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# Fracture - PowerPoint PPT Presentation

Fracture mechanic, analysis, Griffith equation

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METE 327

Fall 2008

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

●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

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

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

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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

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

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

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

Influence of Jacobson 5/16/08

corrosive

environment

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

Notch

Root

Extent

of flat

fracture

METE 327 Physical Metallurgy Copyright 2008 Loren A. 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