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Home , Elasticity (physics), Plasticity (physics)

University of Liège Aerospace & Mechanical Engineering

Fracture , Damage and Non Linear Mechanics: J-Integral

Ludovic Noels

Computational & Multiscale Mechanics of Materials – CM3 http://www.ltas-cm3.ulg.ac.be/ Chemin des Chevreuils 1, B4000 Liège [email protected]

Fracture Mechanics – NLFM – J-Integral Linear Elastic (LEFM) y • Cracked body: summary s – 3 failure modes ∞ Mode I Mode II Mode III s s (opening) (sliding) () yy yy

2a x

s∞

– Asymptotic solution governed by intensity factors

syy 퐾 Asymptotic s 흈mode 푖 = 푖 푓mode 푖 휃 + ℴ 푟0 Zone of asymptotic yy 2휋r solution (in terms of 1/r1/2) dominance True syy mode 푖 푟 mode 푖 0 풖 = 퐾푖 푔 휃 + ℴ 푟 2휋 x Structural response

2021-2022 Fracture Mechanics – NLFM – J-Integral 2 Linear Elastic Fracture Mechanics (LEFM)

• Cracked body: summary

– PT = Eint - Qu before – Crack closure integral • Energy required to close crack tip 풖+횫풖 ′ ′ ΔΠ푇 = 풕 ⋅ 풖 푑풖 푑퐴 Da Δ퐴 풖 after

– Energy release rate ti 0 • Variation of potential energy in case of crack growth t i

퐺 = −휕A 퐸int − 푊ext = −휕퐴ΔΠ푇 • In linear 0 . 1 1 ퟎ t Du/2 퐺 = −휕퐴ΔΠ푇 = − lim 풕 ⋅ 횫풖 푑퐴 Δ퐴→0 Δ퐴 Δ퐴 2 0 u0 u0 +Du u – In & if crack grows straight ahead i i i i

퐾2 퐾2 퐾2 퐺 = 퐼 + 퐼퐼 + 퐼퐼퐼 퐸′ 퐸′ 2휇

2021-2022 Fracture Mechanics – NLFM – J-Integral 3 Linear Elastic Fracture Mechanics (LEFM)

• Cracked body: summary B – J-integral y r D q • Strain energy flow x a G – Exists if an internal potential exists • Is path independent if the contour G embeds a straight crack tip

• No assumption on subsequent growth direction

• Can be extended to plasticity if no unloading (see later)

– If crack grows straight ahead G=J

– In linear elasticity (independently of crack growth direction):

2021-2022 Fracture Mechanics – NLFM – J-Integral 4 Linear Elastic Fracture Mechanics (LEFM)

• Analytical y y y – SIF from full- solution s∞ • Limited cases t∞ t∞ s x x s∞ x ∞ 2a 2a 2a t∞ t∞

s∞ – From energetic consideration 2 2 2 퐾퐼 퐾퐼퐼 퐾퐼퐼퐼 • Growing straight ahead crack 퐺 = −휕 퐸 − 푊 퐺 = + + A int ext 퐸′ 퐸′ 2휇 • From J-integral • Numerical (e.g. FEM)

– bi depends on geometry & crack length • Tabulated solutions (handbooks) – http://ebooks.asmedigitalcollection.asme.org/book.aspx?bookid=230 2021-2022 Fracture Mechanics – NLFM – J-Integral 5 Linear Elastic Fracture Mechanics (LEFM)

• Small Scale Yielding assumption – LEFM: we have assumed the existence of a K-dominance zone

syy Does not always syy K-dominance zone Asymptotic syy exist LEFM asymptotic ( asymptotic -1/2 syy in r solution in 1/r1/2) True syy True syy

x x Plasticity

• This holds of if the process zone (in which irreversible process occurs) – Is a small region compared to the specimen size & – Is localized at the crack tip – Validity of this approach? • We check the dimensions ∝Process 2 퐾퐼 zone size 푎, 푊 − 푎 > 2.5 0 휎푝 • Non-linear fracture mechanics – Derivation of the LEFM validity criterion – Providing solutions when LEFM criterion is not met

2021-2022 Fracture Mechanics – NLFM – J-Integral 6 Elastoplastic behavior • Power law – This law can be rewritten in terms of the total deformations • stress is replaced by 2

equivalent stress 1.5

• strain is replaced by

p

0

0 p s s

/ 1 equivalent strain /  = 2; n = 1 s s  = 4; n = 1  = 2; n = 4 • The governing law becomes 0.5  = 4; n = 4  = 2; n = 11  = 4; n = 11 or 0 0 2 4 6 8 10 0  E/s 0 • Parameter n  E / sp p – n→∞: perfect plasticity – n→1: “elasticity” – Doing so requires 2 assumptions • There is no unloading • As elastic strains are assimilated to plastic strains, the material is incompressible – Which are satisfied if • We are interested only in crack initiation and not in crack propagation • The stress components remain proportional with the loading • Elastic deformations are negligible compared to plastic ones 2021-2022 Fracture Mechanics – NLFM – J-Integral 7 HRR field

• Summary 2.5 s~ ; n = 3 rr – Assumptions Plane  ~s ; n = 3 2 qq ~s ; n = 3 • J2-plasticity with power law description rq ~s ; n = 3 • Small deformations 1.5 e

~ s • There is no unloading and loading is 1 proportional in all the directions (ok for crack initiation and not for crack propagation) 0.5

• Elastic strains are assimilated to plastic strain 0 0 50 100 150 (material is incompressible) q [degq [°].] • Semi-infinite crack 7 – HRR results for semi-infinite mode I crack Plane  • Asymptotic stress, strain and fields 6 Plane s 5

4

n I 3

2

1

0 0 5 10 15 n n

• J is a “plastic strain intensity factor” 2021-2022 Fracture Mechanics – NLFM – J-Integral 8 HRR field • Summary (2) 0.2 n = 3; Plane  – HRR results for semi-infinite mode I crack (2) n = 3; Plane s

