73666136 introductory fracture mechanics

8/22/2019 73666136 Introductory Fracture Mechanics

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FRACTURE MECHANICS AND INSTRUMENTED

IMPACT TEST

An Informal Introduction with Worked Examples


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CONTENTSPage No.

FRACTURE MECHANICS - An Informal Introduction with Worked Examples 0 CONTENTS i-iii

PREFACE iv IMPORTANT NOTE ACKNOWLEDGEMENT OF SOURCES v

............................................................................................................................................. Chapter 1-Introduction to fracture Mechanics 1

............................................................................................................................................. General 1

1.2 Historic Failures/Developments that Spurred the Emergence of Fracture Mechanics 21.3 Some aspects of fracture in tension/impact/fatigue tests 3

1.3.1 Ductile vs Brittle - behaviour under the conventional, slow tensile test. 3

1.3.2. Microscopic Aspects of Ductile and Brittle fractures in Steel 41.3.3. Fatigue of Ductile materials 4

1.3.4 IMPACT TOUGHNESS and IMPACT TESTS 4

Impact tests 5

Pendulm Impact Test Details 6 Charpy Data 6 Ductile- Brittle Transition Temperature (DBTT) 7 1.4 Conventional versus Fracture Mechanics based Design 7 1.5 A Note on Crack Plane Orientation (CPO) in Fracture Test Specimens 8 1.6 Different Regimes of Fracture Mechanics 8 Example 1.1. DBTT Example: T I T A N I C failure 9

............................................................................................................................................. Chapter 2-Linear Elastic Fracture Mechanics (LEFM) 10

2.1 GRIFFITH THEORY OF BRITTLE FRACTURE 10

2.1.1 Background 10

2.1.2. Expression for the Critical Fracture Stress of a Brittle Solid 11 Example 2.1.1: Glass Fracture and Griffith Theory 13 2.2 IRWINS STRAIN ENERGY RELEASE RATE AND STRESS INTENSITY FACTOR 132.3 COMPLAIANCE AND STRAIN ENERGY RELEASE RATE 16

PREFACE TO GRIFFITH EQUATION PROBLEMS AND PROBLEMS 182.4 MODES OF LOADING, SIF AND LEFM 212.4.1 Three Modes of Loading based on crack surface displacements 21

2.4.2 The stress intensity factor (SIF) and stress distribution 22 Salient points of the stress distributions 24 Stress Triaxiality, Plane Strain and Plane Stress 252.5 CRACK TIP PLASTICITY 26

2.6. LEFM FRACTURE TOUGHNESS (KIC) TESTING 28

Relevant features of the ASTM E 399 Standard 30 2.7. TRIANGLE OF INTEGRITY 32 ............................................................................................................................................. S

pecimen SIFs and Typical Fracture Toughness Values 33

................................................................................................................................................................2.8. LEAK BEFORE BREAK (LBB) CONCEPT 34
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WORKED EXAMPLES/PROBLEMS IN LEFM 35

............................................................................................................................................. Chapter 3 - APPLICATIONS OF LEFM 53

................................................................................................................................................................3.1. FATIGUE CRACK GROWTH (FCG) 53

3.1.1. Introduction 53

3.1.2. Three Stages of Fatigue Crack Growth (FCG) 55

3.1.3. Stage II FCG: PARIS LAW 56

3.1.4. FCG Testing 59

Paris Constants for Some Common Steels 62

WORKED EXAMPLES IN FCG 63

3.2. STRESS CORROSION CRACKING (SCC) 733.2.1. General 733.2.2. Treatment of Crack Growth Rate 74

WORKED EXAMPLES IN SCC 77

3.3. ASME CODE APPROACH FOR DESIGN AGAINST BRITTLE FRACTURE:

RTNDT-KIR CURVE APPROACH 84

3.3.1. Ductile-Brittle Transition Temperature: Different Approaches 84

3.3.2KIR curve determination for a 9Cr-1Mo steel Illustration 85

3.3.2.1. Specimen Fabrication 85

3.3.2.2. Drop-Weight specimen fabrication and Drop-Weight Test (DWT) 86

3.3.2.3. Charpy specimen 89

3.3.3. Determination ofRTNDT and ASME KIR curve for 9Cr-1Mo Base Material 903.3.3.1. Results from Drop-Weight Test 90

3.3.3.2. Results From Charpy Test 91

3.3.3.3. RTNDT and KIR Curve 92

............................................................................................................................................. Chapter 4 - Elastic-Plastic Fracture Mechanics-EPFM 94

............................................................................................................................................. 4.1. Crack Tip Opening Displacement (CTOD) and J-Integral Approach 94

4.2. EPFM in Practical Terms 96

4.3. J-Measurement 974.4. Begley-Landes Multi-Specimen JIC method 100 4.5. Standard Method for crack-tip opening displacement (CTOD) Determination 101 EPFM PROBLEMS 105 Appendix - Chapter 4: A Note on CREEP CRACK GROWTH (CCG) 125

............................................................................................................................................. Chapter 5 - An EPFM Application: ASTM E 1921 Master Curve (MC) 132

5.1. Reference Temperature (T0) and Master Curve Approach 132

5.2. Statistical basis of the Master Curve 133

5.3. Validity limits 135


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5.4. Multi temperature equation for T0 136

5.5. Master Curve equations 137

5.6. DETERMINATION OF T0 AND MC FOR A 9Cr-1Mo STEEL EXAMPLE 1375.6.1. T0

dyfrom Pre-Cracked Charpy Test (PCVN) 137

Typical Calculation ofKJd for Specimen No. 19 Tested at -50 C (Fig. 5.3) 1435.7. APPLICATION OF THE ASME CODE CASES N-629 AND N-631 144

5.8. A NEW UPPER-SHELF FRACTURE TOUGHNESS MASTER CURVE(USFTMC or USMC) FOR FERRITIC STEELS 145

UPPER-SHELF FRACTURE TOUGHNESS MASTER CURVE (USFTMC or USMC)CALCULATION PROCEDURE 146

5.9. Effect of strain rate or stress intensity factor rate (SIF) rate on T0 146

............................................................................................................................................. Chapter 6 Instrumented Charpy Impact Test, Charpy-Fracture Toughness Correlationsand Reference Temperature Prediction 148

6.1. Instrumented Charpy Impact Test 148

6.1.1. Introduction 148

Additional strength and toughness values from IIT 152

A NOTE ON TD (brittleness transition temperature) 1526.1.2. Calculation Procedures in Instrumented Charpy Impact Test 153Load-time data processing 153

Loading or Strain Rates in Some Fracture Tests 154KId and JId/KJd estimation by conventional methods 154

KJd and J1d estimation by the Modified Schindler Procedure 157

6.2. Fracture Toughness Correlations with Charpy Energy and other Parameters 1606.2.1. Direct Charpy energy (CV) temperature - T0 Correlations 160

6.2.2. Older CVN energy (CV) - KIC Correlations and T0 estimates 160

Rolfe, Novak and Barsom (RNB correlation 160

Sailors and Corten (SC correlation 161

Roberts lower-bound correlation (RLB correlation 161

Barsom and Rolfes Kd-Kc (dynamic to static fracture toughness) temperature shift

(Barsom-Rolfe Shift - BRS) procedure 161Marandez-Sanz Procedure 161

Another lower bound correlation for the lower-shelf and lower transition region 162

6.2.3. New Reference Temperature Correlations 162

Schindler-Sreenivasan Procedure (SSP)162

RNB, SC and BLB estimates 163

Note on Size Correction 163PRS-Parameter Correlations 163

Mean-8 Procedure (M8P) 164

6.2.4. FATT-Master Curve (FATT-MC) approach for lower-bound fracture toughness 164

6.2.5. Lower-bound estimate for upper-shelf fracture toughness 166

6.3. Dynamic Fracture Toughness from Instrumented Drop-Weight Test 168

Illustrative Example 6.1:Complete IIT data for a service exposed 2.25Cr-1Mo Steel 169 Illustrative Example 6.2: Comparison of actual fracture toughness data for

a modified 403SS 176

Illustrative Example 6.3: Comparison of KId from IIT and Drop-Weight NDTwith RTNDT based KIR Curve for a 403 SS martensitic Stainless Steel 403SS-IGC 177

Test Standards and Hand Books (Reference T) 178

Bibliography and Additional References (Reference B) 179


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PREFACE

When it was suggested that a book introducing Fracture Mechanics in an informal way with emphasis onworked examples should be prepared, the first thought that came to the mind of the authors was that when so

many scholarly treatises are available, why one more? To our knowledge, the first authoritative book

exclusively on Fracture Mechanics was by Prof. J. F. Knott - "Fundamentals of Fracture Mechanics" (1973). But

it is a scholarly textbook most suited to academic pursuit and, because of the time of its publication, it restricts

itself to mainly Linear Elastic Fracture Mechanics (LEFM). Prof. Knott followed with a sequel, perhaps the first

book of its kind, Worked Examples in Fracture mechanics; it also restricts itself to mainly LEFM and it contains

only worked examples. Subsequently, as listed in the Bibliography, many treatises have been published, the

latest being the ones by Perez (2004), Ramesh (1999) and Saxena (1998), the latter covering non-linear fracture

mechanics and creep crack growth problems. Most of the listed references, which include some freely available

web-resources, contain numerous solved and unsolved problems and seem to be rather heavy on theory or more

detailed in treatment. So, we felt that a book introducing Fracture Mechanics in an informal way, with emphasis

on worked examples, will not be out of place. Hence this venture. Accordingly, this volume has been titled asFracture Mechanics - An Informal Introduction with Worked Examples. Its preliminary model is Prof.

