch b$\ ME.~. r -~~f N~t 0 -f

EO) 0P4r4r4r4fH I H r4u -4 0 00 0O00 8* 0 0 0 0 0 0 *0 0 0 0 0 *0 0 0 0 0 0 *00

DOz

cc% c- -st-N t-~U No OD ~ r N co 4'~V- C- aol\CSrx\:;s0 0 .- 6 N R ;10 NC

0'---t -0% F t icuO40o o Ir-I -1 -4 C4, 00~.N- r.~ -

c 04.~ -s3' C'j.4C1-4t.~

* ~ c R-(~-t r-.C',4 88880 0 0 0 0 9 0 0i.IiIiIl i r i0 00 0 0

* 0 00 0 0 00 0 000080

li q' r! Oic*CoCIJ0Ce1-09. O 1-N A.

5, tf'.N -iC&N04 r-~C' cme Ho u cZr;NNNCMNm mCj l -0 0000 8888

0 0000 0880

t- 9C.I Vm -,xýCjt-00 00 00 0 HHN'~tr' 0 1AU - - = 0

.X~ 0 000000 000000000000000-

WADD TR 60-517 7.4700 0 00 0 30 0-

7.3.2.2 (Cont'd)

In o -- j 0,0 r-4. UN a, 01%0' :V H a_ *o H8

--I -Z- M"N H 0 0 0888

* [email protected]\w000 " 0 H 0 \

IAIA

w M M0 0, 0M 0P cc 0A M U0 _: 0 0 -N \0 --I 0

0'. oM I X\r rClt~ H OD ~0% n-oH M ---- r- 0'.' '-A H

00i M , .0 0 10 0 0 0 00 0 0 00 1 0

[,-,D CrCr -' ~-ZIU\C. C'j 0'. o c 0'~ )~ N~ k0 l 0i

0 0H 'fl g 0080

CDO -aH 0,~c. cC'. c'j g-c ;H

00 C~ CIYYClý0 8808z

00 r-a HC' Hm c IfUt- -a ) HO 00 00 0 r- ~H H 1 0 0 8 C)8 8

00 0008808N ~~ @000000 00 0 0 0 0 *~0 0 0i lil

* Z zs MciX\L-\ ~ \ 0 - C C l IAg'r ~ ~ U R \0 M N

0 rr~-4HHH r-4 00 IA:*C 0000 888§ 8080

tl -mXAEIO %0 Cr-r- co8 y h ý r .

0 -- 9 \0r- HrrH-r00 08

0 0H r4HIH H ri - rs - H rH000088 8 8 8

00 0 0 00 0 0 0 0 0

c cr~r-AliWs. 555 311 131 w 5MA11wC4 Ll - 9 0\ - W M H 2 IA~ Nlo RU N m H C -\0' CV'.r-

`2 r4HCa)C'HcRt`\Om r CO0 H r- NNNNj f-

0 r-mw--cH cc~ccm CJ MAO -J

r! rv amt' ut -%Hr-\ al mau o ta '---- cq sH ~t- -sam

o r.4 I r ( i t,-Cv0'. '.OC 0 ''- H0 888 N0Lc1r

lasli licsatia sa! .99995

.IH O\OOOOOO"O\OOOOHOOOOOO

WADD TR 60-517 7.48

tI7.3.2.2 (Cont'd)

-( __rr .

Ea (T- T1 )

0`0

0 E l .95

0 .02 .04 .06 .08 .1

Kt2

a

FIGURE 7.3.2.2-1 NONDIMENSIONAL RADIAL STRESS AS FUNCTION OFKt FOR THE CASE ah = 52

an.4r/o

.3 .95

.9.2 .85

(1v ~.8Ea (To-T 1 ) .1 .75

.5

.4.2

0*

0 .02 .04 .06 .08 .1

Kt2

a

FIGURE 7.3.2.2-2 NONDIMENSIONAL TANGENTIAL STRESS AS FUNCTION

OF FOR THE CASE a= 52

WADD TR 60-517 7.49

7.3.2.2 (Cont' d)

1.0!U.8 ah:2o

.2

N !I /oh=5E (1-v )a T00 6 __ - oh=21

0 .04 .08 .12 .16 .2

Kt2

a

FIGURE 7.3.2.2-3 NONDIMENSIONAL SURFACES TANGENTIAL STRESS ASKt

FUNCTION OF K WITH ah AS A PARAMETERa

0/b:

.2 8

• . . . .25

-z0 02 .04 .06 .08 .1

Eaa

FIGURE 7.3.2.2-4 NONDIMENSIONAL SURFACES TANGENTIAL STRESSFOR A HOLLOW CYLINDLR WITH OUTER SURFACESAT TEMPERATURE T1 AND INNER SURFACE INSULATED

-WADD Tb 60-517 7.50

.2~

0A .02 .0 .0-0

-. -.- - - - ,Kt-- - -- - - - - - - - - -

7.3.2.2 (Cont' d)

For hollow cylinders, the solution of the thermal stresses is more complicatedthan that of solid cylinders because of the add-.tional (inner) boundary. A parametric studybecomes very unwieldy. As an example, a particular case of a hollow cylinder will be con-sidered in which the outer surface (r=b) is kept at an elevated temperature T1 for t >o and

there is no loss of heat from the inner surface r 0. Figure 7.3.2.2-4 presents(1-)-eej rb2

the ordinate E 01 against Kt/a2 for values 0, 0.25, 0.5, and 0.75 of the ratioEaT1

a/b. It appe -rs that the values for the solid cylinder are reasonable approximations to thosefor the hollow cylinder only in the ca3e of fairly thick-walled cylinders.

7.4 REFERENCES

7-1 Horvay, G., Thermal Stress Problems in Nuclear Reactors, Summer SessionCourse in Stress Analysis at the Massacnusetts Institute of Technology,June 17-28, 1957.

7-2 Reissner, E., Stresses and Small Displacements of Shallow Spherical Shells IU,Journal Mathematics and Physics, Volume 27, 1948, p 240.

7-3 Horvay, G., Linkous, C. and Born, J. S., Analysis of Short Thin AxisymmetricalShells Under Axisymmetric Edge Loading, Journal Applied Mechanics, Volume 23,1956, (a) p 68, (b)p489.

7-4 Timoshenko, S. and Goodier, J. N., Theory of Elasticity, McGraw-Hill BookCompany, Inc., New York.

7-5 Jaeger, J. C., On Thermal Stresses in Circular Cylinders, PhilosophicMagazine, Volume 36, 1945, pp 418-428.

WADD TR 60-517 7.53

SECTION 8

STRUCTURAL ASSEMBLIES

WADD TR 60-517 8.0

SECTION 8

STRUCTURAL ASSEMBLIES

TABLE OF CONTENTS


8 Structural Assemblies 8.2

8.1 Shells of Revolution with Bulkheads 8.3

8.2 Determination of Interaction Forces and Moments 8.6

8.2.1 Solution for the Interior Bulkhead 8.7

8.2.2 Solution for External Bulkhead 8.8

WADD, TR 60-517 8.1

SECTION 8 - STRUCTURAL ASSEMBLIES

The structural elements analyzed in this Manual are, in almost all cases, integralcomponents of a complete structural assembly. The beams analyzed in Section 4, forexample, are usually components of a larger frame or beam-like structure; the purelymonocoque shells analyzed in Section 7 are components of a larger semimonocoque stiffenedstructure.

In all such cases the boundaries or edges of each structural subsystem are subject tothe restraint of adjacent structure. Since the over-all structure is an integral assembly, thedeflections at the boundaries of adjacent structural components must be compatible with eachother. This section considers the mutual interaction of these components, thereby providingthe tools necessary for the over-all structural analysis.


f Influence coefficienth Thicknessr Radial coordinate

3D Flexural rigidity Eh

12(1-y 2

E Young' s modulusH Interaction force per unit of lengthM Interaction moment per unit of lengthP Uniform normal pressure on bulkhead

Q Coefficient of linear thermal expansion6 Radial deflectione Rotation

Angle between the normal to shell surface and axis of revolution

AT Elevated temperature with respect to base

Subscripts

1,2 ShellsB BulkheadH, M Due to edge interaction loads and moments, respectively.

WADD TR 60-517 8.2

8.1 SHELLS OF REVOLUTION WITH BULKHEADS

This sub-section considers the interaction of shells of revolution, under axi-symmetric loading (mechanical and thermal), with transverse circular bulkheads lyingperpendicular to the axis of revolution (Figure 8. 1-1 (a)).

For general shells of revolution, the analysis Is approximate, giving reliableresults within the limitations spelled out in Sub-sections 7.1 and 7.2. For cylindricalshells, the analysis is rigorous according to thin shell theory.

In order to determine the interaction stresses at the junction of shell and bulk-head components, the deflections and slopes at the unjoined shell and bulkhead edges arefirst determined (Figure 8.1-1 (b)).

The deflections and slopes 6', 61 and 01 ,0 at the unjoined shell edges, dueto axisymmetric thermal and mechanical lads, caln be determined by the methods ofSection 7. The bulkhead is considered to be a circular plate (with or without a concentrichole). Its edge slope and deflection, Of and 6 j, for radially symmetric transverse Iokd

may be determined from a plate bending analysis. Thus, as shown by Timoskenko, * for auniformly loaded solid circular plate free to rotate at the outer edges (Figure 8.1-2),

pr 3

B 8D (1)

and 6'B r% 0.

After the unjoined edge deflections have been determined as above, edge momentsand loads are applied (Figure 8. 1-1(c)), such that the radialdeflections and changes in edgeslopes of the shell components and the bulkhead are all equal, and the joint is in equilibrium.These conditions require that

e1 ÷ 6 + 6 ( ,2)

0e' + ew '+ e"f Ofe + Off

andM+M +M =01 B 2 (3)

H1 +H +H = 0

where the double primes refer to the effects of the edge interaction loads (M, H). However,the edge deflections and rotations due to the interaction loads can be written as

61 flH + flM •o1 H1 1 M1

e1;= 1lH H1 +1M M 1 (4)

2 f2H 2+f2M M2H2 +2 M2

STilmoskenko, S., "Theory of Plates and Shells," McGraw-Hill Book Co., Inc.