• Process zone 0.1 y

~ 0 with

-0.1 – If SSY

• CTOD -0.2 – New definition -0.2 -0.1 0 0.1 0.2 ~x y 1 s0/E = 0.001 p ux Plane  s0/E = 0.002 u 0.8 p y s0/E = 0.004 p s0/E = 0.008 x 0.6 p

r* n d 0.4

– 0.2

– Also function of J 0 0 0.1 0.2 0.3 0.4 0.5 – We did not assume SSY !!! 1/n 2021-2022 Fracture Mechanics – NLFM – J-Integral 9 HRR field • Validity in SSY – We have two asymptotic solutions • HRR field is valid in the process zone • LEFM is still valid in the elastic zone close to the crack tip HRR asymptotic HRR asymptotic yy -1/(n+1) s s in r -1/(n+1) yy syy in r syy n+1 LEFM asymptotic Log 1 s in r-1/2 yy 2 True syy LEFM 1 asymptotic x -1/2 Plasticity syy in r Log r

– Conditions • This is the case if all sizes are 25 times larger than the plastic zone

t a L

2021-2022 Fracture Mechanics – NLFM – J-Integral 10 HRR field • Validity in elasto-plastic conditions – Deformations are small – We still have one asymptotic solution valid • HRR field is valid in the process zone – LEFM is NOT valid in the elastic zone close to the crack HRR asymptotic yy -1/(n+1) HRR asymptotic s syy in r s in r-1/(n+1) n+1

yy Log syy 1 LEFM asymptotic -1/2 syy in r 2 True syy LEFM 1 asymptotic -1/2 x syy in r Plasticity Log r

– Conditions • This is the case if all sizes are 25 times larger than CTOD

t a L

2021-2022 Fracture Mechanics – NLFM – J-Integral 11 HRR field • Validity in large yielding – Example: ligament size is too small – Small deformations assumption does not hold – Neither HRR field nor LEFM asymptotic fields are valid HRR asymptotic HRR asymptotic yy -1/(n+1) -1/(n+1) s s in r syy in r yy syy n+1 LEFM asymptotic Log 1 -1/2 syy in r True syy 2 LEFM 1 x asymptotic -1/2 Plasticity syy in r Log r – There is no zone of J-dominance – Plastic strain concentrations depend on the experiment

2021-2022 Fracture Mechanics – NLFM – J-Integral 12 HRR field • Crack initiation criteria – In SSY:

• Criteria based on J or dt are valid: J ≥ JC or dt ≥ dC

– J & dt depend on a, the geometry, the loading, …

• But as the LEFM solution holds, we can still use K(a)≥KC 0 – May be corrected by using the effective length aeff if s∞< 50% of sp – In Elasto-Plastic conditions:

• Criteria based on J or dt are valid: J ≥ JC or dt ≥ dC

– J & dt depend on a, the geometry, the loading, …

• The LEFM solution DOES NOT hold, we CANNOT use K(a)≥KC – In Large Scale Yielding • Plastic strain concentrations depend on the experiment • Zones near free boundaries or other cracks tend to be less stressed – Solution is no longer uniquely governed by J

– Relation between J & dt is dependent on the configuration and on the loading

– The critical JC measured for an experiment might not be valid for another one – A 2-parameter characterization is needed

2021-2022 Fracture Mechanics – NLFM – J-Integral 13 Relation J-G • For elasto-plastic materials – J is a useful concept as it can be used as crack initiation criterion – Relation J-G • J = G if an internal potential is defined & if the crack grows straight ahead • For an elasto-plastic material – Since no internal potential can be defined J ≠ G – But let us consider two specimens 1: Non-linear 2: Elasto- » One with a non-linear elastic material (1) elasticity plasticity » One with an elasto-plastic material (2) with the same loading response J1=G1 J ? G

– If the crack will grow straight ahead 2 2 s » J1 = G1 as there is a potential defined Q Q

» Before crack propagation (which True would involve possible unloading) sTS the stress state is the same for the

two materials J2 = J1 = G1 = G2 • If the crack will grow straight ahead, before crack propagation: J = G for an elasto-plastic material E p e True 

2021-2022 Fracture Mechanics – NLFM – J-Integral 14 Computation of J • Available methods of computing J – Numerical (see previous lectures) • Contour integral • Domain integral – Energetic approach • Numerical • Experimental – Experimental • Multiple specimen testing • Deeply notched SENB • Eta factor approach – Engineering • Use of handbooks • If J expression is known it can be applied to – Fracture toughness test – Fracture criteria and stability prediction

2021-2022 Fracture Mechanics – NLFM – J-Integral 15 Computation of J : Energetic approach • Prescribed displacement – Let us consider a specimen • With a prescribed displacement • A compliance depending on the crack length – Using compliance curves • Energy release rate: Q Q+dQ • :

• Energy release rate in terms of displacements

– We have Q’ Q’ A A Q Q’ du A’ > A

Eint E int u' u' u u

2021-2022 Fracture Mechanics – NLFM – J-Integral 16 Computation of J : Energetic approach Q • Prescribed displacement (2) Q* – Interpretation of Loading Crack A=A Q*+dQ 0 growth for an elastic material

• Body with crack surface A0 loaded up to Q* Unloading • Crack growth dA at constant grip A=A0+dA the specimen becomes more flexible u u* the load decreases by • Unload to zero • The area between the 2 curves is then - G dA – For an elasto-plastic material • Since J = G only before crack growth, this method can be used – Either experimentally with 2 specimens with a different initial crack length – Or by computing the curve u(a, Q) » Analytically » Using FEM

2021-2022 Fracture Mechanics – NLFM – J-Integral 17 Computation of J : Energetic approach • Prescribed loading – Let us consider a specimen • With a prescribed loading (dead load) • A compliance depending on the crack length – Using compliance curves • Energy release rate: du Q • Complementary energy: Q

Q Q’ A A Q* Q • Energy release rate in terms of A’ >A displacements u dQ – We have

Eint Eint

u u' u* 2021-2022 Fracture Mechanics – NLFM – J-Integral 18 Computation of J : Energetic approach • Prescribed loading (2) Q Loading – Interpretation of Q* Crack growth A=A0 for an elastic material Unloading

