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Fracture mechanic, analysis, Griffith equation

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## PowerPoint Slideshow about 'Fracture' - nizamtm

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●What is Fracture?

●The Griffith Equation

●The Orowan Modification

●Statistics—Weibull Distribution

●Examples

– 35 MgO samples

– Wesgo Al95 Alumina

– Plaster

●Transition to Fracture Mechanics

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

●Most Fracture failures are “unexpected” failures

●What is an “unexpected” failure?

●The expected load carrying capacity of a

structural member is its yield strength times its

cross sectional area

●The presence of a flaw can change this

●The following approach was proposed by

Griffith in 1920

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Uniform Stress, σ σ

A plate of unit thickness has a through crack

of length 2c. The crack has a surface energy

of 4cγγ, where γγ is the energy required to

create a unit area of new surface.

The total energy content of the plate, under the

applied stress is σ σ

energy is reduced by the volume of material in

the cylinder of radius c.

σ2/E. Now, if the crack extends

δ δ

2 c

2/E. Due to the crack, this

Uniform Stress, σ σ

This energy is π

a small amount, δ δ , then the strain energy is

further reduced by an amount (π

The additional surface is 4 δ δ , so the additional

surface energy is 4 δ δ γ γ.

π c2σ

(π c2σ

σ2/E)δ δ.

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

We set these two quantities equal to one another

So, 4δ γ

stress at which the crack will propagate is:

δ γ = 2π π cδ σ

δ σ2/E. The deltas cancel, and so the

And that is the Griffith Equation.

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

For ductile materials, Orowan modified the

Griffith equation in 1952 by substituting two

terms for γ γ : γ γ s is the surface energy, and γ γ pis the

plastic energy. γ γ s is about 1-2 J/m2, and γ γ p is 100-

1000 J/m2, so we can write:

Before we use this equation as a lead-in to Fracture Mechanics,

some additional aspects of brittle fracture:

Statistical nature–Weibull Distribution

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

●Probability of failure, S = n/(N+1) where N is the

total number of specimens and n is the order

number in increasing strength

● Then Si = (1-e -Bi) where Bi is called the risk of

rupture, and Bi = ((σi – σu)/σo)m

● Where σi is the ith specimen strength, σu is the

so-called “zero strength” , σo is a normalizing

factor and m is the flaw density exponent.

●Sometimes B is multiplied by a dimensionless

volume to account for a possible size effect.

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

●Take the double logarithm of both sides and plot

log log S vs. log (σi – σu) for different values of

σu.

● The best value of σu will give a straight line of

slope m.

●A computer program was written to find the best

fit using different choices of σu . (minimize the

sum of the deviations squared from a line).

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

35 MgO Strength Values in increasing order

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

23700

23950

25050

25800

25950

26900

26950

27050

27800

28050

28150

28200

28300

28400

28550

28800

29000

29300

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

29350

29400

29450

29600

29700

30400

30500

30850

30950

31000

31200

31400

31450

31700

31750

31900

34400

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Fracture Mechanics Jacobson 5/16/08

Fracture Mechanics

Earlier we derived the Orowan modification to the Griffith

equation, to take into account plastic deformation as an energy

absorbing mechanism in crack propagation:

absorbing mechanism in crack propagation:

Earlier we derived the Orowan modification to the Griffith

equation, to take into account plastic deformation as an energy

George Irwin, in the early 1960's proposed a quantity G, the

strain energy release rate, or the crack extension force, giving an

equation for the fracture stress in terms of G:

equation for the fracture stress in terms of G:

George Irwin, in the early 1960's proposed a quantity G, the

strain energy release rate, or the crack extension force, giving an

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

but, Jacobson 5/16/08γp and Gc are difficult quantities to measure, so Irwin

proposed the stress intensity factor,

stress and a is the half crack length. It is similar to a stress

concentration factor, and can be easily measured. K has

somewhat unusual units of Mpa-m1/2 or KSI-in1/2. Now,

and K2 = GE so, substituting we get:

and K2 = GE so, substituting we get:

and K2 = GE so, substituting we get:

but, γp and Gc are difficult quantities to measure, so Irwin

proposed the stress intensity factor,

stress and a is the half crack length. It is similar to a stress

concentration factor, and can be easily measured. K has

somewhat unusual units of Mpa-m1/2 or KSI-in1/2. Now,

somewhat unusual units of Mpa-m1/2 or KSI-in1/2. Now,

where σ is the

where σ is the

where σ is the

proposed the stress intensity factor,

stress and a is the half crack length. It is similar to a stress

or, more generally, where α includes a number of geometrical

factors:

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

The Critical Stress Intensity Factor Jacobson 5/16/08

● Usually expressed as KIc, and it is a true

material property

●There is a minimum thickness in order to be

certain that plane strain conditions exist

– Thickness, B = 2.5(KIc/σo)2 is the critierion where

σo is the 0.2% offset yield strength

●There can be no assurance that a test will be

valid until it is performed

●An estimate of thickness can be made using an

expected value of KIc

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Crack Opening Modes Jacobson 5/16/08

I

II

III

II

III

I

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Types of Test Specimens Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Types of Test Specimens (cont.) Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Different types of load-displacement Jacobson 5/16/08

curves

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Explanation of Curves Jacobson 5/16/08

●Type I—typical of most ductile metals. Line OP

is at 5% lower slope than tangent OA. PS = PQ

– If x1 is more than ¼ of xS then material is too ductile

●Type II—shows a pop-in, but must meet the

maximum ductility criterion—PQ is max load

●Type III—shows a complete pop-in instability

and is characteristic of a brittle “elastic” material

● PQ is used to calculate KQ and if the thickness is

greater than calculated for plane strain, then KQ

equals KIC.

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Example: Thin-Wall Pressure Vessel Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Thin-Wall Pressure Vessel Jacobson 5/16/08

●Material: Ti 6Al 4V, Hoop Stress = 360 MPa

● KIc = 57 Mpa m1/2

● σo = 900 Mpa

●Crack oriented as shown above

●For this type of loading and geometry:

2 = (1$.21aπσ2)/Q

● KI

●a =Surface crack depth

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Thin-Wall Pressure Vessel Jacobson 5/16/08

● For a wall thickness of 12 mm, and σ/σ0 = 0.4

●If 2c=2a then Q = 2.35

● Then the critical crack size, ac = 15.5 mm

●The critical crack depth is greater than the wall

thickness and the vessel will “leak before burst”

● However if a/2c = 0.05, Q = 1 and ac = 6.6 mm

●This is less than the wall thickness

●So, vessel will burst!

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Steel, compression side Jacobson 5/16/08

Crack growth by fatigue

Pre-existing crack

1"

Steel, tension side

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Influence of Jacobson 5/16/08

corrosive

environment

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Periodic Overloads Jacobson 5/16/08

Notch

Root

Extent

of flat

fracture

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Periodic Overloads Jacobson 5/16/08

●The overload creates a larger plastic zone

●If applied infrequently, the residual stress does

not build up to slow the crack propagation

●If applied to often, it causes a greater rate of

crack propagation

●At an intermediate rate, the plastic zone, and

hence the residual compressive stress, is

present for more of the growth cycles

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

Plastic Zone Size Jacobson 5/16/08

rp = (1/2π)(σ2a/σ0

2)

For a steel with

a = 10 mm,

σ = 400 Mpa

σ0 = 1500 Mpa

rp = 0.113 mm

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

METE 327 Physical Metallurgy Copyright 2008 Loren A. Jacobson 5/16/08

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