LEFM—Linear Elastic Fracture Mechanics

Crack-tip fields for plane stress and plane strain: Mode I and Mode II

33

Linear, isotropic elastic solids with Young's modulus, , and Poisson's ratio, .The in-plane stresses in the crack tip fields are the same in plane stress and plane strain, however, 0 in plan

E ν

σ = 33 11 22e stress and ( ) in plane strainσ ν σ σ= +

Universal behavior ( !) as approaches crack tip:all problems rMode I - -Symmetric stresses & strains fields at tip.

22 12Mode I-- On the plane ahead of the tip: , 0 (standard definition)2

where is called the Mode I stress intensity factor

I

I

Kr

K

σ σπ

= =

The distributions, , are given in most texts on LEFMIαβσ

0

( )2

( )2

II

II

Kr

K ru u uE

αβ αβ

α α α

σ σ θπ

θπ

=

= + 2

in plane stress

in plane strain1

E EEν

=

=−

Universal behavior ( !) as approaches crack tip ( in the texts):IIall problems r αβσ

Mode II - -Anti - symmetric stresses & strains fields at tip.

12 22 On the plane ahead of the tip: , 0 (standard definition)2

where is called the Mode II stress intensity factor

II

II

Kr

K

σ σπ

= =

Universal behavior ( !) as approaches crack tip:all problems rMode III - -Out - of - plane shearing (also called "anti - plane shear").

( )13 23 12

3

, sin ,cos ( / 2 on plane ahead of tip)2 22

2 sin ( /(2(1 )) is shear modulus)2 2

IIIIII

III

K K rr

K ru G EG

θ θσ σ σ ππ

θ νπ

⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞= ≡ +⎜ ⎟⎝ ⎠

Derivation of the Mode III fields given in class

By superposition (linear elasticity), conditions at any crack tip can always berepresented as a sum of the three modes.

0

( )2

( )2

IIII

IIII

Kr

K ru u uE

αβ αβ

α α α

σ σ θπ

θπ

=

= +

Some Basic Solutions

There are several excellent compilations of solutions. We will make use of Tada, Paris & Irwin

1x

2x

122 1 22 2

1

Exact solution:

( , 0)x x a xx aσσ

∞

= > =−

IK aσ π∞=

0

1

2

3

1 2 3 4 5

Tensile crack

Shear crack

112 1 22 2

1

Exact solution:

( , 0)x x a xx aτσ∞

= > =−

IIK aτ π∞=

Exact solution

/or

/

σ σ

τ τ

∞

∞

1 /x a

2K

rπ

Asymptotic crack-tipfield

For these two problems the K-field gives and accurateestimate of the stress ahead of the crack for r/a<1/4

Some Basic Solutions, continued

1.122IK aσ π∞=tensileedge - crack :

1/ 22 tan

2Ib aK aa b

πσ ππ

∞ ⎡ ⎤⎛ ⎞= ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

periodic tensilecracks :

2IK aσ π

π∞=penny shapedcrack :

Mode I at all pointsof crack edge

ASTM compact tensile specimen

1

0.60.275

0.25/ 2

h bh bc D bthickness t b

=== =

≡ =

1IP aK a Fbt b

⎛ ⎞= ⎜ ⎟⎝ ⎠

(see Tada, et al page???

1F

/a b

An example of results in theStress Analysis of Cracks Handbookby H. Tada, P.C. Paris and G.R. IrwinASME Press, New York, NY, 2000

An edge-notched infinite beam under purebending. Pages 55-57 of the Handbook

Results are presented on the this page and thenext for the stress intensity factor (a mode I problem),the crack opening displacement at the surface, and the additional rotation due to the presenceof the crack.

