Autofrettage Technology and Its Applications in Pressured Apparatuses 9782759831128

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Current Natural Sciences

Ruilin ZHU, Guolin ZHU and Zhaoyang LIU

Autofrettage Technology and Its Applications in Pressured Apparatuses

Printed in France

EDP Sciences – ISBN(print): 978-2-7598-3111-1 – ISBN(ebook): 978-2-7598-3112-8 DOI: 10.1051/978-2-7598-3111-1 All rights relative to translation, adaptation and reproduction by any means whatsoever are reserved, worldwide. In accordance with the terms of paragraphs 2 and 3 of Article 41 of the French Act dated March 11, 1957, “copies or reproductions reserved strictly for private use and not intended for collective use” and, on the other hand, analyses and short quotations for example or illustrative purposes, are allowed. Otherwise, “any representation or reproduction – whether in full or in part – without the consent of the author or of his successors or assigns, is unlawful” (Article 40, paragraph 1). Any representation or reproduction, by any means whatsoever, will therefore be deemed an infringement of copyright punishable under Articles 425 and following of the French Penal Code. The printed edition is not for sale in Chinese mainland. Ó Science Press, EDP Sciences, 2023

Foreword

Pressure vessels are one of the pressured apparatuses, which are widely used in many industrial fields, such as mechanical engineering, chemical engineering, petrochemical engineering, pharmaceutical factory, energy industry, material industry, food industry, metallurgical industry, oil industry, architectural engineering, aeronautical engineering, aerospace industry, weaponry industry, etc. This book focuses on the study of pressure vessels as one of the pressured apparatuses. The basic requirement of pressure vessels is safety and economy. Security is the core issue. It is necessary to maximize the economy under the premise of fully guaranteed security. The strength and structure of pressure vessels are the most important factors that influence safety. The stresses in the wall of a pressure vessel must be reduced and limited in a safe range in structural design so as to ensure the strength of the pressure vessel. The circumferential stress on the inner surface is quite great and uneven, using high or ultra-high-strength material or increasing the thickness cannot solve the problem, introducing pre-stresses into the wall is an excellent way to reduce the stresses in the wall, and raising the load-bearing capacity of a pressure vessel. The autofrettage technology is an effectual measure to introduce pre-stresses into the wall. After a pressure vessel is processed with autofrettage technology, the circumferential stress on the inner surface is lowered and the distributions of the stresses in the wall become even because of the superposition of the operation stresses and the pre-stresses (residual stresses or thermal stresses), and all stresses are within the elastic range, the elastic operating range enlarged and the elastic load-bearing capacity raised. Especially, when the operating pressure is greater than the yield pressure of the inner surface or initial yield pressure, processing with autofrettage technology for a pressure vessel becomes more necessary. At present, there is no monograph on autofrettage technology in the world. This book is written based on our research work and absorbs our latest research results. In this book, the systematic theory of autofrettage technology of pressure vessels is established and studied, and a large number of application examples are presented. DOI: 10.1051/978-2-7598-3111-1.c901 Ó Science Press, EDP Sciences, 2023

IV

Foreword

The whole book consists of seven chapters. Chapter 1 develops the theories of elastic-plastic stresses of pressure vessels based on the maximum shear stress theory (Tresca yield criterion) and the maximum distortion strain energy theory (Mises yield criterion). Chapter 2 discusses the mechanical autofrettage technology according to the maximum shear stress theory (Tresca yield criterion). Chapter 3 studies the mechanical autofrettage technology according to the maximum distortion strain energy theory (Mises yield criterion). Chapter 4 investigates the mechanical autofrettage technology by limiting circumferential residual stress based on maximum distortion strain energy theory (Mises yield criterion). Chapter 5 studies the mechanical autofrettage technology when the operating pressure is several times of the initial yield pressure according to the maximum shear stress theory and the maximum distortion strain energy theory. In consideration of that when there is a temperature difference between the inner and outer surfaces, thermal stresses will appear in the wall of a pressure vessel and the thermal stresses can be used as pre-stresses, chapter 6 studies thermal autofrettage technology according to the maximum shear stress theory (Tresca criterion), and chapter 7 studies thermal autofrettage technology according to the maximum distortion strain energy theory (Mises criterion). This book tries to clarify the ideas, introduce the principle, and show the effect of autofrettage technology. In this book, interesting and useful results for engineering applications are presented, the laws contained in the autofrettage theory are revealed, the essential cause and reason for the obtained results are analyzed, the inherent and meaningful relations between various parameters in autofrettage theory are brought to light, and the safe depth of plastic region (overstrain) and the conditions of loading or the optimum operation conditions are found out. This book pays attention to the rigor of the theoretical deduction and the maneuverability of the engineering practice, it serves three purposes. The first purpose is to provide a reference book for the analysis and design of pressured apparatuses that are widely used in industries. The second purpose is to help students and graduate students have a good command of the theory and applications of autofrettage technology of pressure vessels. The third purpose is to aid engineers in understanding the background and cause and effect of some of the design equations in related Codes. Thanks to the authors of the references. The authors are aware that due to the limited level of the authors, there must be some mistakes in the book. Please do not hesitate to point them out.

Ruilin ZHU Hunan Normal University, Hunan University of Information Technology June 29, 2022 Email: [email protected]; [email protected]

Contents Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

III

CHAPTER 1 Study on Elastic-Plastic Mechanical Stresses in Cylindrical Pressure Vessels .

1

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1 2 7 17 20

CHAPTER 2 Mechanical Autofrettage Technology Based on Tresca Yield Criterion . . . . . .

21

1.1 Introduction . . . . . . . . . . . . . . . . . 1.2 Studies of Elastic Stresses . . . . . . . 1.3 Analysis of Elastic-Plastic Stresses 1.4 Chapter Summary . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . .

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56 70 77 81

CHAPTER 3 Mechanical Autofrettage Technology Based on Mises Yield Criterion . . . . . .

83

2.1

General Study on Mechanical Autofrettage Technology . . . . . . . . . . 2.1.1 In General Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 The Critical Radius Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.3 The Optimum Plastic Depth kj* (kj is Written as kj*) . . . . . . 2.1.4 The Results When kj = kj* . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Mechanical Autofrettage Technology Under Entire Yield State . . . . 2.3 Mechanical Autofrettage Technology with Radius of Elastic-Plastic Juncture Being Arithmetic Mean Radius of Inside Radius and Outside Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Mechanical Autofrettage Technology with Radius of Elastic-Plastic Juncture Being Geometrical Mean Radius of Inside Radius and Outside Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Mechanical Autofrettage Technology with Minimum Equivalent Total Stress on Elastic-Plastic Juncture . . . . . . . . . . . . . . . . . . . . . 2.6 Comparison Between Three Cases . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.1

. . . . . .

General Study on Mechanical Autofrettage Technology . . . . . . . . . . . . 3.1.1 In General Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83 83

VI

Contents

3.1.2 The Critical Radius Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 The Optimum Plastic Depth kj* (kj is Written as kj*) . . . . . . 3.1.4 The Results When kj ¼ kj or k 2 lnkj2  k 2  kj2 þ 2 ¼ 0 . . . . . 3.2 Mechanical Autofrettage Technology Under Entire Yield State . . . . 3.2.1 The Residual Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 The Total Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Mechanical Autofrettage Technology with Radius of Elastic-Plastic Juncture Being Arithmetic Mean Radius of Inside Radius and Outside Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 The Solutions with Radius of Elastic-Plastic Juncture Being Geometrical Mean Radius of Inside Radius and Outside Radius . . . 3.5 The Solutions with Minimum Equivalent Total Stress on Elastic-Plastic Juncture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Comparison Between the Three Cases . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 k = 2.5 > kc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 k = 2 < kc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 98 . . 99 . . 100 . . 109 . . 109 . . 111

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133 146 147 150 153 157

CHAPTER 4 Mechanical Autofrettage Technology by Limiting Circumferential Residual Stress Based on Mises Yield Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 4.1

The Optimum Plastic Depth When Circumferential Residual Stress on the Inside Surface Controlled . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Distribution of Residual Stresses When Circumferential Residual Stress on the Inside Surface Controlled . . . . . . . . . . . . . . . . . . . . . . 4.2.1 General Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 The Residual Stresses for Entire Yield . . . . . . . . . . . . . . . . . 4.2.3 The Residual Stresses with Radius of Elastic-Plastic Juncture Being Arithmetic Mean Radius of Inside Radius and Outside Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 The Residual Stresses with Radius of Elastic-Plastic Juncture Being Geometrical Mean Radius of Inside Radius and Outside Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5 The Residual Stresses When Equivalent Total Stress on Elastic-Plastic Juncture is the Minimum . . . . . . . . . . . . . 4.3 The Total Stresses and the Load-Bearing Capacity When Circumferential Residual Stress on the Inside Surface Controlled . . . 4.4 Control Circumferential Total Stress Directly . . . . . . . . . . . . . . . . . 4.5 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . 159 . . 166 . . 166 . . 176

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186 196 198 202

Contents

VII

CHAPTER 5 Mechanical Autofrettage Technology Under Low Load . . . . . . . . . . . . . . . . . 203 5.1 5.2 5.3 5.4

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Optimum Plastic Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . Analysis of Residual Stresses Under the Optimum Plastic Depth . Analysis of the Stresses Caused by Internal Pressure and Total Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Analysis of the Effect of Load Ratio (λ) and Plastic Depth (kiλ) . 5.6 Analysis of Load-Bearing Capacity . . . . . . . . . . . . . . . . . . . . . . . 5.6.1 Based on Tresca Criterion . . . . . . . . . . . . . . . . . . . . . . . . 5.6.2 Based on Mises Criterion . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . 203 . . . . 204 . . . . 208 . . . . . . .

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219 228 234 234 237 240 244

CHAPTER 6 Summary of Implement Methods and Their Characteristics of Mechanical and Thermal Autofrettage Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 6.1

Implement Methods and Characteristics of Mechanical Autofrettage Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 6.2 Implement Methods and Characteristics of Thermal Autofrettage Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 CHAPTER 7 Thermal Autofrettage Technology Based on Tresca Yield Criterion . . . . . . . . 257 7.1 7.2 7.3 7.4

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derivation of Thermal Stresses . . . . . . . . . . . . . . . . . . . . . . . The Characteristics of the Thermal Stresses . . . . . . . . . . . . . The Analysis of Total Stresses and Investigation of Optimum Operation Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 The Total Stresses Under Optimum Operation Conditions . . 7.7 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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273 284 296 302 306

CHAPTER 8 Thermal Autofrettage Technology Based on Mises Yield Criterion . . . . . . . . 307 8.1 8.2

The Analysis of Thermal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Analysis of Total Stresses and Investigation of Optimum Operation Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

307 309 331 345 348

Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349

Chapter 1 Study on Elastic-Plastic Mechanical Stresses in Cylindrical Pressure Vessels 1.1

Introduction

Cylindrical pressure vessels are widely used in industries. Since the distributions of stresses caused by operation pressure in the wall of a cylinder under an elastic state are very uneven, the load-bearing capacity is thus limited seriously. The autofrettage technique is an effective and important measure to even distributions of stresses, improve safety and raise the load-bearing capacity, and extend operating life for pressure vessels. Usually, in the most commonly employed autofrettage process, a pressure vessel is pressurized to a quite high internal hydraulic pressure; as a result, the portion of the cylinder from the inner radius to some intermediate radius becomes plastic, while the remaining portion remains elastic. After releasing the pressure subjected to the pressure vessel during the autofrettage process (called autofrettage pressure, pa), the residual stresses are set up in the wall of the pressure vessel. When the pressure vessel is put into production, the operation stresses caused by the operation load are superimposed on the residual stresses so that advantageous distributions of total stresses in the wall of the pressure vessel are achieved, thus the pressure vessel can endure a higher pressure. We call these methods mechanical autofrettage because the pre-stresses are caused by mechanical forces. Mechanical autofrettage technology has been widely applied in industries, such as high and ultra-pressure reactors, high and ultra-pressure compressors, pipings, chemical equipment, petrochemical equipment, high-pressure pump, battleship, and tank cannon barrels, and fuel injection systems for diesel engines, etc. Since temperature differences in the wall of an apparatus can cause pre-stresses, heating and/or cooling the thick apparatuses to obtain pre-stresses is a promising approach for the autofrettage of thick apparatuses. We call this autofrettage technology thermal autofrettage technology, which will be investigated in the last two chapters. In autofrettage technology of pressure vessels, the variation law of the residual stresses and total stresses (the residual stresses plus operation stresses), the reasonable and the optimum plastic depth or overstrain, the reasonable and the DOI: 10.1051/978-2-7598-3111-1.c001 © Science Press, EDP Sciences, 2023

2

Autofrettage Technology and Its Applications in Pressured Apparatuses

optimum load-bearing capability, the relationship between some key parameters, the optimum temperature difference, etc., are important to the design of high-pressure apparatus. In order to discover the general law in autofrettage theory for the design of high and ultrahigh-pressure vessels, it is important to investigate the autofrettage technology deeply, systematically, and in detail. Engineering practice shows that the mechanical behaviors of the engineering materials tally with the maximum shear stress theory (Tresca’s yield criterion) and the maximum distortion strain energy theory (Mises’s yield criterion), thus the research work is based on these two theories[1]. Engineering conditions vary in thousands of ways. We bypass specific engineering conditions and carry out the research according to the following ideal conditions: (1) the materials of pressure vessels are perfectly elastic-plastic and the Bauschinger effect is temporarily neglected, the compressive yield limit of the material is equal to the tensile yield limit; (2) the strain hardening is ignored; (3) there is no any defect in the materials. The problems with autofrettage under specific engineering conditions can be resolved by reference to the results of this study on the basis of the specific engineering conditions. Elastoplastic stresses are the foundation of autofrettage technology, in this chapter; we study the elastoplastic mechanical stresses of cylindrical pressure vessels.

1.2

Studies of Elastic Stresses

When a cylindrical pressure vessel is subjected to an internal pressure p, in an elastic state, Lamé equations express the elastic stresses in three directions, i.e., the axial elastic stress rpz , the radial elastic stress rpr , and the circumferential elastic stress rph , that are as follows[2, 3]: 8 p rz 1 p > > ¼ 2 > > ry k  1 ry > > >  > < rp
k 2 rp k2  x2 p r ¼ 1 2 z ¼ 2 2 ry x ry x ðk  1Þ ry > > >
 p > p > 2 > rh k r k2 þ x2 p > > : ¼ 1þ 2 z ¼  2 2 ry x ry x ðk  1Þ ry

8 p rz 1 > > ¼ 2 > > > k 1 p > >  > < rp
k 2 rpz r ¼ 1 2 or p x p > > >
 p > p > 2 > rh k rz > > : ¼ 1þ 2 p x p

ð1:1Þ

In order to show the stress characteristics and facilitate the follow-up research, by taking k = 4, Lamé equations are plotted in figure 1.1. When it’s not necessary, we sometimes omit the superscript p, s, and d. From equation (1.1) and figure 1.1, it is seen that the distributions of the elastic stresses along with radius r or x (x = r/ri) are seriously uneven, with rph being the

Study on Elastic-Plastic Mechanical Stresses

3

FIG. 1.1 – Distributions of stresses in the wall of a cylindrical pressure vessel. greatest and rpr being the smallest (algebraic value, not absolute value). Therefore, with the internal pressure p increasing, whether based on the maximum shear stress theory (Tresca criterion) or based on the maximum distortion strain energy theory (Mises’s yield criterion), it is certain that the inside surface of the pressure vessel becomes yielded first, and then the plastic region develops towards the outside surface and retains a plastic cylinder concentric with the elastic cylinder. The equivalent stress based on the maximum distortion strain energy theory (Mises’s yield criterion) is[1] rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i 1h d ðrh  rr Þ2 þ ðrr  rz Þ2 þ ðrz  rh Þ2 re ¼ ð1:2Þ 2 For a cylindrical pressure vessel, the axial stress σz is the arithmetic mean value of circumferential stress σθ and radial stress σr, or rz ¼

rh þ rr 2

Equation (1.3) is satisfied under both elastic and plastic states. Substituting equation (1.3) into equation (1.2) leads to pffiffiffi 3 2 d ðrh  rr Þ or rh  rr ¼ pffiffiffi rde re ¼ 2 3

ð1:3Þ

ð1:4Þ

Normally rph is the first principal stress (σ1) and rpr is the third principal stress (σ3). Therefore, the equivalent stress based on the maximum shear stress theory (Tresca criterion) is rse ¼ r1  r3 ¼ rh  rr

ð1:5Þ

4

Autofrettage Technology and Its Applications in Pressured Apparatuses

From equations (1.4) and (1.5), the relationship between the equivalent stress based on the maximum shear stress theory rse, and the equivalent stress based on the maximum distortion strain energy theory rde is pffiffiffi 3 s d re ¼ ð1:6Þ r 2 e When the equivalent stress reaches the yield limit σy, or σe = σy, the material of the pressure vessel becomes yielded. Therefore, the yield condition based on the maximum distortion strain energy theory is 2 rh  rr ¼ pffiffiffi ry ð1:7Þ 3 and the yield condition based on the maximum shear stress theory is rh  rr ¼ ry

ð1:8Þ

By comparing equations (1.7) and (1.8), it is known that the results based on the maximum distortion strain energy theory can be obtained by replacing σy in the results based on the maximum shear stress theory with p2ffiffi3 ry , and the results based on the maximum shear stress theory can be obtained by replacing σy in the results pffiffiffi based on the maximum distortion strain energy theory with 3ry =2. In fact, this is the right way to transit from the Tresca criterion to the Mises criterion or from the Mises criterion to the Tresca criterion, as long as there is the functional relation as equation (1.3) between the stresses in three directions (axial, radial and circumferential directions). We call the way DS transition rule. According to the maximum shear stress theory, the equivalent stress of the stresses caused by operation pressure p (called operation stresses) rps e is rp rp 2k 2 rp rps 2k 2 p e ¼ h r ¼ 2 z ¼ 2 2 ry ry ry x ry x ðk  1Þ ry

or

rps 2k 2 e ¼ 2 2 p x ðk  1Þ

ð1:9Þ

In accordance with the above DS transition rules, the equivalent operation stresses based on the maximum distortion strain energy theory (Mises yield criterion) rpd e is pffiffiffi
p pffiffiffi 2 pffiffiffi 2  3 rh rpr 3k 3k rpd p rpd e e ¼ 2 2 ð1:10Þ ¼  or ¼ 2 2 ry x ðk  1Þ ry p x ðk  1Þ 2 ry ry rps ei

Equation (1.9) or equation (1.10) shows that when x = 1 or on the inside surface pffiffi 2 ps pd p 3k 2k 2 or rpd ei is the greatest, which is rei ¼ k 2 1 p or rei ¼ k 2 1 p. When rei = σy, the

pd pressure vessel fails to work. Then, letting rps ei = σy or rei = σy, we obtain the maximum elastic load-bearing capability or initial yield pressure of an unautofrettaged cylinder, pe

pe k 2  1 ¼ 2k 2 ry

ðTresca yield criterionÞ

ð1:11Þ

Study on Elastic-Plastic Mechanical Stresses

pe k 2  1 ¼ pffiffiffi ry 3k 2

5

ðMises yield criterionÞ

ð1:12Þ

Although increasing the thickness or increasing k can raise the maximum elastic load-bearing capability, the increase in pe is restricted greatly, especially when k is quite great, even if k is increased rapidly, pe rises very slowly, as is intuitively shown in figure 1.2.

FIG. 1.2 – The maximum elastic load-bearing capability of an unautofrettaged cylinder. From equations (1.11), (1.12), and figure 1.2, the limit of rpye is 12 or p1ffiffi3, in other words, no matter how thick a pressure vessel is, even if it is infinitely thick, the r pressure it can contain is only 0.5σy or pyffiffi3. From the above, it is thus evident that measures must be taken to raise the load-bearing capability of pressure vessel. Autofrettage technology is an effective measure to even the distributions of stresses in the wall of the pressure vessel and to raise the load-bearing capability of the pressure vessel. The key to autofrettage technology is the pre-stresses in the wall of the pressure vessel. The pre-stresses are usually achieved either by applying hydraulic pressure to the bore of the thick-walled pressure vessel or by pushing an oversized mandrel through the bore or by detonating an explosive charge inside the pressure vessel to deform it plastically, thus the compressive residual stresses on the inner surface of the pressure vessel produced, enabling it to withstand higher working pressure. We call these methods mechanical autofrettage since the pre-stresses are caused by mechanical forces. We first study mechanical autofrettage. We will study mechanical autofrettage technology in chapters 2–5. For equations (1.1) and (1.9), setting

rps e ry

rp

¼ rhy obtains x = k. When x < k,

p p ps rps e > rh , or within the whole wall, re is always larger than rh .

6

Autofrettage Technology and Its Applications in Pressured Apparatuses pd

rp

For equations (1.1) and (1.10), the setting rrey ¼ rhy obtains qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi x¼ 3  1k ¼ xa \k

ð1:13Þ

p p pd When x < xa, rpd e > rh ; When x > xa, re < rh . There is a crossover point of pffiffi p pd pd r rp p 3 . curve rrey and curve rhy at x = xa. When x = xa, rrey ¼ rhy = ðpffiffi31Þðk 2 1Þ ry

For k = 4 and rpy = 0.8, Lamé equations in another form are plotted in figures 1.3 and 1.4.

FIG. 1.3 – The distributions of operation stresses based on the Tresca criterion. p In figure 1.3, at x = k, rps e = rh .

FIG. 1.4 – The distributions of operation stresses based on the Mises criterion.

rpd e

p p pd In figure 1.4, at x = xa = 3.4, rpd e = rh ; When x < xa, re > rh ; When x > xa, p < rh .

Study on Elastic-Plastic Mechanical Stresses

1.3

7

Analysis of Elastic-Plastic Stresses

Elastoplastic stresses are the basis of autofrettage technology. In the case to not cause confusion, we omit some superscripts or subscripts. When a pressure vessel is subjected to an internal pressure p, as the internal pressure p increases, after the yielding of the inside surface and before the entire yielding of the whole wall, the inner layer of the pressure vessel is a plastic cylinder (or a plastic region) and the outside layer of the pressure vessel is a homocentric elastic cylinder (or an elastic region) as shown in figure 1.5, where ri is the inner radius, ro is the outer radius and rj is the radius of elastic-plastic juncture. The elastic cylinder is separated from the plastic cylinder, as is shown in figure 1.6 which is the elastic cylinder, and figure 1.7 which is the plastic cylinder, where p j is the interaction pressure between the plastic layer and the elastic layer. Then, the elastic region is a cylinder with inside radius rj and outside radius ro, subjected to internal pressure p j; and the plastic region is a cylinder with inside radius ri and outside radius rj, subjected to internal pressure p and external pressure p j. We now do mechanical analysis on these two cylinders.

FIG. 1.5 – A section of an elastic-plastic cylinder.

FIG. 1.6 – The elastic layer of the elastic-plastic cylinder.

8

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 1.7 – The plastic layer of the elastic-plastic cylinder. In whether the elastic region or plastic region, at an arbitrary radius r, cutting the cylinder with two concentric circles which are at a distance of dr, two planes with included angle dθ, and two parallel planes at a distance of dz, we obtain an element as shown in figure 1.8 as a free-body diagram. The equilibrium equation in the radial direction for the differential element is as follows: ðrr þ drr Þðr þ drÞdh  rr rdh  2rh dr sin

dh ¼0 2

FIG. 1.8 – A differential element from the wall of the cylinder. dh Ignoring higher order infinitesimal and taking sin dh 2  2 into consideration, we then have drr ð1:14Þ rh  rr ¼ r dr

Equation (1.14) is tenable in both the elastic state and plastic state.

Study on Elastic-Plastic Mechanical Stresses

9

In the plastic region, according to the maximum distortion strain energy theory, substituting equation (1.7) into equation (1.14) obtains 2 dr drr ¼ pffiffiffi ry r 3

ð1:15Þ

Integrating both sides of equation (1.15) gives[4] 2 rr ¼ pffiffiffi ry ln r þ C 3

ð1:16Þ

where C is an integration constant, to be determined by the following boundary conditions  rr ¼ p; r ¼ ri ð1:17Þ rr ¼ pj ; r ¼ rj Substituting the first one of equation (1.17) into equation (1.16) obtains the integration constant C, then substituting C back into equation (1.16) results in 2 r rr ¼ pffiffiffi ry ln  p ri 3

or

rr 2 p ¼ pffiffiffi ln x  ry ry 3

Substituting equation (1.18) into equation (1.7) obtains
 2 r rh 2 p rh ¼ pffiffiffi ry ln þ 1  p or ¼ pffiffiffi ðln x þ 1Þ  r r r 3 3 i y y

ð1:18Þ

ð1:19Þ

The axial stress σz can be obtained by equation (1.3). Therefore, the stresses within the plastic region are summarized as follows:

 8 rz 2 1 r p 2 1 p > > þ ln þ ln x  ¼ pffiffiffi  ¼ pffiffiffi > > > ri ry ry ry 3 2 3 2 > > > > > r 2 p > > r ¼ pffiffiffi ln x  < ry ry 3 ð1:20Þ > rh 2 p > > ¼ pffiffiffi ð1 þ ln x Þ  > > > ry ry 3 > > > > r > e > : 1 ry Seemingly, the elastic-plastic stresses within the plastic region are not related to the plastic depth kj, but in fact, kj affects the elastic-plastic stresses within the plastic region. Substituting the second one of equation (1.17) into the second one of equation (1.20) obtains 2 rj pj ¼  pffiffiffi ry ln þ p or ri 3

pj 2 p ¼  pffiffiffi ln kj þ ry ry 3

ð1:21Þ

10

Autofrettage Technology and Its Applications in Pressured Apparatuses

The inner surface of the elastic layer is just the outer surface of the plastic layer (see figures 1.5–1.7). So, on the inner surface of the elastic layer, there is also rh  rr ¼ p2ffiffi3 ry . According to Lamé formula, on the inner surface of the elastic layer
2 ro þ1 rj rh ¼ pj
2 ro 1 rj

and

rr ¼ pj

ð1:22Þ

Substituting equation (1.22) into equation (1.7) obtains ry ro2  rj2 pj ¼ pffiffiffi 3 ro2

ð1:23Þ

Letting equation (1.23) equal equation (1.21) obtains the relationship between the internal pressure p, the depth of plastic region kj (kj = rj/ri), the radius ratio k (k = ro/ri) and yield strength σy: kj2 k 2  kj2 p 2 1 2 ¼ pffiffiffi ln kj þ pffiffiffi  pffiffiffi ¼ pffiffiffi ln kj þ pffiffiffi ry 3 3 3 3k 2 3k 2 In the plastic region, letting

re ry



rh ry

ð1:24Þ

gives

pffiffiffi p 2 2 3  pffiffiffi ln x  pffiffiffi ry 3 3 When x = kj, equation (1.25) becomes rpy  p2ffiffi3 ln kj  pffiffi k 2 k 2 k 2 k 2 p p2ffiffi ln kj ¼ pffiffi j . Letting pffiffi j  2 pffiffi 3 results in  2 2 ry 3 3k 3k 3 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi kj  3  1k ¼ xa

ð1:25Þ pffiffi 2 pffiffi 3. 3

From equation (1.24),

ð1:26Þ

Therefore, when the load is determined by equation (1.24), in the plastic region, if kj ≤ xa, rrye  rryh ; if kj ≥ xa, rrye  rryh . On the other hand, from equation (1.25) we obtain pffiffiffi
pffiffiffi 3 p 2 3 x  exp  ð1:27Þ ¼ xz 2 ry 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k ¼ xa , xz does not When x ≤ xz, rrye  rryh ; when x ≥ xz, rrye  rryh . If kj  exist within 1 ≤ x ≤ kj. xz and xa do not exist at the same time. The elastic layer has an inner radius ri, outer radius ro, and is subjected to internal pressure p j. Therefore, according to the Lamé formula, the elastic stresses within the elastic region are as follows:

Study on Elastic-Plastic Mechanical Stresses

11

8 rz > > > > > ry > > > > > rr > > < ry > rh > > > > > ry > > > > re > > : ry

kj2 ¼ pffiffiffi 3k 2

 kj2 k2 rz k2 ¼ pffiffiffi 1 2 ¼ 1 2 x ry x 3k 2
ð1:28Þ 
 kj2 k2 rz k2 p ffiffi ffi ¼ 1þ 2 ¼ 1þ 2 x ry x 3k 2 2 kj ¼ 2 x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi In the elastic region, setting rrye  rryh leads to x  3  1k ¼ xa = 0.8556k < k. re rh re rh Then, when x ≤ xa, ry  ry ; when x ≥ xa, ry  ry . pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k = 0.8556k < k, rryh  rrye certainly occurs at some point in Since xa ¼ the elastic region. In the plastic region, rryh  rrye is conditional, that is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k ¼ xa or equation (1.26). If there exists rryh  rrye in the plastic region, kj  there certainly exists rryh  rrye in the whole elastic region. If based on the maximum shear stress theory, according to the DS transition rule, the corresponding results are listed as follows. The plastic stresses within the plastic region 8 rz r p p ¼ 0:5 þ ln  ¼ 0:5 þ ln x  > > > r r r r y i y y > > > > rr p > > > < r ¼ ln x  r y y ð1:29Þ rh p > > ¼ 1 þ ln x  > > ry ry > > > > > r > : e1 ry The relationship between the internal pressure p, the depth of plastic region kj (kj = rj/ri), the radius ratio k (k = ro/ri), and yield strength σy is as follows[3]. k 2  kj2 p 1 kj2 ¼ ln kj þ  2 ¼ ln kj þ ry 2 2k 2k 2

ð1:30Þ

In fact, the first term of equation (1.24) or equation (1.30) is the pressure to cause the plastic layer to become entirely yielded, and the second term is the initial yield pressure of the elastic layer. When based on the maximum shear stress theory, in the plastic region, letting p p p rh re ry  ry results in ry  ln x  0. When x = kj, ry  ln x  0 becomes ry  ln kj  0. From k 2 k 2

equation (1.30), rpy  ln kj ¼ 2k 2 j  0. Therefore, when based on the maximum shear stress theory, in the plastic region, we always have rrye  rryh . This is because that

12

Autofrettage Technology and Its Applications in Pressured Apparatuses

rse = σθ − σr, and σr < 0, then, rse > σθ whether in elastic or plastic zone. However, pffiffi since rde ¼ 23 ðrh  rr Þ, obviously, it is possible that rde < σθ. The elastic stresses within the elastic region 8 k2 r > > > z ¼ j2 > > ry 2k > >

 > > kj2 > rr k2 rz k2 > > ¼ 1  1  ¼ < ry 2k 2 x2 ry x2 ð1:31Þ 
 2
kj > rh k2 rz k2 > > ¼ 1 þ 1 þ ¼ > > > ry 2k 2 x2 ry x2 > > 2 > > re kj > > : ¼ 2 ry x In the elastic region, we always have rrye  rryh . In equations (1.24) and (1.30), letting kj = 1 obtains the initial yield pressure of the unautofrettaged pressure vessel, i.e. equations (1.11) and (1.12) respectively. When kj = 1, the terms expressing the entire yield pressure in equations (1.24) and (1.30) disappear. In equation (1.24), letting kj = k obtains the entire yield pressure, or equation (1.28). When kj = k, the term expressing initial yield pressure in equation (1.24) disappears. py 2 ¼ pffiffiffi ln k ry 3

ð1:32Þ

When k = ∞, equation (1.24) becomes p1 2 1 ¼ pffiffiffi ln kj þ pffiffiffi ry 3 3 p p1 1 ¼ ¼ pffiffiffi. ry ry 3 Letting k = ∞ in equation (1.20) results in 8 rz 2 2 2 x > > ¼ pffiffiffi ln x  pffiffiffi ln kj ¼ pffiffiffi ln > > ry > 3 3 3 kj > > > > > > rr 2 2 1 2 x 1 > > > < ry ¼ pffiffi3ffi ln x  pffiffi3ffi ln kj  pffiffi3ffi ¼ pffiffi3ffi ln kj  pffiffi3ffi

ð1:33Þ

When k = ∞, if kj = e0.5, then,

> > rh 2 2 1 2 x 1 > > ¼ pffiffiffi ln x  pffiffiffi ln kj þ pffiffiffi ¼ pffiffiffi ln þ pffiffiffi > > ry > 3 3 3 3 kj 3 > > > > > re > > : 1 ry

ðplastic regionÞ

ð1:34Þ

Study on Elastic-Plastic Mechanical Stresses

13

Letting k = ∞ in equation (1.28) results in 8 rz > ¼0 > > > ry > > > kj2 > rr > > > < ry ¼  pffiffi3ffix 2 kj2 rh rr ðelastic regionÞ > > p ffiffi ffi ¼ ¼  > > > ry ry 3x 2 > > > 2 > k r > > : e ¼ j2 ry x

ð1:35Þ

If based on the maximum shear stress theory, in view of the DS transition rule, the corresponding results are listed as follows py ¼ ln k ð1:36Þ ry p1 1 ¼ ln kj þ 2 ry 8 rz > > > ry > > > > rr > > < ry rh > > > > > ry > > r > > : e ry

¼ ln x  ln kj ¼ ln

ð1:37Þ

x kj

1 x 1 ¼ ln x  ln kj  ¼ ln  2 kj 2 1 x 1 ¼ ln x  ln kj þ ¼ ln þ 2 kj 2

ðplastic regionÞ

ð1:38Þ

1 8 rz > > > ry > > > > > > rr > > < ry rh > > > > > r y > > > > r > > : e ry

¼0 ¼

kj2 2x 2

kj2 rr ¼ 2x 2 ry kj2 ¼ 2 x ¼

ðelastic regionÞ

ð1:39Þ

It is noted that no matter within the elastic region or plastic region, the equivalent stress remains unchanged when based on the maximum shear stress theory and the maximum distortion strain energy theory. The reason for this is as follows: Under the same operation pressure p, the relationship between the equivalent stress based on the maximum shear stress theory rse and the equivalent stress based pffiffi on the maximum distortion strain energy theory rde is rde ¼ 23 rse (equation (1.6)), but the stress components based on the maximum distortion strain energy theory

14

Autofrettage Technology and Its Applications in Pressured Apparatuses

are p2ffiffi3 times as much as those based on the maximum shear stress theory, therefore, rde = rse . To intuitively display the variation of elastoplastic stresses, figure 1.9a–d shows the distributions of elastoplastic stresses based on the maximum distortion strain energy theory, where Figure 1.9a is for k = 4, kj = 1.5, xa = 3.422399 > kj, and p/σy = 0.964351… which is calculated from equation (1.24). Figure 1.9b is for k = 4, kj = 1.694172…, xa = 3.422399 > kj and p/σy = 1.082532… which is calculated from equation (1.24). Figure 1.9c is for k = kj = kc = 2.2184574899167… (entirely yielded), p/σy = 0.920081… which is calculated from equation (1.24) or equation (1.32), xz = 1.94029 from equation (1.27). xa = 1.898112 < kj, so, xa does not exist and xz appears. Figure 1.9d is for k = 4, kj = 3.7, xz = 3.478307 from equation (1.27), xa = 3.422399 < kj and p/σy = 1.594088… which is calculated from equation (1.24). Figure 1.10a–d shows the distributions of elastoplastic stresses based on the maximum shear stress theory, where Figure 1.10a is for k = 4, kj = 1.5, p/σy = 0.835153… which is determined from equation (1.30). Figure 1.10b is for k = 4, kj = 1.694172…, p/σy = 0.9375… which is determined from equation (1.30). Figure 1.10c is for k = kj = kc = 2.2184574899167… (entirely yielded), p/σy = 0.796812… which is determined from equation (1.30) or equation (1.36). Figure 1.10d is for k = 4, kj = 3.7, p/σy = 1.38052… which is calculated from equation (1.30).

FIG. 1.9 – The distributions of elastoplastic stresses based on the Mises yield criterion.

Study on Elastic-Plastic Mechanical Stresses

FIG. 1.9 – (continued).

15

16

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 1.10 – The distributions of elastoplastic stresses based on the Tresca yield criterion.

Study on Elastic-Plastic Mechanical Stresses

17

FIG. 1.10 – (continued). Pressure vessels play a key role in the pillar industries in the national economy, emerging industries of strategic importance, aviation, military space technology, and so on. In recent years, with the rapid change of the world economic situation, resource quality deterioration, and energy structure adjustment, in addition to the large-scale, harsh medium environment, extreme service temperature, etc. development direction, pressure vessels are also showing a development situation of lightweight of heavy pressure vessels. Autofrettage is one of the effective ways of lightweight heavy-pressure vessels. The lightweighting, green design, and manufacturing of pressure vessels under the premise of ensuring their inherent safety, and the reduction of environmental pollution caused by the failure of pressure vessels subjected to disasters, are outstanding problems that need to be solved focally. It is hoped that autofrettage will play an important role in the lightweighting, green design, and manufacturing of pressure vessels[5–16].

1.4

Chapter Summary

The main equations and conclusion are listed in table 1.1.

Within plastic region

Tresca criterion

Relationship between k, kj, p, and σy

Mises criterion

= σθ − σr pffiffiffi pffiffiffi 3 3 s d ðrh  rr Þ ¼ r re ¼ 2 2 e

Equivalent stress σe

Elastoplastic stresses

Elastoplastic stresses

Within elastic region rse

rz p ¼ 0:5 þ ln x  ry ry rr p ¼ ln x  ry ry rh p ¼ 1 þ ln x  ry ry re 1 ry

kj2 rz ¼ 2 ry 2k 
 2
kj rr k2 rz k2 ¼ 2 1 2 ¼ 1 2 ry 2k x ry x

 kj2 rh k2 rz k2 ¼ 2 1þ 2 ¼ 1þ 2 ry 2k x ry x 2 re kj ¼ ry x 2

k 2  kj2 p 1 kj2 ¼ ln kj þ  2 ¼ ln kj þ 2k 2 ry 2 2k kj = 1: maximum elastic load-bearing capability or initial yield pressure, pe pe k 2  1 ¼ 2k 2 ry kj = k: entire yield pressure, py py 2 ¼ pffiffiffi ln k ry 3 kj2 rz
 ¼ pffiffiffi rz 2 1 p ry 3k 2 ¼ pffiffiffi þ ln x  
 ry ry 2
3 2 2 k rr k rz k2 j ¼ pffiffiffi 1 2 ¼ 1 2 rr 2 p ry x ry x ¼ pffiffiffi ln x  3k 2 ry ry 3 
 2
2 k r k r k2 j h z rh 2 p p ffiffi ffi ¼ ¼ 1 þ 1 þ ¼ pffiffiffi ð1 þ ln x Þ  ry x2 ry x2 3k 2 ry ry 3 2 re re kj 1 ¼ ry ry x 2

Autofrettage Technology and Its Applications in Pressured Apparatuses

Criterion Tresca criterion Mises criterion

18

TAB. 1.1 – Collection of the main equations conclusion.

Relationship between k, kj, p, and σy

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pp ffiffiffi In elastic region, when x  3  1k ¼ xa re rh re rh = 0.8556k,  ; when x ≥ xa,  ry ry ry ry

kj2 k 2  kj2 p 2 1 2 ¼ pffiffiffi ln kj þ pffiffiffi  pffiffiffi ¼ pffiffiffi ln kj þ pffiffiffi ry 3k 2 3k 2 3 3 3 2 pe k  1 py 2 ¼ pffiffiffi ; ¼ pffiffiffi ln k ry ry 3k 2 3

Study on Elastic-Plastic Mechanical Stresses

TAB. 1.1 – (continued).

19

20

Autofrettage Technology and Its Applications in Pressured Apparatuses

References [1] Liu H.W. (2019) Mechanics of materials (in Chinese). Higher Education Press, Beijing. [2] Wang Z.W. (1990) Design of chemical pressure vessels (in Chinese). Chemical Industrial Press, Beijing. [3] Yu G.C. (1980) Chemical pressure vessels and equipment (in Chinese). Chemical Industrial Press, Beijing. [4] The compiling group of “handbook of mathematics” (1979) A handbook of mathematics (in Chinese). Higher Education Press, Beijing. [5] Chen X.D., Fan Z.C., Chen Y.D., et al. (2017) Green and intelligent design, manufacturing and maintenance of pressure vessels in China, CPVT 34 (11), 12. [6] Zhan Y., Wen J.F., Tu S.D., et al. (2016) Combination rules for multiple coplanar cracks based on fatigue life, CPVT 33 (2), 1. [7] Bu F., Qian C.F. (2013) On reasonable design of tank wall of large storage tank under static loads, CPVT 30 (12), 31. [8] Chen G.H., Chen L.J. (2012) Solution-dependent path crack propagation simulation and its application, CPVT 29 (10), 36. [9] Liu W., Li T., Zheng J.Y., et al. (2014) Recent development of super-high pressure vessel codes and standards, CPVT 31 (12), 47. [10] Qiu P., Shi J.F., Zheng J.Y., et al. (2017) Brief introduction of 2017 ASME PVP Conference, CPVT 34 (9), 73. [11] Gao Z.L., Li Y.B., Lei Y.B. (2015) Progress and case study on probabilistic assessment of reactor pressure vessels under pressurized thermal shock, J. Mech. Eng. 51 (20), 67. [12] Lai H.S., Fan D.S., Jia Y.F., et al. (2017) Key technical challenges of high density polyethylene (HDPE) pipe in nuclear equipment, CPVT 34 (12), 45. [13] Kabanova O.E. (2015) Determining the optimal yield radius and optimal pressure of autofrettage of curved elements of pipelines and pipes, Chem. Pet. Eng. 50 (9–10), 638. [14] Lee M.J., Lee B.C., Lim J.G., et al. (2014) Residual stress analysis of the thermal barrier coating system by considering the plasma spraying process, J. Mech. Sci. Technol. 28 (6), 2161. [15] Horst B., Dirk B. (2014) Full exploitation of lightweight design potentials by generating pronounced compressive residual stress fields with hydraulic autofrettage, Adv. Mater. Res. 907, 17. [16] Jiang G.F., Sun L., Chen G. (2014) Experimental study of 304 stainless steel fatigue life considering material re-strain hardening, J. Mech. Strength 36 (6), 850.

Chapter 2 Mechanical Autofrettage Technology Based on Tresca Yield Criterion The pressure applied to the cylindrical pressure vessel, or the pressure to achieve a plastic depth kj is called autofrettage pressure pa. For mechanical autofrettage, the key to autofrettage technology is the residual prestresses remaining in the wall of the pressure vessels after eliminating autofrettage pressure pa. Based on the maximum shear stress theory (Tresca yield criterion)[1], to achieve a plastic depth kj, the pressure exerted on a pressure vessel is evidently equation (1.30), this is just autofrettage pressure pa, or kj2 k 2  kj2 k 2 ln kj2 þ k 2  kj2 pa ¼ ln kj þ 0:5  2 ¼ ln kj þ ¼ ry 2k 2k 2 2k 2

ð2:1Þ

Obviously, the greater kj is, the greater pa is. It’s not difficult to foresee that kj is bound to affect residual stresses, load-bearing capability, etc., and under different strength theories, the extent and the way of its influence will be different. Therefore, it is necessary to investigate autofrettage technology based on the maximum shear stress theory (Tresca yield criterion). We proceed from the general situation.

2.1

General Study on Mechanical Autofrettage Technology

In our research work, we have obtained some research results about mechanical autofrettage technology based on the maximum shear stress theory (Tresca criterion), which are the foundation of this research and are summed up as follows.

DOI: 10.1051/978-2-7598-3111-1.c002 © Science Press, EDP Sciences, 2023

22

Autofrettage Technology and Its Applications in Pressured Apparatuses

2.1.1

In General Forms

1. The components of residual stresses When a pressure vessel is subjected to an autofrettage pressure pa which is greater than its initial yield pressure (pe) and smaller than the entire yield pressure (py), its inner layer becomes plastic and its outer layer remains elastic. After eliminating autofrettage pressure pa, the residual stresses remain in the wall of the pressure vessels. Based on the maximum shear stress theory (Tresca criterion), we obtained the residual stresses as follows[2]. Within the plastic region (1 ≤ x ≤ kj) " r0z 1 kj2 x2 ¼ þ ln 2  2 ry 2 k kj

! # kj2 1 2 1  2 þ ln kj k2  1 k

" r0r 1 kj2 x2 ¼  1 þ ln 2  2 ry 2 k kj

!  # kj2 1 k2 2 1 2 1  2 þ ln kj k2  1 k x

" r0h 1 kj2 x2 ¼ þ 1 þ ln  ry 2 k 2 kj2

!  # 2 kj2 1 k 1þ 2 1  2 þ ln kj2 k2  1 k x

k 2  kj2 þ k 2 ln kj2 1 r0e ¼1 x2 ry k2  1

ð2:2Þ

ð2:3Þ

ð2:4Þ

ð2:5Þ

It is not difficult to learn that as x increases, r0z , r0h , and r0e increase. When qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 k 2 kj2 þ k 2 ln kj2 0 x decreases as x increases; When x  , r , r0r increases r k 2 1 k 2 1 as x increases. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 \kj [3]. From ln kj2 \kj2  1, we can prove that k 2 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 In fact, is x0 in the following section. k 2 1 Within the elastic region (kj ≤ x ≤ k) " r0z 1 kj2 ¼  ry 2 k 2

! # kj2 kj2  1  ln kj2 1 2 ¼ constant [ 0 1  2 þ ln kj ¼ k2  1 k 2ðk 2  1Þ r0r ¼ ry

  k 2 r0z k 2  x 2 kj2  1  ln kj2 \0 1 2 ¼ x ry x2 2ðk 2  1Þ

ð2:6Þ

ð2:7Þ

Mechanical Autofrettage Technology Based on Tresca Yield Criterion r0h ¼ ry

  k 2 r0z k 2 þ x 2 kj2  1  ln kj2 [0 1þ 2 ¼ x ry x2 2ðk 2  1Þ r0e kj2  1  ln kj2 k 2 ¼ [0 ry k2  1 x2

23

ð2:8Þ

ð2:9Þ

Obviously, as x increases, r0h and r0e decrease, r0r increases, and r0z is kept constant. Figure 2.1 is the typical distribution of the residual stresses where k = 4 and kj = 1.6 are taken as an example. The meaning of each curve and main parameters have been marked in the figure.

FIG. 2.1 – The typical distributions of the residual stresses. At the inside surface, x = r/ri = 1, then, from equations (2.2) to (2.5), the residual stresses on the inside surface are as follows. k 2 ln kj2  kj2 þ 1 r0zi 0 ¼ 2ðk 2  1Þ ry

ð2:10Þ

r0ri ¼0 ry

ð2:11Þ

k 2 ln kj2  kj2 þ 1 r0hi r0 ¼ 2 zi  0 ¼ 2 k 1 ry ry

ð2:12Þ

24

Autofrettage Technology and Its Applications in Pressured Apparatuses

r0ei ¼ r0hi  r0ri ¼ r0hi  0

ð2:13Þ

At the inside surface, the absolute value of r0ei or r0hi is the maximum. At the elastic-plastic juncture, x = kj, the residual stresses obtained by equations (2.2)–(2.5) are equal to those obtained by equations (2.6)–(2.9). Therefore, the residual stresses in the whole wall are continuous. Substituting x = kj into equations (2.2)–(2.5) or equations (2.6)–(2.9) leads to the residual stresses at the elastic-plastic juncture: r0zj kj2  1  ln kj2 0 ¼ ry 2ðk 2  1Þ

ð2:14Þ

! k 2 r0zj 1 2 0 kj ry

ð2:15Þ

! r0zj k 2 r0zj 1þ 2 [ 0 ry kj ry

ð2:16Þ

r0ri ¼ ry r0hj ry

0

r0hj ry

\

¼

r0ej k 2 r0zj k 2 ðkj2  1  ln kj2 Þ ¼2 2 ¼ 1 ry kj ry ðk 2  1Þkj2

ð2:17Þ

It can be proved that r0h , r0e , and r0z monotonically increase with x increasing in the plastic region, r0h and r0e monotonically decrease with x increasing in the elastic region, r0z remains constant with x increasing in the elastic region, r0r monotonically increases with x increasing in the elastic region[3]. At the elastic-plastic juncture, r0hj , r0ej , and r0zj are the maximum within the whole wall (algebraic value, not absolute value).  0  rhi  r0hj k 2 ln kj2  kj2 þ 1 kj2  1  ln kj2 k 2 þ kj2    r  r ¼ k2  1 2ðk 2  1Þ kj2 y y ð2k 2 kj2 þ k 2 þ kj2 Þ ln kj2  ðk 2 þ 3kj2 Þðkj2  1Þ ¼ 2kj2 ðk 2  1Þ k 2 1

Since ln kj2 [ 2 k 2j þ 1, ð2k 2 kj2 þ k 2 þ kj2 Þ ln kj2 > j

It is easy to prove that  0  r0 r  Therefore,  rhiy > rhjy .

4k 2 kj2 þ 2k 2 þ 2kj2 kj2 þ 1

4k 2 kj2 þ 2k 2 þ 2kj2 kj2 þ 1

ðkj2  1Þ

[ k 2 þ 3kj2 .

At the outside surface, x = k, then, from equations (2.6) to (2.9), the residual stresses on the outside surface are as follows.

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

Within the whole wall, the whole wall,

r0e ry

[

r0h ry ,

25

r0zo kj2  1  ln kj2 r0zj ¼ ¼ [0 ry 2ðk 2  1Þ ry

ð2:18Þ

r0ro ¼ 0

ð2:19Þ

0\r0ho ¼ 2r0zo \r0hj [ 0

ð2:20Þ

r0eo ¼ r0ho

ð2:21Þ

r0r ry

\0. Because

r0e ry

r0

r0

¼ ryh  ryr ,

r0e ry

r0

r0

 ryh ¼  ryr [ 0, or within

except r0ei = r0hi and r0eo = r0ho .

2. Autofrettage pressure pa Autofrettage pressure is the internal pressure applied to a pressure vessel to cause a plastic depth kj before it is put into production, which is k 2  kj2 pa ¼ ln kj þ ry 2k 2 That is just equation (2.1). 3. The location of the intersection of the three residual stress curves or the abscissa at which r0e = 0 (defined as x0) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2  kj2 þ k 2 ln kj2 r x0 ¼ ¼ ð2:22Þ \kj ri k2  1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 In figure 2.1, x0 ¼ = 1.377924. k 2 1 The value of radial residual stresses at x = x0 (r0rx0 ) is the minimum, which is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2  kj2 þ k 2 ln kj2 kj2  1  ln kj2 r0rx0 r0rmin r0zx0 r0hx0 x0 r0zj ¼ y0s þ ¼ ¼ ¼ ¼ ln þ ¼ ln ry ry ry ry kj ry 2ðk 2  1Þ kj2 ðk 2  1Þ ð2:23Þ In figure 2.1, y0s = −0.128759223 4. The location where r0z = 0 (defined as x1) rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x1 ¼

e

1k 2 þ k 2 ln k 2 j j k 2 1

ð2:24Þ

Autofrettage Technology and Its Applications in Pressured Apparatuses

26

x1 is obtained by letting r0z = 0 in equation (2.2). From ln kj2 \kj2  1, we can prove that x1 < kj. On the other hand, since r0zi < 0, r0zj > 0, and r0z > 0 within the whole elastic region, it can be concluded that r0z = 0 occurs in the plastic region, or x1 < kj. 5. The total stresses The total stresses include the residual stresses and the elastic stresses caused by internal pressure p

p

p

rT r0 rp z ¼ z þ z ry ry ry

ð2:25Þ

rT r0 rp r ¼ r þ r ry ry ry

ð2:26Þ

rp rT r0 h ¼ hþ h ry ry ry

ð2:27Þ

 0   0  rT rh r rph rT rT rr rpr e r h ¼  ¼ þ þ  ry ry ry ry ry ry ry  0    p 0 p 0 r rh rr r r rp ¼  þ h r ¼ e þ e ry ry ry ry ry ry

ð2:28Þ

rp

p

where rrzy , rrry and rhy are expressed by equation (1.1), rrey is equation (1.9). Within the plastic region (1 ≤ x ≤ kj) ! " # kj2 rT 1 kj2 x2 1 1 p 2 z ¼ þ ln 2  1  2 þ ln kj þ 2 2 2 k 1 k  1 ry ry 2 k k kj " rT 1 kj2 x2 r ¼  1 þ ln 2  2 ry 2 k kj

ð2:29Þ

!  # kj2 1 k2 k2  x2 p 2 1 2 1  2 þ ln kj  2 2 2 k 1 k x x ðk  1Þ ry ð2:30Þ

" rT 1 kj2 x2 h ¼ þ 1 þ ln  ry 2 k 2 kj2

!  # kj2 1 k2 k2 þ x2 p 2 1 þ 1  2 þ ln kj þ k2  1 k x2 x 2 ðk 2  1Þ ry ð2:31Þ

k 2  kj2 þ k 2 ln kj2 rT 2k 2 p e þ ¼1 ry x 2 ðk 2  1Þ x 2 ðk 2  1Þ ry

ð2:32Þ

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

Setting

rT e ry

27

 1 obtains the load-bearing capability k 2  kj2 pa p  ln kj þ ¼ ry 2k 2 ry

ð2:33Þ

 T re ry

d

k 2 kj2 þ k 2 ln kj2 2 4k 2 1 p x 3  k 2 1 x 3 ry  0 also obtains equation (2.33). k 2 1 k 2 k 2 rT When rpy ¼ ln kj þ 2k 2 j ¼ rpay , rey  1 = constant within the whole plastic region. T  rT r rT rT rT rT T Since rey ¼ rhy  rry = constant and rzy ¼ 12 rhy þ rry , the three curves for rT z , rh , and rT to each other. r are parallel  

Setting

dx

rT h ry

d

Setting

dx

¼

 0 obtains p x 2 ðk 2  1Þ k 2  kj2 þ k 2 ln kj2 pa  þ [ ry 2k 2 2k 2 ry

ð2:34Þ

 T d

Setting

rr ry

dx

 0 obtains k 2  kj2 þ k 2 ln kj2 x 2 ðk 2  1Þ pr pa p   ¼ \ ry 2k 2 2k 2 ry ry

ð2:35Þ

When x = 1, equation (2.35) becomes kj2  1 pri p  ln kj  ¼ ry 2k 2 ry

ð2:36Þ

When x = kj, equation (2.35) becomes kj2  1 pro p ¼  ln kj  \0 ry 2 ry

ð2:37Þ

Consequently, as long as the operating pressure p does not exceed autofrettage pressure pa,

rT e ry

increases with the increase of x. If rpy  rT z ry

increases with the increase of x. Clearly, x. When

p ry

satisfies equation (2.35),

x, otherwise,

rT r ry

x 2 ðk 2 1Þ 2k 2

þ

k 2 kj2 þ k 2 ln kj2 2k 2

[

T p a rh ry , ry

always increases with the increase of

monotonously increases with the increase of

rT r ry

does not increase monotonously with the increase of x in the plastic qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 rT r0r r region. Especially, when p = 0, ry is just ry , therefore, when x  x0 ¼ , k 2 1 rT r ry

decreases as x increases; when x ≥ x0,

rT r ry

equation (2.35) and equation (2.32), when

increases as x increases. Besides, from

p ry

¼ rpyr ,

rT e ry

= 0. Since

rT e ry

=

rT h ry

−

rT r ry ,

when

Autofrettage Technology and Its Applications in Pressured Apparatuses

28 rT e ry rT h ry

= 0, there must be rT r

=

=

ry

rT z

ry .

rT h ry

=

rT r ry .

Since rT z ¼

Therefore, if p ≤ pr,

rT r ry

rT þ rT r h 2 ,

rT h ry

when

intersection in the plastic region, and at the intersection, be proved that at the intersection,

ry

rT r ry ,

there must be

does not increase monotonously with the

increase of x in the plastic region, the three curves for rT r

=

rT rT r h ry , ry

and

rT z ry

have an

rT e ry

= 0. In addition, it can  T  rr  gets the minimum, or  ry  gets the maximum.

The abscissa (x0T ) of the intersection point is solved out from equation (2.35) as follows sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2  kj2 þ k 2 ln kj2 2k 2 p x0T ¼  ð2:38Þ k2  1 k 2  1 ry Within the elastic region (kj ≤ x ≤ k) kj2  1  ln kj2 rT 1 p z þ 2 ¼ ¼ constant [ 0 k  1 ry ry 2ðk 2  1Þ

ð2:39Þ

rT k 2  x 2 kj2  1  ln kj2 k2  x2 p r  ¼ \0 ry x2 2ðk 2  1Þ x 2 ðk 2  1Þ ry

ð2:40Þ

rT k 2 þ x 2 kj2  1  ln kj2 k2 þ x2 p h þ ¼ [0 ry x2 2ðk 2  1Þ x 2 ðk 2  1Þ ry

ð2:41Þ

kj2  1  ln kj2 k 2 rT 2k 2 p e ¼ þ 2 2 [0 2 2 ry k 1 x x ðk  1Þ ry

ð2:42Þ

Since within the whole wall, rT h ry rT eo ry

rT

¼  rry [ ¼

rT eh ry

¼

0,

kj2 1ln kj2 k 2 1 rT

then, þ

r0r ry

within

\0 and the

rpr ry

\0,

whole

2 p k 2 1 ry . rT h ry decrease

rT r ry

rT rT rT rT e e r h ry ¼ ry  ry , or ry  rT rT e h except that ry [ ry ,

\0, while

wall,

rT

rT

Evidently, rey and as x increases, rry increases as x increases, and rzy is kept constant within the elastic region. In order to summarize and confirm the above conclusions, several figures for total stresses are given according to equations (2.29)–(2.32) and (2.39)–(2.42), as shown in figures 2.2–2.8. (1) Add operation stresses to figure 2.1 to illustrate the total stresses as shown in figure 2.2. When k = 4 and kj = 1.6,

p ry

¼ rpay ¼ ln kj þ

k 2 kj2 2k 2

= 0.890004 >

pri ry .

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

29

FIG. 2.2 – The total stresses when k = 4, kj = 1.6 and p = pa. rT

rT

rT

rT

Within the whole wall, rey [ rhy except reoy ¼ rhoy . Within the plastic region, curves 1, 2, and 3 are parallel to each other because p = p a. Because p = pa, σe ≡ σy within the whole plastic region, and σe < σy within the elastic region. rT r ry

increases monotonously with the increase of x within the whole wall.

(2) When k = 4 and kj = 1.6, 0.421254 < rpye = 0.46875 < rpay .

pa ry

= 0.890004,

p ry

¼ ln kj 

FIG. 2.3 – The total stresses when k = 4, kj = 1.6 and p = pri.

kj2 1 2k 2

¼ prriy =

Autofrettage Technology and Its Applications in Pressured Apparatuses

30

rT

rT

rT

rT

Within the whole wall, rey [ rhy except reoy ¼ rhoy . Within the plastic region, curves 1, 2, and 3 are not parallel to each other because p ≠ pa. When p < pa, all stresses are lower than σy within the whole wall. rT r ry

increases monotonously with the increase of x in the plastic region, but it is in

a critical state because p = pri. Therefore, x0T = 0. (3) When k = 4 pa ry = 0.890004.

kj = 1.6,

and

p ry

=

pr ry

= 0.270082


pri ry

= 0.421254.

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

31

FIG. 2.5 – The total stresses when k = 4, kj = 1.6 and p = pe > pri.

rT

rT

rT

rT

Within the whole wall, rey [ rhy except reoy ¼ rhoy . When p < pa, all stresses are lower than σy within the whole wall. Within the whole wall, curves 1, 2, and 3 are not parallel to each other because σe is not constant. rT r ry

increases monotonously with the increase of x within the whole wall because p > pri. (5) When k = 4 and kj = 1.6,

p ry

= 0.95 >

pa ry

>

pri ry .

FIG. 2.6 – The total stresses when k = 4, kj = 1.6 and p > pa.

Autofrettage Technology and Its Applications in Pressured Apparatuses

32

rT

rT

rT

rT

Within the whole wall, rey [ rhy except reoy ¼ rhoy . Within the whole wall, curves 1, 2 and 3 are not parallel to each other because σe is not constant. rT r ry

increases monotonously with the increase of x within the whole wall because p > pri. Since p > pa, σe > σy within the whole plastic region. (6) When k = 4 and kj = 2,

p ry

=

pe ry

= 0.46875
lnk2,

ð2:94Þ

ð2:95Þ

42

Autofrettage Technology and Its Applications in Pressured Apparatuses r0

k Setting reiy ¼ 1  kk 2ln1k   1 just obtains kk 2ln 1  1, i.e. equation (2.44). Thus, under entire yield, or kj = k, to avoid reverse yielding or r0ei ≥ −σy, there must be k ≤ kc. Figures 2.13 and 2.14 show the distributions of the residual stresses for k = kc and k = 2, respectively. To show the reverse yielding, figure 2.15 shows the distributions of the residual stresses for k = 2.5 (>kc). The meaning of each curve and coordinates of key points as well as the main parameters are marked in the figures. 2

2

2

FIG. 2.13 – The distributions of the residual stresses for k = kc and kj = k.

FIG. 2.14 – The distributions of the residual stresses for k = 2 and kj = k.

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

43

FIG. 2.15 – The distributions of the residual stresses for k = 2.5 and kj = k. The total stresses within the whole wall (1 ≤ x ≤ k)   rT 1 x2 ln k 2 1 p z 1 þ ln 2  2 ¼ þ 2 k  1 ry ry 2 k k 1

ð2:96Þ

  rT 1 x2 ln k 2 k 2 ln k 2 k2  x2 p r ln 2  2 þ 2 ¼  ry 2 k k  1 ðk  1Þx 2 x 2 ðk 2  1Þ ry

ð2:97Þ

  rT 1 x2 ln k 2 k 2 ln k 2 k2 þ x2 p h 2 þ ln 2  2  2 ¼ þ ry 2 k k  1 ðk  1Þx 2 x 2 ðk 2  1Þ ry

ð2:98Þ

  rT k 2 ln k 2 2k 2 p 2k 2 p k 2 ln k 2 1 e ¼1 2 þ ¼ 1þ 2  ry ðk  1Þx 2 x 2 ðk 2  1Þ ry k  1 ry k 2  1 x 2 p

From equation (2.99), when rpy  ln k ¼ ryy , the larger x is, the larger rT e ry

p ry

rT e ry

ð2:99Þ is. Setting

py ry .

 1 leads to  ln k ¼ That is to say, when kj = k and k ≤ kc, the maximum load is the entire yield pressure. When a pressure vessel is subjected to entire yield pressure py = σylnk, Lamé equations become rpz ln k ¼ 2 ry k  1

ð2:100Þ

rpr k2  x2 ln k ¼ 2 2 ry x ðk  1Þ

ð2:101Þ

Autofrettage Technology and Its Applications in Pressured Apparatuses

44

At the inner surface,

rpei ry

rph k2 þ x2 ln k ¼ 2 2 ry x ðk  1Þ

ð2:102Þ

rpe k 2 ln k 2 ¼ 2 ry ðk  1Þx 2

ð2:103Þ

¼ kk 2ln1k  1. It is seen that when kj = k and p = σylnk, 2

2

even if k ≤ kc, rpei ≥ σy if autofrettage process is not carried out. When kj = k and p = py = σylnk, the total stresses are as follows

as x increases,

rT

rT rT z r ry , ry

and

rT h ry

rT 1 z ¼ þ ln x  ln k ry 2

ð2:104Þ

rT r ¼ ln x  ln k ry

ð2:105Þ

rT h ¼ 1 þ ln x  ln k ry

ð2:106Þ

rT e 1 ry

ð2:107Þ

increase. At the inner surface, rT 1 zi ¼  ln k ry 2

ð2:108Þ

rT ri ¼  ln k ry

ð2:109Þ

rT hi ¼ 1  ln k ry

ð2:110Þ

rT ei ¼1 ry

ð2:111Þ

For rziy   1, k ≤ e1.5 (kc < e1.5); for k ≤ e (kc < e). At the outer surface,

rT hi ry

  1, k ≤ e2 (kc < e2); for

rT zo ¼ 0:5ry

rT ri ry

  1, ð2:112Þ

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

45

rT ro ¼ 0

ð2:113Þ

rT ho ¼ ry

ð2:114Þ

rT eo ¼ ry

ð2:115Þ

T Seemingly, as long as k ≤ e (kc < e), there certainly are |rT z | ≤ σy, |rr | ≤ σy, T ≤ σy, and |re | ≤ σy, but if k ≥ kc and kj = k, reverse yield will occur. When kj = k, py = σylnk and k ≤ kc, the distributions of the total stresses are plotted in figures 2.16 and 2.17 for k = kc and k = 2, respectively. The meaning of each curve and the main parameters are marked in the figures.

|rT h|

FIG. 2.16 – The distributions of the total stresses when k = kc and kj = k.

FIG. 2.17 – The distributions of the total stresses when k = 2 and kj = k.

Autofrettage Technology and Its Applications in Pressured Apparatuses

46

All in all, the premise of kj = k is k ≤ kc. In addition, when kj = k, (1) Autofrettage pressure or equation (2.1) becomes mum load when kj = k; qffiffiffiffiffiffiffiffiffiffiffi 2 2 (2) x0 ¼ kk 2ln1k \k; qffiffiffiffiffiffiffiffi r0 r0 r0rx0 ln k 2 k 2 1ln k 2 ¼ ¼ ln (3) rzx0y ¼ rhx0 r k 2 1 þ 2ðk 2 1Þ ; y y qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (4) x1 ¼

2.3

e

1k 2 þ k 2 ln k 2 k 2 1

pa ry

¼ ln k, which is the maxi-

.

Mechanical Autofrettage Technology with Radius of Elastic-Plastic Juncture Being Arithmetic Mean Radius of Inside Radius and Outside Radius

Sometimes, some designers will be inclined to use the average radius as the radius of the plastic region. When the radius of the plastic region is an arithmetic mean value of the inner and outer radius of the pressure vessel, there is rj ¼

ri þ ro 2

or kj ¼

rj k þ 1 ¼ 2 ri

e ¼ 0:5

or

ð2:116Þ

Substituting equation (2.116) into equations (2.2)–(2.5) obtains the residual stresses within the plastic region " ! # r0z 1 ðk þ 1Þ2 4x 2 ð k þ 1Þ 2 ðk þ 1Þ2 1 ¼ þ ln  1 þ ln ð2:117Þ k2  1 ry 2 4k 2 4k 2 4 ð k þ 1Þ 2 " r0r 1 ðk þ 1Þ2 4x 2 ¼  1 þ ln  ry 2 4k 2 ð k þ 1Þ 2

!  # ð k þ 1Þ 2 ð k þ 1Þ 2 1 k2 1 2 1 þ ln k2  1 4k 2 4 x ð2:118Þ

" r0h 1 ðk þ 1Þ2 4x 2 ¼ þ 1 þ ln  ry 2 4k 2 ð k þ 1Þ 2

!  # ð k þ 1Þ 2 ð k þ 1Þ 2 1 k2 1þ 2 1 þ ln k2  1 4k 2 4 x ð2:119Þ

k2  r0e ¼1 ry

ðk þ 1Þ2 þ k2 4 k2  1

2

ln ðk þ4 1Þ 1 x2

ð2:120Þ

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

47

Substituting equation (2.116) into equation (2.6) into equation (2.9) obtains the residual stresses within the elastic region r0z ¼ ry

ðk þ 1Þ2 4



r0e ¼ ry

ð2:121Þ

 k 2 r0z \0 x 2 ry

ð2:122Þ

  k 2 r0z 1þ 2 [0 x ry

ð2:123Þ

r0r ¼ ry r0h ¼ ry

2

 1  ln ðk þ4 1Þ [0 2ðk 2  1Þ 1

ðk þ 1Þ2 4

2

 1  ln ðk þ4 1Þ k 2 [0 k2  1 x2

ð2:124Þ

When k = 2, kj ¼ k þ2 1 = 1.5. The residual stresses within the whole wall are plotted in figure 2.18. The meaning of each curve and the main parameters are marked in the figure.

FIG. 2.18 – The residual stresses within the whole wall when k = 2 and kj ¼ k þ2 1.

When k = 3, kj ¼ k þ2 1 = 2. The residual stresses within the whole wall are plotted in figure 2.19. The meaning of each curve and the main parameters are marked in the figure.

48

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 2.19 – The residual stresses within the whole wall when k = 3 and kj ¼ k þ2 1. At the inner surface (x = 1), r0ei r0hi ¼ ¼ ry ry

ðk1Þðk þ 3Þ 4

2

 k 2 ln ðk þ4 1Þ k2  1

ð2:125Þ

ðk þ 1Þ2 ¼ c0 4

ð2:126Þ

Letting r0ei ≥ −σy obtains 5k 2 þ 2k  7  4k 2 ln

The image of c = 0 is plotted in figure 2.20.

2

FIG. 2.20 – The image of 5k 2 þ 2k  7  4k 2 ln ðk þ4 1Þ ¼ 0.

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

49

From figure 2.20, it is seen that when k is greater than a certain value, c < 0, or r0ei /σy < −1. Solving equation (2.126) obtains k  2:618253. . . ¼ ka

ð2:127Þ

When k ≤ ka, |r0ei | ≤ σy. When k = ka, kj ¼ k þ2 1 = 1.809126…. The residual stresses within the whole wall are plotted in figure 2.21. The meaning of each curve and the main parameters are marked in the figure.

FIG. 2.21 – The residual stresses within the whole wall when k = ka and kj ¼ k þ2 1. When k ≥ kc, equation (2.46) or k 2 ln kj2  k 2  kj2 þ 2 ¼ 0 can guarantee a pressure vessel against reverse yield when processed with autofrettage. The graph of k 2 ln kj2  k 2  kj2 þ 2 ¼ 0 is shown as a curve ad in figure 2.22. As mentioned above, when k ≤ kc, kj can be k. In figure 2.22, the straight line oa is for kj = k.

FIG. 2.22 – The relationship between kj and k.

Autofrettage Technology and Its Applications in Pressured Apparatuses

50

Figure 2.22 is applicable for both the maximum shear stress theory (Tresca yield criterion) and the maximum distortion strain energy theory (Mises yield criterion). 2 2 Substituting kj ¼ k þ2 1 into k 2 ln kj  k 2  kj þ 2 ¼ 0 just results in 2

5k 2 þ 2k  7  4k 2 ln ðk þ4 1Þ ¼ 0, or equation (2.126). kj ¼ k þ2 1 is also plotted in figure 2.22 as straight-line ob. The coordinates of the intersection point of the straight line ob and the curve ad are f (ka, 1.809127). In figure 2.22, if k ≥ ka, such as in section fb, the pressure vessel will undergo reverse yield when processed with autofrettage. From figure 2.22, it is known that when k < ka, the plastic depth kj ¼ k þ2 1 is smaller than kj*. However, this may be at the expense of reducing load-bearing capacity. The equivalent stress of the total stresses k2  rT r0 rp e ¼ e þ e ¼1 ry ry ry Setting

rT e ry

ðk þ 1Þ2 þ k2 4 k2  1

2

ln ðk þ4 1Þ 1 2k 2 p þ 2 2 2 x x ðk  1Þ ry

ð2:128Þ

 1 gives p 1 ðk þ 1Þ2 k þ1   þ ln ry 2 2 8k 2

ð2:129Þ

Accordingly, when kj ¼ k þ2 1, the maximum load-bearing capacity of a pressure vessel is p 1 ðk þ 1Þ2 k þ1 ¼  þ ln ry 2 2 8k 2 Effectively, substituting equation (2.130). Letting

p ry

kj ¼ k þ2 1

into

equation (2.1)

ð2:130Þ also

results

in

2

þ 1Þ ¼ 12  ðk 8k þ ln k þ2 1 ≤ 1 obtains 2

k  3:102921. . . ¼ kz

ð2:131Þ

When k ≤ 3.102921… = kz, p ≤ σy and rT ri ≥ −σy; When k ≥ kz, p ≥ σy and ≤ −σy. When k ≤ ka, there must be k ≤ kz. As mentioned above, when k ≤ kc, kj = k, and the maximum load-bearing capacity of a pressure vessel is the entire yield load rpy ¼ ln k, that is plotted in rT ri

2 figure 2.23 as curve oa; and when k ≥ kc, kj is determined by k 2 ln kj k 2  kj2 þ 2 ¼ 0, and the maximum load-bearing capacity of a pressure vessel

is

p ry

¼ k k1 2 , that is plotted in figure 2.23 as curve ad. 2

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

51

FIG. 2.23 – The load-bearing capacity under various conditions.

