Institute of Mechanics and Fluid Dynamics (IMFD)Chair of Applied Mechanics – Solid Mechanics

Prof. Dr. rer. nat. habil. Meinhard Kuna

ExercisesFracture Mechanics Computations

Summer Term 2016

The present exercise book is published as supplementary material for the lecture "‘Frac-ture Mechanics Computations"’ taught at the Institute of Mechanics and Fluid Dynamicsat TU Bergakademie Freiberg.

All rights of this publication are reserved. Duplication and distribution are only allowedwith the prior written permission of the publisher.

Lecturer in charge: Prof. Dr. Meinhard KunaEditors: Dr. Matthias Scherzer,

Dr. Geralf Hütter,Lutz Zybell,Dr. Frank Rabold,Dr. Michael Budnitzki

1. edition entitled "Bruch- und Schädigungsmechanik"2. upgraded and extended edition March 20113. english edition March 20124. edition with new corporate design April 20145. error-corrected edition March 2015

3

Part I.Problems

1. Linear elastic fracture mechanics

1.1. A long metal strip of width b containing a centered through-thickness crack of length2a is symmetrically loaded at infinity with the traction σ∞. The strip breaks alongthe crack ligaments when σ∞ = σ∞c is reached. Calculate the critical stress intensityfactor KIc for the given values σ∞c = 215.28 MPa, a = 5 cm and b = 20 cm. Thestress intensity factor KI is defined by

KI = σ∞√πaF (ζ) , F (ζ) =

(1− 0.5ζ + 0.37ζ2 − 0.044ζ3)√1− ζ

, ζ =2a

b.

1.2. The strength of a silicon wafer strip (rectangular cross-section with widthB = 40 mmand height h = 0.6 mm) is tested in a four point bending experiment (Figure 1a,L = 37, 5 mm). The specimen breaks at a load of F = Fc = 90 N.

a) Determine the length c of the surface defect leading to fracture. Use the modelof a surface flaw (Figure 1b), with the stress intensity factor being given by

KI = σ√πc · Y , Y = 1.12 .

The fracture toughness of silicon is given by KIc = 1.2 MPa√

m.

b) Calculate the critical load for fracture, assuming that by means of previousmechanical polishing there are not any surface flaws larger than cpol = 50 nm.

F

4L

h2L

(a)

c

σ σ

(b)

Figure 1: (a) Four point bending specimen (b) surface flaw

4 Part I. Problems

1.3. A cylindrical pressure vessel with radius R contains a through thickness crack oflength 2a (2a� R) far away from the hemisphere closings. The crack orientation isslanted by the angle β with respect to the circumferential direction of the cylinder(see Figure 2). Furthermore, the wall thickness of the pressure vessel is substantiallysmaller than its radius (t� R).

Calculate the stress intensity factors KI and KII under the given assumptions, if thevessel is loaded by an internal pressure p.

σᵩ

σᵩ

β

p

R

t

β

2a

σzσz

σz

Figure 2: Cylindrical pressure vessel with crack

1.4. A three point bending specimen with dimensions W = B = 10 mm and a crack oflength a = 3 mm is symmetrically loaded by the concentrated force F (see Figure3). The specimen breaks at a load of F = Fc = 1466 N.

Calculate the fracture toughness KIc of the tested material for the given definitionof the stress intensity factor for a three point bending specimen

KI =3S F

2BW 2

√π a · Φ

( aW

)

Φ( aW

)=

1.99− α(1− α)(2.15− 3.93α + 2.7α2)

(1 + 2α)(1− α)3/2.

F

a

S=4W

Wthickness B

Figure 3: Three point bending specimen with crack

1. Linear elastic fracture mechanics 5

1.5. The stress intensity factor KI for a Griffith crack of length 2a in an infinite elasticplate under tension loading σ∞ is calculated by KI = σ∞

√π a. If the same crack

is symmetrically loaded by a pair of concentrated forces F (dimension: force perthickness) instead of tractions σ∞ (see Figure 4), the stress intensity factor KI isgiven by

KI =2F√π a

.