WADD TR 60-517 8.3

8.1 (Cont'd)

hB

(a) Integral Shells and Bulkhead

}•"51 B 2

t t I1 2

(b) Deflections of Unjoined Shell and Bulkhead Edges

H1 HB H2

M 2

(c) Interaction Loads at Joined Shell and Bulkhead Edges

FIGURE 8. 1-1 DEFLECTIONS, LOADS AND MOMENTS AT A SHELL ANDBULKHEAD JUNCTION

WADD TR 60-517 6.4

8.1 (Cont'd)

26 H 2 + 2•M M2

6"f *HB fBH B (4)

eB fB *MB cont' dB 'BM 3'

where, from Paragraph 7. 1. 1, for v = .30

f 2.571 1/2 r 3/21H E (sin@1

f = ft 3.305im I.H Eh 2

8.496 1 (5)

Eh1m 3 -sin 01

f~2H =2.571 (s~n /2 (hr )3/2 2r

f2M 52H2

8.496 rh

3 I i

f.M =- W. _/ _

4.-

S~FIGURE 8.1-2 EDGE ROTATION OF SOLID CIRCULARS~BULKHEAD DUE TO UNIFORM PRESSURE•, LOADING

S• WADD TRt 60-517 8.5 I

2 1-~ C:!---- -- - - - - - - --- ---- -- N

8.1 (Cont'd)

and from Timoshenko* and Figure 8.1-1(a),

[r 2 +r + -3BH Lr2r•2 EhB

rb(6)

f__-30] 12rfBM r2 r2 3h

Inserting Eqs. (4) into (2) and (3) and combining equations, results in

(f1 H + H1 + (BH) H2 + (flM) M1 + 0 =

(fBH Hi + (f2 H + fSH) H2 + 0 + (f2M) M2 = (6B- 62)

.7)H + 0 + (film +fB M1 + (fM BM O -0E)~iH~1 iMBM 1 B 12 =%.{

+ 0 + (f2.H) 112 + f MI *(f 2M~ BM)M2 B 2

and

M (MB -(M1 + M2 )

HB -(H 1 + H2)

Thus, once the deflections and slopes at the unjoined shell and bulkhead edges(corresponding to the right side of Eqs. (7)) are determined, (7) and (8) provide sixalgebraic simultaneous equations for the determination of the three interaction forces(Hi, H2, HB) and moments (M1 , M2 , MB).

8.2 DETERMINATION OF INTERACTION FORCES AND MOMENTS

The long cylinder shown in Figure 8.2-1 is heated to a uniform temperature ofAT, while the bulkheads remain unheated (ATE = 0). Find the restraining forces at theinterior and end bulkheads.

2 * Timoshenko, S., "Theory of Plates and Shells," McGraw-Hill Book Co., Inc.

WADD TR 60-517 8. 6

S- -- '.--.---- - - -

ý.e- -'

8.2 (Cont'd)

Interiorh Bulkhead

I-EndrBulkhead

hB hBB4

A

FIGURE 8.2-1 LONG CYLINDER WITH INTERIOR AND END BULKHEADS

8.2.1 Solution ior the interior Bulkhead (Figure 8.2.1-1)

The free expansions due to the uniform temperature elevation are

A' 6 0 - ruAT1 2

B 1 2 BAlso from the symmetry of the problem

M = -M 2 = M

M 0B

H1 = H2 = -HB/2 = H

Substituting the above in Eqs. (7) of Paragraph 8.1 yields the independent equations

(fH + 2 fBH) H + (1lM)M=-raAT

(1H) H + (f1M) M =0

Solving the above equations resuits in

WADD TR 60-517 8.7

8.2.1 (Cont' d)

-r a AT

If H + 2fBH - t1H f'iM

andM =flM H ,

where the influence coefficients f, f" are determined from Eqs. (5) and (6) of Parigraph 8. 1.

2H

M

2 H

FIGURE 8.2.1-1 INTERIOR BULKHEAD

8.2.2 Solution for the External. Bulahead (Figure 8.2.2-1)

As in the case of the interior bulkhead,

56 = raAT

Also, for equilibrium of the joint (Eqs. (8) of Paragraph 8. 1)

M = -MB = M

H = -H =HB

Substituting the above in Eqs. (7) of Paragraph 8.1 yields the independent equations

WADD TR 60-517 8.8

8.2.2 (Cont'd) I(lH + BH H + (1M) M =-r aAT t

)fH) H +(IfM -:-fbM) M =0

from which

H = -raATSH= [(ffl~l P

ilM "Ill+f)a +f tl~ 1H +BH) ifM BfM)

and

(flM + fMBM)

HH

M M

MI

M

H H

FIGURE 8.2.2-1 EXTERNAL BULKHEAD

WADD TR 60-517 8.9

SECTION 9

STABIUTY OF STRUCTURES

WADD TR 60-517 9.0

SECTION 9

STABILITY OF STRUCTURES

TABLE OF CONTENTS


9 Stability of Structures 9.2

9.1 Stability Criteria 9.4

9.1.1 Stability by Eigenvectors 9.7

9.1.1.1 Euler Equilibrium Equation 9.7

9.1.1.2 Buckling Load 9.11

9.1.2 Stability by Virtual Work 9.14

9.1.3 Properties of Eigenvectors (Deflection Modes) and Eigenvalues(Critical Loads) 9.15

9.1.4 Approximate Methods of Determining Stability 9.18

9.1.4.1 Si'gle Functions 9.18 5

9.1.4.1.1 Upper Bound 9.18

9.1.4.1.2 Lower Bound 9.19

9.1.4.2 Multiple Functions 9.22

9.1.5 Shear Energies 9.26

9.1.6 Plasticity and Eccentricity 9.29

9.1.7 Temperature 9.31

9.2 Non-Dimensional Buckling Curves 9.32

9.2.1 Non-Dimensional Stability 9.32

9.3 Curved Plates and Shells 9.48

9.4 References 9.50

WADD TR 60-517 9.1 !Ii

SECTION 9 - STABILITY OF STRUCTURES

The solution of the structural problem requires the determination of the displacementdue to a given loading. As the magnitude of compressive loading is increased a limitingvalue is approached at which a small lateral force will produce considerable lateral deflec-tion. This limiting value is called a critical (buckling) load at which the structure becomesmetastable. This section is concerned with a discussion of the criteria of stability and thecharacteristic deformation modes of structures (eigenvectors). Methods to calculite thebuckling load from the eigenvectors or to approximate the buckling load by determining upperand lower bounds are presented. The effect of neglecting various flexibilities is discussedand formulations are presented. The effects of eccentricity, plasticity and temperature arediscussed qualitatively. Non-dimensional buckling curves and examples which include theeffects of uniform temperature and creep are also presented.


a, Coefficients of eigenvectors defining the lateral deflection

b Length of half wave of buckle (length of pin-connected column, width of plate)e Distance from shear center to line of action of transverse shear V.f Deflection pattern or eigenvector; Average thickness of sandwich facesh Thickness of plate.; Centroidal thickness of sandwich platehc T'1ickness of sandwich core

k Stability constant1 Length of columnm Virtual internal moment due to a unit virtual forceq Lateral load intensityr Distance from shear centert Thickness of plate; Timeu Displacementu1 Elgenvector

w Lateral displacementw0 1wIxm-am lateral displacement; Initial approximation of laterai displacement

wj Slope ýa j direction

wij Change of slops (negative of curvature)

x, y, z Distances along coordinate axes

A Area; Linear operatorAV Effective transverse shear area

C Constant coefficientD Stiffness of isotropic plateDij Stiffness associated with the moment M1 i

EA Initial (Young' s) modulus

ER Effective stability modulus-4E -oT

SES Secant modulus a

ET Tangent modulus = d d /da

WADD TR 60-517 9.2

F Generalized force or buckling loadFcr Buckiing load

FM Buckling load c;nsidering bending energy orly

FV Buckling load considering transverse shear energy only

FQ Buckling load considering toreoxal energy only

Gs Secant shear modulus

G Shear modulu3 of corecII Moment of inertia of cr-)ss sectionK Non-dimensional stabiltty constantL Potential energyL* Complementary energyMij Moment on ith plane in the x z direction

Ni Load on tth plane and in jth direction

Q Torque acting on cross section

T Temp-rature

SU Change in Ntrain energy due to a 1-A-tual displacementa U* Cirmge in complementary strain energy -Ale to a virtual force

V Transverse sh&Gr load on cross section; V'olumeaW Cbange in potential energy oi external loads due to a virtual displacement6W* Change in potential energy of external displaeements due to a virtual force

AVG Tran3verse she,:r s'iffness of cross section = IEsdAVET Bend.ng stiffness -f cross saction = E Sy dA

EA Axial stiffness of cross seetion = fE.d

STorsional stiffness of cross section fG5 r2zA

1 Coefficien. of linear thermal expansion5ij Kronecker delta = 1 if = j

=Oifi jStrein

f ftrain compcnent associated with stress compon2nt r.

C Strain duet to stresses

7? Ratio of bending stiffnessK Curvature - rate of change. of slow?

Ei;envalue assoviated with eigenvector ui and buckling load F.Constant - Lagrangian mulContan - agr~.nianmukip, ier

V Poisson' s ratio EA Radius of gyratiozr of cross !action /-

SStress t.th.Stress on I plane in j di ractioL

T h-%ear streGscr %bscript denoting critical value

WADD TR 60-17 9.3

9.1 STABILITY CRITERIA

A structure will deform under given loading and boundary conditions in such a manneras to make the potential energy stationary. As the load is increased, the structure deformsmore and more into one of its characteristic buckling patterns. The buckling pattern forwhich the strain energy is a minimum for given boundary displacements is called the lowesteigenvector of the structure and corresponds to the lowest buckling load (or eigenvalue).More generally, the eigenvectors are those deformation patterns for which the potentialenergy is stationary with respect to an arbitrary virtual displacement. Consequently, theeigenvectors also satisfy the Euler Equilibrium Equation. This equation Is obtained by acalculus of variation procedure which makes the potential energy stationary.

Any Initial eccentricities of the structure can be expressed as the sum of eigenvectors.As the load is applied and Increased, the magnitudes of the eigenvectors (which dcine the dis-ulacement of the structure) will increase. The percentage Increase (of the order of 1 )

1-F/F1

is greatest for the smallest eigenvector (corresponding to the smallest characteristic load Fi= F1)

and becomes exceedingly large as F -- F 1. Thus the structure will tend to buckle in the

lowest eigenvector mode no matter what initial eccentricities exist. (The eccentricity cantven be in the form of a amall lateral disturbance to a straight column.)

It is possible to Investigate the stability of a structure by examining the change In thepotential energy (SL = 5 U - 5W), defined in Paragraph 1.7.2. 1, due to an arbitrary virtualdisplacement which satisfies restraining displacement boundary conditions.