Letting

p ry

2

2

þ 1Þ ¼ 12  ðk 8k þ ln k þ2 1 ¼ k k1 leads to 5k 2 þ 2k  7  4k 2 ln ðk þ4 1Þ ¼ 0, 2 2 2

2

þ 1Þ þ ln k þ2 1 is also plotted in figure 2.23 as curve ob. or equation (2.126). rpy ¼ 12  ðk 8k 2 The coordinates of the intersection point of the curve ob and the curve ad are f(ka, 0.854126). In figure 2.23, if k ≥ ka, such as in the section fb, rT e > σy. 2 For comparison, the maximum elastic load or initial yield load rpye ¼ k2k1 2 is also plotted in figure 2.23 as the dash curve. As k increases, point n and point m gradually coincide with one point. ! 2k 1 ðk þ 1Þ2 k þ1 2k 1 ðk þ 1Þ2 k þ1  1   þ ln k  þ ln [ 2 ¼ ln 2k 2 2 k þ1 2 8k 2 8k 2 k þ1 þ 1



1 ðk þ 1Þ2 2ðk  1Þ2 ð7k  1Þ þ 0 ¼ 2 8k 2 ð6k þ 2Þ 8k 2

p ry when kj [ k þ2 1, or

This means that the load-bearing capacity as is seen in figure 2.23. This is because is smaller than that when kj = kj*.

k þ1 2

kj ¼ k þ2 1 is smaller than lnk, the plastic depth when kj ¼ 2

þ 1Þ From figure 2.23, we can also see that when k ≤ ka, 12  ðk 8k þ ln k þ2 1 \ k k1 2 2 ; 2

þ 1Þ þ ln k þ2 1 [ when k ≥ ka, 12  ðk 8k 2 yield occurs due to kj ¼ k þ2 1.

k 2 1 k2 .

2

However, when k ≥ ka, |r0ei | > σy or reverse 2

þ 1Þ þ ln k þ2 1 ¼ rpy , In addition, when kj ¼ k þ2 1, autofrettage pressure rpay ¼ 12  ðk 8k 2 which is just equation (2.130). This is because the general equation of autofrettage

pressure

pa ry

¼ ln kj þ

k 2 kj2 2k 2

is obtained just by setting

rT e ry

¼ 1 in a plastic region.

52

2.4

Autofrettage Technology and Its Applications in Pressured Apparatuses

Mechanical Autofrettage Technology with Radius of Elastic-Plastic Juncture Being Geometrical Mean Radius of Inside Radius and Outside Radius

Taking the geometric mean value of the inner and outer radius as the plastic radius is a measure worthy of consideration for the design of autofrettaged pressure vessels. When the radius of the plastic region is a geometric mean value of the inner and outer radius of the pressure vessel, then pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffi k1 1 ¼ pffiffiffi rj ¼ ro ri or kj ¼ k or e ¼ ð2:132Þ k1 k þ1 Substituting equation (2.132) into equations (2.2)–(2.5) obtains the residual stresses within the plastic region   r0z 1 x 2 k  1  ln k ln þ ¼ ð2:133Þ k2  1 ry 2 k      r0r 1 x2 1 1 1 k2  1  ln k 2 1 2 ¼ ln þ  1 þ k k k 1 ry 2 k x

ð2:134Þ

     r0h 1 x2 1 1 1 k2  1  ln k 2 1þ 2 ¼ ln þ þ 1 þ k k k 1 ry 2 k x

ð2:135Þ

r0e k 2  k þ k 2 ln k 1 ¼1 k2  1 x2 ry

ð2:136Þ

Substituting equation (2.132) into equations (2.6)–(2.9) obtains the residual stresses within the elastic region r0z k  1  ln k ¼ constant [ 0 ¼ 2ðk 2  1Þ ry r0r ¼ ry r0h ¼ ry



ð2:137Þ

 k 2 r0z 1 2 \0 x ry

ð2:138Þ

  k 2 r0z 1þ 2 [0 x ry

ð2:139Þ

r0e k  1  ln k k 2 ¼ [0 k2  1 x2 ry

ð2:140Þ

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

53

pffiffiffi When k = 3, kj ¼ k = 1.732051…. The residual stresses within the whole wall are plotted in figure 2.24, where the meaning of each curve and the main parameters are marked.

pffiffiffi

FIG. 2.24 – The residual stresses within the whole wall when k = 3 and kj ¼ k . pffiffiffi When k = 3.5, kj ¼ k = 1.870829…. The residual stresses within the whole wall are plotted in figure 2.25, in which it is shown that |r0ei | > σy when k = 3.5. The meaning of each curve and the main parameters are marked in the figure.

pffiffiffi

FIG. 2.25 – The residual stresses within the whole wall when k = 3.5 and kj ¼ k .

54

Autofrettage Technology and Its Applications in Pressured Apparatuses At the inner surface (x = 1), from equation (2.136), r0ei r0hi k  1  k 2 ln k ¼ ¼ k2  1 ry ry

ð2:141Þ

k 2 þ k  2  k 2 ln k ¼ d  0

ð2:142Þ

Letting r0ei ≥ −σy obtains

The image of d = 0 is plotted in figure 2.26.

FIG. 2.26 – The image of k2 + k − 2 − k2lnk = 0.

From figure 2.26, it is seen that when k is greater than a certain value, d < 0. This indicates r0ei /σy < −1. Solving equation (2.142) obtains ð2:143Þ k  3:042297. . . ¼ kb pffiffiffi When k ≤ kb, |r0ei /σy| ≤ 1. When k = kb, kj ¼ k = 1.744218…. The residual stresses within the whole wall are plotted in figure 2.27, where the meaning of each curve and the main parameters are marked. pffiffiffi 2 2 Substituting kj ¼ k into k 2 ln kj  k 2  kj þ 2 ¼ 0 just results in k2 + k − 2 p ffiffiffi − k2lnk = 0, or equation (2.142). kj ¼ k has been plotted in figure 2.22 as curve oh. The coordinates of the intersection point of the curve oh and the curve ad are g (kb, 1.744218). In figure 2.22, if k ≥ kb, such as in section gh, the pressure vessel will undergo reverse yield when processed with autofrettage. pffiffiffi From figure 2.22, it is known that when k < kb, the plastic depth kj ¼ k is smaller than kj*. Nevertheless, this may be at the expense of reducing load-bearing capacity.

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

55

pffiffiffi

FIG. 2.27 – The residual stresses within the whole wall when k = kb and kj ¼ k .

The equivalent stress of the total stresses rT r0 rp k 2  k þ k 2 ln k 1 2k 2 p e ¼ e þ e ¼1 þ k2  1 x 2 x 2 ðk 2  1Þ ry ry ry ry Setting

rT e ry

ð2:144Þ

 1 gives pffiffiffi p k 1 þ ln k  ry 2k

ð2:145Þ

Accordingly, the maximum load-bearing capacity of a pressure vessel is rpy ¼ pffiffiffi pffiffiffi k1 k. 2k þ ln k when kj ¼ pffiffiffi Effectively, substituting kj ¼ k into equation (2.1) also results in equation (2.145). pffiffiffi Letting rpy  k1 2k þ ln k ≤ 1 obtains k  3:591121. . . ¼ kx

ð2:146Þ

kx > kb, therefore, when k ≤ kb, there must be k < kx, thus p ≤ σy, accordingly rT ri ry

  1.

pffiffiffi 2 ln k ¼ k k1 leads to k2 + k − 2 − k2lnk = 0, or equation (2.142). Letting k1 2 2kpþ ffiffi ffi p k1 ry ¼ 2k þ ln k is also plotted in figure 2.23 as curve oh. The coordinates of the intersection point of the curve oh and the curve ad are g (kb, 0.891957). In figure 2.23, if k ≥ kb, such as in the section gh, rT e > σy.

Autofrettage Technology and Its Applications in Pressured Apparatuses

56

  pffiffiffi 1 ðk þ 1Þ2 k þ1 k1 1 ðk þ 1Þ2 k  1 k þ1 þ ln    þ ln k ¼  þ ln pffiffiffi 2 2 8k 8k 2 2 2k 2 2k 2 k kp þffiffi1 2 1 1 ðk þ 1Þ k1 2 k [   þ 2 k þ 1 pffiffi þ 1 8k 2 2 2k 2 k pffiffiffi p ffiffiffi pffiffiffi ð k  1Þ3 ð15k k þ 11k þ 5 k þ 1Þ pffiffiffi ¼ 0 8k 2 ðk þ 1 þ 2 k Þ

pffiffiffi This means that the load-bearing capacity when kj ¼ k is smaller than that pffiffiffi when kj ¼ k þ2 1, as is seen in figure 2.23. This is because k þ2 1 [ k , or the plastic pffiffiffi depth when kj ¼ k þ2 1 is larger than that when kj ¼ k . pffiffiffi 2 From figure 2.23, we can also see that when k ≤ kb, k1 þ ln k \ k k1 2 ; when 2k p ffiffi ffi k1 k 2 1 0 k ≥ kb, 2k þ ln k [ k 2 . However, when k ≥ kb, |rei | > σy or reverse yield occur if pffiffiffi kj ¼ k . pffiffiffi In addition, when kj ¼ k , there are conclusions that pffiffiffi p (1) Autofrettage pressure rpay ¼ k1 2k þ ln k ¼ ry , which is just equation (2.145). As mentioned above, this is because the general equation of autofrettage pressure pa ry

k 2 kj2 2k 2 is obtained rT results in rey ¼ 1;

¼ ln kj þ

also (2) x0 ¼

rT e ry

¼ 1 in plastic region, and

p ry

¼ rpay

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

k 2 k þ k 2 ln k \k; k 2 1 0 0 rhx0 rrx0

r0 (3) zx0 ¼ ¼ ¼ ln ry ry ry qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (4) x1 ¼

2.5

just by setting

e

1k þ k 2 ln k k 2 1

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k  1 þ k ln k k  1  ln k ; þ k2  1 2ðk 2  1Þ

.

Mechanical Autofrettage Technology with Minimum Equivalent Total Stress on Elastic-Plastic Juncture

On the elastic-plastic juncture, x = r/ri = kj, from equation (2.32) or equation (2.42), we obtain ! rT 1 k2 2k 2 2k 2 p ej ¼1 2  1 þ ln k ð2:147Þ þ j 2 2 2 2 k  1 kj ry kj ðk  1Þkj ry d

Setting

rT  ej ry

dkj

¼ 0 results in

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

57

  p   exp ry  1 rj p ð2:148Þ kj ¼ ¼ exp or e ¼ ry k1 ri   When kj ¼ exp rpy and 1 ≤ kj ≤ k, the equivalent total stress at the elastic-plastic juncture rT ej is the minimum, the minimum value is   2 exp 2 rpy  1 rT k k 2 kj2  1 ej   ¼ 2 ¼ 2 ry k  1 exp 2 p k  1 kj2

ð2:149Þ

ry

When

p ry

¼ rpye ¼ k2k1 2 , 2

kj ¼ Letting kj ¼ exp





k 2 1 2k 2

 2  rj k 1 ¼ exp 2k 2 ri

ð2:150Þ

 k obtains k 2 ln k 2  k 2  1

ð2:151Þ

Equation (2.151) is natural, because ln k 2 [ 2 kk2 1 þ 1. Therefore, 2

k 2 ln k 2 [ 2k 2

 k2  1 2k 2 2 ¼ 2 k  1 [ k2  1 2 k þ1 k þ1

Thus, when the load is initial yield pressure pe, the plastic depth determined by equation (2.150) cannot exceed the radius ratio k. 2 When rpy ¼ k k1 2 ,  2  rj k 1 kj ¼ ¼ exp ð2:152Þ k2 ri 2  Letting kj ¼ exp k k1  k obtains 2

d

2  k ln k k 2 1

dk

k 2 ln k 1 k2  1

ð2:153Þ

k k 1 k ln k ¼ ðk 2 1 ðk 2  1  ln k 2 Þ [ 0, when k → 1, kk 2ln 1 ! 2, and solving k 2 1 ¼ 1 Þ2 2

2

obtains k ¼ 2:2184574899167. . . ¼ kc

ð2:154Þ

Therefore, when rpy ¼ k k1 2 , the plastic depth determined by equation (2.152) cannot exceed radius ratio k if k > kc. Equation (2.152) is plotted in figure 2.28. 2

Autofrettage Technology and Its Applications in Pressured Apparatuses

58

FIG. 2.28 – Comparison between kj and k.   When k < kc, for equation (2.148), setting kj ¼ exp rpy  k leads to p py  ln k ¼ ry ry Setting k 2 ln k k 2 1

p ry

¼ k k1 ¼ ln k just obtains equation (2.44) or equation (2.153), or 2 2

¼ ln k, when

p ry

p k 2 1 ry ¼ k 2 , kj < k, but the 2 k < kc, ln k\ k k1 2 , the maximum load is k 2 1 = kc, ln k ¼ k 2 , the maximum load is

¼ 1. This indicates that when k > kc, ln k [

maximum load is unlikely p ry

ð2:155Þ

p ry

p ry

¼ k k1 2 ; when 2

¼ ln k, kj = k; when k

k 2 1 k2 ,

if

kc2 1 kc2

¼ ¼ ln kc . Substituting equation (2.148) into equations (2.2)–(2.5) obtains the residual stresses within the plastic region   2 3 p 0 2 exp 2  1 ry rz 1 4 2k p 5 ln x 2 þ  2 ð2:156Þ ¼ ry 2 k  1 ry k2  1     2 3 0 1 p p   2 exp 2 exp 2 ry ry 1 p p 1 k 1 2 5 ¼ 4  1 þ ln x 2  2  @1  þ2 A 2 ry ry k  1 k2 k2 ry 2 x r0r

ð2:157Þ     2 3 0 1 p   2 exp 2 rpy r0h 1 4exp 2 ry p p 1 k 1þ 2 5 ¼ þ 1 þ ln x 2  2  @1  þ2 A 2 ry ry k  1 k2 k2 ry 2 x ð2:158Þ

Mechanical Autofrettage Technology Based on Tresca Yield Criterion   k 2  exp 2 rpy þ 2k 2 rpy 1 ¼1 x2 ry k2  1 r0e

59

ð2:159Þ

Substituting equation (2.148) into equations (2.6)–(2.9) obtains the residual stresses within the elastic region   0 exp 2 rpy  1  2 rpy rz ¼ constant [ 0 ð2:160Þ ¼ ry 2ðk 2  1Þ r0r ¼ ry r0h ¼ ry r0e

When k = 2, rpy

 k 2 r0z \0 x 2 ry

ð2:161Þ

  k 2 r0z 1þ 2 [0 x ry

ð2:162Þ

1

  exp 2 rpy  1  2 rpy k 2

[0 ð2:163Þ k2  1 x2   2 p ¼ k2k1 ¼ 1:454991. . .. The residual stresses 2 ¼ 0:375, kj ¼ exp r y ry

¼



within the whole wall are plotted in figure 2.29, where the meaning of each curve and the main parameters are marked.

FIG. 2.29 – The residual stresses within the whole wall when

p ry

¼ k2k1 2 ¼ 0:375. 2

60

Autofrettage Technology and Its Applications in Pressured Apparatuses

When k = 2, rpy ¼ ln k ¼ 0:693147:::, kj = k = 2. The whole wall is plastic, and the residual stresses within the whole wall are plotted in figure 2.30, where the meaning of each curve and the main parameters are marked.

FIG. 2.30 – The residual stresses within the whole wall when kj = k = 2.

When k = 3,

p ry

¼ k k1 ¼ 89, kj ¼ exp 2 2

  p ry

¼ 2:432425. . .. The residual stresses

within the whole wall are plotted in figure 2.31, where the meaning of each curve and the main parameters are marked.

FIG. 2.31 – The residual stresses within the whole wall when

p ry

¼ k k1 ¼ 89. 2 2

Mechanical Autofrettage Technology Based on Tresca Yield Criterion When k = 3,

p ry

¼ k k1 ¼ 89, 2 2

r0ei ry

< −1. In light of kj ¼ exp

61

  p ry

, we cannot directly

determine whether the equivalent residual stress on the inner surface r0ei exceeds strength limit σy. When k ≥ kc, in order to avoid reverse yield, for equation (2.159), setting

r0ei ry

≥ −1 obtains   p p exp 2 þ k 2  2k 2  2  0 ry ry

or

k2 

  exp 2 rpy  2

ð2:164Þ

2 rpy  1

For a certain k, the load-bearing capacity rpy when k ≥ kc should be determined by equation (2.164), as is shown by curve am in figure 2.23. For equation (2.164), the   domain of k2 is

p ry



1 2

and

p ry

exp 2rpy pffiffiffi  ln 2. The function k 2 ¼ 2 p 1 ry

2

is plotted in

figure 2.32. There are two branches for the graphs of the function: curve og and  p curve bah. In addition, the function kj ¼ exp ry is plotted in figure 2.32 as the solid p ry

¼ k k1 is also plotted in figure 2.32 as the dash curve 2 pffiffiffi ofac. In figure 2.32, curve og corresponds to the domain of rpy  ln 2, and k < 1 on curve og, thus it is of no practical significance; curve bah corresponds to the domain   curve odai and the function

of

p ry

 12. For the function k 2 ¼

2

2rpy p 2ry 1

exp

2

2 , setting dk  ¼ 0 gives

d 2rpy

    p p p exp 2  exp 2 þ1 ¼ 0 ry ry ry

ð2:165Þ

Solving equation (2.165) obtains the horizontal ordinate of the minimum value point a p ¼ 0:796812. . . ð2:166Þ ry   exp 2rpy 2 Substituting rpy ¼ 0:796812. . . into k 2 ¼ leads to k = kc. 2rpy 1     exp 2rpy 2 2 Letting k 2 ¼ ¼ k ¼ exp 2 rpy also obtains equation (2.165). When p j 2ry 1   2 obtains equation (2.165) k = kj, substituting k 2 ¼ kj2 ¼ exp 2 rpy into rpy ¼ k k1 2 again. This shows that the three curves bah, ofac, and odai intersect at one point. This point is just the minimum value point (point a) of curve bah. From figure  2.32,

when rpy \0:796812:::, k > kc, and k is larger than kj determined by kj ¼ exp p ry

p ry

, and

is larger than the value determined by rpy ¼ k k1 2 . This is reasonable and makes sense. When rpy [ 0:796812:::, k > kc, and k is smaller than kj determined by 2

62

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 2.32 – The relationship between k (kj) and rpy . kj ¼ exp

  p ry

. This is unreasonable and makes no sense. Curve odac can ensure

k > kj, but cannot ensure r0ei ≥ −σy. Therefore, when k > kc, the meaningful control curve is curve ba, which is also curve am in figure 2.23. If the design conditions are controlled by curve ba when k > kc, there are both k > kj and r0ei ≥ −σy. For example, when k = 3, from equation (2.164), rpy ¼ 0:558725:::. When k = 3 and   p p ¼ 0:558725:::, k ¼ exp j ry ry ¼ 1:748442. . . = kj*. The residual stresses within the whole wall are plotted in figure 2.33, where the meaning of each curve and the main parameters are marked.

FIG. 2.33 – The residual stresses within the whole wall when k = 3 and rpy ¼ 0:558725:

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

63

Compared with figure 2.31, the stresses are enormously improved.  residual  p In reality, when k ¼ kj ¼ exp ry , equation (2.165) and equation (2.164) are   equivalent. When kj ¼ exp rpy (but k > kj) equation (2.164) becomes equation (2.46), i.e., k 2 ln kj2  k 2  kj2 þ 2  0 i.e. k 2 

kj2 2 . ln kj2 1

Therefore,   curve ad in figure 2.22 is also the relation between k and kj when 2 kj ¼ exp rpy and r0ei = −σy. However, when the load-bearing capacity rpy ¼ k k1 2 , there is rT σ within the whole plastic region; when the load-bearing capacity e ≡  y   p equivalent to kj ¼ exp rpy , is restrained by where kj ry ¼ ln kj k 2 ln kj2  k 2  kj2 þ 2  0, rT ej < σy at the elastic-plastic juncture. Equation (2.164) can ensure kj < k and r0ei ≥ −σy, but under the premise of ensuring kj < k and r0ei ≥ −σy as well as rT being the minimum, the load-bearing   ej capacity rpy calculated by kj ¼ exp rpy may not be the most satisfactory. From

equation (2.164), the greater k is, the smaller kj is, thus the smaller rpy is, because   kj ¼ exp rpy . When k = ∞, kj = e0.5, hence rpy = 0.5, that is merely the maximum   elastic load with k = ∞. Besides, when kj ¼ exp rpy , from figure 2.23, the smaller than that under load-bearing capacity under the case k > kc is unexpectedly  

the case k < kc. So, if it is not necessary, kj ¼ exp rpy is not a satisfactory solution.   When exp 2 rpy þ k 2  2k 2 rpy  2 ¼ 0, which is equation (2.164), equation (2.159) become

  k 2  exp 2 rpy þ 2k 2 rpy 1 r0e 2 ¼1 ¼1 2 2 2 x x ry k 1

ð2:167Þ

The equivalent residual stress expressed by equation (2.167) is exactly that under the condition of k 2 ln kj2  k 2  kj2 þ 2  0.   When kj ¼ exp rpy , autofrettage pressure or equation (2.1) becomes   pa p 1 1 p ¼ þ  exp 2 ð2:168Þ ry ry ry 2 2k 2     dðpa =ry Þ dðpa =ry Þ p 1 1 ¼ 1  exp 2 ¼ 1  exp 2 rpy ¼ 0 gives rpy ¼ ln k. . Setting 2 2 dðp=ry Þ dðp=ry Þ ry k k   d2 ðpa =ry Þ ¼  k22 exp 2 rpy \0. Consequently, when rpy ¼ ln k, rpay reaches its maximum dðp=r Þ2 y

¼ rpy ¼ ln k. It should be noted that only if k ≤ kc can we have   2 ¼ ln k. For k ≥ kc, from equation (2.164), we obtain rpy  2k1 2 exp 2 rpy ¼ k2k2 2 .

value, which is p ry

pa ry

Substituting this equation into equation (2.168) gives

64

Autofrettage Technology and Its Applications in Pressured Apparatuses pa k 2  1 ¼ k2 ry

ð2:169Þ

Equation (2.33) indicates that in general when the load-bearing capacity

p ry

equals autofrettage pressure, or rpy ¼ rpay , it is the optimum, in this case rT e ≡ σy in the p k 1 whole plastic region. So, when rpay ¼ k k1 2 , the load-bearing capacity is also r ¼ k 2 . y   2 p In this case i.e. kj ¼ exp rpy and k ≥ kc, when rpay ¼ k k1 2 , if r is controlled by equay 2

2

tion (2.164), rpy \ rpay , and rT ej < σy. For example, when k = 3, from equation (2.164),

p ry rT ej ry

¼ 0:558724,

and

from

equation (2.169)

pa ry

¼ 89,

from

equation (2.149)

¼ 0.756998. When k ≤ kc, the optimum load-bearing capacity is also rpy ¼ rpay ¼ ln k, and in this

case rT e ≡ σy in the whole plastic region. When k = ∞, from equation (2.169), rpay ¼ 1; from equation (2.164), rpy ¼ 0:5; when rpy ¼ 0:5, from equation (2.168), we also have rpay ¼ 1. When k = ∞, equation (2.168) become pa p 1 ¼ þ ry ry 2

ð2:170Þ

For equation (2.168), when k ≤ kc, rpy = 0 to lnkc; when k ≥ kc, rpy = 0 to the value   determined by equation (2.164). When rpy = 0, kj ¼ exp rpy ¼ 1. The pressure vessel is at the initial yield stage, in which the pressure is the initial yield load rpye ¼ k2k1 2 . For 2

a pressure vessel with a certain radius ratio k ≥ kc, the maximum load for rT ej to reach the minimum is the value determined by equation (2.164). When rpy equals the value determined by equation (2.164), rpay ¼ 2 rpye ¼ k k1 2 . For the convenience of engineering applications, some data calculated by equations (2.164) and (2.168) are listed in table 2.1. Equation (2.168) is plotted in figure 2.34. In figure 2.34, curve ab is the connection of the ends of the curves described by equation (2.168), which shows the maximum load-bearing capacity rpy and the maximum autofrettage pressure rpay for k ≥ kc. From equation (2.164) we can find k2 expressed by rpy, substituting k2 into equation (2.169) we obtain rpay , as a result, the equation of curve ab in figure 2.34 is generated, that is   exp 2 rpy  2 rpy  1 pa   ¼ ð2:171Þ ry exp 2 p  2 2

ry

Mechanical Autofrettage Technology Based on Tresca Yield Criterion TAB. 2.1 – Numerical value for

p ry

and

65

pa ry .

k

p ry

pa ry

k

p ry

pa ry

2.3 2.4 2.5 2.6 2.7 2.8 2.9 3 3.1 3.2

0.67708 0.636288 0.612134 0.595387 0.582876 0.573101 0.565222 0.558724 0.553272 0.54863

0.810964 0.826389 0.84 0.852071 0.862826 0.872449 0.881094 8/9 = 0.888889 0.895942 0.902344

3.3 3.4 3.5 3.6 3.8 4 4.5 5 10 ∞

0.544632 0.541154 0.538103 0.535407 0.530865 0.527194 0.520552 0.51615 0.503692 0.5

0.908173 0.913495 0.918367 0.92284 0.930748 15/16 = 0.9375 0.950617 24/25 = 0.96 0.99 1

FIG. 2.34 – The graph of

pa ry

  ¼ rpy þ 12  2k12 exp 2 rpy .

The coordinates of point a is a(ln kc, ln kc). As mentioned above, the meaningful domain of k2 is rpy  12, and when rpy = 0.5, k2 = ∞, consequently from equation (2.168) or equation (2.169), rpay ¼ 1, or the coordinates of point b is b(0.5, 1). The coordinates of point c is c(0, 0.5). The practically significant curves for relation between rpy and rpay are located in a quadrilateral abco, where the straight line bc is parallel to the straight line ao. For k ≤ kc, the ends of the curves described by equation (2.168) are on the straight line ao where rpay ¼ rpy . Thereupon the included angle between the straight line ao and the abscissa axis or vertical axis is 45°. In figure 2.34, the dash curves have no practical significance, they are an extension of solid curves or only the elongation of the mathematical relation expressed by equation (2.168).

66

Autofrettage Technology and Its Applications in Pressured Apparatuses The total stresses are expressed as follows 8 T rz r0 > > > ¼ z þ > > ry ry > > > > rT r0 > > > > r ¼ r þ < ry ry > rT r0h > h > ¼ þ > > > ry ry > > > > T > r0 > re > ¼ e þ : ry ry

rpz ry rpr ry rpe ry

ð2:172Þ

rpe ry

If the pressure contained in a pressure vessel does not exceed the initial yield pressure, or p ≤ pe, all stresses will not exceed the yield strength σy, then autofrettage is not necessary. However, if autofrettage is performed, the  stresses will be ðwhen p ¼ pe ; evened and reduced. Between 1 and k, if kj ¼ exp rpy 2  kj ¼ exp k2k1 Þ, the equivalent total stress at the elastic-plastic juncture rT 2 ej is the minimum, the minimum value  2 is  equation (2.149). k 1 When p = pe, kj ¼ exp 2k 2 , then, equation (2.149) becomes   2  p k 1 T 2 2 exp 2 exp  1 1 rej ry k2 k k   ¼ 2

k 2 1 ¼ 2 ry k  1 exp 2 p k  1 exp k 2

ð2:173Þ

ry

Several figures are presented to show the above-mentioned conclusions. 2  2 k 1 = 0.444444, k ¼ exp = 1.559623, the operation For k = 3, rpy ¼ k2k1 2 j 2k 2 stresses, residual stresses, and total stresses are shown in figures 2.35–2.37, respectively.

FIG. 2.35 – The distribution of the operation stresses for k = 3, p = pe and kj ¼ exp





k 2 1 2k 2

.

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

67

FIG. 2.36 – The distribution of the residual stresses for k = 3, p = pe, and kj ¼ exp

FIG. 2.37 – The distribution of the total stresses for k = 3, p = pe, and kj ¼ exp

In figure 2.37, For k = 3,

p ry

¼

rT ej ry

k 2 1 2k 2

 .



k 2 1 2k 2

.

= 0.662499. This must be the minimum. 2  ¼ rpye = 0.444444, kj = 1.4 < exp k2k1 , the total stresses are 2

k 2 1 2k 2

shown in figure 2.38.





68

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 2.38 – The distribution of the total stresses for k = 3, p = pe, and kj = 1.4. In figure 2.38, For k = 3,

p ry

rT ej ry

= 0.674968 > 0.662499.

pe ¼ k2k1 2 ¼ r = 0.444444, exp y 2



k 2 1 2k 2



< kj = 1.7 < kj* = 1.748442,

the total stresses are shown in figure 2.39.

FIG. 2.39 – The distribution of the total stresses for k = 3, p = pe, and kj = 1.7.

rT

In figure 2.39, rejy = 0.668629 > 0.662499. For k = 3, from equation (2.164) or table 2.1, rpy = 0.558724 >   exp rpy = kj* = 1.748442, the total stresses are shown in figure 2.40.

pe ry ,

kj ¼

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

69

FIG. 2.40 – The distribution of the total stresses for k = 3, p = 0.558724 and kj = kj*. rT

In figure 2.40, rejy = 0.756998. For k = 3, from equation (2.164) or table 2.1, rpy = 0.558724 > the total stresses are shown in figure 2.41.

pe ry ,

kj = 1.6 < kj*,

FIG. 2.41 – The distribution of the total stresses for k = 3, p = 0.558724, and kj = 1.6.

rT

In figure 2.41, rejy = 0.763525 > 0.756998. For k = 3, from equation (2.164) or table 2.1, rpy = 0.558724 > the total stresses are shown in figure 2.42. In figure 2.42,

rT ej ry

= 0.757596 > 0.756998.

pe ry ,

kj = 1.8 > kj*,

Autofrettage Technology and Its Applications in Pressured Apparatuses

70

FIG. 2.42 – The distribution of the total stresses for k = 3, p = 0.558724 and kj = 1.8.

2.6

Comparison Between Three Cases

pffiffiffi So far, we have investigated autofrettage under four cases: kj = k, kj ¼ k þ2 1, kj ¼ k   and kj ¼ exp rpy . Some of the results obtained here are similar to those in ref.[4–6].   kj = k is for k < kc, and when k < kc, rpy ¼ ln k, that is equivalent to k ¼ exp rpy .   Therefore, when k < kc, kj = k is equivalent to kj ¼ exp rpy . Then, we do a com  pffiffiffi parison between the three cases: kj ¼ k þ2 1, kj ¼ k and kj ¼ exp rpy . I. For k = 2.5 > kc 1. kj ¼ k þ2 1 kj ¼

k þ1 , kj1 ¼ 1:75 2

The equivalent residual stress From equation (2.120), in the plastic region, k2  r0e ¼1 ry

ðk þ 1Þ2 þ k2 4 2 k 1

2

ln ðk þ4 1Þ 1 1:939561 ¼ 1 x2 x2

In the elastic region, from equation (2.88), r0e ¼ ry

ðk þ 1Þ2 4

2

 1  ln ðk þ4 1Þ k 2 1:122939 ¼ x2 k2  1 x2

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

The load-bearing capacity, from equation (2.130), p 1 ðk þ 1Þ2 k þ1 ¼ 0:814616 ¼  þ ln ry 2 2 8k 2 The equivalent operation stress, from equation (1.9), rpe 2k 2 p 1:939561 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The equivalent total stress, from equation (2.15), rT r0 rp e ¼ e þ e ry ry ry Then, in the plastic region, rT e 1 ry In the elastic region, rT 3:0625 e ¼ x2 ry The autofrettage pressure, from equation (2.130), pa p ¼ ry ry

2. kj ¼

pffiffiffi k kj ¼

pffiffiffi k , kj2 ¼ 1:581139

The equivalent residual stress From equation (2.136), in the plastic region, r0e k 2  k þ k 2 ln k 1 1:805108 ¼1 ¼ 1 2 2 k 1 x x2 ry From equation (2.140), in the elastic region, r0e k  1  ln k k 2 0:694892 ¼ ¼ k2  1 x2 x2 ry

71

72

Autofrettage Technology and Its Applications in Pressured Apparatuses

The load-bearing capacity, from equation (2.145), pffiffiffi p k1 þ ln k ¼ 0:758145 ¼ ry 2k The equivalent operation stress, from equation (1.9), rpe 2k 2 p 1:805108 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The equivalent total stress, from equation (2.28), rT r0 rp e ¼ e þ e ry ry ry Thus, in the plastic region, rT e 1 ry In the elastic region, rT 2:5 e ¼ 2 x ry The autofrettage pressure pa p ¼ ry ry

3. kj ¼ exp p ry

  p ry

When k = 2.5, from equation (2.164) or table 2.1, the load-bearing   capacity p = 0.612134. When ry = 0.612134, from equation (2.148), kj ¼ exp rpy , kj3 =

1.844363. The equivalent residual stress From equation (2.159), in the plastic region, r0e 2 ¼1 2 x ry From equation (2.163), in the elastic region, r0e k  1  ln k k 2 1:401675 ¼ ¼ k2  1 x2 x2 ry The equivalent operation stress, from equation (1.9),

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

73

rpe 2k 2 p 1:805108 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The equivalent total stress, from equation (2.15), rT r0 rp e ¼ e þ e ry ry ry Then, in the plastic region, rT 0:194892 e ¼1 x2 ry In the elastic region, rT 3:206783 e ¼ x2 ry The autofrettage pressure, from equation (2.169), pa k 2  1 ¼ ¼ 0:84 k2 ry When k = 2.5, the equivalent residual stress and the equivalent total stress are plotted in figures 2.43 and 2.44, respectively, where curve beh is the equivalent pffiffiffi residual stress for kj ¼ k þ2 1, curve cdl is the equivalent residual stress for kj ¼ k ,   curve afg is the equivalent residual stress for kj ¼ exp rpy ; curve nqv is the equivpffiffiffi alent total stress for kj ¼ k þ2 1, curve nzw is the equivalent total stress for kj ¼ k ,   curve mst is the equivalent total stress for kj ¼ exp rpy . The main parameters are marked in these two figures. Using these figures, we can compare the value and shape of the stress images.

FIG. 2.43 – Comparison of the equivalent residual stress when k = 2.5.

74

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 2.44 – Comparison of the equivalent total stress when k = 2.5. II. For k = 2 < kc 1. kj ¼ k þ2 1 kj ¼

k þ1 , kj1 ¼ 1:5 2

The equivalent residual stress From equation (2.120), in the plastic region, k2  r0e ¼1 ry

ðk þ 1Þ2 þ k2 4 2 k 1

2

ln ðk þ4 1Þ 1 1 ¼ 11:664574 2 x2 x

From equation (2.88), in the elastic region r0e ¼ ry

ðk þ 1Þ2 4

2

 1  ln ðk þ4 1Þ k 2 0:585426 ¼ x2 k2  1 x2

The load-bearing capacity, from equation (2.130), p 1 ðk þ 1Þ2 k þ1 ¼ 0:624215 ¼  þ ln ry 2 2 8k 2 The equivalent operation stress, from equation (1.9), rpe 2k 2 p 1:664574 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The equivalent total stress, from equation (2.15),

Mechanical Autofrettage Technology Based on Tresca Yield Criterion rT r0 rp e ¼ e þ e ry ry ry Thus, in the plastic region, rT e 1 ry In the elastic region, rT 2:25 e ¼ 2 x ry The autofrettage pressure pa p ¼ ry ry

2. kj ¼

pffiffiffi k kj ¼

pffiffiffi k , kj2 ¼ 1:414214

The equivalent residual stress From equation (2.136), in the plastic region, r0e k 2  k þ k 2 ln k 1 1:590863 ¼1 ¼ 1 k2  1 x2 x2 ry From equation (2.140), in the elastic region, r0e k  1  ln k k 2 0:409137 ¼ ¼ k2  1 x2 x2 ry The load-bearing capacity, from equation (2.145), pffiffiffi p k1 þ ln k ¼ 0:596574 ¼ ry 2k The equivalent operation stress, from equation (1.9), rpe 2k 2 p 1:590863 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The equivalent total stress, from equation (2.15), rT r0 rp e ¼ e þ e ry ry ry

75

76

Autofrettage Technology and Its Applications in Pressured Apparatuses

Then, in the plastic region, rT e 1 ry In the elastic region, rT 2 e ¼ ry x 2 The autofrettage pressure pa p ¼ ry ry

3. kj ¼ exp

  p ry

and kj = k

When k = 2 < kc, kj = k, the load-bearing capacity

p ry

= lnk = 0.693147, which   is greater than the load-bearing capacity when k = 2.5 and kj ¼ exp rpy . When   p p = lnk, k ¼ exp j3 ry ry = k. There is no elastic region when kj = k. The equivalent residual stress From equation (2.93), r0e k 2 ln k 2 1:848392 ¼1 2 ¼1 x2 ry ðk  1Þx 2 The equivalent operation stress, from equation (1.9), rpe 1:848392 ¼ x2 ry The equivalent total stress rT e 1 ry The autofrettage pressure pa p ¼ ¼ 0:693147 ry ry When k = 2, the equivalent residual stress and the equivalent total stress are plotted in figures 2.45 and 2.46, respectively, where curve beh is the equivalent pffiffiffi residual stress for kj ¼ k þ2 1, curve cdg is the equivalent residual stress for kj ¼ k ,   curve af is the equivalent residual stress for kj ¼ exp rpy or kj = k; curve nqv is the

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

77

equivalent total stress for kj ¼ k þ2 1, curve nzw is the equivalent total stress for   pffiffiffi kj ¼ k , curve nm is the equivalent total stress for kj ¼ exp rpy or kj = k. The main parameters are marked in these two figures. Using these figures, we can compare the value and shape of the stress images.

FIG. 2.45 – Comparison of the equivalent residual stress when k = 2.

FIG. 2.46 – Comparison of the equivalent total stress when k = 2.

2.7

Chapter Summary

The main equations and conclusion are listed in table 2.2.

78

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 2.2 – The main equations and conclusion of this chapter. The components of residual stresses in general forms Within the plastic region (1 ≤ x ≤ kj) " # ! 0 kj2 rz 1 kj2 x2 1 2 ¼ þ ln 2  1  2 þ ln kj k2  1 ry 2 k 2 k kj ! "  # kj2 r0r 1 kj2 x2 1 k2 2 1  ¼  1 þ ln  1  þ ln k j k2  1 ry 2 k 2 k2 x2 kj2 " !  # kj2 r0h 1 kj2 x2 1 k2 2 1þ 2 ¼ þ 1 þ ln 2  1  2 þ ln kj k2  1 ry 2 k 2 k x kj Within the elastic region (kj ≤ x ≤ k) r0z kj2  1  ln kj2 ¼ constant [ 0 ¼ ry 2ðk 2  1Þ   r0r k 2 r0z ¼ 1 2 \0 ry x ry   r0h k 2 r0z ¼ 1þ 2 [0 ry x ry

General equivalent residual stress k 2  kj2 þ k 2 ln kj2 1 r0e ¼1 (1 ≤ x ≤ kj) x2 ry k2  1 r0e kj2  1  ln kj2 k 2 ¼ [ 0 (kj ≤ x ≤ k) ry k2  1 x2 General autofrettage pressure pa k 2  kj2 pa ¼ ln kj þ ry 2k 2 The abscissa of the intersection of curves of the residual stress or the abscissa at which r0e = 0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2  kj2 þ k 2 ln kj2 r x0 ¼ ¼ \kj k2  1 ri The value of residual stresses when x = x0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 0 0 k 2  kj2 þ k 2 ln kj2 kj2  1  ln kj2 rzx0 rhx0 rrx0 r0rmin x0 r0zj ¼ ¼ ¼ ¼ ln þ ¼ ln þ ry ry ry ry kj ry 2ðk 2  1Þ kj2 ðk 2  1Þ The abscissa at which r0z = 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 1k þ k ln k j j

x1 ¼ e k 2 1 Equivalent operation stress rpe rph rpr 2k 2 rpz 2k 2 p ¼  ¼ 2 ¼ 2 2 ry ry ry x ry x ðk  1Þ ry The critical radius ratio kc = 2.218 457 489 916 7… k 2 ln k k 2 ln k ¼ 1, and 2 ¼ 1 is equivalent to 2pe = py, i.e. kc is the solution of 2 k 1 k 1 2 k 1 2 ¼ ln k 2k 2

Mechanical Autofrettage Technology Based on Tresca Yield Criterion TAB. 2.2 – (continued). The optimum plastic depth kj* 2 k 2 ln kj2  k 2  kj þ 2 ¼ 0 (e0.5 ≤ kj* ≤ kc and k ≥ kc)

1.

2.

3.

4. 5.

When k ≤ kc, kj = k The results when kj = kj* Residual stresses in axial, radial, and circumferential direction 8 8 0 2 > 2 r r0z kj > > z > > ¼ ln x  0:5 ¼ > > > > > > ry 2k 2 ry > > > > > >   < r0 < 0 rr k 2 r0z r ¼ ln x þ x 2  1 (1 ≤ x ≤ kj*) (kj* ≤ x ≤ k) ¼ 1 2 > > ry ry x ry > > > > > >   > > > > r0 > > r0h k 2 r0z > > : h ¼ ln x  x 2 > ¼ 1 þ : ry ry x 2 ry Equivalent residual stress 2 2 r0e 2 r0e kj ¼ 1  2 (1 ≤ x ≤ kj*) ¼ (kj* ≤ x ≤ k) x x2 ry ry The optimum load-bearing capability and autofrettage pressure p pa pe k 2  1 ¼ ¼2 ¼ ry ry k2 ry pffiffiffi pffiffiffi x0  2 = constant x1  e = constant pffiffiffi r0z r0r r0h r0rmin r0rx0 1  ln 2 s ¼ constant ¼ ¼ ¼ ¼  ln 2  0:5 ¼  y0 ¼ 2 ry ry ry ry ry y0s is not related to k and kj

The results when kj = kj*, p = pa = 2pe 1. The components of total stresses 8 T 8 T kj2 rz rz 1 1 > > > > ¼ ¼ constant > > þ ¼ ln x  > > > > ry 2k 2 2 k2 ry > > > >   < < rT kj2 1 1 rT 1 r r (kj* ≤ x ≤ k) ¼ ln x  1 þ 2 (1 ≤ x ≤ kj*) ¼  > > k ry ry 2 x2 k2 > > > >   > > > > rT 1 > > rT k 2 1 1 > > : h ¼ j : h ¼ ln x þ 2 þ k ry ry 2 x2 k2 2. Equivalent residual stress kj2 rT rT e e  1 (1 ≤ x ≤ kj*) ¼ 2 (kj* ≤ x ≤ k) ry ry x The results when kj = k 1. The optimum load-bearing capability and autofrettage pressure p pa py ¼ ¼ ln k ¼ ry ry ry 2. The total stresses when kj = k and p = py = σylnk rT 1 z ¼ þ ln x  ln k ry 2 rT r ¼ ln x  ln k ry rT h ¼ 1 þ ln x  ln k ry

79

80

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 2.2 – (continued).

rT e 1 ry 3. The total stresses on an inner surface when kj = k and p = py = σylnk rT 1 zi ¼  ln k ry 2 rT p ri ¼  ln k ¼  ry ry rT hi ¼ 1  ln k ry rT ei ¼1 ry k þ1 The results when kj ¼ 2 1. When k ≤ ka = 2.618253…, |r0ei | ≤ σy; when k ≥ ka, |r0ei | ≥ σy k þ1 = 1.809126… 2. When k = ka, kj ¼ 2 3. The maximum load-bearing capacity and autofrettage pressure p 1 ðk þ 1Þ2 k þ1 ¼  þ ln 2 ry 2 2 8k ðk þ 1Þ2 k þ 1 k2  1 1 ðk þ 1Þ2 2 \ þ ln ; when k ≥ ka,  þ ln k þ2 1 [ k k1 2 , 2 2 2 k 2 8k 8k 2 0 but when k ≥ ka, |rei | > σy T 5. When k ≤ 3.102921… = kz, p ≤ σy and rT ri ≥ −σy; when k ≥ kz, p ≥ σy and rri ≤ −σy. When k ≤ ka, there must be k ≤ kz 6. When k ≥ ka, rT e > σy

4. When k ≤ ka, 12 

pa 1 ðk þ 1Þ2 k þ1 p ¼ ¼  þ ln 2 ry ry 2 8k 2   pffiffiffi pffiffiffi p 1 ðk þ 1Þ2 k þ1 k þ1 p k1 k > þ ln k (kj ¼ k ) 8. ¼  þ ln ¼ ¼ j ry 2 2 2 ry 2k 8k 2 7. Autofrettage pressure

1 ðk þ 1Þ2 k þ 1 k2  1 1 ðk þ 1Þ2 2 \ þ ln ; when k ≥ ka,  þ ln k þ2 1 [ k k1 9. When k ≤ ka,  2 , 2 2 2 2 k 2 8k 8k 2 0 but when k ≥ ka, |rei | > σy pffiffiffi The results when kj ¼ k 1. When k ≤ kb = 3.042297…, |r0ei | ≤ σy; when k ≥ kb, |r0ei | ≥ σy pffiffiffi 2. When k = kb, kj ¼ k = 1.744218… 3. The maximum load-bearing capacity and autofrettage pressure pffiffiffi p k1 þ ln k ¼ ry 2k   pffiffiffi pffiffiffi p k1 p 1 ðk þ 1Þ2 k þ1 k þ1 ¼ ¼  þ ln 4. þ ln k (kj ¼ k ) < kj ¼ 8k 2 ry 2k ry 2 2 2 pffiffiffi k 2  1 pffiffiffi k 2  1 p k1 p k1 ¼ ; when k ≥ kb, ¼ , 5. When k ≤ kb, þ ln k \ þ ln k [ ry 2k k2 ry 2k k2 but when k ≥ kb, |r0ei | > σy

Mechanical Autofrettage Technology Based on Tresca Yield Criterion

81

TAB. 2.2 – (continued). The results when kj ¼   1. Overstrain e ¼ 2. When kj ¼ exp

exp

p ry

  rj p ¼ exp ry ri

1

k1

  p ry

, the equivalent stress σej is the minimum, that is   p rT k 2 exp 2 ry  1 k 2 kj2  1 ej   ¼ 2 ¼ 2 ry k  1 exp 2 p k  1 kj2 ry

3. The allowable load in order for r0ei /σy ≥ −1 when k ≥ kc 



exp 2 rpy þ k 2  2k 2 rpy  2  0 or k 2 

 

2rpy p 2ry 1

exp

2

4. The relationship between autofrettage pressure and the allowable load   pa p 1 1 p ¼ þ  2 exp 2 ry ry 2 2k ry

References [1] Liu H.W. (2019) Mechanics of materials (in Chinese). Higher Education Press, Beijing. [2] Yu G.C. (1980) Chemical pressure vessels and equipment (in Chinese). Chemical Industrial Press, Beijing. [3] The compiling group of “handbook of mathematics” (1979) A handbook of mathematics (in Chinese). Higher Education Press, Beijing. [4] Zhu R.L. (2008) Results resulting from autofrettage of a cylinder, Chin. J. Mech. Eng. 21(4), 105. [5] Zhu R.L. (2008) Ultimate load-bearing capacity of cylinder derived from autofrettage under ideal condition, Chin. J. Mech. Eng. 21(5), 80. [6] Zhu R.L. (2013) Study on autofrettage for medium-thick pressure vessels, J. Eng. Mech. 139 (12), 1790.

Chapter 3 Mechanical Autofrettage Technology Based on Mises Yield Criterion Based on the maximum distortion strain energy theory (Mises yield criterion), to achieve a plastic depth kj, the pressure exerted on a pressure vessel is evidently equation (1.24), this is just autofrettage pressure pa[1, 2], or kj2 k 2  kj2 k 2 ln kj2 þ k 2  kj2 pa 2 1 2 pffiffiffi ¼ pffiffiffi ln kj þ pffiffiffi  pffiffiffi ¼ pffiffiffi ln kj þ pffiffiffi ¼ ry 3 3 3 3k 2 3k 2 3k 2

ð3:1Þ

Obviously, as in the case of the maximum shear stress theory (Tresca yield criterion), the greater kj is, the greater pa is. It’s not hard to foresee that kj is bound to affect residual stresses, load-bearing capability, etc., and under different strength theories, the extent and the way of its influence will be different. Therefore, it is necessary to investigate autofrettage technology based on the maximum distortion strain energy theory. We also proceed from the general situation.

3.1

General Study on Mechanical Autofrettage Technology

In our research, we obtained some results about mechanical autofrettage technology based on the maximum distortion strain energy theory[3–7], which is the foundation of this research.

3.1.1

In General Forms

1. The components of residual stresses When a pressure vessel is subjected to an autofrettage pressure pa which is greater than its initial yield pressure (pe) and smaller than the entire yield pressure (py), its inner layer becomes plastic and its outer layer remains elastic. After DOI: 10.1051/978-2-7598-3111-1.c003 © Science Press, EDP Sciences, 2023

84

Autofrettage Technology and Its Applications in Pressured Apparatuses

eliminating autofrettage pressure pa, the residual stresses remain in the wall of the pressure vessels. Based on the maximum distortion strain energy theory (Mises yield criterion), the residual stresses can be expressed as follows[2]. Within the plastic region (1 ≤ x ≤ kj) ! " # kj2 r0z 1 kj2 x2 1 2 ¼ pffiffiffi 2 þ ln 2  1  2 þ ln kj ð3:2Þ k2  1 ry k kj 3 k " r0r 1 kj2 x2 ¼ pffiffiffi 2  1 þ ln 2  ry kj 3 k

!  # kj2 1 k2 2 1 2 1  2 þ ln kj k2  1 k x

" r0h 1 kj2 x2 ¼ pffiffiffi 2 þ 1 þ ln 2  ry kj 3 k

!  # kj2 1 k2 2 1þ 2 1  2 þ ln kj k2  1 k x

k 2  kj2 þ k 2 ln kj2 1 r0e ¼1 x2 ry k2  1

ð3:3Þ

ð3:4Þ

ð3:5Þ

Based on the maximum shear stress theory (Tresca yield criterion), it is easy to know[8] that as x increases, r0z , r0h , and r0e increase within the plastic region qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 (1 ≤ x ≤ kj). When x  , r0r decreases as x increases; When k 2 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 x , r0r increases as x increases. From ln kj2 \kj2  1, we can prove that k 2 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 j. k 2 1 q\k ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 k kj2 þ k 2 ln kj2 In fact, is just the relative location of the intersection of the three k 2 1 residual stress curves, i.e., the abscissa at which r0e ¼ 0, or x0 in the following section. Within the elastic region (kj ≤ x ≤ k) " r0z 1 kj2 ¼ pffiffiffi 2  ry 3 k

kj2 1  2 þ ln kj2 k r0r ¼ ry

!

# kj2  1  ln kj2 1 pffiffiffi ¼ constant [ 0 ¼ k2  1 3ðk 2  1Þ

ð3:6Þ

  k 2 r0z k 2  x 2 kj2  1  ln kj2 pffiffiffi \0 1 2 ¼ x ry x2 3ðk 2  1Þ

ð3:7Þ

  k 2 r0z k 2 þ x 2 kj2  1  ln kj2 pffiffiffi [0 1þ 2 ¼ x ry x2 3ðk 2  1Þ

ð3:8Þ

r0h ¼ ry

r0e kj2  1  ln kj2 k 2 ¼ [0 ry k2  1 x2

ð3:9Þ

Mechanical Autofrettage Technology Based on Mises Yield Criterion

85

Clearly, as x increases, r0h and r0e decrease, r0r increases, and r0z is kept as a constant. Figure 3.1 is the typical distribution of the residual stresses, where k = 4 and kj = 1.6. The meaning of each curve and main parameters are marked in the figure.

FIG. 3.1 – The typical distributions of the residual stresses. When based on the maximum distortion strain energy theory, the components of residual stresses are p2ffiffi3 times as much as those based on the maximum shear stress pffiffi theory, and as mentioned above, rde ¼ 23 rse , as a result, the equivalent residual stress based on the maximum distortion strain energy theory (Mises yield criterion) is the same as that based on the maximum shear stress theory (Tresca yield criterion). At the inside surface, x = r/ri = 1, then, from equations (3.2)–(3.5), the residual stresses on the inside surface are as follows. k 2 ln kj2  kj2 þ 1 r0zi 0 ¼  pffiffiffi ry 3ðk 2  1Þ

ð3:10Þ

r0ri ¼ 0

ð3:11Þ

r0hi 2 k 2 ln kj2  kj2 þ 1 r0zi ¼ 2 ¼  pffiffiffi 0 k2  1 ry ry 3

ð3:12Þ

pffiffiffi  0  pffiffiffi 0 k 2 ln kj2  kj2 þ 1 pffiffiffi r0zi 3 rhi r0ri 3 rhi r0ei ¼ 3 ¼  ¼ 0 ¼ k2  1 ry ry 2 ry ry 2 ry

ð3:13Þ

86

Autofrettage Technology and Its Applications in Pressured Apparatuses

0 0 At the inside surface, the absolute   0  value of rhi is the maximum while rri ¼ 0. 0     Therefore, when rei ¼ ry , there is rhi ¼ ry . Therefore, in the next chapter, we will study mechanical autofrettage technology by limiting circumferential residual stress based on the Mises yield criterion. At the elastic-plastic juncture, x = rj/ri = kj, the residual stresses calculated by equations (3.2)–(3.5) are equal to those calculated by equations (3.6)–(3.9). Therefore, the residual stresses in the whole wall are continuous. Substituting x = kj into equations (3.2)–(3.5) or equations (3.6)–(3.9) leads to the residual stresses at the elastic-plastic juncture

r0zj kj2  1  ln kj2 0 ¼ pffiffiffi ry 3ðk 2  1Þ r0rj ¼ ry r0hj ry

0

r0hj ry

\

¼

k2 1 2 kj

!

r0zj 0 ry

! r0zj k 2 r0zj 1þ 2 [ 0 ry kj ry

r0ej pffiffiffi k 2 r0zj k 2 ðkj2  1  ln kj2 Þ ¼ 3 2 ¼ 1 ry kj ry ðk 2  1Þkj2

ð3:14Þ

ð3:15Þ

ð3:16Þ

ð3:17Þ

Based on the maximum shear stress theory (Tresca yield criterion), r0z [ 0, r0h , r0e and r0z monotonically increase as x increases in the plastic zone, r0h and r0e monotonically decrease as x increases in the elastic zone, r0z remains a constant as x increases in the elastic zone, r0r monotonically increase as x increases in the elastic zone. At the elastic-plastic juncture, r0hj , r0ej , and r0zj are the maximum within the whole wall (algebraic value, not absolute value). Regardless of the value of kj, the equivalent residual stress at an elastic-plastic juncture (r0ej ) is certainly tension and smaller than yield strength σy. Therefore, it is quite necessary to pay close attention to the residual stresses at the inside surface. At the outside surface, x = ro/ri = k, then, from equations (3.6)–(3.9), r0zo kj2  1  ln kj2 r0zj ¼ ¼ pffiffiffi [0 ry ry 3ðk 2  1Þ

ð3:18Þ

r0ro ¼ 0

ð3:19Þ

0 \ r0ho ¼ 2r0zo \ r0hj [ 0

ð3:20Þ

Mechanical Autofrettage Technology Based on Mises Yield Criterion r0eo kj2  1  ln kj2 pffiffiffi r0zo ¼ 3 ¼ [0 ry k2  1 ry Within the whole wall, r0r  0; r0eo \r0ho . r0

r0

In the elastic region, letting rye ¼ ryh gives x ¼ r0 \ ryh .

ð3:21Þ

pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k = 0.8556 k = xa, or xa is

the abscissa at which r0e ¼ r0h . Therefore, when x < xa, r0e ry

87

r0e ry

[

r0h ry ;

when x > xa,

2. Autofrettage pressure pa Autofrettage pressure is the internal pressure applied to a pressure vessel to cause a plastic depth kj before it is put into production, which is just equation (3.1), or k 2  kj2 pa 2 ¼ pffiffiffi ln kj þ pffiffiffi ry 3 3k 2 3. The location (defined as x0) of the intersection of the three residual stress curves or the abscissa at which r0e ¼ 0. X0 is the same as that based on the maximum shear stress theory (Tresca yield criterion), which is as follows sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2  kj2 þ k 2 ln kj2 r x0 ¼ ¼ ð3:22Þ \kj ri k2  1 The value of radial residual stresses at x = x0 (r0rxo ) is the minimum, which is r0rx0 r0rmin r0zx0 r0hx0 2 x0 r0zj ¼ ¼ ¼ ¼ pffiffiffi ln þ ry ry sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ry ry ry 3ffi kj 2 2  k 2 þ k 2 ln k 2 k k  1  ln kj2 2 j j j p ffiffi ffi ¼ y0d ¼ pffiffiffi ln þ 2 kj ðk 2  1Þ 3 3ðk 2  1Þ

ð3:23Þ

y0d is p2ffiffi3 times of y0s in chapter 2. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 In figures 3.1 and 2.1, x0 ¼ = 1.377924 < kj = 1.6. In figure 3.1, k 2 1 d s y0 ¼ 0:148678344, and in figure 2.1, y0 ¼ 0:128759223, y0d ¼ p2ffiffi3 y0s . 4. The location (defined as x1) where r0z ¼ 0 Although the residual stresses based on the maximum distortion strain energy theory are p2ffiffi3 times of the residual stresses based on the maximum shear stress theory, the abscissa, x1, at which r0z ¼ 0, is the same based on both strength theories, is as follows,

88

Autofrettage Technology and Its Applications in Pressured Apparatuses rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x1 ¼

e

1k 2 þ k 2 ln k 2 j j k 2 1

ð3:24Þ

In figures 3.1 and 2.1, x1 = 1.567273. 5. Lamé equations The components of the operation stresses in three directions, or the axial elastic stress rpz , the radial elastic stress rpr and the circumferential elastic stress rph , are the same as those based on the maximum shear stress theory (Tresca criterion), and the equivalent operation stress is as follows: pffiffiffi  p pffiffiffi 2  pffiffiffi 2 p 3 rh rpr 3k rz 3k rpe p ¼  ¼ ¼ 2 ry x ry x 2 ðk 2  1Þ ry 2 ry ry

ð3:25Þ

6. The total stresses The total stresses include the residual stresses and the elastic stresses caused by internal pressure p. The total stresses in general form within the plastic region (1 ≤ x ≤ kj) ! " # 2 2 2 k k rT 1 x 1 1 p j j z ¼ pffiffiffi 2 þ ln 2  1  2 þ ln kj2 ð3:26Þ þ 2 21 k k ry k k  1 ry k 3 j " rT 1 kj2 x2 r ¼ pffiffiffi 2  1 þ ln 2  ry kj 3 k

!  # kj2 1 k2 k2  x2 p 2 1  1  2 þ ln kj  k2  1 k x2 x 2 ðk 2  1Þ ry ð3:27Þ

" kj2 rT 1 x2 h ¼ pffiffiffi 2 þ 1 þ ln 2  ry kj 3 k

!  # 2 kj2 1 k k2 þ x2 p 1 þ 1  2 þ ln kj2 þ k2  1 k x2 x 2 ðk 2  1Þ ry ð3:28Þ

pffiffiffi 2 k 2  kj2 þ k 2 ln kj2 3k rT p e þ ¼1 ry x 2 ðk 2  1Þ x 2 ðk 2  1Þ ry Setting

rT e ry

ð3:29Þ

 1 obtains the load-bearing capability k 2  kj2 pa p 2  pffiffiffi ln kj þ pffiffiffi ¼ ry ry 3 3k 2

ð3:30Þ

Mechanical Autofrettage Technology Based on Mises Yield Criterion  T re ry

d

Setting

dx

¼

k 2 kj2 þ k 2 ln kj2 2 k 2 1 x3

89

pffiffi 2 3k 1 p  2k 2 1 x 3 ry  0 also obtains equation (3.30). k 2 k 2

r Therefore, when rpy ¼ p2ffiffi3 ln kj þ pffiffi3k 2j ¼ rpay , rey  1 = constant within the plastic   T  pffiffi rT r rT rT rT rT region. Since rey ¼ 23 rhy  rry = constant and rzy ¼ 12 rhy þ rry , the three curves for T

T T parallel to each other within the plastic region. rT z ; rh ; rr are  T r

h ry

d

Setting

dx

 T d

Setting

rr ry

dx

 0 obtains p x 2 ðk 2  1Þ k 2  kj2 þ k 2 ln kj2 pa pffiffiffi  pffiffiffi þ [ 2 2 ry r 3k 3k y  0 obtains k 2  kj2 þ k 2 ln kj2 x 2 ðk 2  1Þ pr pa p pffiffiffi   pffiffiffi ¼ \ ry ry ry 3k 2 3k 2

ð3:31Þ

When x = 1, equation (3.31) becomes kj2  1 pri p 2  pffiffiffi ln kj  pffiffiffi ¼ ry ry 3 3k 2

ð3:32Þ

When x = kj, equation (3.31) becomes kj2  1 pro p 2  pffiffiffi ln kj  pffiffiffi ¼ \0 ry ry 3 3k 2

ð3:33Þ

As long as the operation pressure p does not exceed autofrettage pressure pa, rT

rT e ry

rT

and rhy increase with the increase of x. Clearly, rzy always increases with the increase of x. When

p ry

satisfies equation (3.31),

x, otherwise,

rT r ry

monotonously increases with the increase of

rT r

does not increase monotonously with the increase of x in the plastic qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 rT rT r0 region. Especially, when p = 0, rry is just ryr , hence, when x  x0 ¼ , rry k 2 1 ry

decreases as x increases; when x ≥ x0, equations (3.29) and (3.31), when rT e ry rT h ry

= 0, there must be =

rT r ry

=

rT z ry .

rT h ry

=

rT r ry .

p ry

rT r ry

increases as x increases. Besides, from  pffiffi  T r rT rT rT ¼ rpyr , rey = 0. Since rey = 23 rhy  rry , when

Since rT z ¼

Therefore, if p ≤ pr,

rT r ry

rT þ rT r h 2 ,

when

rT h ry

=

rT r ry ,

there must be

does not increase monotonously with the

increase of x in the plastic region, the three curves for section in the plastic region, at the intersection,

rT e ry

rT rT r h ry , ry

and

rT z ry

have an inter-

= 0. In addition, it can be proved

Autofrettage Technology and Its Applications in Pressured Apparatuses

90

that at the intersection,

rT r ry

 T r  gets the minimum, or  rry  gets the maximum. The

abscissa (x0T ) of the intersection point of curves of total stresses is solved from equation (3.31) as follows. x0T

ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi k 2  kj2 þ k 2 ln kj2 3k 2 p  2 ¼ k2  1 k  1 ry

ð3:34Þ

However, in engineering practice, x0T usually does not exist between 1 and k. Usually, x0T \1. The total stresses in general form within the elastic region (kj ≤ x ≤ k) kj2  1  ln kj2 rT 1 p z þ 2 ¼ pffiffiffi ¼ constant [ 0 2 k ry  1 ry 3ðk  1Þ

ð3:35Þ

rT k 2  x 2 kj2  1  ln kj2 k2  x2 p r p ffiffi ffi  ¼ \0 ry x2 x 2 ðk 2  1Þ ry 3ðk 2  1Þ

ð3:36Þ

rT k 2 þ x 2 kj2  1  ln kj2 k2 þ x2 p h pffiffiffi þ 2 2 ¼ [0 2 ry x x ðk  1Þ ry 3ðk 2  1Þ

ð3:37Þ

pffiffiffi 2 kj2  1  ln kj2 k 2 3k rT p e ¼ þ 2 2 [0 2 2 ry k 1 x x ðk  1Þ ry

ð3:38Þ

Within the whole wall, In the elastic region, plastic region,

rT e ry

rT e ry

r0r ry

\0 and

rpr ry

\0, then,

rT r ry

\0 within the whole wall.

decreases as x increases. Therefore, if

rT e ry

 1 within the

< 1 within the elastic region.

In the elastic region, setting

rT e ry



rT h ry ,

we obtain

pffiffiffi p  kj2  1  ln kj2 pffiffiffi 2 pffiffiffi 3k 2  k 2  x 2 3k  k 2  x 2  ry 3 pffiffiffi 2 Since kj2−1 > lnkj, kj2−1−lnkj > 0. When 3k  k 2  x 2  0, or pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ffiffiffi kj 1ln kj p x 3  1k = xa, meanwhile ry   pffiffi3 (that’s of course in practice), p ffiffi ffi T T k 2 1ln k 2 rT rh rT re 3k 2  k 2  x 2  0, or x ≥ xa, meanwhile rpy   j pffiffi3 j , rey  rhy , ry  ry ; When but

p ry



kj2 1ln kj2 pffiffi 3

is impossible in practice, as a result, when x ≥ xa,

Therefore, when x = k,

rT eo ry



rT ho ry

is absolute.

rT e ry



rT h ry .

Mechanical Autofrettage Technology Based on Mises Yield Criterion

91

T Obviously, xa is also the abscissa at which rT e ¼ rh .     T T pffiffiffi 2 r r k 2 1ln k 2 pffiffiffi 2 3k  k 2  kj2 . 3k  k 2  kj2 rpy   j pffiffi3 j Setting rejy  rhjy results in pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 2 pffiffiffi k 2 1ln k 2 When 3k  k 2  kj2  0, or kj  3  1k = xa, meanwhile rpy   j pffiffi3 j pffiffiffi rT rT (that's of course in practice), rejy  rhjy ; When 3k 2  k 2  kj2  0, or kj ≥ xa, mean-

while rpy  

kj2 1ln kj2 pffiffi 3

(minus pressure),

practice, as a result, when kj ≥ xa, rT ej ry

rT ej ry





rT hj ry ,

rT hj ry .

but rpy  

kj2 1ln kj2 pffiffi 3

is impossible in

Therefore, when x = kj,

rT ej ry



rT hj ry

or

rT hj ry

 is conditional and depends on kj ≤ xa or kj ≥ xa. In fact, we can also make the following inference: When x < xa,

r0e ry

[

When x < xa,

rpe ry

[

rT e ry rT e ry

rT ej ry

[

r0

¼ rye þ rT h ry ;

rpe ry

and

when x =

rT h ry

r0h ry , rph ry ,

r0e ry

\ ryh .

rp

rpe ry

\ rhy .

r0e ry

¼ ryh , when x > xa,

when x = xa,

rpe ry

¼ rhy , when x > xa,

r0

¼ ryh þ

rT xa, rey

r0

when x = xa,

¼

rT h ry ;

rph ry .

r0

rp

Therefore, there must be that when x < xa,

when x > xa, rT rT rT

rT e ry

rT

\ rhy .

rT

The variation tendencies of rzy , rhy , rry and rey are the same as that based on the maximum shear stress theory. In order to summarize and validate the above-mentioned results, we present several figures for total stresses according to equations (3.26)–(3.29) and equations (3.35)–(3.38) as shown in figures 3.2–3.11. (1) Add operation stresses to figure 3.1 to illustrate the total stresses as shown in figure 3.2. When k = 4 and kj = 1.6,

p ry

¼ rpay ¼ p2ffiffi3 ln kj þ

k 2 kj2 pffiffi 3k 2

= 1.027688.

FIG. 3.2 – The total stresses when k = 4, kj = 1.6 and p = pa.

Autofrettage Technology and Its Applications in Pressured Apparatuses

92 xa =

pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k = 3.422399 > kj. When x < xa, rT

rT e ry

[

rT h ry ;

When x > xa,

rT e ry

rT

\ rhy ;

rT

At x = xa, rey ¼ rhy . Within the plastic region, curves 1, 2, and 3 are parallel to each other. T When p = pa, rT e ≡ σy within the whole plastic region, and re \ry within the elastic region. There is no x0T between 1 and k. T T Because p = pa, the three curves for rT z ; rh ; rr are parallel to each other within the plastic region. (2) When k = 4 and kj = 3.422399 = xa,

p ry

k 2 kj2 pffiffi 3k 2

¼ rpay ¼ p2ffiffi3 ln kj þ

= 1.575377.

FIG. 3.3 – The total stresses when k = 4, kj = xa = 3.422399 and p = pa.

xa =

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k = 3.422399 = kj. At x = xa = kj,

rT e ry

¼

rT h ry .

T When p = pa, rT e ≡ σy within the whole plastic region, and re < σy within the elastic region. There is no x0T between 1 and k. T T Because p = pa, the three curves for rT z ; rh ; rr are parallel to each other within the plastic region.

(3) When k = 4 and kj = 3.6 > xa,

p ry

¼ rpay ¼ p2ffiffi3 ln kj þ

k 2 kj2 pffiffi 3k 2

= 1.588792.

When kj > xa, the point at which σe = σθ is within the plastic region, represented by xa0 , and xa does not exist. When x < xa0 , rT e ry

¼

rT h ry .

At x = kj,

rT e ry

rT

rT e ry

[

rT h ry ;

When x > xa0 ,

\ rhy . For figure 3.4, xa0 = 3.46239 > xa.

rT e ry

rT

\ rhy ; At x = xa0 ,

Mechanical Autofrettage Technology Based on Mises Yield Criterion

93

FIG. 3.4 – The total stresses when k = 4, kj = 3.6 and p = pa. T When p = pa, rT e ≡ σy within the whole plastic region, and rh [ ry near kj. T There is no x0 between 1 and k. T T Because p = pa, the three curves for rT z ; rh ; rr are parallel to each other within the plastic region.

(4) When k = 4 and kj = 1.6,

p ry

= 0.9
kj. When x < xa,

At x = xa,

rT e ry

¼

rT h ry .

rT e ry

[

rT h ry ;

When x > xa,

rT e ry

rT

\ rhy ;

Autofrettage Technology and Its Applications in Pressured Apparatuses

94

Within the whole wall, curves 1, 2, and 3 are not parallel to each other because is not constant. rT e Since p < pa, rT e < σy within the whole wall. T There is no x0 between 1 and k. T T Because p ≠ pa, the three curves for rT z ; rh ; rr are not parallel to each other within the plastic region. p ry

(5) When k = 4 and kj = 1.6,

= 1.1 >

pa ry .

FIG. 3.6 – The total stresses when k = 4, kj = 1.6 and p > pa.

xa =

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k = 3.422399 > kj. When x < xa, rT

rT e ry

[

rT h ry ;

When x > xa,

rT e ry

rT

\ rhy ;

rT

At x = xa, rey ¼ rhy . Within the whole wall, curves 1, 2, and 3 are not parallel to each other because is not constant. rT e Since p [ pa ; rT e [ ry within the whole plastic region. There is no x0T between 1 and k. T T Because p ≠ pa, the three curves for rT z ; rh ; rr are not parallel to each other within the plastic region. (6) When k = 4 and kj = 3.6 > xa,

p ry

= 1.3
xa, xa0 is within the plastic region, and xa does not exist. When x < xa0 , rT e ry

[

rT h ry ;

When x > xa0 ,

rT e ry

rT

\ rhy ; At x = xa0 ,

xa0 = 3.463761 > xa. At x = kj,

rT h ry

rT e ry

¼

= 1.001982 and

rT h ry

rT e ry

= 0.955529. For figure 3.7,

= 0.958831.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

95

FIG. 3.7 – The total stresses when k = 4, kj = 3.6 and p < pa.

