Numerical Simulation of Fracture of Materials · 2017. 1. 3. · Numerical Simulation of Fracture of Materials Allgemeines Wahlfach / Optional Course WS 2009/2010 Dr. Steffen Brinckmann,

Numerical Simulation of Fracture of Materials

Allgemeines Wahlfach / Optional Course WS 2009/2010

Dr. Steffen Brinckmann, Dr. Rebecca Janisch

Contact information: [email protected],[email protected]

Why should we care?

What will you learn here?

Classical fracture mechanics

Fracture & FEM

Density Functional Theory

Quantum mechanics

Application of DFT: ’abinit’

Application of FEM: ’Abaqus’

Apply to two fracture cases

Location normally: IA 1/21; CIP: UHW 1219

FAQ??

Q: Do I have enough computer knowledge?A: Programming part will be very easy.

Q: Why are the notes not “complete”?A: To increase active thinking and participation,

certain parts are left blank.

The formal things:

Time of classes: Tuesday 8AM - 9:30AM

and??

QUESTIONS ???


H.L. Ewalds and R.J.H. Wanhill: Fracture Mechanics(hard to get)

D. Gross and T. Seelig: Bruchmechanik, Springer

T.L. Anderson: Fracture Mechanics: Fundamentals andApplications, CRC PR Inc.

or search for Fracture Mechanics at Amazon, Google, ...


Overview of Fracture

Overview Continuum Mechanics

Classical Fracture Mechanics


Let’s look at it!

What is fracture?

Are there different types?

What are we doing here?

Fracture of concrete

Fracture of metals

Fracture of steel

Fracture of wood

www.ecometry.biz/PicturesPatterns

Fracture of a polymere

Fracture of ???

Fracture of explosive

Rae et al. (2002) Proc. R. Soc. Lond.


Let’s look at it!

What is fracture?

Are there different types?

What are we doing here?

What is failure?

Yielding dominated Fracture dominated- plasticity - localized- ductile - brittle- material dominates - crack dominated

Defects: Defects:

Some definitions ...

crack tip

crack process zone

crac

k fro

nt

crack wake

crack surface

Tangential and normal cracks

Different modes

mode I mode II mode III

Intergranular and Transgranularfracture cannot always be differentiated. Why?

Fracture of materials with precipitates

void nucleation void growth void coallescenceinitial state

Especially in polymers ...

Some metals ...

slow crackfast crack

At temperature ...

stress to fracture

1

10

meltingtemperature

Question: Why?

energy to fracture

meltingtemperature

In these lectures we concentrate on

mode I loading

brittle cracks

single cracks

metals

without precipitates





Overview of Continuum Mechanics

Stress & Strain

Equilibrium

Elasticity and Plasticity

The stress is a tensor.The traction is a vector.

SV

appliedt

nt1

2

traction: ti = σijnj

normal stress: σ11, σ22

shear stress: σ12, σ23

When do traction & stress have the same value?

What are the units of traction & stress?

The strain is a tensor.

If Xi is the original position,

and xi is the current position of a particle

then it has moved by ui = xi − Xi.

The strain is εij = 12 (ui,j + uj,i)

What is the unit of strain?

ui,j: ith displacement component differentiated in jth direction.ui,j = ∂ui

∂Xj

Plane stress and strain are 2D approximations.

PLANE STRESS PLANE STRAIN

picture: sheet of paper levee / dike

tickness: t→ 0 t→∞σ33 = 0 ε33 = 0

What other stress and strain components arezero?

What constrains can lead to plane strain?


Stress & Strain

Equilibrium

Elasticity and plasticity

Equilibrium: Tug-of-war

traction:∫

S ti dS = 0

stress: σij,j = 0

divergence of stress: divσ = 0

gradient: ∂∂xσ = 0

What is the difference?


Stress & Strain

Equilibrium

Elasticity and plasticity

Elastic behavior:σ

E

ε

Hooke’s law: σij = Cijklεkl

linear elastic: Cijkl = const

How does the material unloaded?

What are the units?

Elastic constants:

Young’s modulus: E = 2µ(1 + ν)

Poisson’s ratio: ν = E2µ − 1

Shear modulus: µ = E2(1+ν)

What are the units?

Plastic behavior:

y

σ

E

ε

σH

Yield: σmises = σflow

Flow stress: σflow = σy + Hε

Mises stress: σmises = 12

(σij − σkk

3 δij) (σij − σkk

3 δij)

How does the material unloaded?

How much energy is dissipated?





Overview of Classical Fracture Mechanics

Pre-Griffith models

Energy Balance models

Current engineering models

Pre-Griffith models

Principal Stress model

Principal strain model

Mohr-Coulomb model

Drucker-Prager model

Stress space

σI

σII

Principal stress = largest Eigenvalue of stress tensor.

2D: σI = 12(σx + σy) +

√[12(σx − σy)

]2 + σ2xy

Principal Stress modelRankine, Lamé, Navier (∼1800)

Fracture occurs,if principal stress reaches strength.σcompression ≤ σI ≤ σtension

Tσ

Tσ

σC

σC

σI

σII

Principal Stress model (continuation)

Advantages: Disadvantages:

Principal Strain modelSaint-Venant, Bach (∼1880)

Fracture occurs,if principal strain reaches critical value.