Knotts book, but it covers more topics starting from Griffiths Theory to Non-Linear Fracture Mechanics and

some Applications in ASME Codes. It also introduces the ASTM E 1921 Reference Temperature Approach. No

attempt has been made to derive the relations from first principles, nor a micro-mechanistic nor a metallurgical

view-point adopted. An informal Strength of Materials textbook approach with ample explanation of the

necessary terms involved has been the line adopted during the writing of this monograph. Any engineering

student who has completed a Strength of Materials course would benefit in following this book. Of course, it

can be used as a companion or prescribed book, for an introductory course in Engineering Fracture Mechanics.

On completion of this book, one will be in a position to appreciate the relative importance of Fracture

Mechanics in safety analysis and assurance of structural integrity. Most of the Fracture testing Techniques havebeen covered with a coverage deep enough to impart an overview of the test. But no attempt has been made to

cover the mechanical or instrumentation aspects. Hence, after mastering this book, one is not expected to walk

into a Test Lab equipped with the wherewithals to perform a test; however, with help from the appendedreferences and standards, one would well be on the way to equip oneself for such a task; in short, this book is

not a manual on test techniques.

No originality is claimed for the material presented in this book (see Note and Acknowledgement in the next

page). About 50 worked examples have been provided. Many problems whose answers are only available, have

been worked out. The material on KIR and Reference Temperature determination has been provided based on the

test results from our laboratory at Indira Gandhi Centre for Atomic Research. Many glaring omissions may be

there: like, Stress Intensity Factor Measurement or Computational Techniques, many Advanced Defect

Tolerance Procedures, Creep-Fatigue Crack Growth, Variable Amplitude Fatigue Crack Growth, OverloadEffects on Fatigue, etc. They are intentional as they are perceived to be outside the ambit and aim of the present

book. Hence this book is neither comprehensive nor exhaustive, but sufficiently introductory and informative.

Unusual for a Fracture Mechanics Book, a chapter (Chapter 6 the final Chapter) on Instrumented Impact


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Testing, Charpy correlations and Prediction of Reference Temperature has been added, mostly based on work at

IGCAR, since it has practical implications in many Engineering Applications such as Quality Control, Test

Temperature Decisions and Preliminary Design, especially with availability of some new correlations. Thus wehope the material we have provided will serve not only as an appetizer but also as an enhancer and nourisher of

knowledge of Fracture Mechanics, strictly within the Strength of Materials viewpoint (hence no Mathematics

beyond a knowledge of Preliminary Calculus is assumed). In short, this book will prepare the reader for a

pleasant first encounter with the subject of Fracture Mechanics and the cited Standards and References will givedirections for further progress. If that is the result, then the aim of this book would have been more than served.

IMPORTANT NOTE ACKNOWLEDGEMENT OF SOURCES

Two sets of references are appended at the end: (i) Test Standards and Hand Books and(ii) Bibliography and Additional References. Item (i) pertains to test standards like those

of American Society for Testing and Materials (ASTM), British Standards (BS), ASME

Code, ASM Metals Handbook, IAEA Technical Report etc. Item (ii) pertains to general

bibliographical references, monographs and journal articles etc. relating to Fracture

Mechanics. In the text, references to material under Item (i) are preceded by Letter T: for

example Ref. 1 under Item (i) is referred as T1 in the text. Similarly, references to material

under Item (ii) are preceded by letter B: for example Ref. 1 under Item (ii) is referred as B1 in

the text. Most of the problems and material have been adapted from References B1, B3,

B6, B7, B9, B10, B12, B13, B19, B20, B21, B23, B24, B26, B27, B28, B30, B31, B32, B36,

B41, B43, B44, B45, B48 to B55, B56, B59, B60 to B62 and B65 (Item (ii)), where thebold underlined references indicate the source of problems or worked examples.

Sources for figures, tables etc. are cited (following the above notation) at theappropriate places in the text.


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Chapter 1-Introduction to fracture Mechanics

1.1. General

Fracture involves breaking up of a body into two or more parts with the creation of newsurfaces. Thus it is destructive and, often, as shown in Fig. 1, catastrophic, resulting in loss

of men, materials and money.

Hence Understanding how things break can avoid having to explain why they broke

later on! [B24] That is the realm of Fracture Mechanics: understanding fracture of

materials and structures and, thereby, predict, prevent and mitigate fracture failures.

(a) Boeing 737-200, Aloha Flight243, 1988 T2 Tanker, The Schenectady, 1941

Fig. 1.1. Catastrophic fracture failures [B10]

OUTLINEWhat is fracture? Historical introduction. How fracture was

assessed qualitatively in pre-fracture mechanics days, using

tension and impact tests, fracture appearance, ductile-brittle

transition temperature (DBTT) etc. Need and usefulness offracture mechanics. Different Fracture mechanics Regimes.

Fracture Specimen Orientation in Steel Plates and an example of

DBTT limited failure.


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1.2 Historic Failures/Developments that Spurred the Emergence of Fracture Mechanics

Enormous increase in the use of metals (mainly irons and steels) for structural applications in

the late 19th

and early 20th

centuries and later led to many accidents, with loss of life andwealth, owing to failure of these structures. In particular, there were numerous accidentsinvolving steam boiler explosions and railway equipment. Some of these accidents were dueto poor design, but it was also discovered that material deficiencies in the form of pre-existing

flaws could initiate cracking and fracture.

A new era of accident-prone structures was ushered in by the advent of all-welded designs,notably the Liberty ships and T-2 tankers of World War II. Out of more than 2500 Liberty

ships built during the war, 145 broke in two and almost 700 experienced serious failures. Thefailures often occurred under very low stresses, for example even when a ship was docked,and this anomaly led to extensive investigations which revealed that the fractures were brittle

and that flaws and stress concentrations were responsible. It was also discovered that brittle

fracture in the types of steel used was promoted by low temperatures. As clarified in the lastdecade, the failure of the TITANIC (a completely riveted structure unlike the all-weldedstructures of World War-II) in 1912 resulted from the use of poor steel at low temperature.

Indian Connection to Failure and Fracture: On 2 May 1953 exactly one year to the dayafter their introduction, a Comet aircraft (the de Havilland Comet was the first jet-propelledairliner) broke up in flight near Calcutta. In all, 20 Comets crashed between 1952 and 1971taking the lives of almost 500 people. It was eventually discovered that their fuselages had

exploded whilst climbing up to cruising height, weakened by the fatigue of repeatedpressurization and depressurization. The aircraft whose wreckage was discovered had begun

to crack at the corner of one of the automatic direction finder (ADF) aerial cut-outs, andanother tested on the ground burst open at the corner of a window. Hatch corners and window

corners had to be modified to reduce the stress concentration.

In modern times, the emergence of aerospace technology (with use of high-strength structuralmaterials at high stresses for weight reduction) and nuclear power and the attendant concernabout the structural integrity of aero-space structures and nuclear reactors has greatly

contributed to the development of Fracture Mechanics.

To summarise, in the above cases, failures could be attributed to :

the all-welded construction which eliminated crack-arresting plate boundaries present inriveted joints

the presence of crack-like flaws/metallurgical defects in welded joints - like inclusions, lackof fusion, weld cracks etc.

the use of materials whose low resistance to crack advance (toughness) was further reduced bylow temperatures for example, Titanic. Poor design - window hatch corners in Comet resulted in stress concentration and fatigue

cracks.


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1.3 Some aspects of fracture in tension/impact/fatigue tests [B65]

1.3.1 Ductile vs Brittle - behaviour under the conventional, slow tensile test .

ductile

Fig. 1.2

Slow controlled extension; final instabilitydue to gross area reduction. Cup-and-conefailure surface characterised by 45

oshear

lip.

brittle

Fig. 1.3

Fast catastrophic failure with no warning.Transverse granular cleavage surface in

ferritic steels - no shear lip.

Fig. 1.4 Fig. 1.5

Relatively high energy absorption capacity

(stress-strain area) - " tough". Localised

yielding at high stress concentrationsredistributes stresses advantageously. Shear

stresses cause failure, triaxiality effect is

relatively benign.

No significant yielding, so no reduction of

high stresses. Low energy absorption

capacity. Triaxial stresses cause failure.


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1.3.2. Microscopic Aspects of Ductile and Brittle fractures in Steel

(a) TransgranularCleavage

(b) Intergranular

FractureFig. 1.6 [B26]

(c) (d)

1.3.3. Fatigue of Ductile materials [B65]: Fatigue is recognised as a mechanism of crack growthterminated by catastrophic fracture - the S-N diagram, shown in Fig. 1.7, may be used to predict

failure. S-N diagram approach is mostly empirical. Hence a more basic approach involving an

understanding of the fundamental fracture mechanisms, processes and factors involved can helppredict and assess fracture failure more confidently. Fracture Mechanics provides one such

approach.

Fig. 1.7a Fig. 1.7b

1.3.4 IMPACT TOUGHNESS and IMPACT TESTS

Toughness

Measure of the amount of energy a material can absorb before fracture

Low loading rates as in normal tension testsArea under the stress-strain curve up to fracture

Dynamic loadingImpact energy


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Impact tests:

Charpy (simple three-point bend:TPB) and

Izod (cantilever bend and done only at room temperature) Tests

Fig. 1.8. Pendulum Impact Testing [B26]


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Pendulm Impact Test Details

Toughness quantified in terms of the amount of energy required to fracture a notched specimen of thematerial struck by a hammer or pendulum. Notch introduces three factors: (a) Stress concentration;(b) Triaxial stresses which reduce shear stresses favouring brittle fracture; and (c) increased strain

rate.