F

F

a0 a0

Figure 4: Griffith crack loaded by two concentrated forces

a) Calculate the critical crack lengths ac for a given critical stress intensity factorKIc for both loading conditions.

b) What can be stated regarding crack stability after KI has exceeded KIc forboth loading conditions ?

1.6. A Griffith crack of length 2a is loaded by a couple of concentrated forces F(dimension: force per thickness) located at x = b (see Figure 5a).

2a

y

x

bF

F

(a)

2a

y

x

bF

F

bF

F

(b)

Figure 5: Griffith crack loaded by concentrated forces

6 Part I. Problems

a) Verify that the Westergaard stress function

Z(z) =F

π (z − b)

(a2 − b2

z2 − a2

)1/2

can be used to solve the corresponding elastic boundary value problem.

b) Prove that the stress intensity factor at the crack tip x = a is given by

KI = limr→0

√2πr σ22(x = a+ r, y = 0) =

F

(π a)1/2

(a+ b

a− b

)1/2

.

c) Show that in case of two symmetrical concentrated force couples (see Figure5b) the Westergaard stress function and the stress intensity factor can becalculated by

Z(z) =2F z

π (z2 − b2)

(a2 − b2

z2 − a2

)1/2

, KI =2F

(a2 − b2)1/2

(aπ

)1/2

.

Use the results obtained in part b).

Hint : The complex Westergaard stress function is related to the Kolosov-Muschelischwili stress function by:

Φ′(z) =1

2Z(z), χ′′(z) = −1

2z Z ′(z) .

1.7. The linear elastic boundary value problem at a crack tip in polar coordinates (r, θ)(see Figure 6) is to be evaluated using series expansion . A separation ansatz

F = rλ+1h(θ)

with a still unknown function h(θ) is chosen for the Airy stress function.

Pr

θ

y

x

Figure 6: Polar coordinates at the crack tip

1. Linear elastic fracture mechanics 7

Calculate the exponent λ by following these steps:

a) Formulate the boundary conditions for the crack surfaces.

b) Introduce the given separation ansatz in the differential equation of the bound-ary value problem ∆∆F = 0. The Laplace operator ∆(·) in polar coordinatesis given by

∆(·) =∂2(·)∂r2

+1

r

∂(·)∂r

+1

r2

∂2(·)∂θ2

.

Solve the resulting ordinary differential equation for the unknown functionsh(θ). Regard only the even part h(−θ) = h(θ) of the function (correspondingto mode I) in the following.

c) Calculate the corresponding stresses

σrr =1

r

∂F

∂r+

1

r2

∂2F

∂θ2

σθθ =∂2F

∂r2

τrθ = − ∂

∂r

(1

r

∂F

∂θ

).

d) Determine the exponent λ from the condition that the obtained stresses fulfillthe boundary conditions of the problem and that they are nontrivial, i.e. dif-ferent from zero (→ eigenvalue problem).

e) Addition: Repeat c) and d) for the uneven part h(−θ) = −h(θ) of the functioncorresponding to mode II.

Hint: In order to obtain finite displacements as well as a finite stored strain energy,only solutions satisfying λ ≥ 0 are admissible.

1.8. The eigenfunctions at the crack tip can be determined by using the complex Kolosov-Muschelischwili stress functions (equivalent to problem 1.7). The correspondingansatz is

Φ(z) = A · zλ+2, χ′(z) = B · zλ+2, z = x+ iy .

a) Formulate the boundary conditions for the crack surfaces.

8 Part I. Problems

b) Determine the stresses corresponding to the given ansatz by using the followingrelation:

σyy + iτxy = 2Re [Φ′(z)] + zΦ′′(z) + χ′′(z) .

c) Use the Euler relation z = r eiθ in order to introduce polar coordinates.

d) Calculate the exponent λ from the condition, that the calculated stresses fulfillthe boundary conditions from part a) and that they are nevertheless nontrivial,i.e. different from zero (→ eigenvalue problem).

The hint in problem 1.7 applies.