Within the whole wall, curves 1, 2, and 3 are not parallel to each other because rT e is not constant. Since p < pa, σe < σy is within the whole plastic region. There is no x0T between 1 and k. T T Because p ≠ pa, the three curves for rT z ; rh ; rr are not parallel to each other within the plastic region. (7) When k = 4 and kj = 3.6 > xa,

p ry

= 1.7 >

pa ry .

When kj > xa, xa0 is within the plastic region, and xa does not exist. When x < xa0 , rT e ry

[

rT h ry ;

When x > xa0 ,

rT e ry

rT

\ rhy ; At x = xa0 , rT h

rT e ry

¼

rT h ry

= 1.0171435. For figure 3.8,

rT e

xa0 = 3.461886 > xa. At x = kj, ry = 1.061571 and ry = 1.015853. Within the whole wall, curves 1, 2, and 3 are not parallel to each other because is not constant. rT e Since p > pa, rT e [ ry within the whole plastic region. T There is no r0 between 1 and k. T T Because p ≠ pa, the three curves for rT z ; rh ; rr are not parallel to each other within the plastic region. rT

In the above examples, since p > pri, rry increases monotonously with the increase of x in the plastic region. (8) When k = 4 and kj = 1.65, rpy = prriy = 0.5160901 < rpye = 0.541266 < rpay = 1.057356. pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi rT rT rT rT 3  1k = 3.422399 > kj. When x < xa, rey [ rhy ; When x > xa, rey \ rhy ; xa =

Autofrettage Technology and Its Applications in Pressured Apparatuses

96

FIG. 3.8 – The total stresses when k = 4, kj = 3.6 and p > pa.

At x = xa,

rT e ry

¼

rT h ry .

rT r ry

increases monotonously with the increase of x in the plastic region, but it is in a critical state, the intersection point of the three curves for total stresses is just on rT

rT

the vertical axis where x0T = 1. This is because p = pri. reiy = 0 and rry gets the minimum. The coordinates of the intersection point c are c (1,  rpy ) = c (1,  prriy ) = c (1, −0.5160901). Within the whole wall, curves 1, 2, and 3 are not parallel to each other because p ≠ pa, thus rT e is not constant.

FIG. 3.9 – The total stresses when k = 4, kj = 1.65 and p = pri.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

97

Since p < pa, rT e < σy is within the whole wall. T T Because p ≠ pa, the three curves for rT z ; rh ; and rr are not parallel to each other within the plastic region. (9) When k = 4 and < rpay = 1.057356.

kj = 1.65,

p ry

= 0.3
xa,

rT e ry

rT

\ rhy ;

rT

rT

At x = xa, rey ¼ rhy . Since p < pri, the three curves for total stresses have an intersection point in the plastic region, and the coordinates of the intersection point a are a (x0T ,  rpy ) = a rT

rT

(1.182891, −0.3). When x ¼ x0T ¼ 1:182891, rey = 0 and rry gets the minimum (−0.3). Within the whole wall, curves 1, 2, and 3 are not parallel to each other because p ≠ pa, thus rT e is not constant. Since p < pa, σe < σy within the whole wall. Because p ≠ pa, the three curves for T T rT z ; rh ; rr are not parallel to each other within the plastic region. (10) When k = 4 and kj = 2, rpye = 0.541266 < rpy = 0.6 < prriy = 0.692124 < rpay = 1.23339. pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi rT rT rT rT 3  1k = 3.422399 > kj. When x < xa, rey [ rhy ; When x > xa, rey \ rhy ; xa = At x = xa,

rT e ry

¼

rT h ry .

Autofrettage Technology and Its Applications in Pressured Apparatuses

98

FIG. 3.11 – The total stresses when k = 4, kj = 2, and pri.

Since p < pri, the three curves for total stresses have an intersection point in the plastic region, and the coordinates of the intersection point b are b (x0T ,  rpy ) = b rT

rT

(1.182891, −0.6). When x = x0T = 1.081759, rey = 0 and rry gets the minimum (−0.6). Within the whole wall, curves 1, 2, and 3 are not parallel to each other because p ≠ pa, thus rT e is not constant. Since p < pa, σe < σy within the whole wall. From the above we see that some stresses exceed the strength limit σy, therefore, it is necessary to seek more reasonable control methods and conditions.

3.1.2

The Critical Radius Ratio

For equation (3.13), setting kj = k leads to r0ei k 2 ln k 2 ¼1 2 ry k 1 For equation (3.27), letting

r0ei ry

ð3:39Þ

 1 k 2 ln k 1 k2  1

ð3:40Þ

The solution of equation (3.40) is k  kc ¼ 2:218 457 489 916 7. . . Therefore, when k ≤ kc, irrespective of kj, even if kj = k, ry  r0ei  0.

ð3:41Þ

Mechanical Autofrettage Technology Based on Mises Yield Criterion

99

The results are the same as those based on the maximum shear stress theory (Tresca criterion), see equations (2.32) and (2.33). This is because the equivalent residual stress based on the maximum shear stress theory is the same as that based on the maximum distortion strain energy theory, although the components of residual stresses based on the two strength theories are different.

3.1.3

The Optimum Plastic Depth kj* (kj is Written as kj*)

For equation (3.13), setting plastic depth kj* 2 2 k 2 lnkj  k 2  kj þ20

or

r0ei ry

  1 and writing kj as kj* obtain the optimum

k2 

kj2  2 ln kj2  1

ðe0:5  kj  kc and k  kc Þ ð3:42Þ

Equation (3.42) is just equation (2.46). Thus, it can be seen that although the components of residual stresses based on the maximum distortion strain energy theory (Mises’s yield criterion) are different from those based on the maximum shear stress theory (Tresca yield criterion), the optimum plastic depth kj* is the same. pffiffiffi From equation (3.42), when k → ∞, kj ¼ e; letting kj* = k just results in equation (3.40). When k ≥ kc, if kj ≤ kj*, i.e., the plastic depth determined by equation (3.42), ry  r0ei  0. When k ≤ kc, even if kj = k, there must be that ry  r0ei  0. The optimum plastic depth kj* is illustrated in figure 3.12, which is the same as figure 2.9. From figure 3.12 and equation (3.42), we know that when k ≥ kc, as k increases, kj* decreases, this means that for r0ei   ry , thicker pressure vessels need shallower plastic depth.

FIG. 3.12 – The optimum plastic depth kj*.

100

Autofrettage Technology and Its Applications in Pressured Apparatuses

In figure 3.12, point a is the intersection of straight line oa (kj* = k) and curve ab ðk 2 ln kj2  k 2  kj2 þ 2  0Þ, setting kj* = k in the equation ðk 2 ln kj2  k 2  k kj2 þ 2  0Þ just obtains kk 2ln 1  1. The coordinates of point a are a (kc, kc); point b is at pffiffiffi infinity; thus, the coordinates of point b are b (∞, e). 2 k In fact, kk 2ln 1 ¼ 1 is equivalent to 2pe = py, i.e., 2

2

k2  1 k 2  1 2 ln k pffiffiffi ¼ pffiffiffi ; ¼ ln k or 2 2k 2 3k 2 3

where pe is the maximum elastic load-bearing capability or initial yield pressure, and py is the entire yield pressure. k 2 1

k ffiffi1 just results in equaInterestingly enough, letting prriy ¼ p2ffiffi3 ln kj  pj ffiffi3k 2 ¼ rpye ¼ p 3k 2 pri pe pri pe tion (3.42). When kj < kj*, ry \ ry ; when kj > kj*, ry [ ry . Therefore, when kj ≤ kj*,

if rpy [ rpye , there must be rpy [ of x in the plastic region.

3.1.4

pri ry ,

thus,

rT r ry

2

increases monotonously with the increase

The Results When kj ¼ kj  or k 2 lnkj2  k 2  kj2 þ 2 ¼ 0

1. The components of residual stresses Within the plastic region (1 ≤ x ≤ kj*) r0z 2 1 ¼ pffiffiffi ln x  pffiffiffi ry 3 3

ð3:43Þ

r0r 2  ¼ pffiffiffi ln x þ x 2  1 ry 3

ð3:44Þ

r0h 2  ¼ pffiffiffi ln x  x 2 ry 3

ð3:45Þ

r0e 2 ¼1 2 x ry

ð3:46Þ

Within the elastic region (kj* ≤ x ≤ k) 2 2 r0z kj ¼ pffiffiffi ry 3k 2

ð3:47Þ

Mechanical Autofrettage Technology Based on Mises Yield Criterion

101

r0r ¼ ry

  k 2 r0z 1 2 x ry

ð3:48Þ

r0h ¼ ry

  k 2 r0z 1þ 2 x ry

ð3:49Þ

2 2 r0e kj ¼ [0 x2 ry

ð3:50Þ

The complex formulas, or equations (3.2)–(3.9), are significantly simplified. On the inner surface, equations (3.43)–(3.46) become: r0zi 1 ¼  pffiffiffi ry 3

ð3:51Þ

r0ri ¼ 0

ð3:52Þ

r0hi 2 ¼  pffiffiffi \  1 ry 3

ð3:53Þ

r0ei ¼ 1 ry

ð3:54Þ

Therefore, if plastic depth kj = kj*, the residual stresses at the inner surface are always compressive and constant, which have nothing to do with k and kj, or have nothing to do with the thickness and overstrain of a pressure vessel. The equivalent residual stress at inner surface just reaches −σy. However, jr0hi j exceeds yield strength pffiffi σy. This is inevitable, because re ¼ 23 ðrh  rr Þ, at the inner surface, r0ri ¼ 0, then pffiffi r0 r0 r0ei ¼ 23 r0hi , when reiy ¼ 1, there must be rhiy ¼  p2ffiffi3 \  1. We will present the research results when r0h is controlled in the next chapter (chapter 4). On the elastic-plastic juncture, equations (3.43)–(3.46) become: r0zj kj2  2 2 1 ¼ pffiffiffi ln kj  pffiffiffi ¼ pffiffiffi ry 3 3 3k 2

ð3:55Þ

 r0rj 2  ¼ pffiffiffi ln kj þ kj2  1 ry 3

ð3:56Þ

 2  2 ¼ pffiffiffi ln kj  kj ry 3

ð3:57Þ

r0hj

102

Autofrettage Technology and Its Applications in Pressured Apparatuses r0ej 2 ¼1 2 ry kj

ð3:58Þ

2. The optimum load-bearing capability When the load exerted on the pressure vessel is just autofrettage pressure pa, r0ei  ry within the whole plastic region irrespective of k and kj. When k ≥ kc and 2 kj = kj*, autofrettage pressure pa (see equation (3.1)) becomes p2ffiffi3 k k1 2 . Then, when k ≥ kc and kj = kj*, the optimum load-bearing capability is p pa pe 2 k2  1 ¼ ¼ 2 ¼ pffiffiffi ry ry ry 3 k2

p ry

ðk  kc Þ

ð3:59Þ

2 2 When k → ∞, rpy ¼ rpay ¼ 2 rpye ¼ p2ffiffi3 k k1 → p2ffiffi3, and rpye ¼ p1ffiffi3 k k1 → p1ffiffi3. 2 2 When k ≤ kc, kj = kj* = k, autofrettage pressure pa or equation (3.1) becomes ¼ rpay ¼ p2ffiffi3 ln k. Then, the optimum load-bearing capability is

p pa 2 py ¼ ¼ pffiffiffi ln k ¼ ry ry r 3 y

ðk  kc Þ

ð3:60Þ

The optimum load-bearing capability is illustrated as curve oad in figure 3.13, where the coordinates of some key points are marked.

FIG. 3.13 – The optimum load-bearing capacity of an autofrettaged pressure vessel.

p

In figure 3.13, point a is the crossover point of curve oa (rpy ¼ p2ffiffi3 ln k ¼ ryy ) and

2 2 p2ffiffi ln k = p2ffiffi k 1 results in curve ad (rpy ¼ 2 rpye ¼ p2ffiffi3 k k1 2 ), then, letting 3 3 k2

k 2 ln k k 2 1

¼ 1, as is

Mechanical Autofrettage Technology Based on Mises Yield Criterion

103

just equation (3.40). Therefore, the abscissa of point a is kc, and when k = kc, 2 2 = p2ffiffi3 k k1 = 0.920079…. When k < kc, p2ffiffi3 ln k < p2ffiffi3 k k1 When k > kc, 2 2 ;

p2ffiffi ln k 3 p2ffiffi ln k 3

2 > p2ffiffi3 k k1 2 .

Points b and d are at infinity; thus, the coordinates of points b and d are b (∞, p1ffiffi3) and d (∞, p2ffiffi), respectively. 3

3. The location (defined as x0) of the intersection of the three residual stress curves or the abscissa at which r0e ¼ 0, or x0 When plastic depth kj=kj*, equation (3.22) becomes pffiffiffi x0  2

ð3:61Þ

pffiffiffi In fact, for equation (3.46), letting r0e ¼ 0 just results in x = x0 ≡ 2. x0 is a constant and is not related to k and kj. The value of residual stresses when x = x0 or the minimum radial residual stress is y0d ¼

r0r r0z r0h r0rmin r0rx0 1  ln 2 ¼ ¼ ¼ ¼   pffiffiffi ¼ constant ry ry ry ry ry 3

ð3:62Þ

y0d is a constant and not related to k and kj. y0d ¼ p2ffiffi3 y0s . That is to say, when plastic depth kj = kj*, for any k, the three curves for the residual stress collect at the same point within the plastic region and the intersection is:   pffiffiffi 1  ln 2 2;  pffiffiffi ¼ ð1:414214. . .; 0:17716. . .Þ 3 4. The location (x, defined as x1) where r0z ¼ 0, or x1 When plastic depth kj = kj*, equation (3.24) becomes pffiffiffi x1  e

ð3:63Þ

pffiffiffi In fact, for equation (3.43), letting r0z ¼ 0 just results in x = x1 ≡ e. x1 is a constant and is not related to k and kj. When k = 4, from equation (3.42), kj* = 1.694172, the residual stresses are illustrated in figure 3.14 to summarize and validate the above-mentioned results. The meaning of each curve and main parameters have been marked in the figure. 5. The components of total stresses when kj ≥ kc, plastic depth kj = kj*, and load p = pa = 2pe.

104

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 3.14 – The residual stresses when kj = kj*. Within the plastic region (1 ≤ x ≤ kj*) rT ln x 2 1 2 z ¼ pffiffiffi  pffiffiffi þ pffiffiffi ry 3 3k 2 3

ð3:64Þ

rT ln x 2 2 2 r ¼ pffiffiffi  pffiffiffi þ pffiffiffi ry 3 3k 2 3

ð3:65Þ

rT ln x 2 2 h ¼ pffiffiffi þ pffiffiffi ry 3k 2 3

ð3:66Þ

rT e 1 ry

ð3:67Þ

T T Clearly, rT z ; rr ; and rh increase as x increases.

Setting

rT ri ry

2 ¼  p2ffiffi3 þ pffiffi32k 2 ¼  p2ffiffi3 k k1 ¼  rpy   1 results in 2

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi ffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi k  2ð2 þ 3Þ ¼ ð1 þ 3Þ2 ¼ 1 þ 3 ¼ 2:732051 ¼ kr When k ≤ kr = 2.732051, jrT ri j can not reach σy. 2 p When k = kr = 2.732051, ry ¼ 2 rpye ¼ p2ffiffi3 k k1 ¼ 1. 2

ð3:68Þ

Mechanical Autofrettage Technology Based on Mises Yield Criterion

Setting

rT hi ry

¼ pffiffi32k 2  1 results in sffiffiffiffiffiffiffi 2 k  pffiffiffi ¼ 1:07457 ¼ kh 3

105

ð3:69Þ

Therefore, usually rT h can not reach σy. T For equations (3.66) and (3.67), setting rT e  rh obtains pffiffiffi  2 pffiffiffi 2  pffiffiffi  3k  2 3 1 k 1 2 3  x  exp  ¼ exp ¼ exp 2 2k 2 k2 k2 2 ð3:70Þ pffiffiffi pffiffiffi 3 p 2 3 ¼ exp  ¼ xz 2 ry 2 pffiffi  When k = 1, xz ¼ exp 23  1 = 0.874612; When k = 2.6, xz ¼ pffiffi  pffiffi 1 exp 23  2:62:6 = 2.050526; When k = ∞, xz ¼ exp 23 = 2.377443. Within the elastic region (kj* ≤ x ≤ k) kj2 rT z ¼ pffiffiffi ¼ constant ry 3k 2

ð3:71Þ

  kj2 1 rT 1 r ¼  pffiffiffi 2  2 k ry 3 x

ð3:72Þ

  kj2 1 rT 1 h ¼ pffiffiffi 2 þ 2 k ry 3 x

ð3:73Þ

kj2 rT e ¼ 2 ry x

ð3:74Þ

T T Clearly, rT e and rh decrease as x increases, and rr increases as x increases. p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T 2 pffiffiffi r k 3  1k = xa. This shows again that if Letting rhjy ¼ p1ffiffi3 þ pffiffi3jk 2  1 leads to kj 

T kj ≤ xa, rT hj  rej . 2 þ20 In order for r0ei to be greater than σy, we have gotten k 2 ln kj2  k 2  kj 0.5 (e ≤ kj* ≤ kc and k ≥ kc). If kj > kj*, or the actual value of plastic depth kj is larger than the value calculated by k 2 ln kj2  k 2  kj2 þ 2 ¼ 0, there will be pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k into k 2 ln kj2  k 2  k 2 ln kj2  k 2  kj2 þ 2 [ 0. Then, substituting kj ¼ kj2 þ 2 obtains pffiffiffi pffiffiffi ð3:75Þ k 2 ln½ð 3  1Þk 2   3k 2 þ 2

Equation (3.75) is plotted in figure 3.15.

106

Autofrettage Technology and Its Applications in Pressured Apparatuses

pffiffiffi

pffiffiffi

FIG. 3.15 – The graph of k 2 ln½ð 3  1Þk 2   3k 2 þ 2. pffiffiffi pffiffiffi Setting k 2 ln½ð 3  1Þk 2   3k 2 þ 2  0 (see figure 3.27) results in k  2:300122. . . ¼ kf ð3:76Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k = kj, thus rT That is to say, when k > kf, kj* < hj \ry ; when k < kf, pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi T 3  1k = kj, thus rhj [ ry . For example, when k = 2.25, kj* = 2.046308, kj* > pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi rT 3  1k = 1.925099 < kj* = 2.046308, rhjy = 1.054898 > 1; when k = 2.5, kj* = pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi rT 1.844363, 3  1k = 2.138999 > kj* = 1.844363, rhjy = 0.891584 < 1. pffiffiffi pffiffiffi When k 2 ln½ð 3  1Þk 2   3k 2 þ 2 ¼ 0, or k = kf = 2.300122…, equation (3.70) becomes xz = xa = kj* = 1.967984. qffiffiffiffiffi rT Letting rhi ¼ pffiffi2 2  1 leads to k  p2ffiffi = 1.07457… = kθ. Therefore, usually rT hi y

3k

3

can not reach σy. To confirm the above conclusions, we present several total stress figures as follows. The total stresses for k = 4 > kr, kj = kj* and p = pa = 2pe are illustrated in figure 3.16. The total stresses for k = 2.25 < kf, kj = kj* and p = pa = 2pe are illustrated in figure 3.17. The total stresses for k = kr, kj = kj* and p = pa = 2pe are illustrated in figure 3.18. The total stresses for k = kf, kj = kj* and p = pa = 2pe are illustrated in figure 3.19. The meaning of each curve and main parameters have been marked in the figures.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

107

When k = kr = 2.732051, from equation (3.42) or equation (2.34), kj* = 1.785136; When k = kf = 2.300122, from (3.42) or equation (2.34), kj* = 1.967984.

FIG. 3.16 – The total stresses with k = 4, kj = kj* and p = pa = 2pe. pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k = 0.8556 k = 3.422399 > kj*. ThereIn figures 3.14 and 3.16, xa ¼ T fore, xa exists and xz does not exist. In other words, if kj < xa, there is rT h = r e = σy T in the elastic region, and when rT h = re = σy, x = xa, see figures 3.14 and 3.16; if T T T kj > xa, there is rh = re = σy in the plastic region, and when rT h = re = σy, x = xz, T T see figure 3.17; if kj = xa, there is rhj ¼ rej ¼ ry at the elastic-plastic juncture where x = xa = xz, see figure 3.19. Because k = 4 > kr, p > σy, and rT ri < −σy. Because pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi T 3  1k, thus rhi < σy. k > kf, kj*
xa = 1.925099, so, xz exists and xa does not exist. Because k = 2.25 < kr, p < σy and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k, thus rT rT ri > −σy. Because k < kf, kj* > hj > σy.

FIG. 3.18 – The total stresses with k = kr, kj = kj* and p = pa = 2pe. In figure 3.18, xa = 2.337542; kj* = 1.785136 < xa, so, xa exists and xz does not pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k, = −σ . Because k > k , k < exist. Because k = kr, p = σy and rT y f j* ri rT < σ . y ri

FIG. 3.19 – The total stresses with k = kf, kj = kj* and p = pa = 2pe. T In figure 3.19, since k = kf, at kj = kj*, rT hj is just rej , and they are just σy, and xz = xa = kj*.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

109

When based on the maximum distortion strain energy theory (Mises’s yield r0 criterion), sometimes rhi ¼  p2ffiffi \  1 and rT h > σy, we will solve the problem in the y

3

next chapter (chapter 4).

3.2

Mechanical Autofrettage Technology Under Entire Yield State

3.2.1

The Residual Stresses

As mentioned above, when k ≤ kc, the plastic depth kj can be k. When kj = k, or overstrain ε = 100%, there is only a plastic zone and no elastic zone within the whole wall of a pressure vessel, then, the components of residual stresses, or equations (3.2)–(3.5) become equations (3.77)–(3.80). Within the whole wall (1 ≤ x ≤ k)   r0z 1 x2 ln k 2 ¼ pffiffiffi 1 þ ln 2  2 ð3:77Þ ry k k 1 3   r0r 1 x2 ln k 2 k 2 ln k 2 þ ¼ pffiffiffi ln 2  2 ry k k  1 ðk 2  1Þx 2 3

ð3:78Þ

  r0h 1 x2 ln k 2 k 2 ln k 2  ¼ pffiffiffi 2 þ ln 2  2 ry k k  1 ðk 2  1Þx 2 3

ð3:79Þ

r0e k 2 ln k 2 ¼1 2 ry ðk  1Þx 2

ð3:80Þ

Clearly, r0z , r0e and r0h increase as x increases. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 2 2 2 2 r x0 ¼ ri ¼ becomes x0 ¼ kk 2ln1k . When x < x0 ¼ kk 2ln1k , r0r k 2 1 qffiffiffiffiffiffiffiffiffiffiffi 2 2 decreases as x increases; When x > x0 ¼ kk 2ln1k , r0r increases as x increases. At the inner surface, or x = 1, r0ei ry

r0ei ry

¼ 1  kk 2ln1k , which is equation (3.27). 2

2

k 0 ¼ 1, kk 2ln 1 ¼ 1, therefore, under entire yield, or kj = k, for rei   ry ,  0 0 2 2 rhi k   p2ffiffi k ln k there must be k ≤ kc; and when kk 2ln 1 ¼ 1, ry ¼  3 k 2 1 \  1, or when rei ¼ ry ,  0  r  [ ry . We will solve the problem in the next chapter (chapter 4).

When

hi

Clearly,

2

r0eo ry

[ 0.

Autofrettage Technology and Its Applications in Pressured Apparatuses

110

Figures 3.20 and 3.21 show the distributions of the residual stresses for k = kc and k = 2, respectively. The meaning of each curve and main parameters have been marked in the figures.

FIG. 3.20 – The distributions of the residual stresses for k = kc and kj = k.

FIG. 3.21 – The distributions of the residual stresses for k = 2 and kj = k. qffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 2 2 2 k Since x0 ¼ kk 2ln1k when kj = k, if k = kc, kk 2ln 2 and 1 ¼ 1, then, x0 ¼ r0zx0 ry

r0hx0 ry

r0rx0 ry

pffiffi 2. Therefore, the coordinate of the intersections of the ¼  1ln 3 pffiffiffi  pffiffi 2 . The ordinate of three curves for the residual stresses in figure 3.20 is a 2;  1ln 3

y0d ¼

¼

¼

the intersections of the three curves for the residual stresses in figure 3.21 (k ≠ kc) is 2 r0 r0 k2 ffiffi 2 k 2 = −0.13514…. The abscissa of the intersec¼ rhx0 ¼ rrx0 ¼ p1ffiffi ln kln2 1 þ kp1ln

r0zx0 ry

y

y

3

3ðk 1Þ

tions of the three curves for the residual stresses in figure 3.21 (k ≠ kc) is qffiffiffiffiffiffiffiffiffiffiffi 2 2 x0 ¼ kk 2ln1k = 1.359556….

Mechanical Autofrettage Technology Based on Mises Yield Criterion

3.2.2

111

The Total Stresses

The total stresses within the whole wall (1 ≤ x ≤ k) when the plastic depth kj = k.   rT 1 x2 ln k 2 1 p z p ffiffi ffi ¼ ð3:81Þ 1 þ ln 2  2 þ 2 k ry k k  1  1 ry 3   rT 1 x2 ln k 2 k 2 ln k 2 k2  x2 p r þ 2 ¼ pffiffiffi ln 2  2  2 2 2 ry k k  1 ðk  1Þx x ðk  1Þ ry 3

ð3:82Þ

  rT 1 x2 ln k 2 k 2 ln k 2 k2 þ x2 p h  2 ¼ pffiffiffi 2 þ ln 2  2 þ 2 2 2 ry k k  1 ðk  1Þx x ðk  1Þ ry 3

ð3:83Þ

pffiffiffi 2  pffiffiffi 2  3k 3k p k 2 ln k 2 1 rT k 2 ln k 2 p e ¼1 2 þ ¼ 1 þ  ry ðk  1Þx 2 x 2 ðk 2  1Þ ry k 2  1 ry k 2  1 x 2 From equation (3.84), when rT e

p ry

p  p2ffiffi3 ln k ¼ ryy , the larger x is, the larger

ð3:84Þ rT e ry

is.

p

Setting ry  1 just obtains rpy  p2ffiffi3 ln k ¼ ryy . That is to say, when kj = k and k ≤ kc, the maximum load is entire yield pressure. When a pressure vessel is subjected to p entire yield pressure rpy ¼ p2ffiffi3 ln k ¼ ryy , Lamé equations become rpz ln k 2 ¼ pffiffiffi ry 3ðk 2  1Þ

ð3:85Þ

rpr k2  x2 ln k 2 ¼  pffiffiffi ry 3x 2 ðk 2  1Þ

ð3:86Þ

rph k2 þ x2 ln k 2 ¼ pffiffiffi ry 3x 2 ðk 2  1Þ

ð3:87Þ

rpe k 2 ln k 2 ¼ 2 ry ðk  1Þx

ð3:88Þ

p

rp

k Clearly, when kj = k and rpy ¼ p2ffiffi3 ln k ¼ ryy , reiy ¼ kðk 2ln1Þ , even if k ≤ kc, rpei may exceed σy if autofrettage process is not performed. When plastic depth kj = k and p = py, the total stresses are as follows   rT 2 1 z þ ln x  ln k ð3:89Þ ¼ pffiffiffi ry 3 2 2

2

112

Autofrettage Technology and Its Applications in Pressured Apparatuses rT 2 r ¼ pffiffiffi ðln x  ln k Þ ry 3

ð3:90Þ

rT 2 h ¼ pffiffiffi ð1 þ ln x  ln k Þ ry 3

ð3:91Þ

rT e 1 ry

ð3:92Þ

rT h ry

rT

¼ p2ffiffi3 ð1 þ ln x  ln k Þ ¼ rey ¼ 1 leads to pffiffiffi pffiffiffi   pffiffiffi 3 3 p 2 3 x ¼ exp ln k þ  1 ¼ xb ¼ exp  ¼ xz 2 2 ry 2

Setting

rT

rT

rT

ð3:93Þ

rT

When x < xb, rhy \ rey ¼ 1; When x > xb, rhy [ rey ¼ 1. When p = py, equation (3.93) is just equation (1.25), or xb = xz.  pffiffi pffiffi xb ¼ exp ln k þ 23  1 = exp(lnk)exp 23  1 =  k pffi3 < k. This shows exp 1 2

rT h ry

rT e ry ,

that the curve is always intersected with the curve and the intersection point is between 1 and k. When k = 1, x = 0.866025; When k = 1, xb = kc, xb = b  pffiffi 3 2.084483 < kc. xb ¼ exp ln k þ 2  1 is plotted in figure 3.22. As x increases,

rT rT z r ry , ry ,

and

rT h ry

increase.

FIG. 3.22 – The image between k and xb.

Mechanical Autofrettage Technology Based on Mises Yield Criterion At the inner surface,

113

  rT 2 1 zi  ln k ¼ pffiffiffi ry 3 2

ð3:94Þ

rT 2 p ri ¼  pffiffiffi ln k ¼  ry ry 3

ð3:95Þ

rT 2 hi ¼ pffiffiffi ð1  ln k Þ ry 3

ð3:96Þ

rT e 1 ð3:97Þ ry  pffiffi rT rT For rziy   1, there must be k  exp 1 þ2 3 = 3.91974; for rhiy   1, there must  pffiffi rT ri be k  exp 1 þ 23 = 6.462559; there must be for ry   1, pffiffi T r k  exp 23 = 2.377443. The premise of kj = k is k ≤ kc, therefore, rziy   1, rT hi ry

rT

  1 and rriy   1 are natural. At the outer surface, rT 1 zo ¼ pffiffiffi ry 3

ð3:98Þ

rT ro ¼0 ry

ð3:99Þ

rT 2 ho ¼ pffiffiffi [ 1 ry 3

ð3:100Þ

rT e 1 ry

ð3:101Þ

From equation (3.83), at the outer surface,   rT 1 2 ln k 2 2 p ho ¼ pffiffiffi 2  2 þ 2 k  1 ry ry k 1 3 Setting

rT ho ry

ð3:102Þ

 1 obtains pffiffiffi p p 2 2 3 2 py y  pffiffiffi ln k  pffiffiffi k  1 ¼ \ ry ry ry 3 2 3

ð3:103Þ

Autofrettage Technology and Its Applications in Pressured Apparatuses

114

Equation (3.103) is the maximum load for rT ho ≤ σy.

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ffi From equation (3.103), it is known that when k = kr = 2ð2 þ 3Þ ¼ pffiffiffi p p 1 þ 3 > kc, ryy reaches the maximum, ryy = 0.660535…, and when k = kr,

py ry

= 1.160535. When k = kc, p y ry

p y ry

= 0.616746… and

is plotted in figure 3.23 along with

py ry

and

pe ry .

py ry

= 0.920079…. >

p y ry .

It is seen that p y [ pe .

FIG. 3.23 – The allowable load for rTho ≤ σy. When plastic depth kj = k, p = py, and k ≤ kc, the distributions of the total stresses are plotted in figures 3.24 and 3.25 for k = kc and k = 2, respectively. The meaning of each curve and coordinates of key points as well as the main parameters are marked in the figures.

FIG. 3.24 – The distributions of the total stresses when kj = k, p = py and k = kc.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

115

T In figures 3.24 and 3.25, after x ≥ xb, rT h ≥ re = 1.

FIG. 3.25 – The distributions of the total stresses when kj = k, p = py and k = 2. When kj = k, p = p−y = 0.616746σy and k = kc, the distributions of the total stresses are plotted in figure 3.26.

FIG. 3.26 – The distributions of the total stresses when kj = k, p = py− and k = kc. − T In figure 3.26, rT ho = σy, re < σy because p = p y < py. When kj = k, p = p−y = 0.568327σy and k = 2, the distributions of the total stresses are plotted in figure 3.27. T In figure 3.27, rT ho ¼ ry , re \ry . When kj = k, p = 0.65σy > p−y and k = kc, the distributions of the total stresses are plotted in figure 3.28. T In figure 3.27, rT ho = 1.01696σy, re \ry . − When kj = k, p = 0.6σy < py and k = kc, the distributions of the total stresses are plotted in figure 3.29. T In figure 3.29, rT ho = 0.991459σy, re \ry .

116

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 3.27 – The distributions of the total stresses when kj = k, p = py− and k = 2.

FIG. 3.28 – The distributions of the total stresses when kj = k, p > py− and k = kc.

Whether it is necessary that rT ho \ry in engineering practice can be determined by the engineers according to the results of this research. All in all, the premise of kj = k is k ≤ kc. In addition, when kj = k, we have the following results. (1) Autofrettage pressure pa 2 ¼ pffiffiffi ln k ry 3 This is just the maximum load under the condition kj = k;

Mechanical Autofrettage Technology Based on Mises Yield Criterion

117

FIG. 3.29 – The distributions of the total stresses when kj = k, p < py− and k = kc.

(2) The abscissa of the intersection of the three residual stress curves or the abscissa at which r0e ¼ 0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 ln k 2 \k x0 ¼ k2  1 When k = kc, x0 ¼

pffiffiffi 2.

(3) The value of residual stresses when x = x0 or the minimum stresses are r0zx0 r0hx0 r0rx0 1 ln k 2 k 2  1  ln k 2 þ pffiffiffi ¼ ¼ ¼ pffiffiffi ln 2 ry ry ry 3 k 1 3ðk 2  1Þ (4) The abscissa at which r0z ¼ 0, or x1 is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x1 ¼

3.3

e

1k 2 þ k 2 ln k 2 k 2 1

Mechanical Autofrettage Technology with Radius of Elastic-Plastic Juncture Being Arithmetic Mean Radius of Inside Radius and Outside Radius

The overstrain is defined as e¼

rj  ri ro  ri

ð3:104Þ

118

Autofrettage Technology and Its Applications in Pressured Apparatuses

When the radius of the plastic region is an arithmetic mean value of the internal and external radius of the cylindrical pressure vessel, or rj ¼

ri þ ro 2

or

kj ¼

k þ1 2

ð3:105Þ

the overstrain ε = 50%. Figure 3.12 is the relationship between kj and k, so we can add kj ¼ k þ2 1 into this figure as straight line ofb, as is shown in figure 3.30. In figure 3.30, (1) curve afgd is for k2lnkj2−k2−kj2+2 = 0, point a is an intersection of curve afgd, straight line oac (kj = k) and curve oan; point f is an intersection of curve afgd and straight line ofb; point g is an intersection of curve afgd and curve ogh; point pffiffiffi d is at infinity, its coordinates are d (∞, e). (2) straight line oac is for kj = k (entirely yielded), point c is at infinity, and its coordinates are c (∞, ∞). (3) straight line ofb is for kj ¼ k þ2 1, point b is at infinity, its coordinates are b (∞, ∞). pffiffiffi (4) curve ogh is for kj ¼ k (see below), point h is at infinity, and its coordinates are h (∞, ∞).   (5) curve oan is for kj ¼ exp

k 2 1 k2

(see below), point n is at infinity, and its coor-

dinates are n (∞, e).

2  (see below), point m is at infinity, and its (6) curve om is for kj ¼ exp k2k1 2 pffiffiffi coordinates are m (∞, e).

FIG. 3.30 – The relationship between kj and k.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

119

Substituting equation (3.105) into equation (3.2)–(3.5) obtains the residual stresses within the plastic region " r0z 1 ð k þ 1Þ 2 4x 2 ¼ pffiffiffi þ ln  2 ry 4k 3 ð k þ 1Þ 2

! # ð k þ 1Þ 2 ð k þ 1Þ 2 1 1 þ ln k2  1 4k 2 4

" r0r 1 ðk þ 1Þ2 4x 2 ¼ pffiffiffi  1 þ ln  2 ry 4k 3 ð k þ 1Þ 2

ð3:106Þ

!  # ð k þ 1Þ 2 ð k þ 1Þ 2 1 k2 1 2 1 þ ln k2  1 4k 2 4 x ð3:107Þ

" r0h 1 ðk þ 1Þ2 4x 2 ¼ pffiffiffi þ 1 þ ln  2 ry 4k 3 ðk þ 1Þ2

!  # ð k þ 1Þ 2 ð k þ 1Þ 2 1 k2 1þ 2 1 þ ln k2  1 4k 2 4 x ð3:108Þ

k2  r0e ¼1 ry

ðk þ 1Þ2 þ k2 4 k2  1

2

ln ðk þ4 1Þ 1 x2

ð3:109Þ

Substituting equation (3.105) into equations (3.6)–(3.9) obtains the residual stresses within the elastic region r0z ¼ ry

ðk þ 1Þ2 4



r0e ¼ ry

ð3:110Þ

 k 2 r0z \0 x 2 ry

ð3:111Þ

  k 2 r0z 1þ 2 [0 x ry

ð3:112Þ

r0r ¼ ry r0h ¼ ry

2

 1  ln ðk þ4 1Þ pffiffiffi [0 3ðk 2  1Þ 1

ðk þ 1Þ2 4

2

 1  ln ðk þ4 1Þ k 2 [0 k2  1 x2

ð3:113Þ

When k = 2, kj = 1.5. The residual stresses within the whole wall are plotted in figure 3.31, where the meaning of each curve and the main parameters are marked.

120

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 3.31 – The residual stresses within the whole wall when k = 2, kj = 1.5. When k = 3, kj = 2. The residual stresses within the whole wall are plotted in figure 3.32, where the meaning of each curve and the main parameters are marked.

FIG. 3.32 – The residual stresses within the whole wall when k = 3, kj = 2. Based on the maximum shear stress theory, letting r0ei   ry obtain 5k 2 þ 2k  7  4k 2 ln

ðk þ 1Þ2 ¼ c0 4

ð3:114Þ

Equation (3.114) is just equation (2.90), therefore, when k ≤ ka = 2.618253,  ei  ry . ka is just the abscissa of the intersection point (f) of curves ob and ad in figure 3.30.

 0 r

Mechanical Autofrettage Technology Based on Mises Yield Criterion

121

When k = ka, kj = 1.809126…, the distribution of residual stresses within the whole wall is plotted in figure 3.33. The meaning of each curve and the main parameters are marked in the figure.

FIG. 3.33 – The residual stresses within the whole wall when k = ka. k

Converting the longitudinal coordinates of figure 3.30 into kj results in figure 3.34 so as to analyze the ratio of kj to k. When the plastic depth kj = k, pffiffiffi k þ1 kj ¼ k þ2 1, kj ¼ k 2k ; when kj ¼ k , kj ¼ p1ffiffik .

FIG. 3.34 – The relationship between

kj k

and k.

kj k

¼ 1; when

Autofrettage Technology and Its Applications in Pressured Apparatuses

122

pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1 k = x a, As mentioned above, when kj  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffi ffi 3  1k results in Then, letting kj ¼ k þ2 1 

rT ej ry



rT hj ry ;

when kj ≥ xa,

1 k  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1:406076 ¼ kd pffiffiffi 311

rT ej ry



rT hj ry .

ð3:115Þ

kd is marked in figure 3.34. That is to say, when kj ¼ k þ2 1, if k ≥ kd, pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi rT rT 3  1k, accordingly rejy  rhjy . kj  The equivalent stress of the total stresses k2  rT r0 rp e ¼ e þ e ¼1 ry ry ry Setting

rT e ry

ðk þ 1Þ2 þ k2 4 2 k 1

2 pffiffiffi 2 ln ðk þ4 1Þ 1 3k p þ x 2 x 2 ðk 2  1Þ ry

ð3:116Þ

 1 gives p 1 ðk þ 1Þ2 2 k þ1  pffiffiffi  pffiffiffi þ pffiffiffi ln 2 ry 2 3 4 3k 3

ð3:117Þ

Accordingly, the maximum load-bearing capacity of a pressure vessel is p1ffiffi 3



2

ðk þ 1Þ pffiffi 4 3k 2

p ry

¼

þ p2ffiffi3 ln k þ2 1 when kj ¼ k þ2 1.

Substituting kj ¼ k þ2 1 into equation (3.1) also results in equation (3.117). Letting

p ry

2

¼ p1ffiffi3  ðk4pþffiffi31Þ þ p2ffiffi3 ln k þ2 1 ≤ 1 obtains k2 k  2:654911. . . ¼ kz

ð3:118Þ

T When k ≤ kz, p ≤ σy and rT ri ≥ −σy; When k ≥ kz, p ≥ σy and rri ≤ −σy. When k ≤ ka, there must be k ≤ kz. When based on the maximum distortion strain energy theory, if k ≤ kc, kj = k, and the maximum load-bearing capacity of a pressure vessel is the entire yield load p p2ffiffi ry ¼ 3 ln k, that is plotted in figure 3.35 as curve oa; and if k ≥ kc, kj is calculated by

k 2 ln kj2  k 2  kj2 þ 2 ¼ 0, and the maximum load-bearing capacity of a pressure 2 vessel is rp ¼ p2ffiffi k k1 2 , that is plotted in figure 3.35 as curve ad. Point a is the intery

3

2 section point of the curve rpy ¼ p2ffiffi3 ln k and curve rpy ¼ p2ffiffi3 k k1 2 . The coordinates of point a are a (kc, 0.920079). 2 ðk þ 1Þ2 p p1ffiffi  pffiffi Letting þ p2ffiffi ln k þ2 1 ¼ p2ffiffi k k1 just leads to 2 2 r ¼

3 4 3k 3 3 ðk þ 1Þ2 ln 4 ¼ 0, which is just equation (3.114), or p2ffiffi ln k þ 1 is also plotted in figure 3.35 as curve ob. 2 3

y

5k þ 2k  7  4k 2

p ry

p1ffiffi 3

ðk þ 1Þ2 pffiffi 4 3k 2

2

equation (2.90).

¼  þ The coordinates of the intersection point of the curve ob and the curve ad are f (ka, 0.986264). In figure 3.35, if k ≥ ka, such as in the section fb, rT e > σy.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

123

FIG. 3.35 – Comparison of load-bearing capacity. For comparison, the maximum elastic load or initial yield load rpye ¼ p1ffiffi3 k k1 is also 2 plotted in figure 3.35 as dash curve on. As k increases, point n and m gradually coincide to one point (the meanings of point m and curve am will be explained below.). ! 2 1 ðk þ 1Þ2 2 k þ1 ðk  1Þ2 ð7k  1Þ pffiffiffi ln k  pffiffiffi  pffiffiffi 0 þ pffiffiffi ln ¼ pffiffiffi 2 3 3 4 3k 2 3 2 3k 2 ð6k þ 2Þ 2

py ry

This means that the load-bearing capacity at kj ¼ k þ2 1 is smaller than ¼ p2ffiffi3 ln k, as is seen from figure 3.35. This is because kj [ k þ2 1, or the plastic

depth when kj ¼ k þ2 1 is smaller than that when kj = kj*. From figure 3.35, we can also see that when k ≤ ka, 2 2 2 2 ðk þ 1Þ ðk þ 1Þ p1ffiffi  pffiffi þ p2ffiffi3 ln k þ2 1 \ p2ffiffi3 k k1 when k ≥ ka, p1ffiffi3  4pffiffi3k 2 þ p2ffiffi3 ln k þ2 1 [ p2ffiffi3 k k1 2 ; 2 . 3 4 3k 2  0 However, when k ≥ ka, r  [ ry or reverse yield will occur due to kj ¼ k þ 1. ei

In

addition,

p2ffiffi ln k þ 1 2 3

when

2

kj ¼

k þ1 2 ,

autofrettage

pressure

pa ry

¼

p1ffiffi 3

2

 ðk4pþffiffi31Þ þ k2

¼ rpy , which is just equation (3.117). This is because the general equation of k 2 k 2

r autofrettage pressure rpay ¼ p2ffiffi3 ln kj þ pffiffi3k 2j is obtained just by setting rey ¼ 1 in a plastic region. 0 p The total stresses rT ¼ r0 þ rP , rT e ¼ re þ re . pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi k þ1 3  1k = xa, or k = kd, equation (3.70) becomes When kj ¼ 2 ¼ xz = xa = kj. The total stresses for k = 1.406076 = kd, kj ¼ k þ2 1 = 1.203038 and rp ¼ p1ffiffi  T

y

ðk þ 1Þ2 pffiffi 4 3k 2

þ p2ffiffi3 ln k þ2 1 are illustrated in figure 3.36.

3

Autofrettage Technology and Its Applications in Pressured Apparatuses

124

The ðk þ 1Þ2 pffiffi 4 3k 2

total

stresses

for

kj ¼ k þ2 1 = 1.1

and

p ry

¼ p1ffiffi3 

kj ¼ k þ2 1 = 1.25

and

p ry

¼ p1ffiffi3 

k = 1.2 < kd,

þ p2ffiffi3 ln k þ2 1 are illustrated in figure 3.37.

The

total

stresses

for

k = 1.5 > kd,

2

ðk þ 1Þ pffiffi 4 3k 2

þ p2ffiffi3 ln k þ2 1 are illustrated in figure 3.38. The meaning of each curve and mweain parameters are marked in the figures.

2

FIG. 3.36 – The total stresses for k = kd, kj ¼ k þ2 1 and rpy ¼ p1ffiffi3  ðk4pþffiffi31Þk 2 þ p2ffiffi3 ln k þ2 1. T In figure 3.36, since k = kd, at x = kj, rT hj is just re , and is just σy, and xz = xa = kj. In figure 3.37, from equation (3.70), xz = 1.042057; kj = 1.1 > xa = 1.02672, so, xz exists and xa does not exist. Because k = 1.2 < kz, p < σy and rT ri > −σy. Because pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi T k < kd, kj > 3  1k, thus rhj > σy. In figure 3.38, xa = 1.2834; kj = 1.25 < xa, so, xa exists and xz does not exist. pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k, thus Because k < kz, p < σy and rT ri > −σy. Because k > kd, kj < rT < σ . y hj

The total stresses for k = 2.8 > kz, kj ¼ k þ2 1 = 1.9, and p2ffiffi ln k þ 1 2 3

p ry

= 1.052653 are illustrated in figure 3.39.

The total stresses for k = 2.654911 = kz, kj ¼ k þ2 1 = 1.8274555 and 2

ðk þ 1Þ pffiffi 4 3k 2

2

¼ p1ffiffi3  ðk4pþffiffi31Þ þ k2 p ry

¼ p1ffiffi3 

þ p2ffiffi3 ln k þ2 1 = 1 are illustrated in figure 3.40. The meaning of each curve and main parameters have been marked in the figures.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

125

2

FIG. 3.37 – The total stresses for k = 1.2 < kd, kj ¼ k þ2 1 and rpy ¼ p1ffiffi3  ðk4pþffiffi31Þk 2 þ p2ffiffi3 ln k þ2 1.

2

FIG. 3.38 – The total stresses with k = 1.5 > kd, kj ¼ k þ2 1 and rpy ¼ p1ffiffi3  ðk4pþffiffi31Þ þ p2ffiffi3 ln k þ2 1. k2 In figure 3.39, xa = 2.395679; kj = 1.9 < xa, so, xa exists and xz does not exist. Because k > kz, p > σy and rT ri = −1.05265 < −σy. Because k > kd, kj < pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi T 3  1k, thus rhj < σy.

126

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 3.39 – The total stresses with k = 2.8 > kz, kj ¼ k þ2 1 and

p ry

2

¼ p1ffiffi3  ðk4pþffiffi31Þ þ p2ffiffi3 ln k þ2 1. k2

In figure 3.40, xa = 2.271541; kj = 1.8274555 < xa, so, xa exists and xz does not pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k, rT hj < σy. Since k = kz, p = σy and

exist. Because k > kd, kj < T rT ri σri = −σy.

FIG. 3.40 – The total stresses with k = 2.654911 = kz, kj ¼ k þ2 1 = 1.8274555 and p = σy.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

3.4

127

The Solutions with Radius of Elastic-Plastic Juncture Being Geometrical Mean Radius of Inside Radius and Outside Radius

If the radius of the plastic region is a geometric mean value of the inner radius ri and outer radius rj of the pressure vessel, or pffiffiffiffiffiffiffiffi ð3:119Þ rj ¼ ri ro then kj ¼

pffiffiffi k

ð3:120Þ

pffiffiffi Adding kj ¼ k into figure  03.30  obtains curve ogh. It is seen that only section og  on curve oh can ensure that rei   ry . Substituting equation (3.120) into equations (3.2)–(3.5) obtains the residual stresses within the plastic region   r0z 1 x 2 k  1  ln k ¼ pffiffiffi ln þ ð3:121Þ k2  1 ry k 3  

  r0r 1 x2 1 1 1 k2  1  ln k 2 1 2 ¼ pffiffiffi ln þ  1 þ k k k 1 ry k x 3

ð3:122Þ

 

  r0h 1 x2 1 1 1 k2  1  ln k 2 1þ 2 ¼ pffiffiffi ln þ þ 1 þ k k k 1 ry k x 3

ð3:123Þ

r0e k 2  k þ k 2 ln k 1 ¼1 k2  1 x2 ry

ð3:124Þ

Substituting equation (3.120) into equations (3.6)–(3.9) obtains the residual stresses within the elastic region r0z k  1  ln k ¼a ¼ pffiffiffi ry 3ðk 2  1Þ r0r ¼ ry r0h ¼ ry



constant [ 0

ð3:125Þ

 k 2 r0z \0 x 2 ry

ð3:126Þ

  k 2 r0z 1þ 2 [0 x ry

ð3:127Þ

1

128

Autofrettage Technology and Its Applications in Pressured Apparatuses r0e k  1  ln k k 2 ¼ [0 k2  1 x2 ry

ð3:128Þ

pffiffiffi When k = 3, kj ¼ k = 1.732051…. The residual stresses within the whole wall are plotted in figure 3.41, where xa = 2.566799. The meaning of each curve and the main parameters are marked in the figure.

FIG. 3.41 – The residual stresses within the whole wall when k = 3. pffiffiffi When k = 3.5, kj ¼ k = 1.870829…, xa = 2.994599. The residual stresses  within the whole wall are plotted in figure 3.42, in which it is shown that r0ei  [ ry when k = 3.5.

FIG. 3.42 – The residual stresses within the whole wall when k = 3.5.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

129

  Based on the maximum shear stress theory, when k ≤ kb = 3.042297, r0ei   ry . When k = kb, kj = 1.744218…, xa = 2.602988…, the residual stresses within the whole wall are plotted in figure 3.43. The meaning of each curve and the main parameters are marked in the figure.

FIG. 3.43 – The residual stresses within the whole wall when k = kb. In figure 3.43, r0ei is just −σy. kb is just the abscissa of the intersection point (g) of curves oh and ad. pffiffiffi As based on the maximum shear stress theory, in figure 3.30, when kj ¼ k , if k ≥ kb, such as in the gh section, the pressure vessel will undergo reverse yield when processed with autofrettage. From figure 3.30, it is known that when k < kb, the pffiffiffi plastic depth kj ¼ k is smaller than kj*. Nevertheless, this may be at the expense of reducing load-bearing capacity. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi rT rT rT rT 3  1k = xa, rejy  rhjy ; when kj ≥ xa, rejy  rhjy . Letting kj ¼ When kj  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pp ffiffiffi 3  1k results in k 1 k  pffiffiffi ¼ 1:366025 ¼ kg 31 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k, accordingly kg is plotted in figure 3.34. When k ≥ kg, kj  The equivalent stress of the total stresses pffiffiffi 2 3k rT r0e rpe k 2  k þ k 2 ln k 1 p e ¼ þ ¼1 þ 2 2 2 2 k 1 x ry ry ry x ðk  1Þ ry Setting

rT e ry

ð3:129Þ rT ej ry



rT hj ry .

ð3:130Þ

 1 gives p k1 1  pffiffiffi þ pffiffiffi ln k ry 3k 3

ð3:131Þ

Autofrettage Technology and Its Applications in Pressured Apparatuses

130

Accordingly, the maximum load-bearing capacity of a pressure vessel is rpy ¼ pffiffiffi k1 ffiffi þ p1ffiffi ln k when kj ¼ k . p 3k 3 pffiffiffi Effectively, substituting kj ¼ k into equation (3.1) also results in equation (3.131). k1 ffiffi þ p1ffiffi ln k ≤ 1 obtains Letting rpy ¼ p 3k 3 k  2:926388. . . ¼ kx1 rT ri ry  2 p2ffiffi k 1 3 k2

When k ≤ kx1, p ≤ σy, thus

k1 ffiffi þ p1ffiffi ln k ¼ p 3k 3 k1 ffiffi þ p1ffiffi ln k (2.106). rpy ¼ p 3k 3

Letting

ð3:132Þ

 1. just leads to k2+k−2−k2lnk = 0, or equa-

is also plotted in figure 3.35 as curve oh. The tion coordinates of the intersection point of the curve oh and the curve ad are g (kb, 1.029943228). In figure 3.35, if k ≥ kb, such as in the section gh, rT e > σy.   1 ðk þ 1Þ2 2 kþ1 k1 1 pffiffiffi  pffiffiffi  pffiffiffi þ pffiffiffi ln k þ pffiffiffi ln 2 3 4 3k 2 3 3k 3 pffiffiffi pffiffiffi pffiffiffi ð k  1Þ3 ð15k k þ 11k þ 5 k þ 1Þ pffiffiffi pffiffiffi ¼ 0 4 3k 2 ðk þ 1 þ 2 k Þ pffiffiffi This means that the load-bearing capacity when kj ¼ k is smaller than that pffiffiffi when kj ¼ k þ2 1, as is seen in figure 3.35. This is because k þ2 1 [ k , or the plastic p ffiffiffi depth when kj ¼ k þ2 1 is larger than that when kj ¼ k . pffiffiffi k1 ffiffi þ ln k \ p2ffiffi k 2k1 From figure 3.35, we can also see that when k ≤ kb, p ; when 3k 3 2   p ffiffiffi 2 k1  0 ffiffi þ ln k [ p2ffiffi k k1 k ≥ kb, p 2 . However, when k ≥ kb, rei [ ry or reverse yield will 3k pffiffiffi 3 occur due to kj ¼ k . pffiffiffi The total stresses when k = 1.366025 = kg, kj ¼ k = 1.1687707 and rpy ¼ k1 ffiffi þ p1ffiffi ln k are illustrated in figure 3.44. p 3k 3 pffiffiffi The total stresses when k = 1.2 < kg, kj ¼ k = 1.0954451 and rpy ¼ k1 ffiffi þ p1ffiffi ln k are illustrated in figure 3.45. p 3k 3 pffiffiffi The total stresses when k = 1.5 > kg, kj ¼ k = 1.2247449 and rpy ¼ k1 ffiffi þ p1ffiffi ln k are illustrated in figure 3.46. p 3k

3

The meaning of each curve and main parameters have been marked in the figures. T In figure 3.44, since k = kg, at x = kj, rT hj is just re , and they are just σy, and xz = xa = kj.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

131

pffiffiffi

k1 ffiffi þ p1ffiffi ln k. FIG. 3.44 – The total stresses with k = kg, kj ¼ k and rpy ¼ p 3k 3

In figure 3.45, from equation (3.70), xz = 1.041352; kj = 1.0954451 > xa = 1.02672, so, xz exists and xa does not exist. Because k = 1.2 < kx1, p < σy and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3  1k, thus rT r0ri > −σy. Because k < kg, kj > hj > σy.

pffiffiffi

k1 ffiffi þ p1ffiffi ln k. FIG. 3.45 – The total stresses with k = 1.2 < kg, kj ¼ k = 1.0954451 and rpy ¼ p 3k 3

In figure 3.46, xa = 1.2834; kj = 1.2247449 < xa, so, xa exists and xz does not pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k, thus exist. Because k < kx1, p < σy and rT ri > −σy. Because k > kd, kj < rT < σ . y hj

132

Autofrettage Technology and Its Applications in Pressured Apparatuses

pffiffiffi

k1 ffiffi þ p1ffiffi ln k. FIG. 3.46 – The total stresses with k = 1.5 > kg, kj ¼ k = 1.2247449 and rpy ¼ p 3k 3

In addition, when kj ¼

pffiffiffi k , we have the following results.

(1) Autofrettage pressure pa k  1 1 p ¼ pffiffiffi þ pffiffiffi ln k ¼ ry ry 3k 3

which is just equation (3.131). As mentioned above, this is because the general equation of autofrettage pressure rpay ¼ p2ffiffi3 ln kj þ 1 in the plastic region, and

p ry

¼

pa ry

also results

k 2 kj2 pffiffi is obtained 3k 2 rT in rey ¼ 1.

just by setting

rT e ry

¼

(2) The abscissa of the intersection of the three residual stress curves or the abscissa at which r0e ¼ 0, x0, is the same as based on the maximum shear stress theory, or rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2  k þ k 2 ln k \k x0 ¼ k2  1 (3) The value of residual stresses when x = x0 or the minimum stresses are y0d ¼ y0d ¼

r0zx0 r0hx0 r0rx0 1 k  1 þ k ln k k  1  ln k þ pffiffiffi : ¼ ¼ ¼ pffiffiffi ln 2 k 1 ry ry ry 3 3ðk 2  1Þ

(4) The abscissa at which r0z ¼ 0, or x1 is the same as based on the maximum shear stress theory, which is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x1 ¼

e

1k þ k 2 ln k k 2 1

Mechanical Autofrettage Technology Based on Mises Yield Criterion

3.5

133

The Solutions with Minimum Equivalent Total Stress on Elastic-Plastic Juncture

On the elastic-plastic juncture, x = r/ri = kj, rT 1 k2 2k 2 ej ¼1 2  1 þ 2 ln kj 2 k  1 kj ry kj d

Setting

!

pffiffiffi 2 3k p þ 2 ðk  1Þkj2 ry

ð3:133Þ

rT  ej ry

dkj

¼ 0 results in

pffiffi  pffiffiffi  exp 23 rpy  1 3p rj kj ¼ ¼ exp or e ¼ ð3:134Þ ri 2 ry k1 pffiffi  If kj ¼ exp 23 rpy , the equivalent total stress at the elastic-plastic juncture rT ej is the minimum when kj is between 1 and k, and the minimum is pffiffiffi  2 exp 3 rpy  1 rT k k 2 kj2  1 ej pffiffiffi  ¼ 2 ¼ 2 ry k  1 exp 3 p k  1 kj2 ry When

p ry

k ffiffi1 ¼ rpye ¼ p , from equation (3.134), 3k 2  2  k 1 kj ¼ exp 2k 2

ð3:135Þ

2

ð3:136Þ

Therefore, as based on the maximum shear stress theory in the last chapter, when the load is initial yield pressure pe, the plastic depth determined by equations (3.136) or (3.134) cannot exceed radius ratio k. 2 When rpy ¼ p2ffiffi3 ¼ k k1 2 , from equation (3.134),  2  k 1 kj ¼ exp ð3:137Þ k2 2  2 k Letting kj ¼ exp k k1  k also obtains kk 2ln 2 1  1. We have known that when k ≥ kc = 2.2184574899167…, the equation

k 2 ln k k 2 1

 1 is tenable. Therefore, based on 2 the maximum distortion strain energy theory, when rp ¼ p2ffiffi ¼ k k1 2 , the plastic depth y

3

calculated by equations (3.134) or (3.137) can not exceed radius ratio k if k > kc; 2 otherwise, or k < kc, when rpy ¼ p2ffiffi3 ¼ k k1 2 , the plastic depth calculated by equations (3.134) or (3.137) is larger than radius ratio k.  pffiffi  For k < kc, in equation (3.134), the setting kj ¼ exp 23 rpy  k leads to p 2 py  pffiffiffi ln k ¼ ry r 3 y

134

Autofrettage Technology and Its Applications in Pressured Apparatuses

Setting

p ry

¼ p2ffiffi3 k k1 ¼ p2ffiffi3 ln k 2 2

just

obtains

equation (3.40)

but

“≤”

k k 1 changed into “ = ”, or kk 2ln 1 ¼ 1. This indicates that when k > kc, ln k [ k 2 , when 2 2 p p 2 k 1 2 k pffiffi pffiffi 1 ry ¼ 3 k 2 , kj < k, but the maximum load is unlikely ry ¼ 3 k 2 ; when 2

2

the maximum load is rpy ¼ p2ffiffi3 ln k, when rpy ¼ p2ffiffi3 ln k, k < kc, ln k\ k k1 2 ,   2  2 , when k = kc, ln k ¼ k k1 kj = k and kj is smaller than exp k k1 2 2 , kj = k = kc, the 2

maximum load is

p ry

¼ p2ffiffi3

kc2 1 kc2

¼ p2ffiffi3 ln kc . k 2 ln k

Equation (3.137) is equivalent to k 2 1j ¼ 1. Clearly, when k = kc, there is that kj = k = kc, that is to say, when k = kc, the plastic depth calculated by equation (3.137) is equal to the plastic depth calculated by equation (3.42). According to equation (3.137), with the increase of k, kj increase, and according to equation (3.42), with the increase of k, kj decreases. Therefore, when k > kc, the plastic depth calculated by equation (3.137) is certainly larger than the plastic depth calculated by equation (3.42). pffiffiffi According to equations (3.136) and (3.42), when k = ∞, kj ¼ e, kj calculated by equation (3.136) increases with the increase of k, and kj calculated by equation equation (3.42) decreases with the increase of k. Therefore, kj calculated by equation (3.42) is  always  larger than kj calculated by equation (3.136).   kj ¼ exp

k 2 1 2k 2

is plotted in figure 3.30 as curve om, and kj ¼ exp

k 2 1 k2

is

plotted in figure 3.30 as dash curve oan. Substituting equation (3.134) into equation (3.2) into equation (3.5) obtains the residual stresses within the plastic region pffiffiffi " pffiffiffi 2 # expð 3 rpy Þ  1 3k p r0z 1 2  2 ¼ pffiffiffi ln x þ ð3:138Þ 2 k 1 ry k  1 ry 3 pffiffiffi p " pffiffiffi p r0r 1 expð 3 ry Þ ¼ pffiffiffi  1 þ ln x 2  3  2 ry ry k 3

1

pffiffiffi expð 3 rpy Þ k2

pffiffiffi p þ 3 ry

!

# 1  k2  1 2 k2  1 x

ð3:139Þ pffiffiffi p " pffiffiffi p r0h 1 expð 3 ry Þ ¼ pffiffiffi þ 1 þ ln x 2  3  2 ry k ry 3

1

pffiffiffi expð 3 rpy Þ k2

pffiffiffi p þ 3 ry

!

1  k2  1 þ k2  1 x2

#

ð3:140Þ pffiffiffi  pffiffiffi k 2  exp 3 rpy þ 3k 2 rpy 1 ¼1 x2 ry k2  1 r0e

ð3:141Þ

Mechanical Autofrettage Technology Based on Mises Yield Criterion

135

Substituting equation (3.134) into equation (3.6) into equation (3.9) obtains the residual stresses within the elastic region pffiffiffi  pffiffiffi p p r0z exp 3 ry  1  3 ry pffiffiffi ¼ a constant [ 0 ð3:142Þ ¼ ry 3ðk 2  1Þ r0r ¼ ry r0h ¼ ry r0e

ry When

k = 2,

p ry

¼

exp



 k 2 r0z \0 x 2 ry

ð3:143Þ

  k 2 r0z 1þ 2 [0 x ry

ð3:144Þ

1

pffiffiffi  pffiffiffi 3 rpy  1  3 rpy k 2 k2  1

k ffiffi1 ¼p = 0.433012702, 3k 2 2

x2

[0

kj ¼ exp

ð3:145Þ

pffiffi  3 p 2 ry

¼ 1:454991,

xa =

1.711199. The residual stresses within the whole wall are plotted in figure 3.47. The meaning of each curve and the main parameters are marked in the figure.

FIG. 3.47 – The residual stresses within the whole wall when

p ry

k ffiffi1 ¼p = 0.433012702. 3k 2 2

When k = 2, rpy ¼ p2ffiffi3 ln k ¼ 0.8003774:::, kj = k = 2. The whole wall is plastic, the residual stresses within the whole wall are plotted in figure 3.48. The meaning of each curve and the main parameters are marked in the figure.

136

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 3.48 – The residual stresses within the whole wall when kj = k = 2. When k = 3,

p ry

¼ p2ffiffi3 k k1 = 1.0264005…, kj ¼ exp 2 2

pffiffi  3 p 2 ry

= 2.432425…, xa =

2.566799…. The residual stresses within the whole wall are plotted in figure 3.49. The meaning of each curve and the main parameters are marked in the figure.

FIG. 3.49 – The residual stresses within the whole wall when

When k = 3,

p ry

¼ p2ffiffi3 k k1 = 1.0264005, 2 2

r0ei ry

p ry

¼ p2ffiffi3 k k1 ¼ 1.0264005. 2 2

< −1. According to kj ¼ exp

pffiffi  3 p 2 ry ,

we can not directly determine whether the equivalent residual stress on the inner surface r0ei is greater than strength limit σy. When k ≥ kc, in order to avoid reversed yield, for equation (3.141), setting

r0ei ry

≥ −1 obtains

Mechanical Autofrettage Technology Based on Mises Yield Criterion pffiffiffi  exp 3 rpy  2 pffiffiffi p k2  3 ry  1

  pffiffiffi p pffiffiffi p exp 3 þ k 2  3k 2  2  0 or ry ry

137

ð3:146Þ

According to equations (3.126) and (2.128), for the same k, the load-bearing capacity based on the maximum distortion strain energy theory is p2ffiffi3 times of that based on the maximum shear stress When based on the maximum distortion ptheory. ffiffi  rj 3 p strain energy theory, kj ¼ ri ¼ exp 2 ry . Therefore, according to equation (3.134), the plastic depths are the same based on both the theories pffiffi for the same k. For 2 p 2 k 1 example, when k = 4, ry ¼ pffiffi3 k 2 = 1.082532, kj ¼ exp 23 rpy = 2.553589, this is the plastic depth based on the maximum distortion strain energy theory; when   p p k 2 1 k = 4, ry ¼ k 2 = 0.9375, kj ¼ exp ry = 2.553589, this is the plastic depth based on the maximum shear stress theory. It can be seen that the plastic depth based on both theories are the same. When k ≥ kc, the load-bearing capacity rpy is determined by equation (3.146), as is shown by curve am in figure 3.35. From equation (3.146), thedefinition domain of  pffiffi p exp 3ry 2 k2 is rp  p1ffiffi ¼ pb and rp  p1ffiffi ln 2 ¼ pc . The function k 2 ¼ pffiffi p is plotted in y

3

y

3ry 1

3

figure 3.50. There are two branches for the graphs of the function: curve og and pffiffi  3 p curve bah. In addition, the function kj ¼ exp 2 ry is plotted in figure 3.50 as the solid curve odai and the function rpy ¼ p2ffiffi3 k k1 is plotted in figure 3.50 as the dash 2 curve ofac. In figure 3.50, curve og corresponds to the definition domain of p p1ffiffi ry  3 ln 2. Since k < 1 on curve og, it is of no practical significance; curve bah   pffiffi p exp 3ry 2 , corresponds to the definition domain of rp [ p1ffiffi. For the function k 2 ¼ pffiffi p 2

3

y

setting

 dk 2  pffiffi p d 3ry

3ry 1

¼ 0 gives pffiffiffi     pffiffiffi p pffiffiffi p 3p exp 3  exp 3 þ1 ¼ 0 ry ry 2 ry

ð3:147Þ

Solving equation (3.147) obtains the horizontal ordinate of point a, that is the minimum value of bah. p ¼ 0:920079 ð3:148Þ ry   pffiffi p exp 3ry 2 Substituting rpy = 0.920079 into k 2 ¼ pffiffi3 p 1 leads to k = kc. ry

138

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 3.50 – The relationship between k (kj) and rpy . 



pffiffi p 3ry 2 pffiffi p 3ry 1

pffiffiffi  3 rpy also obtains equation (3.147). pffiffiffi  2 When k = kj, substituting k 2 ¼ kj2 exp 3 rpy into rpy ¼ p2ffiffi3 k k1 obtains equa2 Letting

k2 ¼

exp

¼ kj2 exp

tion (3.147) again. This shows that the three curves bah, ofac and odai intersect at one point. This point is just the minimum value point a of curve bah. From figure 3.50, when rpy \0.920079:::, k > kc, and k is larger than kj calculated by pffiffi  2 kj ¼ exp 23 rpy , and k is larger than the value determined by rpy ¼ p2ffiffi3 k k1 2 . This is

reasonable and makes sense; when rpy [ 0.920079:::, k > kc, and k is smaller than kj pffiffi  determined by kj ¼ exp 23 rpy , this is unreasonable and makes no sense. Curve odac

can ensure k > kj, but cannot ensure r0ei ≥ −σy. Therefore, when k > kc, the meaningful control curve is curve ba, which is also curve am in figures 3.35 or 2.23. If the design conditions are controlled by curve ba when k > kc, there are both kj < k and r0ei ≥ −σy. For example, when k = 3, from equation (3.146), rpy ¼ 0.64516:::. When pffiffi  k = 3 and rpy ¼ 0.64516:::, kj ¼ exp 23 rpy ¼ 1:748442::. The residual stresses within

the whole wall are plotted in figure 3.51. The meaning of each curve and the main parameters are marked in the figure. Compared with figure 3.49, pffiffi the  residual stresses are greatly improved. 3 p In fact, when kj ¼ exp 2 ry , equations (3.147) and (3.146) are equivalent. pffiffi  When kj ¼ exp 23 rpy (but k > kj), equation (3.146) becomes equation (3.42), or k 2 ln kj2  k 2  kj2 þ 2  0. Therefore, curve in figure 3.30 is also the relation between k and kj under the pad ffiffi  3 p case that kj ¼ exp 2 ry and r0ei ¼ ry . However, when the load-bearing capacity

Mechanical Autofrettage Technology Based on Mises Yield Criterion

FIG. 3.51 – The residual stresses when k = 3, p ry p ry

p ry

139

= 0.64516….

¼ p2ffiffi3 k k1 2 , σej ≡ σy within the whole plastic region; when the load-bearing capacity  pffiffi  2 ¼ pffiffi3 ln kj equivalent to kj ¼ exp 23 rpy where k 2 ln kj2  k 2  kj2 þ 2  0, 2

σej < σy. Equation (3.146) can ensure kj < k and r0ei   ry , but under the premise of ensuring kj < k and r0ei   ry as well as σej being the minimum, the load-bearing pffiffi  capacity rpy calculated by kj ¼ exp 23 rpy may not be the most satisfactory. From equations (3.146) or (3.42), the greater k is, the smaller kj is, or the smaller rpy is, pffiffi  because kj ¼ exp 23 rpy . When k = ∞, kj = e0.5, hence rpy ¼ p1ffiffi3, that is merely the pffiffi  maximum elastic load with k = ∞. Besides, when kj ¼ exp 23 rpy , from figure 3.35,

than the load-bearing capacity under the case that k > kc is unexpectedly pffiffismaller  3 p that under the case that k < kc. So, if it is not necessary, kj ¼ exp 2 ry is not a satisfactory solution. pffiffiffi  pffiffiffi When exp 3 rpy þ k 2  3k 2 rpy  2  0 (equation (3.146)), equation (3.141) become r0e

k 2  exp

pffiffiffi  pffiffiffi 3 rpy þ 3k 2 rpy 1 2 ¼1 2 x2 x k2  1

¼1 ry pffiffi  When kj ¼ exp 23 rpy , autofrettage pressure

When

p ry

  pffiffiffi p pa p 1 1 ¼ þ pffiffiffi  pffiffiffi exp 3 ry ry ry 3 3k 2 pffiffiffi   p2ffiffi3 ln k, p1ffiffi3  pffiffi31k 2 exp 3 rpy  0. Therefore, rpay 

ð3:149Þ

ð3:150Þ p ry .

