introduction to fracture mechanics: theory of linear

Introduction to Fracture Mechanics: Theory ofLinear Elastic Fracture

Crack tip stress fields, stress intensity, and energy releaserate with examples

S. J. Grutzik

Material Mechanics and Tribology (1851)

Sandia National Laboratories

Albuquerque, New Mexico USA

October 20, 2016 1

SAND2016-11008PE

Course outlineThis is the first lecture in an informal course covering basic fracture mechanics withlectures held monthly. Currently planned lectures include:

Basic Theory Part I, Scott Grutzik Linear elastic fracture criteria, stress intensityfactors, energy release rate

Experimental Fracture Mechanics, Jay Carroll Fracture testing specimens and fullfield measurements

Basic Theory Part II, Scott Grutzik Elastic plastic fracture, anisotropy, propagation

Fracture Resistant Design, Jay Carroll Fracture mechanisms, material selection,propgation

Computational Methods for Brittle Fracture, John Emery Driving force calculationsand meshing considerations with examples

Computational Approaches for Resolving the Driving Force and Resistance Domainintegral for J integral calculations and mtehods for simulationsresistance with case studies

The Materials Science of Fracture Origins of flaws, toughening mechanisms,ductile vs brittle, emerging material classes

Interfacial Fracture, Dave Reedy Features of interfacial fracture, measuringinterfacial toughness, cohesive zone models in FEA

October 20, 2016 2

Sandia fracture examples

Glass to metal seals

Brazed connections

Glass-ceramic microstructure

Photovoltaics

Distinguishing feature of fracture: inherent length scale. If size of detectable cracks

is reduced, fracture strength is increased

October 20, 2016 3

List of texts

Andersion: Fracture mechanics Good general reference, broad coverage

Broberg: Cracks and Fracture Abstract, mathematical approach to fracture

Janssen, Zuidema, Wanhill: Fracture Mechanics Engineering perspective, somediscussion of material mechanisms

Zehnder: Fracture Mechanics Equivalent to a one semester masters level course

Dowling: Mechanical Behavior of Materials Entry level discussion of fatigue

Broek: Elementary Engineering Fracture Mechanics Similar to Anderson, gooddiscussion of damage tolerance

October 20, 2016 4

Fracture of ideal crystal lattice Approximate tensile stress-strain as

half sine wave

σ = σtf sin

(π

2

ε

d/x0

) Impose the two conditions

dσ

dε= E, ε 1

2γ =

∫∞x0

σ(x)dx

Solving for d and σtf we get

σtf =

√Eγ

x0

Large E and γ and small x0 implieslarge σtf

Using the approximation d ≈ 2x0, weget σtf ≈ E

π and γ ≈ Ex0π2

σ

ϵ

σtf

d/x0-1

σ

x0

V

xx0

October 20, 2016 5

Inglis solution for elliptic hole in sheet (1913)1

Inglis computed the stress intensityfrom an elliptic hole in a plate

As a/b→∞ geometry becomescrack-like

Maximum stresses are:

σmax = σ0

(1 + 2

a

b

)σmax = σ0

(1 + 2

√a

ρ

)

where ρ = b2

a is the radius ofcurvature at the end of the ellipse

ab

ρ

σ0

1C.E. Inglis. “Stresses in a plate due to the presence of cracks and sharp corners”. In: Transactions of the Institutionof Naval Architects 55 (1913), pp. 219–230.

October 20, 2016 6

Griffith fracture criterion (1921)2

Criterion: Crack growth will occur if the potential energy decrease produced by anincrement of crack extension equals the energy needed to extend the crack that sameincrement.

σ = σf , if∂Π

∂a=∂W

∂a

For the elliptical crack

∂Π

∂a=πa2σ2

0

E∗, E∗ =

E, if plane stressE

1−ν2 , if plane strain

Apply to Griffith criterion

∂

∂a(Π−W) =

∂

∂a

(πa2σ2

0

E∗− 4γa

)= 0→ σf =

√2γE∗

πa

Using the earlier approximation γ ≈ Ex0π2 ,

σf ≈√

2E2x0

π3a≈ E

4

√x0

a

2A.A. Griffith. “The phenomena of rupture and flow in solids”. In: Philosophical Transactions of the Royal Society A221.163 (1921), pp. 582–593.

