Chapter 2 - General Fracture Mechanics

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Chapter 2 – An Introduction to Fracture Mechanics

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• Definition• Stress Concentration• Effects of cracks on strength of

materials• Strain Energy Release Rate, G • Stress Intensity Factor, K• Application of K to Design and

Analysis

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http://en.wikipedia.org/wiki/Fracture_mechanics

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INTRODUCTION• Most materials show a tendency to fracture when stressed beyond

some critical level.• Usually, fracture is caused by a structural flaw or a crack: under the

loading conditions, a crack may develop (starting from a flaw or stress concentration) and grow slowly in size.

• During the continuing development of the cracks, the material strength decreases until it becomes so low that the service loads cannot be carried any more, and fracture occurs.

• This may occur at stresses (caused by external loads) below the material’s yield strength where fracture would not normally expected.

• In additional to cracks, other types of flaws that are crack-like in form may easily develop into cracks, such as deep surface scratches or gouges, voids in welds, inclusions of foreign substances in cast and forged materials, delaminations in layered materials and so on.

• Cracks or crack-like flaws occur more frequently than we might think about.

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Cracks or crack-like flaws commonly occur in ship structures, in bridge structures, in pressure vessels and piping, in heavy machinery, and in ground vehicles. They are also a source of concern for various parts of nuclear reactors.

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QUESTIONS TO BE ASKED

• What is the residual strength as a function of crack size?• What size of crack can be tolerated at the expected

service load; i.e., what is the critical crack size?• How long does it take for a crack to grow from a certain

initial size to the critical size?• What size of pre-existing flaw can be permitted at the

moment the structure/material starts its service life?• How often should the structure/material be inspected for

crack?

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STRESS CONCENTRATION

Consider above diagram, what will happen when the left bar is cut into two?

Now, for above diagram, if the left side of the bar is cut, then what will happen?

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STRESS CONCENTRATION

It is helpful to consider load-path (load-flow) lines: imaginary lines indicating e.g., how one unit of load is transferred from one loading point to the other. What will happen to the load-flow lines when no cut in the material? What will happen if there is a cut in the material?

Every discontinuity forms an interruption of the load-path, will therefore deviate the load-flow lines and hence cause a stress concentration.

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STRESS CONCENTRATION FACTOR

In general, blunt notches produce lower local stresses, sharp notches cause higher local stresses. The highest local stress σl is a number if times higher than the nominal stress, σnom.

l t nomkσ σ=kt is called theoretical stress concentration factor (elastic stresses)

For an elliptical notch:

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1 2 1 2tb b aka b

ρρ

⎛ ⎞= + = + =⎜ ⎟

⎝ ⎠

For a circle, a=b, kt = 3

For an ellipse of b/a=3, kt = 7

For an ellipse of a/b=3, kt = 1.67

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Elliptical hole in a wide plate under remote uniform tension, and the stress distribution along the x-axis near the hole for one particular case.

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EFFECTS OF CRACKS ON STRENGTH

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• If a load applied to a member containing a crack is too high, the crack may suddenly grow and cause the member to fail by fracturing in a brittle manner, that is with little plastic deformation.

• From the theory of fracture mechanicsfracture mechanics, a quantity called stress intensity factor,stress intensity factor, K, can be defined that characterizes the severity of the crack situation as affected by crack size, stress, and geometry.

• A given material can resist a crack without brittle fracture occurring as long as this K is below a critical value, Kc, which is a property of the material called the fracture fracture toughnesstoughness..

•• Fracture toughnessFracture toughness is a material property which is affected by temperature and loading rate, and secondarily by the thickness of the members.

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Consider two materials, one with low σ0 and high Kc, and the other with an opposite combination, namely high σ0 and low Kc. How do these combinations of properties cause crack sizes in these two materials?

What does the crack size affect the materials in above two cases?

In general, low strength in a tension test is usually accompanied by high ductility and also by high fracture toughness, Conversely, high strength is usually associated with low ductility and low fracture toughness.

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BASIC MODES OF FRACTURE

Mode I: opening mode: caused by tensile loading, crack faces simply moving apart;

Mode II: sliding mode: caused by shear loading, crack faces slide relative to one another in a direction normal to the crack front.

Mode III: tearing mode: caused also by shear loading, crack faces slide relative to one another but in the direction parallel to the crack front.