Consider an eccentric column (Figure 9. 1-1(a)) which is in stable equilibrium Linder theforce aad displecement syetems FV, u1. For an arbitrary virtual displacement 5t1 , the re-

ciprocal theorem. (Pa-lagraph 1. 7. 1) yields

"5L=O . (la)

"Further, since the column is in a state of stable equilibrium, then according to the minimumpotential energy theorem (Paragraph 1. 7. 2. 3),

S2L> 0. (ib)

Wheu Zhe lead on the column becomes nqual to the so-called buckling load Fcr (Figure 9.1-1(b)),+he column daflectione can vary over a wide range with only small variations of this load. Toall Intents and purposes, the deflections u2 are arbitrary and the column is considered to be

in a state ef metastable eautlibrium, for which

24L=53 L= = (ic)

Another. orm of metastable equilibrium, involving the complementary potential, occursif a stop is plac-,d under one end of the column (Figure 9. 1-1(c). I. this case, additional loadla absorbed ry the stop, and the foccea F3 for a given deflection u3 are abitrary, with the re-sult tiat

6L*= 2 L*= O. (1d)

If the column is initially straight (actually, this can never be achieved) and a load F4 , greater

WADD Th 60-517 9.4

9.1 (Cont d)

than F is applied (Figure 9. 1-1(d)!, a small lateral displacement 5u 4 will cause thecr4

column to buckle. The straight colum.n, in this case, is considered to be in a state of un-stable equilibrium, for which

6L =0(le)

52L <0.

The criteria of stability are summarized below,

6L =0;5 2L>0 (2)

Stable Equilibrium 2

Metastable Equilibrium: I(a) Arbitrary Displacement 5 L = 5 L 0 (4)

2(b) Arbitrary Forces 5L* = 5 L*= 0 (5)

Unstable Equilibrium 2- Deflections 6 L = 0; 5 L< 0 , (6)

2 2where 5 L, 6 L are variatiwas due to virtual displacements 6u, and 6 L*, S L* are variationsdue to virtual forces (or stresses) S F.

Employing Eq. (4) will result in the solution for the buckling load provided u is the

eigenvector. If u is not the eigenvector, then employing Eq. (4) results in a buckling loadfor a stiffer structure (6 U expressed in terms of displacements) which is higher than thebuckling load of the actual structure. Employing Eq. (5) results in a displacement for a morefiexdble structure (5 U* expressed in terms of the forces) which can be utilized to obtain abuckling load lower than that of the actual structure.

The first variation (6 L) can be viewed as the difference in potential energy between asystem defined by the true forces and displacements and a system defined by the true forcesbut assumed displacement (which is the true one modified by a virtual displacement). Thevariation (6 L) is zero (and the potential is stationary) when the assumed displacement happens

to be the true displacement. The second variation (6 2 L) can similarly be envisioned as thechange in potential for the system of the true loads and the assumed displacements when sub-jected to a similar virtual displacement. Thus the stability of a structure can be investigatedby assuming a deflection pattern and evaluating the change in potential of such a system whensubjected to a virtual displacement having the same pattern as the assumed displacement. Insuch a case, this change in potential energy 6 L is in effect a second variation upon the truepotential energy. Similar arguments can be applied to the complementary potential (L*).

The conditions for metastable equilibrium as defined by Eqs. (4) and (5) can be redefined

6U =6W (7)

6U* =W*, (8)

where S U is in terms of the assumed strain pattern and 6 U* is in terms of the assumed stress

pattern.

WADD TR 60-517 9.5

9.1 (Cont' d)

F2 =Fcr

Fu

/- Ou

6u1

Lt

u3 L 5 u4 +- u

u +6u.,F..t

u1 , Fu

u u

(a) Stable Equilibrium 0 Le d 0 (b) Metastable Equilibrium-tUndefined Dis-

f6 2 L. > placements 2{0L L 0}

FF 3 +6F 3 F F3P 3 3F4 > cr

u Z

31 1

F U.

(c) Metastable Equilibrium - Undefined Forces (d) Unstable Equilibrium-Straight Column

{ L* 6 0L* = 0} r 6L =0}

FIGURE 9.1-1 STABILITY OF COLUMNS

WADD TR 60-517 9.6

, -- - --- I--- -i--- ~- ~ ..a.

iI

9. 1. 1 Stability By Eigenvectors

The Euler Equilibrium Equation, which expresses the internal equilibrium of thestructure in terms of the displacements, is obtained by determining the condition that thedisplacements must satisfy in order to make the potential energy stationary. The methodof obtaining the Euler Equation is shown in Paragraph 9. 1. 1. 1. The solutions to these equa-tions are the possible deflection modes of the structure and are called eigenvectors.

The solution of the Euler Equilibrium Equation for the eigenvectors "u"11 contains an

eigenvalue "I." which is a function of the buckling load. Techniques for obtaining the buck-ling load are 'shown in Paragraph 9. 1. 1. 2. The buckling load depends upon the boundaryconditions (types of supports, aspect ratio) and the loading conditions. The exact solution forthe eigenvalues can only be obtained in a limited number of cases in which the exact solutionof the Euler Equilibrium Equation can be obtained. Solutions for rectangv'.ar plates withvarious types of boundary conditions and loadings are shown in Figures 2 and 14 to 26 of Re-ference 9-1. These solutions as well as solutions for other geometries and loadings can befound throughout the literature; for example References 9-2 and 9-3. This information shouldbe utilized in a manner described in Sub-section 9. 2 to determine the buckling stress.

In general the solutions are quite complex, and approximate methods are attemptedwhich place upper and lower bounds upon the buckling load. Properties of the elgenvectors andthe eigenvalues are described in Paragraph 9.1.3. Some approximate methods and the effectsof these approximations will be discussed in Paragraph 9.1.4 These methods utilize the energyprinciples discussed previously.

9. 1. 1. 1 Euler Equilibrium Equation

The method for obtaining the equilibrium equation for a rectangular plate, with trans-verse shear energy neglected, is indicated in the following. The potential is minimized by afunction defined by the equilibrium equation. Symbols are defined below and in Figure 9. 1. 1.1-].

th t

Stress component acting on i plane (plane perpendicular to ith direction) andacting in j direction

OSXX Normal stress on x plane acting in x direction

oShear stress on x plane acting in y direction

Cij Strain component associated with stress component oijijth tj

Nij Load per unit of length acting on I plane and in jth direction

N Normal load per unit of length on x plane and in x direction

N Shear load per unit of length on x plane and in y directionxyw Deflection in z directionq Lateral load per unit of area in z directionMij Moment per unit of length action on ith plane in the j x z direction

SMxx Moment per unit of length acting on x plane and in x x z = -y direction,causes rotation (bending) about the y axisMoment per unit of length acting on x plane and in y x z = x direction;

MXy causes rotation (twist) about the x axis

W, w Negative of curvature -

WADD TR 60-517 9.7

9.1.1.1 (Cont' d)

bz

- y

a orl

Nij = Load on i plane in j direction

y Plane

kN

X

a. yPlane

& y Plane

XY F

FIGURE 9. 1.1. 1-1 LOADS ON PLATE

WADD TR 60-517 9.8

9.1.1.1 (Cont d)

D.. Stiffness associated with the moment M and curvature w

Mxx

w + vfw,xx ,yy

MD Yy For isotropic plate

,yy xx E 2D =D =D =D S z dz

D =xyxy 11• ) ,xy

V Poisson' s ratioES Secant modulus

The potential is defined for non-linear elastic material in order to obtain a moregeneral equilibrium equation:

Iij

L=U-W=J ij diE dV - F•FU (la)

rffw'lj f f wd ,-qw} dxdy, (Ib)L= 1

or

L= ,xx Dxx(Wxx+ Y W,yy) d(w, xx)+ 2 ,VD3 (1-v)w ydW

1 2w yWyD y + W,xx)dc ) Nxx W,x (1c)

-0

1 2 1+ N w w + - N w, qw dxdy.iXY ,x ,y 2yy ,y

From the calculus of variations, if

L J b dxdy is stationary, then

S(2)$0

WADD TR 60-51? 9.9

9.1.1. 1 (Cont' d)

a2 . +_q' ( __ + (-1)n -- (2)

+ a 0w,11 " kw, ^""" 8 n 0V xx...x Cont'd

Therefore,

O= -q-(N= wx,, -(Nw), (N w ) - 3 w , ), +[ D+v Jw, +V w, ,

(3a'y+ [2(1-y)Dyw, xy],y + [Dyy(w yy+ w, )] ,YY a

Let

I = =lateral load

H = -(Nxxwx),x Ny y) Wy) - (N yW

= lateral contribution of in-plane loads when plate deforms (first derivative c' shear),

where

Ila -NxxW,xx - w,xy yy w,yy

= contribution for non-changing in-plane loads

I1b= -w, (N~ + N )w, (Ny + N )-,x ,x+ xy,y ,y ( ,y xy,x

= contribution for changing In-plane loads

1= Mxxxx y, + M

= effects of lateral contribution of moment, (second derivative of moment),

where

ffla= D w c + 2 D wx, +D wS=,xxxx + xy wxxyy yy ,yy

= moment effects for isotropic plate

IDb = VW D -2D +D]XXY Ixy xx xy Y

= effects due to Y (zero for isotropic plate)iic = [2D~xwx, W +DD w +D w + 2D w ]BI 2Dxx ,xy x ýwxyy + Dyy,y ,yyy

= effects of changes in stiffness and curvature

IHd= [Dmxwyyx-Dxy,y wxy-Dxyx wxyy +D Y, 2

= effects of v for changes in stiffness and curvature

WADD TR 60-517 9.10

9. 1. 1. 1 (Cont' d)

me = xx[W,xx+ 2D , W + D W,

= effects due to change in stiffness

f v[DmDwyy -2 DXYw , xy+ wyy,yy,xx

effects due to Y for changes in stiffness.

Then

0 =I + 1+11M (3b,

0 =I+ Ha + Ub +Ila + IIIb + Mc + Hid + IIe÷ +lf. (3c)

The linear elastic Isotropic plate with constant loads results in the simple form

0=I+ Ha+ IIa (4a)

-2N w -H w +DV 4 w0Y I 2 XN y yy ,yy

where

V~w= w + 2w +W= xxx ,xxyy W,yyyy

Eh3D 12(1--')

The problerm of a non-linear anisotropic plate results in an exceedingly complex equationwith non-linear coefficients and recourse must be made to approximate procedures.

9. 1. 1. 2 Buckling Load

The buckling load of a structure can be found by determining the eigenvalue whichpermits non-trivial solutions to the Euler Equilibrium Equation. The lowest load which satis-fies an eigenvalue is the buckling load of the structure.

The technique is illustrated for a pin-ended column of constant stiffness. More com-

plex illustrations can be found in various texts, including References 9-2, 9-3, and 9-4.

Exampl Simply supported column of constant stiffness (neglecting shear energy)

y and w

F - - F

FIGURE 9.1.1.2-1 PIN-ENDED COLUMN

6

• WADD TiR 60-517 9.11

9.1.1.2 (Cont'd)

Equation (4b) of Paragraph 9. 1. 1. 1 is used where D =EI, -Nx F & q = Nx=N =0 and deflection w is independent of y.

yy

The Euler Equilibrium Equation becomesd4 d2

El w dw =SEI 44 + F =~ 0 ,(la) •

dx dx2 a

or

d4w d2w =0 (lb)

&4 + 0. d2 bdx dx

where

__ __ __ _= _ (ic)

General Solution: A

w = (CI + C2 x) + C3 cosAx+ c4 sin J x (2)

Boundary Conditions (Simple Supports):

w(O)=w 0l=

d2w(J) =Wd~w(1) 0dx2 dx2

C .C1 = C2 3= and/X = nw/l (n is an integer)

for a non-trivial solution.