Autofrettage Technology and Its Applications in Pressured Apparatuses

140

When

kj = k = kc,

kj2 ¼ exp

pa ry

pffiffiffi  3 rpy ¼ k 2 ,

then,

from

equation (3.150),

¼ rpy ¼ p2ffiffi3 ln kc = 0.920079. pffiffiffi  dðpa =ry Þ dðpa =ry Þ py p 1 p2ffiffi ¼ 1  exp 3 rpy . gives Setting 2 dðp=ry Þ dðp=ry Þ ¼ 0 ry ¼ 3 ln k ¼ ry . k   ffiffi p pffiffiffi d2 ðpa =ry Þ ¼  k 23 exp 3 rpy \0. Consequently, when rpy ¼ p2ffiffi3 ln k, rpay reaches its maxidðp=ry Þ2 mum value, which is rpay ¼ rpy ¼ p2ffiffi3 ln k. It should be noted that only if k ≤ kc, can we p p2ffiffi For k ≥ kc, from equation (3.146), we obtain have ry ¼ 3 ln k. pffiffiffi  2 p k ffiffi2 pffiffi1 3 rpy ¼ p . Substituting this equation into equation (3.150) gives ry  3k 2 exp 3k 2 pa 2 k2  1 ¼ pffiffiffi ry 3 k2

ð3:151Þ

Equation (3.30) indicates that in general, when the load-bearing capacity

p ry

equals autofrettage pressure, or rpy ¼ rpay , it is the optimum. In this case, rT e ≡ σy in the p ry

pa ry

¼ p2ffiffi3 k k1 2 , the load-bearing capacity is also pffiffi  2 2 p 3 p ¼ p2ffiffi3 k k1 and k ≥ kc, when rpay ¼ p2ffiffi3 k k1 2 . In the case that kj ¼ exp 2 , if r is 2 ry y

whole plastic region. So, when

2

controlled by equation (3.146), rpy \ rpay , and rT ej < σy. For example, when k = 3, from p equation (3.146) ry ¼ 0.64516, and from equation (3.151) rpay ¼ 1:0264, from equation (3.135)

rT ej ry

¼ 0.756998.

When k ≤ kc, the optimum load-bearing capacity is

p ry

¼ rpay ¼ p2ffiffi3 ln k, and in this

T case rT e re ≡ σy within the whole plastic region. When k = ∞, from equation (3.151), rpay ¼ p2ffiffi3; from equation (3.146), rpy ¼ p1ffiffi3, when rp ¼ p1ffiffi, from equation (3.150), we also have rpa ¼ p2ffiffi. When k = ∞, equay

3

y

3

tion (3.150) become pa p 1 ¼ þ pffiffiffi ry ry 3

ð3:152Þ

For equation (3.150), when k ≤ kc, rpy is from 0 to p2ffiffi3 ln kc ; when k ≥ kc, rpy is from

p2ffiffi ln kc to the value calculated by 3  pffiffi  exp 23 rpy ¼ 1. When the pressure vessel

equation (3.146). When

p ry

= 0, kj ¼

is at the initial yield stage, the pressure

k 2ffiffi1 . For a pressure vessel with applied to the pressure vessel is initial yield load rpye ¼ p 3k 2 a certain radius ratio k (≥kc), the maximum load for σej to reach the minimum is the value calculated by equation (3.126). When rpy equals the value calculated by 2 equation (3.146), rpay ¼ 2 rpye ¼ p2ffiffi3 k k1 2 . For the convenience of engineering applications, some data calculated by equation (3.146) are listed in table 3.1. Equation (3.150) is plotted in figure 3.52. In figure 3.52, curve ab is the connection of the ends of the curves expressed by equation (3.150), which shows the

Mechanical Autofrettage Technology Based on Mises Yield Criterion

141

maximum load-bearing capacity rpy and the maximum autofrettage pressure rpay for k ≥ kc. Curve ab is derived as follows. From equation (3.146) we can find k2 expressed by rpy, substituting k2 into equation (3.151) we obtain rpay , as a result the equation of curve ab in figure 3.52 is generated, that is equation (3.153). pffiffiffi  pffiffiffi p p pa 2 exp 3 ry  3 ry  1 pffiffiffi  ¼ pffiffiffi ð3:153Þ ry 3 exp 3 rpy  2 TAB. 3.1 – Numerical value from exp

pffiffiffi  pffiffiffi 3 rpy þ k 2  3k 2 rpy  2  0.

k

p ry

pa ry

k

p ry

pa ry

2.3 2.4 2.5 2.6 2.7 2.8 2.9 3 3.1 3.2

0.781825 0.734722 0.706831 0.687494 0.673047 0.66176 0.652662 0.645159 0.638863 0.633503

0.936421 0.954232 0.969948 0.983887 0.996306 1.007417 1.0174 1.026401 1.034545 1.041937

3.3 3.4 3.5 3.6 3.8 4 4.5 5 10 ∞

0.628887 0.624871 0.621348 0.618235 0.61299 0.608751 0.601082 0.595999 0.581613 0.57735

1.048668 1.054813 1.060439 1.065604 1.074735 1.082532 1.097678 1.108513 1.143154 1.154701

FIG. 3.52 – The graph of

pa ry

¼ rpy þ p1ffiffi3  pffiffi31k 2 exp

pffiffiffi  3 rpy .

142

Autofrettage Technology and Its Applications in Pressured Apparatuses

We have known that when k = kc, kc2−1 = kc2lnkc.

or p2ffiffi 3

p ry

kc2

ln kc 2 ln kc kc2 2

¼ p2ffiffi3

Equation (3.153)

pa ry

kc2 ln kc kc2 1 ¼ 1 2 k 2 ln k 1 p2ffiffi c 2 c ¼ kc 2 3

¼ rpy ¼ p2ffiffi3 ln k ¼ p2ffiffi3 ln kc and

is

equivalent

to

pa ry

¼

kc2 2 kc2 2

ln kc ¼ p2ffiffi3 ln kc . This confirms the above conclusion. The coordinate of point a in figure 3.52 is a (p2ffiffi3 ln kc , p2ffiffi3 ln kc ). As mentioned above, the meaningful definition domain of k2 is rp  p1ffiffi, and when

¼ p1ffiffi3, k2 = ∞. Consequently from equation (3.150)–(3.153), coordinate of point b in figure 3.52 is b (p1ffiffi, p2ffiffi). 3

y

pa ry

3

¼ p2ffiffi3, or the

3

The coordinate of point c in figure 3.52 is c (0, p1ffiffi3). The practically significant curves for a relationship between rpy and rpay are located in a quadrilateral abco in figure 3.52, where the straight line bc is parallel to the straight line ao. For k ≤ kc, the ends of the curves described by equation (3.150) are on the straight line ao in figure 3.52, where rpay ¼ rpy . Thereupon the included angle between the straight line ao and the abscissa axis or vertical axis is “45°”. In figure 3.52, the dash curves have no practical significance, they are merely an extension of solid curves, or only the elongation of the mathematical relation expressed by equation (3.150). The total stresses are expressed as follows 8 T rz r0z rpz > > > ¼ þ > > ry ry ry > > > T > > rr r0r rpr > > >

rh rh re > > ¼ þ > > > r ry ry > > y > > > rT r0 rp > > : e ¼ e þ e ry ry ry If the pressure contained in a pressure vessel does not exceed the initial yield pressure, or p ≤ pe, all stresses will not exceed the yield strength σy. Generally, autofrettage is not necessary. However, if autofrettage is performed,the stresses will pffiffi  3 p be evened and cut down. Between 1 and k, if kj ¼ exp 2 ry p = pe,   2  when p ¼ pe ; kj ¼ exp k2k1 , the equivalent total stress at the elastic-plastic 2 juncture rT ej is the minimum, the minimum is equation (3.135). 2  When p = pe, kj ¼ exp k2k1 , then, equation (3.135) becomes 2 pffiffiffi  2  p k 1 2 2 exp 3 exp  1 1 rT ry k2 k k ej pffiffiffi  ¼ 2 k 2 1 ¼ 2 ry k  1 exp 3 p k  1 exp k 2 ry

ð3:155Þ

Mechanical Autofrettage Technology Based on Mises Yield Criterion

143

Several figures are presented to show the above-mentioned conclusions. 2  p pe k 2ffiffi1 k 1 p ¼ r = 0.5132, kj ¼ exp 2k 2 For k = 3, r ¼ = 1.559623, the operation 2 y

3k

y

stresses, residual stresses and total stresses are shown in figures 3.53–3.55, respectively.

FIG. 3.53 – The distribution of the operation stresses for k = 3, p = pe and kj ¼ exp

FIG. 3.54 – The distribution of the residual stresses for k = 3, p = pe and kj ¼ exp







k 2 1 2k 2

.



k 2 1 2k 2

.

144

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 3.55 – The distribution of the total stresses for k = 3, p = pe and kj ¼ exp

In figure 3.55, For k = 3,

p ry

rT ej ry

= 0.662499.

k ffiffi1 ¼p ¼ rpye = 0.5132, kj = 1.4 < exp 3k 2 2



k 2 1 2k 2



k 2 1 2k 2

 .

 , the total stresses are

shown in figure 3.56.

FIG. 3.56 – The distribution of the total stresses for k = 3, p = pe and kj = 1.4.

In figure 3.56, For k = 3,

p ry

rT ej ry

= 0.674968 > 0.662499. 2  pe k ffiffi1 k 1 ¼p ¼ = 0.5132, exp < kj = 1.7 < kj* = 1.748442, the ry 2k 2 3k 2 2

total stresses are shown in figure 3.57.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

145

FIG. 3.57 – The distribution of the total stresses for k = 3, p = pe and kj = 1.7.

In figure 3.57,

rT ej ry

= 0.668629 > 0.662499.

For k = 3, from equation (3.146),

p ry

= 0.64516 >

pe ry ,

kj ¼ exp

pffiffi  3 p 2 ry

= kj* =

1.748442, the total stresses are shown in figure 3.58.

FIG. 3.58 – The distribution of the total stresses for k = 3, p = 0.64516 and kj = kj*. rT

In figure 3.58, rejy = 0.756998. For k = 3, from equation (3.146), stresses are shown in figure 3.59.

p ry

= 0.64516 >

pe ry ,

kj = 1.6 < kj*, the total

146

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 3.59 – The distribution of the total stresses for k = 3, p = 0.64516 and kj = 1.6. rT

In figure 3.59, rejy = 0.763525 > 0.756998. For k = 3, from equation (3.146), rpy = 0.64516 > stresses are shown in figure 3.60.

pe ry ,

kj = 1.8 > kj*, the total

FIG. 3.60 – The distribution of the total stresses for k = 3, p = 0.64516 and kj = 1.8. In figure 3.60,

3.6

rT ej ry

= 0.757596 > 0.756998.

Comparison Between the Three Cases

pffiffiffi So far, we have investigated autofrettage under four cases: kj = k, kj ¼ k þ2 1, kj ¼ k pffiffi  and kj ¼ exp 23 rpy . kj = k is for k < kc, and when k < kc, rpy ¼ p2ffiffi3 ln k, that is

Mechanical Autofrettage Technology Based on Mises Yield Criterion

147

pffiffi  equivalent to k ¼ exp 23 rpy . Therefore, when k < kc, kj = k is equivalent to pffiffi  pffiffiffi kj ¼ exp 23 rpy . Then, we do a comparison between three cases: kj ¼ k þ2 1, kj ¼ k pffiffi  and kj ¼ exp 23 rpy .

3.6.1

k = 2.5 > kc

1. kj ¼ k þ2 1 kj ¼

k þ1 , kj1 ¼ 1:75 2

The equivalent residual stress is as follows. From equation (3.109), in the plastic region, k2  r0e ¼1 ry

ðk þ 1Þ2 þ k2 4 k2  1

2

ln ðk þ4 1Þ 1 1.939561 ¼1 x2 x2

This is the same as that based on the maximum shear stress theory (see last chapter). From equation (3.113), in the elastic region, r0e ¼ ry

ðk þ 1Þ2 4

2

 1  ln ðk þ4 1Þ k 2 1.122939 ¼ x2 k2  1 x2

This is the same as that based on the maximum shear stress theory. From equation (3.117), the load-bearing capacity, p 1 ðk þ 1Þ2 2 k þ1 ¼ 0:940637 ¼ pffiffiffi  pffiffiffi þ pffiffiffi ln ry 2 3 4 3k 2 3 The load-bearing capacity is p2ffiffi3 times of that based on the maximum shear stress theory. From equation (1.10), the equivalent operation stress, pffiffiffi 2 3k rpe p 1.939561 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The load-bearing capacity is different from that based on the maximum shear stress theory, but the equivalent operation stress is the same as that based on the maximum shear stress theory. From equations (3.29) or (3.154), the equivalent total stress, rT r0 rp e ¼ e þ e ry ry ry

148

Autofrettage Technology and Its Applications in Pressured Apparatuses

Hence, in the plastic region, rT e 1 ry In elastic region rT 3.0625 e ¼ x2 ry The equivalent total stress is the same as that based on the maximum shear stress theory. The autofrettage pressure pa p ¼ ry ry

2. kj ¼

pffiffiffi k kj ¼ k 0:5 , kj2 ¼ 1:581139

The equivalent residual stress is as follows. From equation (3.124), in the plastic region, r0e k 2  k þ k 2 ln k 1 1.805108 ¼1 ¼1 2 2 k 1 x x2 ry From equation (3.128), in the elastic region r0e k  1  ln k k 2 0.694892 ¼ ¼ k2  1 x2 x2 ry The equivalent residual stress is the same as that based on the maximum shear stress theory. From equation (3.131), the load-bearing capacity. p k 1 1 ¼ pffiffiffi þ pffiffiffi ln k ¼ 0:875431 ry 3k 3 The load-bearing capacity is p2ffiffi3 times of that based on the maximum shear stress theory. From equation (1.10), the equivalent operation stress pffiffiffi 2 3k rpe p 1.805108 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The load-bearing capacity is different from that based on the maximum shear stress theory, but the equivalent operation stress is the same as that based on the maximum shear stress theory.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

149

From equations (3.29) or (3.154), the equivalent total stress rT r0 rp e ¼ e þ e ry ry ry Then, in the plastic region, rT e 1 ry In the elastic region rT 2.5 e ¼ 2 x ry The equivalent total stress is the same as that based on the maximum shear stress theory. The autofrettage pressure pa p ¼ ry ry 3. kj ¼ exp

pffiffi  3 p 2 ry

When k = 2.5, from equation (3.146) or table 3.1, the load-bearing capacity = 0.706831. The load-bearing capacity is p2ffiffi3 times of that based on the maximum shear stress theory. When rpy = 0.706831, from equation (3.134), kj ¼ pffiffi  exp 23 rpy , kj3 = 1.844363.

p ry

The equivalent residual stress is as follows. From equation (3.141), in the plastic region, r0e 2 ¼1 2 x ry From equation (3.145), in the elastic region r0e k  1  ln k k 2 1.401675 ¼ ¼ k2  1 x2 x2 ry The equivalent residual stress is the same as that based on the maximum shear stress theory. From equation (1.10), the equivalent operation stress is as follows. pffiffiffi 2 3k rpe p 1.805108 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The load-bearing capacity is different from that based on the maximum shear stress theory, but the equivalent operation stress is the same as that based on the maximum shear stress theory.

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Autofrettage Technology and Its Applications in Pressured Apparatuses

From equations (3.29) or (3.154), the equivalent total stress rT r0 rp e ¼ e þ e ry ry ry Thus, in the plastic region, rT 0.194892 e ¼1 x2 ry In the elastic region rT 3.206783 e ¼ x2 ry The equivalent total stress is the same as that based on the maximum shear stress theory. The autofrettage pressure pa 2 k2  1 ¼ pffiffiffi ¼ 0:969948 ry 3 k2 When k = 2.5, the figures for the distributions of the equivalent residual stress and the equivalent total stress are the same as figures 2.43 and 2.44, respectively (see last chapter).

3.6.2

k = 2 < kc

1. kj ¼ k þ2 1 kj ¼

k þ1 , kj1 ¼ 1:5 2

The equivalent residual stress is as follows. From equation (3.109), in plastic region, k2  r0e ¼1 ry

ðk þ 1Þ2 þ k2 4 k2  1

2

ln ðk þ4 1Þ 1 1 ¼ 1  1.664574 2 2 x x

From equation (3.113), in elastic region r0e ¼ ry

ðk þ 1Þ2 4

2

 1  ln ðk þ4 1Þ k 2 0.585426 ¼ x2 k2  1 x2

The equivalent residual stress is the same as that based on the maximum shear stress theory.

Mechanical Autofrettage Technology Based on Mises Yield Criterion

151

From equation (3.117), the load-bearing capacity. p 1 ðk þ 1Þ2 2 k þ1 ¼ 0:720781 ¼ pffiffiffi  pffiffiffi þ pffiffiffi ln 2 ry 2 3 4 3k 3 The load-bearing capacity is p2ffiffi3 times of that based on the maximum shear stress theory. From equation (1.10), the equivalent operation stress pffiffiffi 2 3k rpe p 1.664574 ¼ 2 2 ¼ x2 ry x ðk  1Þ ry The load-bearing capacity is different from that based on the maximum shear stress theory, but the equivalent operation stress is the same as that based on the maximum shear stress theory. From equations (3.29) or (3.154), the equivalent total stress rT r0 rp e ¼ e þ e ry ry ry Then, in the plastic region, rT e 1 ry In elastic region rT 2.25 e ¼ 2 x ry The equivalent total stress is the same as that based on the maximum shear stress theory. The autofrettage pressure pa p ¼ ry ry

2. kj ¼

pffiffiffi k kj ¼

pffiffiffi k , kj2 ¼ 1:414214

The equivalent residual stress is as follows. From equation (3.124), in the plastic region, r0e k 2  k þ k 2 ln k 1 1.590863 ¼1 ¼1 k2  1 x2 x2 ry

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Autofrettage Technology and Its Applications in Pressured Apparatuses

From equation (3.128), in the elastic region r0e k  1  ln k k 2 0.409137 ¼ ¼ k2  1 x2 x2 ry The equivalent residual stress is the same as that based on the maximum shear stress theory. From equation (3.131), the load-bearing capacity. p k 1 1 ¼ pffiffiffi þ pffiffiffi ln k ¼ 0:688864 ry 3k 3 The load-bearing capacity is p2ffiffi3 times of that based on the maximum shear stress theory. From equation (1.10), the equivalent operation stress pffiffiffi 2 3k rpe p 1.590863 ¼ ¼ x2 ry x 2 ðk 2  1Þ ry The load-bearing capacity is different from that based on the maximum shear stress theory, but the equivalent operation stress is the same as that based on the maximum shear stress theory. From equations (3.29) or (3.154), the equivalent total stress rT r0 rp e ¼ e þ e ry ry ry Hence, in the plastic region, rT e 1 ry In the elastic region rT 2 e ¼ ry x 2 The equivalent total stress is the same as that based on the maximum shear stress theory. The autofrettage pressure pa p ¼ ry ry pffiffi  3. kj ¼ exp 23 rpy and kj = k When k = 2 < kc, kj = k, the load-bearing capacity rpy ¼ p2ffiffi3 ln k = 0.800377, The load-bearing capacity is p2ffiffi times of that based on the maximum shear stress theory, 3

Mechanical Autofrettage Technology Based on Mises Yield Criterion

153

pffiffi  and it is greater than the load-bearing capacity when k = 2.5 and kj ¼ exp 23 rpy . pffiffi  When rpy ¼ p2ffiffi3 ln k, kj ¼ exp 23 rpy = k. There is no elastic region when kj = k. From equation (3.80), the equivalent residual stress r0e k 2 ln k 2 1.848392 ¼1 2 ¼1 x2 ry ðk  1Þx 2 This is the same as that based on the maximum shear stress theory. From equation (1.10), the equivalent operation stress rpe 1:848392 ¼ x2 ry This is the same as that based on the maximum shear stress theory. The equivalent total stress rT e 1 ry This is the same as that based on the maximum shear stress theory. The autofrettage pressure pa p ¼ ¼ 0:800377 ry ry When k = 2, the figures for the distributions of the equivalent residual stress and the equivalent total stress are the same as figures 2.45 and 2.46, respectively (see last chapter).

3.7

Chapter Summary

The main equations and conclusion are listed in table 3.2. TAB. 3.2 – The main equations and conclusion of this chapter. The components of residual stresses in general forms Within the plastic region (1 ≤ x ≤ kj) " # ! kj2 r0z 1 kj2 x2 1 2 ¼ pffiffiffi 2 þ ln 2  1  2 þ ln kj k2  1 ry k kj 3 k ! "  # kj2 r0r 1 kj2 x2 1 k2 1  ¼ pffiffiffi 2  1 þ ln 2  1  2 þ ln kj2 k2  1 ry k x2 kj 3 k " !  # kj2 r0h 1 kj2 x2 1 k2 2 1þ 2 ¼ pffiffiffi 2 þ 1 þ ln 2  1  2 þ ln kj k2  1 ry k x kj 3 k k 2  kj2 þ k 2 ln kj2 1 r0e ¼1 ry k2  1 x2

154

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 3.2 – (continued). Within the elastic region (kj ≤ x ≤ k) r0z kj2  1  ln kj2 ¼ pffiffiffi ¼ constant [ 0 ry 3ðk 2  1Þ   r0r k 2 r0z k 2  x 2 kj2  1  ln kj2 pffiffiffi \0 ¼ 1 2 ¼ ry x ry x2 3ðk 2  1Þ   r0h k 2 r0z k 2 þ x 2 kj2  1  ln kj2 pffiffiffi [0 ¼ 1þ 2 ¼ ry x ry x2 3ðk 2  1Þ

r0e kj2  1  ln kj2 k 2 ¼ [0 ry k2  1 x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi r0 r0 Within the elastic region, when x < xa (= 3  1k = 0.8556k), e [ h ; when x > xa, ry ry r0e r0h \ ry ry General autofrettage pressure pa k 2  kj2 pa p 2 ¼ pffiffiffi ln kj þ pffiffiffi ¼ ry ry 3k 2 3 The abscissa of the intersection of curves of the residual stress or the abscissa at which r0e = 0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2  kj2 þ k 2 ln kj2 r \kj x0 ¼ ¼ k2  1 ri The value of residual stresses at x = x0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 0 0 k 2  kj2 þ k 2 ln kj2 kj2  1  ln kj2 rzx0 rhx0 rrx0 r0rmin 2 x0 r0zj 2 d p ffiffi y0 ¼ ¼ ¼ ¼ ¼ pffiffiffi ln þ ¼ 3 ln þ pffiffiffi ry ry ry ry ry kj2 ðk 2  1Þ 3ðk 2  1Þ 3 kj The abscissa at which r0z = 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 1k þ k ln k j j

x1 ¼ e k 2 1 The critical radius ratio kc = 2.218 457 489 916 7… k 2 ln k k 2 ln k ¼ 1, and 2 ¼ 1 is equivalent to 2pe = py, i.e kc is the solution of 2 k 1 k 1 k2  1 2 2 pffiffiffi ¼ pffiffiffi ln k 3k 2 3 The optimum plastic depth kj* kj2  2 2 2 k 2 ln kj  k 2  kj þ 2  0 or k 2 ¼ (e0.5 ≤ kj* ≤ kc and k ≥ kc) ln kj2  1 When k ≤ kc, kj = k The results under the case that kj = kj* 1. Residual stresses in axial, radial, and circumferential direction 8 8 0 kj2  2 rz 2 1 r0 > > > > > z¼ p ffiffiffi ¼ pffiffiffi ln x  pffiffiffi > > > > ry ry  3k 2  > 3 3 > > > < r0 < 2 2 2 r0r k 2 r0z r ¼ pffiffiffi ln x þ pffiffiffi  pffiffiffi (1 ≤ x ≤ kj*) ¼ 1 2 2 > > r r x 3 3 x 3 y y > > > >   ry0 > > 2 2 2 > r0h > r0h k rz > > : ¼ pffiffiffi ln x  pffiffiffi > : ¼ 1þ 2 ry 3 3x 2 ry x ry

(kj* ≤ x ≤ k)

Mechanical Autofrettage Technology Based on Mises Yield Criterion

155

TAB. 3.2 – (continued). 2. Equivalent residual stress r0e 2 r0e kj2  2 ¼1 2 (1 ≤ x ≤ kj*) ¼ (kj* ≤ x ≤ k) ry ry x2 x 3. The optimum load-bearing capability and autofrettage pressure p pa pe 2 k2  1 ¼ ¼ 2 ¼ pffiffiffi ry ry ry 3 k2 pffiffiffi 4. x0  2 = constant pffiffiffi x1  e = constant r0 r0 r0 r0 r0 1  ln 2 5. y0d ¼ z ¼ r ¼ h ¼ rmin ¼ rx0   pffiffiffi ¼ constant ry ry ry ry ry 3 y0d is not related to k and kj The results under the case that kj = kj*, p = pa = 2pe 1. The components of total stresses 8 T 8 kj2 rz ln x 2 1 2 > > rT > > z > > p ffiffi ffi p ffiffi ffi p ffiffi ffi ¼  þ ¼ pffiffiffi ¼ constant >r > 2 > > ry 3 3 3k > > y 3k 2 > > > > > >   < T < T kj2 1 rr ln x 2 2 2 rr 1 (1 ≤ x ≤ kj*) (kj* ≤ x ≤ k) ¼ pffiffiffi  pffiffiffi þ pffiffiffi p ffiffi ffi ¼  > ry > ry 3 3 3k 2 > > 3 x2 k2 > > > >   > > > > > rT > kj2 1 ln x 2 2 rT 1 > > h h > > p ffiffi ffi p ffiffi ffi þ ¼ p ffiffi ffi ¼ þ : : ry ry 3 3k 2 3 x2 k2 2. Equivalent total stress kj2 rT rT e e  1 (1 ≤ x ≤ kj*) ¼ 2 (kj* ≤ x ≤ k) ry ry x pffiffiffi (1) When k ≤ kr = 1 þ 3 = 2.732051, p ≤ σy, hence rT ri > −σy pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi (2) When k ≥ kf = 2.300122…, kj* < 3  1k(=xa), thus rT hj < σy; pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi T 3  1k, thus rhj > σy when k < kf, kj* > (3) When x  exp

pffiffiffi pffiffiffi pffiffiffi 2   2 pffiffiffi k 1 2 3 3k  2 3 p 2 3 T ¼ exp   ¼ exp ¼ xz , rT e  rh 2k 2 k2 2 2 2 ry

pffiffiffi pffiffiffi (4) When kj = kj*, if k 2 ln½ð 3  1Þk 2   3k 2 þ 2 ¼ 0 or k = kf, xz = xa = kj* = 1.967984… qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi k þ1 ¼ (When kj ¼ 3  1k, or 3  1k, or k = kd, xz = xa = kj; When kj ¼ k ¼ 2 k = kg, xz = xa = kj.) pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi (5) When kj < 3  1k(=xa), xa exists and xz does not exist; When kj > xa, xa does not exists and xz exist The results under the case that kj = k 1. The optimum load-bearing capability and autofrettage pressure p pa 2 py ¼ ¼ pffiffiffi ln k ¼ ry ry ry 3 2. The total stresses under the case that kj = k and p = py = σylnk   rT 2 1 z ¼ pffiffiffi þ ln x  ln k ry 3 2

156

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 3.2 – (continued).

rT 2 r ¼ pffiffiffi ðln x  ln k Þ ry 3 rT 2 h ¼ pffiffiffi ð1 þ ln x  ln k Þ ry 3 rT e 1 ry p ffiffi ffi pffiffiffi    pffiffi 3 p 2 3 3. x ¼ exp ln k þ 23  1 ¼ xb ¼ exp  ¼ xz 2 ry 2 rT rT rT rT When x < xb, h \ e ¼ 1; When x > xb, h [ e ¼ 1. When p = py, xb = xz ry ry ry ry 4. The total stresses on inner surface under the case that kj = k and p = py = σylnk   rT 2 1 zi  ln k ¼ pffiffiffi ry 3 2 T rri 2 p ¼  pffiffiffi ln k ¼  ry ry 3 rT 2 hi ¼ pffiffiffi ð1  ln k Þ ry 3 rT ei ¼1 ry k þ1 The results under the case that kj ¼ 2  0  0 1. When k ≤ ka = 2.618253…, rei   ry ; When k ≥ ka, rei   ry k þ1 = 1.809126… 2. When k = ka, kj ¼ 2 3. The maximum load-bearing capacity and autofrettage pressure p 1 ðk þ 1Þ2 2 k þ1 2 py ≤ pffiffiffi ln k ¼ ¼ pffiffiffi  pffiffiffi þ pffiffiffi ln 2 ry 2 r 3 4 3k 3 3 y 1 ðk þ 1Þ2 2 k þ1 2 k2  1 4. When k ≤ ka, pffiffiffi  pffiffiffi þ pffiffiffi ln ; \ pffiffiffi 2 3 4 3k 2 3 3 k2

  1 ðk þ 1Þ2 2 k þ1 2 k2  1 [ pffiffiffi þ pffiffiffi ln , but when k ≥ ka, r0ei  [ ry When k ≥ ka, pffiffiffi  pffiffiffi 2 2 2 k 3 4 3k 3 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi rT rT 1 hj ej 5. When k  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3  1k = xa, and  = 1.406076 = kd, kj  pffiffiffi r r y y 311 T 6. When k ≤ kz = 2.654911…, p ≤ σy and rri ≥ −σy; When k ≥ kz, p ≥ σy and rT ri ≤ −σy; When k = kz, p = σy and rT ri = −σy When k ≤ ka, there must be k ≤ kz 7. When k ≥ ka, rT e > σy qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi k þ1 ¼ 8. When kj ¼ 3  1k, or k = kd, xz = xa = kj 2 pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi (When kj ¼ k ¼ 3  1k, or k = kg, xz = xa = kj; When k = kf, xz = xa = kj*.) pa 1 ðk þ 1Þ2 2 k þ1 p ¼ pffiffiffi  pffiffiffi þ pffiffiffi ln ¼ ry 2 ry 3 4 3k 2 3   pffiffiffi p 1 ðk þ 1Þ2 2 k þ1 k þ1 p k1 1 10. ¼ pffiffiffi  pffiffiffi þ pffiffiffi ln ¼ pffiffiffi þ pffiffiffi ln k (kj ¼ k ) kj ¼ > 2 ry 2 2 ry 3 4 3k 3 3k 3 9. Autofrettage pressure

Mechanical Autofrettage Technology Based on Mises Yield Criterion

157

TAB. 3.2 – (continued).

pffiffiffi The results under the case that kj ¼ k  0  0 1. When k ≤ kb = 3.042297…, rei   ry ; When k ≥ kb, rei   ry pffiffiffi 2. When k = kb, kj ¼ k = 1.744218… 3. The maximum load-bearing capacity and autofrettage pressure p k1 1 ¼ pffiffiffi þ pffiffiffi ln k ry 3k 3   pffiffiffi p k1 1 p 1 ðk þ 1Þ2 2 k þ1 k þ1 kj ¼ ¼ pffiffiffi þ pffiffiffi ln k (kj ¼ k ) < ¼ pffiffiffi  pffiffiffi þ pffiffiffi ln 4. ry ry 2 2 3k 3 3 4 3k 2 3 pffiffiffi pffiffiffi k1 2 k2  1 k1 2 k2  1 5. When k ≤ kb, pffiffiffi þ ln k \ pffiffiffi ; when k ≥ kb, pffiffiffi þ ln k [ pffiffiffi , but 3 k2 3k 3 k2  0 3k when k ≥ kb, rei  [ ry ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pp ffiffiffi rT rT hj ej 6. When k ≥ kg, kj  3  1k, accordingly  ry ry ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pp ffiffiffi 7. When kj ¼ k ¼ 3  1k, or k = kg, xz = xa = kj qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi k þ1 (When kj ¼ ¼ 3  1k, or k = kd, xz = xa = kj; When k = kf, xz = xa = kj*.) 2 pffiffiffi  3p rj The results under the case that kj ¼ exp 2 ry ri pffi  1. Overstrain e ¼ 2. When kj ¼

exp

3p 2 ry

1

pffiffi k1  exp 23 rpy ,

the equivalent stress σej is the minimum, that is pffiffiffi  2 exp 3 rpy  1 rT k k 2 kj2  1 ej pffiffiffi  ¼ 2 ¼ 2 ry k  1 exp 3 p k  1 kj2 ry

3. The allowable load for r0ei =ry   1 when k ≥ kc pffiffiffi  pffiffiffi exp 3 rpy þ k 2  3k 2 rpy  2  0 or k 2 

 exp



pffiffi p 3ry 2 pffiffi p 3ry 1

4. The relationship between autofrettage pressure and the allowable load   pffiffiffi p pa p 1 1 ¼ þ pffiffiffi  pffiffiffi exp 3 ry ry ry 3 3k 2 pa p  ry ry

References [1] Liu H.W. (2019) Mechanics of materials (in Chinese). Higher Education Press, Beijing. [2] Yu G.C. (1980) Chemical pressure vessels and equipment (in Chinese). Chemical Industrial Press. [3] Zhu R.L., Yang J.L. (1998) Autofrettage of thick cylinders, Int. J. Press. Vessels Pip. 75(6), 443. [4] Zhu R.L. (2008) Results resulting from autofrettage of a cylinder, Chin. J. Mech. Eng. 21(4), 105.

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Autofrettage Technology and Its Applications in Pressured Apparatuses

[5] Zhu R.L. (2008) Ultimate load-bearing capacity of cylinder derived from autofrettage under ideal condition, Chin. J. Mech. Eng. 21(5), 80. [6] Zhu R.L. (2013) Study on autofrettage for medium-thick pressure vessels, J. Eng. Mech. 139 (12), 1790. [7] Zhu R.L., Zhu G.L., Mao A.F. (2016) Plastic depth and load-bearing capacity of autofrettaged cylinders, J. Mech. Sci. Tech. 30(6), 2627. [8] (1979) The compiling group of “handbook of mathematics”. A handbook of mathematics (in Chinese). Higher Education Press, Beijing.

Chapter 4 Mechanical Autofrettage Technology by Limiting Circumferential Residual Stress Based on Mises Yield Criterion When the maximum distortion strain energy theory (Mises’s yield criterion) is applied to study the autofrettage for pressure vessels, there are some problems needed to be studied further. Based on our previous work[1–3], this chapter is intended to do relevant research with the help of the mathematical analysis of ref.[4].

4.1

The Optimum Plastic Depth When Circumferential Residual Stress on the Inside Surface Controlled

The residual stresses based on the maximum distortion strain energy theory (Mises’s criterion) are given as equations (3.2)–(3.9). The depth of plastic region (kj) affects the residual stresses, the greater kj is, the greater the equivalent residual stress r0e is, so it is inadvisable to simply raise kj, otherwise, compressive yield may occur after removing the autofrettage pressure, pa. However, raising kj can raise the load-bearing capacity of a pressure vessel, thereupon, in our previous studies, we put forward an equation for kj*, the maximum plastic depth (kj) for a certain k to ensure r0ei ≥ −σy: 2 2 k 2 ln kj  k 2  kj þ2 ¼ 0

ð4:1Þ

where e0.5 ≤ kj* ≤ kc = 2.2184574899167… and k ≥ kc. When k ≤ kc, r0ei ≤ −σy never 2 2 occurs irrespective of kj. Equation (4.1) is equation (2.34). k 2 ln kj  k 2  kj þ2 ¼ 0 is applicable for both the maximum shear stress theory (Tresca yield criterion) and the maximum distortion strain energy theory (Mises’s yield criterion). When kj is determined by equation (4.1)—we write kj = kj*, and according to the maximum distortion strain energy theory, the equations for residual stresses are found in equations (3.43)–(3.50). DOI: 10.1051/978-2-7598-3111-1.c004 © Science Press, EDP Sciences, 2023

160

Autofrettage Technology and Its Applications in Pressured Apparatuses

By taking k = 3, kj = 1.5 < kj*, kj = 1.748442 = kj* and kj = 2 > kj*, the residual stresses based on the maximum distortion strain energy theory are illustrated as shown in figure 4.1a–c, respectively, to show the effect of kj on residual stresses. In figure 4.1, points a, b, and c are intersections of the three curves for the qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2 kj2 þ k 2 ln kj2 r residual stresses, their abscissas are calculated by x0 ¼ ri ¼ \kj k 2 1 (equation (2.22) or equation (3.22)), their coordinates are a (1.325159, −0.11142), b pffiffiffi (x ¼ 2, −0.17717) and c (1.478033, −0.23276). When k = 3, kj = 1.5 < kj*, both |r0ei | and |r0hi | are below σy. When k = 3, kj = 1.748442 = kj*, |r0ei | just reach σy, but |r0hi | exceeds σy. When k = 3, kj = 2 > kj*, both |r0ei | and |r0hi | exceed σy. Small kj causes small residual stresses and the effect of autofrettage is not obvious; Great kj causes great residual stresses and the effect of autofrettage is obvious, but there will be reversed yielding after releasing autofrettage pressure pa if the plastic depth is too deep. For this reason, we presented a formula to control kj so that |r0ei | ≤ σy, that is equation (4.1). Nevertheless, we have found that when kj = kj*, at the inside surface, |r0ei | and |r0hi | just simultaneously reach σy according to the maximum shear stress theory (Tresca yield criterion), and |r0hi | exceeds σy when |r0ei | just reaches σy according to the maximum distortion strain energy theory (Mises’s yield criterion) although kj = kj*. The reasons are as follows. When based on the maximum distortion strain energy theory, with the help of ref.[5], we obtain the equivalent stress rde as follows, pffiffiffi 3 2 d ð4:2Þ ðrh  rr Þ or rh  rr ¼ pffiffiffi rde re ¼ 2 3 When rde reaches the strength σy, the material becomes yielded. At the inside surface (r = ri), the radial residual stress r0ri = 0, then, circumferential residual stress r0hi ¼ p2ffiffi3 ry , which exceeds the strength σy. This may lower the safety of a pressure vessel. So, it is necessary to limit |r0hi | in order to find the maximum plastic depth kj for |r0hi | to be below σy, such kj is written as kjθ. Therefore, letting  pffiffi3ðk22 1Þ ð1



kj2

þk

2

ln kj2 Þ 

r0hi ry

¼

 1 obtains:

pffiffiffi pffiffiffi 2 2k 2 ln kjh  2kjh2  3k 2 þ ð2 þ 3Þ  0 or vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  u  u2 k 2  1  pffiffi3ffi pffiffi u jh pffiffiffi ðe 3=4  kjh  kch ; k  kch Þ k¼t 2 ln kjh2  3

ð4:3Þ

pffiffiffi Replacing 3 with 2 just leads to k 2 ln kj2  k 2  kj2 þ 2 ¼ 0. This is the result of letting r0ei = −σy. Limiting circumferential stress is essentially the maximum principal stress theory.

Mechanical Autofrettage Technology by Limiting Circumferential

161

FIG. 4.1 – Distributions of the residual stresses based on the maximum distortion strain energy theory.

162

Autofrettage Technology and Its Applications in Pressured Apparatuses

From equation (4.3), k ! 1; kjh ¼ e

pffiffi 3=4

¼ 1:541896. . . ¼ kjh1

Letting kjθ = k in equation (4.3) obtains: pffiffiffi pffiffiffi k 2 ln k 2 2 þ 3 k 2 ln k 2 þ 3 ¼ ¼ or k2  1 k2  1 2 4

ð4:4Þ

ð4:5Þ

The solution of equation (4.5) is k ¼ 2:024678965. . . ¼ kch \kc ¼ 2:2184574899167

ð4:6Þ

Therefore, when k ≤ kcθ, no matter how deep the depth of the plastic region is, even if the whole cylinder is yielded, or kj = k, |r0hi | can not exceed σy after removing pa from the cylinder; when k > kcθ, if kj is not greater than the value determined by equation (4.3), or kj ≤ kjθ, |r0hi | will not exceed σy after removing pa, otherwise, |r0hi | will exceed σy after removing pa. Clearly, when k = kcθ, kjθ = k = kcθ. pffiffi From equation (4.3), the domain of k is kjh  e 3=4 = kjθ∞. and kjh  qffiffiffiffiffiffiffiffiffiffi pffiffi 2þ 3 = 1.366025…. 2 The relation between kjθ and k or equation (4.3) is shown in figure 4.2. The graph of equation (4.3) has two branches: curve od and curve abc. Curve od corresponds to qffiffiffiffiffiffiffiffiffiffi pffiffi pffiffi kjh  2 þ2 3 = 1.366025; Curve abc corresponds to kjh  e 3=4 = 1.541896…. For curve od, k < 1. This is unreasonable and, thus, it is of no significance in practice. For section ab of curve abc, kj > k, therefore, this section is of no significance in practice. The meaningful control curve is section bc of curve abc. At point b, kj = k = kcθ = 2.024678965….

FIG. 4.2 – The graph of relation between kjθ and k.

Mechanical Autofrettage Technology by Limiting Circumferential

163

Limiting r0ei ≥ −σy, we get k 2 ln kj2  k 2  kj2 þ 2 ¼ 0 (e0.5 ≤ kj* ≤ kc and k ≥ kc). pffiffiffi pffiffiffi 2 2  2kjh  3k 2 þ ð2 þ 3Þ ¼ 0 is For comparison, the meaningful part of 2k 2 ln kjh plotted along with the meaningful part of k 2 ln kj2  k 2  kj2 þ 2 ¼ 0 in figure 4.3.

FIG. 4.3 – The plastic depth. From figure 4.3, it is seen that for the same k, kj* > kjθ. To avoid r0hi ≤ −σy, kjθ should be smaller than kj*. pffiffiffi In figure 4.3, kj ¼ k þ2 1 and kj ¼ k are plotted along with p ffiffi ffi p ffiffi ffi 2  3k 2 þ ð2 þ 3Þ ¼ 0 and k 2 ln kj2  k 2  kj2 þ 2 ¼ 0, where curve 2k 2 ln kjh2  2kjh pffiffiffi pffiffiffi obwmnc is for 2k 2 ln kjh2  2kjh2  3k 2 þ ð2 þ 3Þ ¼ 0, curve obafgd is for pffiffiffi k 2 ln kj2  k 2  kj2 þ 2 ¼ 0, curve omfs is for kj ¼ k þ2 1 and curve ongh is for kj ¼ k .  pffiffi  pffiffiffi Points c and d are at infinity, their coordinates are c 1, e 3=4 and d ð1; eÞ. Point a is the intersection point of straight line oba (kj = k) and curve afgd (k 2 ln kj2  k 2  kj2 þ 2 ¼ 0), its coordinate is a (kc, kc). Point b is the intersection point of straight line oba (kj = k) and curve bwmnc pffiffiffi pffiffiffi 2 (2k ln kjh2  2kjh2  3k 2 þ ð2 þ 3Þ ¼ 0), its coordinate is b (kcθ, kcθ). Point f is the intersection point of straight line omfs (kj ¼ k þ2 1) and curve afgd (k 2 ln kj2  k 2  kj2 þ 2 ¼ 0), its coordinate is f (ka, 1.80913) = f (2.618253, 1.80913). pffiffiffi Point g is the intersection point of curve ongh (kj ¼ k ) and curve afgd (k 2 ln kj2  k 2  kj2 þ 2 ¼ 0), its coordinate is g (kb, 1.74422) = g (3.042297, 1.74422). When based on the maximum distortion strain energy theory and kj ¼ k þ2 1, from equation (3.108),   r0hi 1 k þ3 4k 2 2 þ ln ¼ pffiffiffi ð4:7Þ ry 3 2ðk þ 1Þ k 2  1 k þ 1

164

Autofrettage Technology and Its Applications in Pressured Apparatuses

It is easy to prove that when k → 1, figure 4.4.

r0hi ry

FIG. 4.4 – The relationship between

Letting

r0hi ry

! 0. Equation (4.7) is plotted in

r0hi ry

and k when kj ¼ k þ2 1.

¼ 1 obtains 8k 2 ln

pffiffiffi pffiffiffi k þ1 ¼ ð2 3 þ 1Þk 2 þ 2k  3  2 3 2

ð4:8Þ

The solution of equation (4.8) is k  2:360344. . . ¼ kah \ka

ð4:9Þ

When based on the maximum distortion strain energy theory and kj ¼ k þ2 1, for r0hi ≥ −σy, k ≤ kaθ or k ≤ 2.36035…. pffiffiffi 2  3k 2 þ On the other hand, substituting kj ¼ k þ2 1 into 2k 2 ln kjh2  2kjh pffiffiffi ð2 þ 3Þ ¼ 0 also results in equation (4.8). When k = kaθ, kj ¼ k þ2 1 = 1.680175…. Therefore, in figure 4.3, point m is the intersection point of straight line omfs pffiffiffi pffiffiffi (kj ¼ k þ2 1) and curve bwmnc (2k 2 ln kjh2  2kjh2  3k 2 þ ð2 þ 3Þ ¼ 0), its coordinate is m (kaθ, 1.680175) = m (2.360344, 1.680175). pffiffiffi When based on the maximum distortion strain energy theory and kj ¼ k , from equation (3.103), r0hi 2 k  1  k 2 ln k ¼ pffiffiffi k2  1 ry 3

ð4:10Þ

Mechanical Autofrettage Technology by Limiting Circumferential It is easy to prove that when k → 1, figure 4.5.

r0hi ry

! 0. Equation (4.10) is plotted in

FIG. 4.5 – The relationship between

Letting

r0hi ry

165

r0hi ry

and k when kj ¼

pffiffiffi k.

¼ 1 obtains 2k 2 ln k  2k 

pffiffiffi 2 pffiffiffi 3k þ 2 þ 3 ¼ 0

ð4:11Þ

The solution of equation (4.11) is k ¼ 2:65972. . . ¼ kbh \kb

ð4:12Þ pffiffiffi When based on the maximum distortion strain energy theory and kj ¼ k , for r0hi ≥ −σy, k ≤ kbθ or k ≤ 2.65972…. pffiffiffi pffiffiffi 2 On the other hand, substituting kj ¼ k into 2k 2 ln kjh2  2kjh  3k 2 þ pffiffiffi pffiffiffi ð2 þ 3Þ ¼ 0 also results in equation (4.11). When k = kbθ, kj ¼ k = 1.630865…. pffiffiffi Therefore, in figure 4.3, point n is the intersection point of curve ongh (kj ¼ k ) and p ffiffi ffi p ffiffi ffi curve bwmnc (2k 2 ln kjh2  2kjh2  3k 2 þ ð2 þ 3Þ ¼ 0), and its coordinate is n (kbθ, 1.630865) = n (2.65972, 1.630865). k Converting the relation of kj  k into the relation of kj  k, we obtain k

2 2 figure 4.6. In figure 4.6, curve abg is for kj obtained from 2k 2 ln kjh  2kjh  pffiffiffi 2 pffiffiffi kj 2 2 2 2 3k þ ð2 þ 3Þ ¼ 0; curve abcf is for k obtained from k ln kj  k  kj þ 2 ¼ 0; pffiffiffi k k curve ad is for kj obtained from kj ¼ k þ2 1 and curve ae is for kj obtained from kj ¼ k .

Autofrettage Technology and Its Applications in Pressured Apparatuses

166

FIG. 4.6 – The relation of

kj k

 k.

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 3  1k into 2k 2 ln kjh  2kjh2  3k 2 þ ð2 þ 3Þ ¼ 0 obtains pffiffiffi pffiffiffi pffiffiffi ð4:13Þ 2k 2 ln½ð 3  1Þk 2   ð3 3  2Þk 2 þ 2 þ 3 ¼ 0

Substituting kj ¼

Solving equation (4.13) results in k ¼ 2:105406. . . ¼ kfh \kf

ð4:14Þ

kfθ is marked in figures 4.3 and 4.6. That is to say when kj is restrained by pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 2 pffiffiffi ffiffiffi 3k þ ð2 þ 3Þ ¼ 0, if k ≥ kfθ, kj  3  1k, accordingly

2  2k 2 ln kjh2  2kjh rT ej ry

rT

 rhjy . In order to facilitate the engineering application, some data calculated by 2  equation (4.3) are listed in table 4.1 along with some data calculated by k 2 ln kj 2 2 k  kj þ 2 ¼ 0 for comparison.

4.2

4.2.1

The Distribution of Residual Stresses When Circumferential Residual Stress on the Inside Surface Controlled General Discussion

When kj = kjθ, the residual stresses within the plastic region are as follows.  2 r0z 1 r 1 1 1 p ffiffi ffi ð4:15Þ ¼  ¼ pffiffiffi ln x 2  ln r 2 2 ry 3 3 i

kjθ 2.024678965 = kcθ 1.882849 1.837525 1.806908 1.764911 1.735954 1.72725 1.71415 1.696894 1.682783 1.670971 1.660907 1.652212 1.644614 1.637911 1.63195 1.626613 1.621806 1.617453 1.613492 1.609874 1.606556 1.603502 1.591296 1.582618 1.576167 1.571211

kjθ/k 1 0.918463 0.885554 0.860432 0.820889 0.78907 0.778582 0.761844 0.73778 0.716078 0.696238 0.677921 0.660885 0.644947 0.629966 0.61583 0.602449 0.589748 0.577662 0.566138 0.555129 0.544595 0.534501 0.48963 0.452177 0.420311 0.392803

kj* – – – – – – 2.2184574899167 = kc 2.046308 1.968122 1.92208 1.889454 1.864437 1.844363 1.827762 1.813731 1.801676 1.791183 1.781952 1.773758 1.766432 1.759838 1.75387 1.748442 1.7273 1.712755 1.702177 1.694172

kj*/k – – – – – – 1 0.90947 0.855705 0.817906 0.787273 0.760995 0.737745 0.716769 0.697589 0.679878 0.663401 0.647983 0.633485 0.619801 0.606841 0.594532 0.582814 0.531477 0.489359 0.453914 0.423543

kjθ/kj* – – – – – – 0.778582 0.837679 0.862189 0.875501 0.884367 0.890836 0.895817 0.899797 0.903062 0.905795 0.908122 0.910129 0.911879 0.913419 0.914785 0.916006 0.917103 0.921262 0.924019 0.925971 0.927421

167

k 2.024678965 = kcθ 2.05 2.075 2.1 2.15 2.2 2.2184574899167 = kc 2.25 2.3 2.35 2.4 2.45 2.5 2.55 2.6 2.65 2.7 2.75 2.8 2.85 2.9 2.95 3 3.25 3.5 3.75 4

Mechanical Autofrettage Technology by Limiting Circumferential

TAB. 4.1 – Some data of kjθ and kj*.

∞

kjθ 1.564158 1.559444 1.556118 1.554108 1.553673 pffiffiffi 3 exp 4

kjθ/k 0.347591 0.311889 0.282931 0.263408 0.258946

kj* 1.682956 1.675565 1.670397 1.667294 1.666625

kj*/k 0.37399 0.335113 0.303709 0.282592 0.277771

0

pffiffiffi e

0

kjθ/kj* 0.929411 0.930697 0.931586 0.932114 0.932227 pffiffiffi pffiffiffi 3 e/exp = 0.935207 4

Autofrettage Technology and Its Applications in Pressured Apparatuses

k 4.5 5 5.5 5.9 6

168

TAB. 4.1 – (continued).

Mechanical Autofrettage Technology by Limiting Circumferential

169

pffiffiffi pffiffiffi  r0r 1 2þ 3 2þ 3 2 ¼ pffiffiffi ln x þ  ry 2x 2 2 3

ð4:16Þ

pffiffiffi pffiffiffi  r0h 1 2þ 3 2 3 2 ¼ pffiffiffi ln x  þ ry 2x 2 2 3

ð4:17Þ

pffiffiffi pffiffiffi 3þ2 3þ2 r0e ¼1 ¼1 2 2x 2 ry 2ðr=ri Þ

ð4:18Þ

It is seen that equations (4.15)–(4.18) are unconcerned with kj and k, this means that when r0hi is controlled, or kj = kjθ, the residual stresses, and their equivalent stress are determined only by radial relative location (r/ri = x) and independent of kj and k, or at a certain x, the residual stresses and their equivalent stress are identical for any kj and k. Equation (4.18) shows the following conclusion. qffiffiffiffiffiffiffiffiffiffi pffiffi 3þ2 When x  = 1.366025, r0e ≤ 0, at the inside surface where x = 1, 2 pffiffi 0 rei dr0e 3 ry ¼  2 , that is the minimum within the whole wall because dx [ 0; when ffiffiffiffiffiffiffiffiffiffi qp ffiffi 3þ2 0 0 x 2 , re ≥ 0; within the whole wall of a pressure vessel, jre j\ry ; at the ffiffi p 0 r 3þ2 0\ rejy ¼ 1  2k since elastic-plastic juncture where x = kj, 2 \1, j pffiffi p ffiffi ffi kj2  exp 23 [ 1 þ 3=2, this means that r0ej is tension and can not reach σy. qffiffiffiffiffiffiffiffiffiffi pffiffi 3þ2 0 0 0 0 At x ¼ rri ¼ 2 , re = 0, and this is just the location where rh = rr = rz . That is to say, when kj = kjθ, no matter how great k is, the three curves for the residual stresses collect at the same qffiffiffiffiffiffiffiffiffiffi  point within the plastic region and the interpffiffi pffiffi 3 þ 2 p1ffiffi ln 32þ 2  12 = (1.366025…, −0.13984…). section is: 2 ; 3 We once presented the location (x0) of the intersection of the three residual stress curves or the abscissa at which r0e = 0 in general form, that is as follows r x0 ¼ ¼ ri

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2  kj2 þ k 2 ln kj2 \kj k2  1

ð4:19Þ

Equation (4.19) is equation (2.10). When k and kj meet the relation expressed by equation (4.3), x0 just becomes x0h

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3þ2 ¼ 2

ð4:20Þ

170

Autofrettage Technology and Its Applications in Pressured Apparatuses

Setting r0z = 0 obtains

pffi 3 x1h ¼ e 4 ¼ kjh1

ð4:21Þ

When kj = kjθ, the residual stresses within the elastic region are as follows, pffiffi 2þ 3 r0z kj2  2 ¼ pffiffiffi ð4:22Þ ry 3k 2 r0r ¼ ry r0h ¼ ry

1

1þ

k2

!

ðr=ri Þ2 k2 ðr=ri Þ2

!

r0z ¼ ry r0z ¼ ry

 1

 k 2 r0z x 2 ry

ð4:23Þ

1þ

 k 2 r0z x 2 ry

ð4:24Þ



pffiffiffi r0e kj2  ð 3 þ 2Þ=2 ¼ x2 ry

ð4:25Þ

The distribution of equivalent residual stress within the whole wall is demonstrated in figure 4.7, the related parameters are shown in the figure.

FIG. 4.7 – The distribution of equivalent residual stress within the whole wall for various kjθ. For figure 4.7, we make the following explanation. (1) Curve baa: This is the curve of the equivalent residual stress when k = kjθ = kcθ = 2.024678965…. In this case, the cylinder is entirely yielded, or there is no elastic zone in the wall, x varies from 1 to kjθ (=kcθ) within the plastic region (from point b to point a). When the plastic depth is controlled by

Mechanical Autofrettage Technology by Limiting Circumferential

(2)

(3)

(4)

(5)

(6)

(7)

171

equation (4.3), or kj = kjθ, all curves of the equivalent residual stresses in the plastic zone for any k and kjθ are located on curve ba, starting from point b and ending at some point before point a. Curve bcd: k = 2.101833…, kjθ = 1.805…. The curve of the equivalent residual stress within the plastic region is curve bc and the curve of the equivalent residual stress within the elastic zone is curve cd. x varies from 1 to kjθ = 1.805… within the plastic region (point b to c), and from kjθ to k (point c to d) within the elastic zone. Curve bef: k = 2.2, kjθ = 1.7359545…. The curve of the equivalent residual stress within the plastic region is curve be and the curve of the equivalent residual stress within the elastic zone is curve ef. x varies from 1 to kjθ = 1.7359545… within the plastic zone (point b to e), and from kjθ to k (point e to f) within the elastic zone. Curve bgh: k = 2.5, kjθ = 1.652212…. The curve of the equivalent residual stress within the plastic zone is curve bg and the curve of the equivalent residual stress within the elastic zone is curve gh. x varies from 1 to kjθ = 1.652212… within the plastic zone (point b to g), and from kjθ to k (point g to h) within the elastic zone. Curve bij: k = 3, kjθ = 1.603502…. The curve of the equivalent residual stress within the plastic zone is curve bi and the curve of the equivalent residual stress within the elastic zone is curve ij. x varies from 1 to kjθ = 1.603502… within the plastic zone (point b to i), and from kjθ to k (point i to j) within the elastic zone. Curve bkl: k = 4, kjθ = 1.571211…. The curve of the equivalent residual stress within the plastic zone is curve bk and the curve of the equivalent residual stress within the elastic zone is curve kl. x varies from 1 to kjθ = 1.571211… within the plastic zone (point b to k), and from kjθ to k (point k to l) within the elastic zone. pffiffi Curve bmn: k = ∞, kjh ¼ e 3=4 ¼ 1.541896:::. The curve of the equivalent residual stress within the plastic zone is curve bm and the curve of the equivalent residual stress within the elastic zone is curve mn. x varies from 1 to pffiffi kjh ¼ e 3=4 ¼ 1.541896::: within the plastic zone (point b to m), and from kjθ to ∞ (point m to n) within the elastic zone.

The coordinates  pffi of some key points in figure 4.7 are a (kcθ, 0.544797…),  pffiffi pffiffi 3 3 3 þ 2 b 1;  2 , m e 4 ; 1  pffi3 = m (1.541896, 0.215112…), n (∞, 0), qffiffiffiffiffiffiffiffiffiffi  2e 2 pffiffi 3þ2 v ð x 0 ; 0Þ = v 2 ; 0 . From figure 4.7 and equation (4.18), it is known that all curves of the equivalent residual stresses for any k and kjθ within the plastic are qregion  located on the ffiffiffiffiffiffiffiffiffiffi pffiffi 3þ2 identical curve ba and pass through the same point v 2 ; 0 , except that the different curve for different k and kjθ is located on a different section of curve ba. Saying, the above curves for the plastic zone, ba (k = kjθ = kcθ), bc (k = 2.101833…, kjθ = 1.805…), be (k = 2.2, kjθ = 1.7359545…), bg (k = 2.5, kjθ = 1.652212…), bi pffiffi (k = 3, kjθ = 1.603502…), bk (k = 4, kjθ = 1.571211…) and bm (k = ∞, kjh ¼ e 3=4 ) are all on curve ba, or they coincide with each other.

172

Autofrettage Technology and Its Applications in Pressured Apparatuses

Corresponding to figures 4.7–4.10 are axial residual stress, radial residual stress, and circumferential residual stress respectively.

FIG. 4.8 – The distribution of axial residual stress within the whole wall for various kjθ. r0

When k = ∞, in the elastic region, ryz ¼ 0. The coordinates of some key points in  pffi3  figure 4.8 are a (kcθ, 0.314539…), b (1, −0.5), m e 4 ; 0 = m (1.541896, 0), n (∞, 0).

FIG. 4.9 – The distribution of radial residual stress within the whole wall for various kjθ. pffiffi pffiffi k 2 ð 3 þ 2Þ=2 r0 When k = ∞, kj ¼ exp 43 , in the elastic region, ryr ¼  j pffiffi3x 2 ¼ pffi pffiffi 3 exp 2 ð 3 þ 2Þ=2 pffiffi  . The coordinates of some key points in figure 4.9 are a (kcθ, 0), b 2 3x pffi3  (1, 0), m e 4 ; 0.124195::: = m (1.541896, −0.124195), n (∞, 0).

Mechanical Autofrettage Technology by Limiting Circumferential

173

FIG. 4.10 – The distribution of circumferential residual stress within the whole wall for various kjθ.

pffiffi 3 4 , in the elastic region, pffiffi pffiffiffi pffiffiffi 3 2 0 exp  ð 3 þ 2Þ=2 2 rh kj  ð 3 þ 2Þ=2 pffiffiffi pffiffiffi ¼ ¼ ry 3x 2 3x 2

When k = ∞, kj ¼ exp

coordinates  The  of some key points in figure 4.10 are a (kcθ, 0.629077), b (1, −1), pffi 3 4 m e ; 0.124195::: = m (1.541896, 0.124195), n (∞, 0) and s (xc, 0) = s r0

(1.4540151, 0). xc is the solution of equation (4.26) that is obtained by letting ryh ¼ 0 for equation (4.24).  pffiffiffi pffiffiffi ð4:26Þ 2x 2 ln x 2 þ 2  3 x 2 þ 2 þ 3 ¼ 0 In figures 4.8–4.10, the curves with the same labeled letters represent the residual stresses under the same k and kjθ, saying curves bij in figures 4.8–4.10 represent the residual stresses under k = 3 and kjθ = 1.603502, i.e. curve bij in figure 4.8 is the axial residual stress when k = 3 and kjθ = 1.603502; curve bij in figure 4.9 is the radial residual stress when k = 3 and kjθ = 1.603502; curve bij in figure 4.10 is the circumferential residual stress when k = 3 and kjθ = 1.603502, and so on. About the terminal connection curve (dotted curve) in figures 4.8–4.10: For a r0zo r0ho r0eo ry , ry and ry , r0 curve, rroy  0 for any

certain kjθ, from equation (4.3) we get k, letting x = k, we get connecting all k and kjθ.

r0zo ry

(or

r0ho r0eo ry , ry )

we get the terminal connection

174

Autofrettage Technology and Its Applications in Pressured Apparatuses

Figure 4.11 shows the axial residual stress together with radial residual stress and circumferential residual stress for k = 3 and kjθ = 1.603502.

FIG. 4.11 – The axial, radial, and circumferential residual stress for k = 3 and kjθ = 1.603502.

The coordinate  of point a in figure 4.11 is a (x0, −0.13984) = q ffiffiffiffiffiffiffiffiffiffi ffiffi p 3þ2 = a (1.366025, −0.13984), the coordinate of point c in a 2 ; 0:13984 figure 4.11 is c (xa, 0.107035) = c (2.566799, 0.107035), where xa is the abscissa of pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi r0 r0 the intersection point of curves ryh and rye , xa ¼ 3  1k. However, if kj ≠ kjθ, or the correlation between kj and k does   0  not meet equar0hi  r  tion (4.3), the above features do not exist, or even  ry >1 or  reiy >1. This case is illustrated in figure 4.12, where curves 1 and 2 coincide with each other in the plastic region and both pass through the point o (xc, 0) because kj = kjθ, but curves 3 and 4 do not coincide with each other in the plastic region and do not pass through the point o (xc, 0), and they do not coincide with curve 1 and 2 either because kj ≠ kjθ. In figure 4.12, curve 1 is curve bij in figure 4.10; curve 2 is curve bkl in figure 4.10; curve 3 is for k = 4 and kj = 1.7 > kjθ; curve 4 is for k = 4 and kj = 1.5 < kjθ. Curves 3 and 4 are plotted from equations (3.4) and (3.8). When k = 4, kj = 1.7 > kjθ, |r0hi | > σy; When k = 4, kj = 1.5 < kjθ, jr0hi j\ry , but the load-bearing capacity will be smaller than that under the case of kj = kjθ. When kj = kjθ, within the plastic region, all curves of the residual stresses for any k and kjθ are located on the same curve, the greater kjθ is, the longer the curve of the residual stress for this kjθ is. Because the deepest plastic depth is kjθ = kcθ, the longest curve of the residual stress is the one with kjθ = kcθ, see curve ba in figures 4.7–4.10, where x varies from 1 to kjθ = kcθ = 2.024678965. The distributions of the equivalent residual stress, axial residual stress, radial residual stress, and

Mechanical Autofrettage Technology by Limiting Circumferential

175

FIG. 4.12 – Comparison between the circumferential residual stresses under different k and kj.

circumferential residual stress within the whole wall for different k and kjθ are illustrated in figures 4.7–4.10, respectively. For comparison, when k = 3, kj* = 1.748442… and k = 3, kjθ = 1.603502…, the residual stresses are plotted in figure 4.13a; when k = 4, kj* = 1.694172… and k = 4, kjθ = 1.571211…, the residual stresses are plotted in figure 4.13b. In figure 4.13, solid curves, which ensures jr0hi j\ry , are for kj = kjθ; and dash curves, which ensures jr0ei j  ry , are for kj = kj*. qffiffiffiffiffiffiffiffiffiffi   pffiffi pffiffi 0 r0hx0 r0rx0 r0rmin 3 þ 2 rzx0 3þ2 1 p1ffiffi When kj = kjθ, x0 ¼ , as a 2 , ry ¼ ry ¼ ry ¼ ry ¼  2  3 ln 2 result, irrespective of k, the three for residual stress qcurves  collect at the fixed point ffiffiffiffiffiffiffiffiffiffi ffiffi p pffiffi pffiffiffi 3 þ 2 p1ffiffi ln 32þ 2  12 . When kj = kj*, x0 ¼ 2, within the plastic region: o 2 ; 3 r0zx0 ry

r0hx0 ry

r0rx0 ry

r0rmin ry

pffiffi 2, as a result, irrespective of k, the three curves for ¼  1ln 3   pffiffiffi pffiffi 2 . 2;  1ln residual stress collect at the fixed point within the plastic region: o0 3

¼

¼

¼

Since kjθ < kj* for the same k, the residual stresses under the case kj = kjθ are smaller than those under the case kj = kj*. pffiffiffi r0 r0 When kj = kj*, rhiy ¼  p2ffiffi3 \  1. To make rhiy = −1, just replace 3 with 2. In pffiffiffi fact, on many occasions, replacing 3 in the results from the case kj = kjθ with 2 can pffiffiffi obtain the results in the case kj = kj*. For examples, replacing 3 in pffiffiffi 2 pffiffiffi 2 2 2 2k ln kjh  2kjh  3k þ ð2 þ 3Þ ¼ 0 (equation (4.3)) with 2 just obtains qp ffiffiffiffiffiffiffiffiffiffi ffiffi pffiffiffi 3þ2 2 2 2 2 k ln kj  k  kj þ 2 ¼ 0 (equation (2.34)); replacing 3 in x0h ¼ (equa2 pffi pffiffiffi pffiffiffi 3 tion (4.20)) with 2 just obtains x0 ¼ 2; replacing 3 in x1h ¼ e 4 (equation (4.21)) pffiffiffi with 2 just obtains x0 ¼ e.

Autofrettage Technology and Its Applications in Pressured Apparatuses

176

FIG. 4.13 – Comparison between residual stresses when kj = kjθ and kj = kj*.

4.2.2

The Residual Stresses for Entire Yield

Under the entire yield state, or if kj = k, when r0ei = −σy (in this case kj = k = kc), r0

hi r0hi < −σy (see figure 3.23). To control r0hi   ry , setting ry ¼   2 2 2 p1ffiffi 2 þ ln 12  ln2 k  k 2ln k   1 (see equation (3.59)) results in equation (4.5). k 1 k 1 k 3

Thereupon, when kj = k, to control r0hi   ry , we only need to limit k ≤ kcθ, if

Mechanical Autofrettage Technology by Limiting Circumferential

177

k ≤ kcθ, r0hi   ry when kj = k. Figures 4.14 and 4.15 show the distributions of the residual stresses for k = kcθ = 2.024678965 and k = 1.5 < kcθ, respectively. Compared with figure 3.20, |r0hi | has been restrained within σy.

FIG. 4.14 – The distributions of the residual stresses for kj = k and k = kcθ.

FIG. 4.15 – The distributions of the residual stresses for kj = k and k = 1.5 < kcθ.

4.2.3

The Residual Stresses with Radius of Elastic-Plastic Juncture Being Arithmetic Mean Radius of Inside Radius and Outside Radius

When kj ¼ k þ2 1 and k = ka = 2.618253, r0ei ¼ ry , but r0hi \  ry (see figure 3.32). h   2 i 2 r0 þ 1Þ2 þ 1Þ2 1 From equation (3.88), rhiy ¼ p1ffiffi3 ðk 4k þ 1 þ 2 ln k þ2 1  1  ðk 4k þ ln ðk þ4 1Þ kk 2 þ 2 2 1 , h i 0 2 r 2 which can be arranged as rhiy ¼ p1ffiffi3 2ðkkþþ31Þ þ k4k 2 1 ln k þ 1 , that is just equation (4.7), setting

r0hi ry

  1 must result in k ≤ kaθ.

178

Autofrettage Technology and Its Applications in Pressured Apparatuses

Therefore, if kj ¼ k þ2 1, when k ≤ kaθ, r0hi ≥ −σy. Figure 4.16 shows the distributions of the residual stresses for k = kaθ = 2.360344, kj = (k + 1)/2 = 1.680172.

FIG. 4.16 – The distributions of the residual stresses for k = kaθ and kj = (k + 1)/2. Compared with figure 3.32, |r0hi | has been restrained within σy. Figure 4.17 shows the distributions of the residual stresses for k = 2.2 < kaθ and kj = (k + 1)/2 = 1.6.

FIG. 4.17 – The distributions of the residual stresses for k < kaθ and kj = (k + 1)/2. r0

In the case of k = 2.2 < kaθ, rhiy = −0.89899 > −1. Figure 4.18 shows the distributions of the residual stresses for k = 2.5 < kaθ and kj = (k + 1)/2 = 1.75.

Mechanical Autofrettage Technology by Limiting Circumferential

179

FIG. 4.18 – The distributions of the residual stresses for k > kaθ and kj = (k + 1)/2. In the case of k = 2.5 > kaθ,

4.2.4

r0hi ry

= −1.08491 < −1.

The Residual Stresses with Radius of Elastic-Plastic Juncture Being Geometrical Mean Radius of Inside Radius and Outside Radius

pffiffiffi k and k = kb = 3.042297, r0ei = −σy, but r0hi < −σy (see figure 3.42). 2 r0 k1   1 obtains equation (4.10). To control r0hi ≥ −σy, setting rhiy ¼ p2ffiffi3 kkk 2 ln 1 pffiffiffi Therefore, if kj ¼ k , when k ≤ kbθ, r0hi ≥ −σy. Figure 4.19 shows the distribupffiffiffi tions of the residual stresses for k = kbθ = 2.659725, kj ¼ k = 1.630866. Compared with figure 3.42, |r0hi | has been restrained within σy.

When kj ¼

pffiffiffi

FIG. 4.19 – The distributions of the residual stresses for k = kbθ and kj ¼ k .

