Liberty ship SS Schenectady was a tanker built during WWII. It underwent brittle hull fracture in ice cold water before leaving the shipyard.

Tensile fracture of a brass rod showingcharacteristic cup-cone.

Fracture of Solids

Ductile fracture surface of brass showing so-called “void coalescence”.

Fracture of SolidsTheoretical tensile strength of an elastic solid

max max

max

max

max

2sin @ / 4;

2for small / ; ; where /

; for / 22 2/

xx

xx eE e x a

xE E a

x a aE

ps s l s sl

pl s sl

l ls lp p

s p

æ ö= = =ç ÷è ø

æ ö= = =ç ÷è ø

æ öæ ö æ ö= = »ç ÷ç ÷ ç ÷è øè ø è ø

=

a

s

s

U(r)

r

F(r)

r

l/4

F(r) = − ddrU (r)

λ

Fracture of Solids

Work of fracture

2g a

s

s

gg

σ

e = (x / ao )

E

σ max

Utotal = 2γ = σ max sin2π xλ0

λ /2

∫ dx =λσ max

π

λ = 2πγσ max

σ max =Eγao

⎛

⎝⎜⎞

⎠⎟

1/2

Theoretical strength of a defect-free solid

σ max =Eγao

⎛

⎝⎜⎞

⎠⎟

1/2

Work of fracture for a defect-free solid

Take E = 10 G Pa = 1010 Pa, g = 2 J m-2, ao = 3 x 10-10 m.

σ max =1010 × 23×10−10

⎛⎝⎜

⎞⎠⎟

1/2

≈ 8×109Pa = 8 GPa

Stress Concentrations

¥s

¥s2 4 2

2 4 2

2 4

2 4

4 2

4 2

1 1 3 4 cos22

1 1 3 cos22

3 21 sin 22

rr

r

a a ar r r

a ar r

a ar r

qq

q

ss q

ss q

ss q

¥

¥

¥

é ùæ ö æ ö= - + + -ê úç ÷ ç ÷

è ø è øë ûé ùæ ö æ ö

= + - +ê úç ÷ ç ÷è ø è øë ûæ ö

= - - +ç ÷è ø

Airy stress function2 4 2

2 cos24 4 2r a a

rj s q

æ ö= - - +ç ÷

è ø

qqs

qqqs

rrs

a

max

@ ; = /2 and 3 /3

2

0r rr

r a

qq

q

q p p

s ss s s=

=

= =

=

Stress Concentrations

For the circular flaw 3tK =

Def: Stress Concentration factor, Kt

maxt

nom

Kss

=

Elliptical flaw

2a2b

( )max 1 2 /a bs s= +

2 4 2

2 4 2

2 4

2 4

4 2

4 2

1 1 3 4 cos22

1 1 3 cos22

3 21 sin 22

rr

r

a a ar r r

a ar r

a ar r

qq

q

ss q

ss q

ss q

¥

¥

¥

é ùæ ö æ ö= - + + -ê úç ÷ ç ÷

è ø è øë ûé ùæ ö æ ö

= + - +ê úç ÷ ç ÷è ø è øë ûæ ö

= - - +ç ÷è ø

@r = a; θ = π / 2 and 3π / 2σθθ = 3σ nom

Kt =3σ nom

σ nom

= 3

Kt = (1+ 2a / b)The radius of curvature at a is,

2 /b ar =

( )1/ 2max 1 2 /as s ré ù= +ë ûso,

for

a >> b

σ max = 2σ a / ρ( )1/2

Stress field of an Elliptical flaw

x

y

ab

( )( )( )2

0

sinh 2 cosh 2 cosh 2 2

cosh 2 1

yy

yy o

along ybbs s

s a a as a

= =

+ -=

-ao

ao is the ellipse defined by the flaw.

bo

The Cartesian coordinates, x, y areconnected to the elliptical coordinatesa, b by

( )coshx iy a ia b+ = +

2a

( ) ( )( )

( ) ( )

2

2 22 2

/1 / 11 / 1 /1 / 1 /

yy b ar a

r a b ar a b a

ss

é ù+ ê ú= +

ê úé ù+ - -é ù+ - - ê úë ûë ûë û

Stress field of an Elliptical flaw

r is the distance along y = 0. The relative values of r/a and b/a compared to1 determine the behavior of and define regions of interest.

In the region r/a < 1 the leading term (dominant) including the bluntnesscontribution of b/a is,

( ) ( )( )

( ) ( )

2

22

/1 1/ 2 // 2 /

yy b a

b a r ab a r a

ss

é ù» +ê ú

+ê ú+ ë ûr < a

As b/aÞ 0 (a sharp “crack”) the stress field decays as (a/r)1/2.

