Fm Lecture 2

M. Mirzaei, Fracture Mechanics

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Fracture Mechanics Lecture Notes: 2 Majid Mirzaei, PhD, Associate Professor, Dept. of Mechanical Eng., TMU [email protected] http://www.modares.ac.ir/eng/mmirzaei/FM.htm

Fundamentals of Fracture Mechanics Fracture is the separation of a component into, at least, two parts. This separation can also occur locally due to formation and growth of cracks. Let us investigate the force required for such a separation in a very basic way. A material fractures when sufficient stress and work are applied on the atomic level to break the bonds that hold atoms together. Figure 1 shows a schematic plot of the potential energy and force versus separation distance between atoms.

Repulsion

Attraction

PotentialEnergy

Distance

Bond Energy

EquilibriumSpacing

Tension

Compression

BondEnergy

Cohesive Force

AppliedForce

λ

K

--

-

--

-

--

---

-- -

--

--

--X0

Figure 1


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The bond energy is given by,

0

bx

E Pdx∞

= ∫ (2-1)

where x0 is the equilibrium spacing and P is the applied force. A reasonable estimate of the cohesive strength at the atomic level can be obtained by idealizing the interatomic force-displacement relationship as one half the period of a sine wave, so we may write:

sincxP P πλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

(2-2)

where λ is defined in Fig.1. For small displacements, we may consider further simplification by assuming:

cxP P πλ

⎛ ⎞= ⎜ ⎟⎝ ⎠

(2-3)

Hence, the bond stiffness (i.e., the spring constant) can be defined by:

cPk πλ

= (2-4) If both sides of this equation are multiplied by the number of bonds per unit area and the equilibrium spacing x0 (gage length), then k can be converted to Young’s modulus E and P to the cohesive stress cσ . Solving for cσ gives:

0c

Exλσ

π= (2-5)

Assuming 0xλ ≈ , we may write Eqn (2-5) as:

cEσπ

≈ (2-6) For steels with the Young’s modulus of 210 GPa, the above equation estimates a fracture stress of 70000MPa, which is almost 25 times the strength of the most strong steels!! The reason behind the above huge discrepancy is the existence of numerous defects in ordinary materials. These defects can be quite diverse by nature. Starting from the atomic scale, they may include point defects (for example vacancies: atoms missing), line defects, extra atomic planes (dislocations). On the microstructural level we may consider defects due to grain boundaries, porosity, etc. Some of these defects may also evolve


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during the processes of deformation and fracture. For instance, plastic deformation involves various movements of dislocations which may interact and eventually result in local damages. Plastic deformation may also lead to the formation of microvoids which may coalesce and evolve to microcracks. On the other hand, the term fracture mechanics has a special meaning: description of fractures which occur by propagation of an existing sharp crack. Hence, the assumption of a preexisting crack is essential in fracture mechanics. As shown above, it is evident that numerous microscopic defects and/or microcracks naturally exist in ordinary materials. However, the scope of Engineering Fracture Mechanics is almost entirely concerned with macrocracks which are either present in the components, as a result of manufacturing processes like welding, or develop during the service by various failure mechanisms such as fatigue or creep. The crack propagation can occur in many ways. For instance we may have fast-unstable and slow-stable crack growth under monotonic loading, or a cycle by cycle growth under alternating loads. In general, the resistance to crack growth can be defined by a special term called the toughness of the material. Ductile Versus Brittle By definition, ductile fracture is always accompanied by a significant amount of plastic deformation, while brittle fracture is characterized by very little plastic deformation (see Fig.2). Both types of fracture have distinctive features on macro and micro levels.

BrittleDuctile

Figure 2 To a large extent, ductility and brittleness depend on the intrinsic characteristics of materials such as chemical composition and microstructure. Nevertheless, extrinsic parameters like temperature, state of stress, and loading rate may have substantial influences on the fracture properties of materials. In general, materials show brittleness at low temperatures, high strain rates, and triaxial state of tensile stress. Let us consider the deformation and fracture mechanisms of a ductile material subject to a simple tension test during which several strength levels can be defined (see Fig.3). The first one is the proportional limit below which there is a linear relationship between the stress and strain (point A). The second one is the elastic limit which defines the stress level below which the deformation is totally reversible (point B).


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Strain

StressTest

Specimen Gauge length

AB

C

D

E

F

Figure 3

The third level is the yield strength which marks the beginning of irreversible plastic deformation (point C). Some materials show a clear yield point and also a lower yield point like point D. For others the yield strength is a point that is difficult to define and, in practice, it is defined as the intersection of the σ-ε curve and a line parallel to the elastic portion of the curve but offset from the origin by 0.2% strain. Beyond this point and up to the next level, which is the ultimate tensile strength (point E), the plastic deformation is uniform along the gage length of the specimen. The inflection point in the σ-ε curve is due to the onset of localized plastic flow or necking as depicted in the Figure 4(A). Finally, the point F represents the final fracture.

A B C D

Figure 4

The occurrence of necking results in a triaxial state of stress. Accordingly the plastic deformation becomes more difficult and small particles within the microstructure start to fracture or separate from the matrix causing microvoids, as depicted in Figure 4(B). The microscopic appearance of microvoid formation is depicted in Fig.5.


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Figure 5 The resulting microvoids will eventually coalesce and form an internal disk-shaped crack. The final fracture occurs by a shearing-off process of the internal crack towards the specimen surface. The result is a typical cup-cone fracture depicted in Figures (4D) and (6a).

Figure 6

As mentioned before, brittle fracture is characterized by very little plastic deformation which usually results in flat fracture surfaces as depicted in Figure 6b. The most important type of brittle fracture is called Cleavage. In this type of fracture, which is also called transgranular, cracking occurs through separation of certain crystalloghraphic planes within grains. The result is a very shiny and flat fracture surface. Another type of fracture, called intergranular, occurs through separation of grains from each other and can be attributed to those mechanisms which weaken the grain boundaries. Creep fracture is a typical example.


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Linear Elastic Fracture Mechanics, Energy Approach The aim of this section is to present the fundamental aspects of Linear Elastic Fracture Mechanics (LEFM) using an Energy approach. Let us start with a brief review of some of the relevant material from the theory of elasticity. Consider a deformable body in an equilibrium state under the influence of surface tractions and body forces as depicted in Fig.7. The virtual work can be defined as “the work done on a deformable body, by all the forces acting on it, as the body is given a small hypothetical displacement which is consistent with the constraints present”. The virtual displacements are represented by the symbol “δ”, known as the variation operator. In general, the loadings consist of body forces and surface tractions. The later are prescribed over a part of the boundary designated by Sσ . Over the remaining boundary, designated by uS , the displacement field u is prescribed. However, it must be ensured that

0δ =u on uS to avoid violating the constraints.