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STRAIN ENERGY RELEASE RATEConsider a cracked member under a Mode I load, P, where the crack has length a as shown in the above drawing, and assume that the behavior of the material is linear-elastic (load vs displacement being linear).

U is potential energy stored in the member and the rate of change of potential energy with increase in crack area is defined as the strain energy release rate, G.

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STRESS INTENSITY FACTOR, K

Consider a coordinate system for describing the stresses in the vicinity of a crack is shown in above drawing with Mode I loading.

K, characterizes the magnitude (intensity) of the stresses in the vicinity of an ideally sharp crack tip in a linear-elastic and isotropic material.

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STRESSES NEAR THE CRACK TIP

( )2

Iij ij

K fr

σ θπ

=

For any case of Mode I loading, the stresses near the crack tip depends on r and θ :

KI is a measure of the severity of the crack, its definition in a formal mathematical sense is:

( ), 0lim 2I yr

K rθ

σ π→

=

It is generally convenient to express KI as:

( ); /IK YS a Y F a Lπ= =

Y is geometry constant depending on crack size (a) and a size parameter of the body (L), S is remotely applied stress

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( ); /I gK YS a Y F a Lπ= =

Sg is gross section nominal stress, calculated under the assumption that no cracks is present.

Case (a) is called a central crack, it is a through-thickness crack, crack length is defined as half of the total crack length

Case (b) is called a double edge crack, or simply double surface crack, again, it is through-thickness crack

Case (c) is called a edge crack, or a surface crack, again, it is through-thickness crack, in both cases, crack length is defined as total crack length

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Values for small a/b and limits for 10% accuracy:

( ) ( ) ( )( ) ( ) ( )

; 1.12 ; 1.12

/ 0.4 / 0.6 / 0.13g g ga K S a b K S a c K S a

a b a b a b

π π π= = =

≤ ≤ ≤

Stress intensity factors for

three cases of cracked plates under tension

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Values for small a/b and limits for 10% accuracy:

( ) ( ), 1.12 / 0.4I ga b K S a a bπ= ≤

Stress intensity factors for

various cases of bending

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Stress intensities for a

round shaft with a

circumferential crack.

(a) Axial load P: ( )2 ; 1.12; 10%, / 0.21gPS Y a bbπ

= = ≤

(b) Bending moment M: ( )3

4 ; 1.12; 10%, / 0.12gMS Y a bbπ

= = ≤

(c) Torsion T, K=KIII: ( )3

2 ; 1.00; 10%, / 0.09gTS Y a bbπ

= = ≤

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FRACTURE MECHANICS ANALYSIS

I IcK K S aπ= =

Materials Selection

Design StressAllowable flaw size

Applied Stress

Crack sizeFracture toughness

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The plane strain fracture toughness KIC is a fundamental material property that depends on many factors, the most influential of which are temperature, strain rate, and microstructure. The magnitude of KIC diminishes with increasing strain rate and decreasing temperature.

According to this equation, three variables must be considered relative to the possibilities for fracture of some structural component – namely, the fracture toughness (Kc or KIC), the imposed stress (S), and the flaw size (a) –assuming that Y has been determined.

21orIC ICc c

K KS aSYY a ππ

⎛ ⎞≤ = ⎜ ⎟⎝ ⎠

If the KIC and the magnitude of crack size are specified by

application constraint, it gives how to determine the design (or

critical) stress.

If the KIC and the level of stress are fixed by design situation, it gives the maximum allowable

flaw size.

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FRACTURE MECHANICS ANALYSIS• The strength of materials approach – traditional approach

to structural design and material selection

Applied Stress Yield or tensile strength

• The fracture mechanical approach – has three important variables

Applied Stress

Fracture ToughnessCrack Size

Fracture mechanics quantifies the critical combinations of these three variables.

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FRACTURE MECHANICS ANALYSIS

I IcK K YS aπ= =

• This equation shows the design criterion for material selection in engineering applications.

• You must first decide what is most important about you component design. Once such a priority list is established, certain critical decision can be made. If one of the above three parameters is fixed, then other two can only change according to fracture mechanics relation. If two of them are fixed, basically you have also fixed third one.

• Increasing strength, such as yield strength, usually leads to smaller critical crack length.

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Example 2.1A high-strength steel has a yield strength of 1,460 MPa and a KIC of 98 MPa√m. Calculate the size of a surface crack that will lead to catastrophic failure at an applied stress of ½ Y.S.