• Body with crack surface A0 loaded up to Q* A=A0+dA • Crack growth dA at constant load the specimen becomes more flexible u displacement increment u* u*+du • Unload to zero • The area between the 2 curves is then G dA – For an elasto-plastic material • Since J = G only before crack growth, this method can be used – Either experimentally with 2 specimens with a different initial crack length – Or by computing the curve Q(a, u) » Analytically » Using FEM

2021-2022 Fracture Mechanics – NLFM – J-Integral 19 Experimental determination of J

• Multiple specimen testing (Begley & Landes, 1972) u,Q – Consider 4 specimens with

• 4 different crack lengths a1 < a2 < a3 < a4 Thickness t • Under displacement control W • No fracture taking place a

– This leads to compliance curves L=4W • Integration to obtain the internal energies

Q Eint/t u4 a1

a2 u3

a3 u2 a4 u1

u a

u1 u2 u3 u4 a1 a2 a3 a4

2021-2022 Fracture Mechanics – NLFM – J-Integral 20 Experimental determination of J • Multiple specimen testing (2) – Since The area between the curves u(a, Q) and u(a+da, Q) is equal to - G t da

The slopes of the extrapolated curves Eint(u, a) lead to J

Eint/t J u4 a1

u3 a2 u2 a3 u1 a J 4 1 a u

a1 a2 a3 a4 u1 u2 u3 u4 • Small displacements: LEFM holds • For large displacements

– This technique is very time consuming but it avoids unloading at the crack

2021-2022 Fracture Mechanics – NLFM – J-Integral 21 Experimental determination of J

• Deeply notched specimen testing Q, u – Consider 1 Single Edge Notch Bend specimen • Prescribed loading Q Thickness t – Non-dimensional analysis W • 9 input variables: a 0 . -2. -1 – [sp ] = kg s m , [W-a] = m, [t] = m, L [E] = kg.s-2.m-1, [n] = -, [n] = -, [L] = m, [W] = m, [Q] = kg.m.s-2 • 3 dimensions 9-3 = 6 independent non-dimensional inputs 0 2 – n, n, E / sp , QL / Et(W-a) , L/(W-a), W/(W-a) • 1 output variable: [u] = m 1 relation in terms of the non-dimensional inputs

– Prescribed loading • Before crack propagation:

2021-2022 Fracture Mechanics – NLFM – J-Integral 22 Experimental determination of J • Deeply notched specimen testing (2) – J-integral

with

– We have

•

푄퐿 퐿 푊 휕 휕 휕 퐸푡 푤 − 푎 2 푤 − 푎 푤 − 푎 휕푎 휕푎 휕푎

•

푄퐿 휕 퐸푡 푤 − 푎 2 휕푄

2021-2022 Fracture Mechanics – NLFM – J-Integral 23 Experimental determination of J • Deeply notched specimen testing (3) – J-integral

with

– So we have

Q a

– Can we neglect 퐼 and 퐼 ? 1 2 Jt(W-a)/2 • This would be convenient since this would involve the load-displacement curve only u

2021-2022 Fracture Mechanics – NLFM – J-Integral 24 Experimental determination of J • Deeply notched specimen testing (4) q – If the Single Edge Notch Bend specimen u • Is deeply notched • Has a remaining ligament W-a fully plastic L – There is a plastic hinge • The curve Q – u represents essentially the plastic deformations • Crack is deep enough so that the plastic hinge is localized between the applied load and the crack tip y • Example: perfectly plastic material x 푊−푎 푊−푎 2 2 푀 = 푡 휎푦푑푦 = 2푡 휎0푦푑푦 푊−푎 푝 − 0 2 푊 − 푎 2 푀 = 푡휎0 푝 2 0 푡휎푝 푀 = 푊 − 푎 2 4

2021-2022 Fracture Mechanics – NLFM – J-Integral 25 Experimental determination of J • Deeply notched specimen testing (5) q – If the Single Edge Notch Bend specimen u • Is deeply notched • Has a remaining ligament W-a fully plastic L – There is a plastic hinge y • Perfectly plastic material 0 푡휎푝 푀 = 푊 − 푎 2 x 4 • Static equilibrium: 푄퐿 Q/2 q 푀 = u 4 M • And we have L/2 0 0 푡휎푝 휕푄 푡휎푝 2 푄 = 푊 − 푎 2 = −2 푊 − 푎 = − 푄 퐿 휕푎 퐿 푊 − 푎 • The J-integral becomes

퐼1 = 0 and 퐼2 = 0 2021-2022 Fracture Mechanics – NLFM – J-Integral 26 Experimental determination of J

• Eta factor approach Q a – Expression Jt(W-a)/2 • Is convenient since it involves only the load-displacement curve • Is valid for SENB with plastic hinge u q u

L • Elastic materials

– SENB with L/W = 4 & a/W > 0.5: I1 and I2 can be neglected* • Elastic-plastic materials

– SENB at low temperature, for L/W = 4 & 0.6> a/W > 0.4: I1 and I2 can be neglected (Comparison with multiple specimen testing**) – How can this expression be generalized to • Other geometries, loadings • Elastic-plastic materials *Srawley JE (1976), On the relation of J to work done per unit uncracked area: total, or component due to crack, International Journal of Fracture 12, 470–474 **Castro P, Spurrier P, and Hancock J (1984), Comparison of J testing techniques and correlation J-COD using structural steel specimens, International Journal of Fracture 17,83–95. 2021-2022 Fracture Mechanics – NLFM – J-Integral 27 Experimental determination of J • Eta factor approach (2) – How can be generalized?

– For other specimens and through-cracked structures • Assuming deformations are largely plastic

• Eta factor:

• With hJ depending on – Geometry & loading

– Crack length (a/W)

– Material hardening (n) – But • Materials are not rigidly plastic, so there is an elastic-plastic response

• How to determine hJ?

2021-2022 Fracture Mechanics – NLFM – J-Integral 28 Experimental determination of J • Eta factor approach (3) – How can be generalized (2) ?

– For elastic-plastic behavior • Split of elastic and plastic parts of – The displacement

– The J-integral

– The Crack Mouth-Opening Displacement Q Q, u a

Thickness t W Jpt(W-a)/hp v/2 v/2 u L=4W up ue • Factor hp still remains to be evaluated

2021-2022 Fracture Mechanics – NLFM – J-Integral 29 Experimental determination of J • Eta factor approach (4) – How can be generalized (3) ?