Su

Su

S

T

f Figure 7

Accordingly, the virtual work may be defined by:

virtS V

i i i iS V

W

T u dS f u dV

δ δ δ

δ δ

= +

= +

∫ ∫

∫ ∫

T. u f . u

(2-7)

in which T and f are the surface-traction and body-force vectors, respectively. Now we may invoke the stress boundary relations and implement the divergence theorem to obtain:


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( )

( ) ( )

,

0

, ,

i i i i ij j i i iS V S V

ij i i ijV

ij j i i ij i jV

T u dS f u dV n u dS f u dV

u f u dV

f u u dV

δ δ σ δ δ

σ δ δ

σ δ σ δ=

+ = +

⎡ ⎤= +⎣ ⎦

⎡ ⎤= + +⎢ ⎥

⎣ ⎦

∫ ∫ ∫ ∫

∫

∫

(2-8)

In the above, ijσ are the components of the Cauchy stress tensor and jn are the components of the outward unit normal to the surface. The first grouping of terms within the last integral equals zero because of the equilibrium. The product of the symmetric stress tensor with the skew-symmetric part of ,i juδ is also zero. Since the symmetric part of ,i juδ is nothing but a variation in the strain tensor, we may write the following expression known as the principle of virtual work, PVW:

i i i i ij ijS V V

T u dS f u dV dVδ δ σ δε+ =∫ ∫ ∫ (2-9)

The general form of the constitutive expressions for an elastic continuum can be written as:

ijij

ψσ ρε∂

=∂

(2-10)

in which ρ is the density, andψ is the strain energy density. If we substitute the above expression for stresses in Eq. (2-9), we will have:

i i i i ijijS V V

i i i iS V V V

T u dS f u dV dV

T u dS f u dV dV dV U

ψδ δ ρ δεε

δ δ ρδψ δ ρψ δ

∂+ =

∂

+ = = =

∫ ∫ ∫

∫ ∫ ∫ ∫

(2-11)

where U is the total strain energy stored in the body. The left side of Eq. (2-11) may be defined as a variation in the potential energy, V− , so we may write:

, ( ) 0, 0V U U Vδ δ δ δ− = + = Π = (2-12) in which Π is the total potential energy of the body. The above expression states that during the elastic deformation the external work is converted to stored elastic strain energy, and vice versa, so that the variation of the total potential energy is zero.


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In Fracture Mechanics, however, the total potential energy is the only source for crack growth. Accordingly, an energy criterion for the onset of crack growth can be defined in the following general form:

dG RdAΠ

= − ≥ (2-13)

in which G is called the energy release rate (also known as the crack driving force), A is the cracked area, and R is the resistance of the material to crack growth. The energy release rate, G, can be considered as the energy source for the crack growth and may be obtained from the stress analysis of the cracked geometry. On the other hand, the resistance to crack growth, R, can be considered as the energy sink and depends on the operating fracture mechanism. It should be mentioned that the latter depends on many factors including: the chemical composition and microstructure of the material, temperature, environment, loading rate, and the state of stress. Fixed Displacement Condition In continuation of our discussions concerning the energy approach, we will investigate the behavior of cracked components under two distinct loading conditions. First, suppose that we have stretched a cracked component by the amount ∆ and fixed it as depicted in Figure 8.

aa

P

a+da

-dP

Δ

A

B

C

D

Δ

Figure 8


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The amount of elastic strain energy stored in the component is equal to the triangle ABD and the slope of the load-displacement curve represents the stiffness of the component. Let us initially assume that the stored energy is sufficient to maintain an incremental crack growth, da, under the fixed displacement condition. Since the component with a longer crack has a lower stiffness, the stored elastic energy decreases to a new level equal to the triangle ACD. Since there is no externally applied load in the system, the total potential energy is equal to the strain energy, the only source to provide the required energy for the crack growth. Hence, we may write:

0

( 0), ( ),2

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PV U U Pd

dU dPGB da B da

Δ

Δ Δ

Δ= Π = = Δ =

Δ⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ (2-14)

where B is the thickness of the component.

Constant Load Condition: Here we consider a cracked component under a constant external load P as depicted in Fig. 9.

aa

P

P

a+da

dΔ

A

B C

DE

Δ

F

Figure 9 The amount of elastic strain energy stored in the component is equal to the triangle ABE. Now we assume that the available energy is sufficient to maintain an incremental crack growth, da, under the constant load condition. The component with a longer crack has a lower stiffness but, in this case, the stored elastic energy increases to a new level equal to


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the triangle ACD. The reason is that an excess amount of energy provided by moving the constant load P through the distance d∆ (equal to rectangle BCDE) has now been added to the system. Hence, we may write:

0 2

2

21

2P P

V P

PU Pd

PU V P

P U

dU P dGB da B da

Δ

= Δ

Δ= Δ =

ΔΠ = − = − Δ

Δ⇒Π = − = −

Δ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫

(2-15)

Note that in both cases the energy release rate is provided by the stored energy U and is equal to:

1 dUGB da⎛ ⎞= ⎜ ⎟⎝ ⎠

(2-16)

Moreover, for both cases we may write:

2

2

P

CPP CGB a

dU dUda da Δ

Δ=

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(2-17)

in which C is the compliance of the component. The above equation can be used to obtain G provided that the variation of compliance with the crack length is available. In practice, various analytical, numerical, and experimental techniques are available for this purpose. Assignment 1: Find the critical load for the Component Shown in the figure in terms of the resistance R. Assume a >> b , a >> h.


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Example Consider a large plate under remote uniaxial tensile stress. The plate has a central through-thickness crack of the length 2a, as depicted in Fig 10.

X

av

σ

σ

Figure 10 The strain energy of the above system consists of two parts: the elastic energy of the plate without crack, plus the strain energy required to introduce the crack. The latter is equal in magnitude to the work required to close the crack by the stresses acting in its position.

0

00

14 v( )2

C

a

U U U

B x dx Uσ

= +

= +∫ (2-18)

The expression for v can be obtained from a complete stress analysis of this cracked geometry. As will be shown later, this expression is:

2 22v a xEσ

= − (2-19) which shows that the crack-opening is maximum at the center of the crack and zero at its tip. Substituting for v in Eq.(2-18) we have:

2 20

BU a UE

π σ= + (2-20) which in combination with Eq.(2-15) results in:


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2

1

2

dUGB da

aE

σ

π σ

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

(2-21)

The above equation was derived for two crack tips. Accordingly, the G expression for each crack tip is:

2aGE

πσ= (2-22)

The above equation is remarkable as it shows how the energy release rate increases with increasing the far-field stress and the crack length. We may generalize the above equation for different components by writing it as:

2aGE

πσβ= (2-23) in which, β is a parameter that depends on the geometry and loading condition. Now suppose that we have an experimental specimen, similar to the one shown in Fig 10 but with finite dimensions, made of a specific alloy. Also consider a real component, made of the same material, but quite different with respect to geometry, size, loading, and crack length (see Fig 11). Also assume that we have managed to obtain the β parameter for both components and call them β1 and β2 respectively. We may perform a fracture toughness test on the experimental specimen by gradually increasing the stress and noting the critical stress level σc at which the crack starts to grow. Accordingly, we may obtain the critical energy release rate as:

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1c

caG R

Eπσβ= = (2-24)

which, in fact, represents a material property called fracture toughness. On the other hand, if the applied loading on the real component creates a far-field tensile stress around the crack tip, there exists a similitude condition between the two components. Since the two components are made of the same material, we may calculate the fracture stress for the real component as follows:

22

2

2 2

fc

cf

aG R

EEG

a

πσβ

σπβ

= =

⇒ = (2-25)


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Finally, the obtained fσ can be used to calculate the amount of the external load and/or moment associated with the onset of crack growth.

a1

σ

σMP

P

a2

σ

σ

Figure 11

In practice, however, crack growth may occur in very complicated stress fields. In general, we consider three basic modes for crack growth, although mixed-mode growth is also possible. Mode I is the opening or tensile mode where the crack faces separate symmetrically with respect to the x1-x2 and x1-x3 planes. In Mode II, the sliding or in-plane shearing mode, the crack faces slide relative to each other symmetrically about the x1-x2 plane but anti-symmetrically with respect to the x1-x3 plane. In the tearing or anti-plane mode, Mode III, the crack faces also slide relative to each other but anti-symmetrically with respect to the x1-x2 and x1-x3 planes. The energy release rates related to these modes are termed GI, GII, and GIII respectively. In mixed mode problems we simply add the energy release rates of different contributing modes to obtain the total energy release rate.

III

mode III

II

mode II

I

mode I

Figure 12: Three basic loading modes for a cracked body: (a) Mode I, opening mode; (b) Mode II, sliding mode; (c) Mode III, tearing mode.


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Assignment 2: A cylindrical pressure vessel, with a diameter of 6.1 m and a wall thickness of 25.4 mm, underwent catastrophic fracture when the internal pressure reached 17.5 MPa. The steel of the pressure vessel had E = 210 GPa, a yield strength of 2450 MPa, a value of GC = 131 kJ/m2.

a) Show that failure would not have been expected if the Von Mises yield criterion had served for design purposes. b) Using the energy approach determine the size of crack that might have caused this failure.

Crack Growth Instability Analysis In the previous section we used the energy approach to define a criterion for the onset of crack growth. The energy approach can also be used for the analysis of different aspects of further crack growth such as stability, dynamic crack growth, and crack arrest. For this purpose we use the energy diagrams as depicted in Fig 13.

σ1 σ1 σ1

σ2

a1 Δa1 Δa2a2a3

R

G1

G3

G2

Figure 13 In these diagrams, the left horizontal axis is used to define the original crack lengths from which we may draw different lines whose slopes are related to the far-field stress. The intersections of these lines with the vertical axis represent the energy release rate, G. The vertical axis also represents the crack growth resistance, GC or R. In general, this


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resistance may vary as the crack grows because of different reasons such as the formation of shear lips in the plane stress condition or non-homogeneities in material. However, for a component made of homogenous isotropic material, with no thermal gradient, under plane strain condition, the R-Curve may be represented by a horizontal line. Figure 13 shows that the crack with the length a1 under the far field stress σ1 and the corresponding energy release rate G1 can not grow, because G1 < R . In order to make it grow, we have to increase the stress level to σ2. Nevertheless, longer cracks like a2 or a3 can grow even under σ1. In the above figure, a2 represents a crack under constant load condition whose energy release rate increases with further growth. In this case, the excess energy may accelerate the crack and the resulting unstable growth might cause a catastrophic failure. On the other hand, a3 represents a crack under fixed displacement condition whose energy release rate decreases with further growth. In this case the crack may stop after a stable growth equal to ∆a1. In practice the crack arrest may occur at some further distance like ∆a2 since the crack has some stored energy which may provide the crack driving force even if the apparent G is less than R. Figure 14 shows the energy diagram with a rising R curve which usually occurs in plane stress conditions.

σ1 σ1 σ1

a1 Δa1 Δa2a2a3

R

G1

G3

G2

Figure 14 The increase in resistance can be attributed to the formation of shear lips, which in turn results from plastic deformations at the crack tip. We will elaborate on this issue later when we discuss the crack tip plasticity. As depicted in Fig.14, the criteria for unstable crack growth under constant load for plane stress can be defined as:

, G RG Ra a

∂ ∂≥ ≥

∂ ∂ (2-26)


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Crack Speed In this section we will obtain an estimation of the crack speed in an unstable growth condition. The modeling is considered for an idealized situation of an infinite sheet with a central crack of length 2a under remote tensile stress σ. The idea is that the surplus of energy, represented by triangle ABC in Fig. 15, can be converted to the kinetic energy of the material elements in the crack path as they move apart from each other. As will be shown later, the horizontal and vertical displacements of these elements can be written as:

2 ( )

2 ( )

u

v

u ar fE

v ar fE

σ θ

σ θ

⎧ =⎪⎪⎨⎪ =⎪⎩

(2-27)

σ

ac

a

r

R

A

B C

θr

Y

X

v

u

Figure 15 in which ( )uf θ and ( )vf θ are geometric parameters. As the crack grows, its tip would be further from the considered elements. Hence, we may assume r a∝ and combine different constants to form C1 and C2 and write:

1 1

2 1

a au C u CE Ea av C v C

E E

σ σ

σ σ

⎧ = ⇒ =⎪⎪⎨⎪ = ⇒ =⎪⎩

(2-28)


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in which the dot means differentiation with respect to time. Accordingly, the kinetic energy for the plate can be defined and calculated as follows:

( )

( )

2 2

22 2 2

1 22

22 2

2

121212

KE u v dxdy

a C C dxdyE

k a aE

ρ

σρ

σρ

= +

= +

=

∫∫

∫∫ (2-29)

In the above equations ρ is the density and we have combined all the constants in a single term k. Moreover, as “a” is the only characteristic length in the infinite plate, the value of the integral was considered proportional to 2a . On the other hand the surplus of energy, which can be converted to kinetic energy for two crack tips, can be defined and calculated as follows:

( )

2

22

2

2 ( ) 2

( )

c

c

a

Sa

a

ca

c

E G R da

aR a a daE

a aE

πσ

πσ

= −

= − − +

= −

∫

∫ (2-30)

in which ac is the initial crack length and a is the crack length at every instant. Equating the two energies we may find the crack growth rate as:

2 1 caEak aπ

ρ⎛ ⎞= −⎜ ⎟⎝ ⎠

(2-31)

A more detailed analysis of the crack tip stress field has given an estimation of 0.38 for the first term on the right hand side of the above equation. The second term on the right is the speed of propagation of longitudinal waves in the material, so we have:

0.38 1 cS

aa Va

⎛ ⎞= −⎜ ⎟⎝ ⎠

(2-32)

Based on the above expression, it is clear that there is a limit to the crack speed in every material. Nevertheless, the speed of unstable crack growth is comparable with the speed of propagation of sound waves in the material. This means that an unstable growth of an initial crack with a few millimeters length may destroy several kilometers of a pipeline in a few minutes!


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Crack Branching Another interesting aspect of a growing crack is branching. As depicted in Fig. 16, under constant load, where the energy release rate increases with further crack growth, there might be a point where the available energy becomes twice the energy required to grow a single crack. As mentioned before, this surplus of energy usually accelerates the crack, but if the material permits, the situation may change in favor of crack branching. In general, when we observe that a component has been shattered into numerous pieces, we may think of too much energy available and/or too little energy required for crack to grow. The examples may include the fracture caused by an explosion and/or a glass of water slipping from your hand!