εcompression ≤ εI ≤ εtension

σC

σI

σII

σC

Tσ

Tσ

Assume: σIII = 0

Principal Strain model (continuation)


Principal strain = largest Eigenvalue of strain tensor.EεI = σI − ν(σII + σIII)

Mohr-Coulomb modelMohr, Coulomb (end 1800s)

Fracture occurs,if Mohr-circle touches boundary.

σI

σII

Tσ

Tσ

σC

σCσIσIII

σII

σ

τ

Assume: σIII = 0

Mohr-Coulomb model (continuation)


Drucker-Prager modelDrucker, Prager (∼1920)

Fracture occurs,if shear stress reaches value whichdepends on hydrostatic pressure.

σI

σII

σC

Tσ

TσσC

Assume: σIII = 0

Drucker-Prager model (continuation)


Disadvantage of Pre-Griffith models

Only maximum stress important (no fatigue)

Stress concentrations always lead to failure

Plasticity not included

Not for complex loading conditions


Pre-Griffith models



Griffith Energy Balance ApproachGriffith (1920)

w

h

2a

Utotal = U0 + Ua + Uγ − UF

U0: elast. energy of loadeduncracked plate

Ua: change in elast. energydue to crack formation πσ2a2

E

Uγ: cleavage energy

UF: work by external forces

a� h ∼ w; unit thickness

What are the units?

Griffith Energy Balance Approach (continuation)

if γs is the surface formation energy per area:

U =

equilibrium: dUda = 0 =

4aγs

πσ2a2

E

a

U

instable equilibrium

Griffith Energy Balance Approach (continuation)

σ√

a =√

2γsEπ

Material properties:

Configuration:


Irwin’s extensionIrwin (1948)

Fracture occurs,if πσ

2aE ≥ πσ2

c aE = Gc = R

≥ 2(γs + γp)

σc:

Gc: critical energy release rate

R:

γp: plastic strain work per surface area

γs:

Irwin’s extension (continuation)


What are the units of R and γs?


Pre-Griffith models



A convenient coordinate system

σ11

σ22

rϕ

2

1

Plane strain: κ = 3− 4ν, σ33 = ν(σ11 + σ22)Plane stress: κ = 3−ν

1+ν , σ33 = 0

Mode I elastic fields

σ11

σ22

σ12

=KI√2πr

cosϕ1− sin ϕ

2 sin 3ϕ2

1 + sin ϕ2 sin 3ϕ

2sin ϕ

2 cos 3ϕ2

u1

u2=

KI

2G

√r

2π(κ− cosϕ)

cos φ2

sin φ2

What is stress at the crack tip?

What is the shape at the crack tip?

Irwin’s Stress Intensity Factor ModelIrwin (1950s)

Fracture occurs,if KI = σ

√πa ≥ KIc

KI: MPa√

m

KIc:

plane stress: G = K2I

E

plane strain: G = K2I

E (1− ν2)

Stress Intensity Factors

Gross, Seelig; Bruchmechanik

Convenient: Stress Intensity Factors are added up..

KI shear stress + KI point load + KI external load ≥ KIc

However, crack interaction is not possible:KI center−cracked + KI edge−cracked

However, different modes add up:KI + KII≥ KIc

Irwin’s Stress Intensity Factor ModelIrwin (1950s)

Fracture occurs,if KI = σ

√πa ≥ KIc

The critical values are determined experimentally.

Ewalds, Wanhill; Fracture mechanics

The critical values is a material parameter.

Courtney; Mechanical behavior of Materials

Why are not all on the diagonal?

The critical values is temperature dependent.

30CrNiMo8 (20◦): KIc = 3650MPa√

mm

30CrNiMo8 (−20◦): KIc = 2000MPa√

mm

Why??

Stress Intensity Factors (continuation)


The J-Integral

The J-Integral

ds ds2

U

ti

U =∫ εij

0 σkl dεkl

J =∫

S

[U ds2 − ti ui,1 dS

]ui,1: strain

The J-Integral

Fracture occurs, if J ≥ Jc

J = G =K2

I

E(1− ν2)

contour-independent !

applicable also for plastic cases

The J-Integral (continuation)


Plastic zone

Plasticity at the crack tip

yσ

22σ

r

Plane strain

2rp = 13π

(KIσy

)2

Plane stress

2rp = 1π

(KIσy

)2

only for: small-scale plasticity!

What is the effective crack-length?

Plasticity at the crack tip

thickn

ess

Test your understanding

The R-curve

Definition of R-curve

Griffith and Irwin had said:πσ2a

E = G(Fapplied, a) ≥ Gc(a) = R(a)

1G(F ,a)

2G(F ,a)

R,G

a

R(a)G(F ,a)c

a0

Stable crack growth:∂G∂a

∣∣F=const. ≤

dRda

Why does R-curve increase? Why that criterium?Why ∂a?

Application of R-curve

a

2h

b

applied force

dGda = +24F2a

EBh3

applied displacement

dGda = −48F2a

EBh3

What is the difference? Why? Explain!

Why is it an analytical expression?

Documents

Numerical Simulation of Fracture of Materials · 2017. 1. 3. · Numerical Simulation of Fracture of Materials Allgemeines Wahlfach / Optional Course WS 2009/2010 Dr. Steffen Brinckmann,