Example 2.1 Solution:We are assuming an ideal case of plane strain conditions and assume surface crack, therefore F=1.12. Within these limitations, we can use:

ICK YS aπ=

With Y=1.12 and S=0.5*Yield Strength (S)., then

( )( )

( )

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1.12 0.5* *or

98MPa m1 1 4.573 10 m 4.573mm1.12 0.5* 1.12 0.5 1460MPa

IC

IC

K S a

Ka xS

π

π π−

= ∗

= = = =∗ ∗ ∗⎡ ⎤⎣ ⎦

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Example 2.2Given that a quality-control inspection can ensure that a structural ceramic part will have no flaws greater than 25 μmin size, calculate the maximum service stress available with (a) SiC and (b) partially stabilized zirconia.

Example 2.2 Solution:We can treat this problem as a general fracture mechanics problem using fracture mechanic relation with assuming surface crack, therefore Y=1.12, in which case:

ICf

KSY aπ

=

This problem assumes that the maximum service stress will be thefracture stress for a part with flaw size = a = 25 μm. The KIC for SiC and PS-Zirconia (PSZ) is 3 and 9 MPa√m, respectively.

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(a) For SiC:

-6

3MPa m 302MPa1.12* 25 10 m

fSπ

= =× ×

(b) For PSZ:

-6

9MPa m 907MPa1.12* 25 10 m

fSπ

= =× ×

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EXAMPLE 2.3

A center-cracked plate has dimensions b=50 mm, t = 5 mm, and large h, and a load of P=50 kN is applied.

(a) What is the stress intensity factor K for a crack length of a=10 mm?

(b) For a=30 mm?(c) What is the critical crack length ac for fracture if the

material is 2014-T651 aluminum?

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Example 2.3 Solution:(a)To calculate K, we need to know stress:

( )( )0.050MN 100MPa

2 2 0.05m 0.005m10mm= 0.250mm

gPSbt

ab

α

= = =

= =

Since α ≤ 0.4, it is within 10% to use Y=1.

( ) ( )100MPa 0.010m 17.7MPa mgK FS aπ π= = =(b) For a=30 mm:

30mm 0.650mm

ab

α = = =

This does not satisfy α ≤ 0.4, so that the more general expression for Yis needed:

( ) ( )

21 0.5 0.326 1.2921

1.292 100MPa 0.030m 39.7MPa mg

Y

K YS a

α αα

π π

− += =

−

= = =

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Example 2.3 Solution (cont.)(c) For 2014-T651 Al, KIC = 24 MPa√m. Since ac is not known, Y cannot be determined directly. We may need to use trial and error method.

First, assume that α ≤ 0.4 is satisfied, in which case Y=1.

Solving for ac gives:2 2

1 1 24MPa m 0.0183m=18.3mm100MPa

ICc

g

KaSπ π

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

This corresponds to:

18.3 0.3750

cab

α = = =

which satisfies α ≤ 0.4, so that the estimated Y=1 is acceptable and the result obtained can be accepted.

However, it is better to use trial and error to make more accurate estimation. This method is described as below:

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Example 2.3 Solution (Cont.)Trial and error method:

Use the ac value from above as a rough estimate, and calculate K for several crack lengths in this neighborhood. Use the sample procedure as in (b) above for each crack length. When K = KIC is obtained, the crack length is the desired value.

Calc. No Trial a α = a/b Y K=YSg√πamm MPa√m

1 10 0.20 1.021 18.12 15 0.30 1.051 22.83 20 0.40 1.100 27.64 16.5 0.33 1.063 24.25 16.3 0.326 1.0617 24.0

The desired value of KIC=24MPa√m was between calculated values No 2 and 3, and the second of two additional guesses gave the desired K=KIC at a crack length of ac=16.3 mm.

One can also use graphical procedure to obtained the same results.

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EXAMPLE 2.4• In a thin-walled pressure Vessel with a crack growing in

the wall, two possibilities exist: (1) the crack may gradually extend and penetrate the wall, causing a leak before sudden brittle fracture can occur;(2) Sudden brittle fracture may occur prior to the vessel leaking. Since a brittle fracture in a pressure vessel may involve explosive release of the vessel contents, a leak is by far the more preferable of the two undesirable possibilities.

• Note that a leak is easily detected from a pressure drop or from the presence of vessel contents where they do not belong. Hence, pressure vessels should be designed to leak before they fracture. This is so-called leak-before-break design of pressure vessels.