σ1

a1 Δa1

R

R

Figure 16

Crack Arrest As mentioned before, unstable crack growth may lead to catastrophic failure and must be prevented at all cost. One practical remedy is to use riveted patches or other types of stiffeners to simulate a fixed-displacement condition and arrest the crack. The location of the arrester must be chosen properly by considering the kinetic energy of the crack. As depicted in Fig. 17, the patch may decrease the energy down to the point C where the crack is expected to stop after a growth equal to ∆a2. In practice, however, the crack may grow further until its kinetic energy is consumed. This point has been specified in Fig. 17 by considering the area CDE roughly equal to ABC.


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σ1

a1 Δa1

Δa2

Δa3

RG1

A

BC D

E

Figure 17

LEFM, Stress Approach Investigation of crack-tip stress and displacement fields is important because these fields govern the fracture process occurring at the crack tip. In subsequent sections the stress analyses of the three major modes of crack growth will be presented.

The Mode III Problem The analysis of Mode III is relatively simple because we may assume that u1 = u2 = 0 and u3 = u3(x1 , x2). Accordingly, we may only consider the following nonzero strain components:

3 3,12

uα αε = (2-33)

where, the Greek subscripts have the range 1, 2. Therefore, the only nontrivial stress components are: 3 32α ασ με= (2-34)


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where μ is the shear modulus. Accordingly, the only relevant equation of equilibrium in the absence of body forces is: 3 , 0α ασ = (2-35)

The above three sets of equations can be combined to yield Laplace's equation in terms of

displacements:

23, 3 0u uαα = ∇ = (2-36)

In order to solve the above equation we use the complex variable method which provides a powerful technique for establishing the solutions of plane elasticity problems. The complex variable z is defined by z = x1 + ix2 or, equivalently, in polar coordinates z = reiθ. The overbar denotes the complex conjugate, z = x1 − ix2. It can be shown that,

1

2

( ) ( ) / 2( ) ( ) / 2

x e z z zx m z z z i=ℜ = += ℑ = −

(2-37)

where eℜ and mℑ denote the real and imaginary parts respectively. Let f(z) be a holomorphic (analytic) function of the complex variable z. A complex function is holomorphic in a region if it is single valued and its complex derivative exists in the region. For such a function the Cauchy-Riemann equations can be written as:

1 2 1 2

,u v v ux x x x∂ ∂ ∂ ∂

= = −∂ ∂ ∂ ∂

(2-38)

Differentiating the Cauchy-Riemann equations twice and combining them, we have: 2 2 0u v∇ =∇ = (2-39)

Thus, the real and imaginary parts of any holomorphic function are solutions to Laplace's equation. Therefore, the solution of Eq. (2-36) can be written as,

31 ( ) ( )u f z f zμ⎡ ⎤= +⎣ ⎦ (2-40)

Introducing Eq. (2-40) into Eq. (2-33) we may write:

31

32

1 ( ) ( )2

( ) ( )2

f z f z

i f z f z

εμ

εμ

⎡ ⎤′ ′= +⎣ ⎦

⎡ ⎤′ ′= −⎣ ⎦

(2-41)

Combining Eqns (2-34) and (2-41) we have:


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31 32 2 ( )i f zσ σ ′− = (2-42)

Now, let the origin of the coordinate system be located at the tip of a crack lying along the negative x1 axis as shown in Figure 18.

σx

σy

θr

Y

X

τxy

Figure 18 Crack-tip region and coordinate system Next, we will focus our attention upon a small region D containing the crack tip and consider the following holomorphic function: 1( ) ,f z Cz C A iBλ+= = + (2-43)

where A, B, and λ are real undetermined constants. For finite displacements at the crack tip we must have: ( 0), 1z r λ= = − ). The substitution of Eq. (2-43) into Eq. (2-42) yields:

31 32 2( 1) 2( 1) ( )(cos sin )i Cz r A iB iλ λσ σ λ λ λθ λθ− = + = + + + (2-44) whence,

31

32

2( 1) ( cos sin )2( 1) ( sin cos )

r A Br A B

λ

λ

σ λ λθ λθσ λ λθ λθ

= + −= − + +

(2-45)

The boundary condition that the crack surfaces remain traction free requires that 32 0σ = on θ π= ± . Consequently, we have:

sin cos 0sin cos 0

A BA B

λπ λπλπ λπ

+ =− =

(2-46)

To avoid the trivial solution, the determinant of the coefficients of the above equations must vanish. This requires that sin 2 0λπ = , which for λ> −1 has the following roots:


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1 , / 2, 0,1,2,...2

n nλ = − = (2-47) Of the infinite set of functions of the form of Eq. (2-43) that yield traction-free crack surfaces within D, the function with λ = –1/2 for which A = 0, provides the most significant contribution to the crack-tip fields. For this case Eq. (2-45) becomes,

311/ 2

32

sin( / 2)cos( / 2)(2 )

IIIKr

σ θσ θ

−⎧ ⎫ ⎧ ⎫=⎨ ⎬ ⎨ ⎬

⎩ ⎭⎩ ⎭ (2-48)

which is usually written as:

311/ 2

32

sin( / 2)cos( / 2)(2 )

IIIKr

σ θσ θπ

−⎧ ⎫ ⎧ ⎫=⎨ ⎬ ⎨ ⎬

⎩ ⎭⎩ ⎭ (2-49)

It should be noted that the origin of the above, rather unfortunate, modification is that K was originally considered as a way to calculate G, so the π term was artificially added to the denominator to facilitate the calculations. Also note that B has been chosen such that: { }1/ 2

32 00lim (2 )III r

K r θπ σ =→= (2-50)

We will also have:

12

32 sin( / 2)

2IIIK ru θ

μ π⎛ ⎞= ⎜ ⎟⎝ ⎠

(2-51)

The quantity KIII is referred to as the Mode III stress intensity factor, which is established by the far field boundary conditions and is a function of the applied loading and the geometry of the cracked body. Whereas the stresses associated with the other values of λ are finite at the crack tip, the stress components of Eq. (2-49) have an inverse square root singularity at the crack tip. It is clear that the latter components will dominate as the crack tip is approached. In this sense, Eqs (2-49) and (2-51) represent the asymptotic forms of the elastic stress and displacement fields.

The Mode I and Mode II Problems For the Mode I problem, we assume the displacement field as u1 = u1(x1, x2), u2 = u2(x1, x2), and u3 = 0. Accordingly we may write:

3

3 33

10,

0,

i Eαβ αβ αβ γγ

α αα

νε ε σ νδ σ

σ σ νσ

+ ⎡ ⎤= = −⎣ ⎦

= = (2-52)


23

In the absence of body forces the equilibrium equations reduce to , 0αβ ασ = , and the nontrivial compatibility equation becomes:

, , 0αβ αβ αα ββε ε− = (2-53)

The equilibrium equations will be identically satisfied if the stress components are expressed in terms of the Airy stress function, 1 2( , )x xΨ = Ψ , such that

, ,αβ αβ γγ αβσ δ= −Ψ +Ψ (2-54)

After the introduction of Eq. (2-54) into Eq. (2-52), the compatibility equation requires that the Airy function satisfy the biharmonic equation: 2 2, ( ) 0ααββΨ = ∇ ∇ Ψ = (2-55)

Noting that 2∇ Ψ satisfies Laplace's equation, we can write the following expression (analogous to the antiplane problem): 2 ( ) ( )f z f z∇ Ψ = + (2-56)

where f(z) is a holomorphic function. Equation (2-56) can be integrated to yield the following real function. [More details can be found in my lecture notes on the theory of elasticity, http://www.modares.ac.ir/eng/mmirzaei/elasticity.htm (3. 2D Static Boundary Value Problems: Plane Elasticity)].