– For elastic-plastic behavior (2) • It is more accurate to measure the CMOD v than the displacement u

• But v and Q are not work conjugated ’ the new hp factor has to be evaluated with FE Q Q, u a

Thickness t CMOD, W ’ v v/2 v/2 Jpt(W-a)/hp

L=4W v

vp ve

2021-2022 Fracture Mechanics – NLFM – J-Integral 30 Experimental determination of J

’ Q, u • Evaluate hp for a given geometry – Characterize non-linear material response Thickness t • From uniaxial tensile test W – Linear FE analysis v/2 v/2

• Extract KI and then Je(Q) L=4W a Q – Non-linear analysis

• Load the specimen incrementally p • At each increment W – Extract J (domain integral) v 퐽 푄 = 퐽 − 퐽 푄 푝 푒 a J-J vp ve – Compute 푊푝 e

hp’ – Factor h ’ is the slope of p 1

Wp/t(W-a) 2021-2022 Fracture Mechanics – NLFM – J-Integral 31 Engineering determination of J • Use of handbooks – Split of J:

– Use of handbooks to determine Je (see SIF lecture) and Jp:

• HRR field

• Since this solution holds for fully plastic solutions, we need to adapt it

– J becomes Jp – r becomes W-a – Since » s becomes Q 0 » sp becomes yielding load Q0 • Tabulation of the values for different geometries

–

or depending on the test

– Similar relations for np, up and dt

2021-2022 Fracture Mechanics – NLFM – J-Integral 32 Engineering determination of J

y • Determination of Jp s – Example of centered crack plate

2W h

• x 2a

s

• With 1 1 0 s0/E = 0.001 s /E = 0.001 p p 0 s0/E = 0.002 Plane  s /E = 0.002 0.8 p 0.8 p 0 s0/E = 0.004 s /E = 0.004 p p 0 Other hi tabulated s0/E = 0.008 s /E = 0.008

0.6 p 0.6 p

n

n

d d 0.4 0.4

0.2 0.2 Plane s

0 0 0 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4 0.5 1/n 1/n1/n

2021-2022 Fracture Mechanics – NLFM – J-Integral 33 Engineering determination of J

• Determination of J (2) y p s – Example of centered crack plate (2)

Plane  n=1 n=2 n=3 n=5 n=7

a/W h1 2.80 3.61 4.06 4.35 4.33 2W h = h2 3.05 3.62 3.91 4.06 3.93 1/8 x h3 0.303 0.574 0.840 1.300 1.630 2a

a/W h1 2.54 2.62 2.65 2.51 2.28 = h2 2.68 2.99 3.01 2.85 2.61 1/4 s h3 0.536 0.911 1.220 1.640 1.840

2021-2022 Fracture Mechanics – NLFM – J-Integral 34 Engineering determination of J • Effective crack length – Evaluation of is required in the engineering method

• Effective crack length aeff=a + h rp (h not to be mistaken for the previous one)

• Recall in SSY, h =1/2 &

• For Large Scale Yielding this would lead to lengths larger than the ligament

– Correction of h:

– Use of the same plastic zone expression

2021-2022 Fracture Mechanics – NLFM – J-Integral 35 Fracture toughness testing for elastic-plastic materials • Procedure detailed in the ASTM E1820 norm • What is measured?

– Plane strain value of J prior to significant crack growth: JC – Plane strain value of J near the onset of stable crack growth: JIC – J vs Da resistance curve for stable crack growth: JR • Pre-cracked specimens that can be used – Either SENB or Compact Tension specimen Q, u

Thickness t W a Q, u v/2 v/2

L=4W h v/2 W

a v/2

Thickness t Q, u

2021-2022 Fracture Mechanics – NLFM – J-Integral 36 Fracture toughness testing for elastic-plastic materials

st • 1 step: Determine crack length a0 after Q, u fatigue loading Thickness t – 3 cycles of loading-unloading W a • Between and /2 with the Qfat Qfat Qfat v/2 v/2 maximal loading during pre-cracking L • Crack length is obtained using the tabulated compliance curves Ce(a) Q

• See lecture on Energetic Approach Qfat

1

Ce(a0 )

v

2021-2022 Fracture Mechanics – NLFM – J-Integral 37 Fracture toughness testing for elastic-plastic materials

st • 1 step: Determine crack length a0 after Q, u fatigue loading (2) Thickness t – Example: SENB W a • If v/2 v/2

L • Then

• In elasticity: C(a) = Ce(a) = v/Q

–

– In terms of Load Line displacement:

2021-2022 Fracture Mechanics – NLFM – J-Integral 38 Fracture toughness testing for elastic-plastic materials

• 2nd step: Proceed with the testing: increase loading – At regular intervals i: unload Q • In order to determine ai with the compliance a0 Qi • Do not unload too much in order to avoid reverse plastic loading 1 • At least 8 unloadings are required C(ai )

– After the final loading step • Unload to zero and • Mark final crack length v • Break open (cool down if required to have brittle fracture) • Measure final crack length

2021-2022 Fracture Mechanics – NLFM – J-Integral 39 Fracture toughness testing for elastic-plastic materials

• 3rd step: Data reduction

– For each pair (ai, Qi)

• Calculate Je,i : 2 퐾퐼 푎푖 – Using crack length 퐽 = 푒,푖 퐸′

– E.g. SENB:

• Calculate Jp,i : – Plastic displacement Q a0 Qi 1 – Increment in plastic work

CLLE(ai+1 )

– Since for SENB h-factor is equal to 2:

u Average ligament size up, i+1

2021-2022 Fracture Mechanics – NLFM – J-Integral 40 Fracture toughness testing for elastic-plastic materials

• 3rd step: Data reduction (2) Q a – For each pair (a , Q ) 0 i i Qi • Calculate J : e,i 1 2 퐾퐼 푎effi 퐽푒,푖 = 퐸′ C(ai )