180

Autofrettage Technology and Its Applications in Pressured Apparatuses

4.2.5

The Residual Stresses When Equivalent Total Stress on Elastic-Plastic Juncture is the Minimum

When kj ¼ exp

pffiffi  3 p 2 ry , the circumferential residual stress is equation (3.120), at the

inner surface pffiffiffi  pffiffiffi  2 3 0 1 p 0 exp 3 exp 3 rpy p ffiffi ffi pffiffiffi p k 2 þ 1 ry rhi 1 4 p @ 5 ð4:27Þ ¼ pffiffiffi þ1  3  1  þ 3 A 2 ry ry k  1 k2 k2 ry 3 Setting

r0hi ry

  1 obtains

pffiffiffi   pffiffiffi pffiffiffi p 3 2 pffiffiffi 2 p 2 þ 3  k  3k 0 exp 3 þ ry ry 2 2

or

k2 

exp

pffiffiffi  pffiffi 3 rpy  2 þ2 3 pffiffiffi p pffiffi3 3 ry  2 ð4:28Þ

Equation (4.28) is the load-bearing capacity (rpy ) of a cylinder with a certain pffiffi  k greater than kcθ when kj ¼ exp 23 rpy and |r0hi | = σy. pffiffi For equation (4.28), the domain of k2 is rpy  12 ¼ p1 and rpy  p1ffiffi3 ln 2 þ2 3 ¼ p2 .   pffi pffiffi p 2þ 3 The function k 2 

exp

3ry  pffiffi p pffi3 3ry  2

2

is plotted in figure 4.20. There are two branches for

the graphs of the function: curve og and curve bafh. In addition, the function kj ¼ pffiffi  2 exp 23 rpy is plotted in figure 4.20 as the solid curve oadi and the function rpy ¼ p2ffiffi3 k k1 2 is plotted in figure 4.20 as the dash curve ofdc. In figure 4.20, curve og corresponds to pffiffi the domain of rpy  p1ffiffi3 ln 2 þ2 3, and k < 1 on curve og, thus it is of no practical sigpffiffi  nificance; curve bafh corresponds to the domain of rpy  12. For kj ¼ exp 23 rpy , letting pffiffi  pffiffi 2 2 kj = k and substituting k ¼ exp 23 rpy into equation (4.28) gives kk 2ln1k ¼ 2 þ2 3, that is just equation (4.3). Thus, the ordinate of the intersection (point a) of curve bah and curve oadi is kcθ = 2.024678965. When k = kcθ, from equation (4.28), the corresponding load-bearing capacity rpy = 0.814539…. Therefore, the coordinate of point a pffiffi  2 2 leads to p2ffiffi3 k k1 ¼ is a (0.814539, kcθ). Substituting k ¼ exp 23 rpy into rpy ¼ p2ffiffi3 k k1 2 2 p2ffiffi ln k 3

k or kk 2ln 1 ¼ 1, that is just equation (2.28). Therefore, the coordinate of the intersection (point d) of curve ofdc and curve kc). Substituting  oadi is d(0.920081, pffiffi pffiffi 2 2 p 3 3 2 k 1 k 1 2 pffiffi 2 into equation (4.28) obtains exp 2 2 ¼ þ 1   2  k r 2 2 ¼ 0, the k k y

2

3

solution of which is k = 2.0336493…. The corresponding load-bearing capacity p ry = 0.8754994…. Therefore, the coordinate of the intersection (point f) of curve ofdc and curve bah is f (0.8754994, 2.0336493).

Mechanical Autofrettage Technology by Limiting Circumferential

181

FIG. 4.20 – The relationship between k (kj) and rpy .  For the function k 2 

exp



pffi pffiffi p 3ry 2 þ2 3 pffiffi p pffi3 , 3ry  2

dk 2  setting p ffiffi p ¼ 0 gives d

3ry

pffiffiffi pffiffiffi     pffiffiffi p pffiffiffi p pffiffiffi p 2þ 3 2þ 3 3 exp 3 exp 3 ¼0  þ ry ry ry 2 2

ð4:29Þ

Solving equation (4.29) obtains the horizontal ordinate of the minimum value point a p ¼ 0:814539. . . ð4:30Þ ry   pffi pffiffi p exp 3ry 2 þ2 3 pffiffi p pffi3 Substituting rp = 0.814539… into k 2 ¼ leads to k = kcθ. It is thus y

3ry  2

evident that point a is the minimum value point. For curve bah, pffiffi on  section ba, 3 p k > kj; on section ah, k < kj, where kj is determined by kj ¼ exp 2 ry .   pffi pffiffi p pffiffiffi  exp 3ry 2 þ2 3 pffiffi p pffi3 Letting k 2 ¼ ¼ kj2 ¼ exp 3 rpy also obtains equation (4.29), and 3ry  2 p ffiffiffi  when k 2 ¼ kj2 ¼ exp 3 rpy , equation (4.29) is just equation (4.3). When k = kj, pffiffiffi  2 2 k substituting k 2 ¼ kj2 ¼ exp 3 rpy obtains kk 2ln into rpy ¼ p2ffiffi3 k k1 2 1 ¼ 1. From

figure 4.20, when rpy \0.814539 (corresponding to section ba), k > kcθ, and k is larger pffiffi  than kj determined by kj ¼ exp 23 rpy , and k is larger than the value determined by p ry

p ¼ p2ffiffi3 k k1 2 . This is reasonable and makes sense; when r [ 0.814539 (corresponding y 2

Autofrettage Technology and Its Applications in Pressured Apparatuses

182

to section ha), k > kcθ, and k is smaller than kj determined by kj ¼ exp

pffiffi  3 p 2 ry , this is

unreasonable and makes no sense. Curve odac can ensure k > kj, but can not ensure r0hi ≥ −σy. When k > kcθ, the meaningful control curve is curve ba. Besides, when p 1 ry ! 2, k → ∞. Curve ba is re-plotted in figure 4.21.

FIG. 4.21 – The reasonable relationship between k and rpy . Since kjθ obtained by ensuring r0hi ≥ −σy is smaller than kj* obtained by ensuring ≥ −σy, the load-bearing capacity rpy in the case of kj = kjθ is certainly smaller than that in the of kj = kj*. When k < kcθ, there is kj = k, thus rp ¼ p2ffiffi ln k, that is r0ei

y

3

expressed by curve oa in figure 4.21. If the design conditions are controlled by curve ba when k > kcθ, there are both kj < k and r0hi ≥ −σy. For example, when k = 3, from equation (4.28), pffiffi  p p 3 p = 0.5452382…. When k = 3, = 0.5452382, k ¼ exp j ry ry 2 ry = 1.603502… = kjθ. The residual stresses (expressed by equations (3.118)–(3.125)) within the whole wall are plotted in figure 4.22. The meaning of each curve and the main parameters are marked in the figure. Compared with figure 3.48, the residual stresses are reduced, because when r0hi ≥ −σy, kj = kjθ, and when r0ei ≥ −σy, kj = kj*, and kjθ is smaller than kj*. However, the load-bearing capacityis reduced. pffiffi  In fact, when kj ¼ exp 23 rpy , equations (4.29) and (4.28) are equivalent. When pffiffi  kj ¼ exp 23 rpy (but k > kj), equation (4.28) becomes equation (4.3). Therefore, curve pffiffi bc  in figure 4.2 or figure 4.3 is also the relation between k and kj 3 p when kj ¼ exp 2 ry and r0hi = −σy. However, when the load-bearing capacity p ry

¼ p2ffiffi3 k k1 2 , σej ≡ σy within the whole plastic region; when the load-bearing capacity 2

Mechanical Autofrettage Technology by Limiting Circumferential

FIG. 4.22 – The residual stresses when k = 3, p p2ffiffi ry ¼ 3 ln kj k 2 ln kj2  k 2

(equivalent

to

kj ¼ exp

pffiffi  3 p 2 ry

),

p ry

where

183

= 0.5452382.

kj

is

restrained

by

kj2

 þ 2  0, σej < σy at the elastic-plastic juncture. Equation (4.28) or equation (4.29) can ensure kj < k and r0hi ≥ −σy, but under the premise of ensuring kj < k and r0hi ≥ −σy as well as σej being the minimum, the pffiffi  load-bearing capacity rpy calculated by kj ¼ exp 23 rpy may not the most satisfactory. From equation (4.28) or equation (4.29), pffiffithe greater k is, the smaller kj is, or p the smaller ry is. When k = ∞, kj ¼ exp 43 , hence rpy ¼ 12, that is merely the

maximum elastic load with k = ∞ based on the maximum shear stress theory pffiffi  3 p (Tresca yield criterion). Besides, when kj ¼ exp 2 ry , from figure 4.21, the than that load-bearing capacity under the case of k > kcθ is unexpectedly pffiffi smaller  3 p under the case of k < kcθ. So, if it is not necessary, kj ¼ exp 2 ry is not a satisfactory solution, the load-bearing capacity is needed to be further explored. phigher ffiffi  3 p When kj ¼ exp 2 ry , autofrettage pressure   pffiffiffi p pa p 1 1 p ffiffi ffi p ffiffi ffi ¼ þ exp 3  ð4:31Þ ry ry ry 3 3k 2 pffiffiffi  When rpy  p2ffiffi3 ln k, p1ffiffi3  pffiffi31k 2 exp 3 rpy  0. Therefore, rpay  rpy . pffiffiffi  When kjθ = k = kcθ, kj2 ¼ exp 3 rpy ¼ k 2 , then, from equation (4.31),

pa ry

¼ rpy ¼ p2ffiffi3 ln kch = 0.814539. When rpy ¼ p2ffiffi3 ln k, rpay reaches the maximum value, which is rpay ¼ rpy ¼ p2ffiffi3 ln k. It is known that only when k ≤ kcθ, rp ¼ p2ffiffi ln k. For k ≥ kcθ, from equation (4.28), y

3

184

we obtain

Autofrettage Technology and Its Applications in Pressured Apparatuses p ry

 pffiffi31k 2 exp

pffiffiffi  pffiffi þffiffi 3 3 rpy ¼ 12  22p . Substituting this equation into equa3k 2

tion (4.31) gives pffiffiffi pa 2 þ 3 k 2  1 ¼ pffiffiffi k2 ry 2 3

ð4:32Þ

Equation (3.18) indicates that in general when the load-bearing capacity

p ry

equals autofrettage pressure, or rpy ¼ rpay , it is the optimum, in this case rT e ≡ σy in the pffiffi pa 2 þpffiffi 3 k 2 1 whole plastic region. So, when ry ¼ 2 3 k 2 (k ≥ kcθ), the load-bearing capacity is pffiffi  pffiffi pffiffi 2 pa 3 p 2 þpffiffi 3 k 2 1 p also rpy ¼ 2 2þpffiffi3 3 k k1 2 . In the case that kj ¼ exp 2 ry , if k ≥ kcθ, when ry ¼ 2 3 k 2 , ry

0 is controlled by equation (4.28), rpy \ rpay , rT ej < σy and rhi = −σy. For example, when k = 3, kjθ = 1.603502, rpy = 0.5452382 and rpay = 0.957645… (from equation (4.31)), rT ej ry

= 0.687464 (equation (3.115)). When k ≤ kcθ, the optimum load-bearing capacity is rpy ¼ rpay ¼ p2ffiffi3 ln k, and in this

case rT e ≡ σy in the whole plastic region.

pffiffi ¼ 2 2þpffiffi3 3; from equation (4.28), rpy ¼ 12, pffiffi when rpy ¼ 12, from equation (4.31), we also have rpay ¼ 2 2þpffiffi3 3. When k = ∞, equation (4.31) become

When k = ∞, from equation (4.31),

pa ry

pa p 1 ¼ þ pffiffiffi ry ry 3

ð4:33Þ

For equation (4.31), when k ≤ kcθ, rpy = 0 to p2ffiffi3 ln kch ; when k ≥ kcθ, p p2ffiffi ry = 3 ln kch to the value determined by equation (4.28). For a pressure vessel with a certain radius ratio k ≥ kcθ, the maximum load for σej to reach the minimum is the value determined by equation (4.28). When rpy equals the value pffiffi pffiffi 2 determined by equation (4.28), rpay ¼ 2 2þpffiffi3 3 k k1 ¼ 2 þ2 3 rpye . For the convenience of 2 engineering applications, some data calculated by equation (4.28) are listed in table 4.2. Equation (4.31) is plotted in figure 4.23. Equation (4.31) is just equation (3.120). Therefore, when k ≤ kcθ, the curves for equation (4.31) is just those in figure 3.51. When k ≥ kcθ, the curves for equation (4.31) is those in figure 3.34 except that the ends are different from each other. The connection of the ends of the curves in figure 4.23 is curve cd, which is lower than ab in figure 3.34 of the last chapter because the maximum load-bearing capacity rpy and the maximum autofrettage pressure rpay are lower than those in chapter 3. In figure 4.23, the curve cd is the connection of the ends of the curves described by equation (4.31), which shows the maximum load-bearing capacity rpy and the maximum autofrettage pressure rpay for k ≥ kcθ. Substituting k2 in equation (4.28) into

Mechanical Autofrettage Technology by Limiting Circumferential TAB. 4.2 – Numerical value from exp

185

pffiffiffi  pffiffi pffiffi pffiffiffi 3 rpy þ 23 k 2  3k 2 rpy  2 þ2 3 ¼ 0.

k

p ry

pa ry

k

p ry

pa ry

kcθ 2.05 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3 3.1

0.814539 0.730678 0.683141 0.636884 0.610605 0.592829 0.579793 0.569754 0.561762 0.555241 0.549817 0.5452382 0.541323

0.814539 0.820991 0.833053 0.854757 0.873692 0.89031 0.904974 0.917979 0.929566 0.939933 0.949247 0.957645 0.965243

3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4 4.5 5 10 ∞

0.537939 0.534988 0.532395 0.5301 0.528057 0.526228 0.524583 0.523096 0.521747 0.516553 0.513068 0.503025 0.5

0.97214 0.97842 0.984154 0.989403 0.994221 0.998654 1.002742 1.006519 1.010016 1.024148 1.034256 1.066577 1.07735

FIG. 4.23 – The graph of equation (4.32) obtains generated, that is

pa ry ,

pa ry

¼ rpy þ p1ffiffi3  pffiffi31k 2 exp

pffiffiffi  3 rpy .

as a result, the equation of curve cd in figure 4.23 is

pffiffiffi  pffiffiffi pffiffiffi p p pa 2 exp 3 ry  3 ry  1 2 þ 3 pffiffiffi  ¼ pffiffiffi pffiffi ry 4 3 exp 3 p  2 þ 3 ry

ð4:34Þ

2

When k = kcθ, rpay ¼ rpy ¼ p2ffiffi3 ln k ¼ p2ffiffi3 ln kch and pffiffiffi 2 kch ln kch 2 þ 3 4 2 2 pffiffiffi kch 1¼ ln kch or kch ¼ 2 4 kch  1 2þ 3

or

2 ln kch

pffiffiffi 2 1 2 þ 3 kch ¼ 2 2 kch ð4:35Þ

186

Autofrettage Technology and Its Applications in Pressured Apparatuses

pa ry

Equation (4.34) is equivalent to tion (4.35), we obtain

2 2 kch ln kch 1 p 2þ 3 2 kch  2

ffi ¼

¼ p2ffiffi3

2 kch 1 2 kch

2 2 2 kch ln kch 1 kch ln kch pffi . 2 k 2 2 þ 3 kch 1

. Then,

ch

pa ry

Applying equa-

2

¼ rpy ¼ p2ffiffi3 ln k ¼ p2ffiffi3 ln kch . This

confirms the above conclusion. The coordinate of point c is c (p2ffiffi3 ln kch ,p2ffiffi3 ln kch ). As mentioned above, the meaningful domain of k2 is

p ry

p ry

¼ 12, k2 = ∞. Consequently, from pffiffi equation (4.31) or equation (4.32) or equation (4.33), rpay ¼ 2 2þpffiffi3 3, or the coordinate pffiffi of point d is d (12, 2 2þpffiffi3 3). The coordinate of point c is f (0, p1ffiffi3). The practically significant curves for a relation between rpy and rpay are located in a quadrilateral cdfo, where the straight line bdf is parallel to the straight line aco. For k ≤ kc, the ends of the curves described by equation (4.31) are on the straight-line segment oc where rpay ¼ rpy . Thereupon the included angle between the straight line oa and the abscissa axis or vertical axis is 45°. It is similar to the previous chapter, in figure 4.23, the dash curves have no practical significance, they are merely an extension of solid curves or only the elongation of the mathematical relation expressed by equation (4.31).

4.3

 12, and when

The Total Stresses and the Load-Bearing Capacity When Circumferential Residual Stress on the Inside Surface Controlled

The equivalent total stress rT r0 rp e ¼ e þ e ry ry ry

ð4:36Þ

The equivalent stress of the stresses caused by the internal pressure p is: pffiffiffi  p pffiffiffi 2  3 rh rpr 3k p 1 rpe ¼  ð4:37Þ ¼ 2 ry k  1 ry x 2 2 ry ry In plastic region pffiffiffi 2 k 2  kj2 þ k 2 ln kj2 1 3k rT p e ¼1 þ 2 2 2 2 x ry k 1 x ðk  1Þ ry

ð4:38Þ

In elastic region pffiffiffi 2 kj2  1  ln kj2 k 2 3k rT p e ¼ þ ry k2  1 x 2 x 2 ðk 2  1Þ ry

ð4:39Þ

Mechanical Autofrettage Technology by Limiting Circumferential

187

At the elastic-plastic juncture (x = r/ri = kj), equations (4.38) and (4.39) both become: pffiffiffi 2 k 2 ðkj2  1  ln kj2 Þ 3k p=ry rT e ¼ þ 2 ð4:40Þ ry ðk 2  1Þkj2 ðk  1Þkj2 When k 2  kj2 pa p 2 ¼ pffiffiffi ln kj þ pffiffiffi ¼ ry ry 3 3k 2

ð4:41Þ

rT e ry

 1 within the whole plastic region. In this case, the load-bearing capacity rpy is the most ideal. When kj is determined by equation (4.3), or kj = kjθ, equation (4.41) becomes pffiffiffi pffiffiffi p 2 þ 3 k 2  1 2 þ 3 pe ph ¼ pffiffiffi ¼ ¼ ð4:42Þ ry k2 2 ry ry 2 3 Equation (4.42) is the ultimate allowable loading of a pressure vessel with kj = kjθ under the condition r0hi = –σy (|r0ei | < σy) and based on the maximum distortion strain energy theory (Mises’s yield criterion), denoted by pθ. When k ≥ kcθ and under the condition r0hi = –σy (|r0ei | < σy), the load-bearing capacity of an autofrettaged pressure vessel should be determined by equation (4.42). In fact, equation (4.42) is just equation (4.32). The value determined by equation (4.42) is smaller than that determined by another equation proposed in chapter 3 based on the maximum distortion strain 2 energy theory and kj = kj*: prmy ¼ p2ffiffi3 k k1 2 , but greater than that proposed in chapter 2 based on the maximum shear stress theory and kj = kj*: rpyt ¼ k k1 2 , and it is greater than the maximum elastic load-bearing capability (initial yield pressure) of an pffiffi k 2ffiffi1 unautofrettaged pressure vessel: rpye ¼ p , or it is 1 þ 23 (=1.866025) times of the 3k 2 2

initial yield pressure of an unautofrettaged pressure vessel. It is seen that ph ¼ pm 2þ pt , this is wonderful. Another interesting thing is that letting equation (4.42) equal the pffiffi pffiffi 2 2 2 p p ¼ ryy ¼ p2ffiffi3 ln k obtains kk 2ln1k ¼ 2 þ2 3, or entire yield pressure ryy , i.e. rpy ¼ 23pþffiffi3 2 k k1 2 equation (4.5). Setting rphy ≤ 1 results in pffiffiffi ð4:43Þ k  krh ¼ 2 þ 3 ¼ 3:732051. . . When k ≤ krθ, p ≤ σy, hence rT ri > −σy.

pffiffi 2 p and ryy are plotted in figure 4.24, where curve ad is for rphy ¼ 2 2þpffiffi3 3 k k1 2 ; py py 2ffiffi p curve oab is for ry ¼ 3 ln k (based on Mises yield criterion); curve oc is for ry ¼ ln k ph pt pm pe ry , ry , ry , ry

(based on Tresca yield criterion); curve cg is for

pt ry

¼ k k1 (based on yield Tresca 2 2

(based on Mises yield criterion); curve on is for criterion); curve bf is for prmy ¼ p2ffiffi3 k k1 2 2

188

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 4.24 – Comparison between load-bearing capacity. pe ry

2 pe k 2ffiffi1 p ¼ k2k1 (based on Mises 2 (based on Tresca yield criterion); curve oh is for r ¼ y 3k 2 yield criterion). The application conditions for curve ad are: (1) kj = kjθ; (2) r0hi = –σy 0 (|rei | < σy); (3) based on the maximum distortion strain energy theory (Mises’s yield pffiffi 2 criterion); (4) k ≥ kcθ; (5) rp ¼ rph ¼ 2 þpffiffi 3 k k1 2 . y

y

2 3

The application conditions for curve oa are: (1) kj = k; (2) |r0hi | ≤ σy, |r0ei | < σy; (3) based on the maximum distortion strain energy theory (Mises’s yield criterion); p (4) k ≤ kcθ; (5) rpy ¼ ryy ¼ p2ffiffi3 ln k. The application conditions for curve oab are: (1) kj = k; (2) |r0ei | ≤ σy; (3) based on the maximum distortion strain energy theory (Mises’s yield criterion); (4) k ≤ kc; p (5) rpy ¼ ryy ¼ p2ffiffi3 ln k. The application conditions for curve oc are: (1) kj = k; (2) |r0ei | ≤ σy; (3) based on the maximum shear stress theory (Tresca yield criterion); (4) k ≤ kc; p (5) rpy ¼ ryy ¼ ln k. The application conditions for curve cg are: (1) kj = kj*; (2) r0hi = r0ei = –σy; (3) based on the maximum shear stress theory (Tresca yield criterion); (4) k ≥ kc; 2 (5) rpy ¼ rpyt ¼ k k1 2 . The application conditions for curve bf are: (1) kj = kj*; (2) r0hi = –σy and –σy < r0ei < 0; (3) based on the maximum distortion strain energy theory (Mises’s 2 yield criterion); (4) k ≥ kc; (5) rpy ¼ prmy ¼ p2ffiffi3 k k1 2 . pffiffi 2 Point a is the intersection of curve ad and curve oa. Setting rphy ¼ 2 2þpffiffi3 3 k k1 ¼ 2 pffiffi 2 2 py 2 þ 3 2 k ln k pffiffi ry ¼ 3 ln k just results in k 2 1 ¼ 2 , that is equation (4.5). Therefore, the horizontal ordinate of point a is kcθ. When k ≤ kcθ, curve oa works; When k ≥ kcθ, curve ad works.

Mechanical Autofrettage Technology by Limiting Circumferential Point b is the intersection of curve bf and curve oab. Setting p2ffiffi ln k 3

pm ry

189 p

¼ p2ffiffi3 k k1 ¼ ryy ¼ 2 2

just results in kk 2ln1k ¼ 1, that is equation (2.28). Therefore, the horizontal ordinate of point b is kc. When k ≤ kc, curve oab works; When k ≥ kc, curve bf works. 2 p ¼ ryy ¼ ln k Point c is the intersection of curve cg and curve oc. Setting rpyt ¼ k k1 2 2

2

just results in the same equation kk 2ln1k ¼ 1. Therefore, the horizontal ordinate of point c is also kc. When k ≤ kc, curve oc works; When k ≥ kc, curve cg works. Point s is the intersection of curve ad and the extension of curve oc. Setting pffiffi py ph 2 þpffiffi 3 k 2 1 ry ¼ 2 3 k 2 ¼ ry ¼ ln k obtains pffiffiffi k 2 ln k 2 2 þ 3 p ffiffiffi ¼ ð4:44Þ k2  1 3 2

2

The solution of equation (4.44) is k ¼ 2:456797. . . ¼ ks

ð4:45Þ

p

When k = ks, rphy ¼ ryy = 0.898858…. Therefore, the coordinate of point s is s (ks, 0.898858). Further, if kj = kjθ, equations (4.38) and (4.39) become respectively: pffiffiffi pffiffiffi 2 3þ2 3k p=ry rT e ¼1 þ ð4:46Þ 2x 2 ry k2  1 x2 pffiffiffi pffiffiffi 2 kj2  ð 3 þ 2Þ=2 3k p=ry rT e ¼ þ 2 2 x ry k  1 x2

ð4:47Þ

From equation (4.46), it is known that: (1) Provided rpy  pffiffi 2 2 rT e  4x ð2pffiffi33 þ 2Þ k k1 2 (negative), r   1, this is definitely feasible for p > 0 in engiy pffiffi pffiffi 2 2 ð 3 þ 2Þ2x 2 k 2 1 pe k 1 T ffiffi pffiffi neering; (2) As long as rpy  ð 3 þ2p2Þ2x k 2 , re > 0, while k 2 \ ry , so 3 2 3 pffiffi pffiffi p 3pþffiffi 2 k 2 1 3 þ 2 pe T when p > pe, rT e > 0; (3) So long as ry  2 3 k 2 ¼ 2 ry , re ≤ σy. When x = 1, pffiffi 2 3 pe T if rpy ¼ k2k1 2 ¼ 2 r , then rei = 0. y From equation (4.47), it is known that: (1) Provided rpy [ pffiffi k 2 ð 3 þ 2Þ=2 k 2 1 T  j pffiffi3 k 2 (negative), re > 0, this is definitely feasible for p > 0 in engineering, so the equivalent residual stress within the elastic zone is always tension; pffiffi pffiffi pffiffi x 2 kj2 þ ð 3 þ 2Þ=2 k 2 1 p 3pþffiffi 2 k 2 1 3 þ 2 pe T pffiffi (2) As long as rpy \ , r < σ , so when  ¼ 2 2 y e ry 2 ry , k 3 2 3 k rT e ≤ σy. At inside surface, x = r/ri = 1, then, from equation (4.46): pffiffiffi 2 pffiffiffi 3k p 3 rT ei ¼  ry k 2  1 ry 2

ð4:48Þ

190

Autofrettage Technology and Its Applications in Pressured Apparatuses

pffiffi 2 pffiffi3 k 1 \  2 (negative), rT ei cannot be lower than −σy; Unless p > 2pe, 2 3 k2 pffiffi p 3 þ 2 pe T rT cannot be higher than σ . So, when 0   y ei 2 ry , −σy ≤ rei ≤ σy. Especially, ry pffiffi when rpy ¼ 32þ 2 rpye , rT e ≡ σy within the whole plastic zone. At the elastic-plastic juncture, x = r/ri = kj, from (4.46) or (4.47): pffiffiffi pffiffiffi 2 rT kj2  ð 3 þ 2Þ=2 3k p=ry ej ¼ þ 2 ð4:49Þ 2 ry k  1 kj2 kj

Unless

p ry

pffiffi p 3 þ 2 pe T Clearly, rT ej > 0 within the whole elastic zone. If ry  2 ry , re can not be pffiffi higher than σy. So, when 0  rpy  32þ 2 rpye , 0 < rT ej ≤ σy. Especially, when pffiffi kj2 rT p 3 þ 2 pe T e 2 ry , rej = σy at the elastic-plastic juncture and ry ¼ x 2 at a general location ry ¼ within the elastic zone. pffiffi When rpy ¼ 32þ 2 rpye , or the load-bearing capacity is determined by equations (4.42), (4.46) and (4.47) become respectively:

rT e 1 ry

ð4:50Þ

kj2 rT e ¼ 2 ry x

ð4:51Þ

Figure 4.25 shows the distribution of the equivalent stress of total stress.

FIG. 4.25 – The distribution of the equivalent stress of total stress for kj = kjθ.

p ry

¼

pffiffi 3 þ 2 pe 2 ry

and

Mechanical Autofrettage Technology by Limiting Circumferential

191

The additional remarks for figure 4.25 are as follows. (1) Horizontal line baa: k = 2.024678965… = kcθ, kj = kjθ = k = kcθ. Within the rT

rT

plastic zone, rey is a horizontal line: rey = 1, x varies from 1 to kjθ (=kcθ = k) or from point b to point a, and then from 2.024678965… (kjθ) to 2.024678965… (k) or from point a to point a within the elastic zone (no elastic zone, the “curve” of the equivalent total stress is actually a point within the elastic zone). (2) Curve bcd: k = 2.1…, kjθ = 1.806908…. Within the plastic zone,

rT e ry

is a hori-

rT e ry

zontal line: = 1, x varies from 1 to kjθ, or from point b to point c, and then from 1.806908… to 2.1…, or from point c to point d within the elastic zone. (3) Curve bef: k = 2.5…, kjθ = 1.6522121…. Within the plastic zone,

rT e ry

is a hori-

rT e ry

zontal line: = 1, x varies from 1 to kjθ, or from point b to point e, and then from 1.6522121… to 2.5…, or from point e to point f within the elastic zone. (4) Curve bgh: k = 3, kjθ = 1.60350225…. Within the plastic zone,

rT e ry

is a hori-

rT e

zontal line: ry = 1, x varies from 1 to kjθ, or from point b to point g, and then from 1.844363… to 3, or from point g to point h within the elastic zone. (5) Curve bkl: k = 4, kjθ = 1.57121054…. Within the plastic zone,

rT e ry

is a horizontal

rT e

line: ry = 1, x varies from 1 to kjθ, or from point b to point k, and then from 1.57121054… to 4, or from point k to point l within the elastic zone. pffiffi rT (6) Curve bmn: k = ∞, kjh ¼ e 3=4 . Within the plastic zone, rey is a horizontal line: rT e ry

= 1, x varies from 1 to kjθ, or from point b to point m, and then from kjθ to ∞, or from point m to point n within the elastic zone.

The prerequisite to the above arguments is that k and kj meet equation (4.3) or pffiffi pffiffi 2 kj = kjθ and rpy ¼ 23pþffiffi3 2 k k1 ¼ 32þ 2 rpye . Grasping these laws is helpful to the design of 2 pffiffi 2 high and ultrahigh pressure vessels. If kj ≠ kjθ or rpy 6¼ 23pþffiffi3 2 k k1 2 , the above facts are | may exceed σy. untenable, and |rT pffiffi e When rpy ¼ 32þ 2 rpye , the stresses caused by operation pressure are as follows: pffiffiffi 3þ2 rpz ¼ pffiffiffi ð4:52Þ ry 2 3k 2 pffiffiffi   pffiffiffi 3þ2 3þ2 rpr rpz k2 ¼ 1  2 ¼ pffiffiffi  pffiffiffi 2 ry ry x 2 3k 2 3x 2

ð4:53Þ

pffiffiffi   pffiffiffi rph rpz 3þ2 3þ2 k2 ¼ 1 þ 2 ¼ pffiffiffi þ pffiffiffi 2 ry ry x 2 3k 2 3x 2

ð4:54Þ

192

Autofrettage Technology and Its Applications in Pressured Apparatuses

pffiffiffi  p  pffiffiffi 3 rh rpr 3þ2 rpe ¼  ¼ 2x 2 ry 2 ry ry

ð4:55Þ

pffiffi When rpy ¼ 32þ 2 rpye and kj = kjθ, the components of the total stresses are: Within the plastic zone: pffiffiffi 3þ2 rT r0z rpz ln x 2 1 z ¼ þ ¼ pffiffiffi  þ pffiffiffi ð4:56Þ 2 ry ry ry 2 3k 2 3

Letting

rT h ry

ð4:57Þ

pffiffiffi pffiffiffi rT 3þ2 2  3 ln x 2 h ¼ pffiffiffi þ pffiffiffi þ pffiffiffi ry 2 3k 2 3 2 3

ð4:58Þ

rT e 1 ry

ð4:59Þ

pffiffi pffiffi3  1 gives þ 2 2 3 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi  pffiffiffi 2 þ 3 2  3 x  exp 3  ¼ xzh 2k 2 2

¼ lnpxffiffi3 þ 2

pffiffiffi pffiffiffi 3þ2 3þ2 rT ln x 2 r ¼ pffiffiffi þ pffiffiffi  pffiffiffi 2 ry 2 3k 2 3 3

pffiffi 3 ffiffiþ 2 p 2 3k 2

rT e ry

From equations (4.57) and (4.58),

¼

pffiffi  T 3 rh ry 2

ð4:60Þ

 rT  rry  1, the same as

equation (4.50). Within the elastic zone: 2 kjh rT r0 rp z ¼ z þ z ¼ pffiffiffi ry ry ry 3k 2

ð4:61Þ

    kjh2 1 rT r0 rp rT k2 1 r ¼ r þ r ¼ z 1  2 ¼  pffiffiffi 2  2 k ry ry ry ry x 3 x

ð4:62Þ

    kjh2 1 rT r0h rph rT k2 1 z h ¼ þ ¼ 1  2 ¼ pffiffiffi 2 þ 2 k ry ry ry ry x 3 x

ð4:63Þ

rT

k2

From equations (4.62) and (4.63), rey ¼ xj2 , the same as equation (4.51). Figure 4.26 is a comparison between the equivalent stresses of total stresses under different internal pressure and kj = kjθ, from which it is known that only when

Mechanical Autofrettage Technology by Limiting Circumferential p ry p ry

¼ 6¼

pffiffi pffiffi 3pþffiffi 2 k 2 1 3 þ 2 pe ¼ 2 2 ry and k p2ffiffi 3 3 þ 2 pe 2 ry and/or kj ≠ kjθ,

193

kj = kjθ, the operation state is optimum. Otherwise, if

either rT e > σy or the load-bearing capacity is lowered. pffiffi In figure 4.26, curve 1 is just curve bgh in figure 4.25, in this case, rpy ¼ 32þ 2 rpye , pffiffi p 3 þ 2:1 pe T rT e ≡ σy within the whole plastic zone; Curve 2: ry ¼ ry , rei > σy; Curve 3: 2 pffiffi p 3 þ 1:9 pe 0 < rT Curve 4: p = pe, 0 < rT Curve 5: ei < σy; ei σy; ry , 2 ry ¼ pffiffi 2 p 3 pe k 1 T T ry ¼ 2k 2 ¼ 2 ry , rei /σy = 0; Curve 6: p = 0.8pe, rei /σy < 0.

FIG. 4.26 – Comparison between the equivalent total stresses under different p and kj = kjθ. Besides, for certain k, when kj < kjθ, though the residual stresses are smaller than those when kj = kjθ, the load-bearing capacity is dropped. For example, for k = 3, if kj = kjθ (=1.60350225…), from equation (4.41) or (4.42), rpy = 0.9576…; while if kj = 1.5, from equation (4.41), rpy = 0.9012… krθ and kjθ = 1.571211. In this case, p = pθ > σy and rT r < −σy. In figure 4.27, k = 3 < krθ, so p = pθ < σy and rT > −σ . y r

FIG. 4.29 – The total stresses for k = krθ, kjθ = 1.576574 and p = pθ = σy.

FIG. 4.30 – The total stresses for k = 4 > krθ, kjθ = 1.571211 and p = pθ > σy. Figure 4.31 shows the total stresses in three directions for k = kfθ = 2.105406 pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi rT rT 3  1k ¼ xa and rejy ¼ rhjy ; When x < xa, and kjθ = 1.801385. In this case, kjh ¼ rT e ry

[

rT h ry ;

When x > xa,

rT e ry

rT

\ rhy ; When x = xa = kjθ,

rT ej ry

¼

rT hj ry .

Autofrettage Technology and Its Applications in Pressured Apparatuses

196

FIG. 4.31 – The total stresses for k = kfθ, kjθ = 1.801385 and p = pθ.

4.4

Control Circumferential Total Stress Directly

Circumferential total stress rT h is expressed as pffiffiffi pffiffiffi    p r rT r0 1 2þ 3 2 3 k2 1 p h ¼ h þ h ¼ pffiffiffi ln x 2  þ þ 1 þ ry ry ry x 2 k 2  1 ry 2x 2 2 3 At the elastic-plastic juncture,

rT h ry

is maximum. ! pffiffiffi pffiffiffi! rT 1 2þ 3 2  3 k2 1 p hj 2 ¼ pffiffiffi ln kj  þ þ 1þ 2 2 2 ry 2 2kj kj k  1 ry 3

Setting

rT hj ry

ð4:65Þ

ð4:66Þ

≤ 1 results in

pffiffiffi pffiffiffi kj2 ðk 2  1Þ 2kj2 ln kj2  ð2 þ 3Þ þ ð2  3Þkj2 k 2  1 p p0h p ffiffi ffi   ¼ ry k 2 þ kj2 ry k 2 þ kj2 2 3

ð4:67Þ

In equation (4.67), kj and k are related by equation (4.3), or kj = kjθ. For the above k = 2.05 pffiffi p0h 2 þpffiffi 3 k 2 1 = 0.820991… > . 2 k r y 2 3

and

kjθ = 1.882849,

p0h ry

= 0.726638…,

ph ry

¼

p0

For the above k = 2.05 and kjθ = 1.882849 and rpy ¼ rhy = 0.726638, according to rT ry

0

¼ rry þ

rp ry ,

the total stresses are shown in figure 4.32, from which it is seen that rT h

T is reduced below σy within the whole wall and rT hj = σy at kjθ, re < σy within the whole wall.

Mechanical Autofrettage Technology by Limiting Circumferential

197

FIG. 4.32 – The total stresses for k = 2.05, kjθ = 1.882849 and p = p0h .

ph ry

In general, using equation (4.3), pffiffi 2 ¼ 2 2þpffiffi3 3 k k1 2 .

p0h ry

FIG. 4.33 – In figure 4.33, curve aoc is for curve bod is for

p0h ry ;

ph ry

is plotted in figure 4.33 along with

p0h ry

and

ph ry .

pffiffi 2 ¼ 2 2þpffiffi3 3 k k1 2 , the limit of

ph ry

is

pffiffi 2 þpffiffi 3 2 3

= 1.07735;

curve aoc ensures rT e ≡ σy within the whole wall, curve bod

ensures rT hj = σy; point o is the intersection of curve aoc and curve bod, the

198

Autofrettage Technology and Its Applications in Pressured Apparatuses

coordinate of point o is o (kh, 0.834306) = o (2.105406, 0.834306); when k ≤ kh, T p0h ≤ pθ, when k ≥ kh, p0h ≥ pθ; on section oa, rT e ≡ σy and rhj > σy; on section oc, T T T T rT e ≡ σy and rhj < σy; on section ob, re < σy and rhj = σy; on section od, re σy because p0h is very great and rT hj = σy; sections oa and ob are quite short. p0

For k = 3 (k > kh) and kjθ = 1.603502, rhy = 1.41576, the total stresses are shown in figure 4.34. The meaning of each curve and the main parameters are marked in the figure.

FIG. 4.34 – The total stresses for k = 3 k > kh, kjθ = 1.603502 and

p0h ry

= 1.41576.

rT

When k > kh and rhjy = 1, the equivalent stress is quite great and the distribution of the total equivalent is unfavorable. Therefore, it is reasonable to limit equivalent stress.

4.5

Chapter Summary

2 (1) For |r0ei | ≤ σy, the depth of plastic zone is k 2 ln kj2  k 2  kj þ2 ¼ 0 0.5 where e ≤ kj* ≤ kc = 2.2184574899167…, when k ≥ 2.2184574899167…; 2 ¼ 2 rpye . load-bearing capacity is rp4y ¼ p2ffiffi3 k k1 2 pffiffiffi pffiffiffi 2  3k 2 þ ð2 þ 3Þ (2) For |r0hi | ≤ σy, the depth of plastic zone is 2k 2 ln kjh2  2kjh pffiffi ¼ 0 where e 3=4 ≤ kjθ ≤ kcθ = 2.024678965…, when k ≥ 2.024678965…; pffiffi pffiffi 2 load-bearing capacity is rpy ¼ 23pþffiffi3 2 k k1 ¼ 32þ 2 rpye . The data 2.024678965… is 2 pffiffi 2 2 the solution of kk 2ln1k ¼ 2 þ2 3. pffiffi pffiffi 2 ¼ 32þ 2 rpe is the (3) For an autofrettaged cylinder, kj = kjθ with rp ¼ 3pþffiffi 2 k k1 2 y

2 3

y

optimum operation state—not only safe but also highly capable, under the

Mechanical Autofrettage Technology by Limiting Circumferential

199

T 2 2 state, rT e ≡ σy within the plastic zone and re /σy = kj /x < 1 within the elastic zone. (4) For the same k, kjθ < kj*, this means the depth of plastic zone to ensure |r0hi | ≤ σy is less than that to ensure |r0ei | ≤ σy. (5) Irrespective of k, the three curves for residual stress at a general radial location collect at the same point   within the plastic zone and the intersection is ffiffiffiffiffiffiffiffiffiffi qp ffiffi pffiffi 3 þ 2 p1ffiffi 3þ2 1 ln 2  2 = (1.931852…, −0.13984…) when kj = kjθ, or 2 ; 3 p ffiffi ffi pffiffiffi

2; ðln 2  1Þ= 3 = (1.414214…, −0.17716…) when kj = kj*. (6) When r0hi controlled, or kj = kjθ, the residual stresses, and their equivalent stress are determined only by the relative location (r/ri) within the plastic zone and independent of kj and k, or once the location, r/ri is determined, the residual stresses and their equivalent stress within the plastic zone are identical for any kj and k.

The main equations and conclusion are listed in table 4.3. TAB. 4.3 – The main equations and conclusion of this chapter. The critical radius ratio When the plastic depth covers the whole wall, for |r0ei | ≤ σy, k ≤ kcθ kcθ = 2.024678965… < kc = 2.218 457 489 916 7… pffiffiffi k 2 ln k 2 þ 3 ¼ kcθ is the solution of 2 , k 1 4 pffiffiffi p ffiffi ffi k 2 ln k 2 þ 3 2 2 þ 3 k2  1 pffiffiffi , i.e. is equivalent to pffiffiffi ln k ¼ ¼ and 2 k 1 4 2 3 3k 2 pffiffiffi pffiffiffi 2 p ph 2 2þ 3 3 þ 2 k  1 py pe or ¼ ¼ pffiffiffi ¼ ¼ pffiffiffi ln k or py ¼ 2 ry ry ry k2 3 2 3 pffiffiffi 3 in the molecule replaced with 2, then pffiffiffi k 2 ln k 2 þ 3 k 2 ln k becomes 2 ¼ ¼ 1; 2 k  1 pffiffiffi 4 k 1 2 2 þ 3 k2  1 2 k2  1 pffiffiffi ln k ¼ pffiffiffi becomes pffiffiffi ln k ¼ 2 pffiffiffi ; 2 2 3 3k 3 3k 2 pffiffiffi 2 2 p ph 2 p 2 k  1 py 2 3 þ 2 k  1 py ¼ ¼ pffiffiffi ¼ ¼ pffiffiffi ln k becomes ¼ pffiffiffi ¼ ¼ pffiffiffi ln k; ry ry ry ry k2 ry 3 3 k2 3 2 3 pffiffiffi 2þ 3 pe becomes py = 2pe py ¼ 2 The optimum plastic depth kjθ pffiffi pffiffiffi pffiffiffi 2 2k 2 ln kjh2  2kjh  3k 2 þ ð2 þ 3Þ  0 (e 3=4  kjh  kch ; k  kch ) pffiffiffi Replacing 3 with 2 results in k 2 ln kj2  k 2  kj2 þ 2 ¼ 0 (e0.5 ≤ kj* ≤ kc and k ≥ kc) For the same k, kj* > kjθ, or to avoid r0hi ≤ −σy, kjθ should be smaller than kj* When k ≤ kc, kj = k If

200

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 4.3 – (continued).

The results when kj = kjθ 1. Residual stresses in axial, radial, and circumferential direction pffiffi pffiffiffi 8 0 8  2þ 3 rz 1 1 1 3 > > r0z kj2  2 2 2 > > p ffiffi ffi p ffiffi ffi ¼   ln x ¼ ln x > > ¼ pffiffiffi > > > > ry 2 2 3 3 ffiffiffi > > r p pffiffi2ffi > >  3k 2  0 < 0 < y0 rr 1 2þ 3 2þ 3 r k rz (k ≤ x ≤ k) 2 r (1 ≤ x ≤ kjθ) ¼ pffiffiffi ln x þ  jθ ¼ 1 2 2 > > 2x 2 r 3 r x r y > > y pffiffiffi pffiffiffi > >    y0 > 0 > 0 2 > > r 1 2þ 3 2  3 r k rz > > > > : h ¼ pffiffiffi ln x 2  : h ¼ 1þ 2 þ 2x 2 2 ry ry x ry 3 pffiffiffi Replacing 3 in the molecule with 2 obtains the results when kj = kj* 2. Equivalent residual stress pffiffiffi pffiffiffi pffiffiffi r0e r0e kj2  ð 3 þ 2Þ=2 3þ2 3þ2 ¼1 ¼ 1  (1 ≤ x ≤ k ) ¼ (kjθ ≤ x ≤ k) jθ 2x 2 ry ry x2 2ðr=ri Þ2 pffiffiffi Replacing 3 with 2 obtains the results when kj = kj* 3. The optimum load-bearing capability and autofrettage pressure pffiffiffi pffiffiffi p ph pa 2 þ 3 k 2  1 2 þ 3 pe p ffiffi ffi ¼ ¼ ¼ ¼ ry ry ry k2 2 ry 2 3 pffiffiffi p pa pe 2 k2  1 Replacing 3 in the molecule with 2 obtains ¼ ¼ 2 ¼ pffiffiffi ry ry ry 3 k2 pffiffiffi When 3 in the molecule replaced with 2, then, ph 2 k 2  1 pm ¼ pffiffiffi ¼ ry ry 3 k2 which is the optimum load-bearing capacity based on the maximum distortion strain energy theory and kj = kj* When based on the maximum shear stress theory and kj = kj*, the optimum load-bearing capacity is pt k 2  1 ¼ k2 ry pm þ pt ph ¼ 2 pffiffiffi 2 2 k ln k 2þ 3 , or k = kcθ, pθ = py ¼ When 2 2 1 rkp ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi ffi 3þ2 = a constant 4. x0h  2pffiffiffi pffiffiffi Replacing 3 in the molecule with 2 results in x0  2 pffi 3 5. x1h ¼ e 4 = kjθ∞ = constant pffiffiffi pffiffiffi Replacing 3 with 2 results in x1  e pffiffiffi pffiffiffi  pffiffiffi 3þ2 1 3þ2 3 r0 r0 r0 r0 r0 1 1  ¼ pffiffiffi ln  ¼ a constant 6. y0d ¼ z ¼ r ¼ h ¼ rmin ¼ rx0  pffiffiffi ln 2 2 2 ry ry ry ry ry 2 3 3 pffiffiffi 1  ln 2 Replacing 3 in the molecule with 2 obtains y0d =  pffiffiffi , the result when kj = kj* 3

Mechanical Autofrettage Technology by Limiting Circumferential

201

TAB. 4.3 – (continued). The results when kj = kjθ, p = pθ 1. The components of total stresses pffiffiffi pffiffiffi pffiffiffi  9 3þ2 3 3þ2 > rT ln x 2 1 1 z > þ ¼ pffiffiffi  þ pffiffiffi ¼ pffiffiffi ln x 2  > > 2 2 > 2 2k 2pffiffiffi 2 3k p ry 3 3 > ffiffi ffi > = 2 rT ln x 3 þ 2 3 þ 2 r (1 ≤ x ≤ kjθ) ¼ pffiffiffi þ pffiffiffi  pffiffiffi > ry 3 2 ffiffiffi 3k 2 2 3pffiffiffi > p > > > rT ln x 2 3þ2 2  3 > h > ; ¼ pffiffiffi þ pffiffiffi þ pffiffiffi 2 ry 2 3 3 2 3k 9 kj2 rT > z > ¼ pffiffiffi ¼ constant > > > ry 3 k2 >  > = 2  T T 2 kj rr rz k 1 1 (kjθ ≤ x ≤ k) ¼ 1  2 ¼  pffiffiffi 2  2 x k > ry ry x 3 > >     > kj2 1 > rT rT k2 1 > h ; ¼ z 1 þ 2 ¼ pffiffiffi 2 þ 2 > x k ry ry x 3 pffiffiffi Replacing 3 in the molecule with 2 obtains the results when kj = kj* 2. Equivalent residual stress kj2 rT rT e e  1 (1 ≤ x ≤ kjθ) ¼ 2 (kjθ ≤ x ≤ k) ry ry x pffiffiffi (1) When k ≤ krθ = 2 þ 3 = 3.732051, p ≤ σy, hence rT ri > −σy pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi rT rT ej hj (2) When k ≥ kfθ = 2.105406…, kjθ < 3  1k(=xa), accordingly  ; when k < kfθ, ry ry pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi rT rT ej hj 3  1k, accordingly  kjθ > ry ry rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffi pffiffi T (3) When x  exp 3  2 þ2k 2 3  22 3 = xzθ, rT e ≤ rh pffiffiffi pffiffiffi pffiffiffi (4) When kj = kjθ, if 2k 2 ln½ð 3  1Þk 2   ð3 3  2Þk 2 þ 2 þ 3 ¼ 0 or k = kfθ, xzθ = xa = kjθ = 1.801385… pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi (5) When kj < 3  1k(=xa), xa exists and xz does not exist; When kj > xa, xa does not exists and xz exist k þ1 The results when kj ¼ 2 1. When k ≤ kaθ = 2.360344…, |r0hi | ≤ σy; When k ≥ kaθ, |r0hi | ≥ σy k þ1 = 1.680172… 2. When k = ka, kj ¼ 2 3. The maximum load-bearing capacity and autofrettage pressure p 1 ðk þ 1Þ2 2 k þ1 py ¼ pffiffiffi  pffiffiffi þ pffiffiffi ln ≤ p2ffiffi3 ln k ¼ ry ry 2 3 4 3k 2 3 1 ðk þ 1Þ2 2 k þ1 2 k2  1 \ pffiffiffi 4. When k ≤ kaθ, pffiffiffi  pffiffiffi þ pffiffiffi ln ; 2 3 4 3k 2 3 3 k2 1 ðk þ 1Þ2 2 k þ1 2 k2  1 þ pffiffiffi ln , but when k ≥ kaθ, |r0hi | > σy [ pffiffiffi When k ≥ kaθ, pffiffiffi  pffiffiffi 2 2 3 4 3k 3 3 k2 See also table 3.2 of chapter 3

202

Autofrettage Technology and Its Applications in Pressured Apparatuses

TAB. 4.3 – (Continued). pffiffiffi The results when kj ¼ k 1. When k ≤ kbθ = 2.65972…, |r0hi | ≤ σy; When k ≥ kbθ, |r0hi | ≥ σy pffiffiffi 2. When k = kbθ, kj ¼ k = 1.630865… See also table 3.2 of chapter 3 pffiffiffi  rj 3p The results when kj ¼ ¼ exp 2 ry ri 1. The allowable load in order for r0hi /σy ≥ −1 when k ≥ kcθ pffiffiffi  pffiffi pffiffiffi pffiffiffi  pffiffi3ffi exp 3 rpy  2 þ2 3 p ffiffi ffi p 2 þ 3  0 or k 2  k 2  3k 2  exp 3 rpy þ pffiffiffi p pffiffi3 ry 2 2 3 ry  2 pffiffiffi  pffiffiffi p ffiffi ffi p Replacing 3 in the molecule with 2 obtains exp 3 ry þ k 2  3k 2 rpy  2  0, the result when kj = kj* See also table 3.2 of chapter 3

References [1] Zhu R.L., Yang J.L. (1998) Autofrettage of thick cylinders, Int. J. Press. Vessels Pip. 75 (6), 443. [2] Zhu R.L. (2013) Study on autofrettage for medium-thick pressure vessels, J. Eng. Mech. 139 (12), 1790. [3] Zhu R.L., Zhu G.L., Mao A.F. (2016) Plastic depth and load-bearing capacity of autofrettaged cylinders, J. Mech. Sci. Technol. 30 (6), 2627. [4] The compiling group of “handbook of mathematics” (1979) A handbook of mathematics (in Chinese). Higher Education Press, Beijing. [5] Liu H.W. (2019) Mechanics of materials (in Chinese). Higher Education Press, Beijing.

Chapter 5 Mechanical Autofrettage Technology Under Low Load 5.1

Introduction

When a pressure vessel is processed with autofrettage, it must be ensured that the pressure vessel is not yielded compressively. Based on the ultimate load bearing criterion, we found the suitable kj for a pressure vessel with k not to be yielded when processed with autofrettage, which is k 2 ln kj2
k 2
kj2 þ 2 ¼ 0, or equation (2.46), and which is fit for both the maximum shear stress theory (Tresca yield criterion) and the maximum distortion strain energy theory (Mises’s yield criterion). If kj is determined by this equation, the ultimate load-bearing capacity is 2pe. It is well known that the greater kj is, the greater the load-bearing capacity is, but the more complex the autofrettage process is. Therefore, if it is not necessary for a cylinder to bear 2pe, kj can be decreased to reduce the complexity of the autofrettage process. Accordingly, the following questions should be resolved: How to determine an appropriate plastic depth kj for a certain p and k? What is the relationship between kj and k for a certain p when p ≠ 2pe? What results will be obtained under new conditions? How is p for a certain kj and k determined? What are the characteristics of residual and total stresses under new conditions? In this chapter, we study the aforementioned problems and do the comparison of the results based on the maximum shear stress theory (Tresca yield criterion) and the maximum distortion strain energy theory (Mises’s yield criterion) in order to bring to light essential correlation and laws contained in the results. On the basis of the previous work[1–4], this chapter is intended to study mechanical autofrettage technology under low load with the help of the mathematical analysis of reference[5].

DOI: 10.1051/978-2-7598-3111-1.c005 © Science Press, EDP Sciences, 2023

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204

5.2

The Optimum Plastic Depth

Equation (2.21) indicates that, in general, when the load-bearing capacity rpy equals

autofrettage pressure pa, or rpy ¼ rpay , it is optimum, in this case, rT e ≡ σy in the whole plastic region. Based on the maximum shear stress theory (Tresca yield criterion)[6, 7], autofrettage pressure is equation (2.1), or k 2
kj2 k 2 ln kj2 þ k 2
kj2 pa ¼ ln kj þ ¼ ry 2k 2 2k 2

ð5:1Þ

Based on the maximum distortion strain energy theory (Mises’s yield criterion)[6, 7], autofrettage pressure is equation (3.1), or k 2
kj2 k 2 ln kj2 þ k 2
kj2 pa 2 pffiffiffi ¼ pffiffiffi ln kj þ pffiffiffi ¼ ry 3 3k 2 3k 2

ð5:2Þ

p pe pa ¼k ¼ ry ry ry

ð5:3Þ

Set

λ is called the reinforcing coefficient for load-bearing capacity. 2 For the maximum shear stress theory, rpye ¼ k2k
1 2 , therefore, p pe k2
1 ¼k ¼k ry 2k 2 ry For the maximum distortion strain energy theory,

ð5:4Þ pe ry

k ffiffi
1 ¼p , therefore, 3k 2 2

p pe k2
1 ¼ k ¼ k pffiffiffi ry ry 3k 2

ð5:5Þ

Substituting equation (5.4) into equation (5.1), substituting equation (5.5) into equation (5.2), and writing kj as kjλ, we obtain the same result, that is 2 2 k 2 ln kjk
ðk
1Þk 2
kjk þk ¼ 0

ðe0:5  kjk  kck and k  kck Þ

ð5:6Þ

That is to say, the maximum shear stress theory and the maximum distortion strain energy theory result in the same equation, i.e., equations (5.6). Equation (5.6) presents the optimum plastic depth, kjλ, for fixed k and λ. In other words, for fixed k and λ, if the plastic depth is determined by equation (5.6), then the plastic depth is the minimum; thus, the complexity of the autofrettage process is reduced, and the load-bearing capacity p = λpe can be ensured. If the actual plastic region depth kj < kjλ, that is, the pressure that a vessel can contain is lower than λpe and can be calculated by equations (5.1) or (5.2), then the pressure vessel is safe. If kjλ < kj < kj*, i.e., the pressure that a vessel can contain is higher than λpe and can

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205

be calculated by equations (5.1) or (5.2), then the pressure vessel is safe, but the autofrettage process is more complex. Thus, equation (5.6) is a key formula. When λ = 2, or p = 2pe, equation (5.6) is just equation (2.34), or k 2 ln kj2
k 2
kj2 þ 2 ¼ 0. If kj and k are related by equation (5.6) and a pressure vessel contains the 0 pressure p = pa, then, rT e ≡ σy in the plastic region, jrei j  ry ; When λ = 2, 0 rei ¼
ry . According to ref.[6], the equivalent residual stress at a general location in the plastic region is k 2
kj2 þ k 2 ln kj2 r0e r0h r0r ¼
¼1
ry ry ry ðk 2
1Þx 2

ð5:7Þ

The equivalent residual stress at a general location in an elastic region is r0e r0h r0r k 2 ðkj2
1
ln kj2 Þ ¼
¼ ðk 2
1Þx 2 ry ry ry

ð5:8Þ

Although the components of residual stresses in three directions (r0h , r0r , and r0z ) based on the maximum shear stress theory are different from those based on the maximum distortion strain energy theory, the equivalent residual stress based on both strength theories is the same. In fact, by combining equations (5.1) and (5.7) or combining equations (5.2) and (5.7), we can determine that if p ≤ 2pe, or λ ≤ 2, then jr0ei j  ry . If the practical plastic depth is larger than kj*, then jr0ei j [ ry , which is not allowed. If the practical plastic depth is larger than kjλ and smaller than kj*, then jr0ei j\ry , and the load-bearing capacity can be improved to k rpye \ rpy \2 rpye , but the autofrettage process is more complex. If the practical plastic depth is smaller than kjλ, then the autofrettage process is easier, but the load-bearing capacity is reduced to rpy \k rpye . When k ≥ kc and kj ≤ kj*, jr0ei j  ry ; when k ≥ kc and kj = kj*, r0ei ¼
ry , and p = 2pe; when k ≤ kc, kj = k (entirely yielded), jr0ei j\ry and a cylinder can bear entire yield loading, py. We focus on the case of k ≥ kc. rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 2
k From equation (5.6), we obtain k ¼ ln k 2jk
k þ 1. Therefore, the definition jk

domain of kjλ is that kjλ > e(λ−1)/2 and kjλ ≤ λ0.5. kjλ, i.e., equation (5.6) is plotted in figure 5.1, in which the intersection points between the straight line (kjλ = k) and each curve are o (1, 1), a (1.2071…, 1.2071…), b (1.43005…, 1.43005…), c (1.671272…, 1.671272…), d (1.93322…, 1.93322…), m (kc, kc), i.e. m (2.218457…, 2.218457…), f (2.52973…, 2.52973…) …. Furthermore, points n (∞, e0.5) and v (∞, 0) are at infinity, and the curve mn corresponds to λ = 2, the curve gov corresponds to λ = 1. The straight line kjλ = k divides the curves into two parts. The above parts of the straight line kjλ = k have no significance. Figure 5.1 is applicable to both the maximum shear stress theory and the maximum distortion strain energy theory. In figure 5.1,

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Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 5.1 – Curves of k 2 ln kjk2
ðk
1Þk 2
kjk2 þ k ¼ 0 for different λ. Curve 1, which has one branch, is for λ = 0. Curve 2, which has one branch, is for λ = 0.2. Curve 3 is for λ = 1, which includes the curve og and the horizontal coordinate axis. Curve 4 is for λ = 1.2, which has two parts. The intersection point of the straight line (kjλ = k) and curve 4 is point a. Coordinate of point a is a (kcλ, kcλ), or a (1.2071…, 1.2071…). Curve 5 is for λ = 1.4, which has two parts. The intersection point of the straight line (kjλ = k) and curve 5 is point b. Coordinate of point b is b (kcλ, kcλ), or b (1.43005…, 1.43005…). Curve 6 is for λ = 1.6, which has two parts. The intersection point of the straight line (kjλ = k) and curve 6 is point c. Coordinate of point c is c (kcλ, kcλ), or c (1.671272…, 1.671272…). Curve 7 is for λ = 1.8, which has two parts. The intersection point of the straight line (kjλ = k) and curve 7 is point d. Coordinate of point d is d (kcλ, kcλ), or d (1.93322…, 1.93322…). Curve 8 is for λ = 2, which has two parts. The intersection point of the straight line (kjλ = k) and curve 8 is point m. Coordinate of point m is m (kcλ, kcλ), or m (kc, kc), i.e. m (2.218 457 489 916 7…, 2.218 457 489 916 7…). Curve 9 is for λ = 2.2, which has two parts. The intersection point of the straight line (kjλ = k) and curve 9 is point f. Coordinate of point f is f (kcλ, kcλ), or f (2.52973…, 2.52973…). In conclusion, from equation (5.6) and figure 5.1, we can draw the following conclusions. (1) When λ ≤ 1 (curves 1 − 3), kjλ is always greater than k irrespective of k. This is meaningless in practice.

Mechanical Autofrettage Technology Under Low Load

207

(2) When λ > 1 (curves 4 − 9), the curves for equation (5.6) are divided into two branches, on the left branch, k < 1, and kjλ > k, this is meaningless in practice; on the right branch, the straight line (kjλ = k) divides the curves into two parts: above the straight line (kjλ = k), kjλ > k, this is meaningless in practice; below the straight line (kjλ = k), kjλ < k. Letting kjλ = k in equation (5.6) obtains the critical radius ratio corresponding to a certain λ, or kcλ, which is 2 2 2 kck ln kck
kðkck
1Þ ¼ 0 or

k¼

2 2 kck ln kck 2 kck
1

ð5:9Þ

Equation (5.9) is the maximum radius ratio when the whole wall is yielded while jr0ei j  ry under a certain λ. In this case, kjλ = k = kcλ. When k ≤ kcλ, jr0ei ry j\1 irrespective of kj even if kj = k. When kjλ = k, p = py = σylnk < λpe. Because 1 ≤ λ ≤ 2, 1 ≤ kcλ ≤ kc and 1 ≤ kjλ ≤ kj*. λ indicates the multiple of pe. Equation (5.9) is applicable for both the maximum shear stress theory (Tresca yield criterion) and the maximum distortion strain energy theory (Mises’s yield criterion). Equation (5.9) is plotted in figure 5.2.

FIG. 5.2 – Correlation between λ and kcλ. From equation (5.9) and figure 5.2, it is seen that the greater kcλ is, the greater λ is. (3) When k ≥ kc, the maximum of λ is 2 for any k. When λ = 2, kcλ becomes kc, kjλ becomes kj*, equation (5.6) becomes k 2 ln kj2
k 2
kj2 þ 2 ¼ 0 or equak or equation (2.33). kc = tions (2.34), and (5.9) becomes kk 2ln
1 ¼ 1 2.2184574899167…, called critical radius ratio, which is the solution of kcλ in equation (5.9) when λ = 2. (4) For the right of the two branches, kjλ decreases as k increases and increases as λ increases. k → ∞, kjλ = e(λ−1)/2 = kjλ∞. When k = ∞, 2

k ¼ lnkj2 þ 1

ð5:10Þ

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Autofrettage Technology and Its Applications in Pressured Apparatuses

Thus, only when λ > 1, the discussion about autofrettage is significant. Because compressive yield occurs when λ > 2 (if λ > 2, kjλ is higher than the value on curve 8 where r0ei ¼
ry ), curve 9 in figure 5.1 is meaningless. The horizontal axis can be regarded as the curve with λ = 1. Then, the meaningful part of figure 5.1, or the meaningful and possible plastic depth lies in the quasi-infinite area surrounded by the horizontal axis (λ = 1), the straight line kj = k, and the curve k 2 ln kj2
k 2
kj2 þ 2 ¼ 0 (curve 8 in figure 5.1). The coordinate of fourth vertexes of this quasi-infinite area (m, o, v, n) are shown in figure 5.1. The significant part of figure 5.1, or the significant and possible optimum plastic depth is re-plotted as shown in figure 5.3, which is applicable for both the maximum shear stress theory and the maximum distortion strain energy theory.

FIG. 5.3 – Significant and possible optimum plastic depth with p = λpe. From equation (5.6) and figure 5.3, it is seen that the greater k is, the smaller kjλ is. The application condition for equation (5.6) is k ≥ kcλ.

5.3

Analysis of Residual Stresses Under the Optimum Plastic Depth

When plastic depth kj = kjλ, whether based on the maximum shear stress theory or the maximum distortion strain energy theory, the relative location of the intersection of the three residual stress curves or the abscissa at which r0e ¼ 0 is

Mechanical Autofrettage Technology Under Low Load

x0 ¼

pffiffiffi k

209

ð5:11Þ

Based on the maximum shear stress theory, when x = x0, r0zx0 r0hx0 r0rx0 r0rmin k
1
ln k ¼ ¼ ¼ ¼ y0s ¼
2 ry ry ry ry

ð5:12Þ

Based on the maximum distortion strain energy theory, when x = x0, r0zx0 r0hx0 r0rx0 r0rmin k
1
ln k pffiffiffi ¼ ¼ ¼ ¼ y0d ¼
ry ry ry ry 3 2 s d y0 ¼ pffiffiffi y0 3

ð5:13Þ

Whether based on the maximum shear stress theory or the maximum distortion pffiffiffiffiffiffiffiffiffi strain energy theory, the relative location where r0z ¼ 0 is x1 ¼ ek
1 . X0, x1, y0s , and y0d are not related to k or kj, only related to λ. When kj = kjλ and k ≥ kcλ, by means of equation (5.6), the complex expressions of residual stresses in general condition, or equations (2.2)–(2.9), can be simplified and re-expressed as follows. The results based on the maximum shear stress theory are as follows. In the plastic region: r0z r k
1 k
1 ¼ ln x
ð5:14Þ ¼ ln
ri 2 2 ry r0r k k ¼ ln x þ 2
2x 2 ry

ð5:15Þ

r0h k k ¼ ln x
2
þ 1 2x 2 ry

ð5:16Þ

r0e k ¼1
2 x ry

ð5:17Þ

r0z , r0r , r0h , and r0e are related with λ but not with kj, which means that for some λ, the curves of residual stresses in whichever direction (axial, radial and hoop direction) and equivalent residual stress under different kjλ and k coincide in the plastic region. In the elastic region: 2 r0z kjk
k  2k 2 ry

r0r ¼ ry

  2 2 k 2 r0z kjk
k kjk
k 1
2 ¼
x ry 2k 2 2x 2

ð5:18Þ

ð5:19Þ

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Autofrettage Technology and Its Applications in Pressured Apparatuses

r0h ¼ ry



 2 2 k 2 r0z kjk
k kjk
k 1þ 2 ¼ þ x ry 2k 2 2x 2

ð5:20Þ

2 r0e kjk
k ¼ ry x2

ð5:21Þ

r0zi k
1 ¼
2 ry

ð5:22Þ

r0ri ¼0 ry

ð5:23Þ

r0hi r0 r0 ¼ 1
k ¼ 2 zi ¼ ei ry ry ry

ð5:24Þ

r0ei ¼1
k ry

ð5:25Þ

2 r0zo kjk
k ¼ ry 2k 2

ð5:26Þ

r0ro ¼ 0

ð5:27Þ

2 kjk
k r0ho r0 ¼ 2 zo ¼ ry ry k2

ð5:28Þ

2 r0e kjk
k ¼ ry k2

ð5:29Þ

On the inner surface,

On the outer surface,

According to the DS transition rule in chapter 1, if based on the maximum distortion strain energy theory (Mises’s yield criterion), the residual stresses when kj = kjλ and k ≥ kcλ are corresponding as follows: r0z 2 r k
1 ln x 2 k
1 ¼ pffiffiffi ln
pffiffiffi ¼ pffiffiffi
pffiffiffi ry 3 ri 3 3 3

ð5:30Þ

Mechanical Autofrettage Technology Under Low Load

211

r0r ln x 2 k k ¼ pffiffiffi þ pffiffiffi
pffiffiffi 2 ry 3x 3 3

ð5:31Þ

r0h ln x 2 k 2
k ¼ pffiffiffi
pffiffiffi þ pffiffiffi 2 ry 3x 3 3

ð5:32Þ

r0e k ¼1
2 x ry

ð5:33Þ

r0z kjk
k  pffiffiffi ry 3k 2

ð5:34Þ

2

r0z ¼ ry r0h ¼ ry

  2 2 k 2 r0z kjk
k kjk
k 1
2 ¼ pffiffiffi
pffiffiffi x ry 3k 2 3x 2  1þ

 2 2 k 2 r0z kjk
k kjk
k p ffiffi ffi p ffiffiffi ¼ þ x 2 ry 3k 2 3x 2

ð5:35Þ

ð5:36Þ

2 r0e kjk
k ¼ ry x2

ð5:37Þ

r0zi k
1 ¼
pffiffiffi ry 3

ð5:38Þ

r0ri ¼0 ry

ð5:39Þ

r0hi 2 r0 2 r0 ¼
pffiffiffi ðk
1Þ ¼ 2 zi ¼ pffiffiffi ei ry ry 3 3 ry

ð5:40Þ

r0ei ¼1
k ry

ð5:41Þ

2 r0zo kjk
k ¼ pffiffiffi ry 3k 2

ð5:42Þ

On the inner surface,

On the outer surface,

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Autofrettage Technology and Its Applications in Pressured Apparatuses

r0ro ¼ 0

ð5:43Þ

2 r0ho r0 2 kjk
k ¼ 2 zo ¼ pffiffiffi ry ry 3 k2

ð5:44Þ

r0e kjk
k ¼ ry k2

ð5:45Þ

2

In both the plastic and elastic region, the components of residual stresses in three pffiffi directions based on the maximum shear stress theory (Tresca yield criterion) are 23 times of those based on the maximum distortion strain energy theory, and the equivalent residual stress based on both strength theories is the same. The reason for this situation has been explained in chapters 2 and 3. In the case of the maximum shear stress theory (Tresca yield criterion), when plastic depth kj = kjλ, the residual stresses are illustrated figures 5.4–5.6. In figure 5.4, curves 1–5 are circumferential residual stresses in the plastic zone under the state of entire yielding with different λ, curves 4a–4d are circumferential residual stresses in an elastic zone with different k and λ = 1.8, where curve 1 is for λ = 1.2 and kcλ = 1.207094; curve 2 is for λ = 1.4 and kcλ = 1.430048; curve 3 is for λ = 1.6 and kcλ = 1.671272; curve 4 is for λ = 1.8 and kcλ = 1.93322; curve 5 is for λ = 2 and kcλ = kc = 2.2184574899167; curve 4a is for λ = 1.8, k = 2 and kjλ = 1.736906; curve 4b is for λ = 1.8, k = 3 and kjλ = 1.539944; curve 4c is for λ = 1.8, k = 4 and kjλ = 1.515107; curve 4d is for λ = 1.8, k = ∞ and kjλ = e0.4. The parameters (kjλ, kcλ, etc.) are calculated by equations (5.6) and (5.9).

FIG. 5.4 – Circumferential residual stresses with different k and k under Tresca yield criterion.

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213

Figure 5.5 is circumferential residual stresses in the whole wall with different λ when k = 4 and kj = kjλ, where curve 1 is for λ = 1.2 and kjλ = 1.105972; curve 2 is for λ = 1.4 and kjλ = 1.225275; curve 3 is for λ = 1.6 and kjλ = 1.360485; curve 4 is for λ = 1.8 and kjλ = 1.515107; curve 5 is for λ = 2 and kjλ = kj* = 1.694172.

FIG. 5.5 – Circumferential residual stresses with different k and k = 4 under Tresca yield criterion.

Figure 5.6 is equivalent residual stresses in the whole wall with different λ when k = 4 and kj = kjλ, where curve 1 is for λ = 1.2 and kjλ = 1.105972; curve 2 is for λ = 1.4 and kjλ = 1.225275; curve 3 is for λ = 1.6 and kjλ = 1.360485; curve 4 is for λ = 1.8 and kjλ = 1.515107; curve 5 is for λ = 2 and kjλ = kj* = 1.694172.

FIG. 5.6 – Equivalent residual stresses with different k and k = 4 under Tresca yield criterion.