In the region, r/a >1 ( )( )

2

2

1 /12 /

yy b a

r a

ss

+= +

+

and the stress field scales as (a/r)2

σ yy ≈σ (a / 2r)1/2 r < a: near field

r > a: far field

Sharp Cracks

2a

r

q

Central crack of length 2a in an infinite plate under uniform tension. Theleading terms for r <<a,

( ) ( ) ( ){ } ( )( ) ( ) ( ){ } ( )( ) ( ) ( ) ( ) ( )

1/ 2 1/ 2

1/ 2 1/ 2

1/ 2 1/ 2

/ 2 cos / 2 1 sin / 2 sin 3 / 2

/ 2 cos / 2 1 sin / 2 sin 3 / 2

/ 2 sin / 2 cos / 2 cos 3 / 2

xx

yy

xy

a r O r

a r O r

a r O r

s s q q q

s s q q q

s s q q q

é ù= - +ë û

é ù= + +ë û

é ù= +ë û

Def: Stress Intensity Factor, K

K as p=

( )( ) ( ){ }

( )( ) ( ){ }

( )( ) ( ) ( )

1/ 2

1/ 2

1/ 2

cos / 2 1 sin / 2 sin 3 / 22

cos / 2 1 sin / 2 sin 3 / 22

sin / 2 cos / 2 cos 3 / 22

xx

yy

xy

Kr

Kr

Kr

s q q qp

s q q qp

s q q qp

é ù= -ë û

é ù= +ë û

é ù= ë û

Sharp Cracks

The stress intensity factor, K, defines the strength of the crack in much the same way as the Burgers vector defines the strength of a dislocation.

These equations are often written in short-hand as:

σ ij =K2πr

fij (θ )

Crack Loading Modes

Mode I:Opening

Mode II:In-plane shear

Mode III:Out-of plane or longitudinal shear

IKIIK IIIK

Units of K

K Y as p= Y = numerical factor depending on geometry and loading

MPa m

Sharp Cracks

Mode I stress field

Cartesian coordinates

( )( ) ( ){ }

( )( ) ( ){ }

( )( ) ( ) ( )

1/ 2

1/ 2

1/ 2

cos / 2 1 sin / 2 sin 3 / 22

cos / 2 1 sin / 2 sin 3 / 22

sin / 2 cos / 2 cos 3 / 22

xx

yy

xy

K

r

K

r

K

r

s q q qp

s q q qp

s q q qp

= -é ùë û

= +é ùë û

= é ùë û

Cylindrical polar coordinates

( ) ( )( )( ) ( )

2

3

2

cos / 2 1 sin / 2

cos / 22

sin / 2 cos / 2

rrI

r

Krqq

q

q qss q

ps q q

ì é ù+ë ûü ïï ï=ý íï ïþ ïî

Mode II stress field

Sharp Cracks

Cartesian coordinates

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

sin / 2 2 cos / 2 cos 3 / 2

sin / 2 cos / 2 cos 3 / 22

cos / 2 1 sin / 2 sin 3 / 2

ì- +é ùü ë ûïï ï=ý íï ï

-é ùþ ï ë ûî

xxII

yy

xy

Kr

q q qss q q q

ps q q q

Mode III stress field

Cartesian coordinates

( )( )sin / 2

cos / 22xz III

yz

Kr

qss qp

-ìü ï=ý íïþ î

Griffith theory of fracture

Mechanical work during crack extension

2a

dada

2(a+da)

iT

1

2 3T

u1u 2u

Apply tractions loading the crack system

1 21 11/ 2elU Tu® =

1T

During crack extension the work done is

( )2 31 2 1LoadU T u u® = -

Griffith theory of fracture

Mechanical work during crack extension

The elastic energy at location 3 is just

1 31 21/ 2elU Tu® =

The change in elastic energy is just

( )1 3 1 21 2 11/ 2el elU U T u u® ®- = -

Then

2Load elU U= D

Griffith theory of fracture

2a

dada

2(a+da)

iTConsider a crack of length 2ain a 2D plate of infinite extentunder external boundary tractions.

The total energy, UT of thesystem is composed of 3 terms.

T el surf loadU U U U= + -

Here U load is the work doneby the applied loads Ti on thesystem.