Energy Release Rate, Prescribed Load vs. Prescribed Displacement, andRelation to Stress Intensity Factors

2

Prescribed load/thickness, . Load-point displacement energy release rate ( / )

PG J m

= ∆

=

strain energy of the system/thickness potential energy/thickness of load energy of the system/thickness

SEPPE

=∆ ==

Note that for any linear elastic system, / 2,and thusfor :

/ 2The energy release rate for is defined as

1 ( )2 2P P P

SE P

PE SE P P

PE P PG Ga a a

= ∆

= − ∆ = − ∆

∂ ∂ ∆ ∂∆⎛ ⎞ ⎛ ⎞= − ⇒ = =⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠

prescribed load

prescribed load

2

Define ,

which depends only on geometry, including, , and . Thus,12P

CP

a EdC dCP G P

a da da

ν

∆=

∂∆⎛ ⎞ = ⇒ =⎜ ⎟∂⎝ ⎠

compliance

Generic cracked body

Graphical interpretationof energy release ratedue to crack advance

Energy Release Rate, continued: Role of compliance and loading conditions

As indicated in the figure, introduce a linear spring with compliancein series with the cracked body. Let be the total displacement

through which works. Now, let .(

M T

T

T M M

CP

C P C

∆∆

∆ = ∆ + = ∆ +be prescribed

( )22 1 2 1

2

/ )1 1 12 2 2

and a straight-forward calculation (see pg.8of notes)again gives12

M M T

C

PE SE C P C C

dCG Pda

− −

∆

= + = ∆ + ∆ −∆

=

This is not intutive.

MNote! The energy release rate does not depend on C .In particular, G is the same for both prescribed load and prescribed displacement.

Most results for G are obtained either from analytical or numerical calculations (see later).However, the above formula permits experimental evaluation of G by experimentallymeasuring the compliance at two nearly equal crack lengths, a and a+da.

a

Energy Release Rate, continued: Relation between G and K’s

Since G is independent of loading, consider a body under prescribed .Consider two configurations of the cracked body one with a and the other with a+ a.Because the system is elastic, the energy relea

∆∆

sed in advancing the crack a is the sameas the work done in closing the crack from a+ a to a (see figure). Because is prescribed,no work is done by applied loads in closing crack. The work to cl

∆∆ ∆

ose the crack is

Mode I situation:symmetric body andsymmetric loading.

22 2 20

0

2

1 ( ,0) ( ,0 ) ( ,0 )22 ( ) ( )

1 ( ) ( )

1 (Irwin's universal relation!)

a

a

I I

I I

I

W G a x u x u x dx

a xK a K a a dxE x

K a K a a aE

G KE

σ

π

∆ + −

∆

⎡ ⎤∆ = ∆ = −⎣ ⎦

∆ −= + ∆

= + ∆ ∆

⇒ =

∫

∫

22

2 2

2

( ,0) ( ) / 2

8( ,0 ) ( ,0 ) ( )2

in plane stress, /(1 ) in plane strain

I

I

x K a x

a xu x u x K a aE

E E E E

σ π

πν

+ −

=

∆ −− = + ∆

= = −

( )2 2 2

Under all three modes, one finds1 1

2I II IIIG K K KE G

= + +

Discuss units

Energy Methods for Determining Energy Release Rate

Double cantilever beam specimenCompute compliance of specimen treating each arm as a cantilever beam,

and consider the specimen to have unit thickness and P is force/thickness

3 3 3

3 3

2 22

3

4 82 3

1 122

Pa Pa aCEI Eb P Eb

dC P aG Pda Eb

∆ ∆= = ⇒ ≡ =

= =

3/ 2

Since this is a mode I problem (by symmetry), Irwin's relation gives

2 3

IPaKb

=

See notes pg. 10 for the problem of a thin strip (plane stress)with a semi-Infinite crack subject to rigid grips. This is an exact solution. Note it is independent of crack length.

2

2 2

1,4(1 ) 2 1

IE EG K

b bν ν∆ ∆

= =− −

Energy Methods for Determining Energy Release Rate, continued

Delamination of stressed thin film on elastic substrate

1D analysis (uniaxial stress in film)

h

σ

delamination crack

2

2

1strain energy/area in film far ahead of tip = 2

strain energy/area in film far behind tip = 0energy released/area due to crack advance:

12

which is independent of crack length.

hE

hGE

σ

σ=

σ

This problem does not have symmetry and it is an example where the crack is a combinationof mode I and II. Later in the course we will determine the two stress intensity factors.

The simple result for G is valid when the crack is long enough such that steady-state conditionsapply. In practice this means the crack length has to be long compared to the film thickness.