Introduction to Fracture Mechanics: Theory of Linear ...
Transcript of Introduction to Fracture Mechanics: Theory of Linear ...
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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
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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
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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
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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
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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
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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.
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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.
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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
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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
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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
)
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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
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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.
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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)
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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.
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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
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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.
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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.
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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.
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K1 as a fracture criterion (Irwin/Rice)
In the examples below, in which mode is each crack loaded? Whichcrack affects structural integrity the most?
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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
= +
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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.
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Calculating K with tables: formulae5
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
)
5W. C. Young, R. G. Budynas, and A. M. Sadegh. Roark’s Formulas for Stress and Strain. 8th ed. McGraw-Hill,2012.
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Calculating K with tables: formulae6
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)
6W. C. Young, R. G. Budynas, and A. M. Sadegh. Roark’s Formulas for Stress and Strain. 8th ed. McGraw-Hill,2012.
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Calculating K with tables: formulae7
Interior tensileloaded pennycrack
KI =2
πσ0
√πa
Tensile loadedhalf penny edgecrack
KI = 1.1222
πσ0
√πa
7W. C. Young, R. G. Budynas, and A. M. Sadegh. Roark’s Formulas for Stress and Strain. 8th ed. McGraw-Hill,2012.
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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