1

Fracture and fatigue

Key point:

Preexisting surface flaws and preexisting internal cracks play a central role in the failure of materials.

• How do flaws in a material initiate failure?• How is fracture resistance quantified; how do different

material classes compare?

• How do we estimate the stress to fracture?• How do loading rate, loading history, and temperature

affect the failure stress?

ISSUES TO ADDRESS...

Fracture mechanisms

• Ductile fracture– Occurs with plastic deformation and with high energy

absorption before fracture

• Brittle fracture– Little or no plastic deformation with low energy

absorption – Catastrophic

2

An oil tanker that fractured in a brittle manner by crack propagation around its girth

Ductile vs Brittle Failure

Very Ductile Moderately Ductile BrittleFracture behavior

Large Moderate%AR or %EL Small

• Classification:

• Ductile fracture is usually desirable!• Ductile: warning before fracture• Brittle: no warning

3

• Ductile failure:--one piece--large deformation

Figures from V.J. Colangelo and F.A. Heiser, Analysis of Metallurgical Failures(2nd ed.), Fig. 4.1(a) and (b), p. 66 John Wiley and Sons, Inc., 1987. Used with permission.

Example: Failure of a Pipe

• Brittle failure:--many pieces--small deformation

• Evolution to failure:

• Resultingfracturesurfaces(steel)

50 mm

particlesserve as voidnucleationsites.

50 mm

From V.J. Colangelo and F.A. Heiser, Analysis of Metallurgical Failures (2nd ed.), Fig. 11.28, p. 294, John Wiley and Sons, Inc., 1987. (Orig. source: P. Thornton, J. Mater. Sci., Vol. 6, 1971, pp. 347-56.)

100 mmFracture surface of tire cord wire loaded in tension. Courtesy of F. Roehrig, CC Technologies, Dublin, OH. Used with permission.

Moderately Ductile Failure

necking

s

void nucleation

void growth and linkage

shearing at surface fracture

4

Ductile vs. Brittle Failure

Adapted from Fig. 8.3, Callister 7e.

cup-and-cone fracture brittle fracture

Fractured aluminium alloy I, with a dimpled texture

Fractured aluminium alloy II, with a typical cleavage texture

Fractured tensile test bars

”Cup and cone”

5

Scanning electron micrographs (a) spherical dimples characteristic of ductile fracture, (b) parabolic-shaped dimples characteristic of ductile fracture.

Brittle Failure

Arrows indicate origin of crack

6

Transgranular fracture Intergranular fracture

Brittle Failure

Flaws are Stress Concentrators or stress raisers

Results from crack propagation• Griffith Crack

where rt = radius of curvature of the crack tipso = applied stresssm = stress at crack tip

Kt : stress concentration factor

ot

/

tom K

as=⎟⎟

⎠

⎞⎜⎜⎝

⎛r

s=s

21

2

rt

Principles of fracture mechanics

7

Concentration of Stress at Crack Tip

Adapted from Fig. 8.8(b), Callister 7e.

Engineering Fracture Design

r/h

sharper fillet radius

increasing w/h

0 0.5 1.01.0

1.5

2.0

2.5

Stress Conc. Factor, K tsmaxso

=

• Avoid sharp corners!s

Adapted from Fig. 8.2W(c), Callister 6e.(Fig. 8.2W(c) is from G.H. Neugebauer, Prod. Eng.(NY), Vol. 14, pp. 82-87 1943.)

r , fillet

radius

w

h

o

smax

ot

/

tom K

as=⎟⎟

⎠

⎞⎜⎜⎝

⎛r

s=s

21

2

8

Crack Propagation

Cracks propagate due to sharpness of crack tip • A plastic material deforms at the tip, “blunting” the crack.

Energy balance on the crack• Elastic strain energy

• energy stored in material as it is elastically deformed• this energy is released when the crack propagates• creation of new surfaces requires energy

plasticbittle

Deformed region

When Does a Crack Propagate?

• Critical stress for crack propagates in a brittle material

where• E = modulus of elasticity• gs = specific surface energy• a = one half length of internal crack• Kc = sc/s0

• For ductile => replace gs by gs + gp

where gp is plastic deformation energy

212

/

sc a

E⎟⎠⎞

⎜⎝⎛p

g=s

i.e., sm > sc

or Kt > Kc

Example problem 8.1, p217

9

Fracture toughness

KC: fracture toughness

the critical (c) value of the stress intensity factor at a crack tip necessary to produce failure

KC = Ysc ( a)1/2

1. Y: geometry factor, on the order of 1

2. sc the overall applied stress at failure

3. a: the length of a surface crack (one-half the length of an internal crack)

4. Unit of KIC: MPa • m1/2

p

a

a

Critical stress for crack propagation (sc) and crack length (a)

The three modes of crack surface displacement

(a) mode I, opening or tensile mode

plane strain fracture toughness

(b) mode II, sliding mode

(c) mode III, tearing mode

KIC = Ysc ( a)1/2p

10

Room-temperature mechanical properties for selected engineering materials

Fracture ToughnessGraphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibersPolymers

5

KIc

(MP

a·m

0.5)

1

Mg alloysAl alloys

Ti alloys

Steels

Si crystalGlass -soda

Concrete

Si carbide

PC

Glass 6

0.5

0.7

2

4

3

10

20

30

<100>

<111>

Diamond

PVC

PP

Polyester

PS

PET

C-C(|| fibers) 1

0.6

67

40506070

100

Al oxideSi nitride

C/C( fibers) 1

Al/Al oxide(sf) 2

Al oxid/SiC(w) 3

Al oxid/ZrO 2(p)4Si nitr/SiC(w) 5

Glass/SiC(w) 6

Y2O3/ZrO 2(p)4

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• Crack growth condition:

• Largest, most stressed cracks grow first!