Autofrettage Technology and Its Applications in Pressured Apparatuses

214

r0

The dash curve in figure 5.4 is the connection of rhoy for λ = 1.8 and all k. This dash curve is obtained as follows: for some λ, from equation (5.9), kcλ is obtained; kcλ is the maximum kjλ under this λ; for kjλ = e(λ−1)/2 (corresponding to k = ∞) to kcλ, from equation (5.6), k2 is obtained for this λ; substituting k2 into equation (5.28) r0

r0

obtains rhoy ; taking k (=kcλ to ∞) as the horizontal coordinate and rhoy as the vertical coordinate, this dash curve is formed. For example, for λ = 1.8, from equation (5.9), kcλ = 1.93322; for kjλ = e(λ−1)/2 = e0.4 = 1.491825 (corresponding to k = ∞) to 1.93322, from equation (5.6), k2 is obtained for this λ = 1.8; substituting k2 into equation (5.28) obtains r0ho ry

r0ho ry ;

taking k = kcλ = 1.93322 to ∞ as the horizontal coor-

dinate and as the vertical coordinate, the dash curve in figure 5.4 is formed. Figure 5.7 shows the axial residual stress together with radial residual stress and circumferential residual stress for k = 4, λ = 1.8 and kjλ = 1.515107, that is curve 4 in figures 5.4–5.6. The coordinate point of the three curves for pofffiffiffi the crossover  s k
1
ln k residual stresses is w (x0, y0 ) = w k;
2 = w (1.341641…, −0.10611…).

FIG. 5.7 – The distribution of residual stresses in three directions for k = 4 and k = 1.8 under Tresca yield criterion. In the case of the maximum distortion strain energy theory (Mises’s yield criterion), when plastic depth kj = kjλ, the residual stresses are illustrated figures 5.8 and 5.10. In figure 5.8, curves 1 to 5 are circumferential residual stresses in a plastic zone under the state of entire yielding with different λ, curves 4a–4d are circumferential residual stresses in an elastic zone with different k and λ = 1.8, where curve 1 is for λ = 1.2 and kcλ = 1.207094; curve 2 is for λ = 1.4 and kcλ = 1.430048; curve 3 is for λ = 1.6 and kcλ = 1.671272; curve 4 is for λ = 1.8 and kcλ = 1.93322; curve 5 is for λ = 2 and kcλ = kc = 2.2184574899167; curve 4a is for λ = 1.8, k = 2 and

Mechanical Autofrettage Technology Under Low Load

215

kjλ = 1.736906; curve 4b is for λ = 1.8, k = 3 and kjλ = 1.539944; curve 4c is for λ = 1.8, k = 4 and kjλ = 1.515107; curve 4d is for λ = 1.8, k = ∞ and kjλ = e0.4. The parameters (kjλ, kcλ, etc.) are calculated by equations (5.6) and (5.9). Figure 5.9 is circumferential residual stresses in the whole wall with different λ when k = 4 and kj = kjλ, where curve 1 is for λ = 1.2 and kjλ = 1.105972; curve 2 is for λ = 1.4 and kjλ = 1.225275; curve 3 is for λ = 1.6 and kcλ = 1.360485; curve 4 is for λ = 1.8 and kjλ = 1.515107; curve 5 is for λ = 2 and kjλ = kj* = 1.694172. Figure 5.10 is equivalent residual stresses in the whole wall with different λ when k = 4 and kj = kjλ, where curve 1 is for λ = 1.2 and kjλ = 1.105972; curve 2 is for λ = 1.4 and kjλ = 1.225275; curve 3 is for λ = 1.6 and kcλ = 1.360485; curve 4 is for λ = 1.8 and kjλ = 1.515107; curve 5 is for λ = 2 and kjλ = kj* = 1.694172.

FIG. 5.8 – Circumferential residual stresses with different k and k under Mises yield criterion. r0

The dash curve in figure 5.8 is the connection of rhoy for λ = 1.8 and all k. This dash curve is made out as follows: for some λ, from equation (5.9), kcλ is obtained; kcλ is the maximum kjλ under this λ; for kjλ = e(λ−1)/2 (corresponding to k = ∞) to kcλ, from equation (5.6), k2 is obtained for this λ; substituting k2 into equation (5.44) r0

r0

obtains rhoy ; taking k (=kcλ to ∞) as the horizontal coordinate and rhoy as the vertical coordinate, the dash curve is formed. For example, for λ = 1.8, from equation (5.9), kcλ = 1.93322; for kjλ = e(λ−1)/2 = e0.4 (corresponding to k = ∞) to 1.93322, from equation (5.6), k2 is obtained for this λ = 1.8; substituting k2 into equation (5.44) obtains

r0ho ry ;

taking k = kcλ = 1.93322 to ∞ as the horizontal coordinate and

vertical coordinate, the dash curve in figure 5.8 is formed. distortion strain energy theory is theory.

p2ffiffi 3

times of

r0ho ry

r0ho ry

r0ho ry

as the

based on the maximum

based on the maximum shear stress

216

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 5.9 – Circumferential residual stresses with different k and k = 4 under Mises yield criterion.

FIG. 5.10 – Equivalent residual stresses with different k and k = 4 under Mises yield criterion. Figure 5.11 shows the axial residual stress together with radial residual stress and circumferential residual stress for k = 4, λ = 1.8 and kjλ = 1.515107, that is curve 4 in figures 5.8–5.10. The coordinate point of the three curves pffiffiffi of the crossover  d k
1
ln k = s (1.341641…, −0.12252…). for residual stresses is s (x0, y0 ) = s k;
pffiffi 3

r0

r0

It is easy to prove that, if λ is fixed, the three curves for residual stress (ryz , ryr and pffiffiffi  0 rh k k;
k
1
ln (based on the ry ) collect at a fixed point within the plastic zone: 2 pffiffiffi  pffiffi k (based on the maximum distormaximum shear stress theory) or k;
k
1
ln 3 tion strain energy theory) for any k, kj and λ, and the coordinate of the intersection is

Mechanical Autofrettage Technology Under Low Load

217

FIG. 5.11 – The distribution of residual stresses in three directions for k = 4 and k = 1.8 under Mises yield criterion. not related with k and kj but only with λ. If plastic kj ≠ kjλ, this situation does not happen, the coordinate of the intersection is related not only with λ but also with k and kj. r0

According to equations (5.25) or (5.41), at the inner surface, reiy ¼ 1
k and r0ei is the most dangerous. When λ ≥ 1, r0ei  0, this indicates compressive stress; when λ ≤ 2, r0ei 
ry . Thus when 1 ≤ λ ≤ 2, r0ei is safe. Since kj ≥ λ0.5 (kj ≥ e(λ−1)/2 at the same time), when λ ≤ 2, kj2 –λ > 0 in the elastic zone, i.e. r0e [ 0 (tension) in an elastic zone. At an elastic-plastic juncture where x = kjλ, equivalent residual stress is the maximum (algebraic value, not absolute value) in the whole elastic zone, or r0ej ry

¼

2 kjk
k 2 kjk

¼ 1
kk2 . Clearly, 0 < r0ej < σy. jk

Since k = ∞, kjλ = e(λ−1)/2 = kjλ∞, then from equations (5.18)–(5.21), in the whole elastic region, r0z  0; for x ¼ eðk
1Þ=2  1

ð5:46Þ

r0r ek
1
k ¼
; for x ¼ eðk
1Þ=2  1 2x 2 ry

ð5:47Þ

r0h ek
1
k r0 ¼ ¼
r ; for x ¼ eðk
1Þ=2  1 2 2x ry ry

ð5:48Þ

r0e ek
1
k k
1Þ=2 ¼ ; for x ¼ eð 1 x2 ry

ð5:49Þ

218

Autofrettage Technology and Its Applications in Pressured Apparatuses

In elastic zone, x = e(λ−1)/2  ∞, therefore, when k = ∞, in elastic zone, with r0

k
1

r0

k
1

r0

k
1


k
k x = e(λ−1)/2 to ∞, r0z  0; ryr ¼
e2ek
1 to 0; ryh ¼ e2ek
1 to 0; rye ¼ e ek
1
k to 0. When k = ∞, based on the maximum distortion strain energy theory (Mises’s yield criterion), the residual stresses in the elastic region are as follows.

r0z  0; for x ¼ eð

k
1Þ=2

1

ð5:50Þ

r0r ek
1
k k
1Þ=2 ¼
pffiffiffi ; for x ¼ eð 1 2 ry 3x

ð5:51Þ

r0h ek
1
k r0 k
1Þ=2 ¼ pffiffiffi ¼
r ; for x ¼ eð 1 2 ry ry 3x

ð5:52Þ

r0e ek
1
k k
1Þ=2 ¼ ; for x ¼ eð 1 2 x ry

ð5:53Þ

In elastic zone, x = e(λ−1)/2 to ∞, therefore, when k = ∞, in elastic zone, from k
1 r0 r0 r0 ek
1
k ek
1
k ffiffi k
1 ffiffi k
1 x = e(λ−1)/2 to ∞, r0z  0; ryr ¼
p to 0; rhy ¼ p to 0; rye ¼ e ek
1
k to 0. 3e 3e The residual stresses in the plastic region are still equations (5.14)–(5.17) and equations (5.30)–(5.33) when k = ∞. When r0e ¼ 0, x = λ0.5, which is just the abscissa of the intersection of the three curves for the residual stress and depends on λ only. Further description of curve 4 and curves 4a–4d in figure 5.8 is as follows, the other curves are the same. When λ = 1.8, from equation (5.13), we obtain kcλ = 1.93322. kcλ is the maximum plastic depth kj for λ = 1.8. If plastic depth kj = k, (entirely yielded), then k = kcλ = kj = kjλ. In this case, there is only a plastic region and no elastic region, or the whole wall is the plastic region, the curve for residual stresses is curve 4 or curve fo, which is the longest for all k under λ = 1.8. The greater k is, the smaller kjλ is. If k > kcλ = 1.93322, the corresponding kjλ < 1.93322. (1) k = 2 (λ = 1.8), from equation (5.6), we obtain kjλ = 1.736906…. In this case, the curve for residual stresses in the plastic region is curve fn and the curve for residual stresses in the elastic region is curve ns, the section no is dropped out. Within the plastic region, x varies from 1 to kjλ = 1.7369065 (point f to n), within the elastic region, x varies from kjλ to k (=2) (point n to s). (2) k = 3 (λ = 1.8), from equation (5.6), we obtain kjλ = 1.539944…. In this case, the curve for residual stresses in the plastic region is curve fm and the curve for residual stresses in the elastic region is curve mt, the section mo is dropped out. Within the plastic region, x varies from 1 to kjλ = 1.539944 (point f to m), within the elastic region, x varies from kjλ to k (=3) (point m to t). (3) k = 4 (λ = 1.8), from equation (5.6), we obtain kjλ = 1.515107…. In this case, the curve for residual stresses in the plastic region is curve fh and the curve for

Mechanical Autofrettage Technology Under Low Load

219

residual stresses in the elastic region is curve hu, the section ho is dropped out. Within the plastic region, x varies from 1 to kjλ = 1.515107 (point f to h), within the elastic region, x varies from kjλ to k (=4) (point h to u). (4) k = ∞ (λ = 1.8), from equation (5.6), we obtain kjλ = e0.4 = 1.491825…. In this case, the curve for residual stresses in the plastic region is curve fg and the curve for residual stresses in the elastic region is curve gv, the section go is dropped out. Within the plastic region, x varies from 1 to kjλ = e0.4 (point f to g), within the elastic region, x varies from kjλ to k (=∞) (point g to v). Point v is at infinity, the coordinate is v (∞, 0). From figures 5.4 and 5.8. Equations (5.16), and (5.32), it is known that all curves of the circumferential residual stresses under a fixed λ for any k and kjλ within the plastic region are located on the identical curve, such as, when λ = 1.8, all curves of the circumferential residual stresses under a fixed λ = 1.8 for any k and kjλ is fo, except that the different curve for different k and kjλ is located on a different section of curve fo. Saying, the above curves in the plastic zone, fo (k = kjλ = kcλ), fn (k = 2, kjλ = 1.736906), fm (k = 3, kjλ = 1.539944), fh (k = 4, kjλ = 1.515107), fg (k = ∞, kjλ = e0.4) are all on curve fo, or they partly coincide with each other. The greater k is, the shorter the curves for the circumferential residual stresses within the plastic region are, with the curve under k = kcλ = kj = kjλ being the longest, such as, when k = 4 and λ = 1.8, the longest curve for the circumferential residual stresses within the plastic region is curve fo, where k = kcλ = kj = kjλ = 1.93322. When based on the maximum shear stress theory (Tresca yield criterion), if λ ≤ 1, r0hi 
ry and r0ei 
ry can be guaranteed meanwhile. When based on the maximum distortion strain energy theory (Mises’s yield criterion), if λ ≤ 1, r0ei 
ry can be guaranteed but r0hi \
ry . In order for r0hi 
ry , from pffiffi equation (5.40), k  2 þ2 3. This confirms the fact that the load-bearing capacity is pffiffi p pe 3 þ 2 pe 0 ry ¼ 2 ry ¼ k ry when the circumferential residual stress rhi is controlled in chapter 4. When x < xa,

5.4

r0e ry

[

r0h ry ;

when x > xa,

r0e ry

r0

\ ryh .

Analysis of the Stresses Caused by Internal Pressure and Total Stresses

When p = λpe, the stresses caused by internal pressure p at a general location, called operation stresses, are: rpz 1 p k ¼ 2 ¼ 2 ry k
1 ry 2k rpr ¼ ry



   k 2 rpz k 1 1 1
2 ¼
x ry 2 k 2 x 2

ð5:54Þ

ð5:55Þ

220

Autofrettage Technology and Its Applications in Pressured Apparatuses

rph ¼ ry

    k 2 rpz k 1 1 1þ 2 ¼ þ x ry 2 k 2 x 2

ð5:56Þ

Equivalent operation stress based on the maximum shear stress theory (Tresca criterion) is rp rp rps k e ¼ h¼ r ¼ 2 ry ry ry x

ð5:57Þ

Total stresses σT include residual stresses and operation stress. When p = λpe and kj = kjλ, the total stresses based on the maximum shear stress theory (Tresca yield criterion) are as follows. In the plastic region, rT k
1 k z þ 2 ¼ ln x
2 2k ry

ð5:58Þ

rT k k r ¼ ln x
þ 2 2 2k ry

ð5:59Þ

rT k k h ¼ ln x
þ 2 þ 1 2 2k ry

ð5:60Þ

rT rT rT r0 rp e ¼ h
r ¼ e þ e 1 ry ry ry ry ry

ð5:61Þ

Therefore, when a pressure vessel is subject to a pressure p = λpe and its plastic depth is determined by equation (5.6), then, equivalent total stress everywhere in plastic zone σe ≡ σy. In the elastic region, 2 kjk rT z ¼ 2 ð5:62Þ ry 2k  2  kjk rT 1 1 r ¼
ry 2 k2 x2

ð5:63Þ

 2  kjk rT 1 1 h ¼ þ ry 2 k2 x2

ð5:64Þ

2 rT rT r0 rT rp kjk e ¼ h
r ¼ e þ e ¼ 2 ry ry ry ry ry x

ð5:65Þ

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221

On the other hand, when based on the maximum shear stress theory (Tresca yield criterion), if

p ry

¼ rpay ¼

k 2 ln kj2 þ k 2
kj2 2k 2

(equation (5.1)), regardless of kj, there is

k 2 ln kj2 þ k 2
kj2 rpz 1 p ¼ 2 ¼ ry k
1 ry 2k 2 ðk 2
1Þ

ð5:66Þ

Using equations (5.66), the general equations of residual stresses or equations (2.2)–(2.9) are rewritten as follows: In the plastic region, ! r0z 1 kj2 rp 2 2 ¼ þ ln x
ln kj
z ð5:67Þ 2 ry 2 k ry r0r 1 kj2 ¼
1 þ ln x 2
ln kj2 ry 2 k 2 r0h 1 kj2 ¼ þ 1 þ ln x 2
ln kj2 ry 2 k 2

!


rpr ry

ð5:68Þ




rph ry

ð5:69Þ

!

 p  rh rpr r0e r0h r0r ¼
¼1

ry ry ry ry ry

ð5:70Þ

r0z 1 kj2 rpz ¼
ry 2 k 2 ry

ð5:71Þ

In the elastic region,

r0r 1 kj2 kj2 ¼
ry 2 k 2 x 2

!

r0h 1 kj2 kj2 ¼ þ ry 2 k 2 x 2




rpr ry

ð5:72Þ




rph ry

ð5:73Þ

!

 p  rh rpr r0e r0h r0r kj2 ¼
¼ 2

ry ry ry x ry ry Therefore, when p = pa, irrespective of kj, total stresses are as follows.

ð5:74Þ

222

Autofrettage Technology and Its Applications in Pressured Apparatuses

In the plastic region, rT r0 rp 1 kj2 z ¼ z þ z ¼ þ ln x 2
ln kj2 ry ry ry 2 k 2

!

rT r0 rp 1 kj2 r ¼ r þ r ¼
1 þ ln x 2
ln kj2 ry ry ry 2 k 2 rp 1 kj2 rT r0 h ¼ hþ h¼ þ 1 þ ln x 2
ln kj2 ry ry ry 2 k 2

ð5:75Þ ! ð5:76Þ ! ð5:77Þ

rT rT rT e ¼ h
r 1 ry ry ry

ð5:78Þ

rT r0 rp 1 kj2 z ¼ z þ z ¼ 2 ry ry ry 2 k

ð5:79Þ

In the elastic region,

rT r0 rp 1 kj2 kj2 r ¼ r þ r ¼
ry ry ry 2 k 2 x 2

!

rp 1 kj2 kj2 rT r0 h ¼ hþ h¼ þ ry ry ry 2 k 2 x 2 rT rT rT kj2 e ¼ h
r ¼ 2 ry ry ry x

ð5:80Þ ! ð5:81Þ

ð5:82Þ

When p = pa, irrespective of kj, the equivalent total stress is always in a safe range. If plastic depth kj = kjλ, the components of the total stresses and the equivalent total stress are in a safe range. If kj ≠ kjλ, the equivalent total stress is in a safe range, but the components of the total stresses are not necessarily within a safe range. When plastic depth kj = kjλ, equations (5.75)–(5.82) just becomes equations (5.58)–(5.65), respectively. rT

rT

rT

rT

When x < xa, rey [ rhy ; when x > xa, rey \ rhy . When based on the maximum distortion strain energy theory (Mises’s yield criterion), accordingly we have the following results by replacing σy in the results based on the maximum shear stress theory with p2ffiffi3 ry . When p = λpe, the stresses caused by internal pressure p at a general location, called operation stresses, are:

Mechanical Autofrettage Technology Under Low Load rpz 1 p k ¼ ¼ pffiffiffi ry k 2
1 ry 3k 2

223

ð5:83Þ

rpr ¼ ry

    k 2 rp k 1 1 1
2 z ¼ pffiffiffi 2
2 x x ry 3 k

ð5:84Þ

rph ¼ ry

    k 2 rp k 1 1 1 þ 2 z ¼ pffiffiffi 2 þ 2 x x ry 3 k

ð5:85Þ

Equivalent operation stress

pffiffiffi  p  3 rh rpr rpd k rps z ¼
¼ 2¼ e x ry ry 2 ry ry

ð5:86Þ

When p = λpe and kj = kjλ, the total stresses are as follows. In the plastic region, rT ln x 2 k
1 k z ¼ pffiffiffi
pffiffiffi þ pffiffiffi ry 3 3k 2 3

ð5:87Þ

rT ln x 2 k k r ¼ pffiffiffi
pffiffiffi þ pffiffiffi ry 3 3k 2 3

ð5:88Þ

rT ln x 2 k k 2 h ¼ pffiffiffi
pffiffiffi þ pffiffiffi þ pffiffiffi 2 ry 3 3k 3 3

ð5:89Þ

pffiffiffi  T  3 rh rT rT r0 rp e ¼
r ¼ e þ e 1 ry ry ry 2 ry ry

ð5:90Þ

Therefore, similarly, when a pressure vessel is subject to a pressure p = λpe and its plastic depth is determined by equation (5.6), then, equivalent total stress everywhere in plastic zone σe ≡ σy. In the elastic region, 2 kjk rT z ¼ pffiffiffi ð5:91Þ ry 3k 2  2  kjk rT 1 1 r ¼ pffiffiffi 2
2 x ry 3 k

ð5:92Þ

 2  kjk rT 1 1 h ¼ pffiffiffi 2 þ 2 x ry 3 k

ð5:93Þ

224

Autofrettage Technology and Its Applications in Pressured Apparatuses

pffiffiffi  T  2 3 rh rT rT r0 rp kjk e r ¼
¼ e þ e ¼ 2 ry ry ry x 2 ry ry

ð5:94Þ

On the other hand, when based on the maximum distortion strain energy theory (Mises’s criterion), if rpy ¼ rpay ¼

k 2 ln kj2 þ k 2
kj2 pffiffi 3k 2

(equation (5.2)), regardless of kj, there is

k 2 ln kj2 þ k 2
kj2 rpz 1 p ¼ 2 ¼ pffiffiffi ry k
1 ry 3k 2 ðk 2
1Þ

ð5:95Þ

Using equations (5.66), the general equations of residual stresses or equations (2.2)–(2.9) are rewritten as follows: In the plastic region, ! r0z 1 kj2 rp ¼ pffiffiffi 2 þ ln x 2
ln kj2
z ð5:96Þ ry ry 3 k r0r 1 kj2 ¼ pffiffiffi 2
1 þ ln x 2
ln kj2 ry 3 k r0h 1 kj2 ¼ pffiffiffi 2 þ 1 þ ln x 2
ln kj2 ry 3 k

!


rpr ry

ð5:97Þ




rph ry

ð5:98Þ

!

pffiffiffi  0 pffiffiffi  p   3 rh r0r 3 rh rpr r0e rp ¼

¼1
¼1
e ry ry 2 ry ry 2 ry ry

ð5:99Þ

In the elastic region, r0z 1 kj2 rp ¼ pffiffiffi 2
z ry ry 3k r0r 1 kj2 kj2 ¼ pffiffiffi 2
2 ry x 3 k

ð5:100Þ

!

r0h 1 kj2 kj2 ¼ pffiffiffi 2 þ 2 ry x 3 k




rpr ry

ð5:101Þ




rph ry

ð5:102Þ

!

pffiffiffi  0 pffiffiffi  p   kj2 kj2 rp 3 rh r0r 3 rh rpr r0e ¼

¼ 2
¼ 2
e ry x x ry 2 ry ry 2 ry ry

ð5:103Þ

Mechanical Autofrettage Technology Under Low Load

225

Therefore, when p = pa, irrespective of kj, total stresses are as follows. In the plastic region, ! 2 0 p k rT r r 1 j z ¼ z þ z ¼ pffiffiffi 2 þ ln x 2
ln kj2 ð5:104Þ ry ry ry 3 k rT r0 rp 1 kj2 r ¼ r þ r ¼ pffiffiffi 2
1 þ ln x 2
ln kj2 ry ry ry 3 k rp rT r0 1 kj2 h ¼ h þ h ¼ pffiffiffi 2 þ 1 þ ln x 2
ln kj2 ry ry ry 3 k

! ð5:105Þ ! ð5:106Þ

pffiffiffi  T  3 rh rT rT e r ¼
1 ry 2 ry ry

ð5:107Þ

rT r0 rp 1 kj2 z ¼ z þ z ¼ pffiffiffi 2 ry ry ry 3k

ð5:108Þ

In the elastic region,

rT r0 rp 1 kj2 kj2 r ¼ r þ r ¼ pffiffiffi 2
2 ry ry ry x 3 k

!

rp rT r0 1 kj2 kj2 h ¼ h þ h ¼ pffiffiffi 2 þ 2 ry ry ry x 3 k pffiffiffi  T  kj2 3 rh rT rT e r ¼
¼ 2 ry x 2 ry ry

ð5:109Þ ! ð5:110Þ

ð5:111Þ

When p = pa, irrespective of kj, the equivalent total stress is always in a safe range. If plastic depth kj = kjλ, the components of the total stresses and the equivalent total stress are in a safe range. If kj ≠ kjλ, the equivalent total stress is in a safe range, but the components of the total stresses are not necessarily within a safe range. When kj = kjλ, equations (5.104)–(5.111) just becomes equations (5.87)–(5.94), respectively. rT

rT

rT

rT

When x < xa, rey [ rhy ; when x > xa, rey \ rhy . When kj = kjλ (p = pa) and λ = 1.2, total stresses based on the maximum shear stress theory (Tresca yield criterion) are plotted in figure 5.12; When kj = kjλ

226

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 5.12 – The total stresses when kj = kjk and k = 3, k = 1.2 under Tresca yield criterion.

FIG. 5.13 – The total stresses when kj = kjλ and k = 3, k = 1.8 under Tresca yield criterion. (p = pa) and λ = 1.8, total stresses based on the maximum shear stress theory (Tresca yield criterion) are plotted in figure 5.13. When kj = kjλ (p = pa) and λ = 1.2, total stresses based on the maximum distortion strain energy theory (Mises’s yield criterion) are plotted in figure 5.14; When kj = kjλ (p = pa) and λ = 1.8, total stresses based on the maximum distortion strain energy theory (Mises’s yield criterion) are plotted in figure 5.15. The components of the total stresses based on the maximum distortion strain energy theory (Mises yield criterion) are p2ffiffi3 times of those based on the maximum shear stress theory (Tresca yield criterion), the equivalent total stress is unchanged, kjλ and xa are unchanged.

Mechanical Autofrettage Technology Under Low Load

227

FIG. 5.14 – The total stresses when kj = kjλ and k = 3, k = 1.2 under Mises yield criterion.

FIG. 5.15 – The total stresses when kj = kjλ and k = 3, k = 1.8 under Mises yield criterion. From equations (5.62)–(5.65) and equations (5.91)–(5.94), seemingly, total stresses in the elastic zone have nothing to do with λ. However, kjλ depends on λ as seen in equation (5.6). It is thus obvious that it is equations (5.1) or (5.2) that ensures rT e /σy ≡ 1 2 2 everywhere in the plastic zone and rT /σ = k /x in elastic zone irrespective of kj y j e and k. Substituting equation (5.1) into the equation of the equivalent operation stresses, or equation (1.9), obtains the equivalent operation stresses based on the maximum shear stress theory (Tresca yield criterion), rps e

Autofrettage Technology and Its Applications in Pressured Apparatuses

228

rps 2k 2 p
2 k 2
kj2 þ k 2 ln kj2 e ¼ 2 x ¼ ry k
1 ry ðk 2
1Þx 2

ð5:112Þ

Substituting equation (5.2) into the equation of the equivalent operation stresses, or equation (1.10), obtains the equivalent operation stresses based on the maximum distortion strain energy theory (Mises’s yield criterion), rpd e pffiffiffi  p pffiffiffi 2  2 2 2 2 pd p k
kj þ k ln kj 3 rh rr 3k re p ¼
¼ ð5:113Þ ¼ 2 2 ry x ðk
1Þ ry ðk 2
1Þx 2 2 ry ry Equations (5.112) and (5.113) are the same, the reason is that pda ry

rpd e ry

¼

pffiffi ps 3 re 2 ry ,

but

ps p2ffiffi a . 3 ry

¼ Substituting equations (5.112) and (5.7) into equation (5.61) or substituting equations (5.113) and (5.7) into equation (5.90) just obtains rT e  ry ; Substituting equations (5.112) and (5.8) into equation (5.65) or substituting equations (5.113)

and (5.8) into equation (5.94) just obtains rT

rT e ry

k2

¼ xj2 . So, as long as p = pa, rT e ≡ σy in

k2

the plastic zone and rey ¼ xj2 in the elastic zone are inexorable law irrespective of kj and k. In other words, providing p = pa, for any kj and k, which need not be related rT

k2

j 2 2 e by equation (5.6), rT e  ry in the plastic zone and ry ¼ x 2 (0 < kj /x < 1, for kj ≤ x ≤ k) in the elastic zone are inevitable. However, kj and k affect residual stresses. Inadequate kj for a certain k may cause compressive yield, and it is necessary for kj to be smaller than the value determined by equation (5.6), otherwise, compressive yield occurs.

5.5

Analysis of the Effect of Load Ratio (λ) and Plastic Depth (kjλ)

According to equation (5.6), the effect of λ on kjλ is concluded as follows. When k ≤ kc, kjλmax = k = kcλ, and λmax ≤ 2; When k ≥ kc, kjλmax < k, kjλmax is 2 2
k 2
kjk þ 2 ¼ 0 (This equation is corresponding to λ = 2), determined by k 2 ln kjk and λmax = 2. For a fixed k, there are two values of kjλ corresponding to the same λ according to equation (5.6): one is greater than k, which is of no practical significance; another is smaller than k.

dkjk dk

k ðk 2
1Þ

jk ¼ 2ðk 2
k 2 Þ. When kjλ = k, jk

dk

dkjk dk

¼ 1; when kjλ > k,

dkjk dk

\0, which is

of no significance; when kjλ < k, dkjk [ 0, which is of great significance. Thus, for a certain k, when kjλ < k, the greater λ is, the greater kjλ is. When kjλ = k, equa2 k tion (5.6) becomes equation (5.7). When λ = 2, equation (5.7) becomes kk 2ln
1 ¼ 1, the solution of which is k = kc = 2.2184574899167….

Mechanical Autofrettage Technology Under Low Load

229

When k > kjλ, as k continuously increases, kjλ corresponding to various k becomes increasingly closer, or when k is sufficiently large, kjλ is almost unchangeable. For k = ∞, if λ = 2, kj1 = e0.5. Equation (5.6) applies to both the maximum shear stress theory and the maximum distortion strain energy theory. In order to intuitively understand the effect of λ on kjλ, we plot equation (5.6) to show the correlation between λ and kjλ. To plot equation (5.6), we first rewrite the equation as follows. k¼ Setting

dk dkjk

2 2 k 2 ln kjk þ k 2
kjk

k2
1

ð5:114Þ

¼ 0 leads to kjλ = k. That is to say, when kjλ = k, λ reaches a

maximum. Equation (5.114) is plotted as shown in figures 5.16 and 5.17.

FIG. 5.16 – The correlation between λ and kjλ under a certain k.

In figure 5.16, curve 1: k = 1.1, λmax = 1.098336; curve 2: k = 1.2, λmax = 1.193377; curve 3: k = 1.3, λmax = 1.285205; curve 4: k = 1.4, λmax = 1.373928; curve 5: k = 1.5, λmax = 1.459674; curve 6: k = 1.6, λmax = 1.542576; curve 7: k = 1.7, λmax = 1.622768; curve 8: k = 1.8, λmax = 1.700383; curve 9: k = 1.9, λmax = 1.77555; curve 10: k = 2, λmax = 1.848392;

230

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 5.17 – The correlation between λ and k under a certain kjλ.

curve 11: k = 2.1, λmax = 1.919029; curve 12: k = kc, λmax = 2; curve 13: k = 2.5, λmax = 2; curve 14: k = 3, λmax = 2; curve 15: k = 4, λmax = 2; curve 16: k = 5, λmax = 2; curve 17: k = ∞, λmax = 2. Points 1–12 and a are peak points, points 13, 14, …, b are not peak points. The dash curve oa is equation (5.9), or k ¼ kk 2ln
1k ¼ 2

When k ≤ kc, kjλmax = k = kcλ, kmax ¼ kk 2ln
1k ¼ 2

2

2

2 2 kck ln kck 2
1 kck

2 2 kck ln kck 2
1 kck

= λmax.

, such as curves 1–12 in

figure 5.16. When k ≥ kc, kjλmax = kj* < k, λmax = 2, such as curves 12–17 in figure 5.16. 2 (based on When k ≤ kc, equation (5.6) still holds because k rpye ¼ rpay \ ln k\ k k
1 2 2 (based on the the maximum shear stress theory) or k rpye ¼ rpay \ p2ffiffi3 ln k\ p2ffiffi3 k k
1 2 maximum distortion strain energy theory). This is proved as follows. When based on the maximum distortion strain energy theory, substituting equation (5.6) into equations (5.1) or (5.2) results in k rpye ¼ rpay . Substituting equation (5.114) into k rpye also results in k rpye ¼ rpay .

From k rpye ¼ k k2k
1 2 \ ln k, we obtain 2

k 2 ln k 2 [k k2
1

ð5:115Þ

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231

Substituting equation (5.114) into equation (5.115) obtains k 2 ln k 2 [ k 2 ln kj2 þ k 2
kj2 ¼ a  da dkj

¼ 2kj

k2 kj2

ð5:116Þ


1 [ 0. When kj = 1, a = k2−1 1.2. k rpye ¼ 1:2 1:6 21:62 =

1:62
1:1141082 21:62

2

= 0.365625. lnk = ln1.6 =

When k = 1.6, kj = kjλ = 1.114108, p = pa and λ = 1.2, 0.470004, total stresses based on the maximum shear stress theory (Tresca yield criterion) are plotted in figure 5.18. If based on the maximum distortion strain energy theory (Mises’s yield criterion), when k = 1.6 and λ = 1.2, from equation (5.6), kjλ = 1.114108; from

232

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 5.18 – The total stresses when kj = kjλ and k = 1.6, k = 1.2 under Tresca yield criterion. ffiffi
12 = 0.422187. equation (5.9), kcλ = 1.207098 < k. k rpye ¼ 1:2 p1:6 31:6 2

pa ry

¼ p2ffiffi3 ln kj þ

k 2
kj2 pffiffi 3k 2

2 ffiffi 2 2 = 0.422187. p2ffiffi ln k = 0.542713, p2ffiffi k 2k
1 ¼ p2ffiffi3 ln 1:114108 þ 1:6 p
1:114108 = 31:6 3 3 2 0.703646. When k = 1.6, kj = kjλ = 1.114108, p = pa and λ = 1.2, total stresses based on the maximum distortion strain energy theory (Mises’s yield criterion) are plotted in figure 5.19.

FIG. 5.19 – The total stresses when kj = kjλ and k = 1.6, k = 1.2 under Mises yield criterion. If based on the maximum shear stress theory (Tresca yield criterion), when k = 2 and λ = 1.2, from equation (5.6), kjλ = 1.109443; from equation (5.9), pa 22
1 lnk = ln2 = 0.693147, kcλ = 1.207098 < k. k rpye ¼ 1:2 22 2 = 0.45. ry = 0.45.

Mechanical Autofrettage Technology Under Low Load

233

k 2
1 k2

= 0.75. When k = 2, kj = kjλ = 1.109443, p = pa and λ = 1.2, total stresses based on the maximum shear stress theory (Tresca yield criterion) are plotted in figure 5.20.

FIG. 5.20 – The total stresses when kj = kjλ and k = 2, k = 1.2 under Tresca yield criterion. If based on the maximum distortion strain energy theory (Mises’s yield criterion), when k = 2 and λ = 1.2, from equation (5.6), kjλ = 1.109443; from equa2
1 = 0.519615. rpay = 0.519615. p2ffiffi3 ln k = tion (5.9), kcλ = 1.207098 < k. k rpye ¼ 1:2 p2ffiffi32 2

= 0.866025. When k = 2, kj = kjλ = 1.109443, p = pa and λ = 1.2, 0.800377, p2ffiffi3 k k
1 2 total stresses based on the maximum s distortion strain energy theory (Mises’s yield criterion) are plotted in figure 5.21. 2

FIG. 5.21 – The total stresses when kj = kjλ and k = 2, k = 1.2 under Mises yield criterion.

234

5.6 5.6.1

Autofrettage Technology and Its Applications in Pressured Apparatuses

Analysis of Load-Bearing Capacity Based on Tresca Criterion

For a pressure vessel with k and kjλ, its load-bearing capacity p = λpe. When based on the maximum shear stress theory (Tresca criterion), the load-bearing capacity is plotted in figure 5.22, where the coordinates of relevant points are o (1, 0), a (1.2071…, 0.18822…), b (1.43005…, 0.357709…), c (1.671272…, 0.513585…), d (1.93322…, 0.659187…), e (2.2184574899167…, 0.796812…).

FIG. 5.22 – Load-bearing capacity for a pressure vessel with k and kjλ based on the maximum shear stress theory. Given the data in figure 5.22, the following may be drawn to show the significance and applications of figure 5.22. When λ = 1.2, kcλ = 1.2071…, if k ≤ kcλ, |r0ei |≤σy irrespective of kj even if kj = k; if kj = k < kcλ, p = py = σylnk ( 1.2. Then, rpy ¼ k rpye ¼ k k2k
1 2 = 0.365625 < r ¼ ln k y = 0.470004. When k = 1.6, λ = 1.2, from equation (5.6), kjλ = 1.114108; When λ = 1.2, from equation (5.9), kcλ = 1.207098 < k. When k = 1.6, kj = kjλ = 1.114108, p = pa and λ = 1.2, total stresses based on the maximum shear stress theory (Tresca yield criterion) has been plotted in figure 5.18. (II) k = 1.6 < kc, λ = 1.55 2 p 1:62
1 k k2k
1 λmax = 1.542576 < 1.55. 2 ¼ 1:55 21:62 = ry ¼ ln k = 0.470004; 0.472266. Letting k 2 ln k 2 k 2
1

p ry

¼ ln k = 0.470004 = k k2k
1 obtains λ = kmax ¼ 2 2

= 1.542576. When λ = 1.55, from equation (5.9), kcλ = 1.60912 > k.

236

Autofrettage Technology and Its Applications in Pressured Apparatuses When kj = k and p = py = σylnk, the total stresses are equations (2.70)– ≡ σy in the plastic region. Since (2.73), where rT e 2 p k 2
1 1:62
1 T k 2k 2 ¼ 1:55 21:62 > ry ¼ ln k, if k k2k
1 2 is taken as the load, re > σy. In fact, when λ > λmax, the maximum plastic depth is kj = k. When kj = k, from equation (2.60), the equivalent residual stress on inner surr0

from equaface is reiy ¼ 1
kk 2ln
1k = −0.542576. When rpy ¼ k k2k
1 2 , tion (5.57), the equivalent operation stress on inner surface is rpei ry rT ei ry

2

2

2

= λ = 1.55. Then, the equivalent total stress on inner surface is

= 1.55–0.542576 = 1.007424 > 1. (III) k = 2.5 > kc. 2 2 kmax ¼ kk 2ln
1k = 2.181645. When k > kc, λmax = 2. When k = 2.5, λ = λmax = 2, from equation (5.6), kjλ = kj* = 1.844363. When kj = kj*, the 2 pa optimum load is rpy ¼ k rpye ¼ k k2k
1 2 ¼ r = 0.84 > 0.796812. y (2) Given k If k > kc, λmax = 2. If k < kc, from equation (5.9), kmax ¼ kk 2ln
1k ¼ 2

2

2 2 kck ln kck 2
1 kck

(When

k > kc, λmax > 2; When k < kc, λmax < 2.). Between 1 < λ < λmax, select a λ to meet design requirement. When k > kc, if the selected λ > 2, r0ei < − σy. When k < kc, if λ > λmax is selected, the load-bearing capacity can not reach rpy ¼ k rpye ¼ k k2k
1 2 under the given k, because when k < kcλ, the maximum allowable 2

load is

p ry

¼ ln k, and when k < kc, ln k  k k2k
1 2 . 2

(I) k = 1.6 < kc λmax = 1.542576. (A) Select λ = λmax = 1.542576, ln k ¼ k k2k
1 2 = 0.470004. λ = λmax indicates kj = kjλ = k. 2 (B) Select λ = 1.4, lnk = 0.470004 > k k2k
1 2 = 0.426563. 2

(C) Select λ = 1.6, lnk = 0.470004 < k k2k
1 2 = 0.4875. When λ > λmax, the maximum plastic depth is kj = k. When kj = k, from equation (2.60), the equivalent residual stress on inner surface is 2

r0ei ry

¼ 1
kk 2ln
1k = −0.542576. When 2

2

p ry

¼ k k2k
1 2 , from equation (5.57), 2

the equivalent operation stress on inner surface is the equivalent total stress 0.542576 = 1.057424 > 1.

on

inner

rpei ry

= λ = 1.6. Then,

surface

is

rT ei ry

= 1.6–

(II) k = 2.5 > kc 2 2 kmax ¼ kk 2ln
1k = 2.181645. When k > kc, λmax = 2. When k = 2.5, λ = λmax = 2, from equation (5.6), kjλ = kj* = 1.844363. When kj = kj*, the 2 pa optimum load is rpy ¼ k rpye ¼ k k2k
1 2 ¼ r = 0.84 > 0.796812. y

Mechanical Autofrettage Technology Under Low Load

237

If λ = 1.6 is selected, from equation (5.6), kjλ = 1.384516 < kj* = 2 1.844363. rpy ¼ k k2k
1 2 = 0.672 < 0.84. (3) Given p and λ e2p=ry 2p From equation (5.117), kmax ¼ e2p=r . If the given λ > λmax, there is no y
1 ry

solution; if the given λ < λmax, use rpy ¼ k rpye ¼ k k2k
1 2 to determine k. When the given p = 0.796812σy, λmax = 2; When the given p < 0.796812σy, λmax < 2; When the given p > 0.796812σy, λmax > 2. (4) Given p e2p=ry 2p From equation (5.117), kmax ¼ e2p=r . Between 1 < λ < λmax, select a λ to y
1 ry meet design requirement. 2

5.6.2

Based on Mises Criterion

When based on the maximum distortion strain energy theory (Mises’s yield criterion), the load-bearing capacity is plotted in figure 5.23, where the coordinates of relevant points are o (1, 0), a (1.2071…, 0.217338…), b (1.43005…, 0.413047…), c (1.671272…, 0.593037…), d (1.93322…, 0.761164…), e (2.2184574899167…, 0.920079…). Given the data in figure 5.23, the following may be drawn to show the significance and applications of figure 5.23.

FIG. 5.23 – Load-bearing capacity for a pressure vessel with k and kjλ based on the maximum distortion strain energy theory.

238

Autofrettage Technology and Its Applications in Pressured Apparatuses

When λ = 1.2, kcλ = 1.2071…, if k ≤ 1.2071… = kcλ, jr0ei j  ry | irrespective of kj even if kj = k; if kj = k < kcλ, p = py = σylnk ( λmax, then, optimum load is rpy ¼ k rpye ¼ k p 3k 2 k < kcλ. If rp ¼ p2ffiffi ln k is taken as the load (kj = k at the same time), rT e  ry in y

3

k ffiffi
1 k ffiffi
1 ; If k p is taken as the load, rT the plastic region, but rpy ¼ p2ffiffi3 ln k\k p e [ ry . 3k 2 3k 2 2

2

Mechanical Autofrettage Technology Under Low Load

239

(I) k = 1.6 < kc, λ = 1.2 p pe p k 2ffiffi
1 p Then, λmax = 1.542576 > 1.2. ry ¼ k ry ¼ k 3k 2 = 0.422187 < ry ¼ p2ffiffi ln k = 0.542713. When k = 1.6, λ = 1.2, from equation (5.6), kjλ = 3 1.114108; when λ = 1.2, from equation (5.9), kcλ = 1.207098 < k. When k = 1.6, kj = kjλ = 1.114108, p = pa and λ = 1.2, total stresses based on the maximum distortion strain energy theory (Mises’s yield criterion) has been plotted in figure 5.19. (II) k = 1.6 < kc, λ = 1.55 k 2ffiffi
1 ffiffi 2
12 = ¼ 1:55 p1:6 λmax = 1.542576 < 1.55. rpy ¼ p2ffiffi3 ln k = 0.542713; k p 3k 2 31:6 0.545325.

Letting k 2 ln k 2 k 2
1

p ry

k ffiffi
1 ¼ p2ffiffi3 ln k = 0.542713 = k p 3k 2 2

just

obtains

= 1.542576. λ = kmax ¼ When λ = 1.55, from equation (5.9), kcλ = 1.60912 > k. When kj = k and p = py = σylnk, the total stresses are equations (2.70)– in the plastic region. Since (2.73), where rT e  ry 2 kp2ffiffi
1 1:6
1 k 2ffiffi
1 T k 3k 2 ¼ 1:55 pffiffi31:62 > rpy ¼ p2ffiffi3 ln k, if k p is taken as the load, r [ ry . e 3k 2 In fact, when λ > λmax, the maximum plastic depth is kj = k. When kj = k, from equation (3.46), the equivalent residual stress on inner sur2 2 r0 k 2ffiffi
1 , from equaface is rei ¼ 1
kk 2ln
1k = −0.542576. When rp ¼ k p 2 y

3k

y

tion (5.86), the equivalent operation stress on inner surface is rpei ry rT ei ry

= λ = 1.55. Then, the equivalent total stress on inner surface is

= 1.55–0.542576 = 1.007424 > 1. (III) k = 2.5 > kc 2 2 kmax ¼ kk 2ln
1k = 2.181645. When k > kc, λmax = 2. When k = 2.5, λ = λmax = 2, from equation (5.6), kjλ = kj* = 1.844363. When kj = kj*, the k 2ffiffi
1 ¼ rpay = 0.969948 > 0.920079. optimum load is rpy ¼ k rpye ¼ k p 3k 2 (2) Given k If k > kc, λmax = 2. If k < kc, from equation (5.9), kmax ¼ kk 2ln
1k ¼ 2

2

2 2 kck ln kck 2
1 kck

(When

k > kc, λmax > 2; When k < kc, λmax < 2.). Between 1 < λ < λmax, select a λ to meet design requirement. When k > kc, if the selected λ > 2, r0ei \
ry . When k < kc, if λ > λmax is selected, the load-bearing capacity can not reach rpy ¼ k ffiffi
1 under the given k, because when k < kcλ, the maximum allowable k rpye ¼ k p 3k 2 2

load is

p ry

k 2ffiffi
1 ¼ p2ffiffi3 ln k, and when k < kc, p2ffiffi3 ln k  k p . 3k 2

(I) k = 1.6 < kc λmax = 1.542576. k ffiffi
1 = 0.542713. λ = λmax (A) Select λ = λmax = 1.542576, p2ffiffi3 ln k ¼ k p 3k 2 indicates kj = kjλ = k. k 2ffiffi
1 = 0.492552. (B) Select λ = 1.4, p2ffiffi3 ln k [ k p 3k 2 2

240

Autofrettage Technology and Its Applications in Pressured Apparatuses k 2ffiffi
1 (C) Select λ = 1.6, p2ffiffi3 ln k\k p = 0.562917. When λ > λ max, the 3k 2 maximum plastic depth is kj = k. When kj = k, from equation (3.46), the equivalent residual stress on inner surface is r0ei p k 2 ln k 2 kp2ffiffi
1 , from equation (5.86), 2 r ¼ 1
k 2
1 = −0.542576. When r ¼ k y

y

3k

the equivalent operation stress on inner surface is the equivalent total stress 0.542576 = 1.057424 > 1.

on

inner

rpei ry

= λ = 1.6. Then,

surface

is

rT e ry

= 1.6–

(II) k = 2.5 > kc 2 2 kmax ¼ kk 2ln
1k = 2.181645. When k > kc, λmax = 2. When k = 2.5, λ = λmax = 2, from equation (5.6), kjλ = kj* = 1.844363. When kj = kj*, the k 2ffiffi
1 ¼ rpay = 0.969948 > 0.920079. optimum load is rpy ¼ k rpye ¼ k p 3k 2 If λ = 1.6 is selected, from equation (5.6), kjλ = 1.384516 < kj* = k 2ffiffi
1 = 0.775959 < 0.969948. 1.844363. rpy ¼ k p 3k 2 (3) Given p and λ pffi pffiffi 3p=ry 3p From equation (5.119), kmax ¼ peffi3p=r r . If the given λ > λmax, there is no e

(4)

y
1

y

y
1

y

k ffiffi
1 to determine k. When solution; if the given λ < λmax, then, use rpy ¼ k rpye ¼ k p 3k 2 the given p = 0.920079σy, λmax = 2; When the given p < 0.920079σy, λmax < 2; When the given p > 0.920079σy, λmax > 2. Given p pffi pffiffi 3p=ry 3p From equation (5.119), kmax ¼ peffi3p=r r . Between 1 < λ < λmax, select a λ to e

2

meet design requirement.

5.7

Chapter Summary

(1) λ ≤ 2 is required, otherwise, jr0ei =ry j [ 1. (2) When k ≤ kcλ, there is always |r0ei σy| > > r y > > > > r0r > > < ry r0h > > > > > ry > > > > r0e > > : ry

Residual stresses Based on the maximum shear stress theory (Tresca criterion) 8 0 2 rz kjk
k > k
1 > >  ¼ constant ¼ ln x
> > ry 2k 2 > 2 >   > 0 2 2 2 > > k k rr k r0z kjk
k kjk
k > > ¼ ln x þ 2
¼
< ¼ 1
2 ry x ry 2k 2 2x 2 2x 2 (kjλ ≤ x ≤ k) (1 ≤ x ≤ kjλ)   0 2 2 0 2 k k k
k k > r k r jk jk
k > z h ¼ ln x
2
þ 1 > ¼ 1 þ ¼ þ > > ry 2x 2 > x 2 ry 2k 2 2x 2 > > 0 > 2 k > k
k r > ¼1
2 > : e ¼ jk 2 x ry x

242

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 5.1 – (continued).

On the inner surface, r0hi r0zi k
1 r0ri r0 r0 r0ei , ¼
¼ 0, ¼ 1
k ¼ 2 zi ¼ ei , ¼1
k 2 ry ry ry ry ry ry On the outer surface, 2 2 2
k kjk
k
k r0ho r0zo kjk r0 r0e kjk ¼ , r0ro ¼ 0, ¼ 2 zo ¼ , ¼ 2 2 2k k k2 ry ry ry ry Based on the maximum distortion strain energy theory (Mises criterion) 8 8 0 2 2
k > r0z kjk r ln x k
1 > > z > > p ffiffiffi ¼ constant ¼ p ffiffi ffi p ffiffi ffi ¼
> > > > > > ry  3k 2  ry 3 3 > > > > > r0 > r0 2 > > k 2 r0z ln x k k > > r > > < r ¼ pffiffiffi þ pffiffiffi 2
pffiffiffi < ¼ 1
2 r x ry y 3 3x 3 (1 ≤ x ≤ k )   ry0 (kjλ ≤ x ≤ k) jλ 0 2 0 2 r k rz > > r ln x k k
2 h > > h > > ¼ 1 þ p ffiffi ffi p ffiffi ffi p ffiffi ffi ¼

> > > ry > ry x 2 ry > > 3 3x 2 3 > > > r0 > 0 2 > > k k
k > > r e jk > > e : ¼1
2 > : ¼ x ry x2 ry On the inner surface, r0zi k
1 r0ri r0hi 2 r0 2 r0 r0ei ¼
pffiffiffi , ¼ 0, ¼
pffiffiffi ðk
1Þ ¼ 2 zi ¼ pffiffiffi ei , ¼1
k ry r r r r ry 3 3 3 y y y y On the outer surface, 2 2 2 r0ho r0zo kjk
k r0 2 kjk
k r0e kjk
k ¼ pffiffiffi , r0ro ¼ 0, ¼ 2 zo ¼ pffiffiffi , ¼ 2 k2 ry ry ry ry 3k 2 3 k The elastic stresses caused by internal pressure p = λpe Based on the maximum shear stress theory (Tresca criterion) rpz 1 p k ¼ ¼ ry k 2
1 ry 2k 2     rpr k 2 rpz k 1 1 ¼ 1
2 ¼
ry x ry 2 k 2 x 2     p rh k 2 rpz k 1 1 ¼ 1þ 2 ¼ þ ry x ry 2 k 2 x 2 ps rp rp re k ¼ h¼ r ¼ 2 ry ry ry x Based on the maximum distortion strain energy theory (Mises’s criterion) rpz 1 p k ¼ ¼ pffiffiffi ry k 2
1 ry 3k 2     rpr k 2 rpz k 1 1 ¼ 1
2 ¼
pffiffiffi 2
2 k ry x ry 3 x   p   p 2 rh k rz k 1 1 ¼ 1þ 2 ¼ pffiffiffi 2 þ 2 ry x ry x 3 k pffiffiffi  p  p ps r rpd r k r 3 e h ¼
r ¼ 2¼ e 2 ry ry ry ry x

Mechanical Autofrettage Technology Under Low Load TAB. 5.1 – (continued). The total stresses when p = λpe and kj = kjλ Based on the maximum shear stress theory (Tresca yield criterion) Within the plastic region, rT k
1 k z þ 2 ¼ ln x
2 2k ry rT k k r ¼ ln x
þ 2 2 2k ry rT k k h ¼ ln x
þ 2 þ 1 2 2k ry rT rT rT r0 rp e ¼ h ¼ r ¼ e þ e 1 ry ry ry ry ry Within the elastic region, 2 kjk rT z ¼ 2 ry 2k  2  T kjk rr 1 1 ¼
ry 2 k2 x2  2  T kjk rh 1 1 ¼ þ ry 2 k2 x2 2 T T re r rT r0 rp kjk ¼ h ¼ r ¼ e þ e ¼ 2 ry ry ry ry ry x Based on the maximum distortion strain energy theory (Mises’s yield criterion) Within the plastic region, rT ln x 2 k
1 k z ¼ pffiffiffi
pffiffiffi þ pffiffiffi ry 3 3 3k 2 T 2 rr ln x k k ¼ pffiffiffi
pffiffiffi þ pffiffiffi ry 3 3 3k 2 rT ln x 2 k k 2 h ¼ pffiffiffi
pffiffiffi þ pffiffiffi þ pffiffiffi ry 3 3 3k 2 3 pffiffiffi  T  T 0 p 3 rT r r r r e h ¼
r ¼ e þ e 1 2 ry ry ry ry ry Within the elastic region, 2 kjk rT z ¼ pffiffiffi ry 3k 2  2  T k rr 1 1 jk ¼
pffiffiffi 2
2 ry k 3 x  2  k rT 1 1 jk h ¼ pffiffiffi 2 þ 2 x ry 3 k pffiffiffi  T  2 T T re r0 rp kjk 3 rh rr ¼ e þ e ¼ 2 ¼
2 ry ry ry ry ry x T rT r rT rT When x < xa, e [ h ; when x > xa, e \ h ry ry ry ry

243

244

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 5.1 – (continued).

Based on both the maximum shear stress theory and the maximum distortion strain energy theory pffiffiffi x0 ¼ k = constant Based on both the maximum shear stress theory and the maximum distortion strain energy theory pffiffiffiffiffiffiffiffiffi x1 ¼ ek
1 = constant Based on the maximum shear stress theory r0zxo r0hxo r0rxo r0rmin k
1
ln k = constant ¼ ¼ ¼ ¼ y0s ¼
2 ry ry ry ry Based on the maximum distortion strain energy theory r0zxo r0hxo r0rxo r0rmin k
1
ln k pffiffiffi ¼ ¼ ¼ ¼ y0d ¼
= constant ry ry ry ry 3 2 y0d ¼ pffiffiffi y0s 3 x0, x1, y0s and y0d are not related to k or kj, only related with λ

References [1] Zhu R.L., Zhu G.L. (2012) Analysis on autofrettage of cylinders, Chin. J. Mech. Eng. 25(3), 615. [2] Zhu R.L., Zhu G.L. (2013) Effect of optimum plastic depth on stresses and load-bearing capacity of autofrettaged cylinder, Chin. J. Mech. Eng. 26(2), 365. [3] Zhu R.L. (2013) Study on autofrettage for medium-thick pressure vessels, J. Eng. Mech. 139 (12): 1790. [4] Zhu R.L., Li Q. (2014) Inquire into the marvellousness of autofrettage for mono-layered cylinders, J. Theor. Appl. Mech. 52(2), 359. [5] (1979) The compiling group of “handbook of mathematics”, A handbook of mathematics (in Chinese). Higher Education Press, Beijing. [6] Yu G.C. (1980) Chemical pressure vessels and equipment (in Chinese), Chemical Industrial Press, Beijing. [7] Liu H.W. (2019) Mechanics of materials (in Chinese). Higher Education Press, Beijing.

Chapter 6 Summary of Implement Methods and Their Characteristics of Mechanical and Thermal Autofrettage Technology 6.1

Implement Methods and Characteristics of Mechanical Autofrettage Technology

It is well known that cylindrical pressure vessels have unfavorable stress distributions under an elastic state, and the thicker the wall is, the more unfavorable the stress distribution is, as is seen in figure 1.1. To improve the stress distributions, autofrettage technology is usually adopted. The key to autofrettage technology is to introduce prestresses. Traditionally, the autofrettage is performed by applying very high pressure (such as hydraulic pressure, extrusion force, etc.) to the thick-walled pressure vessel so as to deform it plastically, after removing the very high pressure, the residual stresses on the wall of the pressure vessel are produced, enabling the pressure vessel to withstand higher working pressure. We call this method mechanical autofrettage. Currently, the common methods to obtain prestresses include the mechanical extrusion method, direct static pressure method (direct hydrostatic method), explosive expanding pressure method, and solid autofrettage method. The principles and characteristics of the above methods are as follows[1, 2]. 1. Mechanical extrusion method The principle of the mechanical extrusion method is pulling or pushing a conical interference mandrel through the inner surface of the cylinder for the inner surface to be extruded and deformed plastically to obtain residual stresses. For the mechanical extrusion method, there are currently the following three ways to pull or push a mandrel to slide through the cylinder. (1) Press the mandrel through the cylinder by a punch or hydraulic press, which is called the indirect hydraulic pressing method.

DOI: 10.1051/978-2-7598-3111-1.c006 © Science Press, EDP Sciences, 2023

246

Autofrettage Technology and Its Applications in Pressured Apparatuses

(2) Push the mandrel on its back through the cylinder directly by hydraulic pressure, which is called the direct hydraulic pushing method, which is shown in figure 6.1. (3) Pull the mandrel through the cylinder by mechanical force, which is called the mechanical power traction method, which is shown in figure 6.2. The mechanical extrusion method is mainly used in open cylinders or piping, it is economical and sealing is easy to achieve. In the mechanical extrusion method, residual stresses can reach 1500 to 2000 MPa. For materials with certain yield strength, the mechanical extrusion method requires much lower pressure than the hydrostatic method. 2. Direct hydrostatic method The principle of the direct hydrostatic method is directly applying hydraulic pressure to the cylinder for it to be deformed plastically, after removing the hydraulic pressure, residual stresses are produced. This is the earliest method and the most commonly used method, and it is widely used for autofrettage of various gun barrels and high or ultra-high-pressure vessels and piping. The direct hydrostatic method is basically the same as the hydraulic test process of the pressure vessel, and it is also basically the same as the operation process of the pressure vessel. The direct hydrostatic method is simple and flexible to use, in which no special pressure components are required, and the wall can obtain uniform plastic deformation. The direct hydrostatic method is especially suitable for autofrettage treatment of closed pressure vessels. For open cylinders (such as gun barrels, pipe, compressor cylinders, etc.), an end seal device can be used. In the direct hydrostatic method autofrettage, extra high-pressure power source, extra high-pressure pump and extra high-pressure pipeline, and pipe accessories are needed. Therefore, the sealing problem of ultra-high pressure needs to be solved. 3. Explosive expanding pressure method The principle of the explosive expanding pressure method is to produce high pressure in a very short time (10−5 s order of magnitude) by using high-energy dynamite so that the cylinder or pipe produces plastic deformation rapidly under the action of high-pressure gas and shock wave. Since the plastic deformation of the inner wall is a function of the explosive strength, that is, the weight of the dynamite on the unit length, as long as the proper number of explosives is controlled and the pressure produced by the explosion is consistent with the pressure required for a overstrain, the cylinder or pipe can produce certain plastic deformation. In the explosive expanding pressure method, the influence of the end effect should be considered or the plastic deformation of the end of the cylinder or pipe decreases, and the influence of the radial piston effect should also be considered. 4. Solid autofrettage method The solid autofrettage method is that the solid medium, such as lead, which has good plasticity and low melting point, is melted and poured into the cylinder, then, the solid medium is extruded by the press bar, which causes the solid medium to

Summary of Implement Methods and Their Characteristics

247

FIG. 6.1 – Autofrettage processing equipment for direct hydraulic pushing method. 1. Pressured liquid inlet; 2. Plug; 3. Soft steel filling ring; 4. Rubber “O” shaped sealing ring; 5. Mandrel; 6. Cylinder. deform plastically and transfer the pressure to the cylinder wall, as a result, residual stresses are produced because of the plastic deformation of the cylinder wall. In the solid autofrettage method, liquid medium consolidation should be avoided at extremely high pressure. Solid medium has high viscosity after it is plasticized, so the cylinder end seal is easy to solve. Even if a weak part of the wall breaks down, there will be no danger. This method has been applied to the high-pressure pump head with unequal thickness and special-shaped structure. Cylinders after autofrettaged treatment must undergo low-temperature heat treatment in order to stabilize the metallographic structure of the cylinder after plastic deformation. The temperature for low-temperature heat treatment is generally 250 °C to 300 °C, sometimes 360 °C. After about 3 h of heat preservation, the heated cylinder is cooled with the furnace or cooled in the air. Attention should be paid to avoiding stress concentration due to uneven cooling speed. It was once reported that autofrettage residual stress would be attenuated at a high temperature for a long time, and at the same time, corresponding solutions are put forward for this problem. For example, (1) choosing materials with good creep

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FIG. 6.2 – Autofrettage processing equipment for mechanical power traction method. 1. Cylinder; 2. Mandrel; 3. Pulling rod. resistance properties; (2) reasonable selection of overstrain; (3) strictly observing the operation procedures; (4) choosing suitable heat treatment temperature; (5) re-autofrettage, as is an effective way to solve the autofrettage residual stress attenuation. Therefore, the problem of autofrettage residual stress attenuation will no longer be discussed in this book.

6.2

Implement Methods and Characteristics of Thermal Autofrettage Technology

Since the temperature difference in the wall of a cylinder can cause thermal stresses, we put forward an autofrettage technology by the temperature difference between the inner and outer surfaces of a cylindrical pressure vessel. We call this technology temperature difference autofrettage or thermal autofrettage. In thermal autofrettage, there are no dangerous pressured fluids and no expensive hydrostatic installations needed, consequently, once the temperature difference is too high, catastrophic accidents like pressure vessel explosions cannot be caused as in mechanical autofrettage. In addition, the magnitude and regularities of distribution of temperature difference stresses are closely related to temperature difference Δt, thus, various operating stress states can be obtained by changing the temperature difference Δt according to operation conditions, as a result, we can acquire flexible design projects for one design assignment by using thermal autofrettage. Therefore, thermal autofrettage is safe, convenient, and economical and is a potential autofrettage technology. Moreover, there sometimes naturally is a temperature difference in many industrial production processes.

Summary of Implement Methods and Their Characteristics

249

When the temperature on the inner surface of a cylinder is higher than that on the outer surface, it is called internal heating; on the contrary, when the temperature on the inner surface of a cylinder is lower than that on the outer surface, it is called external heating. There are many ways to generate temperature differences. The temperature difference under the inner heating case can be generated by heating the inner surface and/or cooling the outer surface, as is shown in figure 6.3; the temperature difference under the outer heating case can be generated by cooling the inner surface and/or heating outer surface, as is shown in figure 6.4. Heating can be carried out by means of electric heating or steam heating and cooling can be carried out by means of liquid ammonia, etc.

FIG. 6.3 – Temperature difference generated for inner heating.

FIG. 6.4 – Temperature difference generated for outer heating.

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Autofrettage Technology and Its Applications in Pressured Apparatuses

For the inner heating shown in figure 6.3, the heating element is a kind of flexible heating flat ribbon, that is a resistance wire wrapped by high-temperature resistance cotton rope, as is shown in figure 6.5. The flexible heating flat ribbons are wound around the bushing, as is shown in figure 6.6, and the ends of the flexible heating flat ribbons are connected to the power source and temperature controller, etc.

FIG. 6.5 – A section of flexible heating flat ribbon.

FIG. 6.6 – Bushing wound with flexible heating flat ribbons. The heating and cooling system can be retained or dismantled in the operation of the pressure vessel, especially after the thermal stresses have caused the wall of the pressure vessel to yield and the residual thermal stresses have remained in the wall, the heating and cooling system can be dismantled. When the temperature difference reaches the critical temperature difference, Δtc, the inner surface of the pressure vessel initiates yielding, where Δtc is as follows, Dtc ¼

ðk 2  1Þ ln k 2 ð1  lÞry ð1  lÞry ¼A 2 2 2 k ln k  k þ 1 Ea Ea

ð6:1Þ

Summary of Implement Methods and Their Characteristics

251

According to equation (6.1), the temperature difference does not have to be quite large, the inner surface of the pressure vessel can be yielded under the temperature difference. Therefore, after thermal autofrettage treatment, the metallic material of the pressure vessel will not change or deteriorate the metallographic structure, and reduce its mechanical properties and service life due to heating or cooling during thermal autofrettage treatment. For example, when E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, k = 2.5, σy = 500 MPa, from equation (6.1), Δtc = 226.17 °C. This is just the temperature difference between the inner and outer walls, not the temperature at every point in the wall. Even if the temperature of each point in the wall is 226.17 °C, there will be no change in the metallographic structure under such a low temperature. The temperature difference can also be generated by heating on one side and cooling by another side, saying, heating the inner surface to ti = 126.17 °C and cooling the outer surface to to = −100 °C (this is internal heating, external heating is the other way round). According to equation (6.1), it is known that Δtc decreases with the increase of the wall thickness (reflected by radius ratio k). That is, the thicker the cylinder is, the smaller the critical temperature difference, Δtc is. In equaln k tion (6.1), the part A ¼ k 2ðkln1Þ k 2 k 2 þ 1 decreases as k (reflecting wall thickness) increases, or the critical temperature difference, Δtc decreases as k increases. When k → 1, A = 2; when k → ∞, A = 1. For the metallic material, there is little change in E, μ and α, if σy = 500 MPa (E = 2 × 105 MPa, μ = 0.3, ð1lÞr α = 1.2 × 10–5 °C−1), Ea y = 145.8 °C. Then, under the case of σy = 500 MPa, the maximum critical temperature difference, Δtc will not exceed 292 °C. The degradation of the metallographic structure depends mainly on the temperature of the metal rather than the temperature difference. Furthermore, for the metallographic structure of the pressure vessel to deteriorate, it is necessary for the pressure vessel to undergo high temperatures over a long period of time. Moreover, when there is a temperature difference between the internal surface and the external surface, the temperature inside the wall is not high. In the case of steady heat transfer, the distribution of temperature along the wall thickness is as follows: 2

2

t¼

ti ln xk þ to ln x ln k

ð6:2Þ

When k = 2.5 (E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 500 MPa), it is known from the above that Δtc = 226.17 °C. If the internal surface is heated to ti = 246.17 °C and the external surface is at room temperature 20 °C (then the temperature difference between the internal surface and the external surface, Δt = 226.17 °C), the distribution of temperature along the wall thickness is illustrated in figure 6.7. From figure 6.7, it is seen that the temperature inside the wall of the cylinder is not high. Therefore, metallographic structure change/degradation can not happen, so the mechanical properties of the material and service life of the cylinder can not be reduced. If k is increased, the temperature or the temperature critical

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FIG. 6.7 – The distribution of temperature along the wall thickness with k = 2.5. temperature difference Δtc will be decreased. When k = 3 (still, E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 500 MPa), it is known from the above that Δtc = 217.7 °C. If the internal surface is heated to ti = 237.7 °C and the external surface is at room temperature 20 °C (then the temperature difference between the internal surface and the external surface, Δt = 217.7 °C), the distribution of temperature along the wall thickness is illustrated in figure 6.8.

FIG. 6.8 – The distribution of temperature along the wall thickness with k = 3. When σy = 300 MPa, E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, the variation of the critical temperature difference Δtc with radius ratio k is shown in figure 6.9.

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253

FIG. 6.9 – Variation of Δtc with k when σy = 300 MPa. When σy = 500 MPa, E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, the variation of the critical temperature difference Δtc with radius ratio k is shown in figure 6.10.

FIG. 6.10 – Variation of Δtc with k when σy = 500 MPa.

When σy = 700 MPa, E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, the variation of the critical temperature difference Δtc with radius ratio k is shown in figure 6.11.

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FIG. 6.11 – Variation of Δtc with k when σy = 700 MPa. When k = 2, E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, the variation of the critical temperature difference Δtc with yield strength σy is shown in figure 6.12.

FIG. 6.12 – Variation of Δtc with σy when k = 2.

When k = 3, E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, the variation of the critical temperature difference Δtc with yield strength σy is shown in figure 6.13. When k = 4, E = 2 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, the variation of the critical temperature difference Δtc with yield strength σy is shown in figure 6.14. It can be seen that the temperature in the wall is not high, so there will be no change in metallography. Under the condition of keeping temperature difference, the thermal stress can not decay. As for the attenuation of residual temperature

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255

FIG. 6.13 – Variation of Δtc with σy when k = 3.

FIG. 6.14 – Variation of Δtc with σy when k = 4. difference stresses after removing temperature difference, it can be solved just like solving the attenuation of the above mechanical residual stresses.

References [1] Yu G.C. (1980) Chemical pressure vessels and equipment (in Chinese). Chemical Industrial Press, Beijing. [2] Wang Z.W. (1990) Design of chemical pressure vessels (in Chinese). Chemical Industrial Press, Beijing.

Chapter 7 Thermal Autofrettage Technology Based on Tresca Yield Criterion 7.1

Introduction

The metallic materials have the properties of thermal expansion and contraction (expand with heat and contract with cold). The media contained in high-pressure apparatuses are frequently in high or low temperatures, the materials of the high-pressure apparatuses are thus heated or cooled by the media, and thermal expansion and contraction in the materials are thereupon caused. For the thick-walled high-pressure apparatuses, deformations of the wall are not free but highly restrained. Therefore, if the thick high-pressure apparatuses are heated or cooled (whether by operation media or by artificial means), thermal stresses will be produced due to deformation constraints. Thermal stresses are often useful in the operation of high-pressure apparatuses. The study of the thermal stresses can help us to explore the effects of material characteristics on thermal stresses and to create a unique autofrettage technology for thick high-pressure apparatuses when the thermal stresses are advantageous. It is well known that high-pressure apparatuses have unfavorable stress distributions under an elastic state. To improve stress distributions, autofrettage technology is usually adopted. Traditionally, the autofrettage is achieved by mechanical method, or by applying hydraulic pressure to the bore of the thick-walled high-pressure apparatuses to deform them plastically, the compressive residual stresses on the inner part of the high-pressure apparatuses are thus produced, the operation stresses, therefore, are partially offset by the residual stresses, enabling the high-pressure apparatuses to withstand higher working pressure. This is mechanical autofrettage technology. In the former chapters, we have carried out studies on mechanical autofrettage. The key to autofrettage of the thick high-pressure apparatuses is pre-stresses. Since temperature differences in the wall of a high-pressure apparatus can cause pre-stresses, heating and/or cooling the thick high-pressure apparatuses to obtain pre-stresses is a promising approach for autofrettage of thick apparatuses. For this DOI: 10.1051/978-2-7598-3111-1.c007 © Science Press, EDP Sciences, 2023

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Autofrettage Technology and Its Applications in Pressured Apparatuses

reason, we first need to explore the thermal stresses in metal materials. On the basis of studying thermal stresses, we will study thermal autofrettage technology. Most high-pressure apparatuses are usually shaped into cylinders, so we investigate the thermal stresses in cylindrically shaped pressure vessels.

7.2

Derivation of Thermal Stresses

To draw on the advantages and avoid disadvantages of the thermal stresses, it is necessary and meaningful to investigate the magnitudes and the characteristics of the thermal stresses in the materials quantitatively and in detail. Figure 7.1 is a thick cylindrical pressure vessel with internal radius ri and external radius ro; the temperature on the internal wall is ti, and to on the external wall; the temperature difference, Δt = ti − to, °C; and the temperature at any radius r is t, and t = t(r). This is an axial symmetry problem, the cross-section remains plane under thermal stresses, the radial displacement u is related only to radius r, the axial strain εz is constant, and clearly, there is no hoop displacement. We will derive the expressions of thermal stresses based on such a mechanical and mathematical model.