F 2 2nL E . . (Eigenvalue) (3)"n EI 2

.n2 12 El (Critical Loads) (4a)

2

FI = 1E = Buckling Load = Lowest Critical Load (4b)2

Successive eigenvectors and corresponding critical loads are listed in Figure9.1.1.2-2. Note that the deflection amplitudes wio --- wno are undetermined.

WADD TR 60-517 9.12

• • •-

9.1.1.2 (Cont'd)

W 2

2 Wi0 SfvrX22

F1 = El

21

W2 0 sin 'x

42r 2El IF =n 2 E l

FIGURE 9.1.1.2-2 EIGENVECTORS AND EIGENVALUES OF COLUMN

WADD TR 60-517 9.13

9.1.2 Stability by Virtual Work

It has been shown that when the structure buckles, it buckles in the eigenvector shape.For this situation, changes in virtual strain onergy (6 U) are equal to the changes In virtualwork (6W). The principle is again Illustrated for this column:

6 U = aij (Eij)dV d f1 -M6K dx=f EI d 6 d X

dx (1)6W= F6 u =F 1 2 (2a)L

aW= F6 •dx F 6 d(x .(2b)

2 2 dx dvO dx

f0 El -qj- 6 -•w dx =F f -- 6 dx . (3a)

o dx2 dx2 0 x dLet

w = w0f; 6w = wof

where

w= maximum deflection

andf = elgenvector of the structure.

Thenn

Then /01 f If01d=Fo f 2 dx (3b)

w0lw f2 fdxxx=Fw•0 ,

fH t•xxd=F =buckling load (3c)

Simtlaz1v, expressing the moment and change in curvatures for a pin-ended coluknn(M =-Fw, w(O) = wxp(s) =, we o m)

,xx ,xxThe following is obtained:

Eldw (d'/o 0I) \ )/ 6

_Lf xdx-•. . F = J o , x ?

f0 El dx (4a)

4.WADD TR 60-517 9.14 Jd

ZI

[9.1.2 (Cont' d)

and integrating by parts the following alternate expression is obtained:

Fo f f' ,xxdx ( b0 f1 f2(b

f dx 0(4b)

Note that this is true only if f symbolizes the eigenvectors. For a pin-ended

column sin T-•, which yields the critical loads

2 El12 , (Eq. (4a) of Paragraph 9.1.1.2)2

!lhe same critical loads would result if an initial eccentricity is assumed.

The eigenvector f (which is obtained by solving the Euler Equilibrium Equation)S~ may be exceedingly difficult to obtain. However, one can use the properties of eigenvalues

(Paragraph 9.1.3) or the energy principles to obtain anproximate solutions to the bucklingload.

9.1.3 Properties of Eigenvectors (Deflection Modes) and Eigenvalues (Critical Loads)

The following properties of eigenvectors and eigenvalues are of use in the determination

of the stability of structures by direct or approximate. procedures. The properties are illus-trated for a one-dimensional problem but can be extended to the two-dimensional case.

(1) An eigenvector is a deformation pattern compatible with the boundar3 .onditions of thestructure which makes the potential energy of the structure stationary, i.e., 16 1 u =0.

(2) The eigenvectors of a structure with homogeneous boundary condit'.ons are orthogonalto the curvatures of the eigenvectors. When the eigenvectors are orthonormal (weigh-ing function is unity), then

fui uj dx =

(3) Associated with each eigenvector is an eigenvalue which is associated with a buckling loadof the structure. The eigenvalues are non-decreasing for the structural problenms con-sidered.

(4) Restricting the possible (admissible) classes of functions which can be eigenvectors cannotresult in a reduction of the buckling load. This is equivalent to saying that additional re-straints at the boundary can only increase the stability of the structure. The less re-strained the boundary, the lower the buckling load, which increases as boundary goesfrom free, to simple supported, and to clamped.

(5) The addition of restrictions within the structure is equivalent to stiffening the boundaryconditions (property 4). Thus the addition of another support at the center of a columncan increase the stability four-fold.

(6) If two structures have the same cross-sectional stiffness and boundary conditions but thesize of one is smaller than the other, then the eigenvalue of the first Is never less than thesecond. Thus, the stability of structures with identical boundary conditions is never lessfor the smaller structure. A short column is more stable than a long column and a squareshear panel is more stable than a rectangular shear panel that has the same boundary con-ditions and smaller side equal to that of the square.

WADD TR 60-517 P. 15

9.1.3 (Cont' d)

(7) If two structures have the same geometry and boundary conditions but the stiffnesses ofone structure is greater at each corresponding point than the other structure, then thebuckling load of the first is at least as great as the second. Thus if the bending stiffnessof a column varied, the critical load for this column would lie between the critical loadsfor similar columns with constant bending stiffnesses equal to the largest and smallestvalue of the bending stiffness of the column in question.

(8) The u. eigenvector has J-1 nodes in the interior of the structure. The sub-division of ajstructure Into elements bounded by nodes permits analysis of the sub-structure withoutconsidering compatibility of deflections.

(9) A continuous (compatible) deflection can be taken as a sum of weighted orthonormaleigenvectors.

w = Mai u i where ai = fwui dx

(10) The orthonormality of the eigenvectors simplifies operations upon a function which isexpressc I in terms of the eigenvectors. Continued operations (iterations) causes theresulting function to be heavily weighted in favor of the extreme eigenvalue. If theextreme eigenvector is not known, then any arbitrary smooth function (it is assumed thatthe sought after eigenvector is the extreme component of this function) will, upon iteration,approximate the eigenvector. The eigenvalue, or buckling load, can then be found.

Let A be any non-singular operation where

_1-A ui = Xi 1 ui & Xi u A ui (la)_, d2 ui /F

(e.g., letA i dA7 , A = ui dxdxIN = L

Let w0 = z ai ui ,where w0 = initial approximation of the lateral defPo.ttion.

Aw0 =A Z (aiul) = M ai (Aui) = Z.(ai;i-') ui w,

2 -1 2(1b)A(Aw) A2 0 = a,2) uX = X

0 1aA (Au1 '21AnAw o (aii ui = Wn (n= 0, 1,2....)

wn = ai ,i-n u X _ln Zaifuilaiui(,i-nl-n)

Wn+l E a1 AX-(n+l) ui A1-n-1 1 ai ui + M at ui •i-n_ -n-1 (0c)

For the structural problems considered it is assumed that A1< Xi " Consequeutly,

Urn Wn A1

WADD TR 60-517 9,16

7 77 - ~ ,

S9.2.3 (ConL' d) -

Operating upon an assumed deflection pattern results In a calculated deflection patternwhich la closer to the true buckling patlern. If the assumed deflection •Aleru is tietrue one, then t%,, opeýatioa gives the elgenvector modiied by a constant An. = ; u-

Thus if w0 were exactly ai u, then

rO f• "le)w 1

wThe ratio n will vary along the length of the column (ercept when wA is an eigen-

Wn+ivector). This ratio will have a maximum and minimum value for different values of x.It is noted In paragraph 16 of Reference 9-2 that

"( min Max (f

A lower bound on the buckling load can be found by employing this principle. Thetechiique is to assume an initial eccentricity of the column w0 and determine the deflec-tion w due to an axial load F (w r x Fw dxdx with applicable boundary conditions)tio w de o n xil oa F o Jo -l w0

and determine the maximum and minimum ratio of w n. The ratio at the point of

maximum deflection usually results in a lower oound on the buckling load. Closerapproximations will result if the process is repa-ated with the new deflection pattern.

(11) An upper bound can be obtained by a scalar product method as follows since A- u. =

16ui, the following is obtained:

f (A- 1 up (u,) dx fr' ut1 dx=. X , = i (2 a)

f(ui) (u) dx fu , dx

Similarly, A W) w d. JiP A. a ui) (E a1 up dx

f(w) (w) dx f f(au,) E (ai ui) dx

f f(z a X. u) (Z 9: u.) dx. 1 1 1 .. U

Sf 1:(ai ui) E ai ui)&

2 =2 > X (2b)a ai Z a i

sinceai 2 (;,1 _Y 0 .f(A- w). w x > X(cf (w) (w) d - 1 (2c)

if -'l w d2w

dx XX

WA.DD TR 60-517 9.17

9.1.3 (Cont'd)

tnen from Eq. (2c)J _,x) (w) dx

w2 dx (2d)

Assuming any w, and operating upon the assumed function as shown in Eq. (2c), resultsin a value which is never less than the lowest eigenvalue. The evaluation of the eigen-value (buckling load) is relatively insensitive to the inaccuracy of the assumed deflection.

9 In both cases (10 and 11), equality results when the assumed value w is the eigenvectora 1 u 1 and the difference is small when w is approximately a 1 u1

-- • 9.1.4 Approximate Methods of Determining Stability•

The solution to the differential equilibrium equation may be quite difficult to determinemdirectly. Apprormate solutions are obtain from energy considerations. The physical argu-ments parallel the properties of eigenvalues noted in Paragraph 9.1.3. A deflection pattern

- (w) is chosen, corresponding to the given boundary conditions, and is hoped to be a good approx-imation to the first eigenvector. This corresponds to applying a virtual displacement to theattual structure. The solutions are presented for the column but the technique can be extendedto all structures. The stiffness (EI) is retained inside the integral sign to permit equations to beapplied to structures of variable stiffness such as occur when structures are subjected to variationsS:i temperature.

The deflection can be assumed as a single function or a sum of such functions. Theaisumed deflection can be employed to produce a upper and lower bounds on the stability byapproplate manipulations. A sum of functions with undetermined coefficients provides a means

.z whereby the upper and lower bound solutions may be made to approach each other as closely asdesired. The closer the assumed deflection curve is to the true deflection curve, the closerthe approximation to the buckling load.

9.1.4.1 Single Functions

Assuming a displacement pattern is equivalent to increasing the stiffness of a structureby adding additional restraints. The buckling force calculated by the Method of Virtual Displace-ments, where the strain energy is expressed in terms of the assumed deflection, will be greater

"* than the actual buckling force. Employing the Method of Virtual Forces, where the complementarystrain energy is expressed in terms of the actual force, would result in an overestimate of thedeflection (flexibility) of the structure and a lower bound on the buckling load.

9.1.4.1.1 Upper Bound

Assuming a compatible displacement pattern with the corresponding strain distribu-tion results in a larger estimate of the strain energy when expressed in terms of the displace-

N ments. The buckling load in this case will be an upper bound on the actual buckling load.

6U ? 6W

Let w = w0 f where f need not be the eigenvector.

