Introduction to Fracture Mechanics phần 4 ppt

Many different geometries have been evaluated, either analytically or numerically, and are available in the literature, e.g., Compendium of Stress Intensity Factors, D... Plane Crack P

From Hooke’s law, the strains are linearly related to the stresses so that

Since the strains are calculated from the

2 r

K I

ij







Since the strains are calculated from the

displacement gradient,

2

K r

r

K

i









Plane Crack Problem Stress Intensity Factors

The stress intensity factors for Modes I, II and III are

defined as follows:

lim



r

K

lim



   xy 

r



   xy 

r

K

lim



r

K

The stress intensity factor K depends on loading and geometry

Many different geometries have been evaluated, either

analytically or numerically, and are available in the literature,

e.g., Compendium of Stress Intensity Factors, D P Rooke.

Plane Crack Problem Similitude

For a crack of length 2a1 in an infinite plate, subjected

to an applied stress σ1 the stress intensity factor is

known to be Consider two large plates, one with a center crack of length 2a1, the other with a center crack of length 2a2 A stress σ1 is applied to the first

1

1 a

K I   

crack of length 2a2 A stress σ1 is applied to the first

plate, and a stress σ2 is applied to the second plate If

we choose σ1, σ1, σ2 and σ2 so that then the

fields at the crack tip are identical in both cases This

is the principle of similitude, which is very important in fracture mechanics as it allows results from laboratory scale tests to be applied to large scale fracture problems

) 2 ( )

1

(

I

I K

Plane Crack Problem Stress Intensity Factors

How do we apply this analysis to the failure of actual materials? It has been found experimentally that when

the stress intensity factor K (which depends on the

geometry and loading) attains a critical value K C (a

material property) the crack begins to grow, i.e., the

critical condition for the onset of fracture is

K → K c

The condition can also be expressed in terms of the

energy release rate, i.e., ς → ςc

What are some typical values for K C ?

Si3 N4

Al2O3

Glass

K c (MPa )

1



4

3 



8

4 



Steels

Al alloys

Polymers

2 5

.



100

10 



300

30 



Fracture Mechanics #2:

Role of Crack Tip Plasticity

Role of Crack Tip Plasticity

Plastic Zone Size Estimate

Consider inelastic and permanent deformation at the crack tip (stresses are too high for the material to

remain elastic)

First order estimate of plastic zone size:

Assume: plane stress, and the material behavior is elastic-perfectly plastic Set the stress σyy= σys

(along the line θ = 0)

ys

yy

r

K





* 1 2

2 2

2 1

*

2

p

a

K r





 



Where we have used the result that for a semi-infinite crack in a very large plate K I    a

What about the details of the plastic zone shape? The

What about the details of the plastic zone shape? The

shape of the plastic zone is obtained by examining the yield condition, in conjunction with asymptotic K-field

results, for all angles θ around the crack tip Either the

Mises or the Tresca criterion can be applied

Plastic Zone Shape

Recall that for the Tresca yield condition yielding

occurs when  max   ys / 2

We will use the Mises yield condition The Mises

condition in terms of principal stresses is given as

condition in terms of principal stresses is given as

1 3

2 3 2

2 2

where σ y s is the uniaxial yield stress (For a tension test, σ2= σ3=0, σ1= σys )

On the plane θ = 0, σxy= 0 and thus σxx and σyy are the principal stresses σ1 and σ2 The stresses σ z ≡ σ3; σz =0 for plane stress, σ z = ν(σ xx + σ yy) for plane strain

However, in general the shear stress σ xy is not zero and the principal stresses σ1 and σ2 cannot be determined so easily

The principal stresses σ1 and σ2 are evaluated as follows (can use Mohr’s circle, for example):

2

2 2

1

2 2













