Fracture Mechanics: Worked Examples [2 ed.] 9780901716286, 9781138442771, 9780138756789, 9780429605178, 9780429599651, 9780429610691

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Fracture Mechanics: Worked Examples [2 ed.]
 9780901716286, 9781138442771, 9780138756789, 9780429605178, 9780429599651, 9780429610691

Table of contents :

Introduction, Background, Fracture Toughness Determination Yielding Fracture Mechanics, Fatigue Crack Growth, Concluding Remarks, References

Citation preview

FRACTURE MECHANICS - WORKED EXAMPLES

Fracture Mechanics Worked Examples J.F. Knott FRS, FEog, B.Met, SeD, FIM, FWeldl School of M etallurgy and Materials

The University of Birmingham and

P.A. Wilhey BSc, PhD, CPhys, MlnstP RolIs ·Royce plc Bristol

Book 550 CRC Press Taylor & Fronds Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 First issued in hardback 2017

C The Institute of Materials 1993 C loM Communications Ltd 1998

C RC Press is an imprint ofTaylor & Francis Group, an Infonna business No c laim to original U.S, Governme nt works ISBN-13: 978-0-901716-28-6 (pb k) ISBN- 13: 978-1-138-44277-1 (hbk) This book contains infonnation obtained from authentic and high ly regarded sources. Reasonable efforts have been made to publish reliable data and infonnation, but the author and publisher cannot assume responsibili ty for the validity of al l materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if pennission to publish in this fonn has not been obtained. Ifany copyright material has not been acknowledged please write and let us know so we may rectiry in any future reprint. Except as pennitted under U.S. Copyright Law, no part or this book may be reprinted, reproduced, transmitted, or utilized in any ronn by any e lectronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfi lming, and recording, or in any inronnation storage or retrieval system, without written pennission rrom the publishers. For pennission to photocopy or use material electronically from this work, please access www.copyrighl.com (http://www.copyrighl.com/) or contact the Copyrigh t Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Tradema rk Nollce: Product or corporate names may be trademarks or registered trade marks, and are used only for identification and explanation without intent to infringe.

Briti sh Li bra ry Cataloguing-i n-Pub licati on Data Knolt, John Fracture Mechanics: Wo rked Exampl es I. T itle 11 . With ey, Paul 6 20. 11 26

loM Commu nications Lld is a wholl y owned subsidiary of The Insitute of Materia ls Typeset from the authors' d isks by Inforum, Rowlands Castle Visit the Taylor & Frands Web site a t http://www.tay lora ndfr ancis.com and the C RC Press Web site a t hltp :ffwww.crcpress.co m

Contents

List of Worked Examples

VI

Acknowledgements

viii

Foreword

.ix .. 1

Introduction

2 2

Background

3

Fractu re Toughness Delcnnination

4

Yielding Fracture Mechanics .

5

Fatigue Crack Growth

86

6

Concluding Remarks

100

7

References

103 105

3

.35 5')

Appendix - Derivation of the Displacement Within a Crack

2g

List of Worked Examples

2a

. 4

- Calculation of Minimum Defect Size

2e

- Derivation of Stress Instensity Factor

. 6 8

2d

- T he Virtual Work Thcomm

. 9

2e

- Eqi va1ence of G and K

2b - Calculation of Fnlctu rc Stress

JO

21 - Fracture in a Centre-Cracked P ane l

13

2g - Fracture in an SEN (Single Edge Notched) Bend ll:stpiccc

15

2h

16

Fracture in a

crs (Compact Tension)

Testpiece

2g

- Teslpiece Dimensions (I)

17

2j

- Teslpiece Dimensions (2)

18

2k

- Deri vation of G U isng Compliance Method

21

21

- Compli:mcc and Fmcturc Toughness

22

2m - Deri vation of a Weight Function for a Crack i n a nInfi nite S heet

24

2n

25

- Ca1cul2.S[ K,c )' Oy

18

Background Theory

450

=2.5(39.4)' .. 19.1 mm A tcstpiecc thickness of 50 nun is therefore more than adequate for a valid K/c resu lt. For (b) (Working in m)