1.9. A strip specimen of width b containing a surface crack of length a is loaded by alinear pressure distribution (see Figure 7). Calculate the stress intensity factor KI

at the crack tip for a given ratio of a/b = 0.4.

a

b

σ σ

Figure 7: Strip specimen containing a surface flaw

Hint : The stress intensity factors for a surface flaw in a strip specimen loaded underpure tension or pure bending, respectively, can be calculated for a/b < 0.6 by

KtensI = σ0

√π a

[1.12− 0.23

(ab

)+ 10.55

(ab

)2

− 21.72(ab

)3

+ 30.39(ab

)4],

KbendI =

6M0

b2

√π a

[1.12− 1.40

(ab

)+ 7.33

(ab

)2

− 13.08(ab

)3

+ 14.0(ab

)4],

where σ0 is the tension stress andM0 is the thickness-related bending moment.

1.10. A crack of length 2a in an infinite plate (a so-called Griffith crack) is loaded by adistributed crack surface load with amplitude σ, see Figure 8. Determine the stressintensity factor KI at the crack tip by using the weight function method.

Start from the given displacement solution for this configuration under constant

1. Linear elastic fracture mechanics 9

2a

σx

y

σ

σ σ

Figure 8: Griffith crack with linear crack surface load

tension load σ∞ at infinity:

u(2)x

(x, y = 0±, l = 2a

)= − 1

E ′σ∞x

u(2)y

(x, y = 0±, l = 2a

)= ± 2

E ′σ∞

√l2

4− x2

with E ′ =

E for plane stressE

1− ν2for plane strain

K(2)I = σ∞

√πa .

1.11. The maximum admissible internal pressure of a thick-walled pressure vessel withinner radius ri = 700 mm and wall thickness t = 150 mm (see Figure 9) shall bedetermined.

First, calculate the radial distribution of the circumferential stresses σθθ inducedby the internal pressure p for the pressure vessel not containing any crack. Then,determine the admissible internal pressure by means of the following models:

a) Assessment by the maximum principal stress criterion for an admissible equiv-alent stress of σadm = 400 MPa

b) By legal provision, the strength assessment of pressure vessels requires thefracture mechanical assessment of the structure containing a hypothetic half-elliptical surface flaw in axial direction with depth b = t/4 and length 2a = t.

Evaluate the stress intensity factor at the deepest point C of such a crack by

10 Part I. Problems

A

A

tr

p

2a

b

A-A

i C

Figure 9: Thick-walled pressure vessel containing a half-elliptic surface flaw

using the handbook solution (see Figure 12 on page 13) assuming pure tensionwith an amplitude equal to the circumferential stress at the inner surface of thepressure vessel. The fracture toughness of the material is KIc = 50 MPa

√m.

c) Extend the model from part b) by using a linearized stress function (normalizedby the pressure at the inner surface) as superposition of tension and bendingstresses.

Assess the results critically with respect to the following questions:

1. Is it useful to include further terms of the Taylor-series than the first bendingterms?

2. Is it necessary to calculate the stress intensity factors by FEM?

1.12. During the growth of a void in an elastic medium under tension loading strain energyis dissipated due to surface extension. The growth of a spherical void of radius a bythe extension ∆a in an infinite isotropic elastic body should be investigated in thefollowing, see Figure 10.

Derive the Griffith equation1 for this case by calculating the limit load σ = σc.Take into account:

a) the elastic strain energy density under uniaxial tension (in terms of stress),

b) the difference in strain energy density between an infinite elastic body undertension load, containing a spherical void of radius a and the undamaged body

1The energy release rate is proportional to σ2a

1. Linear elastic fracture mechanics 11

aΔa

σ σ

Figure 10: Growth of a spherical void

(without void),

c) the change in strain energy density due to extension of the void by ∆a� a,

d) the energy balance of this extension, if the dissipated elastic strain energy istransformed into surface energy with the density γ ([γ] = MPa ·m).

Hint: lim∆x→0

(x+ ∆x)n − xn

∆x= n · xn−1

1.13. Calculate the energy release rate during crack propagation in a double cantileverbeam (DCB) specimen as depicted in Figure 11. The given condition h� a allowsthe application of beam theory.

