This book is aimed at those in both industry and academic institutions who require a grounding not only in the basic pri
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Pages [119] Year 1993
Table of contents :
Introduction, Background, Fracture Toughness Determination Yielding Fracture Mechanics, Fatigue Crack Growth, Concluding Remarks, References
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 334872742 First issued in hardback 2017
C The Institute of Materials 1993 C loM Communications Ltd 1998
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Briti sh Li bra ry Cataloguingi nPub licati on Data Knolt, John Fracture Mechanics: Wo rked Exampl es I. T itle 11 . With ey, Paul 6 20. 11 26
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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 CentreCracked 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 (Wa) 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 displacemenlload 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 addi 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 planestrain 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 105 kNl 2
28
(dC) da
v 2 ) where
Fracture Mechanics  Worked Examples 23 =
1582 x2.5xlOs
2x25  12.5 kJm2
From EQuation 6, K 2 _ GleE
"
 1y2 =
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 = Iv
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 2x 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):: (12: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(ax)
{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)