The hammer has potential energy due to its initial height, h1When the hammer is released the potential energy -> kinetic energy

At the bottom of the stroke, the hammer impacts on the test specimen

Kinetic energy is used to break or fracture the specimen

The remaining kinetic energy is not enough to raise the pendulum to its initial height at the end of thestroke : h2 < h1

The impact (fracture) energy is the difference in potential energy: U (h1 - h2)

Charpy Data

Fig. 1.9a. Charpy impact energy (toughness) vs. Fig. 1.9b. Charpy impact energy (CV) vs

Test temperature curve for a typical ferritic steel. test temperature (T) for various alloys

the almost constant low energy region at low temperatures is called the lower-shelfwhile the

almost constant high energy region at high temperatures is called the upper-shelf.

Quantitative results are not directly used in design calculations

These results give a qualitative comparison of the toughness:for different materials at the same test temperature

at different test temperatures for the same materialfor the same material subjected to different heat treatments


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Ductile- Brittle Transition Temperature (DBTT)

The ductile-brittle transition is exhibited by BCC metals (see Fig. 1.9), such as low carbonsteel; they become brittle at low temperature or at very high strain rates. FCC metals (Alalloys upper-shelf energy ~ 30 J; austenitic stainless steels upper-shelf energy ~ 150 to 350

J see, Fig. 1.9b)), however, generally remain ductile at low temperatures without anytransition.

Low temperatures, low energy absorption indicates brittle behaviour. As temperature increases, the energy absorbed increases dramatically, indicating a change of

failure mode from brittle to ductile (for ferritic steels).

The effect of a change of loading rate is to move bodily the toughness curve parallel to thetemperature axis. This, together with the shape of the curve, means that a higher rate or lowertemperature will decrease toughness.

Usually, DBTT is defined as the temperature corresponding to a Charpy (CV) value of 28, 30,41 or 68 J, depending on the design codes and types of steels.

1.4 Conventional versus Fracture Mechanics based Design

The traditional design is based on preventing yielding or allowing only local yielding by restrictingdesign stress to a fraction of the yield or ultimate tensile stress (YS or UTS) using a factor of safety(FOS), that is,

Design Stress = (YS or UTS)/FOS

FOS varies from 2 to 4 and 10 for a lift wire rope [ B23, B24]. This approach proved inadequate toaddress issues like those described in Section 1.2 or situations involving DBT.

Fracture Mechanics presupposes the existence of cracks/crack-like flaws in the material, which maybe microscopic (for example, grain size, inclusions etc.), or large (for example, casting or weld

defects), due to manufacture, to corrosion, fatigue,

Fracture Mechanics correlates three parameters quantitatively . . .(i) - load the background stress, ;(ii) - geometry - the crack size, a (and to a lesser extent, crack shape); and

(ii) - material - its resistance to cracking, i.e., its fracture toughness, measured by special testsand predicts, amongst other things,

- degree of safety, or imminence of catastrophic (brittle) fracture- crack growth rate whilst advancing in a controlled manner - remaining component life.

FRACTURE MECHANICS HAS BEEN CALLED

THE SCIENCE AND ENGINEERING OF LIVING WITH DEFECTS.


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1.5 A Note on Crack Plane Orientation (CPO) in Fracture Test Specimens [B7]

The major CPOs in a rolled plate are indicated in the above figure. L, T and ST or S are thethree principal working directions in the rolled plate, namely, Longitudinal (PrincipalRolling Direction-usually the longest dimension or length), Transverse (Next LongestDimension or Width) and Short Transverse (usually the shortest or thickness direction).

Specimen-1 has its long axis in the T-direction and crack propagates in the T-direction;hence, its CPO is designated as T-L and is referred to as T-L specimen. Similarly for others.

Because of mechanical fibering, inclusion banding etc., ductility and toughness dependsstrongly on the specimen orientation (see Example 1.1 on Titanic steel in the next Box).Usually, the L-T specimen (3 in the above figure) shows the highest and S-T (5) shows thelowest impact property.

1.6 Different Regimes of Fracture Mechanics

1. Linear Elastic Fracture Mechanics (LEFM): Here the body/structure as a whole is in theelastic and plasticity is confined to very small region at the defect/crack tip and failure occursafter very little plastic deformation. The relevant fracture mechanics parameter is linear elasticfracture toughness critical stress-intensity factor (SIF) - KIC.

2. Elastic-Plastic Fracture Mechanics (EPFM): Here, significant/extensive plasticity occurs atthe crack tip and fracture is preceded by significant plastic deformation. The relevant fracturemechanics parameter is elastic-plastic fracture toughness: critical J-integral JIC or criticalcrack tip opening displacement (CTOD), IC.

3. Fatigue Fracture Mechanics: Fatigue crack growth (FCG) is related to (SIF range) Kor (J-

integral range) J.4. Creep Fracture Mechanics: Fracture mechanics applied to creep crack growth (CCG): the

relevant fracture mechanics parameter is C*

- a time dependent J-integral rate parameter andits variants.


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Example 1.1. DBTT Example: TITANICfailure

Metallurgical Cause for the RMS Titanic Failure (on 12 April 1912)

A metallurgical analysis of steel taken from the hull of the wreckage of the 46,000 tons shipTitanic reveals that it had a high DBTT, making it unsuitable for service at low temperatures;

at the time of the collision (with an iceberg that was three to six times larger than its own

mass), the temperature of the sea water was -2C. The analysis also shows, however, that the

steel used was probably the best plain carbon ship plate available at the time of the ship'sconstruction (1912). Presence of massive MnS inclusions and banding resulted in very poor

transverse impact properties. (Adapted from: The Royal Mail Ship Titanic: Did a

Metallurgical Failure Cause a Night to Remember? Katherine Felkins, H. P. Leighly, Jr., andA. Jankovic. J. of Materials, 50(1) (1998) pp.12-18.)

Fig. 1.10


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Chapter 2-Linear Elastic Fracture Mechanics (LEFM)

2.1 GRIFFITH THEORY OF BRITTLE FRACTURE

2.1.1 Background

Fracture strength of a solid material function of the cohesive forces between atoms. Theoretical cohesive strength of brittle elastic solids ~ E/10, where E is the Youngs

modulus of elasticity.

Experimentally observed fracture strengths for most engineering materials ~ E/100 toE/1000,

In the 1920s, A. A.Griffith ascribed the above discrepancy to the presence of very small,microscopic flaws or cracks at the surface and within the interior of a body of material.

Flaws detrimental to the fracture strength because an applied stress gets amplified orconcentrated at the tip, with the magnitude of amplification depending on crack

orientation and geometry.

OUTLINEFracture stress of brittle solids and Griffith Theory; Examples of application of Griffith Theory;

Stress Intensity Factor (SIF - K), Irwins Critical Strain Energy Release Rate (GC) and Linear

Elastic Fracture Mechanics (LEFM); Significance of LEFM Fracture Toughness, critical, K

KIC; Compliance and Strain Energy Release Rate; Different Modes of Fracture; SIF and crack

tip stress distribution; Crack Tip Plasticity; Stress Triaxiality, Plane Strain and Plain Stress;

Linear Elastic Fracture Toughness (KIC), Different Specimens and Testing; Fracture Toughness

and its Relation to Structural Geometry and Quality Factors; Leak Before Break; Worked

Examples in LEFM.


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2.1.2. Expression for the Critical Fracture Stress of a Brittle Solid

(a) (b)Fig. 2.1. (a) Body with central crack, remote stress, ; (b) Body with edge crack [B7]

Fig. 2.2 Energy balance during crack growth in a brittle solid [B26]

Cond. In next BOX


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2.1.2. Expression for the Critical Fracture Stress of a Brittle Solid (cond.)Considering an infinite plate with a through-the-thickness elliptical crack of length, 2a (Fig. 2.1a), or an edge-

crack of length, a (Fig. 2.1b) and remote stress, , Griffith, based on energy balance considerations, obtained anexpression for the fracture stress, F, of a brittle material-glass. Basic premise of Griffith (1920) was that thecrack will grow if the energy available for growth equals or exceeds that required for growth and thereby

the total energy of the system is lowered.

Energy available for growth: An elastically strained (stressed) material contains strain energy. As

the crack propagates inwards from the surface of a stressed material, the area of material in whichstrain energy is relaxed corresponds to the two shaded triangles shown in Fig. 2.1a. The area of thesetriangles is approximately a2 and, hence, the relaxation of elastic strain energy is proportional to

the square of a. This is confirmed by calculation.

Energy required for growth: As the crack grows, surface energy needed to form the new surfaces,and this equals 2a (for unit thickness of the plate), where is the surface energy per unit area of thematerial. The value of this term increases as the first power of the depth of the crack.

There is therefore an energy balance between the formation of new surfaces and the relaxationof strain energy in the material.

From the above, it can be seen that shallow cracks consume more energy as surface energy thanreleased as relaxed strain energy. Therefore conditions are energetically unfavourable for crack

propagation. As the crack increases in length the conditions are reversed and more strain energy isreleased than is needed to form the surfaces of the growing crack. This occurs beyond the ' critical

Griffith crack length', ag (or ac). Under these conditions the crack is able to grow at an everincreasing rate. This is shown in Fig. 2.2.