October 20, 2016 7

Orowan model

Assume very sharp elliptic hole, ρ ≈ x0

Flaw will propagate when σmax = σtf

σf

(1 + 2

√a

ρ

)=

√Eγ

x0

For very sharp flaw, a/ρ 1, assume γ ≈ Ex0π2

σf ≈E

2π

√x0

a

Recall Griffith criterion gave σf ≈√

2E2x0π3a

≈ E4

√x0a

October 20, 2016 8

2D elasticity: governing equations

Equilibrium −→ σij,j = 0

Strain-Displacement −→ εij =1

2(ui,j + uj,i)

Hooke’s Law −→ εij =1 + ν

Eσij −

ν

Eδijσkk

Boundary conditions: Must specify either t or u at every surface pointAlso require:

2D −→ σ33 = 0 or ε33 = 0 and σ13 = σ23 = 0 and ε13 = ε23 = 0

Compatibility −→ ∂2ε11

∂x22

+∂2ε22

∂x21

= 2∂2ε12

∂x1∂x2

October 20, 2016 9

2D elasticity: Airy stress function

Define φ such that

σ11 =∂2φ

∂x22

σ22 =∂2φ

∂x21

σ12 = −∂2φ

∂x1∂x2

Observe this satisfies equilibrium, σ11,1 + σ12,2 = 0 and σ12,1 + σ22,2 = 0.Now insert these and Hooke’s Law into the compatibility equation to get

∇2∇2φ = ∇4φ = 0

where ∇2 is the Laplacian operator. φ is biharmonic.In polar coordinates,

∇2 =1

r

∂

∂r

(r∂

∂r

)+

1

r2

∂2

∂θ2=∂2

∂r2+

1

r

∂

∂r+

1

r2

∂2

∂θ2

and

σrr =1

r

∂φ

∂r+

1

r2

∂2φ

∂θ2σθθ =

∂2φ

∂r2σrθ = −

∂

∂r

(1

r

∂φ

∂θ

)

October 20, 2016 10

Crack tip stress fields

Consider traction free crack at originextending to the left

Stresses should go like rλ, let φ = rλ+2f(θ)

x

y

rσθθ(r,±𝜋)=σrθ(r,±𝜋)=0

θ

∇2φ = (λ+ 2)2 rλf(θ) + rλf ′′(θ)

∇4φ = rλ−2[f(4)(θ) +

((λ+ 2)2 + λ2

)f(2)(θ) + (λ+ 2)2 λ2f(θ)

]= 0

rλ+2 6= 0∀r so we have an ODE in f(θ). Substitute the trial solution f(θ) = eαθ to get

α4 +[(λ+ 2)2 + λ2

]α2 + (λ+ 2)λ2 = 0[

α2 + (λ+ 2)2] (α2 + λ2

)= 0

So α = ±iλ or α = ±i (λ+ 2),

f(θ) = B1eiλθ +B2e

−iλθ +B3ei(λ+2)θ +B4e

−i(λ+2)θ

or, equivalently

f(θ) = C1 cosλθ+C2 sinλθ+C3 cos (λ+ 2)θ+C4 sin (λ+ 2)θ

October 20, 2016 11

Crack tip stress fields

Consider traction free crack at originextending to the left

Stresses should go like rλ, let φ = rλ+2f(θ)

x

y

rσθθ(r,±𝜋)=σrθ(r,±𝜋)=0

θ

∇2φ = (λ+ 2)2 rλf(θ) + rλf ′′(θ)

∇4φ = rλ−2[f(4)(θ) +

((λ+ 2)2 + λ2

)f(2)(θ) + (λ+ 2)2 λ2f(θ)

]= 0

This gives

φ(r,θ) = rλ+2 [C1 cosλθ+C2 sinλθ+C3 cos (λ+ 2)θ+C4 sin (λ+ 2)θ]

Enforcing σθθ(r,±π) = φ,rr = 0 and σrθ(r,±π) = −( 1rφ,θ),r = 0 implies

f(±π) = f ′(±π) = 0.cosλπ sinλπ cos (λ+ 2)π sin (λ+ 2)πcosλπ − sinλπ cos (λ+ 2)π − sin (λ+ 2)πλ sinλπ λ cosλπ (λ+ 2) sin (λ+ 2)π (λ+ 2) cos (λ+ 2)π−λ sinλπ λ cosλπ −(λ+ 2) sin (λ+ 2)π (λ+ 2) cos (λ+ 2)π

C1

C2

C3

C4

=

0000

For non-trivial solution, det [Λ] = 4 sin2 (2λπ) = 0 −→ λ = n/2, n ∈ Z.