We have shown that

2load elW U U= = D

Griffith theory of fracture

For a thin plate of unit thickness under load the excess elastic energy in the system owing to the presence of the crack is,

2 2 /elU a Eps=

The surface energy of the crack system is

4surfU ag=

The total energy may now be written as

( )2 2 / 4

T el surf

T

U U W U

U E aap gs

= -

+-

+

=

U

aa*

Excess elastic energy of a cracked solid

Uelasticno crack = 1

2σεV ;σ = Eε

Uelasticno crack

unit thickness= σ 2

2EA

Uelasticcrack

unit thickness= σ 2

2EA+ πσ 2a2

Eexcess elastic energy!"#

Griffith theory of fracture

We can find the position of the energy for fracture, corresponding to the maximum in UT,

( )

2

1/ 2

1/ 2

2

2 / 4 0

2

21

T

G

G

d U a Eda

Ea

Ea

p s g

gs sp

gsp n

= - + =

é ù= = ê úë û

é ùê ú=

-ê úë û

For the case of plane strain

Griffith theory of fracture

Def: Crack extension force

( )eldG U Wda

= - -

G is the (negative) change of potential energy per unit crack extension per unitwidth.

( )2

2eld aG W Uda E

ps gº - = =

Energy/area = Force/length

Griffith theory of fracture

( )

1/ 2

2

21GGEKa gs pn

é ùê ú= =

-ê úë û

What is a “typical value” of K at which fracture of a brittle solid is predicted

Plane Strain

We can rearrange this to read

( )

1/ 2

2

21GE

ags

p n

é ùê ú=

-ê úë û

E ~ 1011 Pa, g ~ 1Jm-2

Many materials exhibit critical K values 10 -100 times larger then this. Why?

KG = 0.45 MPa m

This is the order of magnitude value we expect for the critical stress intensity for fracture of a metal based on the Griffith theory.

Is there a mathematical limit to the value of r in the equation for the crack-tip stress field?

σ ij =K2πr

fij (θ )

Many materials exhibit critical K values 10 -100 times larger then this. Why?

Is there a physical limit to the value of r in the equation for the crack-tip stress field?

KG = 0.45 MPa m

0.45 MPa mGK

Many materials exhibit critical K values 10 -100 times larger then this. Why?

This is the order of magnitude value we expect for the critical stress intensity for fracture of a metal based on the Griffith theory.

Is there a mathematical limit to the value of r in the equation for the crack-tip stress field?

σ ij =K2πr

fij (θ )

Is there a physical limit to the value of r in the equation for the crack-tip stress field?

Many materials exhibit critical K values 10 -100 times larger then this. Why?

YES! an interatomic spacing

Crack tip plasticity

σ ij =K2πr

fij (θ )

Based on the crack tip stress field, will there (always?) be a region nearthe crack tip where the yield stress of the metal is exceeded.

Let’s ignore the term, and set equal to the yield strength.

fij (θ )σ ij

Estimate of the plastic zone size

fij (θ ) = 1 σ ij =σ yield =σ y

rplastic = rp =K 2

2πσ y2

A typical yield strength of a high strength steel is 300 MPa.

Crack tip plasticityEstimate of the plastic zone size

fij (θ ) = 1 σ ij =σ yield =σ y

rplastic = rp =K 2

2πσ y2

A typical yield strength of a high strength steel is 300 MPa.

Crack tip plasticity

σ ∼ r−1/2σ y

rplastic = rp =K 2

2πσ y2

x

y

rplastic

Crack tip plasticityEstimate of the plastic zone size

fij (θ ) = 1 σ ij =σ yield =σ y

rplastic = rp =K 2

2πσ y2

The smallest possible size of the plastic zone corresponds to about a lattice spacing. Let’s examine what the value of would have to be, to correspond to a lattice spacing at the typical value of the critical Griffith value of the stress intensity factor, K = 0.45 MPa.

σ y

rp ≈ ao = 3×10−10 m

σ y =K2πrp

= K

2πao

4.5×105Pa2π × 3×10−10

≅ 1×1010 Pa

How does this compare to the theoretical fracture stress of a solid?

Crack tip plasticityEstimate of the plastic zone size

fij (θ ) = 1 σ ij =σ yield =σ y

rplastic = rp =K 2

2πσ y2

The smallest possible size of the plastic zone corresponds to about a lattice spacing. Let’s examine what the value of would have to be, to correspond to a lattice spacing at the typical value of the critical Griffith value of the stress intensity factor, K = 0.45 MPa.

σ y

Theoretical strength of an elastic solid

ao : Lattice spacing

Smallest possible rp is about 1 nm.

rp ≈ ao = 3×10−10 m

σ y =K2πrp

= K

2πao

4.5×105Pa2π × 3×10−10

≅ 1×1010 Pa

Crack tip plasticity on a microscopic scale

Dislocations define the plastic zone size

Crack tip dislocation interactions and dislocation–dislocation interactions via the Peach-Kohler equation.

Crack tip plasticity on the atomic scale

First principles based calculations: Density Functional Theory and Molecular Dynamic Simulations.

On the atomic scale. The crack tip serves as a source for dislocations. As the tip emits dislocations it become less sharp or “blunted.”

In general we can see that there are multiple length scales associated with crack tip stress/strain fields and elastic-plastic fracture.