Griffith demonstrated that the critical stress c (equivalently represented byf) required for crackpropagation in a brittle material is described by

1

22 sc

E

a

(2.1.1)

where E is the Youngs modulus, s is the specific surface energy and a is one-half the length of aninternal crack.

In practice, the energy required to produce the two new fracture surfaces is significantly greater than

s. The surface energy, s, is that required to break all the chemical bonds at the fracture surface.During fracture the molecular structure of the material around the crack is also disturbed, to a depth

which is sometimes very considerable. The total energy is known as the work of fracture, W.

Therefore,

1

22c

EW

a

(2.1.2)


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2.2 IRWINS STRAIN ENERGY RELEASE RATE AND STRESS INTENSITY FACTOR

Most metals and many polymers do experience some plastic deformation during fracture; thus,

crack extension involves more than producing just an increase in the surface energy. This

complication may be accommodated by replacing s in Eq. 2.1.1 by s + p, where p represents aplastic deformation energy associated with crack extension. Thus,

1

22 ( )s pc

E

a

(2.2.1)

For highly ductile materials, it may be the case that by p s, such that

1

22 pc

E

a

(2.2.2)

Example 2.1.1. Take a glass with E= 69 GPa and W = 0.3 Jm-

. Draw the graph showing the

dependence of fracture stress on crack length.

For various assumed a values of 10, 20, 30, ..,100 m, plugging in the above values ofEand W in Eq. (2.1.1), the corresponding critical fracture stress values are obtained and are

plotted in the Figure below.

As the crack increases from 10 m to 100 m, the fracture stress decreases from 36 MPa

to 11 MPa. The point to be noted is that1

fa

.


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In the 1950s, G.R.Irwin chose to incorporate both by s and p into a single term, GC, as

2( )c s pG (2.2.3)

GC is known as the critical strain energy release rate.

After incorporating the new term and rearranging, Griffith cracking criterion becomes

2

c

aG

E

(2.2.4)

Thus crack extension occurs when 2a/E exceeds the value of GC for the particular materialunder consideration. Therefore,

2

ca EG (2.2.5)

OR

ca EG (2.2.6)

This equation says:

LHS (Some configuration stuff you can calculate) = RHS (Some material property stuff you

can measure)

LHS = a

Stuff you can calculate from the cracked structure configuration

Loading,

crack length, a

This term is referred to as the Stress Intensity Factor (SIF) and given the symbol K

K a (2.2.7)

Khas unusual units: Stress x length = Pam or MPam

K(quantifies the stress-strain distribution ahead of the crack) is fundamentallydifferent from the concept ofStress Concentration Factor (SCF); SCF describes how much

the stress has been elevated at a point (crack or notch tip) compared to the remote stress.


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RHS = Fracture Toughness = cEG

The right hand side of the critical equation is a combination of material propertiesYoungs modulus

Toughness

It is a measured material property called the critical fracture toughness of the material, Kc.

c cK EG (2.2.8)

Critical Condition

The fracture toughness, Kc, is a constant for a given material and independent of the structural

configuration

The critical condition for fast fracture can be written:

K = Kc (2.2.9)

When the K reaches the critical value Kc, the crack propagates without limit:Fracture occurs, as a consequence of:

Increase in crack length

Increase in load K a The tensile strength of a brittle material is determined by the length of the largest

crack existing prior to loading

SIFs for Cracked Bodies

The SIF defined here is strictly speaking valid only for thin, semi-infinite plates, such that,thickness, t


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2.3 COMPLAIANCE AND STRAIN ENERGY RELEASE RATE

Fig. 2.3(a) Fig. 2.3(b)

Fig. 2.3. Applied load = P; Crack length = a; Displacement = [B7]

In this section, we examine the significance of the strain-energy release rate in a notched/cracked

body subjected to elastic loading and its relation to G in greater detail. Figure 2.3(a) shows how G

can be measured. A single-edge notch specimen is loaded axially through pins. The sharpestpossible notch is produced by introducing a fatigue crack at the root of the machined notch. The

displacement of this crack as a function of the axial force is measured with a strain-gage clip

gage attached at the entrance to the notch. Load vs. displacement curves are determined fordifferent length notches, where P = Md (or , as indicated in Fig. 2.3(b)). M is the stiffness of aspecimen with a crack of length a and reciprocal of stiffness, i. e., (1/M) is called the

compliance, C; therefore, d = CP. The elastic strain energy is given by the area under the curve

to a particular value ofPand d.

The concepts of strain energy release rate and SIF were developed by G. R. Irwin in the

1950s. Thus he can be considered as the father of Modern Fracture Mechanics, while

Griffith can be considered as the father of brittle fracture theory.


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21 1 1= = =

2 2 2U Pd PCP P C (2.3.1)

Consider the case shown in Fig. 2.3 where the specimen is rigidly gripped so that an increment ofcrack growth da results in a drop in load from Px to P2.

1 2 1 1 2 2= =d d C P C P

Since CP is constant,C

+ = 0P

C Pa a

C= -

P P

a C a

(2.3.2)

But the crack extension force, G, is defined as

21= = 2 +

2d

U P CG CP P

a a a

(2.3.3)

Putting Eq. (2.3.2) in Eq. (2.3.3),

21= -

2

CG P

a

(2.3.4)

Thus G can be evaluated by determining Cas a function of crack length. In the above, the load Pwas considered for unit specimen thickness (i. e., P/B). For the fixed grip case, no work is doneon the system by the external forces P dd, while for the fixed load case, external work equal toPddis fed into the system.


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PREFACE TO GRIFFITH EQUATION PROBLEMS

Based on more sophisticated analysis, Eq. 2.2.4 can be rewritten as below:

1

2' Cf

E Ga

where2

'(1 )

EE

.

Therefore, 2

12

(1 )

Cf

EG

a

OR2 2

(1 )

Cc

f

EGa

Hence, for the common value of = 0.3 for metals, the effect of considering or not considering thevalue of , will change the result by about 10%.

Example 2.2.1: MARAGING STEEL FRACTURE

If the fracture stress of a large sheet of maraging steel , which contains a central crack oflength 40 mm, is 480 MPa, calculate the fracture stress of a similar sheet containing a crack of

length 100 mm.12

6

40 480 100.02

C Cf

EG EG

a

;

12

1000.05

C Cf

EG EG

a

Taking the ratios off,

0.5

100 100

40

0.02

480 0.05

f f

f

; therefore, to f100 = 303.4 MPa.

Comments: Maraging steel, unlike glass, is both tough and strong; hence, an increase in the

flaw-length by 2.5 times resulted in a reduction of the critical fracture stress by about 37%.


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Example 2.2.2: GLASS SHEET FRACTURE

A sheet of glass measuring 2 m by 200 mm by 2 mm contains a central slit parallel to the 200 mm

side. The sheet is restrained at one end and loaded in tension with a mass of 500 kg.What is the maximum allowable length of slit before fracture occurs?

Assume the following material property values: E= 60 GPa, surface energy ~ 0.5 J/m2, Poisson's ratio= 0.25 and the fracture stress of sound glass = 170 MPa.

Note on Units: When the critical strain energy release rate is in N/m (=J/m 2), E in N/m2, and a in m,the fracture stress is in N/m2 (Pa), which needs to be divided by 10 6 to convert it to MPa (the standard

SI engineering unit). Mass is multiplied by 9.81 m.s-2

to get the force in N. Also note that twice thesurface energy gives GC (see Eq. (2.2.3)).

Then applied nominal stress, =500 9.81

0.2 0.002

= 12.26 MPa; the question is, for this to be equal

to f, what is the ac?

Substitution into the Griffiths equation, i.e.,2 2(1 )

Cc

f

EGa

, gives

94

2 2 12

60 10 0.5 21.355 10

(1 0.25 ) 12.2625 10ca m

This is a central crack in a plate and the full slit length is then, by definition 2 a (see Fig. 2.1a). Thus

the maximum length of slit which can be supported is 0.271 mm.

Comments: As the applied stress is only 12.26 MPa, this indicates how critical even relatively smalldefects are in brittle materials, as this value of applied stress is much less than the fracture stress of

sound (uncracked) glass of 170 MPa. This example illustrates that even the apparently superfluousinformation helps in finding the right information!


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Example 2.2.3: FAILURE CRITERIA: YIELD vs. FRACTURE

A cylindrical pressure vessel, with a diameter of 6.1 m and a wall thickness of 25.4 mm,

underwent catastrophic fracture when the internal pressure reached 17.5 MPa. The steel of the

pressure vessel had E= 210 GPa, a yield strength of 2450 MPa a value of GC = 131 kJ/m2.a) Show that failure would not have been expected based on von Mises yield criterion,

2 2 2 2

1 2 2 3 3 1. ., ( ) ( ) ( ) 2 YSi e b) Based on Griffith's analysis determine the size of crack that might have caused this failure,stating assumptions that you have made.

Assumptions: (i) Fabrication and the orientation of the fatal defect - vessel is made from

welded plates with welds running perpendicular to both hoop and longitudinal stresses. As thehoop stress is the maximum principal stress, one should assume that the defect is

perpendicular to this stress direction. (ii) 25.4 mm plate is quite thick and hence assume plane

strain, Poisson's ratio ~ 0.3.

Solution:

a) Using thin walled pressure vessel theory we get:

1 2 3

1; ; 0

2 4 20

pD pD tif

t t D

1 2

17.5 6100 17.5 61002101MPa; 1050MPa

2 25.4 4 25.4

Substituting these values into the von Mises yield criterion gives:

2 2 2 6 7 2. ., (2101 1050) (1050 0) (0 2101) 6.62 10 1.2 10 ( 2 )YSi e

Thus failure would NOT have been expected on the basis of yield.

b) Substitute into the Griffith formula the relevant values:

3 9

2 2 6 2 2

131 10 210 10

(1 ) (2101 10 ) (1 0.3 )

Cc

f

EGa

m

Hence: ac = 2.18 mm.


