fracture crack propagation, Study notes of Mechanics of Materials

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Chapter 5

1

Overview

Characterisation of Fatigue

There are three commonlyrecognized forms of fatigue:

High cycle fatigue (HCF),

Low cycle fatigue (LCF),

Thermal mechanicalfatigue (TMF)

Fatigue strength isdetermined by runningmultiple specimen tests at anumber of differentstresses.

The objective is to identifythe highest stress that willproduce a fatigue lifebeyond ten million cycles.

This stress is also known as the material's endurance limit or fatigue limit.

Gas turbines are designed so that the stresses in engine components do notexceed this value including an additional safety factor.

What are the important parameters to characterize agiven cyclic loading history?

Mean stress

Stress range

Stress amplitude

Stress Ratio

K – Range

min

max

σ

σ

σ

−

=

∆

min

max

K

K

K

−

=

∆

min

max

σ

σ

σ

−

=

1 2

a

min

max

σ

σ

σ

=

(^12)

m

min max

min max

K K

R

=

=

σ σ

Fatigue Crack Growth – LEFM approach

In 1961 the ideas of LEFM were applied to fatigue crackgrowth by Paris, Gomez and Anderson.

For given cyclic loading, define

K as K

max

– K

min

which

can be found

∆σ

and the geometry of the crack body.

Say that the crack grows an amount

a during N cycles.

Paris, Gomez and Anderson said that the rate of crackgrowth depends on

K in the following way:

Thus plot of log (da/dN) vs. log (

K) should be straight

line with a slope

m

The actual relationship between crack growth rate and ∆

K is depicted on the following slide. There are three

different regime of fatigue crack growth A, B and C.

Fatigue Crack Growth – LEFM approach

m

K

C

da dN

a N

∆

=

→

∆ ∆

Fatigue Crack Growth – LEFM approach

F

ATIGUE

C

RACK

G

ROWTH

Regime A

Concept of the threshold stress intensity

K

th

When

K is ≈

K

th,

, where

K

th

is the threshold stress

intensity factor, the rate of crack growth is so slow that the crackis often assumed to be dormant or growing at an undetectablerate.

An operational definition for

K

th

often used is that if

the rate of crack growth is 10

mm/cycle or less the

conditions are assumed to be at or below

K

th

An important point is that these extremely slow crackgrowth rates represent an average crack advance of lessthan one atomic spacing percycle. How is this possible?

What actually occurs is that there are many cycles with no crack advance, then the crack advances by 1 atomicspacing in a single cycle, which is followed again bymany cycles with no crack advance.

F

ATIGUE

C

RACK

G

ROWTH

Regime B

When we are in regime B (Paris regime) the followingcalculation can be carried out to determine the number ofcycles to failure.

From Paris Law:

∆

K can be expressed in terms of

∆σ

;

Where Y depends on the specific specimen geometry.

Thus the Paris Law becomes:

Assume that Y is a constant. Solve for d

a

and integrate

both sides:

m

K

C

dN da

∆

=

a

Y

K

π

σ

∆

=

∆

m

a

Y

C

dN da

π

σ

∆

=

(

)

∫

∫

f

f o

N

m

m

m

a a

m

dN

CY

a

da

0

2

2

/

/

π

σ

From those expressions, we need to determine the initialcrack length a

o

and the final crack length

a

f

(some times

called the critical crack length).

How do we determine the initial crack length

a

o

?

Crack can be detected using a variety of techniques, rangingfrom simple visual inspection to more sophisticatedtechniques based on ultrasonics or x-rays. If no cracks aredetectable during inspection, we must assume that a crackjust at the resolution of our detection system exists.

How do we determine the final crack length

a

f

? We know

that eventually the crack can grow to a length at which thematerial fails immediately, i.e.

or

C

K

K

→

max

C

f

K

a

Y

→

π

σ

max

F

ATIGUE

C

RACK

G

ROWTH

Regime B

CON’T

Thus we may solve

a

f

as follows:

A very important idea that comes from this analysisis the following: even if a component has adetectable crack, it need not to be removed fromservice! Using this framework the remaining lifecan be assessed. The component can remain inservice provided it is inspected periodically. This isthe

crack-tolerant

or

damage tolerant

design

approach.

2

2

2

1

max

σ

π

Y

K

a

C

f

=

F

ATIGUE

C

RACK

G

ROWTH

Regime B

CON’T

EXAMPLE 5.

A wide thick plate, K

1c

= 40 ksi-in

, contains an

edge crack 0.1 inch long. The plate is subjected toalternating stresses between 0 and 20 ksi. Data onsimilar material under similar environmentalconditions exhibited Paris law with coefficients m =4 and C = 7x 10

for K in ksi-in

. How many

cycles will the plate support before failure?

Solution

It is first necessary to determine the length of thelongest crack the plate can support withoutfailure. A suitable stress intensity factor is:

K

1c

σ

max

π

a

c

π

a

c

a

c

2 inches

Solution For this case, Paris Law has the form

da/dN = C(

∆

K)

m

= C [1.

∆σ

)

√

(

π

a

c

)]

m

Substituting this material’s Paris law parameters,we obtain;

da/dN = 7 x 10

(1.12)

4

(20)

4

π

2

a

2

Rearranging the forgoing equations yields

da/a

2

= 7 x 10

(1.12)

4

(20)

4

π

2

dN

Thus Integrate from a

i

= 0.1 inch to a

c

= 1.02 inch,

N ≈ 5185 cycles

ans

∫

−

−

02 . 1

1 . 0

2

2

4

4

10

da

a

x

N

π

First calculate C in Paris’ lawC = 10

/

3

= 4.55 x 10

,for crack in m/cycle

Obtain stress range,

∆σ

= 54 MPa

Substitute into the given expression:

  

  

−

∆

−

=

−

−

2 / ) 2 ( 2 / ) 2 ( 2 /

)

(

1

)

(

1 ) ( ) 2 (

2

m f m o m m m f

a

a

CY

m

N

π

σ

Solution

  

  

−

=

−

2 1 2 1 2 3 3 3

11

01

0

1

005

0

1

54

02

1

10

55

4

2

/

/

/

). ( ). ( ) (..

π

x

N

f

Thus N

f

= 195 675 cycles