• Calculate Jp,i (e.g. SENB)

v • Total 퐽 value: J

– This allows drawing

• The J vs. Da=a-a0 curve

• This requires accurate determination of a0

Da

2021-2022 Fracture Mechanics – NLFM – J-Integral 41 Fracture toughness testing for elastic-plastic materials

• 4th step: Analysis y – Even before physical crack growth ux uy there is a blunting of crack tip x • Due to plasticity r* • Which results in an apparent Da • Corrected by plotting the blunting line

JR Jblunting

Da

2021-2022 Fracture Mechanics – NLFM – J-Integral 42 Fracture toughness testing for elastic-plastic materials

• 4th step: Analysis (2) JR J – Using data points in between the blunting Jextrapolated 0.15 and 1.5-mm exclusion lines, 0.2mm • Extrapolate J : Δ푎 J ln 퐽 = ln 퐶 + 퐶 ln IC 1 2 1 푚푚 – Plane strain value of J near the onset of stable crack growth: JIC • Fracture toughness 0.15mm 1.5mm Da • Intersection with the 0.2mm offset line – Plane strain value of J prior to significant crack growth: JC • Intersection with the blunting line – J vs Da resistance curve for stable crack growth: JR

– Are sizes correct?

2021-2022 Fracture Mechanics – NLFM – J-Integral 43 Resistance curves • Experimental curves – Example: steel A533-B at 93°C* • Thickness has an important influence • Slight influence of the initial crack length

• Side grooving suppresses the plane stress state

]

]

2 2

1000 Jblunting a =116 mm m

t =25 mm m 0

/

/ kJ

kJ 1500 [

t =100 mm [ 퐼

800 퐼

퐽 퐽 a0 =134 mm 600 1000 a0 =146 mm t =100 mm, 400 side grooved 500 200

0 0 0 1 2 3 4 0 2 4 6 8 Δ푎 [mm] Δ푎 [mm] • From these curves one can extract

– The toughness JIC crack initiation criterion J (a, Q) = JIC

– The resistance curve JR(Da) crack stability criterion? *Andrews WR and Shih CF (1979), Thickness and side-groove effects on J- and δ-resistance curves for A533-B steel at 93◦C,ASTM STP 668 , 426–450. 2021-2022 Fracture Mechanics – NLFM – J-Integral 44 Reminder: LEFM crack grow stability

• Non-perfectly brittle material: Resistance curve

– For non-perfectly brittle materials GC depends on the crack surface

• Therefore GC will be renamed the resistance Rc (A) – Elastoplastic behavior

• Active plastic zone in Rc front of the crack tip Rc ducile t3<< (Plane s) • Plastic work increases Resistance increases Rc ducile t2

• A steady state can be reached Rc fragile in which case a crack propagates Active Plastic zone Plastic wake in a compressive plastic wake plastic zone 0 DA=tDa

– Stability of the crack also depends on the variation of G with crack length • Stable crack growth if • Unstable crack growth if

2021-2022 Fracture Mechanics – NLFM – J-Integral 45 Reminder: LEFM crack grow stability

• Example: Delamination of composites with initial crack a0 Dead load vs Fixed grip Q u h Thickness t h Thickness t a h a h

Rc, G Rc, G Q2 u3 Q3 u Q 2 1 Rc ducile Rc ducile u1 u increasing Q increasing GC brittle GC brittle

a a a0 a* a** a0

– Dead load Q1 – Fixed grip u1 ,u2 or u3 • Perfectly brittle materials: unstable • The crack is stable for any material • Ductile materials: stable, but if a is larger than a** it turns unstable

2021-2022 Fracture Mechanics – NLFM – J-Integral 46 J-controlled crack growth & NLFM crack grow stability J-controlled region • Under which conditions is the crack growth J-controlled? Elastic unloading – Recall that in order to use J da r** • Unloading is prohibited r* • So the solution is valid before crack growth x

– But we wish to use the JR curve • When is the crack growth still J-controlled? Non-proportional loading – HRR solution holds in an annular region • Since HRR solution is singular, there is a region not controlled by J • There is an annular region between r* and r** controlled by J

When can we use the HRR theory during crack growth?

2021-2022 Fracture Mechanics – NLFM – J-Integral 47 J-controlled crack growth & NLFM crack grow stability J-controlled region • Under which conditions is the crack growth J-controlled (2)? Elastic unloading – HRR solution holds in an annular region da r** • As HRR solution is singular, there is a region not controlled by J r* • There is an annular region between x r* and r** controlled by J – If the crack grows by da • There is an elastic unloading Non-proportional loading in a region that scales with da • For a potential to be defined: no unloading – So for J to remains relevant if the crack grows • One clear condition is therefore 푑푎 << 푟 ∗∗ • Another condition to keep asymptotic field proportionality (see next slides): 퐽 퐷 = 퐼퐶 ≪ 푟∗∗ 푑퐽 푑Δ푎

2021-2022 Fracture Mechanics – NLFM – J-Integral 48 J-controlled crack growth & NLFM crack grow stability • Under which conditions is the crack growth J-controlled (3)?

– HRR strain field (between r* and r**): – Since  depends on J and a: y’ • • The crack is moving to the right: r q x’ x Da – With

&

– So one has

with

2021-2022 Fracture Mechanics – NLFM – J-Integral 49 J-controlled crack growth & NLFM crack grow stability • Under which conditions is the crack growth J-controlled (4)?

– HRR strain field (between r* and r**): – Since  depends on J and a: •

• With

• &

– At the end of the day:

2021-2022 Fracture Mechanics – NLFM – J-Integral 50 J-controlled crack growth & NLFM crack grow stability • Under which conditions is the crack growth J-controlled (5)?