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2.4 MODES OF LOADING, SIF AND LEFM

2.4.1 Three Modes of Loading based on crack surface displacements

Fig. 2.4. Mode-I (opening or tensile mode),

Mode-II (sliding mode), and Mode-III (tearing mode) [B26].

All other loading situations can be expressed as a combination of these three basic modes.

ONLY MODE-I IS CONSIDERED FURTHER


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2.4.2 The stress intensity factor (SIF) and stress distribution

The SIF orKcharacterises the elastic stress field for a stressed element near the tip of a sharpcrack at a distance rand angle from the X-axis (Y is in the normal direction to the crack-plane and Z is along the thickness (B) direction). The situation is illustrated below (Fig. 2.5).

Fig. 2.5. Stress components ahead of a (Mode-I) crack in a finite body of thickness B showinglateral contraction in the thickness direction; ris the distance from crack-tip [B59, B48].

Figure 2.5 helps visualise the coordinates for describing the crack-tip stress field for a crack ofdepth or length, a, situated in a body with width, W, and thickness, B. The crack-front (the

leading crack-tip, assuming to be straight) lies along the Z-direction (parallel to B) and the crack

propagates in the X-direction (along W); thus, the X-Z plane constitutes the crack plane. x, yand z are the three normal stress components, z being the stress component normal to thecrack-plane.

Figure 2.6 gives the analytical expressions describing the normal stress components in terms ofKI and distance r and angle . For an infinite body, KI is given by Eq. (2.2.7), while Eq. (2.2.10)describes the KI for a finite-sized body.


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Fig. 2.6. Expressions for stress distribution ahead of a crack in terms of SIF, KI,subscript I denotes Mode-I loading (For Modes II and III, subscripts IIand III will be used respectivey). is the Poissons ratio [B48].


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Fig. 2.7. Distribution of the three normal stresses (see Figs. 2.5&2.6 also).

In Fig. 2.7b, z = 0 (the mid-thickness position) [B59, B48].

Salient points of the stress distributions (Figs. 2.5 to 2.7) are:

The expressions in Fig. 2.6 are the first terms of a series and hold good for small r.

Crack-tip stresses are 1/rand this causes a stress singularity at the crack tip

(i.e. the stresses go to infinity as r goes to zero). Thus in the near-tip region, which iswhere fracture processes occur, the stress field is dominated by the singularity.

Along the critical plane for cracking ahead of the crack tip (where the angle is zero),the equations for the principal normal stress, that is the Y-direction stress (this applies

to other stresses also) reduces to the simple form of:

2y

K

r

(2.4.1)

The numerator, the stress intensity factor, K, in the above essentially gives ameasure of the magnitude, or intensity, of the near-tip elastic stress field for anycombination of remote stress and crack length. Irwin postulated that fracture would

occur at a critical value of K, KC, a material property, called fracture toughness. KCis dependent on thickness and, at sufficiently high thickness, it attains a thickness

independent constant value called plane strain (described in the next box) fracture

toughness denoted by KI C.


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Stress Triaxiality, Plane Strain and Plane Stress

In a normal tensile test, a simple bar is pulled along and is subjected to a uniaxial

stress (that is, stressing in a single direction) at least up to necking. When a crackedbody is subjected to a remote tensile stress, y, as in Fig. 2.5, stresses develop in theother two perpendicular directions due to Poisson effect (see Figs. 2.5 to 2.7). As

shown in Fig. 2.7a, the stress in the thickness direction, z, reduces to zero at thesurfaces (for the simple reason that free surfaces cannot support normal stresses; for

the same reason, as shown in Fig. 2.7b, x = 0 at the crack tip (r =0)) and is amaximum at the centre (mid-thickness, z = 0). So, as the thickness, B, is reduced, the

central region of the specimen having the maximum stress will be reduced, and

beyond a point, maximum stress itself will reduce, leading to a situation known asplane stress, where , z is very small or ~0 and only a biaxial stress field will exist. So,as B increases triaxiality increases, and beyond a certain thickness maximum stress

triaxiality develops.

Plastic deformation depends on shear stresses ( difference of normal stresses; forexample in a simple uniaxial tensile test, shear stress is y; when triaxiality ispresent, for the same y, shear stresses are (y - z) or (y x) or (z y)). Thushigh triaxiality leads to a reduction in shear stresses and plasticity is reduced orconstrained. So at the highest triaxiality, we have maximum normal stresses leading

to restricted plasticity. Normal stresses promote brittle fracture. This condition ofmaximum triaxiality with restricted plasticity is referred to as plane strain;hence, thin specimens experience a condition known as plane stress with reduced or

almost nil triaxiality.

Hence, plane strain fracture toughness, KIC, described above, is obtained insufficiently thick specimens where maximum stress triaxiality has developed and

larger thicknesses have no further effect on triaxiality. This situation is depicted in Fig.2.8, where KC is plotted as a function of B, the specimen thickness; thinner specimens

below the plane strain limit show higher and, later, lower KC because of increased

plasticity and because of the interplay of different extents of plane stress (surface) and

plane strain (central) regions as in Fig. 2,7a.


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Fig. 2.8. Dependence of fracture toughness on specimen thickness:

plane stress-plane strain transition [B6].

2.5 CRACK TIP PLASTICITY

Fig. 2.9. Development of plastic zone (R orrp) at the crack tip [B6].


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From the foregoing Section, it is obvious that plasticity and its extent and distribution play an

important role in controlling the plane strain and plane stress regimes. From the definition of thestress intensity, based on the elastic stress field near a crack tip with = 0, i.e.:

2y

K

r

(2.5.1)

we can see that, as r tends towards zero, the crack tip stresses become singular. But this is

physically not possible; hence a yielded region will exist in the material ahead of the crack for allreasonable stress values. The shape and size of the plastic zone can be determined, to a first

order, from the simple model first proposed by Irwin. Consider a material with a simple elastic-

perfectly plastic response (i.e. no strain hardening occurs). A first estimate of the plastic zonesize ahead of the crack tip (ry), along the plane of the crack, can be obtained by substituting the

yield strength into the above equation (see Fig. 2.9). This truncates the elastic stress field in the

near-tip region, where yielding occurs. The plastic zone size is obtained as:

2

1

2y

YS

Kr

(2.5.2)

Irwin observed that the presence of significant crack tip plasticity caused the specimen to behave

as though it contained a crack of greater length than was actually the case. That is, the

compliance (ratio of displacement to load: /P d , where P and d are load and load pointdisplacements respectively) of the specimen became greater as plasticity developed at the crack

tip. This observation led him to propose a 'plastic zone correction' to crack length, based on amore accurate model of crack tip plastic zone size. A more accurate estimate of plastic zone size

can be obtained by taking the necessary re-distribution of crack tip stresses (which accompanies

yielding) into account. This leads to a larger plastic zone size as indicated in the Fig. 2.9 R orrp. It was shown that,

R = 2ry with effective crack length, aeff = a + ry (2.5.3)

where ry represents a plasticity correction to crack length which should be applied when crack tip

plasticity is relatively extensive, e.g. under plane strain conditions. Under such cases the stressintensity factor is corrected iteratively through taking account of the effective crack length. The

procedure first calculates K using the actual crack length, then finds ry using this value of K. aeffis then found and the Kvalue recalculated. This iteration can be continued further if necessary.

Plasticity is important in fracture mechanics, as the extent of plasticity, relative to specimen

dimensions and crack size, determines the state of stress (plane strain or plane stress) andwhether LEFM is applicable or not. In turn, stress state affects the direction of planes of

maximum shear stress and hence the fracture plane. Thus fracture proceeds perpendicularly tothe maximum principal stress in plane strain, and at 45

to this direction in plane stress.

As a general rule, the stress state approaches plane strain when the plastic zone is about 1/15 of

the crack length and material thickness. Plane stress occurs when the size of plastic zone tendstowards the material thickness. If the plastic zone is of the same order of size as the crack


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length, LEFM would not be valid and yielding fracture mechanics (YFM or elastic-plastic

fracture mechanics-EPFM) parameters must be used. A schematic view of the plane strain

and plane stress plastic zones with reference to the cracked body in Fig. 2.5 is given in Fig. 2.10.

Fig. 2.10. Schematic plane strain, plane stress plastic zones (Ref also Fig. 2.5) [B6].

2.6. LEFM FRACTURE TOUGHNESS (KIC) TESTING

Standard fracture toughness tests are designed to allow reproducible determination of therelevant fracture characterising parameter like plane strain fracture toughness, K1C, the J-integral

or crack tip opening displacement, CTOD. Hence certain conditions have to be met in the tests

regarding specimen and crack geometry, loading parameters and shape of load-displacement

curve, before a valid result can be reported. Bodies like the British Standards Institution (BSI)and the American Society for Testing and Materials (ASTM) have developed standards for


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testing of metallic and other materials. ASTM, E 399, Standard test method for plane-strainfracture toughness of metallic materials, is discussed in the next BOX.

Fig. 2.11. Bend and C(T) Specimens as per ASTM E 399 Standard [B6, B48].