Then f 2 dxP i 2 _> F. (1) '

J ,x dx

WAID TR 60-517 9.18

9.1.4.1.1 (Cont'd)

This equation is identical to Eq. (2d) of Paragraph 9.1.3 and becomes Eq. (3c) of Paragraph9.1.2 when the assumed deflection pattern is the eigenvector.

Similarly it can be concluded for the usual boundary condition where (f) (f ) isidentically zero at the boundaries (e.g., simple or clamped) that

f f dx ff2dx_ f, x > F. (2a) and (2b)

f lfdx f l dx

The inequality expressed by Eqs. (2a) and (2b) is usually smaller than Eq. (1) andshould be employed in obtaining an upper bound. The approximation of the derivatives of theassumed deflection pattern to the derivatives of the eigenvector pattern becomes more in-accurate as the order of the derivatives increases.

It is important that the assumed deflection shape should have no component of de-flection modes which correspond to the buckling of the structure with some restraints removedThe assumed deflection must satisfy all the boundary conditions (restraints) of the structure.This represents a structure with the same or more restraint than the actual structure and re-sults in an upper bound on the buckling load. Thus, if a column has a center support, then theassumed deflection must similarly have zero deflection at the center of the column.

9.1.4.1.2 Lower Bound

Assuming a force and stress distribution is equivalent to underestimating the stiff-nesses and overestimating the displacements of the structure. Employing the Principle ofVirtual Force and expressing the complementary strain energy in terms of an assumed stressdistribution results in a calculated deflection which is greater than the actual deflection. Thisinequality can be employed to obtain a lower bound on the buckling load. A closer bound isobtained when the maximum deflection is computed.

The hinged column is analyzed by taking a virtual force at the center of the columnand resisting it at the hinged ends.

tt4.

2 2

6 F

FIGURE 9.1.4.1.2-1 VIRT"AL FORCE ON PIN-ENDED COLUMN

WADD TR 60-517 9.19

9.1.4.1.2 (Cont'd)

The increased flexibility due to assuming a stress distribution results in an over-estimate of the deflection.

a U* (F)> 6 W*

K 6 M Ž w06 F1/2 w _ _1 F 5

S x 6F dx + w 6F.f/ EI 2 d x)0-o 1/2

Since w = w0 f where w0 = deflection at the center, the following is obtained:

2

F 2:11/2Ff, >_ I2--dx +S 2f(-x El dx AI

In general

F> 1 d (2)

f EI

where f is the assumed deflection mode = 1 at point of virtual load and where m is the virtualinternal moment on the structure due to virtual unit load.

Equation (1) is the ratio of the deflection at the center before and after the axialload is applied. The resulting deflection pattern is closer to the actual eigenvector than theoriginal assumed value. This deflection pattern can be obtained by integrating the curvaturew )a = w0 f, (i.e., f, = ff f, xx dxdx and fl(I/2) = 1) and employing the new deflection

mode in Eq. (2b) of Paragraph 9.1.4.1.1 and Eq. (2) of this paragraph, to obtain closer bounds.This can be continued until the difference between the upper and lower bounds is within acceptablelimits (the upper bound is usually the closer of the two).

The lower bound technique described above is mathematically similiar to the pro-cedure shown in Paragraph 16 of Reference 9-2.

Example: Pin ended column of constant El

4w0 (zero deflection at ends x = 0, 1 but withet w - 12 x (l-x) curvature at the ends representing some

moment restraint)4.-x

f 1 x (I-x)1 2

f' =(f//1) (l-2x)S~f = -8/l2

'xx'

- WADD TR 60-517 9.20

---- - - --- ----- ,.t

9.1.4.1.2 (Cont'd)

Upper Bound:

Using Eq. (2b) of Paragraph 9.1.4. 1. 1,I df 2(64 16/24 ) 2i^

ff, __x ______ _0-2x)2_ 13 (1 - 4/2 +4/3) 10EI

f2 6/14 1 x2tl-X)2 dx 1 512 Ff-fl- dxft2 dx (16/14 01 2-)2dEl E1l : " + 12 >F.

Lower Bound:

Using Eq. (2),

2 2 9.6EIp1/2 f1 2/2 < F.

4 xg-x)x 4 x(1-x) (1-x)dx 1 1 (2)Sx1xxdx 1--- - 4 64(212 E J 4 2 1

Ell 1 /212 E1

9.6EI 10.0 EI<-F <512 - cr- 12

2The buckling load = 1r El = 9.8697 El

12 12

Closer Bounds:

A loading of Fw1 = F 4x (1-X) , upon integration results in1 0

2 316x (1 2x + x

12 zO w20.w2 51 -- 1- 12 +1 w0

Repeating the above operations with the closer approximation results in

9.836 El <F < 9.871 EI12 - cr- 12

Closer bounds are more probable if the assumed deflection pattern matches theboundary conditions' identically. The upper bound procedure is a Rayleigh-Ritz method inwhich a function with a limited number of undetermined coefficients is assumed. The coeffi-cients can be uniquely determined by satisfying the boundary conditions identically. Thefunction

31 0 6 5 1

satisfies the requirements of zero deflection and moment at the ends x=0, ! and would resultin upper and lower bounds which are quite close

WADD TR 60-517 9.21

FA

9.1.4.2 Multiple Functions

The closer the assumed deflection pattern is to the true deflection pattern, the closeris the approximation (see example in Paragraph 9.1.4.1.2). It is convenient to take the assumeddeflected shape as a sum of functions, with undetermined coefficients, which correspond to knowneigenvector solutions of the equilibrium equation for structures which are similar to the actualstructure (e.g., slightly different boundary conditions). Since the eigenvectors are orthogonal,the energy is easily calculated as a function of the squares of undetermined coefficients (ai 2 ).

The coefficients (relative weight) to be assigned to the functions are determined by making thepotential stationary with respect to the undetermined coefficients )= . If the energy is

expressed in terms of the displacements and the functions satisfy boundary conditions which arethe same or more restrained than the actual structure, then the resulting solution must be anupper bound. (See Eq. (2a) of Paragraph 9.1.3). An adequate number of terms representing thedeflection, each with less restrained boundary conditions would result in a lower bound. If theindividual functions do not satisfy the actual boundary conditions, then a constraint (LagrangianMultiplier) relationship can be imposed among the coefficients (in additional to &_L = 0) so that

Oa.the total assumed deflection (Za u.) satisfies the boundary conditions. It is shownlin Reference9-4 that a termwise matching oft;Ie higher derivatives (e.g., slope w x) with an over-all matchingof the lower derivatives (e.g., deflection w) results in a more rapid c'onvergence (closer bounds).Some methods of obtaining an upper bound by assuming functions which satisfy boundary conditionsof equal or greater restraint are known as the Rayleigh-Ritz and Galerkin Methods. The upperbound is the solution of a structure which is stiffer (more restrained) than the actual structure.

The "Rayleigh-Ritz Method" is an extension of Eqs. (1) and (2) of Paragraph 9. 1.4.1to multiple functions where the coefficients are selected to minimize the potential energy. The"Galerkin Method" is an application of the principle of virtual work. If the lateral displacement

nis w = Z2 ai ui, then the stationary property of the potential (6 L = 0) requires that the virtual

i=1energy of the virtual lateral displacement 6w = ui 6 a acting over the equivalent lateral load as

1 I 4 2expressed by the equilibrium equation, Eq. (3) of paragraph 9.1.1.1, (I. e. DV w - NV w - q = 0)must be zero:

(DV 2 4 awui) - N - (V2 aiui) -6 q )ui dxdy 0. (lb)

Equation (lb) is a set of n simultaneous equations which determines the best value of the coeffi-cients of the n functions for obtaining a close upper bound. The Galerkin Method usually resultsin a closer or more rapidly convergent upper bound.

The Trefftz Method obtains upper and lower bounds by assuming deflection patternswhich represent less restraint at the boundaries. Selecting coefficients which make the potentialstationary and simultaneously satisfy the given boundary (constraint) conditions results in a com-puted upper bound upon the buckling of the structure. Ignoring some of the boundary constraintconditions results in a new computed stability which is less than the originally computed one.Care must be taken with the Trefftz Method since a sufficiently large number of functions must beemployed before the lower value can be assumed to be a lower bound. The deflection pattern mustbe fairly accurate or it is possible that the computed buckling load with the less restrained boundarymay overestimate the stability. Employing only a few terms may overestimate the internal energywithin the boundary to overcompensate for the lower energy in the region of the boundary.

WADD TR 60-517 9.22 2

= IN

9.1.4.2 (Cont'd)

The computational techniques for multiple functions are illustrated in two very simpleexamples. These techniques, though powerful, cannot be shown advantageously in the simpleproblems presented.

Example of Rayleigh-Ritz Method - Column of non-constant stiffness with pin ends:

El ='iD, 0 <x<1/2 where <l1El D, l/2_<x <I

F L x 2

FIGURE 9.1.4.2-1 PIN-ENDED COLUMN WITH VARIABLE STIFFNESS

(2n-1) 7xAssume a deflection w = Z a2n- 1 sm 1

n-i

dw n a( (2n-1)irxThen 2=;W = -- (2n-1) an 1 cos 1n=1

d2w T 2 CoD 2 (2n-1)d W,xx Z (2n-1) a l s 1

n=2 1 I

Each term of this deflection pattern satisfies boundary condition

w(o) = w(1) = ow (o) = w () = o,xx , xx

and is a eigenvector for columns of constant El. They are orthogonal in the region 0 to 1/2.

For an elastic structure

oror 1 4 1 4

4 2 .2 (2n-l)irx 1 4 4(2n-l)4 sxSL=J a 2n-1sm 1 dx 2 1 D 1w an(2n-1) a 2n-1-1) x

0 ~Dj~(~1)f1/2 14a 1 sFp1 2 22 2 (2n-1r

F I 2 (2n-1) 2a2 cos -)a dx- 2n-1I1

00 1

S~1/22L j D= -_1-)4 a si {n -)x dx + (.9_1, D 4 (2_1,4 a sin2 2--- dx

14 a2n_l 1 I af142n-I1

- 1 - (2n-1 2) a_ Cos2 (2n-1)xl dx

WADD TR 60-517 9.23

9.1.4.2 (Cont'd)

Since f (2n-1)rx 01 2 (2n-1),x dx=1/21icesin 2 cods = cos

0 0

1/2f/sin2 (2n-1)x dx=1/44 4

4 1)4 2 41 42SF 2 Z (2n-1) + (77 -1) Da- z (2n- a1

1L 4 V + 21_1) 2 14) 2 2-12 2

2r F I-Z (2n-1) 2a 2

d 2 2n 2212

and 2 81 - D 2 .7 (2n-.1) 4a 1 - 1 F (2n-1) 2a =0;fta2n-1 13 D 2 a2n~l-1 I-- F2n-1 ,

2 I kD 1+77a 2n-= 0 and F =(2n-1) 12 2

F min occurs at n = 1.1r2 17+ and 0 =a3 a5•1 2 D 2 3•ndo =a =a a 7 . . . . .