– We have

– Term is in 1/r it changes the proportionality of the solution

• But & are of the same order

• We have J-controlled crack growth if

– From the JR curves, the length scale

can be extracted as – The J-controlled growth criteria are • & • r** remains to be determined • Stability is not established yet

2021-2022 Fracture Mechanics – NLFM – J-Integral 51 J-controlled crack growth & NLFM crack grow stability • Crack growth stability – The J-controlled crack growth criteria are • & • r** is a fraction – Of the remaining ligament W-a

– Or other characteristic lengths (e.g. up to rp / 4) • For SENB: da < 6% (W-a) and D < 10% (W-a) (obtained by FE analysis) – If these criteria are satisfied the stability of the crack can be assessed • J depends on the crack length and loading • Toughness variation with Da is evaluated

from JR curves • Stable crack if

• In terms of the non-dimensional tearing:

2021-2022 Fracture Mechanics – NLFM – J-Integral 52 J-controlled crack growth & NLFM crack grow stability • Crack growth stability (2) – Tabulated values

0 Material Specimen T° (sp +sTS)/2 JIC dJ/da T D [°C] [MPa] [MPa.m] [MPa] [] [mm] ASTM 470 CT 149 621 0.084 48.5 25.5 1.78 (Cr-Mo-V) ASTM 470 CT 260 621 0.074 49 25.8 1.52 (Cr-Mo-V) ASTM 470 CT 427 592 0.088 84 48.6 1.01 (Cr-Mo-V) ASTM A453 CT 24 931 0.124 141 32.8 Stainless steel ASTM A453 CT 204 820 0.107 84 25.6 1.27 Stainless steel ASTM A453 CT 427 772 0.092 65 22.4 Stainless steel 6061-T651 CT 24 298.2 0.0178 3.45 2.79 5 Aluminum

2021-2022 Fracture Mechanics – NLFM – J-Integral 53 Steady State crack grow • Moving crack and elasto-plastic material R – A moving crack sees a c Initial crack Steady state growth plastic wake R t Blunting c ductile 1 • The nonlinear elasticity assumption (power law) is R not valid anymore c brittle Active • Ex: perfectly plastic material Plastic Plastic wake plastic zone – HRR 푛 → ∞ zone 0 DA=tDa

– Steady state • Singularity is weaker in the steady state, so the apparent J limit is larger

– JSS is the steady state limit of J

• = JIC for brittle materials

• Many times JIC for ductile materials – This is characterized by

2021-2022 Fracture Mechanics – NLFM – J-Integral 54 2-parameter theory • A single parameter cannot always fully described the process – Example 1: Thickness effect • In LEFM, recall T-stress is the 0-order term, 2a which is dominant at radius rc T>0 • In general, if T < 0, the measured fracture K K will be larger than for T > 0 rupture • If thick specimen T > 0 • HRR solution is also an asymptotic behavior K Plane s • Measure of J limit is also thickness-dependant c – Example 2: Large Yielding Plane  t • There is no zone of J-dominance HRR asymptotic yy -1/(n+1) • Plastic strain concentrations depend s syy in r n+1

on the experiment Log 1

2 LEFM 1 asymptotic -1/2 syy in r Log r

2021-2022 Fracture Mechanics – NLFM – J-Integral 55 Exercise 1: SENB specimen • Consider the SENB specimen Q, u

– Made of steel Thickness t 0 W • E = 210 GPa, sp = 700 MPa • Power law: =0.5, n=10 v/2 v/2 – Geometry L=4W • t = 25 mm, W = 75 mm, L = 300 mm • a = 28 mm . – Assuming a toughness of JIC = 200 kPa m, dJ/da = 87 MPa • Compute the maximal loading before fracture using LEFM • Compute the maximal loading before fracture using NLFM • Evaluate the crack growth stability

2021-2022 Fracture Mechanics – NLFM – J-Integral 59 Exercise 1: Solution • LEFM solution – SIF (handbook)

•

with

– Effective crack length •

with ?

& with (handbook)

2021-2022 Fracture Mechanics – NLFM – J-Integral 60 Exercise 1: Solution • LEFM solution

– Example Q=150 kN

• Starting point

• Yield load

2021-2022 Fracture Mechanics – NLFM – J-Integral 61 Exercise 1: Solution • LEFM solution (2) – Example Q=150 kN (2)

• First iteration

• Second iteration

2021-2022 Fracture Mechanics – NLFM – J-Integral 62 Exercise 1: Solution • LEFM solution (3) – Whole Q-range with

300 2 K • Or in terms of J = KI / E’ I K 250 C

] 200

1/2

150

[MPa . m . [MPa I 100

K

50

0 0 20 40 60 80 Q [kN]

• Rupture load in LEFM (with corrected h) is 186 kN

2021-2022 Fracture Mechanics – NLFM – J-Integral 63 Exercise 1: Solution • NLFM solution – Handbook

• Jp:

– Yield load:

– Coefficient:

2021-2022 Fracture Mechanics – NLFM – J-Integral 64 Exercise 1: Solution • NLFM solution (2) – Example Q = 150 kN:

– Whole range of Q: • Limit load 177.71 kN – Validity: • J dominance if

0 a, W-a > 25 J /sp = 4.6 mm OK Plane 휀? 푡 >> 4.6 mm • K dominance if

0 2 a, W-a > 2.5 (K /sp ) = 0.149 m KO

2021-2022 Fracture Mechanics – NLFM – J-Integral 65 Exercise 1: Solution • Crack growth stability – Region of J-controlled crack growth • Material length scale:

• For SENB, crack growth is J-controlled if D < 0.1 (W-a) = 4.7 mm OK

• Then, stability can be studied for Da < 0.06 (W-a) = 2.8 mm

– 2 studies:

• a = 28 mm: done

• a = 30 mm so Δ푎=2 mm< 2.8 mm: to do

2021-2022 Fracture Mechanics – NLFM – J-Integral 66 Exercise 1: Solution • Crack growth stability – 2 studies: • a = 28 mm: done

. Qrupt = 177.75 kN, J = 200 kPa m • Dead load: Q = 177.75 kN

but with a = 30 mm

J = 257.22 KPa.m

– Crack stability

• T for this geometry and dead load

• Tearing of the material

stable crack (for Da < 2.8 mm)

2021-2022 Fracture Mechanics – NLFM – J-Integral 67 Exercise 2

• Toughness evaluation Q, u/2 – Ductile material: steel – Follow the norm ASTM E1820 – Normalized specimen h • Compact Tension Specimen vm /2 W