1 ( ) ( ) ( ) ( )2

z z z z z zω ω⎡ ⎤Ψ = Ω + Ω + +⎣ ⎦ (2-57) where ( )zΩ and ( )zω are holomorphic functions. Now, we may substitute the above expression into Eq. (2-54) to write:

11 22

22 11 12

2 ( ) ( )

2 2 ( ) ( )

z z

i z z z

σ σ

σ σ σ ω

⎡ ⎤′ ′+ = Ω +Ω⎣ ⎦⎡ ⎤′′ ′′− − = Ω +⎣ ⎦

(2-58)

Due to symmetry with respect to the crack plane we choose a solution of the form, 1 1,Az Bzλ λω+ +Ω = = (2-59)

where A, B, and λ are real constants. To avoid singular displacements at the crack tip, we should set λ > –1. The introduction of Eq. (2-59) into Eq. (2-58) yields:


24

[ ]{[ ]

22 12 ( 1) 2cos cos( 2)

cos sin( 2) sin }

i r A

B i A B

λσ σ λ λθ λ λ θ

λθ λ λ θ λθ

− = + + −

+ − − + (2-60)which must vanish for θ = ± π. Consequently, we may write:

(2 )cos cos 0

sin sin 0A B

A Bλ λπ λπ

λ λπ λπ+ + =

+ = (2-61)

The existence of a nontrivial solution for the above set of equations requires that sin 2 0λπ = , which for λ> −1 gives the following roots:

1 , / 2, 0,1,2,...2

n nλ = − =

Again, the dominant contribution to the crack-tip stress and displacement fields occurs with λ = –1/2, for which A = 2B. Similar to the antiplane problem, an inverse square root singularity in the stress field exists at the crack tip. Substituting Eq. (2-59), with A = 2B and λ = –1/2, into Eqs (2-58) and (2-60), we find that:

11

12 1/ 2

22

1 sin( / 2)sin(3 / 2)cos( / 2) sin( / 2)cos(3 / 2)

(2 )1 sin( / 2)sin(3 / 2)

IKr

σ θ θσ θ θ θ

θ θσ

−⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪+⎩ ⎭⎩ ⎭

(2-62)

which is usually written as:

11

12 1/ 2

22

1 sin( / 2)sin(3 / 2)cos( / 2) sin( / 2)cos(3 / 2)

(2 )1 sin( / 2)sin(3 / 2)

IKr

σ θ θσ θ θ θ

πθ θσ

−⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪+⎩ ⎭⎩ ⎭

(2-63)

Accordingly, the Mode I stress intensity factor K, is defined as: { }1/ 2

22 00lim (2 )I r

K r θπ σ =→= (2-64)

The displacements can be written as:

1 221

22

cos( / 2) 1 2sin ( / 2)

2 2 sin( / 2) 1 2cos ( / 2)I

u K ru

θ κ θ

μ π θ κ θ

⎧ ⎫⎡ ⎤− +⎧ ⎫ ⎪ ⎣ ⎦⎪⎛ ⎞=⎨ ⎬ ⎨ ⎬⎜ ⎟⎝ ⎠ ⎡ ⎤+ −⎩ ⎭ ⎪ ⎪⎣ ⎦⎩ ⎭

(2-65)

When the foregoing is repeated with A and B being pure imaginary, the following equations can be obtained for the Mode II problem:


25

[ ][ ]

11

12 1/ 2

22

sin( / 2) 2 cos( / 2)cos(3 / 2

cos( / 2) 1 sin( / 2)sin(3 / 2)(2 )

sin( / 2)cos( / 2)cos(3 / 2)

IIKr

θ θ θσσ θ θ θ

πσ θ θ θ

⎧ ⎫− +⎧ ⎫⎪ ⎪⎪ ⎪ = −⎨ ⎬ ⎨ ⎬

⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

(2-66)

1 221

22

sin( / 2) 1 2cos ( / 2)

2 2 cos( / 2) 1 2sin ( / 2)II

u K ru

θ κ θ

μ π θ κ θ

⎧ ⎫⎡ ⎤+ +⎧ ⎫ ⎪ ⎣ ⎦ ⎪⎛ ⎞=⎨ ⎬ ⎨ ⎬⎜ ⎟⎝ ⎠ ⎡ ⎤− − −⎩ ⎭ ⎪ ⎪⎣ ⎦⎩ ⎭

(2-67)

{ }1/ 2

12 00lim (2 )II r

K r θπ σ =→= (2-68)

Finally, we may summarize the expressions derived for different modes by considering the following general expression for the stresses in a cracked body:

( )2ij ijK f

rσ θ

π= + (2-69)

It is clear that the first term is dominant very near to the crack tip. As we move further from the crack tip the singular term weakens and the additional terms become significant. As depicted in Figure 19, equal stress intensity factors for two different cracks, with different lengths, in different geometries, under different loadings, ensure similar crack tip stress fields. Hence, the critical stress intensity factor Kc, obtained at the onset of crack growth for a specific material and geometry, can be interpreted as a mechanical property named fracture toughness.

Figure 19


26

Williams General Solution The most general solution for cracks under generalized in-plane loading was provided by Williams. The solution starts by considering stresses at the corner of a plate under different boundary conditions and corner angles ψ. Here, a crack is treated as a special case where the angle of the plate corner is 2π and the surfaces are traction free, as depicted in Fig.20. The general stress function proposed by Williams is:

( ) ( ) ( ) ( )( )

11 2 3 4

1

sin 1 * cos 1 * sin 1 * cos 1 *

*,

r c c c c

r F

λ

λ

λ θ λ θ λ θ λ θ

θ λ

+

+

Φ = + + + + − + −⎡ ⎤⎣ ⎦=

(2-70)

where c1 , c2 , c3 , and c4 are constants. Using the relevant expressions in polar coordinates, the following expressions can be obtained for the stresses:

( )

( )

( )

1, ,2

1,

1, ,2

1 1 1

1

1 1

rr r

rr

r r

r F Fr r

r F

r Fr r

λθθ

λθθ

λθ θ θ

σ λ

σ λ λ

σ λ

−

−

−

′′= Φ + Φ = + +⎡ ⎤⎣ ⎦

= Φ = +⎡ ⎤⎣ ⎦

′= − Φ + Φ = −

(2-71)

In the above expressions, primes denote derivatives with respect to θ.