F

F

d

h

a

thickness b

Figure 11: Double cantilever beam specimen

12 Part I. Problems

1.14. Calculate the energy release rate for a crack in an isotropic linear elastic materialloaded in mode III by using the J-integral. The near field solution at the crack tipis given by:

u3 = 2KIII

µ

√r

2πsin

θ

2{τ13

τ23

}=

KIII√2πr

{− sin θ

2

+ cos θ2

}.

Hints :

The transformation from cartesian (x, y) into polar coordinates (r, θ) has to beconsidered during differentation.

∂(·)∂x

=∂(·)∂r

∂r

∂x+∂(·)∂θ

∂θ

∂x

∂(·)∂y

=∂(·)∂r

∂r

∂y+∂(·)∂θ

∂θ

∂y

Useful relations between trigonometric functions:

sin(−φ) = − sinφ

sin2 φ+ cos2 φ = 1

sin(φ− ζ) = sinφ cos ζ − cosφ sin ζ

1. Linear elastic fracture mechanics 13

Figure 12: Extract from Murakami et al. (Ed.): “Stress Intensity Factors”

2. Elastic plastic fracture mechanics 15

2. Elastic plastic fracture mechanics

2.1. Find a functional representation for the two-dimensional extension of the plasticzone rp involving only the yield stress σy, the stress intensity factor KI and dimen-sionless constants.

2.2. The fracture toughness of an engineering steel S235 (yield stress 235 MPa) shall bedetermined at room temperature by using a compact tension (CT) specimen, seeFigure 13. In order to use the stress intensity factor concept the plastic zone mustbe small in comparison to the dimensions of the specimen.

Usual standards for the determination of the fracture toughness (e.g. ASTM) requirethat the size of the plastic zone rp calculated by the Irwin estimation does notexceed the twentieth part of the dimensions of the specimen:

rp .1

20min

{a, (W− a), B

}.

Calculate the minimal specimen size by fulfilling this requirement, taking into ac-count the given ratio of the specimen dimensions B = a and W = 2a. The fracturetoughness is expected to be in the range of KIc ≈ 100 MPa

√m.

a

W

F

F

thickness B

H

Figure 13: CT-specimen

2.3. Analyse the shape of the plastic zone under plane stress and plane strain conditionsin the following way:

a) Evaluate the von Mises yield criterion in terms of principal stresses for planestress and plane strain.

b) Compute the principal stresses at the crack tip under mode-I loading fromthe near field solution (first term of the asymptotic expansion) for plane stress

16 Part I. Problems

and plane strain. Discuss the distribution of the principal stresses and themaximum shear stresses at the crack tip depending on the polar angle θ.

c) Estimate the size of the plastic zone at the crack tip by means of the yieldcriterion (see part a) and the principal stresses calculated in part b).

d) Discuss and compare the results for plane stress and plane strain by creatinga sketch of the shapes of the corresponding plastic zones.

2.4. Estimate the shape and the size of the plastic zone at a crack tip loaded by mode-IIIby using small scale yielding assumptions. Insert the near field solution{

τ13

τ23

}=

KIII√2πr

{− sin θ

2

+ cos θ2

}

into the von Mises yield criterion for a given yield stress σy.

2.5. A semi-infinite crack is loaded by a pair of concentrated forces F (force per thickness)located at a distance c from the crack tip (see Figure 14a). The stress intensity factorfor this configuration in a linear-elastic material is given by

KI =2F√2πc

and the yield stress of the material is σy.

a) Estimate the size of the plastic zone d according to the Irwin relation underplane stress and small scale yielding assumptions.

b) Derive the solution according to the Dugdale model for this crack configu-ration as depiceted in Figure 14b. Assume that there must not be any stresssingularity at the crack tip (c+ d). How large is the plastic zone d?

c

F

F(a)

c

F

F

σF

d

(b)

Figure 14: Semi-infinite crack loaded by a pair of single forces

2. Elastic plastic fracture mechanics 17

c) Compare both models for the limit cases

F

c σF

� 1

and

F

c σF

� 1

Hint :√

1 + z2 ≈ 1 + 12z2 for z � 1

3. Fatigue crack growth 19

3. Fatigue crack growth

3.1. During non-destructive testing a crack of length 2a0 = 4 mm is detected in a cycli-cally loaded airplane shell. The swelling load caused by the internal pressure resultsin a maximum stress of σmax

∞ = 100 MPa in the shell structure.