October 20, 2016 12

Crack tip stress fieldsInsert λ = n/2, n ∈ I into f (±π) = f ′ (±π) = 0 to get

C1 = −C3,λC2 = −(λ+ 2)C4 for λ integer

C2 = −C4,λC1 = −(λ+ 2)C3 for λ half integer

C1 = C3 = 0, for λ = 0

or φ(r,θ;λ) = rλ+2f(θ;λ) with

f(θ;λ) =

C1 [cosλθ− cos (λ+ 2)θ] +C3 [λ sinλθ− (λ+ 2) sin (λ+ 2)θ] , λ integer

C1 [λ cosλθ− (λ+ 2) cos (λ+ 2)θ] +C3 [sinλθ− sin (λ+ 2)θ] , λ half integer

How singular are stress fields? −→ Require λ > −1

Strain energy must be bounded

Displacement must be bounded

Uniqueness of solution lost if λ < −1 terms included

Singularity of elliptic hole as ab →∞ is r−1/2

Crack tip fields must be less singular than with a point load at the crack tip

σij =

∞∑n=−1

Cnijrn/2gnij(θ), ui =

∞∑n=−1

Dni rn/2+1hni (θ)

October 20, 2016 13

K field (λ = −1/2 term)

Cartesianσxxσyyσxy

I

=KI√2πr

cos θ2

(1 − sin θ

2sin 3θ

2

)cos θ

2

(1 + sin θ

2sin 3θ

2

)sin θ

2cos θ

2cos 3θ

2

σxxσyyσxy

II

=KII√2πr

sin θ2

(2 + cos θ

2cos 3θ

2

)sin θ

2cos θ

2cos 3θ

2

cos θ2

(1 − sin θ

2sin 3θ

2

)

(σxzσyz

)III

=KIII√

2πr

(sin θ

2

cos θ2

)

Polarσrrσθθσrθ

I

=KI√2πr

54

cos θ2− 1

4cos 3θ

234

cos θ2+ 1

4cos 3θ

214

sin θ2+ 1

4sin 3θ

2

σrrσθθσrθ

II

=KII√2πr

− 54

sin θ2+ 3

4sin 3θ

2

− 34

sin θ2− 3

4sin 3θ

214

cos θ2+ 3

4cos 3θ

2

(σrzσθz

)III

=KIII√

2πr

(sin θ

2

cos θ2

)

Note: Mode III stress and displacement field are not solved by the Airy method.

Since there is only one displacement term to be found, a functional form

u3 = rλ+1h(θ) is assumed and traction free boundary conditions at the crack faces

are enforced.

October 20, 2016 14

K field stresses

−1

−0.5

0

0.5

1

−π −π2 0 π

2 π −π −π2 0 π

2 π −π −π2 0 π

2 π

σ√r/K

i

θ

Mode I

σxxσyyσxy

θ

Mode II

σxxσyyσxy

θ

Mode III

σxzσxy

Max

shearstress

October 20, 2016 15

K field displacementsComputing displacements can be done by converting stress fields to strains viaHooke’s law and then integrating or from Coker and Filon’s3 relation thatdisplacement is related to the Airy function in polar coordinates by

2µur = −∂φ

∂r+ (1 − ν) r

∂ψ

∂θ

2µuθ = −1

r

∂φ

∂θ+ (1 − ν) r2 ∂ψ

∂r

where ∇2φ = ∂∂r

(r∂ψ∂θ

), µ is shear modulus, and ν = ν for plane strain and

ν = ν/(ν+ 1) for plane stress.(uruθ

)= KI

1 + ν

E

√r

2π

( (52− 4ν

)cos θ

2− 1

2cos 3θ

2

−(

82− 4ν

)sin θ

2+ 1

2sin 3θ

2

)+

KII1 + ν

E

√r

2π

(−(

52− 4ν

)sin θ

2+ 3

2sin 3θ

2

−(

72− 4ν

)cos θ

2+ 3

2cos 3θ

2

)uz =

KIII

µ

√2r

πsinθ

2

3E. G. Coker and L. N. G. Filon. A Treatise on Photo-Elasticity. Cambridge University Press, 1931.

October 20, 2016 16

T field (λ = 0 term)

The next term in the Williams Expansion after the K field is called T stress andλ = 0.

φ (θ; λ = 0) = r2C1 (1 − cos 2θ) − 2C3 sin 2θ

Cartesian

σxx = T

σyy = 0

σxy = 0

Polar

σrr =1

2T (1 − cos 2θ)

σθθ =1

2T (1 + cos 2θ)

σrθ =1

2T sin 2θ

T term is a independent of r and so is a uniform stress. The only component of auniform stress field that doesn’t violate the BCs is a constant σxx which matcheswith the above.