θ*

θψ

r

Figure 20 It can be shown that the displacements vary with rλ . Thus, in order to have finite displacements everywhere, we must have λ > 0. Also, the crack faces are traction free, which implies the following equalities: (0) (2 ) (0) (2 ) 0F F F Fπ π′ ′= = = = (2-72)

To ensure a nontrivial solution we must set sin 2 0λπ = , which for λ> 0 has the roots:


27

/ 2, 1, 2,...n nλ = = (2-73)

Accordingly, the most general form for the stress function and the resulting stresses can be written as:

( )

12

1

2

0

*,2

1*,2 *,

nN

n

mij M

ij ijm

nr F

fr f m

r

θ

θσ θ

+

=

=

⎡ ⎤⎛ ⎞Φ = ⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

⎛ ⎞−⎜ ⎟ ⎡ ⎤⎝ ⎠= + ⎢ ⎥⎣ ⎦

∑

∑

(2-74)

where f is a function of F and its derivatives. Eventually, the above expressions result in general expressions in the form of (2-69) for the Mode I and the Mode II crack problems. Again, it is clear that the first term is dominant very near to the crack tip. As we move further from the crack tip the singular term weakens and the additional terms become significant.

Design Philosophy Based on LEFM So far we have learned that cracks may start growing when the stress intensity factor (SIF) reaches a critical value Kc, called fracture toughness. Later we will show that the SIF can be related to the far-field stress and crack length by the following general expression: CK Y a Kσ π= ≥ (2-75)

in which Y is a geometric factor (see Fig. 26). Thus for the design of a cracked, or potentially cracked, structure we have to decide what design variables can be selected, as only two of these variables can be fixed and the third must be determined. For example we may select a special steel to resist a corrosive liquid, so KC is fixed, and the design stress level may also be fixed due to weight considerations. In this case we may calculate the maximum size of tolerable cracks using Eq. (2-75). Based on the above arguments it is clear that the application of LEFM in design procedures usually involves the following activities: 1. Measurement of the critical stress intensity factors that cause fracture for the material. 2. Determination of the size and location of cracks in the structure or component. 3. Calculation of the stress intensity factors for the cracks in the structure or component for the anticipated loading conditions.


28

Figure 21 The first item will be discussed after the concept of crack tip plasticity is introduced. The second task can be performed using some kind of non-destructive test techniques. Examples of such techniques are ultrasound and x-ray techniques, and inspection with optical microscopy. If the crack is detected, most of these techniques will provide an estimation of the crack length. If not, one should assume for design purposes that the structure contains cracks that are just too short to be detected. The third activity, i.e. calculation of stress intensity factors, can be performed using various techniques, including: 1. Finding the analytic solution to the full linear elastic boundary value problem, and deducing stress intensities from the asymptotic behavior of the stress field near the crack tips. 2. Deducing the stress intensity factors from energy methods (In the next section we will discuss the related relationship). 3. Using experimental techniques 4. Using numerical methods such as boundary integral and finite element methods.

K-G Relationship So far we have discussed two different criteria, based on energy considerations and crack tip stress field, for the onset of crack growth. Naturally, there should be a relationship between the two. In this section we will discuss this relationship. Figure 22 depicts a crack of initial length a subject to Mode I loading with the origin located at a distance ∆ behind the crack tip.


29

Δ

Y

XV

Figure 22 Now assume that we may partially close the crack through application of a compressive stress field to the crack faces between x = 0 and x = ∆. The required work would be:

0

v2

2yW dr

σΔ= ∫

(2-76)

The factor of 2 on work is required because both crack faces are displaced. In the above yσ is the compressive stress distribution and v is the crack opening displacement. As this work will be released as energy, the energy release rate G can be written as:

0

00

lim

v2lim2y

WG

drσ

Δ→

Δ

Δ→

=Δ

=Δ ∫

(2-77)

We may define the stresses and displacements in terms of the stress intensity factor as:

2 2

2 2

2

22v

Iy

I

Kr

K a xa xE E a

σπ

σπ

=

−= − =

(2-78)

Noting that x r a= + −Δ and neglecting the second order terms in our calculations, we may rewrite the above as:


30

( )

22 2v 2 2

2 2

I

I

K r rra aE

K rE

π

π

Δ= Δ − + −

= Δ −

(2-79)

Substituting the above into Eq. (2-77), we have:

2

00

2lim IK rG drE rπ

Δ

Δ→

Δ −=

Δ ∫ (2-80)

The result of integration is:

2IKG

E=

(2-81)

which can be modified for plane strain as follows:

( )2

21 IKGE

ν= − (2-82)

The above expressions are general relationships between K and G for Mode I. However, the analysis procedure can be repeated for other modes of loading. When all three modes of loading are present, the energy release rate is given by:

2 2 22 2(1 ) (1 ) (1 )

I II III

I II III

G G G G

K K Kv v vE E E

= + +

= − + − + + (2-83)

As mentioned before, the three modes are additive with respect to energy release rate because it is a scalar quantity. However, it should be noted that the above equation assumes self similar crack growth; i.e., a planar crack is assumed to remain planar and maintain a constant shape as it grows. This is usually not the case for mixed-mode fractures. Assignment 3: A thin-walled cylinder contains a crack forming an angle 30 degree to the longitudinal axis. Numerical data: radius R = 0.2 m, crack half-length a and wall thickness t are a = t = 0.005m, fracture toughness KC = 50 MPa m . Determine at which torque T crack may start to grow.


31

Mixed Mode Fracture When we are dealing with an angled crack (similar to that depicted in Fig 23), the first question is naturally about the magnitude of the far-field stress σ at which the crack starts to grow. The second question is how to determine the direction of further crack growth. σ

σ

Y

X

θ

r

β

Figure 23

The first question can be answered by the methods described in the previous section, while there are two general approaches for predicting the direction of crack growth. In the first approach, it is assumed that the crack growth occurs in the direction perpendicular to the maximum tangential stress at (or near) the crack tip. The second approach considers crack growth in the direction for which the strain energy density is minimal, on the basis that this corresponds to a maximum in energy release rate. The strain energy density in the vicinity of the crack tip may be written as:

( )

[ ]

[ ]

2 211 12 22

11

12

22

1 ( )2

1 (1 cos )( cos )16

1 sin (2cos 1)16

1 ( 1)(1 cos ) (1 cos )(3cos 1)16

I I II IISa K a K K a K

r r

a k

a k

a k

θψ

θ θμ

θ θμ

θ θ θμ

= + + =

= + −

= − +

= + − + + −

(2-84)

where (3 4 )k ν= − for plane strain, and (3 ) /(1 )k ν ν= − + for plane stress.