Design Against Crack Growth

K ≥ Kc = aY ps

--Result 1: Max. flaw sizedictates design stress.

max

cdesign

aY

K

p<s

s

amax

no fracture

fracture

--Result 2: Design stressdictates max. flaw size.

2

1⎟⎟⎠

⎞⎜⎜⎝

⎛

sp<

design

cmax Y

Ka

amax

sno fracture

fracture

• Two designs to consider...Design A--largest flaw is 9 mm--failure stress = 112 MPa

Design B--use same material--largest flaw is 4 mm--failure stress = ?

• Key point: Y and Kc are the same in both designs.

Answer: MPa 168)( B =sc• Reducing flaw size pays off!

• Material has Kc = 26 MPa-m0.5

Design Example: Aircraft Wing

• Use...max

cc

aY

K

p=s

sc amax( )

A= sc amax( )

B

9 mm112 MPa 4 mm--Result:

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Loading Rate• Increased loading rate...

-- increases sy and TS-- decreases %EL

• Why? An increased rategives less time for dislocations to move past obstacles.

s

e

sy

sy

TS

TS

largere

smallere

Charpy test of impact energy

1. A notched specimen – stress concentrating

2. Loading applied rapidly

3. impact energy - the energy necessary to fracture the test specimen

The swinging pendulum

The intial height h

The final height h’

Impact fracture testing

Izod test

Charpy test

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1040 carbon steel: Fe-0.4C-0.75 Mn

J: joule. 1J = 1N • m

Cu: fcc, ductile

Mg alloy: hcp, relatively brittle

Toughness = , the area under the s - ecurve∫ esd

Fracture

Unit of toughness: N/m2

Or (N/m2) • (mm / mm) = N •m / (mm)3

Toughness: the energy necessary to fracture per unit volume of material

Toughness obtained in a tensile test

14

Temperature dependence of the Charpy V-notch inpact energy and shear fracture for an A283 steel.

Schematic curves for the three general types of impact energy-versus-temperature behavior

More DuctileBrittle

15

Ductile-to-Brittle Transtion Temperature (DBTT)

1. in bcc alloys

2. an abrupt drop in ductility and toughness as the temperature is lowered

Plain-carbon steels with various carbon levels

Fe-Mn-0.05C alloys with various manganese levels

• Pre-WWII: The Titanic • WWII: Liberty ships

• Problem: Used a type of steel with a DBTT ~ Room temp.

Reprinted w/ permission from R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(a), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Dr. Robert D. Ballard, The Discovery of the Titanic.)

Reprinted w/ permission from R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(b), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Earl R. Parker, "Behavior of Engineering Structures", Nat. Acad. Sci., Nat. Res. Council, John Wiley and Sons, Inc., NY, 1957.)

Design Strategy: Stay Above The DBTT!

16

Fatique

• Fatique = failure under cyclic stress

• failure after N cycles

• load < T.S

Fatique-testing apparatusfor making rotating-bending tests

17

• Stress varies with time.-- key parameters are S, sm, and frequency

smax

smin

s

time

smS

• Key points: Fatigue...--can cause part failure, even though smax < sc.--causes ~ 90% of mechanical engineering failures.

Random stress cycle.

S = stress amplitude, sm= mean stress,sm = (smax + smin)/2S = (smax – smin)/2Load ratio R = smin/smax

Typical fatique curve (S-N curve)

or fatigue strength

18

• Fatigue limit, Sfat:--no fatigue if S < Sfat

Fatigue Design Parameters

Sfat

case for steel (typ.)

N = Cycles to failure103 105 107 109

unsafe

safe

S = stress amplitude

• Sometimes, thefatigue limit is zero! (a

material does not display a fatique limit)

Adapted from Fig. 8.19(b), Callister 7e.

case for Al (typ.)

N = Cycles to failure103 105 107 109

unsafe

safe

S = stress amplitude

Fatique S-N probability of failure curves for a 7075-T6 Al alloy

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• Crack grows incrementallytyp. 1 to 6

( ) a~ sD

increase in crack length per loading cycle

• Failed rotating shaft--crack grew even though

Kmax < Kc--crack grows faster as

• Ds increases• crack gets longer• loading freq. increases.

crack origin

Fatigue Mechanism

( )mKdNda

D=

Texture of the fatigue fracture surface – clamshell or beachmark texture

Intrusions and extrusions. SEM.

20

Transmission electron fractograph showing fatique striations in aluminum

Each striationis thougth to represent the advance distance of a crack front during a single load cycle, striation width deponds on and incearses with increasing stress range.

Improving Fatigue Life

2. Impose a compressive surface stress (to suppress surface cracks from growing)

--Method 1: shot peening

put surface

into compression

shot--Method 2: carburizing

C-rich gas

3. Remove stress concentrators.bad

bad

better

better

1. Decrease the mean stress level

21

• Engineering materials don't reach theoretical strength.• Flaws produce stress concentrations that cause

premature failure.

• Sharp corners produce large stress concentrationsand premature failure.

• Failure type depends on T and stress:

- for noncyclic s and T < 0.4Tm, failure stress decreases with:- increased maximum flaw size,- decreased T,- increased rate of loading.

- for cyclic s:- cycles to fail decreases as Ds increases.

- for higher T (T > 0.4Tm):- time to fail decreases as s or T increases.

SUMMARY

Homework 8.6 and 8.12, p246