"1he calculated critical load is the same as that for a column of constant averagestiffness which is an upper bound to the true solution with increasing inaccuracy for smallervalues of 17 . Ignoring the even eigenvectors (for computation simplicity) is justified by thelower internal energy level of the first eigenvector but permits only symmetrical bucklingmodes. The error is an overestimate of the stability due to the fact that the actual deflectionis not fully described by the odd eigenvectors alone (when7l J 1.)

Example of the Trefftz Method - Clamped column with constant El, determination ofupper bound:

SF x F

1/2

FIGURE 9.1.4.2-2 CLAMPED COLUMN

Let w = a Cos (2i-l)irx

Then w. -- (21-) I--- a1 sn (2i-l)wx

WADD TR 60-517 9.24

g• .•- --- _ _, ,_ _ _ _ -- -• . . ' ,2 •'• :" ' --- •7 " :;2 "f • !'••••,• *

9.1.4.2 (Cont'd)

2 7r (2i-1)ixw -Z (2i-1)2 _l. a. cos 1

The boundary condition of the displacements are satisfied identically, i.e., w (/2) = w (-1/2) 0.The requirement of zero slope at the ends can be satisfied for the upper limit as follows:

0 = W,x ( w,x =( (2i-1) a, sin 2 Z (-- 1) (21-1) a1 .

"Z (-1) i (21-1) a i = 0

is the constraint condition on the coefficients which ensures an upper bound.

Determine the coefficient (ai) by making the potential stationary but modify the

potential by a term which is identically zero and represents the constraint condition on thecoefficients. The technique is known as the Lagrangian Multiplier Method. Thus,

f d 2w 2 )2]2LXE I -F dx- L Z (-1)i (2i-1) aiJL f I x /l F )d ]

where p is a Lagrangian Multiplier.

14E 2 i cos2 (2i-1)x d2L= El• Zai (2-1 cs I

2 2r i2(211 ai a- F -"- •a (2i-1) jsin2 (2 )•x dx - Z (-l) (2i-1)12

Er4 F20 = 2 -i-L= E (21-1) a1 - (2 1- 1) a1 - i (-1) (2i-1) 4

Eli 33 Fi

-- (21-1)3 F- 2 (2i-1)13

0 = Z (-1) (21-1) a.

0I

v 2

E r 12 (2i-1) - k1 2

WADD TR 60-517 9.25

9.1.4.2 (Cont'd)

where

k F F = ratio of buckling load to Euler buckling load.2 FEIV E12

Solving the above equation for n = 2, 3, 4, 5, ... gives approximate value F 5 FEt 4.63 FE,4xý2 El

4.45 FE, 4.35 FEl ... which converges to F = 4 FE = 2 EI

More rapid convergence would occur if the slopes --at x /2 were satisfied

identically by employing an odd sine series and a constraint condition on the deflections

a sin (21-1M = 0). This would result Ini I n 1

1=1 (2i-1)2 [(2i-1)2- k]

9.1.5 Shear Energies

Ignoring the strain energy due to shear is equivalent to overestimating the stiffnessof the structure and results in higher calculated buckling load. Ignoring the axial strain energydoes not affect the stability calculations if the axial work of external loads is ignored. Theseconclusions become obvious upon examination of the potential of the column in which the trans-verse shear and torsional energies should be considered.

When a column is axially loaded, it deflects laterally due to initial eccentricities.The slope (wx) of the column gives rise to a transverse shear load as shown in Figure 9.1.5-1.

The equivalent transverse shear load (V) acts with the equivalent axial force F at the centroidof the effective areas of the cross section. Twisting due to a torsional moment Q will occurif the transverse shear vector, acting in the direction of the lateral deflection, doeb not passthrough the shear center of the cross section. The loads acting upon the cross section can bedefined as follows:

M =-Fw (la)V Fx (lb)

Q eFw (1c),X

Examining the change in potential due to a virtual displacement,uL 6U-6W=6UM+uUV+uuQUA-1(rWM Q+ v)-+WA=O, (2a)

Q = A - "M + 6WQ + 6WV )TT- 6OW +'

where

6 UM = change in bending strain energy =JM6 (K) dx

6UQ = change in torsional strain energy =fQ5 J-6-G) dx

6 = change in transverse shear strain energy f iv6 V• ) dx•UVG

OUA = change in axial strain energy = JFC (A- )

WADD TR 60-517 9.26

I

9.1.5 (Cont'd) 1

F

Ix Restraint

11

jwElFI'4

F -•

Ftan E) Fw, xV

Centroid of Area

Bendi~ng L_ C iAxis •x •

Shea >

C e n te r e V ,w

;A

FIGURE 9.1.5-1 SHEAR AND TORSIONAL LOAD ON A PIN-ENDED COLUMN

WADD TR 60-517 9.27

9.1.5 (Cont'd)

6 WMQV = M + 5WQ + 6 WV= change in potential energy of the external loads (assuming in-

extensibility) F 6[f-L (-w) dx6 WA = change in potential energy of external loads due to axial deform-

ation of the cross section F 6 f (x,

ande = distance from shear center to line of action of the transverse

shear VM = moment acting on cross section =-FwV = transverse shear load on cross section = F -

Q = torque acting on cross section = e F

EI = bending stiffness =,f ES z2 dA

JG = torsional stiffness = GS r 2 d

A vG= transverse shear stiffness = Gs times effective shear areaE= secant modulus = - aT

S 0"

The changes in axial energy are exactly equal since

5 W dx- F 6J du dx _ ]F6 f dxdx =0. (2b)

Equation (2a) becomes

6 UM +6UV +6UQ- 6WMQV = 0. (2c)

Substituting the equivalent expression containing F into Eq. (2c) results in

f{Fwo(-- -)+Fwx6 ( P Av--)+eFWx 6 e ,x -

F w (f Iewdx - F W 6 w dx 0.El + w,. 6,x A ,G ,GIf X x

"f x _+x3

F (3a)

c f2dx f e + 24 + 4- dx

Fcr 2/ e d), e2 d 3b)

f 2 dx f dX , x

The bending solution alwa• s overestimates the stability of the structure since itassumes infinite shear stiffness AvG and JG. Comparing stability Eq. (4a) of Paragrah 9.1.2

WADD TR 60-517 9.28

I

9.1.5 (Cont' d)

with Eq. (3b) above indicates that the true stability is smaller since +-. + e is a non-2 AV G

negative term. If 1 + e is constant, andAVG JG

V rf 2dx

= = buckling of column with infinite shearM f2 stiffness,

dEI

then

F (4a)cr 1 1 +e2

F A G JGM V

or 1 +1 + e2 1 + 1 + 14b)Fcr FM AVG JG =FM F F•F

and F

F -M(4c)

cr F FM +4eM

F-V F Q

This Is analagous to springs in series where F is the effective stiffness of the thre- springs andC EI

FM is the stiffness of the bending spring =12 FV is the stiffness of the transverse

shear spring = CvAvG. FQ is the stiffness ot the torsion spring CQ JG.2) e

CM (EI/I 2CMFE=+Cj E (5)

+ +C + CQ JG I

If any spring is very flexible as compared to the other springs then the effective stiffness of thesprings in series is approximately equal to, but slightly less than, the stiffness of the very flex- 4ible spring. Thus when the shear and torsional springs are very stiff, the bending solution pre-sents a good upper bound approximation to the stability of the structure. Similarly, if a structureis relatively weak in shear (e.g., a sandwich construction with a soft core), when the value AvG

represents a good upper bound upon the stability. The torsional effect is usually insignificant for 31closed sections (large J) or sections symmetrical about the axis perpendicular to the bendingaxis (e = o).

9.1.6 Plasticity and Eccentricity

The Euler equation becomes non-linear when the bending stiffness becomes a functionof the applied load N and the deflection w and its derivatives. The lateral deflections w increaseswith load, and the moment acting on the section increases at a much greater rate than the load.Instability occurs when the rate of increase of external moment becomes greater than the struc-ture can generate by an increase in internal moment (bending stiffness times curvature). Theload and moment on the cross section tend to reduce the bending stiffness when the stressesexceed the proportional limit. The computed stability of the structure with elastic propertiesmust be an overestimate. The effect of the plasticity of the material can only reduce the stabi-lity and can only be approximated (References 9-1 through -6) at the present state of the art.

WADD TR 60-517 9.29

9.1.6 (Cont' d)

Preliminary investigations (References 9-1 through -8) have indicated that forstructures of small initial eccentricities, an upper bound on stability is obtained when thestiffness is computed assuming an effective modulus of the material equal to the secantmodulus (Es) of the mean stress on the cross section. A lower bound is obtained when the

effective modulus is the tangent modulus (ET). When the initial eccentricity is large, how-

ever, the stability can be below the tangent modulus load. Rvferences 9-5 and -6 consider thecase of plates and columns of small initial eccentricities where the bending stiffnesses (Dxx,

D ) are assumed to have an effective modulus of ET and the twisting stiffnesses (D = D ) areyy Txy yx

assumed to have an effective modulus of E5 . The relative weight of the moduli is determined by

the boundary conditions and the resulting (assumed) deflection patterns (w, ii and w, ij). The

solutions are dependent upon an assumption of isotropy of the material defined by an octhohedraistress-strain law (Section 3) and a constant Poisson ratio of .5 (which slightly overestimates thestability). The stability criteria obtained in this manner are presented in non-dimensional formin subsection 9.2.

The effect of eccentricity is quite pronounced since it initiates plasticity at an earlierload and causes the bending stiffness to decrease more rapidly. The bending stiffness is reducedby the high stresses and the shifting of the neutral axis. Both these effects reduce the stabilityof the structure. The eccentricity cap. cause large deflections of the structure which can changethe initial geometry and cause rapid buckling towards a lower buckling mode as exemplificd bysome shell structures.

The effect of curvature on the stability is qualitatively described below. The quanti-tative effect of initial conditions upon the stability of the structure remains to be determined.The effect is discussed for a pin-ended column but can be applied to most structural problems.

For a pin-ended column M = EI d-w = El w and M = - Fw. As the load increases,dx2 , xx

the change in internal moment 6 (EI wx) must equal the change in external moment 6 (Fw).

EI6 w + w 6(EI) = F6w + w6F. (la)

When the column buckles, 6 F = 0.

2.FwE ~ XX (EI)FM = EI 5w +,xx ow

6 w 6 w=El -.- x + w 6 (EI) ___

6w xx aw 6wF EElw W 6w,5WIb)S6wFM EI + ,x6EI )5w'xxw

FM ' I•w 5w 6 (ib)•.