32

According to this criterion the crack starts to grow when S reaches a critical value Sc, and the direction of crack growth is given when S is a minimum:

2

20, 0dS d Sd dθ θ

= > (2-85)

For example, for pure mode I we have:

( ) ( ) 2

11min

2

0

2( 1)16

I I c

c c

S S a K

kS K

θ θ

μ

= = =⎡ ⎤⎣ ⎦−

⇒ = (2-86)

For the angled crack shown in Fig 23, the strain energy density can be obtained based on the following expressions for the stresses:

2

2

5 1 3 5 3 3cos cos sin sin cos4 2 4 2 4 2 4 22 23 1 3 3 3 3cos cos sin sin sin4 2 4 2 4 2 4 22 21 1 3sin sin4 2 4 22

I IIrr t

I IIt

Ir

K Kr r

K Kr r

Kr

θθ

θ

θ θ θ θσ σ θπ π

θ θ θ θσ σ θπ π

θ θσπ

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − + − + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= + + − − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞= + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

1 3 3cos cos sin cos4 2 4 22

IIt

Kr

θ θ σ θ θπ

⎛ ⎞⎛ ⎞+ −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(2-87)

where we have;

2

2

sin , sin cos

sinI II

t

K a K aσ π β σ π β β

σ σ β

= =

=(2-88)


33

Stress Intensity Factor A major activity in the design process based on fracture mechanics is the determination of the stress intensity factor for the particular problem. In the following sections we will discuss some of the pertinent analytical, experimental, and numerical methods. Analytical Determination of SIF In this section we drive the stress intensity factor expression for an infinite plate with a central crack of length 2a, under remote stress σ0.

aσx

σy

σ0

σ0

σ0

σ0θ

r

Y

X

τxy

Figure 24

We use the general form of the Westergard stress function; e y mΦ =ℜ Φ+ ℑ Φ (2-89)

for which we can define the integration and differentiation operations with respect to z as:

ddzdzddzdz

ddzdz

ΦΦ = Φ = Φ

ΦΦ = Φ = Φ

Φ′ ′Φ = Φ = Φ

∫

∫

∫

(2-90)


34

Writing the Cauchy-Riemann equations we have:

m e dey x dze m dmy x dz

∂ℑ Φ ∂ℜ Φ Φ= ≡ℜ

∂ ∂∂ℜ Φ ∂ℑ Φ Φ

− = ≡ ℑ∂ ∂

(2-91)

Using Eq. (2-91), we can differentiate Eq. (2-89) with respect to “y” and write: ,y m m y e y eΦ = −ℑ Φ+ℑ Φ + ℜ Φ ≡ ℜ Φ (2-92)

Accordingly, we can calculate the stresses using the following Equations:

,

,xx yy

yy xx

e y m

e y m

σ

σ

′= Φ =ℜ Φ − ℑ Φ

′= Φ =ℜ Φ + ℑ Φ (2-93)

Now, we should choose a stress function which satisfies the local and global boundary conditions of this problem. Let us check the following stress function:

02 2

zz aσ

Φ =−

(2-94)

The global and local boundary conditions are:

0B.C.1 ,

B.C.2 0, ,

B.C.3 0, , 0 0

xx yy

xx yy

xx yy

x y z

y x a

y a x a e

σ σ σ

σ σ

σ σ

→∞⇒ →∞⇒ = =

= = ± ⇒Φ = ∞ ⇒ = = ∞

= − < < ⇒ℜ Φ = ⇒ = =

(2-95)

Note that the proposed stress function satisfies these conditions. Now we translate the origin to the crack tip.

02 2

( )( )

z az a aσ +

Φ =+ −

(2-96)

Since we are interested in very near field stresses where z a , we neglect the z2 term and write,

12

02a zσ

−Φ =

(2-97)

Next we switch to polar coordinates with iz re θ= and write Eq. (2-97) as,


35

0[cos( / 2) sin( / 2)]2a irσ θ θΦ = −

(2-98)

Hence, for the normal stress component on the crack line near the tip we have,

22 0 0 02aerθ θσ σ= == ℜ Φ =

(2-99)

Finally, using the original definition of KI for the Mode I problem, we may write:

{ }1/ 222 00

1/ 200

0

lim (2 )

lim (2 )2

I r

I r

I

K r

aK rr

K a

θπ σ

π σ

σ π

=→

→

=

⎧ ⎫⎪ ⎪= ⎨ ⎬⎪ ⎪⎩ ⎭

=

(2-100)

In practice, the far field stress in the x direction does not have any effect on the crack tip stress field. Hence, the above expression for K is also applicable when the sheet is under uniaxial far-field stress in the Y direction. It is possible to obtain the K-expression using the principle of superposition. Thus, for the geometries shown in Figure 25 we may write:

0Ia Ib

Ia

Ia

K K a

K a

K a

σ π

σ π

σ π

+ =

+ =

=

(2-101)

Figure 25


36

Figure 26


37

Elliptical Cracks For many real components cracking starts at free surfaces. These cracks often have semi-elliptical or quarter-elliptical shapes. There may also be elliptical cracks embedded in components. Because of their importance, significant research has been done on modeling and quantification of the effect of elliptical cracks in different structures under various types of loadings. A solution for an embedded elliptical crack for an infinite domain derived by Irwin is:

12 4

2 22sin cosI

a aKb

σ π θ θ⎛ ⎞

= +⎜ ⎟Φ ⎝ ⎠ (2-102)

in which a and b are defined in Fig 27, θ is measured counterclockwise from the point B, and Φ is an elliptical integral of the second kind defined by;

12 2 2/ 2 2

201 sinb a d

bπ

θ θ⎛ ⎞−

Φ = −⎜ ⎟⎝ ⎠

∫ (2-103)

Values for Φ are reported in the form of tables and graphs. It is also possible to find an approximation to the above integral using a series expansion and write the K expression as;

12 4

2 22 2

2

sin cos38 8

Ia aKa bb

σ π θ θπ π

⎛ ⎞= +⎜ ⎟

⎝ ⎠+ (2-104)

From the above expression it is obvious that IK varies along the crack front. Its magnitude is largest at the end of the minor axis and lowest at the end of the major axis as follows:

2

(max)

(min)

I

I

aK

abK

σ π

πσ

=Φ

=Φ

(2-105)

A correction factor of 1.12 is usually considered for the part-through semi-elliptical cracks. For the quarter-elliptical corner crack the factor is 1.2. We may also use the general expression IK Y bσ π= and find the Y factor from the following diagrams.


38

Figure 27


39

Experimental Determination of SIF The different methods used for experimental determination of stress intensity factors belong to a broader domain called experimental stress analysis. Here, we briefly review the usage of electrical resistance strain gages and the method of photoelasticity. An excellent source for detailed study of these and other methods, such as laser interferometry and the method of Caustics, is: Dally, J. W., Riley, W. F., Experimental Stress Analysis, 3rd edition, 1991. Also are the lecture notes on experimental stress analysis, provided by James W. Phillips, Department of Theoretical and Applied Mechanics, University of Illinois at Urbana-Champaign. Strain Gage Method The stress intensity factor can be determined experimentally by placing one ore more strain gages near the crack tip. However, to avoid sever strain gradients, the gages should not be placed at very near field. On the other hand, as we move further from the crack tip, we need more terms to be able to express the field parameters correctly. The three-term representation of the strain field is:

( ) ( )

( ) ( )

( ) ( )

( ) ( )

12

0 0

122

1

12

0 0

122

1

1 12 2

0 1

3cos 1 1 sin sin 22 2 2

cos 1 1 sin2 2

3cos 1 1 sin sin 22 2 2

cos 1 1 sin2 2

32 sin cos sin cos2 2

xx

yy

xy

E A r B

A r

E A r B

A r

A r A r

θ θ θε ν ν

θ θν ν

θ θ θε ν ν ν

θ θν ν

θ θμγ θ θ

−

−

−

⎡ ⎤= − − + +⎢ ⎥⎣ ⎦

⎡ ⎤+ − − +⎢ ⎥⎣ ⎦

⎡ ⎤= − − + −⎢ ⎥⎣ ⎦

⎡ ⎤+ − − +⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

(2-106)

where A0 , B0 , and A1 are unknown coefficients which depend on loading and the geometry of the specimen. For instance we have:

0 2IKAπ

= (2-107)

In general, we need three strain gages to be able to determine the above three unknowns. However, it can be shown that it is possible to use only one gage oriented at angle α and positioned along the Px′ axis, as shown in Fig. 28.