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Relevant features of the ASTM E 399 Standard [B21] :

Two of the specimen types: namely, the Three-Point Bend (TPB) Specimen and the Compact TensionC(T) specimens are depicted in Fig. 2.11 along with the K-calibration expressions.

Validity of the K1C result depends on the shape of the force vs displacement record, specimen size and

crack geometry, and the 0.2% proof strength and toughness of the material at the test temperature. Avalid result requires that:

2

, , ( ) 2.5 IC

YS

Ka B W a

(2.6.1)

These conditions restrict the plastic zone size to less than 1/15 of the relevant dimension and thereby

ensure that plane strain and LEFM conditions prevail.

During the fatigue pre-cracking (to ensure that a sharp enough defect similar to natural cracks is presentto give a lower bound value of toughness), there are restrictions on: (i) the fatigue loading during the

final 1.3 mm or 50% of pre-crack extension (whichever is lower); (ii) crack length (0.45 < a/W < 0.55 ),

difference between crack length measurements on the specimen surfaces (when measured to

+/-0.05 mm this shall not exceed 15% of the average of the two measurements). Figure 2.12(i) showsthe appearance of a typical fatigue crack. Crack length is measured at 8 equidistant points (three points

are shown in Fig. 2.12(ii)c along the crack-front on the fracture surface (plus mean of two surface

cracks counted as one measurement to give 9 values), the difference between any two measurements

must be less than 10% of the initial crack length ( a0 = average of 9 values, called 9-point average); and

(iii) the plane of the crack must always be within 10 of the plane of crack extension. A 9-point average

crack length measurement on a C(T) specimen is illustrated in Fig. 5.1 (Section 5).

During the fracture toughness test, load is applied at a set rate, corresponding to Kchanging in the range0.5 MPams-1 to 3.0 MPams-1, while a trace of load versus displacement () is recorded (see NOTE*).

The appearance of a trace suitable for K1C determination will conform to one of three types shown inFig. 2.12(ii)a. F i g u r e 2 . 1 2 ( i i ) b shows the c on st ru cti on u se d t o o bt ai n t he

fracture load, PQ, which on substitution in the appropriate K expression givesan apparent (tentative) fracture toughness, KQ. The purpose of the 5% secant

offset construction and the read-out of the corresponding secant offset load, P5,is to ensure that at PQ, maximum crack extension (a) is < 2% of the initial crack

length, a0. In Type-I (Fig. 2.12(ii)a), all loads prior to the displacement corresponding to P =

P5 (i. e., = P=P5) are less than P5 and hence P5 =PQ. In Type-II curve, there

is a load maximum prior to P5 (this called a pop-in: pop-ins giving load changes of B, the wall thickness. This allows the part-through crack to penetrate the wall

(Figure b).

4. Once wall penetration occurs, the part-through crack very quickly grows through the ligaments to become a

through-thickness crack with a length l1 = 2c, where 2c is the surface length of the part-through crack at wall

penetration (Figure c). Hence the aspect ratio of the part-through crack is an important parameter in a leak-

before-break analysis. Remember also that the length l of a through-thickness crack is defined as 2a when you

are substituting fora in stress intensity equations.

5. The value oflcrit must be > l1.6. Calculate the time for the crack to grow from l

1

to lcrit

. If the leak rate of fluid is detectable in this time, then a

leak-before-break design case is established.

Figure a Figure b Figure c


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WORKED EXAMPLES/PROBLEMS IN LEFM

Example 2.1.

Rocket motor casings may be fabricated from either of two steels :(a) low alloy steel yield 1.2 GPa toughness 70 MPam,

(b) maraging steel yield 1.8 GPa toughness 50 MPamThe relevant Code specifies a design stress of yield/1.5. Calculate the minimum defect sizewhich will lead to brittle fracture in service for each material, and comment on the result (thislast is important).

Solution: Assume KI = a with centre crack 2a. Fracture occurs at the design stress and thecorresponding critical defect size to be determined.

(a) Low alloy steel: Design stress, = (1200 MPa/1.5) = 800 MPa.

Therefore, critical a =2 2

IC

2 2

70

800

K

= 2.43.10

-3m = 2.43 mm and 2a = 4.9 mm

(b) Maraging steel: Design stress, = (1800 MPa/1.5) = 1200 MPa.

Therefore, critical a =2 2

2 2

50

1200

ICK

= 5.53.10

-4m = 0.553 mm and 2a = 1.1 mm

Comments: In case (b), the critical defect size of 1.1 mm is dangerously small, may be

even below the level of detectability; hence, use of a higher toughness alloy or reductionof design stress is suggested.


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Example 2.2.

The CTS test piece is from a 1.2 GPa (Yield Stress) steel. If the failure load is 10 kN, what fracture

toughness is indicated? Is the result valid? Specimen dimensions in mm.

Note that width and crack size are reckoned from the load's line of action.

Solution: From Table 2.1, the relevant SIF calibration is that for Finite Plate, Edge Crack, Tensile

Force (Case(c)-CTS) as given below:

I

PK Y a

bW

with

5.23 (5.16 5.88)

1 1.07Y

, where = crack aspect ratio = a/W.

Given that, W= 20 mm, b = B = W/2 = 10 mm.

a = (W- 9) = 20 9 = 11 mm and = 11/20 = 0.55, which gives Y= 8.644.At failure, P= 10 kN and YS = 1200 MPa.

Then,3

3

3 3

10 108.644 11 10

10 10 20 10

IC

PK Y a

BW

= 80.3 MPam

Validity condition is given by Eq. (2.5.1) as:2

, , ( ) 2.5 IC

YS

Ka B W a

2

80.32.5

1200

= 0.011 m = 11.1 mm > a (11 mm), (Wa) (9 mm) and < B (20 mm).

Comments: As the evaluated SIF value of the test does not satisfy the ASTM E 399

validity condition for a and B, the Plane Stress Fracture Toughness, KC = 80.3

MPam is not a valid KIC.This is an instructive problem in the sense that, unlike other mechanical tests, like,

tension, hardness, etc., in a fracture mechanics test, it is not possible a priori to

ensure that the test will result in valid/conclusive results. Based on the result, as in

the present case, we may have to repeat the test using larger specimens and alteredtest parameters to obtain a valid result. Hence, a fracture toughness test is costly,

time consuming and sophisticated because of the stringent requirements on test

conditions/specimen dimensions.


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Example 2.7.4.

The long strip may be made from either of the two materials :(a) tough, weak yield 700 MPa plane strain toughness 100 MPam

(b) brittle, strong yield 1400 MPa plane strain toughness 50 MPamA central crack extends through the strip. Plot, as a function of crack length, the failure stressfor each material due to the separate mechanisms of elastic fracture and plastic collapse.

Comment on the trends of these graphs.

Assumptions:

(i) Elastic Fracture: The governing equation is: KI = a, fracture occurring at KI =KIC and critical fracture stress c = KIC/a.

(ii) Plastic Collapse: Plastic collapse occurs when the net section stress equals the

yield stress, i. e., for the centre cracked panel considered here,

( 2 ) (1 2 ) (1 2 )net YS

P P

a aB W a BWW W

OR (1 2 )

YS

a

W

So we have to plot c and net as a function of 2a or2a/W for the two cases givenabove.

Solution: The calculated results are plotted for the Cases (a) and (b) in Figs. (a) and (b) below

in the next Box: Example.2.7.4b.

It may be noted that the curve showing the brittle fracture stress is a parabola due itsdependence on a, whereas the plastic collapse stress is a straight line varying from YS at2a/W = 0 to zero stress at 2a/W = 1.

Example 2.3.

The toughness of a 700 MPa yield structural steel is estimated to be 140 MPam. What size and massof SEN bend test specimen is necessary, and what capacity of testing machine would be required ?Assume an a/W = 0.5 and steel density = 7.9 gm/cc.

Solution: From Table 2.1 (Case (e)), the SIF expression for the SEN Bend geometry is as givenbelow:

6I

PK Y a

bW

with

1.12 (3.43 1.89)

1 0.55Y

The relative dimensions of the ASTM E 399 SEN Bend specimen are given in Fig. 2.11; where B

varies from 0.25 to 1W. Usually B = WorW/2. Here it is assumed that B = 0.5W, with a view toreduce specimen material.

For an a/W= 0.5 = , Y= 1.424. For a KIC = 140 MPam, from the ASTM validity Eq. (2.6.1),

2

, , ( ) 2.5 IC

YS

Ka B W a

,

with YS = 700 MPa,

2

2.5 IC

YS

K

=

2140

2.5700

= 0.1 m = 100 mm

Hence, for a B = 100 mm, W= 200 mm and a = 100 mm and span, S= 4W= 800 mm and specimentotal length, L > 800 mm.

Specimen Mass:

For the above specimen, Volume = 800 x 200 x 100 mm3

= 16000 cc and this gives

a specimen mass = 16000 x 7.9 x 10-3 kg = 126.4 kg, conservatively as actual L > 800 mm.

Load and Test Machine Capacity:For this we must estimate the fracture load. Recasting the SIF expression for the SEN specimen givenabove, load, P is given by:

6

ICK bW

Y a

=3 3

3

140 100 10 200 10

6 1.424 100 10

= 584685 N ~ 590 kN

So, the minimum load capacity required for the machine is: 590 kN.

F or a 100 mm C(T) specimen (with B = (W/2)), foll owing the proportions of F ig. 2.11, the machine

capacit y requir ed is 2052 kN, th e specimen mass bein g 48 kg.