FIG. 7.1 – A thick cylindrical pressure vessel with temperature difference in the wall. At an arbitrary radius r, cutting the cylinder with two concentric circles at a distance of dr, two planes with included angle dθ, and two parallel planes at a distance of dz, we obtain an infinitesimal volume element as is shown in figure 7.2.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

259

Considering equilibrium of the element in r direction, we obtain the equilibrium differential equation of the cylinder at any point as follows[1], 8 dr r  rh > < rþ r ¼0 dr r ð7:1Þ > : drz ¼ 0 dz where σr, σθ, and σz are radial stress, circumferential stress, and axial (thermal) stress, respectively. Equation (7.1) is fit for the materials in both elastic and plastic states.

FIG. 7.2 – An element cutting out of a cylindrical pressure vessel. When an elastic material is heated to temperature t °C and can expand freely, the thermal strains are etr ¼ eth ¼ etz ¼ at, where etr ; eth ; and etz are radial strain, circumferential strain, and axial thermal strain, respectively; α is the thermal expansion coefficient of the materials, °C−1. If the material is constrained, the thermal stresses are consequently caused in the material. When the material is constrained, at an arbitrary radius r, the total strains are thermal strains plus strains produced by thermal stresses. For the materials at elastic state, according to Hook’s law[1],   8 8 l 1þl t >   t t 1 > rr ¼ 2G er þ e  at > > rt lðrth þ rtz Þ þ at etr ¼ > > 1  2l 1  2l  > > > >  E r > > > > t  l 1þl 1 t < < rh ¼ 2G eth þ et  at rh lðrtz þ rtr Þ þ at eth ¼ or 1  2l 1  2l  ð7:2Þ E  > >   > t > 1 t l 1þl > > > > r lðrtr þ rth Þ þ at e ¼ > > rtz ¼ 2G etz þ et  at > > : zt E z > 1  2l 1  2l > : t czr ¼ 0 szr ¼ 0 Where rtr , rth , and rtz are thermal radial stress, thermal circumferential stress,  t  and thermal axial stress, respectively; et ¼ etr þ eth þ etz ¼ 12l rr þ rth þ rtz þ 3at; μ E is the Poisson ratio of the material; E is Young’s elastic module of the material; G is shearing elastic modules of the material, G ¼ 2ð1Eþ lÞ.

260

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For the axial symmetry problem, we take any circumferential fiber with radius r in the cylinder for study, as is shown in figure 7.3.

FIG. 7.3 – The radial displacement of a circumferential fiber. Before deformation the radius of the circumferential fiber in figure 7.3 is r. After deformation, the circumferential fiber arrives at the dotted circle in figure 7.3. The radius of the circumferential fiber is r + u after deformation, that is, the radial displacement is u, as is shown in figure 7.3. Therefore, after deformation, the length or the circumference of the circumferential fiber is 2π (r + u), and before deformation, the circumference of the circumferential fiber is 2πr, thus, the circumferuÞ2pr ential strain of the circumferential fiber is eth ¼ 2pðr þ2pr ¼ ur . When the radial displacement of the circumferential fiber with radius r is u, the radial displacement of the circumferential fiber with radius r + dr is u + du, as is shown in figure 7.4. Then, the deflection of a radial length dr is du. Therefore, the t dw radial strain is etr ¼ du dr . In a similar way, the axial strain is ez ¼ dz . Since the axial t deformation is uniform, etz ¼ dw dz = a constant. Clearly, czr ¼ 0. Therefore, we have etr ¼

du ; dr

eth ¼

u ; r

etz ¼

dw ¼ a constant; ctzr ¼ 0 dz

FIG. 7.4 – The radial displacements of two circumferential fiber.

ð7:3Þ

Thermal Autofrettage Technology Based on Tresca Yield Criterion

261

where u and w are radial displacement and axial displacement respectively, γzr is the shearing strain in the r−z plane. Then, by using equations (7.3), equation (7.2) can be rewritten as follows    8 du l du u 1þl > t > þ þ þ ez  at < rr ¼ 2G 1  2l dr r dr  1  2l u l du u 1þl > > : rth ¼ 2G þ þ þ ez  at r 1  2l dr r 1  2l

ð7:4Þ

Substituting equation (7.4) into the first one of equation (7.1) obtains d2 u 1 du u 1 þ l dt  ¼ a þ dr 2 r dr r 2 1  l dr

ð7:5Þ

Solving equation (7.5) gives 1þl a u¼ 1  lr

Z

r ri

trdr þ

C1 r C2 þ 2 r

ð7:6Þ

where C1 and C2 are integral constants. Substituting equation (7.6) into equation (7.4) results in Z r 8 Ea B > t > rr ¼  trdr þ A  2 > 2 > ð1  lÞrZ ri r > > > r > Ea Eat B > t > þAþ 2 trdr  < rh ¼ 2 ð1  lÞr ri 1l r > rtz ¼ Eðetz  atÞ þ lðrtr þ rth Þ > > > Eat > > > þ 2lA þ Eetz ¼ > > 1  l > : t szr ¼ 0; strh ¼ 0

ð7:7Þ

EðC þ 2let Þ

z 2 where A ¼ 2ð1 þ1lÞð12lÞ , B ¼ 1EC þ l. When a pressure vessel is not subjected to additional axial loads, there is Z ro rtz 2prdr ¼ 0 ð7:8Þ

ri

Substituting the third one of equation (7.7) into equation (7.8) gives Z ro Ea 1  trdr þ ð2lA þ Eetz Þðro2  ri2 Þ ¼ 0 1  l ri 2 When a pressure vessel is not loaded with other loads but temperature difference, then ðrtr Þr¼ri ¼ ðrtr Þr¼ro ¼ 0 ð7:9Þ

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Substituting equation (7.9) into equation (7.7) gives 8 Z ro Ea > > trdr A ¼ > > < ð1  lÞðro2  ri2 Þ ri B ¼ ri2 A > > > 2ð1  lÞ > : etz ¼ A E

ð7:10Þ

From equations (7.7) and (7.10), we obtain rtz ¼ rth þ rtr ¼ 

Eat þ 2A 1l

ð7:11Þ

For a cylindrical pressure vessel with internal radius ri, external radius ro, and internal wall temperature ti, external wall temperature to as shown in figure 7.5, at any radius r, we take a thin cylinder with differential thickness dr and axial length l. Under steady heat transfer, according to Fourier’s heat transfer law[2–4], the quantity of heat Q conducted through the thin cylinder in unit time is dt ð7:12Þ Q ¼ 2prlk dr where λ is the thermal conductivity of the pressure vessel material, W/m∙°C or W/m∙K. Rewriting equation (7.12) as follows dr Q ¼ 2plkdt ð7:13Þ r and integrating both sides of equation (7.13) from the internal wall to the external wall, i.e. Z ro Z to dr Q ¼ 2plk dt r ri ti we obtain Q ¼ 2plk

ti  to ln rroi

FIG. 7.5 – Heat conduction through a wall of cylinder.

ð7:14Þ

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263

On the other hand, if we integrate equation (7.13) from the inner wall to any radius r, or Z t Z r dr ¼ 2plk Q dt ri r ti we obtain Q ¼ 2plk

ti  t ln rri

ð7:15Þ

Setting equation (7.15) equaling equation (7.14) gives t ¼ tðrÞ ¼

ti ln rro þ to ln rri ln rroi

¼

ti ln xk þ to ln x ti ln k  Dt ln x ln x ¼ ¼ ti  Dt ln k ln k ln k

ð7:16Þ

where k is the radius ratio, or ratio of outside radius to an inside radius, k = ro/ri; x is relative location, x = r/ri. The temperature difference, Δt = ti − to > 0 is called internal heating, Δt < 0 is called external heating. The distribution of wall temperature in the case of internal heating for k = 3 is shown in figure 7.6, where the solid curves are for ti = 200 °C, the dash curves are for ti = 300 °C, and the parameters are as follows: (1) (2) (3) (4) (5) (6)

ti ti ti ti ti ti

= = = = = =

200 200 200 300 300 300

°C, °C, °C, °C, °C, °C,

to to to to to to

= = = = = =

0 °C, Δt = ti−to = 200 °C; 50 °C, Δt = ti−to = 150 °C; 150 °C, Δt = ti−to = 50 °C; 0 °C, Δt = ti−to = 300 °C; 50 °C, Δt = ti−to = 250 °C; 150 °C, Δt = ti−to = 150 °C.

FIG. 7.6 – The distributions of wall temperature of internal heating for k = 3. The distribution of wall temperature in the case of external heating for k = 3 is shown in figure 7.7, where the solid curves are for ti = 0 °C, the dash curves are for ti = −30 °C, and the parameters are as follows:

Autofrettage Technology and Its Applications in Pressured Apparatuses

264 (7) (8) (9) (10) (11) (12)

ti ti ti ti ti ti

= = = = = =

0 °C, to = 0 °C, to = 0 °C, to = −30 °C, to −30 °C, to −30 °C, to

200 °C, Δt = ti−to = −200 °C; 150 °C, Δt = ti−to = −150 °C; 50 °C, Δt = ti−to = −50 °C; = 270 °C, Δt = ti−to = −300 °C; = 220 °C, Δt = ti−to = −250 °C; = 120 °C, Δt = ti−to = −150 °C.

FIG. 7.7 – The distributions of wall temperature of external heating for k = 3. In addition, by integrating equation (7.13) from any radius r to the outside wall, or Z ro Z to dr Q ¼ 2plk dt r r t we obtain Q ¼ 2plk

t  to ln rro

ð7:17Þ

Letting equation (7.17) equal equation (7.14) also gives equation (7.16), which expresses temperature at any location r under steady heat transfer. Equation (7.16) indicates that the temperature of any point in the wall of the cylindrical pressure vessel depends on the temperatures of the inside and outside walls (ti and to) and the internal and external radii (ri and ro), and has nothing to do with material natures. When ti > to, the temperature in the wall of the cylindrical pressure vessel strictly decreases from the internal wall to the external wall, and when ti < to, it strictly increases from the internal wall to the external wall. However, from equation (7.14), the quantity of heat conducted through a wall of a cylinder is associated with the thermal conductivity of the material, the temperature difference, and the radii of the inside and outside walls. Thus, it can be seen that once the inside and outside wall temperatures and the inside and outside radii are determined, heat

Thermal Autofrettage Technology Based on Tresca Yield Criterion

265

transfer is related to the physical properties of the materials; and once the material is determined, heat transfer is related to the boundary temperatures and sizes, and has no relation to the specific distribution of temperature field. Substituting equation (7.16) into equation (7.7) obtains the thermal stresses for elastic materials at any location.   lnðk=xÞ k2  x2 rtr ¼ pt  þ 2 2 ð7:18Þ ln k x k  x2 rth ¼ pt

  1  lnðk=xÞ k2 þ x2  2 2 ln k x k  x2 

rtz

¼ pt

1  2 lnðk=xÞ 2  2 ln k k 1

ð7:19Þ

 ð7:20Þ

EaDt . Equations (7.18) and (7.19) are where pt is called thermal loading, pt ¼ 2ð1lÞ [5–7] . consistent with the results in refs. Therefore, the thermal stresses when ti > to are equal to minus thermal stresses when ti < to; the greater Δt is, the greater the absolute value of thermal stresses are. From equations (7.18)–(7.20) and equation (7.11), it is seen that at any location,

rtz ¼ rtr þ rth

ð7:21Þ

In addition, the thermal stresses increase as elastic modules and the thermal expansion coefficient of the materials increase. The mechanism of this phenomenon is analyzed as follows: The elastic modulus indicates the ability of the connected atoms of the material to resist separation or atomic bonding interactions, thus, greater stresses are needed for the materials with greater elastic modulus to obtain a certain strain. The thermal expansion coefficient is a parameter to characterize the expansion or contraction of the material with temperature changing. For the materials with great thermal expansion coefficients, the free strains are great when heated or cooled. However, if the materials are subject to constraints, the stresses caused because of the constraints must be great. The thermal stresses decrease as the Poisson ratio of the material increases, but the Poisson ratios of most materials are similar. Therefore, adjusting the Poisson ratio to adjust the thermal stresses is not quite effective.

7.3

The Characteristics of the Thermal Stresses

Thermal stresses in the material are the foundation of material selection and design of apparatuses. Therefore, it is necessary to study the principal features of thermal stresses. From equations (7.18)–(7.20), at the inside and outside walls, the thermal stresses are as follows,

266

Autofrettage Technology and Its Applications in Pressured Apparatuses rtri ¼ rtro ¼ 0  rthi ¼ rtzi ¼ pt

   rthi rtzi 2k 2 1 2k 2 1   ¼ ¼  or k 2  1 ln k pt pt k 2  1 ln k  rt 1 2 rt 1 2  2  2 or ho ¼ zo ¼ ln k k  1 ln k k  1 pt pt

ð7:22Þ ð7:23Þ

 rtho ¼ rtzo ¼ pt

ð7:24Þ

where the subscripts i and o represent the quantity at the inside and outside wall respectively. According to equations (7.22)–(7.24), when r = ro, rtro ¼ rtzo  rtho ¼ 0, or rtzo ¼ rtho ; when r = ri, rtri ¼ rtzi  rthi ¼ 0, or rtzi ¼ rthi . This means that at the inside and outside wall, the circumferential thermal stress and axial thermal stress are equal to each other. Within 1 ≤ x ≤ k, rtr ¼ rtz  rth ¼ 0 has qffiffiffiffiffiffiffiffiffiffiffi 2 2 a sole extreme point at x ¼ kk 2ln1k [8], and if Δt > 0 (called internal heating), drtr dx 2

drt

[ 0; if Δt < 0 (called external heating), dx 2r \0. Therefore, if Δt > 0, rtr gets qffiffiffiffiffiffiffiffiffiffiffi 2 2 minimum at x ¼ kk 2ln1k , which means rtr ¼ rtz  rth \0, or rtz \rth within 1 ≤ x ≤ k; qffiffiffiffiffiffiffiffiffiffiffi 2 2 if Δt < 0, rtr gets maximum at x ¼ kk 2ln1k , which means rtr ¼ rtz  rth [ 0, or

rtz [ rth within 1 ≤ x ≤ k. 2 1 2k 2 t t Since ln k 2 [ 2 kk2 1 þ 1, it is easy to obtain ln k \ k 2 1, then rzi ¼ rhi \0; since 2 2 k 1 > lnk , it is easy to obtain ln1k [ k 221, then rtzo ¼ rtho [ 0. Within 1 ≤ x ≤ k, neither rth nor rtz has pole. Therefore, when Δt > 0, rth and rtz increase monotonically from compressive stress on inside wall to tension on the outside wall; when Δt < 0, rth and rtz decrease monotonically from tension on the inside wall to compressive stress on the outside wall. Thus, it can be seen that the variation tendency of thermal stresses and the variation tendency of wall temperature difference are opposite. The reason is that deformation of the material is great where the temperature is high, when the material is constrained, the thermal stresses are certainly great; and vice versa. See figure 7.6 with figures 7.8 and 7.7 with figures 7.9, respectively. From equations (7.22)–(7.24), we know that     2k 2 1 1 2 jrthi j  jrtho j ¼ jrtzi j  jrtzo j ¼ jpt j    2 ln k k 2  1  2 k  1 ln k k þ1 1  ¼ 2jpt j 2 : k  1 ln k [8] 1 k þ1 t t t t Since ln k 2 ¼ 2 ln k [ 2 kk2 1 þ 1 , or ln k \ k 2 1 , jrhi j [ jrho j and jrzi j [ jrzo j, which means that whether Δt > 0 or Δt < 0, the absolute values of thermal stresses on the inside wall are greater than those on an outside wall. Since rtri ¼ 0, as |Δt| increases, the inside wall becomes yielded always earlier than the outside wall. If Δt > 0, when 2

2

Thermal Autofrettage Technology Based on Tresca Yield Criterion

267

t t equivalent thermal stress rts ei ¼ rri  rzi ¼ ry (Tresca criterion), inside wall initiates yielding, where σy is yield limit of the materials. From equation (7.22)–(7.24),   2k 2 1 k 2 ln k 2  k 2 þ 1 ts t t t t  ð7:25Þ ¼ pt rei ¼ rri  rzi ¼ rzi ¼ rhi ¼ pt 2 ðk 2  1Þ ln k k  1 ln k

For equation (7.25), letting rts ei ¼ ry obtains the temperature difference at which the inside wall initiates yielding, called critical temperature difference, Δtc: Dtc ¼

ðk 2  1Þ ln k 2ð1  lÞry ¼ mf k 2 ln k 2  k 2 þ 1 Ea

ð7:26Þ

ln k y where m ¼ k 2ðkln k1Þ 2 k 2 þ 1, reflecting the size of a cylindrical pressure vessel; f ¼ Ea , reflecting material properties. Equation (7.26) shows that the thicker the cylindrical pressure vessel is, the smaller the |Δtc| is, this means that the thicker the cylindrical pressure vessel is, the more easily it becomes yielded under the thermal stresses. 2ð1lÞr ð1lÞr When k → 1, m = 1, Dtc ¼ Ea y ; and k → ∞, m = 0.5, Dtc ¼ Ea y . The critical temperature difference at Δt < 0 is equal to minus equation (7.26). When Δt > 0, the thermal loading at which the inside wall initiates yielding, called critical thermal loading ptc, is obtained from equation (7.26): 2

ptc ¼

2ð1lÞr

ðk 2  1Þ ln k ry ln k 2  k 2 þ 1

k2

or

ptc ðk 2  1Þ ln k ¼m ¼ 2 ry k ln k 2  k 2 þ 1

ð7:27Þ

Equation (7.27) is plotted in figure 7.8 to show the variation tendency of m with the increase of k.

FIG. 7.8 – The variation tendency of m with the increase of k. An example: commonly, elastic modules of steel materials E = 1.95 × 105 MPa, Poisson ratio μ = 0.3, thermal expansion coefficient of the materials α = 1.2 × 10–5 °C−1, if Δt = 50 °C, then, pt = 83.57143 MPa. Setting k = 2.5, the thermal stresses are shown as the solid curves in figure 7.9. If Δt < 0, the trend of variations

268

Autofrettage Technology and Its Applications in Pressured Apparatuses

of thermal stresses is opposite to figure 7.9, as is shown in figure 7.10. The meaning of each curve and the main parameters are marked in the figures.

FIG. 7.9 – Distribution of thermal stresses for elastic cylindrical pressure vessel when Δt > 0.

FIG. 7.10 – Distribution of thermal stresses for elastic cylindrical pressure vessel when

Δt < 0.

From equations (7.23) and (7.24), rtzi pt

→ 2,

rtho pt

and

rtzo pt

rthi pt



rtho pt

¼

rtzi pt



rtzo pt

¼ 2; with k → ∞,

rtho pt

and

rthi pt

and

→ 0.

The variation tendencies of k are shown in figure 7.11.

rthi pt

and

rtzi pt

as well as

rtzo pt

with the increase of

Thermal Autofrettage Technology Based on Tresca Yield Criterion

FIG. 7.11 – The variation tendencies of k when Δt > 0.

t d

r hi pt

dk

 t ¼

d

t

r

zi pt

dk

¼

d

r ho pt

dk

¼

d

rt  zo pt

dk

rthi pt

and

rtzi pt

as well as

k 1Þð2k ln kk ¼ ð2k ln k þk½ðk 2 1Þ ln k2 2

2

rtho pt

and

rtzo pt

269

with the increase of

þ 1Þ

. When k = 1, 2klnk

dð2k ln kk 2 þ 1Þ dk

−k + 1 = 0, while ¼ 2½ln k  ðk  1Þ < 0 because k − 1 > lnk. 2 Therefore, 2klnk−k + 1 < 0, while 2klnk + k2 − 1 > 0, k[(k2−1)lnk]2 > 0, t  t t r r r   rt  2 d phi d pzi d pho d pzo t t t 2k 1 t ¼ ¼ ¼ \0. Besides,   accordingly, 2 ln k \0 and dk dk dk dk k 1

t

t

rhi

rzi 1 2 ln k  k 2 1 [ 0. Thus, the larger k is, the larger pt and pt are; the larger k is, the

t

t

r

r smaller phot and pzot are, as is seen from figure 7.11. 2

Figure 7.11 includes pt > 0 (or Δt > 0, and rtzi \0, rthi \0) and pt < 0 (or Δt < 0, and rtzi [ 0, rthi [ 0). If figure 7.11 represents the case of pt > 0 (Δt > 0), when pt < 0 (Δt < 0), the variation tendencies of

rthi pt

and

rtzi pt

as well as

rtho pt

and

rtzo pt

with

the increase of k are shown in figure 7.12. When pt < 0 (Δt < 0), as k tends to ∞, and

rtzi pt

tend to 2,

rtho pt

and

rtzo pt

tend to 0. In a word, regardless of pt > 0 (Δt > 0) or

pt < 0 (Δt < 0), as the k increases, axis), and rthi pt

=

rtzi pt

rthi pt

and

rtzi pt

rthi pt

rtho pt

and

rtzo pt

get closer to the k axis (horizontal

gradually leave the horizontal axis (k axis) and are stabilized at

= 2.

When Δt > 0 (pt > 0), rtr reaches the minimum at x ¼ qffiffiffiffiffiffiffiffi ln k 2 t ln rrmin 1 1 k 2 1 þ ¼  ln k 2 k 2  1 ln k pt

qffiffiffiffiffiffiffiffiffiffiffi

k 2 ln k 2 k 2 1 ,

the minimum is ð7:28Þ

270

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 7.12 – The variation tendencies of k when Δt < 0.

rthi pt

and

rtzi pt

as well as

rtho pt

and

rtzo pt

with the increase of

qffiffiffiffiffiffiffiffiffiffiffi 2 2 When Δt < 0 (pt < 0), rtr reaches the maximum at x ¼ kk 2ln1k , the maximum is qffiffiffiffiffiffiffiffi ln k 2 t t ln rrmax rrmin 1 1 k 2 1 ð7:29Þ  ¼ ¼ 2  k  1 ln k 2 ln k pt pt rt

rt

Figure 7.13 shows the variation tendencies of rmin and rmax with the increase of k.

t pt

ptt

rrmin

rrmax From figure 7.13, the larger k is, the larger pt and pt are.

FIG. 7.13 – The variation tendencies of

rtrmin pt

and

rtrmax pt

with the increase of k.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

271

In order to show the above conclusions, figure 7.14 illustrates the distributions of rtz pt rtr pt

rt

and pht along x for k = 1.5, 2, 3 and 4, and figure 7.15 illustrates the distributions of along x for k = 1.5, 2, 3 and 4.

FIG. 7.14 – The distributions of rpzt and t

rth pt

along x with k = 1.5, 2, 3 and 4.

FIG. 7.15 – The distributions of rptr along x with k = 1.5, 2, 3 and 4. t

Figures 7.16 and 7.17 is a combination of figures 7.14 and 7.15 for k = 2 and 4, respectively. When Δt < 0 (pt < 0), the thermal stresses are negative, those in figures 7.14– 7.17, or they are symmetric about x-axis.

272

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 7.16 – The distributions of rpzt ,

rth pt

and

rtr pt

along x with k = 2.

FIG. 7.17 – The distributions of rpzt ,

rth pt

and

rtr pt

along x with k = 4.

t

t

If the temperature difference |Δt| exceeds critical temperature difference |Δtc|, the plastic zone (yielded zone) develops further, thus the wall material becomes elastoplastic, part of the material is elastic, and part of the material is plastic. After the temperature difference is released, the plastic part of the material cannot be recovered, and the elastic part is trying to recover, then, the residual stresses (thermal stresses) thereupon remained. These residual stresses (thermal stresses) can be controlled in order to obtain a favorable state of stresses for the cylindrical pressure vessel. In an elastic state, the most dangerous stress under internal pressure operation is tension at the inside wall, and when Δt > 0, the compressive thermal stresses at the internal wall are relatively large. Thus, the thermal stresses when Δt > 0 are

Thermal Autofrettage Technology Based on Tresca Yield Criterion

273

beneficial to decreasing dangerous stress under internal pressure operation, then, this chapter studies the internal heating and internal pressure operation problems.

7.4

The Analysis of Total Stresses and Investigation of Optimum Operation Conditions

The stresses caused by internal operation pressure (p) contained in a pressure vessel are called mechanical stresses, i.e., equation (1.1), thermal stresses plus mechanical stresses are called total stresses. When a pressure vessel is subjected to temperature difference (Δt > 0) and internal operation pressure at the same time, the total stresses are as follows,     lnðk=xÞ k2  x2 p k2 T t p rr ¼ rr þ rr ¼ pt  þ 2 2 1 2 þ 2 ð7:30Þ ln k k 1 x k  x2 x rT h

¼

rth

þ rph

    1  lnðk=xÞ k2 þ x2 p k2  2 2 1þ 2 ¼ pt þ 2 ln k k 1 x k  x2 x

t p rT z ¼ rz þ rz ¼ pt

  1  2 lnðk=xÞ 2 p  2 þ 2 ln k k 1 k 1

ð7:31Þ

ð7:32Þ

T T where rT r , rh and rz are total radial stress, total circumferential stress, and total axial stress, respectively, rpr , rph and rpz are mechanical radial stress, mechanical circumferential stress, and mechanical axial stress, respectively. T T It is easy to prove that the curve for rT h is parallel to the curve for rz  rr , or T T T rh  ðrz  rr Þ is a constant: p T T t t p t p rT h  ðrz  rr Þ ¼ ðrh þ rh Þ  ½ðrz þ rz Þ  ðrr þ rr Þ

¼ ðrth  rtz þ rtr Þ þ ðrph þ rpr Þ  rpz

¼ 0 þ 2rpz  rpz ¼ rpz p ¼ a constant [ 0: ¼ 2 k 1

T Therefore, rT z  rr \rh .   1 2k 2 2k 2 p 2k 2 pt T T  2 2 rh  rr ¼ pt ¼ ðp  pt Þ þ þ 2 2 2 2 2 2 2 ln k x k  x x k x x k x ln k

ð7:33Þ

T T T T T rT h  rr ¼ ðrh  rz Þ þ ðrz  rr Þ

rT hi

T T T T T T rT ro ¼ 0, so if rho  rro  ry , then rho  ry . rhi ¼ p\0, so if rhi  rri  ry , then  ry .

Autofrettage Technology and Its Applications in Pressured Apparatuses

274

T It is known by further analysis that rT h  rr is generally the most dangerous stress in engineering. dðrh rr ÞT dx

ðppt Þ T T ¼  4k x 3 ðk 2 1Þ , it is thus clear that rh  rr has no stationary point 2

T within − ∞ < x < +∞, except p = pt. If p < pt, rT h  rr increase monotonically; pt T T T if p > pt, rT h  rr decrease monotonically; if p = pt, rh  rr is constant: ln k . T Therefore, rT h  rr at the inside and outside walls is the key to ensuring the safety of a pressure vessel. At the inside wall, x = 1, then   1 2k 2 2k 2 p 2k 2 pt T rT  ¼ 2 ðp  pt Þ þ ð7:34Þ  r ¼ p þ 2 t hi ri 2 ln k k  1 k 1 k 1 ln k

If

p ry



k 2 ln k 2 k 2 þ 1 pt ry k 2 ln k 2

T T  k2k1 2 , rhi  rri   ry . Substituting equation (7.26) into 2

T equation (7.34) shows that even if pt = ptc, rT hi  rri   ry provided p > 0. So T T rhi  rri   ry in engineering. At the outside wall, x = k, then   1 2 2p 2 pt T  ¼ ðp  pt Þ þ [0 ð7:35Þ rT  r ¼ p þ 2 t ho ro ln k k 2  1 k  1 k2  1 ln k T Setting rT hi  rri  ry obtains load-bearing capacity or allowable loading, p/σy:

p k2  1 pt k 2 ln k 2  k 2 þ 1 k 2  1 Ea k 2 ln k 2  k 2 þ 1 p1  þ ¼ þ Dt ¼ 2 2 2 2 2 2 ry 2k k ln k 2k 2ð1  lÞry k ln k ry ry ð7:36Þ   k2  1 ln k 2 2ð1  lÞry 2 p or Dt  k ¼ Dt1  ry 2 k 2 ln k 2  k 2 þ 1 Ea The greater Δt is, the greater p1 is. When

p ry

ð7:37Þ

pe ¼ k2k1 2 ¼ r , Δt1 = 0. y 2

T Setting rT ho  rro  ry obtains another load-bearing capacity or allowable loading, p/σy:

p k 2  1 pt k 2  1  ln k 2 k 2  1 Ea k 2  1  ln k 2 p2    ¼ Dt ¼ ð7:38Þ ry 2 2 2ð1  lÞry ry ln k 2 ln k 2 ry or Dt 

 2  k 1 p ln k 2 2ð1  lÞry  ¼ Dt2 2 2 ry k  1  ln k 2 Ea p k 2 1 2 pe ry ¼ 2 ¼ k ry , T T rT ri  ry and rho  rro  ry at

The greater Δt is, the smaller p2 is. When  Therefore, to ensure that should be Δt1 ≤ Δt ≤ Δt2. Setting Δt1 = Δt2 results in rT hi

ð7:39Þ

Δt2 = 0. the same time, there

Thermal Autofrettage Technology Based on Tresca Yield Criterion p py ¼ ln k ¼ ry ry

275

ð7:40Þ

Equation (7.40) is just the mechanical entire yield pressure. If rpy \ ln k, Δt1 < Δt2; if rpy [ ln k, Δt1 > Δt2. This is not allowed in engineering. It happens that there is a similar case that in mechanical autofrettage technology, the load-bearing capacity or allowable loading is also always smaller than the entire yield loading in any case. At the inside wall, x = 1, then,   1 2k 2 k2 þ 1 T T rri ¼ p; rhi ¼ pt  2 p ð7:41Þ þ 2 ln k k  1 k 1 If

p ry



k 2 ln k 2 k 2 þ 1 pt ðk 2 þ 1Þ ln k ry

T  kk2 1 þ 1, rhi   ry . Substituting equation (7.26) into equa2

T tion (7.41) shows that even if pt = ptc, rT hi   ry provided p > 0. So rhi   ry in engineering. At the outside wall, x = k, then,   1 2 2p 2 pt T T T  ¼ ðp  pt Þ þ [0 rT ¼ 0; r ¼ r  r ¼ p þ 2 t ro ho ho ro ln k k 2  1 k  1 k2  1 ln k

ð7:42Þ If Δt = Δtc, substituting equation (7.26) into equation (7.36) obtains: rffiffiffiffiffiffiffiffiffiffiffiffiffi p k 2  1 p3 pe ry  ¼ ¼2 or k ¼ ry k2 ry ry ry  p

ð7:43Þ

where pe is the maximum mechanical elastic load-bearing capability of an unautofrettaged cylindrical pressure vessel or initial mechanical yield pressure, pe k 2 1 ry ¼ 2k 2 . Substituting equation (7.26) into equation (7.38) obtains: p ðk 4  1Þ ln k  ðk 2  1Þ2 p4 ¼  ry k 2 ln k 2  k 2 þ 1 ry In fact, setting Δt1 ≤ Δtc obtains p

p ry



k 2 1 k2

ð7:44Þ

¼ 2 rpye ; setting Δt2 ≤ Δtc results in

p ≥ p4. rp3y , rp4y and ryy are plotted in figure 7.18. The meaning of each curve and the main parameters are marked in the figures. p Figure 7.18 shows that the three curves for rp3y , rp4y and ryy have a common intersection point a, whose coordinates are a (kc, 0.796812…). 2 T ¼ rp3y ¼ 2 rpye , rT As mentioned above, when Δt = Δtc, if rpy  k k1 2 hi  rri  ry can be 2

ln kðk 1Þ T ¼ rp4y , rT guaranteed; if rpy  ðk k1Þ 2 ln k 2 k 2 þ 1 ho  rro ¼ ry can be guaranteed. From figure 7.18, when k < kc, rp3y > rp4y , therefore, when k < kc and Δt = Δtc, only if p ≤ p4, 4

2

T T T rT hi  rri  ry and rho  rro  ry can be guaranteed simultaneously. However, p4 is

276

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 7.18 – The graphs of rp3y ,

p4 ry

and

py ry .

too low. Thus, when k < kc, if Δt = Δtc can not be changed, p can only take the value of p4; if Δt can be changed within 0 < Δt < Δtc, the allowable load p may be raised. The specific method to raise the allowable load p will be presented later. From figure 7.18 again, when k > kc, rp3y < rp4y , therefore, when k > kc and Δt = Δtc, if T T T p ≤ p3, rT hi  rri  ry and rho  rro  ry can be guaranteed simultaneously. T Substituting Δt1 into equation (7.34) must result in rT hi  rri ¼ ry , substituting T T Δt2 into equation (7.35) must result in rho  rro ¼ ry ; Substituting Δt1 into equaT tion (7.35) and setting rT ho  rro  ry , and substituting Δt2 into equation (7.34) and T T setting rhi  rri  ry both result in rpy \ ln k. This result is equivalent to Δt1 < Δt2.

This means that as long as Δt1 < Δt2, or

p ry

\ ln k, it is possible to assure rT hi 

T T rT ri  ry and rho  rro  ry in the meantime. Setting p1 = p2 obtains:

EaDtx ¼ ln k 2 ð1  lÞry

or

Dtx ¼

ð1  lÞry ln k 2 2ð1  lÞry k 2 ln k 2  k 2 þ 1 ¼n ¼ Dtc k2  1 Ea Ea ð7:45Þ

where n = lnk, Δtx is called the ideal temperature difference. The greater k is, the greater Δtx is. When Δt ≤ Δtx, p1 ≤ p2; when Δt ≥ Δtx, p1 ≥ p2. Substituting Δtx into equations (7.36) and (7.38) both obtains p ¼ ln k or k ¼ ep=ry ð7:46Þ ry Setting Δt1 ≤ Δtx results in rpy  ln k, and setting Δt2 ≤ Δtx results in rpy  ln k. In fact, setting Δt = Δtx in equations (7.36) and (7.38) obtains the entire yield pressure p1 p2 p ry ¼ ry ¼ ry ¼ ln k.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

277

Setting Δtx = Δtc or setting p3 = p4 obtains k 2 ln k ¼1 k2  1

or

ln k ¼

k2  1 k2

ð7:47Þ

Equation (7.47) is just equation (2.32), or equation (3.28), etc. The solution for equation (7.47) is k = kc = 2.218 457 489 916 7…, where kc is just the critical radius ratio. kc has been shown in figure 7.18. 2 2 2 2 2 k In fact, kk 2ln 1 ¼ 1 is equivalent to k lnk = k −1, substituting k lnk = k −1 into equation (7.26) and equation (7.45) both gives Δtx = Δtc. n and m reflect magnitudes of Δtx and Δtc, respectively. n and m are shown in figure 7.19. Letting n = m just results in equation (7.47). Then, the coordinate of point b in figure 7.19 is b (kc, 0.796812). When k < kc, n < m or Δtx < Δtc; when k = kc, n = m or Δtx = Δtc, the load-bearing capacity of a pressure vessel reaches the maximum: rp1y ¼ rp2y ¼ rpy ¼ ln k. If Δtx < Δt < Δtc, rpy should be calculated by equation (7.38), the result is rpy \ ln k; if Δt < Δtx rpy should be calculated by equation (7.36), the result is also rpy \ ln k. When k > kc, n > m, or Δtx > Δtc. Under the case of k > kc, if Δt = Δtx, there still is p1 = p2 = σylnk, but when Δt > Δtc, the pressure vessel has been yielded, thermal stresses are not the elastic stresses expressed by equations (7.18)–(7.20). So, when k > kc, the maximum temperature difference is Δtc. When k > kc, Δtc < Δtx, so, when k > kc, rpy is controlled by equation (7.36). When Δt = Δtc, the load-bearing capacity of a pressure vessel when k > kc reaches the maximum. When k > kc and 2 Δt = Δtc, equation (7.36) becomes equation (7.43): rpy ¼ k k1 2 \ ln k. In brief, if k < kc, the maximum load-bearing capacity is rpy ¼ ln k; if k > kc, the p k 1 ¼ ln k just results in maximum load-bearing capacity is rpy ¼ k k1 2 . Letting r ¼ k 2 y equation (7.47). It is interesting and wonderful that in mechanical autofrettage, if 2

2

FIG. 7.19 – The variation tendency of m and n.

278

Autofrettage Technology and Its Applications in Pressured Apparatuses

the plastic region covers the whole wall of a pressure vessel, for compressive yield not to occur after removing autofrettage pressure pa, the expression for ultimate k is also equation (7.47). This fact deserves to be studied further. 2 ¼ ln k we also Setting p3 = p4, we obtain equation (7.47); setting rpy ¼ k k1 2 k obtain equation (7.47); setting m = n, we still obtain equation (7.47). kk 2ln 1 ¼ 1 appeared many times and played an important role in mechanical and thermal autofrettage technology. When k < kc, the larger k is, the smaller Δtc−Δtx is; When k = kc, Δtc = Δtx. When k > kc, the larger k is, the larger Δtx−Δtc is. If p1 and p2, Δt1 and Δt2 are illustrated by taking some typical parameters as examples, the above results will be clearer and more intuitive. The following figures can also be used as a reference for engineering design. For Elastic modulus E = 1.95 × 105 MPa, Poisson ratio μ = 0.3, thermal expansion coefficient of the materials α = 1.2 × 10–5 °C−1, strength limit σy = 350 MPa and radius ratio k = 1.5, k = 2, k = kc, k = 3 and k = 4, Δt1 and Δt2 are plotted in figures 7.20–7.24, respectively. For other parameters, the changing trends and patterns of Δt1 and Δt2 are similar to those of figures 7.20–7.24, because for steel Ea Ea is small, popularly 2ð1lÞ ≈1.5  1.8. For materials, the change of parameters 2ð1lÞ 2

example, here

Ea 2ð1lÞ

= 1.671429….

FIG. 7.20 – The graphs of Δt1 and Δt2 for k = 1.5. In figure 7.20, the abscissa of the intersection point of straight line Δt1 and p straight line Δt2 is ryy . When k = 1.5, Δtc = 184.707 °C, rp3y = 0.555556,

p4 ry

p

= 0.147412, ryy = lnk = 0.405465. When p = p3, Δt1 = Δtc, Δt2 < Δtc; When p = p4, Δt2 = Δtc, Δt1 < Δtc (Δt1 < 0); At the intersection point of straight line Δt1 p and straight line Δt2, or when rpy = rp3y = rp4y = ryy , Δt1 = Δt2 = Δtx = 84.90509 °C pe < Δtc. When rpy ¼ k2k1 2 ¼ r = 0.277778, Δt1 = 0; When y Δt2 = 0. See figure 7.20. 2

p ry

¼ k 21 ¼ k 2 rpye = 0.625, 2

Thermal Autofrettage Technology Based on Tresca Yield Criterion

279

FIG. 7.21 – The graphs of Δt1 and Δt2 for k = 2. In figure 7.21, the abscissa of the intersection point of straight line Δt1 and p straight line Δt2 is ryy . When k = 2, Δtc = 171.0838 °C, rp3y = 0.75, rp4y = 0.548963,

py ry

= lnk = 0.693147. When p = p3, Δt1 = Δtc, Δt2 < Δtc; When p = p4, Δt2 = Δtc, Δt1 < Δtc. At the intersection point of straight line Δt1 and straight line Δt2, or when py p p3 p4 p k 2 1 Δt1 = Δt2 = Δtx = 145.1462 °C < Δtc. When ry = ry = ry = ry , ry ¼ 2k 2 ¼ pe ry

= 0.375, Δt1 = 0; When

p ry

¼ k 21 ¼ k 2 rpye = 1.5, Δt2 = 0. See figure 7.21. 2

FIG. 7.22 – The graphs of Δt1 and Δt2 for k = kc. In figure 7.22, the abscissa of the intersection point of straight line Δt1 and p p straight line Δt2 is ryy . When k = kc, Δtc = 166.8538 °C, rp3y = rp4y = ryy = lnk = 0.796812. At the intersection point of straight line Δt1 and straight line Δt2, or when

Autofrettage Technology and Its Applications in Pressured Apparatuses

280 p ry

=

p3 ry

When

= p ry

p4 ry

¼

py p pe k 2 1 ry , Δt1 = Δt2 = Δtx = Δtc. When ry ¼ 2k 2 ¼ ry k 2 1 2 pe 2 ¼ k ry = 1.960777, Δt2 = 0. See figure 7.22.

=

= 0.398406, Δt1 = 0;

FIG. 7.23 – The graphs of Δt1 and Δt2 for k = 3. In figure 7.23, the abscissa of the intersection point of straight line Δt1 and p straight line Δt2 is ryy . When k = 3, Δtc = 156.2978 °C, rp3y = 0.888889, rp4y = 2.028785,

py ry

= lnk = 1.098612. When p = p3, Δt1 = Δtc, Δt2 > Δtc; When p = p4, Δt2 = Δtc, Δt1 > Δtc. At the intersection point of straight line Δt1 and straight line Δt2, or when py p p3 p4 p k 2 1 Δt1 = Δt2 = Δtx = 230.0513 °C > Δtc. When ry = ry = ry = ry , ry ¼ 2k 2 ¼ pe ry

= 0.444444, Δt1 = 0; When

p ry

¼ k 21 ¼ k 2 rpye = 4, Δt2 = 0. See figure 7.23. 2

FIG. 7.24 – The graphs of Δt1 and Δt2 for k = 4.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

281

In figure 7.24, the abscissa of the intersection point of straight line Δt1 and p straight line Δt2 is ryy . When k = 4, Δtc = 148.303 °C, rp3y = 0.9375, rp4y = 4.376664,

py ry

= lnk = 1.386294. When p = p3, Δt1 = Δtc, Δt2 > Δtc; When p = p4, Δt2 = Δtc, Δt1 > Δtc. At the intersection point of straight line Δt1 and straight line Δt2, or when py p p3 p4 p pe k 2 1 ry = ry = ry = ry , Δt1 = Δt2 = Δtx = 290.2924 °C > Δtc. When ry ¼ 2k 2 ¼ ry = 0.46875, Δt1 = 0; When rpy ¼ k 21 ¼ k 2 rpye = 7.5, Δt2 = 0. See figure 7.24. When k < kc, p4 < py < p3, the smaller k is, the smaller p4 is; and the smaller k is, the greater p3−p4 is. When k = kc, p3 = p4 = py. When k > kc, p3 < py < p4, the greater k is, the greater p4 is, and the greater k is, the greater p4−p3 is. When k > kc, at the intersection point of straight line Δt1 and straight line Δt2, where p Δt1 = Δt2, there is Δtx > Δtc. Therefore, when k > kc, ryy = lnk cannot be taken as the allowable load, the optimum allowable load is equation (7.43), or p p3 pe k 2 1 ry ¼ k 2 ¼ ry ¼ 2 ry . For E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 350 MPa and k = 1.5, k = 2, k = kc, k = 3 and k = 4, rp1y and rp2y are plotted in figures 7.25–7.29, respectively. For other parameters, the changing trends and patterns of rp1y and rp2y are similar to those of figures 7.25–7.29. 2

FIG. 7.25 – The graphs of rp1y and rp2y for k = 1.5.

In figure 7.25, the abscissa of the intersection point of straight line rp1y and straight line rp2y is Δtx ( Δtx, p1 > p2. When k = 1.5, Δtc = 184.707 °C, Δtx = 84.90509 °C < Δtc, rp3y = 0.555556, rp4y = 0.147412,

py ry

= lnk = 0.405465. When Δt = Δtc,

Δt =

Δtx, rp1y

=

p2 ry

=

p3 ry

=

p4 ry

=

py ry .

p1 ry

=

p3 ry

>

See figure 7.25.

p2 ry

=

p4 ry

and

p3 ry

>

py p4 ry , ry


p2. When k = 2, Δtc = 171.0838 °C, Δtx = 145.1462 °C < Δtc, 0.693147. When Δt = Δtc, p1 ry

=

p2 ry

=

p3 ry

=

p4 ry

=

py ry .

p1 ry

=

p3 ry

>

p2 ry

=

p4 ry

p3 ry

py p4 ry = 0.548963, ry = lnk = py p4 py ry , ry < ry . When Δt = Δtx,

= 0.75,

and

p3 ry

>

See figure 7.26.

FIG. 7.27 – The graphs of p1/σy and p2/σy for k = kc. In figure 7.27, the abscissa of the intersection point of straight line rp1y and straight line rp2y is Δtx (=Δtc). When Δt < Δtx, p1 < p2; When Δt > Δtx, p1 > p2. When k = kc, Δtc = Δtx = 166.8538 °C, p1 ry

=

p2 ry

=

p3 ry

=

p4 ry

=

py ry .

p3 ry

=

p4 ry

=

py ry

See figure 7.27.

= lnk = 0.796812. When Δt = Δtx = Δtc,

Thermal Autofrettage Technology Based on Tresca Yield Criterion

283

FIG. 7.28 – The graphs of p1/σy and p2/σy for k = 3. In figure 7.28, the abscissa of the intersection point of straight line rp1y and straight line rp2y is Δtx ( Δtx, p1 > p2. When k = 3, p3 ry = 0.888889, py p2 p4 p3 ry = ry and ry < ry ,

Δtc = 156.2978 °C, Δtx = 230.0513 °C > Δtc, lnk = 1.098612. When Δt = Δtx, rp1y

=

p2 ry

=

p3 ry

=

p4 ry

=

py ry .

Δtc, rp1y

=

p3 ry




py ry .

py ry

=

When Δt =

See figure 7.28.

FIG. 7.29 – The graphs of p1/σy and p2/σy for k = 4. If Elastic modulus E, Poisson ratio μ, the thermal expansion coefficient of the materials α, strength limit σy can be changed, the design projects will be more flexible.

284

7.5

Autofrettage Technology and Its Applications in Pressured Apparatuses

Examples

Example 1. k = 2, E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10−5 °C−1, σy = 300 MPa. This is the case of k < kc. If the temperature difference is not exerted, the 2 load-bearing capacity of the pressure vessel with k = 2 is only rpy ¼ rpye ¼ k2k1 2 = 0.375. (1) From equation (7.45), Δtx = 124.411 °C < Δtc = 146.6433 °C obtained from equation (7.26). When Δt = Δtx = 124.411 °C, equations (7.36) and (7.38) both lead to rpy = 0.693147391 = lnk = ln2 > 0.375. This is the optimum design project; the following projects are available for reference. (2) Taking Δt = 130 °C > Δtx, from equation (7.36) rpy = 0.707439 > ln2, from equation (7.38) rpy = 0.656900602 < ln2. Substituting rpy = 0.707439 into T T T equations (7.34) and (7.35) obtains rT hi  rri = 300 MPa and rho  rro = 310 p MPa > σy, respectively. Substituting ry = 0.656900602 into equations (7.34) T and (7.35) obtains σθi−σri = 260 MPa, rT ho  rro = 300 MPa, respectively. p Therefore, when Δt = 130 °C, ry = 0.656900602 < ln2. (3) Taking Δt = 110 °C < Δtx, from equation (7.36) rpy = 0.656295, from equation (7.38) p/σy = 0.786608202. Substituting rpy = 0.656295 into equaT T T tions (7.36) and (7.38) obtains rT hi  rri = 300 MPa and rho  rro = 274 MPa, p respectively. Substituting ry = 0.786608202 into equations (7.36) and (7.38) T T T obtains rT hi  rri = 404 MPa > σy and rho  rro = 300 MPa, respectively. p Therefore, when Δt = 110 °C, ry = 0.656295 < ln2. (4) Substituting Δt = Δtc = 146.6433 °C into equation (7.36) or substituting k = 2 into equation (7.43) obtains rpy = rp3y = 0.75 > lnk. However, substituting p into equations (7.34) and (7.35) respectively obtains ry = 0.75 T T T rT hi  rri = 300 MPa and rho  rro = 340 MPa > σy. Substituting Δt = Δtc = 146.6433 °C into equation (7.38) or substituting k = 2 into equation (7.44) obtains rpy = rp4y = 0.548962477 < lnk.

Therefore, for a pressure vessel with k = 2, the optimum operation parameters are Δt = Δtx = 124.411 °C, rpy = 0.693147391 = lnk. Under these parameters, the distributions of thermal stresses and total stresses are shown in figures 7.30 and 7.31, respectively. The meaning of each curve and the main parameters are marked in the figures. Example 2. k = 2.5, E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 300 MPa. This is the case of k > kc. If there is no temperature difference, the load-bearing 2 capacity of the pressure vessel with k = 2.5 is only rpy ¼ rpye ¼ k2k1 2 = 0.42.

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285

FIG. 7.30 – Distribution of thermal stresses for k = 2 under optimum operation parameters: Δt = Δtx.

FIG. 7.31 – Distribution of total stresses for k = 2 under optimum operation parameters: Δt = Δtx and p = σylnk. (1) From equation (7.26) Δtc = 139.181 °C; from equation (7.45) Δtx = 164.4624 °C > Δtc. Taking Δt = Δtc = 139.181 °C, from equations (7.36) or (7.43) p p3 pe k 2 1 ry ¼ ry ¼ k 2 ¼ 2 ry = 2 × 0.42 = 0.84. This is the optimum design project; the following projects are available for reference. (2) If taking Δt = 130 °C < Δtc, from equation (7.36) rpy = 0.812295, from equation (7.38) rpy = 1.274343849. Substituting rpy = 0.812295 and Δt = 130 °C into T T T equations (7.34) and (7.35) obtains rT hi  rri = 300 MPa and rho  rro = 247.2 MPa, respectively; Substituting p/σy = 1.274343849 and Δt = 130 °C T into equations (7.34) and (7.35) obtains rT hi  rri = 630 MPa > σy and

Autofrettage Technology and Its Applications in Pressured Apparatuses

286

T rT ho  rro = 300 MPa, respectively. Then, when Δt = 130 °C, < 0.84.

p ry

= 0.812295

Therefore, for a pressure vessel with k = 2.5, the optimum operation parameters are Δt = Δtc = 139.181 °C, p = 2pe. Under these parameters, the distributions of thermal stresses and total stresses are shown in figures 7.32 and 7.33, respectively. The meaning of each curve and the main parameters are marked in the figures.

FIG. 7.32 – Distribution of thermal stresses for k = 2.5 under optimum operation parameters:

Δt = Δtc.

FIG. 7.33 – Distribution of total stresses for k = 2.5 under optimum operation parameters: Δt = Δtc and p = 2pe.

If temperature difference can be chosen within the range below Δtc, then, for a pressure vessel with some k, take the smaller of the values determined by equations (7.36) and (7.38) as allowable loading, this has been shown in examples 1 and 2.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

287

For a certain Δt and p, determine radius ratio k by equations (7.36) and (7.38) or by equations (7.43) and (7.46), and take the greater radius ratio as the real radius ratio. By equations (7.36) and (7.38), we can weigh and adjust the parameters k, p, properties of materials (E, α, μ, and σy) and Δt to construct a pressure vessel that fits practical requirements. Example 3. A pressure vessel is to be subjected to a pressure 225 MPa, determine radius ratio (E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 350 MPa). If there is no temperature difference, under an elastic state, even if k is infinite, the load-bearing capacity of the pressure vessel p = 0.5σy (based on the maximum shear stress theory) or rpy ¼ p1ffiffi3 = 0.577 (based on the maximum distortion strain energy theory). However, in this example, rpy ¼ 225 350 = 0.642857 > 0.577, so even if the wall is infinite, the pressure vessel cannot bear a pressure 225 MPa. qffiffiffiffiffiffiffiffi ry (1) If wall is required to be minimum, then k ¼ ry p ¼ 1:67332 from equa-

tion (7.43) or k ¼ ep=ry ¼ 1:901907 from equation (7.46). Take the greater k = 1.901907 as the real radius ratio. When k = 1.901907, from equation (7.45) Δtx = 134.6154 °C, from equation (7.26) Δtc = 173.259 °C. According to the above conclusions, take Δt = Δtx = 134.6154 °C. Substituting rpy = 0.642857 and Δt = T Δtx = 134.6154 °C into equations (7.34) and (7.35) both obtains rT hi  rri = T T rho  rro = 350 MPa = σy. The distributions of thermal stresses and total stresses are shown in figures 7.34 and 7.35, respectively. The meaning of each curve and the main parameters are marked in the figures.

FIG. 7.34 – Distribution of thermal stresses for k = 1.901907 under optimum operation parameters: Δt = Δtc.

288

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 7.35 – Distribution of total stresses for k = 1.901907 under optimum operation parameters: Δt = Δtc and p = σylnk = 225 MPa. If the wall is not required to be minimum, then choose a suitable k on the basis of practical requirements. (2) Saying k = 2.2, then, from equation (7.45) Δtx = 165.1043 °C, from equation (7.26) Δtc = 167.1825 °C, from equation (7.37) Δt1 = 103.7427 °C, from equation (7.39) Δt2 = 186.349 °C > Δtc, which is of no significance. Substituting p and Δt = Δt1 into equations (7.34) and (7.35) obtains ry = 0.642857 T T T rT hi  rri = 350 MPa = σy and rho  rro = 246.8 MPa < σy, respectively. Substip tuting ry = 0.642857 and Δt = Δtc into equations (7.34) and (7.35) obtains T T T rT hi  rri = 217.2 MPa < σy and rho  rro = 326.1 MPa < σy, respectively. When Δt = Δt1 = 103.7427 °C, k = 2.2, the distributions of thermal stresses and total stresses are shown in figures 7.36 and 7.37, respectively. The meaning of each curve and the main parameters are marked in the figures.

FIG. 7.36 – Distribution of thermal stresses for k = 2.2, Δt = Δt1 = 103.7427 °C.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

289

FIG. 7.37 – Distribution of total stresses for k = 2.2, Δt = Δt1 = 103.7427 °C, and

p = 225 MPa.

When Δt = Δtc = 167.1825 °C, k = 2.2, p = 225 MPa, the distributions of thermal stresses and total stresses are shown in figures 7.38 and 7.39, respectively. The meaning of each curve and the main parameters are marked in the figures. (3) Saying k = 3, from equation (7.26) Δtc = 156.2978 °C, from equation (7.45) Δtx = 230.0513 °C > Δtc, Δtx is of no significance. From equation (7.37) Δt1 = 69.77582 °C, from equation (7.39) Δt2 = 266.1882 °C > Δtc. Δt2 is of no significance. Substituting rpy = 0.642857 and Δt = Δt1 into equations (7.34) and (7.35) T T T obtains rT hi  rri = 350 MPa = σy and rho  rro = 133.3 MPa < σy, respectively. p Substituting ry = 0.642857 and Δt = Δtc into equations (7.34) and (7.35) obtains T T T rT hi  rri = 156.3 MPa < σy and rho  rro = 228.7 MPa < σy, respectively.

FIG. 7.38 – Distribution of thermal stresses for k = 2.2, Δt = Δtc = 167.1825 °C.

290

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 7.39 – Distribution of total stresses for k = 2.2, Δt = Δtc = 167.1825 °C, and

p = 225 MPa.

When Δt = Δt1 = 69.77582 °C, k = 3, the distributions of thermal stresses and total stresses are shown in figures 7.40 and 7.41, respectively. The meaning of each curve and the main parameters are marked in the figures. When Δt = Δtc = 156.2978 °C, k = 3, p = 225 MPa, the distributions of thermal stresses and total stresses are shown in figures 7.42 and 7.43, respectively. The meaning of each curve and the main parameters are marked in the figures. Example 4. A process can supply a temperature difference Δt = 120 °C for a pressure vessel, study the size and the load-bearing capacity of the pressure vessel, E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 350 MPa.

FIG. 7.40 – Distribution of thermal stresses for k = 3, Δt = Δt1 = 69.77582 °C.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

291

FIG. 7.41 – Distribution of total stresses for k = 3, Δt = Δt1 = 69.77582 °C and

p = 225 MPa.

FIG. 7.42 – Distribution of thermal stresses for k = 3, Δt = Δtc = 156.2978 °C.

(1) If the load-bearing capacity is not required, there are many schemes. Such as, I. taking k = 2.5, from equation (7.26), Δtc = 162.3777958 > Δt. From equation (7.36) rpy = 0.730387265. From equation (7.37) rpy = 1.556349. Take rpy = 0.730387265, or p = 255.63554 MPa as real loading. When Δt = 120 °C, k = 2.5, p = 255.63554 MPa, the distributions of thermal stresses and total stresses are shown in figures 7.44 and 7.45, respectively. The meaning of each curve and the main parameters are marked in the figures.

292

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 7.43 – Distribution of total stresses for k = 3, Δt = Δtc = 156.2978 °C and

p = 225 MPa.

FIG. 7.44 – Distribution of thermal stresses for k = 2.5, Δt = 120 °C.

II. taking k = 1.6, from equation (7.26), Δtc = 181.3931336 > Δt. From equation (7.36) rpy = 0.506252, From equation (7.34) rpy = 0.402030791. Take rpy = 0.402030791 or p = 140.71078 MPa as real loading. When Δt = 120 °C, k = 1.6, p = 140.71078 MPa, the distributions of thermal stresses and total stresses are shown in figures 7.46 and 7.47, respectively. The meaning of each curve and the main parameters are marked in the figures. (2) Supposing Δt = 120 °C = Δtx, then, from equation (7.45) k = 1.773688 < kc and rpy = lnk = 0.573060794, or p = 200.57128 MPa.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

293

FIG. 7.45 – Distribution of total stresses for k = 2.5, Δt = 120 °C and p = 255.63554 MPa.

FIG. 7.46 – Distribution of thermal stresses for k = 1.6, Δt = 120 °C.

FIG. 7.47 – Distribution of total stresses for k = 1.6, Δt = 120 °C and p = 140.71078 MPa.

294

Autofrettage Technology and Its Applications in Pressured Apparatuses When Δt = Δtx=120 °C, k = 1.773688, p = 200.57128 MPa, the distributions of thermal stresses and total stresses are shown in figures 7.48 and 7.49, respectively. The meaning of each curve and the main parameters are marked in the figures.

FIG. 7.48 – Distribution of thermal stresses for k = 1.773688, Δt = Δtx = 120 °C.

FIG. 7.49 – Distribution of total stresses for k = 1.773688, Δt = Δtx = 120 °C and p = 200.57128 MPa. (3) Supposing Δt = 120 °C = Δtc, then, from equation (7.26) k = 49.872, it is unrealistic. (4) I. Supposing the pressure vessel is required to contain 200 MPa, or p 200 ry ¼ 350 = 0.571429.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

295

From equation (7.36) by approximation, k = 1.76884; from equation (7.37) by approximation, k = 1.772141. Take k = 1.772141. When Δt = 120 °C, k = 1.772141, p = 200 MPa the distributions of thermal stresses and total stresses are similar to figures 7.48 and 7.49, respectively. II.

Supposing the pressure vessel is required to contain 300 MPa or p 300 ry ¼ 350 = 0.857143. From equation (7.36) by approximation, k = 4.219641; from equation (7.37) by approximation, k = 2.019384. Take k = 4.219641. When Δt = 120 °C, k = 4.219641, p = 300 MPa, the distributions of thermal stresses and total stresses are shown in figures 7.50 and 7.51, respectively. The meaning of each curve and the main parameters are marked in the figures.

FIG. 7.50 – Distribution of thermal stresses for k = 4.219641, Δt = 120 °C.

FIG. 7.51 – Distribution of total stresses for k = 4.219641, Δt = 120 °C and p = 300 MPa.

Autofrettage Technology and Its Applications in Pressured Apparatuses

296

7.6

The Total Stresses Under Optimum Operation Conditions

When k ≤ kc, if Δt = Δtx, then p = py = σy lnk and pt = σylnk = py = p, both are entire yield pressure. Therefore, when k ≤ kc and Δt = Δtx, the total stresses are t p rT r ¼ rr þ rr ¼ py

rT h

rT z

¼

rtz

þ rpz

¼

rth

þ rph

lnðk=xÞ k ¼ ry ln ln k x

ð7:48Þ

  1  lnðk=xÞ k ¼ ry 1  ln ¼ py ln k x

ð7:49Þ

    1  2 lnðk=xÞ 1 k2 ln k  2 ¼ py ¼ ry 1  ln 2  2 ln k k 1 k 1 x

ð7:50Þ

T T T In this case, p−pt = 0, lnptk ¼ ry , then, rT hi  rri ¼ rho  rro ¼ ry . In fact, it is seen T from equation (7.48) to equation (7.48) that, within the whole wall rT h  rr  ry , which can be also referred to in example 1 and figure 7.29. According to equation (7.48) to equation (7.48),     lnðk=xÞ 1 k ln k þ þ rT  r ¼ p ln ¼ r z y y h ln k k2  1 x k2  1

rT z  rr ¼ p y

    1  lnðk=xÞ 1 k ln k  2 ¼ ry 1  ln  2 ln k k 1 x k 1 k2p

y T T T k ln k As x increases, rT h  rz decreases. When x = 1, rhi  rzi ¼ k 2 1 ¼ k 2 1 ry  ry ; 2

r ln k

y T when x = k, rT ho  rzo ¼ k 2 1 \ry . T T As x increases, rz  rr increases.  1ln k  T 1 k 2 ln k ¼ r When x = 1, ry \rT  r ¼ p  1  y ln k y zi ri k 2 1 k 2 1 \ry ;     1 1 ln k T When x = k, ry \rT zo  rro ¼ py ln k  k 2 1 ¼ ry 1  k 2 1 \ry . T T T rT z  rr is parallel with rh and smaller than rh . The total stresses when k = 2 and Δt = Δtx have been shown in example 1, see figure 7.31. When k ≤ kc, if Δt = Δtx, pt = σylnk = py, the equations of thermal stresses or equation (7.18)–(7.20) become

rtr k 2 ln k 1 k 2 ln k ¼ ln x þ 2  k  1 x2 k2  1 ry

ð7:51Þ

rth k 2 ln k 1 k 2 ln k ¼ 1 þ ln x  2  k  1 x2 k2  1 ry

ð7:52Þ

Thermal Autofrettage Technology Based on Tresca Yield Criterion

rtz k 2 ln k 2 ¼ 1 þ ln x 2  2 ry k 1

297

ð7:53Þ

Equation (7.48) to equations (7.48), (7.51)–(7.53) tell us that the total stresses and thermal stresses are not related to elastic modulus E, Poisson ratio μ and thermal expansion coefficient of the materials α under the case that k ≤ kc, Δt = Δtx and pt = σylnk = py. By taking k = 1.5, k = 2 and k = kc as examples, the thermal stresses and the total stresses are illustrated in figures 7.52–7.57, respectively. The meaning of each curve is marked in the figures.

FIG. 7.52 – The thermal stresses with Δt = Δtx, pt = py and k = 1.5.

FIG. 7.53 – The total stresses with Δt = Δtx, pt = p = py and k = 1.5.

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Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 7.54 – The thermal stresses with Δt = Δtx, pt = py and k = 2.

FIG. 7.55 – The total stresses with Δt = Δtx, pt = p = py and k = 2. Under the operation condition that Δt = Δtx and p = pt = σylnk = py, the total stresses are cut down and evened greatly, especially the maximum circumferential tension on the inside surface under internal pressure is cut down considerably. 2 ¼ rp3y ¼ 2 rpye , thermal loading is critical When k ≥ kc, if Δt = Δtc, then rpy ¼ k k1 2 thermal loading ptc, or equation (7.27). Therefore, the equations of thermal stresses or equations (7.18)–(7.20) become ðk 2  1Þ ln xk rtr ðk 2  x 2 Þ ln k þ ¼ 2 ry k ln k 2  k 2 þ 1 x 2 ðk 2 ln k 2  k 2 þ 1Þ

ð7:54Þ

Thermal Autofrettage Technology Based on Tresca Yield Criterion

299

FIG. 7.56 – The thermal stresses with Δt = Δtx, pt = py and k = kc.

FIG. 7.57 – The total stresses with Δt = Δtx, pt = p = py and k = kc. rth ðk 2  1Þð1  ln kx Þ ðk 2 þ x 2 Þ ln k  ¼ 2 k ln k 2  k 2 þ 1 x 2 ðk 2 ln k 2  k 2 þ 1Þ ry

ð7:55Þ

rtz ðk 2  1Þð1  2 ln xk Þ ln k 2 ðk 2  1Þðln x 2 þ 1Þ  k 2 ln k 2  ¼ ð7:56Þ ¼ 2 k ln k 2  k 2 þ 1 ry k 2 ln k 2  k 2 þ 1 k 2 ln k 2  k 2 þ 1 The thermal stresses are irrelevant to the properties of the materials (E, α, μ and σy etc.) and are only related to the sizes of pressure vessels under the case that 2 ¼ rp3y ¼ 2 rpye . k ≥ kc, Δt = Δtc and rpy ¼ k k1 2 When k = kc, (7.51)–(7.53).

k 2 ln k k 2 1

¼ 1, equations (7.54)–(7.56) accordingly become equations

300

Autofrettage Technology and Its Applications in Pressured Apparatuses

For k = 3 and k = 4, the thermal stresses are plotted in figures 7.58 and 7.59, respectively.

FIG. 7.58 – The thermal stresses for Δt = Δtc, k = 3.

FIG. 7.59 – The thermal stresses for Δt = Δtc, k = 4. The total stress is the superposition of thermal stress and operating stress, when k ≥ kc and Δt = Δtc, the total stresses are ðk 2  1Þ ln xk rT rt rp ðk 2  x 2 Þ ln k k2  x2 r þ 2 2  2 2 ¼ r þ r ¼ 2 2 2 2 2 ry ry ry k ln k  k þ 1 x ðk ln k  k þ 1Þ x k

ð7:57Þ

rp ðk 2  1Þð1  ln kx Þ rt rT ðk 2 þ x 2 Þ ln k k2 þ x2 h  2 2 þ 2 2 ¼ hþ h¼ 2 2 2 2 2 k ln k  k þ 1 x ðk ln k  k þ 1Þ ry ry ry x k

ð7:58Þ

Thermal Autofrettage Technology Based on Tresca Yield Criterion

301

rT rt rp ðk 2  1Þð1  2 ln xk Þ ln k 2 1 z  þ ¼ z þ z ¼ 2 k ln k 2  k 2 þ 1 ry ry ry k 2 ln k 2  k 2 þ 1 k 2 rT h ry

rT

 rry ¼ k

ð7:59Þ

ln k 2 þ ðk 2 1Þðx 2 2Þ x 2 ðk 2 ln k 2 k 2 þ 1Þ . It is easy to prove that, when k ≥ kc, within the whole T T T wall there is rT h  rr  ry ; when x = 1, or at inside wall, rhi  rri ¼ ry ; when x = k, T 2 2 2 2 2 rT r 1Þðk 2Þ k when k = kc, kk 2ln or at outside wall, rhoy  rroy ¼ k kln2 ðkk 2þlnðkk 2 k 2 þ 1Þ , 1 ¼ 1, then T rT ho  rro ¼ ry . 2

The thermal stresses have nothing to do with modulus of elasticity E, coefficient of thermal expansion α, Poisson’s ratio μ under the case that k ≥ kc, Δt = Δtc and p p3 pe k 2 1 ry ¼ k 2 ¼ ry ¼ 2 ry . For k = 3 and k = 4, the total stresses are plotted in figures 7.60 and 7.61, respectively.

FIG. 7.60 – The total stresses with Δt = Δtc, k = 3 and

pt ry pt ry

When k = 3, Δt = Δtc = 133.9696 °C (σy = 300 MPa), = 0.746402.

p ry

p ry

When k = 4, Δt = Δtc = 127.1168 °C (σy = 300 MPa), = 0.755437.

¼ k k1 2 . 2

¼ k k1 = 0.888889, 2 2

p ry

¼ k k1 = 0.9375, 2 2

¼ rp3y ¼ 2 rpye , the total Under the operation condition that Δt = Δtc and rpy ¼ k k1 2 stresses are reduced and evened markedly, especially the maximum circumferential tension on the inner surface under internal pressure is reduced enormously. 2

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Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 7.61 – The total stresses with Δt = Δtc, k = 4 and

7.7 1.

2.

p ry

¼ k k1 2 . 2

Chapter Summary Thermal stresses of materials are related to not only the sizes of the materials but also the physical properties of the materials. The materials with a great modulus of elasticity and/or thermal expansion coefficient will have a great thermal stresses, and the materials with great Poisson ratio will have small thermal stresses. The quantity of heat conducted through a wall of a cylindrical pressure vessel in unit time is related to the thermal conductivity of the material, but the distribution of the temperature in the wall of the pressure vessel is irrelevant to the physical properties of the materials, it depends on the temperatures of inside and outside walls and radii of inside and outside walls. When Δt > 0, the wall temperature in the wall of the pressure vessel gradually drops from the inside surface to the outside surface; when Δt < 0, the temperature in the wall of the pressure vessel gradually rises from the inside surface to the outside surface. 3.1 When Δt > 0, rth and rtz monotonically increase from compressive stress on the inside surface to tension on the outside surface; when Δt < 0, rth and rtz monotonically decrease from tension on the inside surface to compressive stress on the outside surface; that is to say, the material is compressed where the temperature is high, and tensioned where the temperature is low. 3.2 No matter Δt > 0 or Δt < 0, the absolute values of stresses on the inside surface are

greater

t than those

t on the outside

t surface. The larger k is, the

rthi

rzi

rrmin

r larger pt and pt as well as pt and rmax pt are; the larger k is, the smaller

t

t

rho

r

pt and pzot are; as |Δt| increases, the inside surface gets yielded always earlier than the outside surface.

Thermal Autofrettage Technology Based on Tresca Yield Criterion

4.

5.

6. 7. 8. 9. 10.

11.

303

3.3 The thicker the pressure vessel is, the smaller the critical temperature difference |Δtc| is. When engineers design thick-wall pressure apparatuses, the parameters k, p, properties of materials (E, α, μ, σy and ri, ro) and Δt can be adjusted to construct a variety of technique proposals for a pressure vessel to fit practical requirements and to draw on the advantages and avoid disadvantages of thermal stresses. Using thermal stresses as pre-stresses in cylindrical high-pressure apparatuses is a potential and promising autofrettage technology for the apparatuses. In the elastic stage, Δt > 0 is beneficial to decreasing dangerous stresses in the cylindrical pressure vessel under internal pressure operation, and Δt < 0 is beneficial to decreasing dangerous stresses in the cylindrical pressure vessel under external pressure operation. If the temperature difference is increased above the critical temperature difference |Δtc|, the materials will enter the elastic-plastic phase, at this time, the residual thermal stresses can be obtained after eliminating the temperature difference. This is another way of autofrettage that adopts residual thermal stresses. For the thermal stress, rtz ¼ rtr þ rth at any location. At the inside surface and outside surface, the thermal circumferential stress and thermal axial stress are equal to each other. qffiffiffiffiffiffiffiffiffiffiffi 2 2 If Δt > 0, rtr gets minimum at x ¼ kk 2ln1k , which means rtr ¼ rtz  rth \0, or rtz \rth within 1 ≤ x ≤ k. p T T p For the total stress, rT h  ðrz  rr Þ ¼ rz ¼ k 2 1 = a constant > 0, thus T rT z  rr  rh . When k ≤ kc, Δtx ≤ Δtc; when k ≥ kc, Δtx ≥ Δtc. When k < kc, the optimum operation conditions are Δt = Δtx, and the load-bearing capacity of a pressure vessel is p = σylnk = pt, rh  rr  ry . When k > kc, the optimum operation conditions are Δt = Δtc, and the load-bearing capacity of a pressure vessel is p pe p k 2 1 k 2 1 ry ¼ k 2 ¼ 2 ry ; when k = kc, ry ¼ k 2 ¼ ln k. Unexpectedly, this result is the same as that obtained in mechanical autofrettage except that Δtx and Δtc are not concerned with mechanical autofrettage. T T T T T Usually, rT hi   ry in engineering. If rho  rro  ry , rho  ry . If rhi  rri  ry , T T T T rhi  ry . If p < pt, rh  rr increase monotonically; if p > pt, rT h  rr pt T T decrease monotonically; if p = pt, rh  rr is a constant: ln k . The main equations and conclusion are listed in table 7.1.