Since the crack is small compared to the dimensions of the structure, the modelof the Griffith crack can be used (see Figure 15). The stress intensity factor forthis configuration is defined by K = σ∞

√π a. The parameters of the Paris crack

growth lawda

dN= C

(∆K(a)

)mare C = 2.5 · 10−12 m · (MPa

√m)−4 and m = 4.

Calculate the remaining lifetime of the structure under the given assumptions. Pro-ceed as follows:

a) Calculate the critical crack length, at which instable crack propagation occurs,assuming a fracture toughness of the used aluminum alloy ofKIc = 35 MPa

√m.

b) Calculate the remaining number of cycles Nc until the structure finally fails.

2a

x

y

σ∞

σ∞

Figure 15: Griffith crack

20 Part I. Problems

3.2. At the circumference of a rotating disc with radius R = 300 mm (see Figure 16)fatigue cracks are detected, which grow subcritically in radial direction. That occursdue to the swelling load, caused by the rotation speed of the disc varying from 0 to10000 rpm.

R

a

ω

Figure 16: Rotating disc containing a radial surface crack

One of the cast iron discs contains a crack of length a0 = 15 mm. The materialparameters are given as follows

- Young’s modulus: E = 210 GPa,- Poisson’s ratio: ν = 0.2,- density: ρ = 7 g/cm3,- fracture toughness: KIc = 65 MPa

√m .

From a handbook the equation

K = Y( aR

) ρω2R2

8

3− 2ν

1− ν√πa, Y

( aR

)' 0.51 + 1.29

a

R

is known for the relation between applied load and stress intensity factor at the cracktip, where ω is the rotation speed of the disc. The crack growth can be describedby the Paris law with

da

dN= 6.9 · 10−12 m

[∆K

(MPa√

m)

]3

.

Calculate

a) the critical crack length of the structure,

b) the remaining lifetime. Use a constant geometry factor Y = Y (a/R) for themean crack length a = 1/2 (a0 + ac) in order to simplify the calculation.

3. Fatigue crack growth 21

3.3. For calculating fatigue crack growth the Paris equation is given by

da

dN= C

[∆K(a)

]m.

A Griffith crack under uniaxial tension load is considered. The crack is loaded inthe range of σ∞min = 0 to σ∞max = σ∞. The cyclic stress intensity can be calculatedby

∆K = ∆σ∞√π a .

Crack growth starts at the initial crack length a0 with ∆K(a0) = 100 MPa√

cm.Calculate the remaining cycles of lifetime N , until the critical value ac = 7.5 mm isreached. Use the following parameters:

- a0 = 5 mm,- C =

√10 · 10−11 MPa−3 mm−1/2,

- m = 3.

3.4. A crack in an infinite body is cyclically loaded by a pair of concentrated forces F(force per thickness) in the center of the crack surfaces (see Figure 17). Calculate theremaining cycles of lifetime N until reaching the critical crack length of ac = 7.5 mmusing the Paris equation and parameters from problem 3.3.

The cyclic stress intensity factor ∆K(a) is given by:

∆K (a) =2∆F√π a

.

Discuss the difference of the fatigue crack growth behavior in contrast to that inproblem 3.3.

ΔF

ΔF

a0 a0

Figure 17: Crack in an infinite body loaded by concentrated forces

23

Part II.Results

1.1 KIc = 101.0 MPa√

m

1.2 a) c = 0.74µm

b) Fc = 346.0 N

1.3 KI = pR

4t

√πa(

3− cos(2β)),

KII = pR

4t

√πa sin(2β)

1.4 KIc =3S Fc

2BW 2

√π a · Φ

( aW

)KIc = 15.8 MPa

√m

1.5 a) loaded at infinity: ac =1

π

(KIc

σ∞

)2

,

loaded with concentrated forces: ac =1

π

(2F

KIc

)2

b) loaded at infinity: instable,loaded with concentrated forces: stable

1.6 a) The given stress function fulfills the bipotential equation. Furthermore, thestress function is the solution of the boundary value problem since it fulfillsthe given boundary conditions.