Comments: The above is a very illustrative and instructive problem that gives an idea of the specimen dimensions andmasses and machine capacity required to successfully measure a valid fracture toughness. In the present case a medium

strength alloy was used, and for low-strength steels specimens with B = 300 to 400 mm have been tested. One disadvantageof the bend specimen is that it requires large amount of material, with comparatively lower machine capacity because of large

S/Wratio. C(T) specimen economises on material, hence the name COMPACT.


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38

Example 2.4.

The long strip may be made from either of the two materials :

(a) tough, weak yield 700 MPa plane strain toughness 100 MPam(b) brittle, strong yield 1400 MPa plane strain toughness 50 MPamA central crack extends through the strip. Plot, as a function of crack length, the failure stress

for each material due to the separate mechanisms of elastic fracture and plastic collapse.

Comment on the trends of these graphs.

Assumptions:

(iii) Elastic Fracture: The governing equation is: KI = a, fracture occurring at KI =KIC and critical fracture stress c = KIC/a.

(iv) Plastic Collapse: Plastic collapse occurs when the net section stress equals theyield stress, i. e., for the centre cracked panel considered here,

( 2 )(1 2 ) (1 2 )

net YS P P

a aB W aBW

W W

OR (1 2 )YS aW

So we have to plot c and net as a function of 2a or2a/W for the two cases givenabove.

Solution: The calculated results are plotted for the Cases (a) and (b) in Figs. (a) and (b) belowin the next Box: Example.2.4a.

It may be noted that the curve showing the brittle fracture stress is a parabola due itsdependence on a, whereas the plastic collapse stress is a straight line varying from YS at2a/W = 0 to zero stress at 2a/W = 1.

(cond. To 2.4a)


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Example 2.4a (continuation of Example 2.4)

2a/W

0.0 0.2 0.4 0.6 0.8 1.0

RemoteStress,

0

200

400

600

800

1000

1200

1400

1600

1800

2000

700MPaYS

ICc

K

a

(1 2 )YS

a

W

Fig. a. Plastc Collapse and Elastic Fracture conditions for Case(a)

Tough, Weak Alloy: KIC

= 100 MPa.m0.5

Fig. b. Plastc Collapse and Elastic Fracture conditions for Case(b)

0.0 0.2 0.4 0.6 0.8 1.0

RemoteStress,

0

200

400

600

800

1000

1200

1400

1600

ICc

K

a

(1 2 )YSa

W

1400MPaYS

2a/W

Brittle, Strong Alloy; KIC = 50 MPa.m0.5

Comments: In the case of the high toughness alloy, Case(a), as shown in Fig-(a), the plasticcollapse stress is below that for brittle fracture. H ence L EF M toughness cannot be measured.

For the case (b), as shown in Fig. (b), critical fracture conditions prevail in the intermediate 2a/Wregions with plastic collapse at low and high stress (high and low crack lengths respectively)

regions.


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Example 2.5. HIGH STRENGTH vs FRACTURE TOUGHNESS

A welded structure is to be fabricated from large sheets of 0.45C-Ni-Cr-Mo steel. The detection limit

of available NDT techniques limits the critical defect size to sizes > 3 mm, as cracks smaller than thisare not detectable. A design stress level of half the tensile strength is proposed.To save weight in the structure it has been suggested that the steel could be heat treated to a highertensile strength level. The current grade has a tensile strength of 1520 MPa, and a candidatereplacement grade has a 2070 MPa strength level. Is this change supportable in fracture mechanics

terms?You may assume plane strain conditions in all computations, and the figure below indicates therelationship between fracture toughness and tensile strength for this steel.Compare the allowable stress levels, and hence weights, in both grades of steel for an allowable initialdefect size of approximately 5 mm.

UTS= where =

2K a

Solution:

From the data in the figure, KIC of the 1520 MPa grade is 66 MPa m, while K1C = 33 MPa m

for the

2070 MPa grade. A through-crack in a large sheet can essentially be treated as one in an infinite plate.

(Cond. To Example 2.5a in the next Box )


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Example 2.5a (continued from Ex. 2.5)

For the alloy heat treated to 1520 MPa, we get:

c

66MPa m 760MPa

2.4mm total flaw size = 4.8mm

a

a

This critical flaw size is larger than the minimum NDT detection limit, and this steel is safe to

use. For the 2070 MPa grade, however, the equation gives:

c

33MPa m 1035MPa

0.33mm total flaw size=0.66 mm

a

a

Thus it is not possible to detect critical defects in this grade before fast fracture occurs. To

allow a critical defect size of 4.8 mm, i. e.,2a, in both grades, for the 2070 MPa grade:

3

33MPa m380MPa

2.4 10 m

Hence for a similar flaw tolerance level, the allowable stress in the higher strength alloy is

half that in the 1570 MPa grade - this would imply a two-fold increase in weight of a

component. Hence, the change is unwise.

Comments: This question illustrates the effect on critical crack length of the loss in fracture

toughness that generally accompanies an increase in tensile strength.


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Example 2.6. QUENCHING and RESIDUAL STRESSES

During water quenching of steel components with a section thickness of 30 mm, heat transfer

calculations indicate that a peak stress of 130 MPa is generated in the section. Prior to heat treatment,the components were ultrasonically inspected to detect defects. The inspection technique has aminimum detection size of 0.5 mm.a) What type of defect will be most critical?

b) Calculate the size of defect which would cause fracture of the component during the quenching

operation, given that the aspect ratio of the crack is 2c/a = 10.c) Would this inspection procedure guarantee integrity of the component if the quenching stressesapproached the proof stress of the steel?Note: K1C = 30 MPa m

and the proof stress = 620 MPa. The stress intensity calibration for thiscomponent and crack geometry is given in the figure below (where the subscript y indicates proof

stress).

Where, for surface flaws:1

2

1.1a

KQ

and for embedded flaws:

1

2

aKQ

Solution Cond. in the Next Box-2.6a


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Example 2.6a (Continuation of 2.6: QUENCHING and RESIDUAL STRESSES)

Solution:

a) From inspection of the stress intensity solutions for surface and embedded flaws it is

obvious that, because of the factor of 1.1 for surface defects, they will become critical at

smaller values ofa than embedded defects.

b) For substitution into the K equation for surface defects, Q must be determined. From the

graph,

For 0.21 and 0.1, 1.12y

aQ

c

2 2

IC

2 2

1.1 1.1 30= = 15.4 mm

1.21 1.21 130c

Ka

This flaw is very much bigger than the NDT detection limit and there should be very little risk

of failure during the quenching.

b) If the quenching stresses approached the proof stress of the material, the situation changes

dramatically. Surface defects are still critical, but the value of Q has changed:

For 1 and 0.1, 0.882y

aQc

2 2

IC

2 2

0.88 0.88 30= = 0.54 mm

1.21 1.21 620c

Ka

As the critical size of defect is around the NDT detection limit, the inspection would not

guarantee integrity. A change to the quenching procedure would have to be implemented, e.g.using a slower quenchant.

Comments: This problem brings out the importance of heat treatment conditions in

controlling residual stresses and their effect on critical defect size.


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Example 2.7: FRACTURE TOUGHNESS TESTS

The figure below shows the load line displacement trace recorded from a standard sized

compact tension fracture toughness specimen. Specimen thickness was 25 mm, the cracklength at fracture was 25 mm and the steel alloy had a yield strength of 650 MPa.

a) Calculate PQ and hence KQ. Apply the required checks on plastic zone size, stress state andplasticity during the test to determine whether KQ is a valid plane strain fracture toughness

value.

b) What is the maximum K1C value that can be determined for this steel using 25 mm thickspecimens?

Note: The K equation for C(T) specimen is given in Fig. 2.11 and Table 2.1 (the former ismore accurate).

(Continued to 2.7a)


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Example 2.7a (Continuation ofExample 2.7: FRACTURE TOUGHNESS TESTS)

a) To find PQ, a line is constructed from the origin with a slope 5% less than that of the tangentto the initial straight line part of the load-displacement record. This line is shown in the figure

above. This line intersects the load-displacement trace at 19 kN, and there is no previoushigher value of load on the trace.

Hence PQ = 19 kN.

To find KQ the function f(a/W) (which is the finite geometry correction factor) must becomputed. From the standard specimen size (see Fig. 2.8), W = 2B = 50 mm, hence a/W =25/50 = 0.5:

-3Q

Q 0.5 0.5

From Fig. 2.8, ( ) = 9.66 and

( )19 10 9.66

= = 32.8 MPa0.025 0.05

af

W

aP f

WK mBW

Validity Check on KQ

(i).

2 232.8

, , ( ) 2.5 = 2.5 , i. e., 6.4 mm650

Q

YS

Ka B W a

(ii). Final constraint to ensure absence too much plasticity:

Pmax/PQ = 21 kN/19 kN = 1.105 < 1.1

As the conditions are satisfied, KQ is KIC = 32.8 MPam.

Comments: Essentially, the first check ensures that the crack tip plastic zone is a small enough

percentage of crack length to ensure LEFM, the constraint on B helps to ensure that plane strain

conditions prevail, while the condition on (W - a) ensures that a plastic hinge does not develop ahead

of the crack.

b) The maximum K1C value that can be determined from this thickness of specimen in steel of

this grade, called K-capacity of the specimen, is found when B fails to satisfy the ValdityCheck Condition (i) above; i.e., when,

2

Q YS

0.0252.5 OR K > = 650 65 MPa m

2.5 2.5

Q

YS

K BB

Comments: This question illustrates the application of constraints for determining valid K1C values

from fracture toughness tests.