304

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 7.1 – The main equations and conclusion of this chapter.

Thermal stresses



 lnðk=xÞ k2  x2 þ 2 2 ¼ pt  ln k x k  x2   1  lnðk=xÞ k2 þ x2  2 2 rth ¼ pt ln k x k  x2   1  2 lnðk=xÞ 2  2 rtz ¼ pt ln k k 1 EaDt pt ¼ 2ð1  lÞ The thermal stresses at the inside and outside walls rtri ¼ rtro ¼ 0   2k 2 1  rthi ¼ rtzi ¼ pt 2 k  1 ln k   1 2 t t rho ¼ rzo ¼ pt  2 ln k k  1 The equivalent thermal stress at the inside wall   2k 2 1 k 2 ln k 2  k 2 þ 1 ts t t t t ¼ pt  rei ¼ rri  rzi ¼ rzi ¼ rhi ¼ pt 2 ðk 2  1Þ ln k k  1 ln k Critical temperature difference, Δtc ðk 2  1Þ ln k 2ð1  lÞry Dtc ¼ 2 k ln k 2  k 2 þ 1 Ea Critical thermal loading ptc ðk 2  1Þ ln k ry ptc ¼ 2 k ln k 2  k 2 þ 1 Total stresses (thermal stresses plus operation stresses)     lnðk=xÞ k2  x2 p k2 t p þ 2 2 1  þ rT r ¼ rr þ rr ¼ pt  ln k k2  1 x k  x2 x2     2 2 1  lnðk=xÞ k þx p k2 p t  1 þ þ ¼ r þ r ¼ p rT t h h h ln k k2  1 x2k2  x2 x2   1  2 lnðk=xÞ 2 p t p  2 rT þ 2 z ¼ rz þ rz ¼ pt ln k k 1 k 1 rtr

T For rT hi  rri  ry , p k2  1 pt k 2 ln k 2  k 2 þ 1 k 2  1 Ea k 2 ln k 2  k 2 þ 1 p1  þ ¼ þ Dt ¼ 2 2 2 2 2 2 ry 2k k ln k 2k 2ð1  lÞry k ln k ry ry   k2  1 ln k 2 2ð1  lÞry 2 p ¼ Dt1 or Dt  k  ry 2 k 2 ln k 2  k 2 þ 1 Ea p1 increases with Δt increasing T For rT ho  rro  ry , p k 2  1 pt k 2  1  ln k 2 k 2  1 Ea k 2  1  ln k 2 p2  ¼ Dt ¼   ry ln k 2 ln k 2 ry ry 2 2 2ð1  lÞry  2  k 1 p ln k 2 2ð1  lÞry  ¼ Dt2 or Dt  2 2 ry k  1  ln k 2 Ea

Thermal Autofrettage Technology Based on Tresca Yield Criterion

305

TAB. 7.1 – (continued). p2 decreases with Δt increasing Min {p1, p2} is the allowable load T T To ensure that rT  rT ri  ry and rho  rro  ry meanwhile, there must be hi Δt1 ≤ Δt ≤ Δt2, p py [ ln k ¼ , Δt1 > Δt2, so, Δt which is larger than Δt1 and smaller than Δt2 does When ry ry not exist When Δt1 = Δt2 p py ¼ ln k ¼ and Δt1 = Δt2 = Δtx; If k ≤ kc, Δtx ≤ Δtc; If k ≥ kc, Δtx ≥ Δtc. So, ry ry p py ¼ ln k ¼ is only applicable to the case of k ≤ kc ry ry where py is just mechanical entire yield pressure If Δt = Δtc, p1 becomes rffiffiffiffiffiffiffiffiffiffiffiffiffi p k 2  1 p3 pe ry  ¼ ¼2 or k ¼ 2 ry k ry ry ry  p where pe is the maximum mechanical elastic load-bearing capability of an unautofrettaged pe k 2  1 ¼ cylindrical pressure vessel, or initial mechanical yield pressure, 2k 2 ry If Δt = Δtc, p2 becomes p ðk 4  1Þ ln k  ðk 2  1Þ2 p4 ¼  ry k 2 ln k 2  k 2 þ 1 ry Ideal temperature difference ð1  lÞry ln k 2 k 2 ln k 2  k 2 þ 1 ¼ Dtc k2  1 Ea Δtx increases as k increases. When Δt ≤ Δtx, p1 ≤ p2; when Δt ≥ Δtx, p1 ≥ p2; when Δt = Δtx, p1 = p2 When k 2 ln k k2  1 ¼ 1 or ln k ¼ k2  1 k2 or k = kc = 2.2184574899167…, Δtx = Δtc p When k < kc, Δtx < Δtc, if Δt = Δtx, the maximum load-bearing capacity is ¼ ln k ry When k = kc, Δtx = Δtc, if Δt = Δtx = Δtc, the maximum load-bearing capacity is p k2  1 ¼ ln k ¼ ry k2 p k2  1 When k > kc, Δtx > Δtc, if Δt = Δtc, the maximum load-bearing capacity is ¼ ry k2 The optimum design When k ≤ kc, if Δt = Δtx, Dtx ¼

p

p = py = σylnk ↔ k ¼ ery pt = σylnk = py = p When k ≤ kc and Δt = Δtx, the thermal stresses are rtr k 2 ln k 1 k 2 ln k ¼ ln x þ 2  k  1 x2 k2  1 ry

306

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 7.1 – (continued).

rth k 2 ln k 1 k 2 ln k ¼ 1 þ ln x  2  ry k  1 x2 k2  1 rtz k 2 ln k 2 ¼ 1 þ ln x 2  2 ry k 1 When k ≤ kc and Δt = Δtx, the total stresses are rT k r ¼  ln x ry rT k h ¼ 1  ln x ry T 2 rz k ln k ¼ 1  ln 2  2 k 1 ry x T within the whole wall rT h  rr  ry When k ≥ kc, if Δt = Δtc,

rffiffiffiffiffiffiffiffiffiffiffiffiffi p k2  1 ry ¼ ↔ k ¼ ry  p ry k2 Thermal loading is critical thermal loading ptc When k ≥ kc and Δt = Δtc, the total stresses are ðk 2  1Þ ln kx rT rt rp ðk 2  x 2 Þ ln k k2  x2 r þ  ¼ r þ r ¼ 2 ry ry ry k ln k 2  k 2 þ 1 x 2 ðk 2 ln k 2  k 2 þ 1Þ x 2k2 p 2 k t T 2 2 2 ðk  1Þð1  ln x Þ r r rh ðk þ x Þ ln k k þ x2  2 2 þ 2 2 ¼ hþ h¼ 2 2 2 2 2 ry ry ry x k k ln k  k þ 1 x ðk ln k  k þ 1Þ 2 k t p 2 ðk  1Þð1  2 ln Þ rT r r ln k 1 z x  2 þ ¼ z þ z ¼ 2 ry ry ry k ln k 2  k 2 þ 1 k 2 k ln k 2  k 2 þ 1

References [1] Liu H.W. (2019) Mechanics of materials (in Chinese). Higher Education Press, Beijing. [2] Nellis G., Klein S. (Dec. 2008) Heat transfer. Cambridge University Press, Cambridge University. [3] Wang Z.X. (Dec., 2014) Heat transfer (in Chinese). Beijing University press, Beijing. [4] Yang S.M., Tao W.Q. (Aug., 2010) Heat transfer (in Chinese). China Higher Education Press, Beijing. [5] Yu G.C. (1980) Chemical pressure vessels and equipment (in Chinese). Chemical Industrial Press, Beijing. [6] Wang Z.W. (1990) Design of chemical pressure vessels (in Chinese). Chemical Industrial Press, Beijing. [7] Zheng J.Y., Sang Z.F. (2019) Process equipment design (in Chinese). Chemical industrial Press, Beijing. [8] The compiling group of “handbook of mathematics”. (1979) A handbook of mathematics (in Chinese). Higher Education Press, Beijing.

Chapter 8 Thermal Autofrettage Technology Based on Mises Yield Criterion 8.1

The Analysis of Thermal Stresses

The thermal stresses in the elastic state are still equations (7.18)–(7.20), or equations (8.1)–(8.3) of this chapter.
 lnðk=xÞ k2  x2 t þ 2 2 rr ¼ pt  ð8:1Þ ln k x k  x2 rth


 1  lnðk=xÞ k2 þ x2  2 2 ¼ pt ln k x k  x2


rtz ¼ pt

1  2 lnðk=xÞ 2  2 ln k k 1

ð8:2Þ

 ð8:3Þ

EaDt . rtz ¼ rtr þ rth at any location. where pt is called thermal loading, MPa, pt ¼ 2ð1lÞ At the inside and outside walls, the thermal stresses are the same as equations (7.22)–(7.24), or equations (8.4)–(8.6) of this chapter:

rtri ¼ rtro ¼ 0
rthi

¼

rtzi

¼ pt


rtho ¼ rtzo ¼ pt

ð8:4Þ

2k 2 1  k 2  1 ln k

1 2  2 ln k k  1

 ð8:5Þ

 ð8:6Þ

When based on the maximum distortion strain energy theory (Mises’s yield criterion)[1], the equivalent thermal stress is DOI: 10.1051/978-2-7598-3111-1.c008 © Science Press, EDP Sciences, 2023

308

Autofrettage Technology and Its Applications in Pressured Apparatuses

rde

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i 1h ¼ ðrh  rz Þ2 þ ðrz  rr Þ2 þ ðrr  rh Þ2 2

On the inner wall, the equivalent thermal stress is rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi iffi      1 h t 2 2 2 td rhi  rtzi þ rtzi  rtri þ rtri  rthi rei ¼ 2 On the outer wall, the equivalent thermal stress is rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  2  2 i 1 h t td rho  rtzo þ rtzo  rtro þ rtro  rtho reo ¼ 2

ð8:7Þ

ð8:8Þ

ð8:9Þ

td When rtd ei = σy, the inner wall initiates yielding; when reo = σy, the outer wall initiates yielding. From equations (8.4) to (8.6),
 2k 2 1 k 2 ln k 2  k 2 þ 1 td t t rei ¼ rzi ¼ rhi ¼ pt 2  ð8:10Þ ¼ pt ðk 2  1Þ ln k k  1 ln k

Equation (8.10) is the same as equation (7.25). From equations (8.4) to (8.6),
 1 2 k 2  1  ln k 2 rtd   ¼ pt 2 \rtd ¼ p t eo ei 2 ln k k  1 ðk  1Þ ln k

ð8:11Þ

For equation (8.10), letting rtd ei ¼ ry obtains the temperature difference at which the inside wall initiates yielding, called critical temperature difference, Δtc: Dtc ¼

ðk 2  1Þ ln k 2ð1  lÞry ¼ mf k 2 ln k 2  k 2 þ 1 Ea

ð8:12Þ

ln k y where m ¼ k 2ðkln k1Þ 2 k 2 þ 1, reflecting the size of a cylindrical pressure vessel; f ¼ Ea , reflecting material properties. Equation (8.12) is the same as equation (7.26) of the last chapter. The critical temperature difference when Δt < 0 is equal to negative equation (8.12). When Δt > 0, the thermal loading at which the inside wall initiates yielding, called critical thermal loading ptc, is obtained from equation (8.12): 2

2ð1lÞr

ptc ¼

ðk 2  1Þ ln k ry ln k 2  k 2 þ 1

k2

ð8:13Þ

An example: commonly, elastic modules of steel materials E = 1.95 × 105 MPa, Poisson ratio μ = 0.3, thermal expansion coefficient of the materials α = 1.2 × 10–5 °C−1, k = 2.5, σy = 300 Mpa, from equation (8.12), Δtc = 139.1809679 °C. Setting Δt = Δtc = 139.1809679 °C, then, pt = 232.631 MPa. The thermal stresses are shown as the solid curves in figure 8.1. If Δt = −139.1809679 °C, the trend of variations of thermal stresses is averse to figure 8.1, as is shown in figure 8.2. The meaning of each curve and the main parameters are marked in the figures.

Thermal Autofrettage Technology Based on Mises Yield Criterion

309

FIG. 8.1 – Distribution of thermal stresses for elastic cylindrical pressure vessel when Δt = Δtc = 139.1809679 °C.

FIG. 8.2 – Distribution of thermal stresses for elastic cylindrical pressure vessel when Δt = Δtc = −139.1809679 °C.

8.2

The Analysis of Total Stresses and Investigation of Optimum Operation Conditions

The total stresses are the same as equations (7.30)–(7.32) or equations (8.14)–(8.16) of this chapter:

 lnðk=xÞ k2  x2 p k2 T t p þ 2 2 1 2 rr ¼ rr þ rr ¼ pt  þ 2 ð8:14Þ ln k k 1 x k  x2 x

Autofrettage Technology and Its Applications in Pressured Apparatuses

310

rT h

¼

rth

þ rph



 1  lnðk=xÞ k2 þ x2 p k2  2 2 1þ 2 ¼ pt þ 2 ln k k 1 x k  x2 x

t p rT z ¼ rz þ rz ¼ pt


 1  2 lnðk=xÞ 2 p  2 þ 2 ln k k 1 k 1

ð8:15Þ

ð8:16Þ

T T where rT r , rh and rz are total radial stress, total circumferential stress, and total axial stress, respectively, rpr , rph and rpz are mechanical radial stress, mechanical circumferential stress, and mechanical axial stress, respectively, which is equation (1.1).
 1 2k 2 2k 2 p 2k 2 pt T T rh  rr ¼ pt  2 2 ð8:17Þ ¼ ðp  pt Þ þ þ ln k x k  x 2 x 2k2  x 2 x 2k 2  x 2 ln k




 lnðk=xÞ k2  x2 k2p  2 2 þ 2 2 2 ln k x k  x2 x k x

ð8:18Þ

 1  lnðk=xÞ k2 þ x2 k2p  2 2 þ 2 2 2 ln k x k  x2 x k x

ð8:19Þ

T rT h  rz ¼ p t


T rT z  rr ¼ p t

ðk 1Þ T According to equation (8.17), if p  k ln kk 2x p t , rT h  rr . When x = 1, ln k 2   2 2 p  1  kk2 ln1k 2 pt , where 1  kk2 ln1k 2 ¼ 1 monotonically increases as k increases[2]. 2

When

k = 1,

1  kk2 ln1k 2 ¼ 0; 2

when

2

2

2

1  kk2 ln1k 2 ¼ 1; 2

k = ∞,

when

k = 4,

T 1  kk2 ln1k 2  0:66. Therefore, in engineering there is generally rT h  rr . 2

þ ln xÞ T According to equation (8.18), if p  k ln kx ðkk 2 lnðk=xÞ pt , r T h  rz . When ln k TT x = 1, p ≥ 0. Therefore, in engineering, there is always rT h  rz . T Thus, rh is the most dangerous. On the outer surface, x = k, then,
 1 2 2p T  ¼ apt þ 2bp rT  r ¼ p ð8:20Þ þ 2 t ro ho ln k k 2  1 k 1 2

T rT ho  rzo ¼


T rT zo  rro ¼ pt

2

2

p ¼ bp k2  1

 1 2 p  2 ¼ apt þ bp þ 2 ln k k  1 k 1

ð8:21Þ

ð8:22Þ

k 1 where a ¼ ln1k  k 221 ¼ kðk1ln 2 1Þ ln k , b ¼ k 2 1. When k increases, a and b decrease simultaneously; when k decreases, a and b increase simultaneously[2]. When k → 1, a → 1, b → ∞; When k → ∞, a → 0, b → 0. a decreases gently. 2

2

Thermal Autofrettage Technology Based on Mises Yield Criterion

Setting

rTd eo

311

ffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h 2  2  2 i 1 T T T T T T ¼ 2 rho  rzo þ rzo  rro þ rro  rho  ry results in 3b2 p2 þ 3abpt p þ a 2 p2t  r2y  0

ð8:23Þ

If p = 0 or the pressure vessel is subjected only to thermal stresses, from equation (8.23), ry ðk 2  1Þ ln k ry ðk 2  1Þ ln k ¼ 2 ¼ 2 ry  pt  ry 2 k  1  ln k k  1  ln k 2 a a

ð8:24Þ

ðk 2  1Þ ln k 2ð1  lÞ ðk 2  1Þ ln k 2ð1  lÞ r ry  Dt  y k 2  1  ln k 2 Ea k 2  1  ln k 2 Ea

ð8:25Þ

 or 

2 2 2 2 2 Since ln k 2 [ 2 kk2 1 þ 1, it is easy to prove that k lnk − k + 1 > k − 1 − lnk . 2

1Þ ln k 2ð1lÞ Consequently, kðk2 1ln k 2 Ea ry [ Dtc . This is natural because when p = 0 and Δt < Δtc, the pressure vessel is of course safe. Finding out p from equation (8.23), we obtain qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3apt  12r2y  3a 2 p2t 3apt þ 12r2y  3a 2 p2t p ¼ p5 ð8:26Þ p7 ¼ 6b 6b 2

When

pt ry

1Þ ln k k ln k k þ 1 ptc 2 k 1  a2 ¼ ðk k 2 1ln k 2 ¼ 2 k 2 1ln k 2 ry , p5 is real. Since ln k [ 2 k 2 þ 1, it is 2

2

2

2

2

2

k k þ 1 also easy to prove that 2 k kln2 1ln k 2 [ 1. Therefore, as long as Δt ≤ Δtc, p5 is real. p7 is totally negative, therefore, it is not taken into account in engineering. As Δt or pt increases, p5 decreases. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 r 1Þ ln k 12r2y  3a 2 p2t  3apt or p5 ≥ 0. Since When pt  ay or pt  kðk2 1ln k 2 ry ¼ pt0 , 2

2

2

ln k 2 [ 2 kk2 1 þ 1, it is easy to prove that pt0 ≥ ptc. Therefore, as long as pt ≤ ptc, pt ≤ pt0, thereby p5 ≥ 0. Therefore, generally p5 ≥ 0. Finding out pt from equation (8.23), we obtain qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3bp þ 4r2y  3b2 p2 ¼ pt3 ð8:27Þ pt  2a 2

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  l  ¼ Dt3 or Dt  3bp þ 4r2y  3b2 p2 Eaa 2 3 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ffi Dt 4 p p 5 1  l ¼ Dt3  3b þ 4  3b2 or ry ry ry Eaa ry

ð8:28Þ

ð8:29Þ

312

Autofrettage Technology and Its Applications in Pressured Apparatuses

When p 2ðk 2  1Þ pffiffiffi ;  ry 3 Δt3 is a real number, and When

2ðk 2 1Þ pffiffi 3

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4r2y  3b2 p2  3bp, or Δt3 ≥ 0.

 2 rpye .

p k 2  1 pf  pffiffiffi ¼ ; ry ry 3

ffiffi ry . Hence, when p ≤ p5, there must be It is easy to prove that p5 \ kp1 3 pffiffiffi 2 2 kp 1 ffiffi ry [ pe ; when k  2, p  ffiffi3 ry . Therefore, generally Δt3 ≥ 0. Besides, kp1 3 pffiffiffi 2 kp 1 ffiffi ry  2pe . That is to say, when k  2, load-bearing capacity or allowable loading 3 cannot reach 2pe. In addition, as a or b increases, Δt3 decreases; as a or b decreases, Δt3 increases. Accordingly, as k increases, Δt3 increases; as k decreases, Δt3 decreases. As p increases, Δt3 decreases; as p decreases, Δt3 increases. When Δt3 = Δtc, p5 is called p5c. On the inner surface, x = 1, then,
 2k 2 1 2k 2 p T T rhi  rri ¼ pt 2  ¼ cpt þ 2dp ð8:30Þ þ 2 k 1 k  1 ln k 2

T rT hi  rzi ¼


rT zi



rT ri

¼ pt

k2p ¼ dp 1

ð8:31Þ

k2

 2k 2 1 k 2p  ¼ cpt þ dp þ k2  1 k 2  1 ln k r

ð8:32Þ

y 1 k ln k k þ 1 2 1 k where c ¼ k2k 2 1  ln k ¼ ðk 2 1Þ ln k ¼ k 2 1  ln k þ 2 ¼ 2  a ¼ ptc , d ¼ k 2 1 ¼ 1 þ b. 2

2

2

2

ðad þ bcÞ ¼

1 ln k

2

ð8:33Þ

When k → 1, a → 1, accordingly c → 1; When k → 1, b → ∞, accordingly d → ∞. When k → ∞, a → 0, accordingly c → 2; When k → ∞, b → 0, accordingly d → 1. a, b, c and d are plotted in figure 8.3. rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  T 2  T 2 i 1 h T Td rhi  rT Setting rei ¼ þ rzi  rT þ rri  rT  ry leads to zi ri hi 2 3d 2 p2  3cdpt p þ c2 p2t  r2y  0

ð8:34Þ

Thermal Autofrettage Technology Based on Mises Yield Criterion

313

FIG. 8.3 – The graph of a, b, c, and d.

If p = 0, or the pressure vessel is subjected only to thermal stresses, from equation (8.30), ry ry ð8:35Þ   pt  c c or Dtc  Dt ¼ Dtc

ð8:36Þ

This is natural because when p = 0 and Δt < Δtc, the pressure vessel is of course safe. Finding out p from equation (8.34), we obtain qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3cpt  12r2y  3c2 p2t 3cpt þ 12r2y  3c2 p2t p ¼ p6 ð8:37Þ p10 ¼ 6d 6d When pt ≤ ptc, or when Δt ≤ Δtc, p10 ≤ 0. Thus, p10 may not be taken into consideration in engineering. Then generally equation (3.31) becomes

p

3cpt þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12r2y  3c2 p2t 6d

¼ p6

ð8:38Þ

When pt 2 ðk 2  1Þ ln k 2 ptc ðk 2  1Þ ln k 2 2ð1  lÞry ¼ 2 ¼ 2Dtc ¼ Dtu ;  ¼ 2 ; or Dt  ry c k ln k 2  k 2 þ 1 ry k 2 ln k 2  k 2 þ 1 Ea

Autofrettage Technology and Its Applications in Pressured Apparatuses

314

p6 is real. Therefore, generally, p6 is real in engineering. When Δt = 2Δtc = Δtu, p6 k 2  1 pffiffiffi pe p6c ¼ ¼ 3 ¼ k2 ry ry ry When Δt = 0 p6 1 k 2  1 pe ps ¼ pffiffiffi ¼ pffiffiffi ¼ ¼ ry ry ry 3d 3k 2 When Δt = Δtc, p6 is also called p6c, because we will see that Δt = Δtc and pffiffiffi 2 ¼ 3 rpye ¼ pr6cy . Δt = 2Δtc both result in rp6y ¼ k k1 2 Finding out pt from equation (8.34), we obtain qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3dp  4r2y  3d 2 p2 3dp þ 4r2y  3d 2 p2 ¼ pt4  pt  ¼ pt0 ð8:39Þ 2c 2c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffii 1  l qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffii 1  l h h ¼ Dt4  Dt  3dp þ 4r2y  3d 2 p2 ¼ Dt0 or 3dp  4r2y  3d 2 p2 Eac Eac ð8:40Þ 2

3 2 3 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2 p p 5 1  l Dt4 Dt 4 p p 51  l ¼ or 43d  4  3d 2   3d þ 4  3d 2 ry ry Eac ry ry ry Eac ry ¼

Dt0 ry

ð8:41Þ

When p = 0, Dt4 ¼ Dtc and Dt0 ¼ Dtc : For Δt0, if we let Δt0 = Δtc, we obtain p = 0 and tuting

p ry

p ry

¼ k k1 into Δt0 obtains Δt0 = 2Δtc. 2

¼ k k1 ¼ 2 2

pffiffiffi pe 3 ry , and substi-

2

For Δt4, if we let Δt4 = Δtc, we obtain

p ry

¼ k k1 ¼ 2 2

pffiffiffi pe 3 ry , and substituting

k 1 k2 2

into Δt4 just obtains Δt0 = Δtc. When

k2  1 p  pffiffiffi ry ¼ pe ðthis is the initial mechanical yield pressureÞ; 3k 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4r2y  3d 2 p2  3dp, thereupon Δt4 ≥ 0.

p ry

¼

Thermal Autofrettage Technology Based on Mises Yield Criterion

315

When k2  1 p  2 pffiffiffi ry ¼ 2pe ¼ pu ðtwo times the initial mechanical yield pressureÞ; 3k 2 Δt4 and Δt0 are real numbers. With the
increase of p,  Δt4 increases; with the decrease of p, Δt4 decreases. dDt0 dp

dp pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 3d 2 2 2 2c 1 

2ð1lÞ Ea .

4ry 3d p

Letting

dDt0 dp

 0 obtains

p 1 k 2  1 pffiffiffi pe p6c  ¼ ¼ 3 ¼ ry d k2 ry ry pffiffiffi Therefore, if p\ 3pe , with the increase of p, Δt0 increases; with the decrease of pffiffiffi p, Δt0 decreases. If p [ 3pe , with the increase of p, Δt0 decreases; with the decrease of p, Δt0 increases.
 dp6 cp c t 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi , letting dp 2 2 2 dpt ¼ 2d 1  dpt  0 leads to 12ry 3c pt

pffiffiffi 2 pffiffiffi ry pffiffiffi 3ðk  1Þ ln k 2 ð1  lÞry pffiffiffi ¼ 3Dtc ¼ Dtm : pt  3 ¼ 3ptc or Dt  2 c k ln k 2  k 2 þ 1 Ea Therefore, generally speaking, as Δt or pt increases, p5 decreases and p6 increases; pffiffiffi as Δt or pt decreases, p5 increases and p6 decreases. When Dt ¼ 3Dtc ¼ Dtm , p6 reaches the maximum, p6m k2  1 pe pu ¼ 2 pffiffiffi ¼ 2 ¼ : 2 ry ry ry 3k In practical applications, for a certain k and Δt, we can take the minimum of p6 and p5 as allowable load; or for a certain load p and Δt, we can take the maximum of equations (8.26) and (8.38) as the radius ratio k. However, we can also seek a better way to determine operation parameters. For equations (8.26) and (8.38), setting p6 = p5 results in 3pt ðad þ bcÞ þ b

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12r2y  3c2 p2t  d 12r2y  3a 2 p2t ¼ 0

ð8:42Þ

The temperature difference to make p6 = p5 is called the optimal temperature difference, Δte. Δte can be obtained from equation (8.42). When Δt < Δte, p5 > p6; When Δt > Δte, p5 < p6. When Δt = Δte, p5 = p6, and p5 and p6 are written as pd. Δte must be smaller than Δtc. Since ðad þ bcÞ ¼ ln1k , then, equation (8.42) becomes 3pt = ln k þ b

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12r2y  3c2 p2t  d 12r2y  3a 2 p2t ¼ 0

ð8:43Þ

Autofrettage Technology and Its Applications in Pressured Apparatuses

316

For equation (8.43), setting pt = ptc leads to sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3ðk 2  1Þ 3 k2 ðk 2  1  ln k 2 Þ2 þ  ¼0 12  3 k 2 ln k 2  k 2 þ 1 k 2  1 k 2  1 ðk 2 ln k 2  k 2 þ 1Þ2

ð8:44Þ

From equation (8.44), we solve the boundary radius ratio ki = 2.0542956825…. When k ≤ ki, Δte ≤ Δtc; when k ≥ ki, Δte ≥ Δtc. ki is not related to material characteristics. When k = ki, Δte = Δtc = 169.95861 °C (elastic modules E = 1.95 × 105 MPa, Poisson ratio μ = 0.3, thermal expansion coefficient α = 1.2 × 10–5 °C−1, yield strength σy = 350 MPa). pffiffiffi 2 ¼ 3 rpye ¼ pr6cy ¼ pr5cy ; when In summary, when Δt = Δtc, if k = ki, rp5y ¼ rp6y ¼ k k1 2 k > ki, p5 > p6; when k < ki, p5 < p6. Δt should be smaller than Δtc. As Δt decreases, p6 decreases, and p5 increases. Therefore, only when k ≤ ki, equation (8.42) will be meaningful, or only when k ≤ ki, Δt can be reduced to obtain the optimal temperature difference Δte that can make p6 = p5, and Δte < Δtc. When k ≥ ki, in the range of Δt ≤ Δtc, there is always p6 ≤ p5, or there is no Δte that is smaller than Δtc and can make p6 = p5, in other words, it is impossible for p5 to be equal to p6. Therefore, when k ≥ ki, in the range of Δt ≤ Δtc, the allowable load is p = p6; if Δt = Δtc is taken, from equation (8.38), the allowable load is rffiffiffiffiffiffiffiffiffiffiffiffiffi p p6 k 2  1 pffiffiffi pe p6c ry ð8:45Þ ¼ ¼ ¼ 3 ¼ or k ¼ 2 ry ry k ry ry ry  p This load is just two times the initial mechanical yield pressure when based on the maximum shear stress theory (Tresca yield criterion). When k = ki, from equation (8.45), rpy ¼ rp6y = 0.763040517. That is to say, only when the load p p6 ry ¼ ry ≥ 0.763041, or k ≥ ki, equation (8.45) can be used to determine the radius ratio k. As long as Δt = Δtc, equation (8.38) becomes equation (8.45). When k ≤ ki, after solving Δte by equation (8.42) or equation (8.43), p6 or p5 (p6 = p5) can be found. As Δt or pt changes, the curves of rp5y  Δt and rp6y  Δt are bound to intersect. The intersection is the optimum design point. As mentioned above, as Δt or pt increases, p5 pffiffiffi decreases and p6 increases, and after Δt exceeds 3Dtc , p6 begins to decrease, until Δt = 2Δtc p6 has no definition. Clearly, the shapes and trends of the curves rp5y  Δt and p6 ry  Δt are affected by the value of k. Therefore, when k increases to a certain extent, the curves of rp5y  Δt and rp6y  Δt will have no intersection. As has been noted, when pffiffiffi k 2ffiffi1 ¼ 2 rpye . Substituting Dt ¼ 3Dtc ¼ Dtm , p6 reaches the maximum rp6y ¼ pr6my ¼ 2 p 3k 2 pffiffiffi k 2ffiffi1 obtains km = Dt ¼ 3Dtc ¼ Dtm into rp5y and setting rp5y ¼ pr6my ¼ 2 p 3k 2 3.181854, which is the maximum radius ratio with p5 = p6m = pu. When 2 ¼ pr6cy . Substituting Δt = 2Δtc = Δtu into rp5y and setting rp5y ¼ Δt = 2Δtc = Δtu, rp6y ¼ k k1 2 p6c ry

¼ k k1 obtains ku = 3.613643, which is the maximum radius ratio with 2 2

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317

p5 = p6c. Curves of rp5y  Δt and rp6y are tangent at Δt = Δtu. If k > ku, the curves of p5 p6 ry  Δt and ry  Δt will have no intersection. However, this doesn’t have an impact on design. In order to confirm the above analysis intuitively, we demonstrate the main results of the analysis by the graphic method. For Elastic modulus E = 1.95 × 105 MPa, Poisson ratio μ = 0.3, thermal expansion coefficient of the materials α = 1.2 × 10–5 °C−1, strength limit σy = 350 MPa and radius ratio k = 1.5, k = 2, k = ki, k = kc, k = km, k = ku, k = 3 and k = 4, rp5y and rp6y are plotted in figures 8.4– 8.11, respectively. For other parameters, the changing trends and morphologies of rp5y and rp6y are similar to those of figures 8.4–8.11, because for steel materials, the change of

Ea 2ð1lÞ

is small, popularly

Ea 2ð1lÞ≈1.5  1.8.

For example, here

Ea 2ð1lÞ

FIG. 8.4 – The graphs of p5/σy and p6/σy for k = 1.5.

= 1.671429….

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Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 8.5 – The graphs of p5/σy and p6/σy for k = 2. The following figures can also be used as a reference for engineering design. In the following figures, the above are partial figures (figure (a)) and the below are corresponding panoramic views (figure (b)). With the increase of k, the difference p5−p6 gets greater and greater. In addition, 2 2 p10 → pr6cy ¼ k k1 when Δt → Δtu = 2Δtc (accordingly, pt → 2pc), rp6y → pr6cy ¼ k k1 2 , r 2 . y Before plotting Δt3, Δt4, and Δt0, we analyze them as follows, 0 1   Dt3 ry

p

b ry C b1  lB B1 þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C\0:   ¼ 3 @   p 2A a Ea d ry 4  3b2 rpy

d

Thermal Autofrettage Technology Based on Mises Yield Criterion

319

FIG. 8.6 – The graphs of p5/σy and p6/σy for k = ki.

It is thus evident that Δt3 monotonically decreases as p increases. In order to show the changing trend of Δt3, the correlation between Δt3 and rpy is shown in figures 8.12–8.14. Figure 8.12 is for k = 1.5 < ki, figure 8.13 is for k = ki and figure 8.14 is for k = 4 >k i. They all confirm the above analysis. Since the d Dt3 ry greater k is, the smaller   is, consequently, the greater k is, the slower Δt3 d rpy decreases.

320

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 8.7 – The graphs of p5/σy and p6/σy for k = kc. Clearly, Δt4 < 0 Δt0. 1   Dt4 p d ry B C 3d 2 ry   ¼ 1l B3d þ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiC   Eac @ A [ 0. Therefore, Δt4 monotonically increases as 2 d rpy p 2 43d ry

k ffiffi1 p increases, until p ¼ pu ¼ 2pe ¼ 2 p r , it begins to be imaginary number. When 3k 2 y p = p u, pffiffiffi 2 pffiffiffi 3ðk  1Þ ln k 2ð1  lÞry : Dt4 ¼ 3Dtc ¼ Dtm ¼ 2 k ln k 2  k 2 þ 1 Ea 2

Thermal Autofrettage Technology Based on Mises Yield Criterion

FIG. 8.8 – The graphs of p5/σy and p6/σy for k = 3. 0 1     Dt0 Dt0 p 2 d ry d B C 3d ry C. Setting ry   0 gives   ¼ 1l B3d  rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi  2 A Eac @ d rpy d rpy 43d 2 p ry

p k 2  1 pffiffiffi pe p6c  ¼ 3 ¼ ry k2 ry ry Therefore, when p = p6c, Δt0 gets the maximum, which is Dt0 ¼ 2Dtc ¼ Dtu ¼

ðk 2  1Þ ln k 2 2ð1  lÞry k 2 ln k 2  k 2 þ 1 Ea

321

322

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 8.9 – The graphs of p5/σy and p6/σy for k = km. After p = p6c and Δt0 = 2Δtc = Δtu, Δt0 decreases as p increases, until k 2ffiffi1 r , then, it begins to be an imaginary number. When p = pu, p ¼ pu ¼ 2pe ¼ 2 p 3k 2 ypffiffi pffiffiffi 2 ln k 2ð1lÞry Dt0 ¼ 3Dtc ¼ Dtm = k 2 3lnðkk 21Þ Ea . k 2 þ 1 pffiffiffi It is shown that when p = pu, Δt0 = Δt4 = 3Dtc ¼ Dtm . In consideration of that Δt0 > Δt4, thus, curves Δt0  rpy and Δt4  rpy are actually the same curve within 0 ≤ p ≤ p u. When rpy = 0, Dt4 ¼ Dtc and Dt0 ¼ Dt4 ¼ Dtc :

Thermal Autofrettage Technology Based on Mises Yield Criterion

323

FIG. 8.10 – The graphs of p5/σy and p6/σy for k = ku. As rpy changes from 0 to rpuy , Δt4 monotonically increases from Δt4 = −Δtc to pffiffiffi Dt ¼ 3Dtc ¼ Dtm . As rpy changes from 0 to rpuy , Δt0 (>Δt4) monotonically increases from Δt0 = Δtc until rpy = pr6cy , where it reaches the maximum Δt0 = 2Δtc (=Δtu), and then Δt0 (>Δt4) start to decrease monotonically from Δt0 = 2Δtc (where rpy = pr6cy ) pffiffiffi until Dt0 ¼ Dt4 ¼ 3Dtc ¼ Dtm (where rpy = rpuy ). Within 0 ≤ p ≤ pu, there always is that Δt0 > Δtc. For example, when k = 2 < ki, k = ki and k = 4 > ki, Δt0  rpy and Δt4  rpy are illustrated in figures 8.15–8.17, respectively. They all confirm the above analysis.

324

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 8.11 – The graphs of p5/σy and p6/σy for k = 4. Ordinarily, curves Δt0  rpy and Δt3  rpy have a crossover point A and curves Δt4  rpy and Δt3  rpy have a crossover point B. Setting Δt0 = Δt3 results in 0 1 0 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2ffi p p 1 p p @3d A ¼ @3b þ 4  3b2 A1 ð8:46Þ þ 4  3d 2 ry ry c ry ry a Using the approximation method, for a certain k, we can find out the coordinates of intersection point A, which is A (rpy , Δt0 = Δt3). Setting Δt4 = Δt3 results in

Thermal Autofrettage Technology Based on Mises Yield Criterion

FIG. 8.12 – The correlation between Δt3 and p/σy for k = 1.5 < ki.

FIG. 8.13 – The correlation between Δt3 and p/σy for k = ki.

FIG. 8.14 – The correlation between Δt3 and p/σy for k = 4 > ki.

325

326

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 8.15 – The graph of Δt0  rpy and Δt4  p/σy with k = 2 < ki.

FIG. 8.16 – The graph of Δt0  p/σy and Δt4  p/σy with k = ki.

FIG. 8.17 – The graph of Δt0  p/σy and Δt4  p/σy with k = 4 > ki.

Thermal Autofrettage Technology Based on Mises Yield Criterion 0 @3d p  ry

1 0 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2ffi p 1 p p A ¼ @3b þ 4  3b2 A1 4  3d 2 ry c ry ry a

327

ð8:47Þ

Similarly, using the approximation method, for a certain k, we can find out the coordinates of intersection point B, which is B (rpy , Δt4 = Δt3). Obviously, the position of the intersection depends on the value of the radius ratio k. In view of the shape of the curves Δt0  rpy , Δt3  rpy and Δt4  rpy as shown typically in figures 8.12–8.17, when curve Δt3  rpy and curve Δt0  rpy or curve Δt4  rpy are tangent, the intersection is about to disappear. The point of tangency must be the highest point of curves Δt0  rpy in the light of the shapes of the curves Δt0  rpy and Δt3  rpy as shown typically in figures 8.12–8.17. Therefore, the point of tangency of the curves Δt0  rpy and Δt3  rpy can be obtained as follows. Substituting 2Δtc to obtain

p ry

¼ pr6cy ¼ k k1 into equation (8.29) and then setting Δt = Δtu = 2 2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4k 4  3  3 4 ¼ k 2 ðk 2  1  ln k 2 Þ k 2 ln k 2  k 2 þ 1

ð8:48Þ

The solution of equation (8.38) is ku ¼ 3:613643 It is the same as above. In consideration of the shapes of the curves Δt0  rpy , Δt4  rpy and Δt3  rpy , intersection point A is certainly above Δtc because Δt0 > Δtc. As for intersection point B, its position depends on the value of radius ratio k. As mentioned above, when Δt = Δtc and p5 = p6, there is k = ki = 2.0542956825…. Therefore, when k = ki, the coordinates of intersection point B is B (rp5y ¼ rp6y , Δt4 = Δt3 = Δtc). When k < ki, the ordinate of intersection point B is Δt4 = Δt3 < Δtc; When k > ki, the ordinate of intersection point B is Δt4 = Δt3 > Δtc. Now, we can plot curves Δt0, Δt3, and Δt4 to intuitively observe their distributions as a function of rpy . The following figures may serve as a reference for engineering design. For the above parameters of material characteristics, or Elastic modulus E = 1.95 × 105 MPa, Poisson ratio μ = 0.3, thermal expansion coefficient of the materials α = 1.2 × 10–5 °C−1 and strength limit σy = 350 MPa and for radius ratio k = 1.5, k = 2, k = ki, k = kc, k = 3, k = ku and k = 4, Δt3  rpy , Δt0  rpy and Δt4  rpy are plotted in figures 8.18–8.24, respectively. For other parameters, the changing trends and morphologies of Δt3  rpy , Δt0  rpy and Δt4  rpy are similar to those of figures Ea is small for steel materials. 8.18–8.24, the reason is also that the change of 2ð1lÞ In fact, if the project application is not considered and only the mathematical relationship is concerned within the definition domain and the co-domain, the

328

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 8.18 – The graphs of Δt3  p/σy, Δt0  p/σy and Δt4  p/σy for k = 1.5.

FIG. 8.19 – The graphs of Δt3  p/σy, Δt0  p/σy and Δt4  p/σy for k = 2.

FIG. 8.20 – The graphs of Δt3  p/σy, Δt0  p/σy and Δt4  p/σy for k = ki. graphs of Δt0  rpy plus Δt4  rpy and rp6y  Δt plus pr10y  Δt are complete ellipses, because equation (8.40) plus equation (8.41) is directly a complete elliptic equation. Figures 8.25 and 8.26 are the graphs of Δt0  rpy plus Δt4  rpy for k = 2 and k = 4,

Thermal Autofrettage Technology Based on Mises Yield Criterion

329

FIG. 8.21 – The graphs of Δt3  p/σy, Δt0  p/σy and Δt4  p/σy for k = kc.

FIG. 8.22 – The graphs of Δt3  p/σy, Δt0  p/σy and Δt4  p/σy for k = 3.

FIG. 8.23 – The graphs of Δt3  p/σy, Δt0  p/σy and Δt4  p/σy for k = ku. respectively. Figures 8.27 and 8.28 are the graphs of rp6y  Δt plus pr10y  Δt for k = 2 and k = 4, respectively. The left side of the vertical axis and the lower side of the lateral axis correspond to “outer heating”.

330

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 8.24 – The graphs of Δt3  p/σy, Δt0  p/σy and Δt4  p/σy for k = 4.

FIG. 8.25 – The graphs of Δt0  p/σy and Δt4  p/σy for k = 2.

FIG. 8.26 – The graphs of Δt0  p/σy and Δt4  p/σy for k = 4.

Thermal Autofrettage Technology Based on Mises Yield Criterion

331

FIG. 8.27 – The graphs of p6/σy  Δt plus p10/σy  Δt for k = 2.

FIG. 8.28 – The graphs of p6/σy  Δt plus p10/σy  Δt for k = 4. For figures 8.25 and 8.26, the definition domain of rpy is rpy =

pu pe ry ¼ 2 ry ðk 2 1Þ ln k 2 2ð1lÞry Ea . k 2 ln k 2 k 2 þ 1

k ffiffi1 ¼ 2p , and 3k 2 2

the codomain of Δt0 (Δt4) is Δt = Δtu = 2Δtc = For figures 8.27 and 8.28, the definition domain of Δt0 (Δt4) is Δt = Δtu = 2Δ 2 ln k 2 2ð1lÞry p p pu pe k 2ffiffi1 p . tc = k 2ðkln1Þ 2 Ea , and the codomain of r is r = r ¼ 2 r ¼ 2 k 2 k 2 þ 1 y

8.3

y

y

y

3k

Examples

Example 1. Elastic modulus E = 1.95 × 105 MPa, Poisson’s ratio μ = 0.3, thermal expansion coefficient α = 1.2 × 10–5 °C−1, strength limit σy = 300 MPa, k = 3 > ki. If the temperature difference is not exerted, the load-bearing capacity of the k 2ffiffi1 r = 153.9601 MPa. pressure vessel with k = 3 is only p ¼ pe ¼ p 3k 2 y

Autofrettage Technology and Its Applications in Pressured Apparatuses

332

From equation (8.12), Δtc = 133.9696 °C; when Δt = Δtc = 133.9696 °C, from equation (8.38) or equation (8.45), p = p6 = 266.6667 MPa; from equation (8.26), p5 = 751.5536 MPa > p6, then, take p6 = 266.6667 MPa as the load; from equation (8.27), pt3 = 376.5421 MPa (p = p6); from equation (8.39), pt4 = 223.9206 MPa (p = p6 and Δt4 = Δtc), pt0 = 447.8411 MPa (p = p6 and Δt4 = Δtc). From equation (8.42), Δte = 218.1078 °C > Δtc. Therefore, for a pressure vessel with k = 3, the optimum operation parameters are p = p6 = 266.6667 MPa, Δt = Δtc = 133.9696 °C, under these parameters, the total stresses along with rTd e and rTs e are shown in figure 8.29. The meaning of each curve and the main parameters are marked in the figure.

FIG. 8.29 – The distribution of total stresses with k = 3, Δt = Δtc and p = p6. If based on the maximum shear stress theory (Tresca yield criterion), the optimum operation parameters are the same. Example 2. E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 300 MPa, k = 1.6 < ki. If the temperature difference is not exerted, the load-bearing capacity of the k 2ffiffi1 r = 105.5468 MPa. pressure vessel with k = 1.6 is p ¼ pe ¼ p 3k 2 y From equation (8.12), Δtc = 155.4798 °C; From equation (8.42), Δte = 96.31247 °C < Δtc. Substituting Δt = Δte = 96.31247 °C into equation (8.26) or equation (8.37) obtains p = p5 = p6 = 156.9785 MPa, σylnk = 141.0011 MPa < p < p2ffiffi3 σylnk = 162.814 MPa. When Δt = Δte, pt = 160.9794 MPa. Then, from equations (8.20) to (8.22), T T T T d  rT rT ho = 337.38 MPa; rho  rzo = 100.6ffi MPa; rzo  rro = 236.8 MPa; reo ¼ qroffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 1 2 ½ðrho  rzo Þ þ ðrzo  rro Þ þ ðrro  rho Þ  = 300 MPa. From equations (8.30) to T T T rT MPa; rT rT ri  rhi = 329.4 MPa; zi  rri = 71.8 MPa; hi  rzi = 257.6 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi T 2 T T 2 T T 2 ¼ 12 ½ðrT hi  rzi Þ þ ðrzi  rri Þ þ ðrri  rhi Þ  = 300 MPa.

(8.32), rTd ei

Thermal Autofrettage Technology Based on Mises Yield Criterion

333

From equations (8.28) and (8.40), Δt3 = Δt4 = 96.31247 °C = Δte, Δt0 = 304.2126 °C. If based on the maximum shear stress theory (Tresca criterion), this is the case of k < kc, from equation (8.40) of the last chapter, Δtx = 84.35963 °C < Δtc = 155.4798 °C obtained from equation (8.27). When Δt = Δtx = 84.35963 °C, equations (8.44) and (8.45) both lead to rpy = 0.47004 = lnk = ln1.6, or p = 141.0011 MPa < p5 or p6. Substituting Δtc = 155.4798 °C into equation (8.26) obtains p5 = 80.0235 MPa; substituting Δtc = 155.4798 °C into equation (8.38) obtains p6 = 182.8125 MPa. Substituting p5 = 80.0235 MPa into equation (8.28) obtains Δt3 = Δtc; Substituting p6 = 182.8125 MPa into equation (8.40) obtains Δt4 = Δtc and Δt0 = 310.9597 °C. When Δt = Δte = 96.31247 °C, p = p5 = p6 = 156.9785 MPa, the total stresses Ts along with rTd e and re are shown in figure 8.30. The meaning of each curve and the main parameters are marked in the figure. According to the maximum shear stress theory (Tresca criterion), rTs e > 1.

FIG. 8.30 – The distribution of total stresses with k = 1.6, Δt = Δte and p = p5 = p6. When Δt = Δtc = 155.4798 °C, p = p5 = 80.0235 MPa, the total stresses along Ts with rTd e and re are shown in figure 8.31. The meaning of each curve and the main parameters are marked in the figure. When p and k are given, the key is to determine the appropriate temperature difference Δt among Δt0, Δt4, Δt3 and Δtc by equations (8.28), (8.40), and (8.12). It is better to select a smaller temperature difference among Δt0, Δt4, Δt3, and Δtc. See example 2. From figures 8.30 and 8.31, it is seen that rT ho [ ry . This is because the pressure is too great. If it is necessary, the pressure should be controlled to control rT ho . From T increases, or r is the maximum equation (8.15), we know that with x increasing, rT h ho along x.

334

Autofrettage Technology and Its Applications in Pressured Apparatuses

FIG. 8.31 – The distribution of total stresses with k = 1.6, Δt = Δtc and p = p5 Δtc; When is Δt fixed, if p2 = p5, usually, k < 1. Therefore, usually, p2 is always greater than p5. When Δt = Δte = 96.31247 °C, p = p2 = 127.8242 MPa, the total stresses along Ts with rTd e and re are shown in figure 8.32. The meaning of each curve and the main parameters are marked in the figure. rT ho  ry .

FIG. 8.32 – The distribution of total stresses with k = 1.6, Δt = Δte and p = p2.

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335

When Δt = Δtc = 155.4798 °C, p = p2 = 62.59748 MPa, the total stresses along Ts with rTd e and re are shown in figure 8.33. The meaning of each curve and the main parameters are marked in the figure. rT ho  ry .

FIG. 8.33 – The distribution of total stresses with k = 1.6, Δt = Δtc and p = p2. Example 3. E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 300 MPa, k = 1.6 < ki, p = 130 MPa. From equation (8.12), Δtc = 155.4798 °C; from equation (8.40), Δt0 = 288.3432 °C and Δt4 = 43.3471 °C; from equation (8.28), Δt3 = 117.5862 °C. Therefore, Δt should be selected between Δt4 and Δt3, or Δt4 ≤ Δt ≤ Δt3. Now we Ts take Δt = 70 °C, the total stresses along with rTd e and re are shown in figure 8.34, where the meaning of each curve and the main parameters are marked. According to the maximum shear stress theory (Tresca criterion), rTs e < 1.

FIG. 8.34 – The distribution of total stresses with k = 1.6, Δt = 70 °C < Δt3 < Δtc, p = 130 MPa and k = 1.6.

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Autofrettage Technology and Its Applications in Pressured Apparatuses

If Δt = 119 °C > Δt3, the total stresses along with rTd and rTs e e are shown in figure 8.35, in which the meaning of each curve and the main parameters are marked. Because Δt > Δt3, rTd eo > 1.

FIG. 8.35 – The distribution of total stresses with k = 1.6, Δt3Δt < Δtc and p = 130 MPa. When Δt = 119 °C, p = p2 = 102.8132 MPa, the total stresses along with rTd e and rTs are shown in figure 8.36. The meaning of each curve and the main e parameters are marked in the figure. rT ho  ry .

FIG. 8.36 – The distribution of total stresses with k = 1.6, Δt = 119 °C and p = p2. Example 4. E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 300 MPa, k = 2.1 > ki, p = 220 MPa. From equation (8.12), Δtc = 144.9011 °C; from equation (8.40), Δt0 = 288.7934 °C and Δt4 = 123.4736 °C; from equation (8.28), Δt3 = 155.5836 °C.

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337

Therefore, Δt should be selected between Δt4 and Δtc, or Δt4 ≤ Δt ≤ Δtc. Now we take Δt = 130 °C, the total stresses along with rTd and rTs e e are shown in figure 8.37, where the meaning of each curve and the main parameters are marked. According to the maximum shear stress theory (Tresca criterion), rTs e < 1.

FIG. 8.37 – The distribution of total stresses with k = 2.1, Δt4 < Δt < Δtc and p = 220 MPa. If based on the maximum shear stress theory (Tresca criterion), this is the case of k < kc, from equation (8.40) of last chapter, Δtx = 133.1682 °C < Δtc = 144.9011 °C obtained from equation (8.27) of the last chapter. When Δt = Δtx = 133.1682 °C, equations (8.44) and (8.45) both lead to rpy = 0.741937 = lnk = ln2.1, or p = 222.5812 MPa > 220 MPa. When Δt = Δtx = 133.1682 °C, p = 222.5812 MPa, the total stresses along with rTd and rTs e e are shown in figure 8.38, in which the meaning of each curve and the main parameters are marked. In this case, rTs e ≡ 1.

FIG. 8.38 – The distribution of total stresses with k = 2.1, Δt = Δtx and p = 222.5812 MPa.

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Autofrettage Technology and Its Applications in Pressured Apparatuses

If p and Δt are given, the key is to determine the appropriate radius ratio k. Set p5 = p to solve k by numerical approximation method from equation (8.26) and set p6 = p to solve k by numerical approximation method from equation (8.38), taking the greater k as real radius ratio. Another way: If the solved k is greater than ki and Δt is not limited or Δt > Δtc, solve k from equation (8.45) and take Δtc as the real temperature difference. See example 5. Example 5. E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 300 MPa, p = 130 Mpa, Δt = 150 °C. Setting p5 = p = 130 Mpa, from equation (8.26), k = 1.770509 < ki; Setting p6 = p = 130 Mpa, from equation (8.38), k = 1.343435 < ki. Take k = 1.770509 as the real radius ratio. When k = 1.770509, Δtc = 151.2867 °C, Δt0 = 270.6439 °C, Δt4 = 18.16004 °C, Δt3 = Δt = 150 °C, p5 = p = 130 MPa, p6 = 203.7163 MPa. When p = 130 Mpa, Δt = 150 °C, and k = 1.770509, the total stresses along Ts with rTd e and re are shown in figure 8.39. The meaning of each curve and the main parameters are marked in the figure.

FIG. 8.39 – The distribution of total stresses with k = 1.770509, Δt = 150 °C and p = 130 MPa.

When Δt = 150 °C < Δtc, k = 1.770509, p = p2 = 102.4869 MPa, the total Ts stresses along with rTd e and re are shown in figure 8.40. The meaning of each curve and the main parameters are marked in the figure. rT ho  ry . If based on the maximum shear stress theory (Tresca criterion), set p1 = p to solve k by numerical approximation method from equation (8.42) of the last chapter and set p2 = p to solve k by numerical approximation method from equation (8.44) of last chapter, take the greater k as real radius ratio.

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339

FIG. 8.40 – The distribution of total stresses with k = 1.770509, Δt = 150 °C and p = p2. Another way: If the solved k is greater than kc and Δt is not limited or Δt > Δtc, qffiffiffiffiffiffiffiffi ry solve k from k ¼ ry p , which is just equation (8.45), take Δtc as real temperature difference. If the solved k is smaller than kc and Δt is not limited or Δt > Δtx, solve k from rpy ¼ ln k and take Δtx as the real temperature difference. Setting p1 = p = 130 MPa, we obtain k = 1.350423; Setting p2 = p = 130 MPa, we obtain k = 1.889912 < kc. Take k = 1.889912 as the real radius ratio. When k = 1.889912, Δtc = 148.7473358 °C obtained from equation (8.27) of the last chapter or equation (8.9) of this chapter. Δt = 150 °C is a little bigger than Δtc. When p = 130 Mpa, Δt = 150 °C and k = 1.889912, the total stresses along with rTd and rTs e e are shown in figure 8.41. The meaning of each curve and the main parameters are marked in the figure.

FIG. 8.41 – The distribution of total stresses with k = 1.889912, Δt = 150 °C and p = 130 MPa.

340

Autofrettage Technology and Its Applications in Pressured Apparatuses

In view of k = 1.889912 < kc, from rpy ¼ ln k, we obtain k = 1.54239. When k = 1.54239, Δtc = 157.077292 °C obtained from equation (8.27) of the last chapter or equation (8.9) of this chapter; from equation (8.40) of the last chapter, Δtx = 77.77775 °C < Δtc. When p = 130 Mpa, Δt = Δtx = 77.77775 °C and k = 1.54239, Ts the total stresses along with rTd e and re are shown in figure 8.42. The meaning of each curve and the main parameters are marked in the figure.

FIG. 8.42 – The distribution of total stresses with Δt = Δtx and p = σylnk. Example 6. E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 300 MPa, p = 250 Mpa, Δt = 110 °C. Setting p5 = p = 250 Mpa, from equation (8.26), k = 1.903431 < ki; Setting p6 = p = 250 Mpa, from equation (8.38), k = 3.005172 > ki. Take k = 3.005172 as the real radius ratio. When k = 3.005172, Δtc = 133.9237816 °C, Δt0 = 266.4988 °C, Δt4 = 110 °C = Δt, Δt3 = 228.5979 °C, p5 = p = 130 MPa, p6 = 203.7163 MPa. When p = 250 Mpa, Δt = 110 °C, and k = 3.005172, the total stresses along Ts with rTd e and re are shown in figure 8.43. The meaning of each curve and the main parameters are marked in the figure. When p = 250 MPa, Δt = Δtc is allowed, from equation (8.45), we obtain k = 2.44949 > ki. When k = 2.44949, Δtc = 139.8116 °C, the total stresses along Ts with rTd e and re are shown in figure 8.44, where the meaning of each curve and the main parameters are marked in the figure. This plan is better. If based on the maximum shear stress theory (Tresca criterion), setting p1 = p = 250 MPa, we obtain k = 3.292849 > kc by numerical approximation method from equation (8.42) of the last chapter; Setting p2 = p = 250 MPa, we obtain k = 2.038807 by numerical approximation method from equation (8.44) of last chapter. Take k = 3.292849 as the real radius ratio. When k = 3.292849, Δtc = 131.5798562 °C obtained from equation (8.27) of the last chapter or equation (8.9) of this chapter. Δt = 110 °C is smaller than Δtc. When p = 250 Mpa, Δt = 110 °C

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341

FIG. 8.43 – The distribution of total stresses with k = 3.005172, Δt = 110 °C and p = 250 MPa.

FIG. 8.44 – The distribution of total stresses with k = 2.44949, Δt = Δtc and p = 250 MPa. Ts and k = 3.292849, the total stresses along with rTd e and re are shown in figure 8.45. The meaning of each curve and the main parameters are marked in the figure. qffiffiffiffiffiffiffiffi ry , we obtain the same results as In view of k = 3.292849 > kc, from k ¼ ry p

figure 8.45. Example 7. A process can supply a temperature difference Δt = 120 °C for a pressure vessel, study the size and the load-bearing capacity of the pressure vessel, E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 350 MPa. (1) If the load-bearing capacity is not required, there are many schemes. Such as,

Autofrettage Technology and Its Applications in Pressured Apparatuses

342

FIG. 8.45 – The distribution of total stresses with k = 3.292849, Δt = 110 °C and p = 250 MPa.

I. Taking k = 2.5, from equation (8.12), Δtc = 162.3777958 > Δt, from equation (8.38) rp6y = 0.761038552, from equation (8.26) rp5y = 1.898979. Take p p6 ry = ry = 0.761038552 as are shown in figure rTs e

real loading. The total stresses along with rTd e and

8.46. The meaning of each curve and the main parameters are marked in the figure.

FIG. 8.46 – The distribution of total stresses with k = 2.5, Δt = 120 °C and

p ry

=

p6 ry

= 0.761038552.

II. Taking k = 1.6, from equation (8.12), Δtc = 181.3931336 > Δt, from equation (8.38) rp6y = 0.533583765, from equation (8.26) rp5y = 0.495861. Take

Thermal Autofrettage Technology Based on Mises Yield Criterion

343

p ry

Ts = 0.495861 as real loading. The total stresses along with rTd e and re are shown in figure 8.47. The meaning of each curve and the main parameters are marked in the figure.

FIG. 8.47 – The distribution of total stresses with k = 1.6, Δt = 120 °C and

p ry

=

p5 ry

= 0.495861.

Such instances are legion. When Δt = 120 °C < Δtc, k = 1.6, p = p2 = 140.7108 MPa, the total stresses Ts along with rTd e and re are shown in figure 8.48. The meaning of each curve and the main parameters are marked in the figure. rT ho  ry .

FIG. 8.48 – The distribution of total stresses with k = 1.6, Δt = 120 °C and p = p2. (2) Supposing Δt = 120 °C = Δte, then, from equation (8.43) by approximation, k = 1.653158 < kc. When k = 1.653158, Δtc = 179.7739059 > Δt. When

344

Autofrettage Technology and Its Applications in Pressured Apparatuses

k = 1.653158, Δt = 120 °C = Δte, from equations (8.26) and (8.38), we obtain p p5 p6 p2ffiffi ry = ry = ry = 0.556731958 < 3 ln k = 0.580453. (3) Supposing Δt = 120 °C = Δtc, then, from equation (8.27) k = 49.872, it is unrealistic, rejected. When E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 350 MPa and k = kc, Δtc = 166.8538 °C. The greater the k is, the greater the Δtc is. If Δtc = 120 °C, k is of course quite great. When E = 1.95 × 105 MPa, μ = 0.3, α = 1.2 × 10–5 °C−1, σy = 300 MPa and k = kc, Δtc = 143.0175618 °C. (4) I. Supposing the pressure vessel is required to contain 200 MPa, or p 200 ry ¼ 350 = 0.571429. From equation (8.26) by approximation, k = 1.665644; from equation (8.38) by approximation, k = 1.689516. Take k = 1.689516. When k = 1.689516, Δtc = 178.7176168 °C > Δt. When p = 200 MPa, Ts Δt = 120 °C and k = 1.689516, the total stresses along with rTd e and re are shown in figure 8.49. The meaning of each curve and the main parameters are marked in the figure.

FIG. 8.49 – The distribution of total stresses with k = 1.689516, Δt = 120 °C and p = 200 MPa.

When Δt = 120 °C < Δtc, k = 1.689516 p = p2 = 170.4853 MPa, the total Ts stresses along with rTd e and re are shown in figure 8.50. The meaning of each curve and the main parameters are marked in the figure. rT ho  ry . II. Supposing the pressure vessel is required to contain 300 MPa, or p 300 ry ¼ 350 = 0.857143. From equation (8.26) by approximation, k = 1.88692; from equation (8.38) by approximation, k = 3.629574. Take k = 3.629574. When k = 3.629574, Δtc = 150.804797 °C > Δt. When p = 300 Mpa, Ts Δt = 120 °C and k = 3.629574, the total stresses along with rTd e and re are shown in figure 8.51. The meaning of each curve and the main parameters are marked in the figure.

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345

FIG. 8.50 – The distribution of total stresses with k = 1.689516, Δt = 120 °C and p = p2 = 170.4853 MPa.

FIG. 8.51 – The distribution of total stresses with k = 3.629574, Δt = 120 °C and p = 300 MPa.

8.4

Chapter Summary

The main equations and conclusion are listed in table 8.1.

346

Autofrettage Technology and Its Applications in Pressured Apparatuses TAB. 8.1 – The main equations and conclusion of this chapter.

Thermal stresses




 lnðk=xÞ k2  x2 þ 2 2 ¼ pt  ln k x k  x2
 1  lnðk=xÞ k2 þ x2  2 2 rth ¼ pt ln k x k  x2
 1  2 lnðk=xÞ 2  2 rtz ¼ pt ln k k 1 EaDt pt ¼ 2ð1  lÞ The thermal stresses at the inside and outside walls rtri ¼ rtro ¼ 0
 2k 2 1  rthi ¼ rtzi ¼ pt 2 k  1 ln k
 1 2 t t rho ¼ rzo ¼ pt  2 ln k k  1 The equivalent thermal stress at the inside wall
 2k 2 1 k 2 ln k 2  k 2 þ 1 ts t t t t ¼ pt  rei ¼ rri  rzi ¼ rzi ¼ rhi ¼ pt 2 ðk 2  1Þ ln k k  1 ln k rtr

Critical temperature difference, Dtc Dtc ¼

ðk 2  1Þ ln k 2ð1  lÞry ln k 2  k 2 þ 1 Ea

k2

Critical thermal loading ptc ptc ¼

ðk 2  1Þ ln k ry k 2 ln k 2  k 2 þ 1

Total stresses (thermal stresses plus operation stresses)

 lnðk=xÞ k2  x2 p k2 t p þ 1  þ ¼ r þ r ¼ p  rT t r r r ln k k2  1 x2k2  x2 x2

 2 2 1  lnðk=xÞ k þx p k2 p t þ 1 þ ¼ r þ r ¼ p rT  t h h h x2k2  x2 x2 ln k k2  1
 1  2 lnðk=xÞ 2 p t p  2 þ 2 rT z ¼ rz þ rz ¼ pt ln k k 1 k 1 On the outer surface, x = k, then,
 1 2 2p T T þ 2 ¼ apt þ 2bp rro  rho ¼ pt  ln k k 2  1 k 1 p T ¼ bp rT ho  rzo ¼ 2 k 
1 1 2 p T þ 2 ¼ apt þ bp  rT zo  rro ¼ pt ln k k 2  1 k 1 2 2 1 2 k  1  ln k 1  ¼ ,b¼ 2 where a ¼ ln k k 2  1 k 1 ðk 2  1Þ ln k On the inner surface, x = 1, then,
 2k 2 1 2k 2 p T þ 2  ¼ cpt þ 2dp  r ¼ p rT t ri hi k 1 k 2  1 ln k

Thermal Autofrettage Technology Based on Mises Yield Criterion

347

TAB. 8.1 – (continued). k2p T ¼ dp rT hi  rzi ¼ 2  k
1 2 2k 1 k2p T þ 2  ¼ cpt þ dp rT zi  rri ¼ pt 2 k 1 k  1 ln k 2 2 2 2 2k 1 k ln k  k þ 1 2 1 k2 ¼ þ 2 ¼ 2  a, d ¼  ¼  ¼ 1þb where c ¼ 2 ðk 2  1Þ ln k k 2  1 ln k k  1 ln k k2  1 For rTd eo ¼

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2  T   T  i 1 h T T 2 T 2 rho  rT  ry þ r  r þ r  r zo zo ro ro ho 2 3b2 p2 þ 3abpt p þ a 2 p2t  r2y  0

or p

3apt þ

6b

or pt 

3bp þ

3bp þ

¼ p5

¼ pt3

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4r2y  3b2 p2 2ð1  lÞ

¼ Dt3 Dt  2a Ea rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  T 2  T 2 i 1 h T rhi  rT ¼ þ rzi  rT þ rri  rT  ry zi ri hi 2 3d 2 p2  3cdpt p þ c2 p2t  r2y  0

or p10 ¼

3cpt 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12r2y  3c2 p2t 6d

or p or 3dp  or

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4r2y  3b2 p2 2a

or

For rTd ei

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12r2y  3a 2 p2t

3cpt þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4r2y  3d 2 p2

2c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3dp  4r2y  3d 2 p2 2ð1  lÞ 2c

p

3cpt þ

6d

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12r2y  3c2 p2t 6d

¼ pt4  pt 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 12r2y  3c2 p2t

3dp þ

¼ p6

¼ p6

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4r2y  3d 2 p2

¼ pt0 2c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3dp þ 4r2y  3d 2 p2 2ð1  lÞ

¼ Dt4  Dt  2c Ea The allowable load = Min {p1, p2}

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi When 3pt ðad þ bcÞ þ b 12r2y  3c2 p2t  d 12r2y  3a 2 p2t ¼ 0 (k ≤ ki ), p6 ¼ p5

Ea

¼ Dt0

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Autofrettage Technology and Its Applications in Pressured Apparatuses

TAB. 8.1 – (continued). qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi The temperature difference solved from 3pt ðad þ bcÞ þ b 12r2y  3c2 p2t  d 12r2y  3a2 p2t ¼ 0 is the optimal temperature difference. Δte The boundary radius ratio ki qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi When pt = ptc, or Δt = Δtc, 3pt ðad þ bcÞ þ b 12r2y  3c2 p2t  d 12r2y  3a2 p2t ¼ 0 becomes sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3ðk 2  1Þ 3 k2 ðk 2  1  ln k 2 Þ2 þ 2  2 12  3 ¼0 2 2 2 k ln k  k þ 1 k  1 k  1 ðk 2 ln k 2  k 2 þ 1Þ2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3ðk 2  1Þ 3 k2 ðk 2  1  ln k 2 Þ2 þ  12  3 ¼ 0 is the The solution of 2 k ln k 2  k 2 þ 1 k 2  1 k 2  1 ðk 2 ln k 2  k 2 þ 1Þ2 boundary radius ratio ki

ki ¼ 2:0542956825. . . When Δt = Δtc, if k = ki, p5 = p6; when k > ki, p5 > p6; when k < ki, p5 < p6 When k ≤ ki, Δte ≤ Δtc; when k ≥ ki, Δte ≥ Δtc When k ≥ ki, in the range of Δt ≤ Δtc, the allowable load is p = p6; if Δt = Δtc, the allowable load is rffiffiffiffiffiffiffiffiffiffiffiffiffi k2  1 ry p ¼ p6 ¼ (k ≥ ki) r or k ¼ y k2 ry  p

p2 k 2  1 pt k 2  1  ln k 2 k 2  1 Ea k 2  1  ln k 2   ¼ ¼ Dt 2 2 2ð1  lÞry ry ry ln k 2 ln k 2 When p ≤ p2, rT ho ≤ σy

References [1] Liu H.W. (2019) Mechanics of materials (in Chinese). Higher Education Press, Beijing. [2] The compiling group of “handbook of mathematics”. (1979) A handbook of mathematics (in Chinese). Higher Education Press, Beijing.

Nomenclature

ri, rj, ro, r: Inside radius, radius of elastic-plastic juncture, outside radius and radius at a general location, respectively. k: Radius ratio or ratio of outside to inside radius, k = ro/ri. kj: Depth of plastic zone, or plastic depth, kj = rj/ri. kj*: The optimum plastic depth, kj* is the value determined by k 2 ln kj
2  k 2  kj
2 þ 2 ¼ 0. kc: Critical radius ratio, kc = 2.218 457 489 916 7…. x: Relative location, x = r/ri. x0: The relative location of the intersection of the three residual stress curves or the abscissa at which r0e ¼ 0, when x = x0, r0z ¼ r0r ¼ r0h . x1: The relative location where r0z ¼ 0. T xa: The relative location or the abscissa where r0e ¼ r0h and/or rT e ¼ rh within the pp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi 3  1k. elastic region, xa ¼

x0T : The relative location of the intersection of the three total stress curves, when T T x ¼ x0T , rT z ¼ rr ¼ rh . xz: The relative location where equivalent elastic-plastic stress is equal to circumpffiffi pffiffi 3 p 2 3 ferential elastic-plastic, i.e. σe = σθ, xz ¼ exp 2 ry  2 . p: Internal pressure. py: Entire yield pressure. pe: Maximum elastic load-bearing capability or initial yield pressure of an unautofrettaged pressure vessel. pa: Autofrettage pressure. p∞: The internal pressure when k→∞. σz: Axial stress. σr: Radial stress.

350

Nomenclature

σθ: Circumferential stress. σy: Yield strength. σe: Equivalent stress. ε: Overstrain. λ: Load ratio, λ = p/pe, λ is chosen between 1*2, which indicates the multiple of initial yield pressure. kjλ: Plastic depth when p= λ.pe, or the optimum plastic depth, or kj corresponding to λ, when λ = 2, kjλ is marked as kj*. ελ: Overstrain when p = λ.pe. kcλ: Critical radius ratio under a certain λ, or critical radius ratio corresponding to λ, 2 2 2 ln kck  kðkck  1Þ ¼ 0. determined by kck kjλ∞: kjλ when k→∞. t: Temperature. α : Thermal expansion coefficient. μ: Poisson’s ratio. E: Young’s elastic modules. Δt: Temperature difference. Δtc: Critical temperature difference. Superscripts: ′: Residual stresses. p: Stresses caused by operation pressure p, or operation stresses. T: Total stresses. s: Quantity based on the maximum shear stress theory (Tresca yield criterion). d: Quantity based on the maximum distortion strain energy (Mises yield criterion). Subscripts: j: Quantity at rj. i: Quantity at ri, saying, r0ei represents the equivalent residual stress on the inside surface. o: Quantity at ro, saying, ti represents the temperature on the outside surface e: Equivalent stress. z, r, θ: Axial, radial and circumferential direction, respectively.