For ta independent of the loading it can be shown that

EI fl dx Ei(x) 6wF - ____ - x

M f2 w (2d)

Thus, for an El which varies with fhe loading, a comparison of Eqs. (ib) and (1c) leads to

WADD TR 60-517 9.30

9.1.6 (Cont' d) El1w f, d

f dFM =

where w w~f 61/

( D Employing D for El generalizes the results. When the structure is linear elastic- the classical result is obtalned. If the structure Is not linear elastlc, then

D mustbe modified by the fcretor AD = W D whiichi is always negative. The value- of AD,xx a(W) 'xxdepends upon the curvature and bonding stiffness, which in turn are dependent upen the load leveland the eccentricity.

D is reduced as the material Is streseed beyond the proportional limit; this rred~aces theeffective modulus and shifts the bending axis. If tbe eccentricities are small, then the expressionin the bracket (D) + AD) can be approximated by moduli corresponding to the averagge stress (e.g.,E7 Is a lower bound for a coluimn), If the eccentricity 's not smiall (this may occur due to thermaldeflections), then the stability may be below the "Tangent Modulus Stability."

7A' Temperma.reS

The -~ipplicatioit of hernt to a structuxu reduces the stiffness of the structure, causes de-formations, end may cause com~patibility forces at the boundaries and internal stresses. 1theseeffects in general will reduce the stability of the structure.

-< For thermal lohds it Is essential to define the axial stiffness, as well as the rotationaland lateral ,Wifnesses. of the salprorts sin;,, the temperature w111 try to Make thE bapports moverelailve tu each o~ther and will generate boundary forces to satisfy compatibility at the bound'ary,these compatibility forcee will affect the stability of the structure.

if the 9xia~l and rotational stiffnesses of the supports are zero (unre3trained), then thetomperature reduces the mioduli of the mvi~e:rtal, chavps the initial eccentricity of the structure,and may generate internal stresses to enforce the interaal compatibility condition that strains Ina cross section be. In a plane.

Calculation of the buckling load is quite simple if it is assumed that the atructure re-wains elast',v up to the buckl~rg load. in taat case the bending stiffness is simply compute-d at

-r each cross section where the modulus varies (date io temperature) at every point in the crossseetioxv (Paragraph 4. 1.! illastrates tecmuaintechniquB.) Th1e over-all stability Is then

