FRACTURE - New Mexico Institute of Mining and Technologyljacobso/Fracture slides.pdf · It is...

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

FRACTURE

METE 327Fall 2008


OUTLINE● What is Fracture?● The Griffith Equation● The Orowan Modification● Statistics—Weibull Distribution● Examples

– 35 MgO samples– Wesgo Al95 Alumina– Plaster

● Transition to Fracture Mechanics


What is Fracture?● 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


Uniform Stress, σ

Uniform Stress, σ

2 cδ

The Griffith EquationA plate of unit thickness has a through crackof length 2c. The crack has a surface energyof 4cγ , where γ is the energy required tocreate a unit area of new surface.The total energy content of the plate, under theapplied stress is σ 2/E. Due to the crack, thisenergy is reduced by the volume of material inthe cylinder of radius c.

This energy is πc2σ2/E. Now, if the crack extendsa small amount, δ, then the strain energy isfurther reduced by an amount (πc2σ2/E)δ.The additional surface is 4 δ, so the additionalsurface energy is 4 δγ.


The Griffith Equation (cont)

So, 4δ γ = 2π cδ σ 2/E. The deltas cancel, and so thestress at which the crack will propagate is:

And that is the Griffith Equation.

We set these two quantities equal to one another


For ductile materials, Orowan modified theGriffith equation in 1952 by substituting twoterms for γ : γ s is the surface energy, and γ pis theplastic energy. γ s is about 1-2 J/m2, and γ p is 100-1000 J/m2, so we can write:

Orowan Modification

Before we use this equation as a lead-in to Fracture Mechanics,some additional aspects of brittle fracture:

Statistical nature–Weibull Distribution


The Weibull Distribution● 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 B

i is called the risk of

rupture, and Bi = ((σ

i – σ

u)/σ

o)m

● Where σi is the ith specimen strength, σ

u 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.


Weibull Distribution (cont.)● Take the double logarithm of both sides and plot

log log S vs. log (σi – σ

u) 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).


1 237002 239503 250504 258005 259506 269007 269508 270509 2780010 2805011 2815012 2820013 2830014 2840015 2855016 2880017 2900018 29300

19 2935020 2940021 2945022 2960023 2970024 3040025 3050026 3085027 3095028 3100029 3120030 3140031 3145032 3170033 3175034 3190035 34400

35 MgO Strength Values in increasing order


Fracture Mechanics

Earlier we derived the Orowan modification to the Griffithequation, to take into account plastic deformation as an energyabsorbing mechanism in crack propagation:

George Irwin, in the early 1960's proposed a quantity G, thestrain energy release rate, or the crack extension force, giving anequation for the fracture stress in terms of G:

Fracture Mechanics

Earlier we derived the Orowan modification to the Griffithequation, to take into account plastic deformation as an energyabsorbing mechanism in crack propagation:

George Irwin, in the early 1960's proposed a quantity G, thestrain energy release rate, or the crack extension force, giving anequation for the fracture stress in terms of G:


but, γ p and Gc are difficult quantities to measure, so Irwinproposed the stress intensity factor, where σ is thestress and a is the half crack length. It is similar to a stressconcentration factor, and can be easily measured. K hassomewhat unusual units of Mpa-m1/2 or KSI-in1/2. Now,

and K2 = GE so, substituting we get:

but, γ p and Gc are difficult quantities to measure, so Irwinproposed the stress intensity factor, where σ is thestress and a is the half crack length. It is similar to a stressconcentration factor, and can be easily measured. K hassomewhat unusual units of Mpa-m1/2 or KSI-in1/2. Now,

and K2 = GE so, substituting we get:

proposed the stress intensity factor, where σ is thestress and a is the half crack length. It is similar to a stress

somewhat unusual units of Mpa-m1/2 or KSI-in1/2. Now, and K2 = GE so, substituting we get:

or, more generally, where α includes a number of geometricalfactors:


The Critical Stress Intensity Factor● Usually expressed as K

Ic, 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(K

Ic/σ

o)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 K

Ic


I

I

II

II

III

III

Crack Opening Modes


Types of Test Specimens


Types of Test Specimens (cont.)


Different types of load-displacement curves


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

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

Q

– If x1 is more than ¼ of x

S then material is too ductile

● Type II—shows a pop-in, but must meet the maximum ductility criterion—P

Q is max load

● Type III—shows a complete pop-in instability and is characteristic of a brittle “elastic” material

● PQ is used to calculate K

Q and if the thickness is

greater than calculated for plane strain, then KQ

equals KIC

.


Example: Thin-Wall Pressure Vessel


Thin-Wall Pressure Vessel● Material: Ti 6Al 4V, Hoop Stress = 360 MPa● K

Ic = 57 Mpa m1/2

● σo = 900 Mpa

● Crack oriented as shown above● For this type of loading and geometry:● K

I2 = (1$.21aπσ2)/Q

● a =Surface crack depth


Thin-Wall Pressure Vessel

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

● If 2c=2a then Q = 2.35● Then the critical crack size, a

c = 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!


1"

Steel, compression side

Steel, tension side

Pre-existing crackCrack growth by fatigue


Influence ofcorrosiveenvironment


Periodic Overloads

Notch Root

Extent of flat fracture


Periodic Overloads● 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


Plastic Zone Size

rp = (1/2π)(σ2a/σ

02)

For a steel with a = 10 mm,σ = 400 Mpaσ

0 = 1500 Mpa

rp = 0.113 mm

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