2퐽퐼퐶 • Thick enough 푊 − 푎, 푡 > 25 0 a 휎푝+휎푇푆 vm /2 – Fracture test Thickness t

Q, u/2

Properties Values Half height ℎ 0.036 m Width 푊 0.06 m Thickness 푡 0.03 m Young 퐸 210 Gpa 0 Yield 휎 푝 600 Mpa Poisson 푣 [-] 0.3 Power law  [-] 1 Power law n [-] 3

2021-2022 Fracture Mechanics – NLFM – J-Integral 68 Exercise 2: Solution

st • 1 step: Determine crack length a0 after fatigue loading Q, u/2 – Compliance before crack growth onset • Check during unloading to have the elastic part

푢푒 1 h = = 0.001772 10−6 m/N 푄 564.32 106 vm /2 W

a vm /2

Thickness t

Q, u/2 Crack growth onset

Before crack growth

2021-2022 Fracture Mechanics – NLFM – J-Integral 69 Exercise 2: Solution

st • 1 step: Determine crack length a0 after fatigue loading Q, u/2 – Compliance • Calibration of the geometry following the norm (a correction for rotation should actually be introduced) h 1 푈 = 퐸푡푢 vm /2 W 1 + ,푒 푄 푎 a = 1.000196 − 4.06391푈 + 11.242푈2 − 106.043푈3 + vm /2 푊 464.335푈4 − 650.677푈5 • For this test Thickness t 푢 1 푒 = = 0.001772 mm/kN Q, u/2 푄 564.32 Properties Values 푈 = 0.23035 Half height ℎ 0.036 m Width 푊 0.06 m 푎0 = 0.2498 Thickness 푡 0.03 m 푊 Young 퐸 210 Gpa 0 Yield 휎 푝 600 Mpa 푎0 = 0.01499 m Poisson 푣 [-] 0.3 • Check size requirements Power law  [-] 1 Power law n [-] 3 푎0 > 1.3 mm, 푎0 > 0.05 푡 OK

2021-2022 Fracture Mechanics – NLFM – J-Integral 70 Exercise 2: Solution

nd • 2 step: Evaluate the couples (푎푖, 푄푖) – Compliance data:

• 푄1 = 172.75 kN 푢푒1 1 = 푎1 = 0.015 m 푄1 564.31 • 푄2 = 187.75 kN 푢푒2 1 = 푎2 = 0.015277 m 푄2 550.73 • …….

푄 푢 풖푒 푎 [mm] [kN] [mm] 푄 Crack growth onset [kN/mm] 172.75 0.3580 564.31 14.988 Before crack growth 187.75 0.4108 550.73 15.277 195.25 0.4411 542.35 15.460 202.75 0.4749 532.50 15.679 210.25 0.5136 520.63 15.950 217.75 0.5596 505.75 16.300 225.25 0.6187 485.56 16.796 232.75 0.7156 449.75 17.740

2021-2022 Fracture Mechanics – NLFM – J-Integral 71 Exercise 2: Solution

nd • 2 step: Evaluate the couples (푎푖, 푄푖) (2) – Elasto-plastic responses:

• 푄1 = 172.75 kN, 푢 = 0.3580 mm 푢푒1 1 = 푢푒 = 0.3061 mm 푢푝 = 푢1 − 푢푒 = 0.05189 mm 푄1 564.31 1 1 1 • 푄2 = 187.75 kN, 푢 = 0.4108 mm 푢푒2 1 = 푢푒 = 0.3409 mm 푢푝 = 푢2 − 푢푒 = 0.06992 mm 푄2 550.73 2 2 2 • …….

풖푒 푄 푢 푎 푢푒 푢푝 [kN] [mm] 푄 [mm] [mm] [kN/mm] [mm] 172.75 0.3580 564.31 14.988 0.3061 0.05189 187.75 0.4108 550.73 15.277 0.3409 0.06992 Crack growth onset 195.25 0.4411 542.35 15.460 0.3600 0.08107 202.75 0.4749 532.50 15.679 0.3808 0.09413 Before crack growth 210.25 0.5136 520.63 15.950 0.4038 0.1098 217.75 0.5596 505.75 16.300 0.4306 0.1291 225.25 0.6187 485.56 16.796 0.4639 0.1548 232.75 0.7156 449.75 17.740 0.5175 0.1981

2021-2022 Fracture Mechanics – NLFM – J-Integral 72 Exercise 2: Solution

• 3rd step: Data reduction – Elastic part – : 푎 푎 푎 2 푎 3 푎 4 2 + 0.886 + 4.64 − 13.32 +14.72 − 5.6 푄 푊 푊 푊 푊 푊 • 퐾퐼 = 1 3 푎 푡푊2 1 − 2 푊 • 푄1 = 172.75 kN, 푎 = 14.988 mm 푎 퐾 2 1 = 0.2498 퐾 = 115.71 MPa 푚 퐽 = 퐼1 = 58016 J ⋅ m−2 푊 퐼1 푒1 퐸′ • 푄2 = 187.75 kN, 푎 = 15.277 mm 푎 퐾 2 2 = 0.25462 퐾 = 127.41 MPa 푚 퐽 = 퐼2 = 70341 J ⋅ m−2 푊 퐼2 푒2 퐸′ • ……. Properties Values 푄 [kN] 푎 푢푒 퐾퐼 퐽푒 [mm] [mm] [Mpa m1/2] [J/m2] Half height ℎ 0.036 m 172.75 14.988 0.3061 115.71 58016 Width 푊 0.06 m 187.75 15.277 0.3409 127.41 70341 Thickness 푡 0.03 m 195.25 15.460 0.3600 133.59 77332 Young 퐸 210 Gpa 0 202.75 15.679 0.3808 140.09 85039 Yield 휎 푝 600 Mpa 210.25 15.950 0.4038 147.03 93679 Poisson 푣 [-] 0.3 217.75 16.300 0.4306 154.66 103649 Power law  [-] 1 225.25 16.796 0.4639 163.51 115856 232.75 17.740 0.5175 176.04 134291 Power law n [-] 3