b) KI =F√πa

√a+ b

a− b

c) Ksym =2F√π

√a

a2 − b2

1.7 a) σθθ|θ=±π = τrθ|θ=±π = 0

24 Part II. Results

b) h = hI + hII

with hI = C1 cos(λ− 1)θ + C2 cos(λ+ 1)θ,

hII = C3 sin(λ− 1)θ + C4 sin(λ+ 1)θ

c) σθθ = (λ+ 1)λrλ−1hI = (λ+ 1)λrλ−1 [C1 cos(λ− 1)θ + C2 cos(λ+ 1)θ] ,

τrθ = −λrλ−1 h′ = λrλ−1 [C1(λ− 1) sin(λ− 1)θ + C2(λ+ 1) sin(λ+ 1)θ]

d) λ =k

2, k = 1, 2, 3, . . .

1.8 a) σyy|θ=±π = τxy|θ=±π = 0

b) σyy + iτxy = 2Re[(λ+ 2)Azλ+1

]+ Az(λ+ 1)(λ+ 2)zλ + (λ+ 2)Bzλ+1

c) transformation into polar coordinates by using z = r eiθ:

σyy + iτxy = rλ+1{

2Re[(λ+ 2)Aei(λ+1)θ

]+

+A(λ+ 1)(λ+ 2)ei(λ−1)θ + (λ+ 2)Bei(λ+1)θ}

d) sin(2λπ) = 0 ⇒ λ =k

2, k = 1, 2, 3, . . .

1.9 KI = KtensionI +Kbending

I

KI = 1.68σ√πa

1.10 K(1)I =

2

πσ√πa

1.11 a) pmax = 76.7 MPa

b) pmax = 30.8 MPa

c) pmax = 32.1 MPa

25

700 750 800 8503.5

4

4.5

5

5.5

r [mm]

σ

/pθθ

crack region

linearized

exact

1.12 a) w =1

2

σ2

E

b) ∆U(a) = −4

3πwa3

c) ∆U(a+ ∆a) = −4

3πw(a+ ∆a)3

d) ∆U(a)−∆U(a+ ∆a) = 4πγ (a+ ∆a)2 − 4πγa2

lim ∆a→ 0: γ =1

4

σ2a

E

1.13 G = − dΠ

b da=F 2a2

bEI= 12

F 2a2

b2Eh3

1.14 J =K2

III

2µ

2.1 The solution is simple.

2.2 B = a = W − a & 580 mm

2.3 a) plane strain:√

12

[(σI − σII)2 + (σII − σIII)2 + (σIII − σI)2

]− σF = 0,

σIII = ν(σI + σII)

plane stress:√σ2

I + σ2II + σIσII − σF = 0, σIII = 0

26 Part II. Results

b)

{σI

σII

}= KI√

2πrcos(θ2

){1 + sin(θ2

)1− sin

(θ2

)}

τI = 12(σII − σIII), τII = 1

2(σI − σIII), τIII = 1

2(σI − σII)

c) rp(θ) =1

2π

(KI

σF

)2

cos2(θ2

){3 sin2(θ2

)+ 1 (plane stress)

3 sin2(θ2

)+ (1− 2ν)2 (plane strain)

27

d) Shape of the plastic zone: ESZ is plane stress state, EVZ is plane strain state

2.4 rp (θ) =3

2π

(KIII

σF

)2

2.5 a) dIrwin = 2 rp =1

π

(KI

σF

)2

=2 c

π2

(F

c σF

)2

b) dDugdale =c

2

√1 +

(F

c σF

)2

− 1

c) dDugdaleSSY =

c

4

(F

c σF

)2

, dDugdaleLSY =

F

2σF

3.1 a) ac = 39 mm

b) Nc = 192247.8 ≈ 192248

3.2 a) ac = 37.8 mm

b) Nc = 29333.3 ≈ 29333

28 Part II. Results

3.3 Nc = 1834.9 ≈ 1835

3.4 Nc = 3510.6 ≈ 3510

problem 3.3: tendency to unstable crack propagationproblem 3.4: tendency to stable crack propagation