T -stress is commonly used in plastic fracture and in crack kinking analysis.

October 20, 2016 17

KI as fracture criterion (Irwin/Rice)

KI, KII, and KIII fully determine the most singular part of the stress field around acrack tip. Linear elastic, isotropic materials have lowest toughness when loaded intension.

→ Use KI as a fracture criterion

Irwin/Rice criterion: Material fractures when KI = KIC.

KIC is fracture toughness and is a material property. KI is a function of far fieldgeometry and loading.

October 20, 2016 18

K1 as a fracture criterion (Irwin/Rice)

In the examples below, in which mode is each crack loaded? Whichcrack affects structural integrity the most?

October 20, 2016 19

Calculating K with tables

Roark’s Formulas for Stress and Strain,Marks Handbook for MechanicalEngineers, etc. all have tables of Kformulae

Very good strategy for simple, “back ofthe envelope” type calculations

Can superpose between multiple crackloading scenarios

= +

October 20, 2016 20

Calculating K with tables: formulae4

Interior crack inplate

KI = σ0

√πa

Edge crack inplate

KI = 1.122σ0

√πa

Concentratedforces on crackface

KI(a) =P√πa

√a+ c

a− c

4W. C. Young, R. G. Budynas, and A. M. Sadegh. Roark’s Formulas for Stress and Strain. 8th ed. McGraw-Hill,2012.

October 20, 2016 21


Arbitrary pres-sure on crackface

KI(a) =1√πa

∫a−a

p(x)

√a+ x

a− xdx

Edge crack inuniformly loadedstrip

KI = σ0

√πaf (a/b) , a/b . 0.6

f (a/b) =1√π

(1.99 − 0.41

a

b+

18.70a2

b2− 38.48

a3

b3+ 53.85

a4

b4

)


October 20, 2016 22


Symmetric edgecracks in strip

KI = σ0

√πaf (a/b) , a/b . 0.7

f (a/b) =1√π

(1.98 + 0.36

a

b−

2.12a2

b2+ 3.42

a3

b3

)

Center crack instrip

KI = σ0

√πa√

cos(πa/2b)


October 20, 2016 23


Interior tensileloaded pennycrack

KI =2

πσ0

√πa

Tensile loadedhalf penny edgecrack

KI = 1.1222

πσ0

√πa


October 20, 2016 24

Calculating K with tables: Example Consider a thin-walled circular pressure vessel of diameter d, wall thickness t, and

pressure P with a crack of length a . 0.6t on inside surface

Stress in vessel wall is σ = Pd2t

KI for edge crack in a strip of width b is KI = σ√πaf(a/b) , f(a/b) =

1√π

(1.99 − 0.41ab + 18.70a

2

b2 − 38.48a3

b3 + 53.85a4

b4

), a/b . 0.6

KI from hoop stress is KI =Pd2t

√πaf(a/t)

KI from pressurized crack faces is KI = P√πaf(a/t)

Total KI is then KI = P(1 + d

2t

)√πaf(a/t)

Using a minimum detectable crack length we can use this to define a maximum

operating pressure Pmax = KIC

(1+ d2t )√πaminf(amin/t)

P

d

a

t

October 20, 2016 25

Energy Release Rate: Crack closure integralEnergy release rate: Strain energy released per unit length of crack advancement

G = −∂Π

∂a

G can be found starting with the principle of virtual work.∫S

tiδui dS =

∫V

σijδεij dV =

∫V

δW dW = δΠ

The elastic solution is given by σ0, ε0, and u0. Now let the crack grow so thatV → V − ∆V and S→ S+ ∆S. The elastic solution goes to σ0 + ∆σ, ε0 + ∆ε, andu0 + ∆u. Π→ Π+ ∆Π where ∆Π can be written as

∆Π =

∫V−∆V

W(ε0 + ∆ε

)dV −

∫V

W(ε0)dV

−

∫S+∆S

(t0i + ∆ti

) (u0i + ∆ui

)dS+

∫S

t0iu

0i dS

For a 2D crack, ∆S = 2∆a and G = lima→0

−∆G

∆aAfter some manipulations, we find

G = −∂Π

∂a= − lim

∆a→0

1

∆a

∫∆a

1

2t0i (∆u

+i − ∆u−

i ) dS

In this context, t0 can be interpreted as the traction required to keep the new crack

surface closed.October 20, 2016 26

Energy Release Rate: Relation to K field

G = −∂Π

∂a= − lim

∆a→0

1

a

∫∆a

1

2t0i (∆u

+i − ∆u−

i ) dS

For a crack growing straight ahead, ∆u+i = ∆u−

i so ∆u+i − ∆u−

i = 2ui (∆a− x1,π)and

G = lim∆a→0

1

∆a

∫∆a0

σi2 (x1, 0)ui (∆a− x1,π) dx1

where σij and ui are now defined in terms of K field. Substituting in K fieldsolution, we find