2021-2022 Fracture Mechanics – NLFM – J-Integral 73 Exercise 2: Solution

• 3rd step: Data reduction – Plastic part – Plastic energy: 푝 • 푊1 =7.3564 J 푝 푄1+푄2 • Δ푊2 = = 3.2504 J 2 푢푝 2−푢푝 1 푝 푝 푝 푊2 = 푊1 + Δ푊2 = 10.6068 J

• ……. 푝 Δ푊2

푝 푝 푄 [kN] 푎 푢풑 Δ푊 푊 푝 [mm] [mm] [J] [J] 푊1 172.75 14.988 0.05189 - 7.3564 187.75 15.277 0.06992 3.2504 10.6068 195.25 15.460 0.08107 2.1350 12.7418 202.75 15.679 0.09413 2.5993 15.3411 210.25 15.950 0.1098 3.2250 18.5660 217.75 16.300 0.1291 4.1380 22.7040 225.25 16.796 0.1548 5.6875 28.3915 232.75 17.740 0.1981 9.9283 38.3198

2021-2022 Fracture Mechanics – NLFM – J-Integral 74 Exercise 2: Solution

• 3rd step: Data reduction – Plastic part (2) – J-integral by 휂: 휂 푊 − 푎0 2 + 0.522 푊푝 퐽 = 푊 1 −2 • 푝1 퐽 = 10776.06 J ⋅ m 푊 − 푎0 푡 푝1

푊 − 푎푖−1 2 + 0.522 Δ푊푝 푊 − 푎 푎 − 푎 퐽 = 퐽 + 푊 푖 1 − 1 + 0.76 푖−1 푖 푖−1 • 푝푖 푝푖−1 푊 − 푎푖−1 푡 푊 푊 − 푎푖−1

푎 = 15.277 mm, Δ푊푝 = 3.2504 J 퐽 = 15753 J ⋅ m−2 2 2 푝2 • …….

푄 [kN] 푎 [mm] 푢 Δ푊푝 푊푝 퐽 Properties Values 풑 풑 Half height ℎ 0.036 m [mm] [J] [J] [J/m2] 172.75 14.988 0.05189 - 7.3564 10776 Width 푊 0.06 m 187.75 15.277 0.06992 3.2504 10.6068 15753 Thickness 푡 0.03 m 195.25 15.460 0.08107 2.1350 12.7418 18978 Young 퐸 210 Gpa 0 202.75 15.679 0.09413 2.5993 15.3411 22866 Yield 휎 푝 600 Mpa 210.25 15.950 0.1098 3.2250 18.5660 27637 Poisson 푣 [-] 0.3 217.75 16.300 0.1291 4.1380 22.7040 33677 Power law  [-] 1 225.25 16.796 0.1548 5.6875 28.3915 41850 Power law n [-] 3 232.75 17.740 0.1981 9.9283 38.3198 55842

2021-2022 Fracture Mechanics – NLFM – J-Integral 75 Exercise 2: Solution

• 3rd step: Data reduction – Elasto-Plastic part – Total J-integral: • 퐽 = 퐽 +퐽 퐽 = 68792 J ⋅ m−2 1 푒1 푝1 1

• …..

2 푄 푎 [mm] Δ푎 퐽푒 퐽풑 퐽 [J/m ] [kN] [mm] [J/m2] [J/m2] 172.75 14.988 0 58016 10776 68792 187.75 15.277 0.2892 70341 15753 86094 195.25 15.460 0.4718 77332 18978 96310 202.75 15.679 0.6910 85039 22866 107906 210.25 15.950 0.9617 93679 27637 121316 217.75 16.300 0.1312 103649 33677 137326 225.25 16.796 0.1808 115856 41850 157706 232.75 17.740 0.2752 134291 55842 190133

2021-2022 Fracture Mechanics – NLFM – J-Integral 76 Exercise 2: Solution

• 4th step: Analysis – Check data: 0.5 mm off-set line

푄 푎 [mm] Δ푎 퐽 [J/m2] At least one data in 0 [kN] [mm] 퐽푏푙푢푛푡푖푛푔 = 2휎푝 Δ푎 each set 172.75 14.988 0 68792 187.75 15.277 0.2892 86094 195.25 15.460 0.4718 96310 202.75 15.679 0.6910 107906 210.25 15.950 0.9617 121316 1.5 mm exclusion line 217.75 16.300 0.1312 137326 225.25 16.796 0.1808 157706 232.75 17.740 0.2752 190133 0.15 mm exclusion line

2021-2022 Fracture Mechanics – NLFM – J-Integral 77 Exercise 2: Solution

• 4th step: Analysis – Extrapolate: • Use date between Δ푎 exclusions lines ln 퐽 = ln 퐶 + 퐶 ln 1 2 1 푚푚 – Plane strain value of J near the onset of stable

crack growth: JIC • Fracture toughness 퐽IC • Intersection with the 0.2 mm off-set line 0.2 mm offset line −2 • 퐽IC = 80 193 J ⋅ m

′ 퐾IC = 퐽IC퐸 = 136 MPa m

– Check validity

퐽퐼퐶 푡, 푊 − 푎 > 25 0 ≃ 3.3 mm 휎푝 휎 + 푇푆 2 2

2021-2022 Fracture Mechanics – NLFM – J-Integral 78 References • Lecture notes – Lecture Notes on Fracture Mechanics, Alan T. Zehnder, Cornell University, Ithaca, http://hdl.handle.net/1813/3075 • Handbook – An engineering approach for elastic-plastic fracture analysis V. Kumar, M.D. German, C.F. Shih, NP-1931 Research project 1237-1, 1981, General electric company, https://inis.iaea.org/search/search.aspx?orig_q=RN:12640849

• Other references – « on-line » • Fracture Mechanics, Piet Schreurs, TUe, http://www.mate.tue.nl/~piet/edu/frm/sht/bmsht.html – Book • Fracture Mechanics: Fundamentals and applications, D. T. Anderson. CRC press, 1991. • An Engineering Approach for Elastic-Plastic Fracture Analysis, V. Kumar, M.D. German, C.F. and Shih CF (1981), Tech. Rep. NP-1931, Electric Power Research Institute. • Fatigue of Materials, S. Suresh, Cambridge press, 1998. 2021-2022 Fracture Mechanics – NLFM – J-Integral 79