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This figure illustrates a crack in the wall of a pressure vessel, showing (a) its growth from a small surface flaw and (b) the minimum critical size of a through-wall crack to provide leak-before-break.

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Example 2.4 Question:

A pressure vessel made of ASTM A517-F steel operates near room temperature and has a wall thickness of t = 50 mm. A surface crack was found in the vessel wall during an inspection. It has an approximately semi-elliptical shape, as illustrated below, with surface length 2c = 40 mm and depth a = 10 mm. The stresses in the region of the crack, as calculated without considering the presence of the crack, are approximately uniform through the thickness and are σz = 300 MPa normal to the crack plane and σx = 150 MPa parallel to the crack plane, where the coordinate system is shown in the figure. What is the safety factor against brittle failure? Would you remove the pressure vessel from service? Discuss your results.

x

yz

σ

tb

2c

aD

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[KIC = 187 MPa√m and σys = 760 MPa for ASTM A517-F steel, the stress intensity factor at Point D for a uniform tension normal to the crack plane is:

For the case of semi-elliptical crack, YD ≈ 1.12, the limits for 10% accuracy is a/t < 0.3, and c/b < 0.2].

( )1.65

; 1 1.464 / 1D Da aK Y Q a c

Q cπσ ⎛ ⎞= ≈ + ≤⎜ ⎟

⎝ ⎠

Example 2.4 Solution:

The material (ASTM A517-F) has a fracture toughness of KIC = 187 MPa√m and a yield strength of σ0 = 760 MPa at room temperature. The stress intensity factor, K, for the given stresses and crack can be estimated. Since c = 20 mm, we have a/c=0.5, also, we have a/t = 0.2, and large b, for which YD = 1.12 is a reasonable approximation. The quantity Q is needed:

1.65

1 1.464 1.466aQc

⎛ ⎞= + =⎜ ⎟⎝ ⎠

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Hence, the maximum K, which occurs at the point of maximum depth of the elliptical crack, is approximately:

The stress-based safety factor against brittle fracture can be calculated directly from K values as these are proportional to σ for any given crack length:

( )0.011.12 300 49.2 MPa m

1.466DaK Y

Qππσ×

= = × =

187 3.8049.2

ICKXK

= = =

This is a reasonably high value, so that it would be safe to continue using the pressure vessel until repairs are convenient. However, the crack should be checked frequently to be sure that it is not growing. In addition, the ASME or other design code for pressure vessels is likely to apply, and it should be consulted in this situation.

Example 2.4 Solution (Cont.):

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EXAMPLE 2.5

A spherical pressure vessel is made of ASTM A517-F steel and operates at room temperature. The inner diameter is 1.5 m; the wall thickness is 10 mm, and the maximum pressure is 6 MPa. Is the leak-before-break condition met? What is the safety factor on K relative to KIC, and what is the safety factor against yielding? [KIC = 187 MPa√m and σys = 760 MPa for ASTM A517-F steel].

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Example 2.5 Solution:

The stresses for a spherical geometry can be seen from the part (b) of above figure. They are approximation within 5% for t/r1 <0.3 and 10% for t/r1 < 0.45.

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The stress in the vessel wall is:

1 6 750 225 MPa2 2 10tpr

tσ ×

= = =×

Combining this value with KIC = 187 MPa√m, the critical crack length is obtained from:

2 21 1 187 0.220 m 220 mm225

ICcrit

t

Kcπ σ π⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠This far exceeds the wall thickness of t=10 mm, so that the leak-before-break condition is met.

When the vessel leaks, the crack length along the surface is c=t, so that c = 10 mm. At this point, the stress intensity factor is:

( )1 225 0.01 39.9 MPa mK FS cπ π= = × × × =

Example 2.5 Solution (Cont.):

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Example 2.5 Solution (Cont.):

Hence, the safety factor on K is:

187 4.6939.9

ICK

KXK

= = =

This is a reasonable value, so that the vessel is safe from brittle fracture. The above value may also be interpreted as a safety factor on stress or pressure for the brittle fracture situation, but it differs from the safety factor against yielding.

Noting that the principal stresses are σ1 = σ2 = 225 MPa and σ3 = 0, the yielding criterion if using the yield stress as:

( )1 2 2 3 3 1, , 225 MPaMAX σ σ σ σ σ σ− − − =

The safety factor against the yielding is therefore:

760 3.38225

yy

a

Xσσ

= = =

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