G =1

E∗

(K2

I + K2II

)+K2

III

2µ

G can be used as an alternate form of the Irwin/Rice fracture criterion:

GC = K2IC/E

∗ , E∗ =

E, if plane stressE

1−ν2 , if plane strain

October 20, 2016 27

Calculating G: compliance method

Principle of mimum potential energy requires that strain energy equals potentialenergy of external loads

Π =1

2

∫Ω

σijSijklσkl dA =1

2FextU =

1

2F2

extΓ(a) , Fext = Γ(a)U

If compliance of external loads are known, use this to find G = ∂Π∂a

= 12F2

ext∂Γ∂a

P

2h

a δP =a3

EI

P

t→ Γ(a) =

a3

EIt=

12a3

Et2h3

G = 2

(1

2P2 ∂Γ

∂a

)=

36a2P2

Et2h3

KI =√EG =

6aP

t√h3

October 20, 2016 28

Calculating G: compliance methodMoment loaded DCB gives G independent of crack length

M

2h

a

θM =a

EI

M

t→ Γ(a) =

a

EIt=

12a

Et2h3

G = 2

(1

2M2 ∂Γ

∂a

)=

12M2

Et2h3

KI =√EG =

2√

3M

t√h3

G independent of a makes the moment loadedDCB a useful experimental specimen

October 20, 2016 29

J Integral

G can be shown to be calculated by the J integral,

Jdef=

∫Γ1

(Wdx2 − ti

∂ui

∂x1dΓ

)=

∫Γ1

(Wn1 − niσij

∂uj

∂x1

)dΓ

x1

x2

Γ1

Γ4

Γ3

Γ2

Consider Γ = Γ1 ∪ Γ2 ∪ Γ3 ∪ Γ4

J =

∫Γ

(Wδ1i − σij

uj

x1

)nj dΓ

=

∫A

∂

∂xi

(Wδ1i − σij

uj

x1

)dA

=

∫A

(∂W

∂xi−∂σij

∂xi

∂uj

∂x1− σij

∂2uj

∂xi∂x1

)dA

∂σij

∂xi= 0 σij

∂2uj

∂xi∂x1= σij

∂εij

∂x1

J =

∫A

(∂W

∂x1− σij

∂εij

∂x1

)dA

σij =∂W

∂εij

∴ J = 0 if Γ is a closed path

October 20, 2016 30

J Integral

J =

∫Γ1

(Wdx2 − ti

∂ui

∂x1dΓ

)=

∫Γ1

(Wn1 − niσij

∂uj

∂x1

)dΓ

x1

x2

Γ1

Γ4

Γ3

Γ2

n1 ds = 0∀ds ∈ Γ2, Γ4

ti ds = σijnj ds = 0∀ds ∈ Γ2, Γ4

→ JΓ2

= JΓ4= 0

JΓ=Γ1∪Γ2∪Γ3∪Γ4= JΓ1

+ JΓ2+ JΓ3

+ JΓ4= 0 and JΓ2

= JΓ4= 0

∴ JΓ1= −JΓ3

J = G for any path starting on the bottom crack face and terminating on the topcrack face

October 20, 2016 31

J Integral: DCB Example

J =

∫Γ1

(Wdx2 − ti

∂ui

∂x1dΓ

)=

∫Γ1

(Wn1 − niσij

∂uj

∂x1

)dΓ

P

2h

a

Γ1 Γ2

Γ3

Γ4Γ5

n1 and ti = 0 on Γ2, Γ4

σij = 0 on Γ3 → W and ti = 0 on Γ3

→ JΓ2

= JΓ3= JΓ4

= 0

October 20, 2016 32

Questions?

Sandia National Laboratories is a multi-mission laboratory managedand operated by Sandia Corporation, a wholly owned subsidiary ofLockheed Martin Corporation, for the U.S. Department of Energy’sNational Nuclear Security Administration under contractDE-AC-4-04AL8500.

October 20, 2016 33

introduction to fracture mechanics: theory of linear

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