Bn5( :~ r

005>2.5( 3:~4r Oy 0:: 279 MPa

The minimum value of proof stress is therefore 279 MPa. Example 2j - Testpiece Dimensions (2) An alloy forging steel has a specified mmimum proof stress, oy = 800 MPa and a guaranteed minimum fracture toughness, KIc= 120 MPa m 1/2 . (3) Calculate the minimum testpiece dimensions needed 10 carry oul valid tests to check the toughness figures . (b) Estimate the weights of sufrlciently large standard SEN bend and crs testpieces. (c) Estimate the test machine capacity required. Solution

r the

Fot(,) B>2.s( : ; the

fmmEqoo"on 10

K/c = 120, 0y = 800

:. BO::2.5(O.15)2 m

=56.25 mm and W = 2B W= 112.5mm

For (b) SEN bend specimen (sec Figure 3. 1)

Fra cture Mechanics - Worked Examples 19 Testpiece length, L = 4W + 10 mm :.[;:;460 mm Ig noring the notch, the volume is B x W xL :. volume = 0.0029 m3 :. mass = 23 kg (since densi ty of steel .. 7900 kg m"3) CTS specimen (see Figure 3.2)

length=2H = 1.2 W = 135 mm total width = 1.25W = 140.6 mm hole diameter, D , = 0.25W= 28.125 mm Ignoring the notch, Volume:::; (2H x B x C) - 2(nIY14 x B) = (0.00 1068 • 0.0000(9) = 0.00 1 ml :. mass = 7.9 kg For (c) in order to obtain the mimmum machine capaci ty required, assume the minimum permissible (aIW) value = 0.45 to calculate the maximum load required to give a Klc of 120 MPam 1n (ef Examples 2g and 2h).

rn'

SEN - bend specimen (sec Figure 3. 1)

From Equation 8, PF =

Now

K BWII2

="c,'--Y,

w= 112.5 mm, B = 56.25 mm and

Y1 = 9.14

(aIW= 0.45)

(Working in MN and m) P _ 120xO. 05625x(0. 1I25)112 F 9. 14 = 0.248 MN

The minimum machine capacity required is therefore 248 kN .

CTS Specimen (see Figure 3.2)

From equation 9,

20

Background Theory _ KIc BWII2 PF-

Y,

for a/W = 0.45, Y2 = 8.34 (see Figure 3.2) Hence

goodgoodgood

Comments Thi s example provides a good jllustration of the practical features involved in the testing o f a reasonably tough f orging s teel. The testpiece size becomes qui te large and , for the S EN bend geometry, uncomfortably heavy for a isngle operator t o handle easily. If the testing machine is of limited capacity, it may be necessary to resort to the bend geometry since the loading span can be increased to facilitatc fracturc. However, this may present difficulties in respect t o the supply of material in suitable form. If we take this example furth er, assuming the et st machi ne is limited to 250 kN c apacity, wc could adopt a higher (aM ,) value within the permitted range (i .e. up to aIW = 0.55) to lower the val ue o f PF. However, since a mi nimum guaranteed K/c value of 120 MPam t12 is specified, in practice a higher Klc will be the case, which in turn will require a higher value o f PF. Thus. B and W must be increased correspondingly. In the p rcsent hypothetical case, B might be set at 6 5 mm ( W = 130 mm) which w ould enable valid K/c results to be obtained up to a avlue of 129 Mpam ll2 . If the et st machine is limited to 250 kN, it would be just possible to est t a tsandard testpiece, provided that 0.54 < aIW 12. 15), which would require exceptional control on the ini tial crack length . It would n ot be poss ible to fracture a CTS testpicce of 'valid' s zi e in the machinc. The use of higher a/W values, outside the S tandard , is clearly attracti ve if the (W-a) requi re ment is, indeed, over stringent. 2.6 Compliance Methods The mathematical expressions for K become complicated f or tsandard test piece geometrics and although the Y functi ons referred in Equation 8, arc tabu lated, the physical sense of what i sbeing calculated is oflen losl. For testpieces, a conceptually more di rect m ethod is to determi ne experimentally the change in stored elast ic energy with increas ing c rack length. measured in terms of the di splacement of the ol ading points, and to ob tain a val ue o f Gfrom the expression:

Fracture Mechanics - Worked Examples 21

G =(~) d~

B da Whilst examples of its usage arc g vi en below, the compliance approach has s all changes in displacement is certain limitations. First, the measurement of m difficult to make with accuracy. Additionally, in many structures, such as a pressure vessel, the position of the loading points is undefin ed. The theoretical methods must then b e used.

Example 2k - Derivation of G Using Compliance Method

If the displacemenl-load relationship for a body of thickness, B, containing a rcack of length a, is g vi en by u = CP, where C is the compliance and is a function of0 ( a l onger crack makes the specimen behave like a w aeker elastic spring), show that the energy release rate per unit thickness, G, is g ven i b y:

G=~( dC)

2 B da if the crack extends under a constant l oad P.

Solut ion The strain energy stored in the body a t cr ack I cngth . a, is g iven by 1/2 Pu = Ih CP (the area under the ol ad-di splacement curve). If the crack extends a tconstant load, the d si placement will increase to (u+ d u) , whilst P remain s constant, because the speci men with crack length (a + (la) behaves like a weaker spring. The new tsrai n e nergy stored i sthen increased t o '2t P(u + tlu) . But, the applied l oad does work of magnitude -P«(1u ), so that the net release of potentia! energy is equal t o -'2t Pdu . i.e . d~ = _1/2 Pdu Now u =

ep

.'. du =C d P + pd C da da da

= P dC since dP = 0 (constant load) da

Now, the change in energy for crack extension is given by: d~ I du - = - - Pda 2 do

22

Backgroulld Theory . d~ _ I p 2 de " da - -'2 da G is defined as the positive value o f

1..( dl;) 8 da

G= p2 (dC)

Hence,

(Ill

28 da

Example 21 • Compliance and Fracture Toughness In the compliance calibration of an edge cracked fracture toughness testpiece of an al uminium alloy, il was observed that a load of 100 kN produced a displacement between the loading pins of 0.3000 mm when the crack length was 24.5 mm and 0.3025 mm when the crack length was 25.5 mm. The fracture load of an identical testpiece, containing a crack of length 25.0 mm is 158 kN. Calculate the cri tical value o f the potential energy release rate at fra cture and hence the plane-strain fracture toughness, KJc ' of the a lloy. All testpieces were 25 mm thick. From Equation 6, for plane strain conditions, G =

,

~(1 E

v has a value of 0.3 and E is 70 GPa for this alloy. Solution AtP=IOOkN, Si nce Then

U

(11

= 24.5 mm and UI = 0.3000 mm

a2

= 25.5 mm and

U2

= 0.3025 mm

= CP Ct = 0.3000)1: 10- 2 mm kN- ' C2 0.3025)1: 10- 2 mm kW' .c C,,, _--,Cl I da a2 - at

=

_dC _•

2

:. de = 0 .(025)1:10 - kN - 1 I

da = 2.5

From Equation 11 ,

Gair =

PF

x 10-5 kN-l 2

28

(dC) da

v 2 ) where

Fracture Mechanics - Worked Examples 23 =

1582 x2.5xlO-s

2x25 - 12.5 kJm-2

From EQuation 6, K 2 _ GleE

"

- -1-y2 =

12.5xlO- 3 x70xlO 3

0.91 =962 .', K/c - 31,0 MPa m lf2

2.7 Weight Functions

We have shown (example IV) that the potential energy release rate per unit thickness. G. is given by (equations 4 and 5) potential potential and that the work done on extending a crack is given (example 2d. p9) by

&;= Jo~Udr.