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Example 2.8: PLASTIC ZONE EFEECT

A thin plate of steel contains a central through-thickness flaw of length 16 mm, which is subjected to a stress of

350 MPa applied perpendicularly to the flaw plane. The 0.2% flow stress of the material is 1400 MPa.

Calculate the plastic zone size and the effective stress intensity level at the crack tip, making reasonable

assumptions about the state of stress.

If, after heat treatment, the flow stress of the steel dropped to 385 MPa, what would the plastic zone size be

under the applied stress of 350 MPa, and what conclusions would you draw about the use of LEFM?

Assumptions: (i) The plate is large compared to the size of the crack so that the simple infinite plate formula for

stress intensity factor applies and a = 8 mm (half the central crack length). That is,

=

= 350 0.008 = 55.49 MPa

K a

K m

(ii) The steel plate is in a state of plane stress as it is stated to be 'thin'. Plane stress prevails if the ratio of plate

thickness to plastic zone size tends towards 1, while plane strain prevails if it tends towards 15. Plane stress is

also a conservative assumption, in that Kvalues are higher (through Irwin's plastic zone correction) when plastic

zones are bigger.

Solution:

Irwin's plastic zone correction factor to crack length is given by:

2 2

-4

p

1 1 55.49= = 2.5 10 m (i.e., 0.25 mm)

2 2 1400YS

Kr

This is small compared with the crack length and its effect on Kwill be correspondingly small:

3eff p= ( ) = 350 (8.25 10 ) = 56.35 MPaK a r m

This is around a 1.5% change and thus a single iteration of the calculation is sufficient.

However, if the flow stress drops to 385 MPa after heat treatment, the plastic zone size now becomes:

2

-3

p

1 55.49= = 3.31 10 m = 3.31 mm

2 385r

3

eff p= ( ) = 350 (11.31 10 ) = 65.97 MPaK a r m

This represents a correction of around 18.9% and the use of LEFM becomes dubious. This is confirmed by thefact that the applied stress (350 MPa) is now some 91% of the flow stress. A yielding fracture mechanics

parameter should be used to characterise the propensity for fracture.


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Example 2.9. SPECIMEN THICKNESS EFECT

Catastrophic fracture occurred in a thick steel plate during proof testing, at an applied stress of

700 MPa. The initiating defect was an embedded sharp penny-shaped flaw with a radius of2.5 cm. Calculate the fracture toughness of this steel.

It is desired to check this value by determining the plane strain fracture toughness fromstandard tests. The yield strength of the steel is 1100 MPa. A sheet of nominally similar steel,7.5 mm thick, is available. Is this sufficiently thick to obtain a valid K1C value? If not, what

thickness of steel should you order?

Given: The stress intensity solution for an embedded circular crack is:

2=K a

Substitution of values into the above formula gives:

2= 700 0.025 = 124.9 MPaK m

If this is assumed to be a valid plane strain fracture toughness value, then the minimumspecimen thickness required is given by:

2 2

IC

YS

124.9> 2.5 = 2.5 = 0.0322m = 32.2 mm

1100

KB

Comments: This is much larger than the plate thickness of 7.5 mm. Hence the thickness ofsteel in stock is insufficient to provide a valid K1C value. A thickness > 35 mm is required.


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Example 2.10. LBB PROBLEM:

a) The stress intensity solution for a semi-elliptic flaw in tension is given below. The K1C

value for a Ti-6Al-4V titanium alloy with a yield strength of 910 MPa, is 115.4 MPam

.Determine the size of the largest stable surface flaw (a/c = 0.4) in a 40 mm thick plate of thisalloy, for a design stress in the plate of 75% of the yield strength. This requires assuming an

initial value ofa/B and iterative calculations of stress intensity, if necessary.

For the semi-elliptic surface flaw, =Y a

K

and the values of Y and are given in

the TABLE below.

b) For the same alloy and design stress, calculate the maximum wall thickness of a pressurevessel which could be designed on a leak-before-break criterion. You may assume that the

aspect ratio (a/c) of the surface flaw remains constant at 0.4, and that for the through-thickness crack:

=K a

What thickness of plate would you order for the vessel?

continued to 2.10a


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Example 2.10a (continuation of 2.10;LBB PROBLEM)

TABLE

a/c Phi Y

a/B0.2 0.4 0.6 0.8

1.051 0.2 0o

45o

90

0.617

0.9901.173

0.724

1.1221.359

0.899

1.3841.642

1.190

1.6571.851

1.151 0.4 0o

45o

90

0.7670.998

1.138

0.8961.075

1.225

1.0801.247

1.370

1.3181.374

1.447

1.277 0.6 0o

45o

90

0.916

1.0241.110

1.015

1.0621.145

1.172

1.1821.230

1.353

1.2431.264

1.571 1.0 0o

45o

90

1.1741.067

1.049

1.2291.104

1.062

1.3551.181

1.107

1.4641.193

1.112

Solution:

a) From the stress intensity solution, the highest K values relate to the maximum depth position (Phi =

90o) and this position is used to find crack depth a. As we do not know the critical crack depth, we willhave to assume a value of a/B to calculate acrit using K1C, and then check whether this gives a value ofa/B close enough to our initial estimate. If not, then we must iterate through the calculation using a

more refined estimate ofa/B.Consider a first estimate ofa/B of 0.2, this gives:

IC

3

c

1.138 0.75 910= = 115.4 MPa

1.151

a 9.3 10 m = or 9.3 mm

caK m

From this, we can find a/B = 9.3/40 = 0.233. This is reasonable close to our estimate of 0.2, but wecould improve the prediction by linearly interpolating for the 'correct' value of a/B. This interpolation

gives a/B = 1.138 + [(0.033/0.2) x (1.225 - 1.138)] = 1.152 - a change of only 1.2%. Hence it is

acceptable to leave the critical crack size as 9.3 mm. If one redoes the calculation, however, acrit = 9.20mm.

(continued to 2.10b)


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Example 2.10b (continuation of 2.10a)

Solution (b):

b) The leak-before-break criterion requires the vessel to be stable (i.e. not to suffer fracture) inthe presence of a surface crack which penetrates the wall (a = B) and very quickly grows to

become a through-thickness crack with athrough-thickness crack = csurface crack. Essentially, this means

that we have to check for stability in the presence of both a surface crack and a through-thickness crack. As the surface length of the semi-elliptic crack is required in calculating K

for the through-thickness crack, we start with the semi-elliptic crack.

Surface crack: a/c = 0.4 and a/B = 0.8 (highest value given in the table, but the ratio of a/Bshould really be 1):

IC

c

1.447 0.75 910

= = 115.4 MPa m1.151

= 5.76 mm

ca

K

a

The value ofcfor this crack is 5.76/0.4 = 14.4 mm

Through-thickness crack:

IC c= 0.75 910 = 115.4 and a 9.1 mmK a

Comments: However, this value of 9.1 mm is less than the value of 14.4 mm found from thesurface crack. Hence the through-thickness defect is critical. The maximum thickness of plate

in which both types of crack would be stable is 9.1 x 0.4 = 3.64 mm. Thus one might specify

that the vessel would be made using plate < 3.5 mm thick, although the safety margin then israther low.

Note: This analysis assumes that such a plate thickness would be adequate to carry the

applied load, i.e. that design stresses would have a maximum value 0.75x910 = 682.5 MPa.This may not be feasible with such a thin wall, or the proposed thickness of 3.5 mm may not

meet other design criteria (corrosion, deflectional stability etc) and the design may have to be

re-assessed. Thus the problem illustrates the possibility of conflict between different design

criteria. Part of the skill of an engineer lies in optimising these types of conflict.


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Example 2.11. Radial Cracks around Cylinders [B31]

A commonly encountered surface crack configuration under a remote applied tension, torsion or

a combined loading system is that shown the Figure next.

The stress intensity factors for the loading system illustrated in the Figure are:

I III= f and = gd d

K a K aD D

For an applied torque T, the torsional shear stress becomes3

16=

T

D

and the correction factors

are: Cond. In Next Box 2.11(a)


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Example 2.11a (continuation of Example 2.11)

2 31 1 3 5 11

f =2 2 8 14 15

d D D d d d

D d d D D D

and

2 2 33 1 3 5 35 16

g =8 2 8 16 128 32

d D D D d d d

D d d d D D D

The crack length (size) is estimated as-

=2

D da .

PROBLEM: Two identical high-strength steel rods are prepared: one for a tension test at 106 MPa andone for torsion at 69 MPa. Calculate KI and and KIII. The rod dimensions are d= 4 mm and D = 8 mm. If

KIIIC = (3/4)*KIC, (a) will the rods fracture? Explain; (b) Calculate the theoretical tensile and torsionfracture stresses, if fracture does not occur in (a). Use KIC = 25 MPam.

Solution:

Given: = 106 MPa; = 69 MPa; KIC = 25 MPam; a = (D - d)/2 = (8 4)/2 = 2 mm; (d/D) = 0.5 and(D/d) = 2. Then from the above equations, the correction factors are: f(d/D) = 1.9 and g(d/D) = 2.91.

Hence the applied SIFs and Mode-III fracture toughness values are:

Thus answer to question (a) is that the rods will not fracture as the applied SIFs in tension and

torsion are less than the respective fracture toughness values; KI

< KIC

and KIII

< KIIIC

.

(b) The fracture stresses are:

IC IIIC25 21.65= = = 167 MPa and = = = 94 MPa1.9 * * 2 /1000 2.91 * 2 /1000

ff f

K K

d da g a

D D

I

III

IIIC IC

=

73666136 introductory fracture mechanics

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