(egFM = M -=M f 2

* The result will be lower than for the structure without temperature since the bending 3tiffne3s Isreduced The effects of additinnal eccentricities and internal stresses cannot be evaluated at thepresent Lime but the probablo eiect will be to further reduce the stability. For small eccentri-cities, the strebs-atrain curve of the maeila emoauecnbemp0ydt poxmeth

* ~~~effect: of plasticity. mtra ep'aue~ eeposdt h

WADD TR 60O-517 9.3i

9.1. 7 (Cont' d)

If the axial and rotational stlffnesses of the supports are not zero, then compatibilityforces and deformation will be generated which may limit the usability of the structure. Ile

determination of the compatibility forces and deformation is quite complex. For example, acolumn might be visualized as an axial and a bending spring in series. Compatibility is satis-fied by a combination of axial strains and lateral deflections. The amount of axial strain (andload) depends upon the eccentricity ratio (wo/p) and the slenderness ratio (l/p). If theseratios are very small (e.g., w0 = 0 for a straight column), compatibility will be satisfied pri-marily by axial strains. In this case the buckling temperature of the structure can be calculatedby computing the temperature which will cause the equivalent buckling strain as defined inParagraph 9. 2. 1, i. e. ,

J 0 T dx ,

where the above parameters are defined in Paragraph 9.2.1. the structure is never perfectlystraight and the bending stiffness is usually less than the axial stiffness. The non-straightfetly

Sstructure satisfies compatibility at the !3oundaries by deflecting laterally (e. g., u f= ý w x

whenever the temperature increases. This is accompanied by an increase of load which is muchsmaller than would occur with a perfectly straight structure. The change in load is a function ofthe initial eccentricity. The non-straight structure never actually buckles but continues to de-fleet. Snap buckles are probably due to a change of the deflection pattern from the thermallyinduced pattern to the buckling (eigenvector) mode.

Depending upon the initial boundary conditions and geometries, there is a temperaturecorresponding to a given maximum deflection but this is essentially different (and higher) than abuckling temperature corresponding to the temperature which will generate the mechanical buck-ling load in a perfectly flat structure. The problem is graphically represented in Figure 9.1.7-1which indicates that the boundary load may never attain the mechanical "stability" load and thatexcessive deflections which violate the original assumptions or destroy the utility of the structuremay occur well before a "stability" load Is attained. The general problem is complex and a ex-position of analytical techniques that consider compatibility forces, deflections, eccentricities,and plasticity must be deferred.

9.2 Non-Dimensional Buckling Curves

The present state of the art indicates that sufficiently accurate stability predictions canhe obtained by utilizing the stress-strain curve of the material at the given temperature with somecombination of the secant modulus (representing twisting) and the tangent modulus (representingbending) as indicated in Paragraph 9.1.6. The immense number of structural materials andoperating temperatures makes it imperative to employ a non-dimensional stress-strain law andstability criteria. The non-dimensional stress-strain law is presented in Section 3. The non-dimensional stability criteria is developed below and could be extended to include, as a firstapproximation, the creep buckling of a structure.

9.2.1 Non-Dimensional Stability

The stability equation can always be presented in the following form:

KE I_ RN = !a-• = b2 I 1

- which defines the condition of stationary potential for adjacant deflection patterns.

WADD TR 60-517 9.32

It I

Fr =F F

-w00 0

11FB

F3

•-wlO

•/" 00=0 Wl0 30

W o 1-12ý T

FT

1wo

NOTES: w 0 = Initial eccentricity

Wcr = Maximum allowable displacement

cr

FIGURE 9.1.7-' DEFLECTION AND LOAD IN A RESTRAINED COLUMN

4WADD mR 60-517 93

9.2.1 (Cont' d)

a= = K ER (2)

E

E E ES (SE 2 - 3R S R R Ab

where

N = axial load per inch of widthK = stability constant defined by linear elastic parametersI = inertia of structure per Inch of widthb length of buckle (length of column, width of plate, etc.)ER = effective stability modulus

ES secant modulus

EA = Apparent initial modulus

A = area per inch of widthD = bending stiffness per inch of width.

The function on the right if, a function of the non-dimensional K, (which is a functionof boundary conlitions, type of loading, and the aspect ratio) and the non-dimensional geometryparameter 1/Ab (which is equivalent to the slenderness ratio for column and thickness ratiofor plates). It Is independent of the material. The non-dimensional expression on the left Isa function of the stress-strain curve of the material, the magnitude of the stresses and theboundary conditions (and to some degree on the geometry of the cross section). Values ofEs,!ER can be found in References 9-1, 9-3, 9-5 and 9-6 and Table 9.2. 1-1.

Since the secant modulus is an upper bound on the stability (with the possible expectionfor creep at low stresses),

ES

ER

E _ K I/Ab2 (4)

Thus, the median strain of the cross section is never greater than a number whichdepends upclý 'oundary conditions, type of loading, aspect ratio, and geometry but not upon thematerial.

The variation 6w can be viewed as a time-induced phenomenon which does not alterthe equilibrium equation or the relationship (Eq. (3)) derived from energy considerations.Creep buckling can be investigated by employing a strain relationship which includes time aswell as stress and temperature. Instability can be approximated by whatever stimulations ofstress, temperature and time applied to the material will cause this critical strain. A strainrelationship which includes the effects of temperature and time as well as stress was presentedin Section 3 and was employed with Eq. (3) to obtain the curves shown in Figures 9.2.1-1 through-5. These curves should give approximate results even when the structure is subjected toelevated temperatures for extended times.

Figures 9.2.1-1 through -5 are obtained in non-dimensional form

WADD TR 60-517 9.34

9. 2.1 (Cont' d)

:4 -4

m0

x~ x A,

0 -rW I,

4-4

00

bibo"i t b

WADD TR 60-51793

9. 2.1 (Coatl d)

C4 0 0

- - - - -- x N -

CA,

, l

bl b> 0 4

-D -R6-1793

9.2.1 (Cont'd)

oL

cc))

I LI

ILJ

4N A

WADD TR 60-517 9.37

9.2.1 (Coattd)

Goo

cQ-

04

I--

L LF-

WADD TR 60-517 9.38

9. 2.1 (Contf d)

U) so X

w_ < t' 0C0

IF-

CzOI CI n t UI c

to1

WADD TR 60-517 9.39

9.2.1 (Contf d)

0I T I

0 --

TIF ° ill - I •ii

X X

-- @ --- i--.

INKI- II

.iq _ co a

- I

'1'N

bl b

WADD TR 60-517 9.40

9.2.1 (Cont'd)

x VI

-~ N0

"I ki

I lk L I t I -

ALz

ix I

'aN Ji

U13 C-2 . 0) m' 1) C CY U)- m~ JN'. N C'J '4 - -. -

WADD TR 60-517 9.41

9.2.1 (Cont'd)j

I 0L I

030

_l 04i 04~b~bo

WADT 05794

9.2.1 (Cont' d)

from E 0 5)

R '0 00 'Ab'and the non-dimersional stress-strain law,

E A = (1 - , a/a0 + 6 sinh a/a (Eq. (2a'of Paragraph 3.2. 1),

010 0 00in the following manner.

(1) The effective modulus ER depends upon the boundary conditions and type of loading

and is expressed as a function of the secant (Es) and tangent moduli (ET) as tab-

ulated in Reference 9-5 and Table 9.2.1-1.

(2) The stress-strain law of the mater!al is assumed to be characterized by the threeparameters EA, o0, and P; which can be obtained from simple unisxial short time

A'E. /\0and long time tests as described in Paragraph 3.2.2. The value ef E.

E R 0

Is computed for various values of o/or0 for a given P; this value is ( 0-0

f3 ( plo of E A

(3) A plot of E versus a,'a for various P is shown in

Figures 9.2.1-1 through -5, The curves are then employed lo obtain the stabilityof a structure as illustrated in the examples shown below. In Figures 9.2. 1-1through -3 two graphs for each effective modulus are presented to improve theaccuracy in reading the curves.

(4) The values of the critical strain parameter IKI/Ab 2) is defined below for varioustypes of constructions.(a) Column 21 - 2c ) Z

a 1- 2(1+i)7r2 C () V + ()

where Cc = end fixity of column

P = radius of gyration = 1 -A1 = l3ngth of column5

(b) Plate or Shell _1-2k2)

12 \_2 bJ2 2 (7)

P I+

where k = stability constant evaluated in various texts,such as References 9-1, -2, -3, and -8.

t plate thicknessb = width of plate

WADD TR 60-517 9.43

9.2.1 (Cont' d)

(c) Sandwich Construction 2 )

- 2 (fEf i 2 (8)\Ab 2 1 x 2k f E h

2(1-Y)2 khc G/

where k = stability constanth = centroidal distance between facesf = average thickness of faces

E modulus of facesh = thickness of core

Gc = shear modulus of core

TABLE 9.2.1-1

EFFECTIVE MODULUS OF PLASTIC STRUCTURES(References 9-5 and 2-6)

Unloaded Ends Type ERt/E A

Unaxial Load

Free - Free Column or long Plate YEA

Short Plate (Es/EA 1/4 + 3/4 (Er/Es)]

Square Plate (Es/EA [.114 +.o88oi (XT/E.s)]

Simple - Free Long Flange ES/EA

Clamped - Free Long Flange (EsEA) [.428 + .572 /1/4+ 3/4 (E/ES)1

Simple - Simple Long Panel (ES/EA) [1/2 + 1/2 /T/•-•4 3/4 (E ]Clamped- Clampled Long Panel (Es/EA) 352 +.6• 4 3/4 (E,/Es)

Shear

Simple - Simple I Shear Panel Es/EA

Clamped - Clamped Clamped &dear Panel Es/EA

S657.

WADD TR 60-517 9.44

9.2.1 (Cont' d)

The use of the non-dimensional buckiing curves is illustrated in the followingexamples

Example I - Column with Shear and Twistint.

A sheet stringer whose cross section ie shown in Figure 9.2.1-6 is assumed to behinged at the ends and forced to bend about the y-y axis due to the planar stiffness of the sheet.

Shear A (1.2+1.2) 1= .24iiCenter =x|AV (1.2) .1 = .12

I (1.2)(.-)(.3) + (1.2)(.1)1.3) + 12 •

Bending = .0108 + .0108 + .0144 = .0360-c•. 1.Y 1 .13

C-jIf . 1 J -9- (1.2 +1.2) (.1) 0008.1--• -- ] 2 .0'>

P I/A = . = .15.24

1 31.62 1 1000

S1.0 .3

FIGURE 9.2.1-6 SHEET STRINGER CROSS SECTION

i •Stati:, Buckling :

Fromi Ea. 6

A + .. Av + -2--

•- •s;- d•/.0015

F1i[(.5 ;L4 P3N .036 t1ec u 0078+.105 "01351 +2.6 ( • • •• .o•:

"The above calLuli/ons a:.e independent of the material and depend up-on the geometry.alone. t n the above etal, ple,dtheh the sanlvers e te r by 1t%. t hi stability pai a, c .a r;•: ~bu;, the torsional shear rF-duces the otability pp-rametp-r by 10%. This is an exceptional case '

• svince the cross section, is usualiy symmatt'zcal (e=0) or closed (lazge J).

;, Te iollowinI ma~terial nropertier, can be assumedi, i h ntr s22S8 l

uminum alloy aý 30C'F. '

E 10 ns 1, ; V --•; (T = 5 (0)• Psi /I= .ooo,

IVA.DB TIR 63-517 9.45

9.2.1 (Cont d)

From Figure 9.2.1-2(a) or -2(b)

a/% = 2.7

The stress Is linear elastic since

S= EA KI

0 o Ab

The buckling stress ar is obtained as

•r • (2.7)5(10) = 13,500 psi.•cr a 0

Creep Buckling:

Equation (6) is modified to Equation (6a) to account for the smaller shear energyunder the creep load. This modification is usually quite small and in view of the inaccuraciesof creep buckling the denominator can usually be assumed to be equal to 1.

Ki ________w (OP.), (6a)

=Ab - A 0 21" •+1) A) -L ] F/Fc.

where F = applied creep loadF = short time buckling load.

Let F 1620F 1620

= .24 6750psi

e= .000675 (from stress-strain data)

and.0001 in/in hr (assumed creep rat6)

.00156 0 ..00 105 6750 = .00143

1+ .0078 +. 13.00

t - .00143 .000675 = 7.55 hours.tcr C .0001

WADID TR 60-517 9.46

9.2.1 (Cont'd)

Example 2 - Simply Supported Sandwich Panel (Figure 9.2.1-7):

f-. 020

- t=.0015

b--10"a=2011

FIGURE 9.2.1-7 SANDWICH PANEL

Assuming 2024ST81 aluminum alloy faces and core at 300 0F,

17 5(03,EA = 10 psi, Z =.3, or= 5 = .0001

Ef d .25

and 2(1+z/) -- 2(1+.3) =0435 (Reference 9-9).c .0015

Compression Loading (k 4):

From Eq. (8),

i k (124 1.3 2

2 b2) 10o;~F KI4(1-7/2 4(1-.-32

2 2 2fEf +2= 2 - 3 2Ab 1+ 7r k 2) _r 4 2(.020)435 /i

4(1- ) hcGc 4(1_.3)2 .3

.0098= 1 + .0098(58) = .64(.0098) = .00626

The shear deformation of the core causes a 36 percent reduction in the stability strain.

EA 107a' \A b 1 5210) (.00626) =12.52.

WADD TR 60-517 9.47

9.2.1 (Cont' d)

From Figure 9.2.1-3(a) or 3(b).

= 10.05 (Plastic since c/o 71% A H2I

ar (al/%) a (10o.05) 5000 = 50,250 psi

N r =(2f) a r =.040) 50, 250 = 2010 ]b/in.

4.0

Shear Loading (k - 5.34 + - = 6.34) :(a/b)

From Eq. (8), 2 (3\2(6. 34)

KI 2 10oS._ = 4(1-.3)2 2 _/ )

Ab 1 + "2(6.34) 435 _ 3)2

4(1-.3)2 . 10A

.0157= .00823

1 + (.0157)(58)

EA KI 7A S -(1.73) .00823 = 28.5% b2 5(10)3

From Figure 9.2.1-1(a) or 1(b),

S/- 12.5 (Plastic)0

(r312.5 (5000) 36- 0 sTcr = 1.73

N = 2f-r= 2(.020)36100 = 1444 lb/in.S~xy

9.3 CURVED PLATES AND SHELLS

The stability of plates assumes that the lateral loads (q) are resisted by bending inthe plate since the boundaries are free to move in the plane of the plate. If the boundaries arerestrained or the geometry of the structure is capable of resisting these lateral loads, thenmembrane type stresses coupled with the curvature of the structure can resist the lateralload. The ability of a structure to provide more than one load path cannot decrease the stabilityof the structure. The load will be distributed in accordance with the stiffnesses of the loadpaths so as to minimize the strain energy. The additional load path is analagous to an addi-tional restraint which cannot decrease the stability. The shell therefore is inherently morestable than the flat plate. Unfortunately the energy functions and the equilibrium equationsare more complicated because of the interaction of the membrane and bending btiffnesses.

WADD TR 60-517 9.48

41I

9.3 (Cont' d)

lThe equilibrium equation for small elastic deflections has been derived by Donnell (citedin Reference 9-8) and can be expressed as

i DM +A +V ( V +q)= 0D -v~w +D _V 4w +V4 (N - V2 w +qEt 3

and can be employed to obtain elgenvalues, where DM = bending stiffness 2- E1 2 (1-v

and PA = axial stiffness ,- Et.

The assumption of small deflections and a stiff load path by membrane action is notusually met by a curved plate due to significant initial eccentricities and changes in geometry 3with loading. The curved panel can buckle at a load significantly below that predicted by themembrane-bending model; and may even fail suddenly if the deflection pattern changes underload and the membrane action becomes limited. This can be visualized as equilibrium existingfor the initial loadings and small deflection eigenvector modes. As loading is continued, theshell deforms along the original eigenvector but apporach a crest similar to Figure 9.1-1 (d)and becomes unstable, deforming to a different eigenvector representing a solution with smallermembrane action. The geometry may be changing simultaneously so that the structure couldsuddenly deflect to the different eigenvector with a lower eigenvalue. The amount of initialeccentricity can significantly axiiet the load at which this cross over can occur. Radiallyoutward pressures should foresta.1 this phenomenon to some degree whereas inward pressureshould precipitate it at an earlier loading.

The stability of the curved uanel and shells will depend upon its bending stiffness DMPits axial stiffness DA, the boundary conditions, the initial condition (geometry, eccentricities,etc.), and the material properties. It is recommended that the present analysis of the curvedplates be based on semi-empirical stability criteria shown in Reference 9-8 (e.g. , C(Et/r) or

•.2 kE 22) (t/r)2 ). The stability coefficients are based on theoretical cosiderations and

12(1-i)modified by the statistics of available experimental data. The stability coefficients should beused in the manner described in Subsection 9.2 except that the correction for shear is notwarranted in view of empiricism of the coefficients. The effective modulus is assumed to beE (which slightly overestimates the stability) since the twisting and axial stiffnesses shouldp~ay a more significant role than for flat plates.

The stability of a shell is illustrated in the following:

Example: A cylinder shown in Figure 9.3-1 is loaded with a uniform compressionloading along the edges. The edges are simply supported.

From Reference 9-8,

ES C t/r

since r_ 10 = 67 C = .42 (Figure 7 of Reference 9-8).

o .42 (15) .0063.E 10

WADD TR 60-517 9.49

9.3 (Cont'd)

t=.15

FIGURE 9.3-1 CYLINDER IN COMPRESSION

Assuming 2024ST81 aluminum alloy at 300 0F,

EA = 10 psi, a = 5(10) 1 = .0001

E7•nd A t =107C-j- (-10)3 (.0063)= 12.6.a r 5

From Figure 9.2. 1-1(a) or -1(b),

- =10.60

= 10.6 (5000) = 53,000 psi

A= t 53,000 (.15) = 7950 lb/in

Fcr = aA = Ncr (27rr) 500,000 lb.

9.4 REFERENCES9-1 Gerard, G. and Becker, H., Buckling of Flat Plates, NACA TN 3781, July

1957.

9-2 Timoshenko, S., Theory of Elastic Stability, McGraw-Hill Book Co., Inc.,1936.

WADD TR 60-517 9.50

9.4 (Cont' d)

9-3 Bleich, F., Buckling Strength of Metal Structures, McGraw-Hill BookCo., Inc., 1952.

9-4 Budiansky, B. and Hu Pai, C., lagrangian Multiplier Method of FindingUpper and Iower Limits to Critical Stresses of Clamped Plate, NACAReport 848, 1946.

9-5 Stowell, E. Z.. A Unified Theory of Plastic Buckling of Column:s and Plates,NACA Report 898, 1948.

9-6 Stowell, E. Z., Critical Shear Stress of an Infinitely Long Plate in the PlasticRegion, NACA TN 1681, 1948.

9-7 Gerard, G., A Creep Buckling Hypothesis, Journal of Aeronautical Sciences,Vol. 23, No. 9, September 1956, p. 879.

9-8 Gerard, G., and Becker, H., Buckling of Curved Plates and Shells, NACATN 3783, August 1957.

9-9 Switzky, H., Design of Stable Structures, Republic Aviation Corporation,Report No. ESAM-30, October 1958.

6lt/

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thermo-structural analysis manual

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