(12)

For a crack growing from zero 10 a crack length of a, r becomes identical to x ; hence dr = dx and we have

Now since neither 0 thickness, then

d~ = J:oudx .

(13)

d~=f."odudx

(14)

da 0 da nor dx vary with a, and as G is defined as dl;jda per unit aK

E

2

=

J."o dadudx 0

(15)

This may be rearranged this to give the form :

Define m(x) as

K=f.°OEdudx oaK da

(16)

E du m(x)= - -

(17)

aK da

hence

24

Background Theory (18)

(18)

function. function. (18)

The term m(x) is known as the weight function. Example 2m - Derivation of a Weight Function for a Crack in an Infinite Sheet Using the equation for u2 (equation A22 in the Appendix) derive the weight function for a crack of length la in an infinite sheet under an applied stress of 0IlPP' Solution From equation A22 we have

2(

u2 =- I-v

E

2)0IJppva·-x ,,---, .

As a = (1 - v2 ) this becomes 2

and

I 2

u2 = EaoQPp'Io

-x

du da

a

-

2a

= -

E

0

app/2

2

2'

va -x Note that we are differentiating with respect to a not with respect 10 x. This can be substituted into the equation for the weight function m(x) to obtain E 20: 0 m(x)=O:KEoapP 12

2

=-0 IlPPl

va - x a

2

K

2 va 2-x From table 2.1 we have K = OIlPP v(na) for an infinite sheet. Substituting for K we obtain

m(x)=

~ J 22

V~

0 _xl

Although for this case it is easier to use the formula found in table 2.1, the weight function method becomes useful for more complex loadings and geometries. The function m(x) is a geometric function which depends simply on the crack length and shape and on the geometry of the test piece or component in which the crack is present. A number of weight functions for different geometries are available in the literature (e.g. fef 20), and it can be of value to

Fracture Mechanics - Worked Examples 25 one geometry to a component with a roughly apply a weight function known for onc comparable geometry but for which the weight weighl function is unknown.

Example 2n - Calculation of the weight function for a cracked bend leslpiece A bend testpiece in pure bending (four point loading) contains a a. The stress across the testpiece varies linearly crack of length Q. zero from a maximum tensile stress (am) at one surface, through :rero centre. 10 in the centre, to a maximum compressive stress (-am) at the other surface. The stress intensity factor is given by K = 1.12IS~(1Ca) 1I2 fa a(x)th 1.1215~(1Ca}1I2 1C Jo (a l2 _X _ x 2 )L12 Jo(a )lIl where x is the distance from the top surface of the testpiece. K =/a =/am (ax )112 Calculate the value of/in the equation m (a1t for the bend lestpiece testpiece if the crack is 0.05 O.OS aIW in length. length .

r"

Soluiion equation for the variati variation on of the stress within the .tIcslpiece cstpiece The equalion must be deri derived. ved. As stress drops linearly from am at x = 0, 10 to zero :rero at x = WI2 and on to -om at x = W, the variation of the stress with distance, x, in the uncracked body, is a(x):: (1-2:cIW) a(x) am (1 -2:cIW) The equation above becomes

=

=

=

2 112 1 "" O -2X' W) W) K = 1.I2IS 1.l21S -(1Ca -( rca)) 0", Om 2 2 112dx 112dx 1t

o(a-x)

{f"J

f"

2 '" _ =I I Ret\ldx]dx l

+x21 Im4ldx

l

(AI2)

The Airy stress equations can be used 10 to obtain values of a(11l and a(12' 2, It is straightforward slraightforward 10 obtain a value for 0"2 as the above equation only needs to be twice wilh respect 10 XI' xl' Hence differentiated Iwice

FraClure Mechanics - Worked Examples 02 =Re$+x2

d(lmo)

107

(Al3)