40

Figure 28

Accordingly, the stress intensity factor can be determined from:

1 3 1 32 cos sin sin cos 2 sin cos sin 22 2 2 2 22

Ix x

KE kr

θ θ θμ ε θ α θ απ′ ′

⎡ ⎤= − +⎢ ⎥⎣ ⎦ (2-108)

where,

11

k νν

−=

+ (2-109)

The choice of the angles α and θ depend on the Poisson’s ratio and can be determined from the table below.

Table 1

For the choice 60α θ= = , the required expression is very simple:

83I x xK E rπ ε ′ ′=

(2-110)


41

Photoelasticity Method Photoelasticity is a whole-field stress analysis technique based on an optical-mechanical property called birefringence, possessed by many transparent polymers. A loaded photoelastic specimen, combined with other optical elements, and illuminated with an ordinary light source exhibits fringe patterns that are related to the difference between the principal stresses in a plane normal to the light propagation direction. A polariscope is needed for viewing the fringes induced by the stresses (see Fig. 29). Two types of pattern can be obtained: isochromatics and isoclinics. The former is related to the principal-stress differences and the latter to the principal stress directions.

Figure 29 The sensitivity of a photoelastic material is characterized by its fringe constant fσ . This constant relates the value “N” associated with a given fringe to the thickness h of the specimen in the light-propagation direction and the difference between the principal stresses in the plane normal to the light-propagation direction as follows:

1 2Nfhσσ σ− =

(2-111)

In practice, the principal stresses are obtained from the stresses defined in Eq. (2-63) and combined with Eqn. (2-111) to give:


42

2211 22

1 2 1222

sin2

2

I

I

Kr

r f NKh

σ

σ σσ σ σ

θπ

π

−⎛ ⎞− = +⎜ ⎟⎝ ⎠

=

⇒ =

(2-112)

For example, let us calculate the value of the stress intensity factor for the specimen shown in Fig 30.

Figure 30 Suppose that the fringe designated with number 5 is located at the distance 0.23 in. from the crack tip, the material fringe value is 43, and the specimen thickness is 0.213 in. The magnitude of the stress intensity factor can be calculated as:

2

2 (0.23)(43)(5)1.2 ksi in

0.213

Ir f NKh

σπ

π

=

= =

(2-113)


43

Numerical Determination of SIF For most practical problems, either there is no analytical solution, or the available solutions are only crude approximations. Hence, numerical techniques play a very important role in determination of SIFs for real components. Numerical methods for fracture mechanics can be categorized in many different ways. We will explore some aspects of the so called “computational fracture mechanics” as we proceed in this course. A brief explanation of different approaches to numerical determination of SIF follows: Ordinary and Extended Finite Element Methods, In the finite element method (FEM), the structure of interest is subdivided into discrete shapes called elements. Different Element types can be used to cover the problem domain. The elements are connected at node points where continuity of the displacement fields is enforced. The displacements at the nodes depend on the element stiffness and the nodal forces. For structural problems, the solution of the problem consists of nodal displacements. The stress and strain distribution throughout the body, as well as crack parameters such as SIF, can be inferred from the nodal displacements. A number of commercial FEM packages have the ability for crack modeling and fracture mechanics calculations. There is also a very useful noncommercial code, called FRANC2D, which is developed in Cornell University. Despite all of its capabilities, the code is surprisingly easy to learn and work with. This is an ideal code for education and research. It can be downloaded at http://www.cfg.cornell.edu/software/software.htm. In the Extended Finite Element Methods (X-FEM), a discontinuous function and the two-dimensional asymptotic crack-tip displacement fields are added to the finite element approximation to account for the crack, using the notion of partition of unity. This enables the domain to be modeled by finite elements with no explicit meshing of the crack surfaces. The initial crack geometry is represented by level set functions, and subsequently signed distance functions are used to compute the enrichment functions that appear in the displacement-based finite element approximation. The method has basically been developed in Northwestern University. Boundary Element Methods, The problems in science and engineering are usually solved mathematically through specifying appropriate boundary conditions. Given these boundary conditions, it is theoretically possible to solve for the tractions and the displacements on the boundary, as well as the stresses, strains, and displacements within the body. The boundary integral equation method (BIE) is a very powerful technique for solving for unknown tractions and displacements on the surface. This approach can also provide solutions for internal field quantities, but finite element analysis is more efficient for this purpose. The BIE formulation leads to a set of integral equations that relate surface displacements to surface tractions. In order to solve for the unknown boundary data, the surface must be subdivided into segments (i.e., elements), and the boundary integrals are approximated by a system of algebraic equations. Once all of the boundary quantities are known,


44

displacements at internal points can be computed. The boundary elements have one less dimension than the body being analyzed, i.e., the boundary of a two-dimensional problem is surrounded by one-dimensional elements, while the surface of a three-dimensional solid is paved with two-dimensional elements. Consequently, boundary element analysis can be very efficient, particularly when the boundary quantities are of primary interest. This method is inefficient, however, when solving for internal field quantities. The noncommercial FRANC3D code, along with OSM (solid modeler) and BES (3D BIM solver) codes provide a powerful set for analyzing fracture mechanics problems. These codes are developed in Cornell University can be downloaded at http://www.cfg.cornell.edu/software/software.htm. Mesh Free Methods, These methods are relatively new and still in the stage of development, but they possess some interesting features that make them strong candidates for computational fracture mechanics in the future. In these methods the problem domain is represented by a set of arbitrarily distributed nodes. There is no need to use meshes or elements for field variable interpolation. The density of nodes can be chosen high near discontinuities like cracks. In contrast to FEM, there is no need for conformation of mesh with cracks and this makes the modeling of evolving cracks easier. Determination of SIF using FEM In general, we may consider two different approaches for determination of SIF using FEM. These are: point matching and energy methods. The former technique is based on inferring the stress intensity factor from the stress or displacement fields in the body. In the latter the energy release rate is first computed and the stress intensity is obtained using Eqs (2-81) and (2-82). Displacement Extrapolation Following a liner elastic analysis, the stress intensity factors can be determined by equating the numerically obtained displacements with their analytical expression in terms of the SIF. For example, the mode-I displacement equations are:

( , )2

Ii i

K ru fG

θ νπ

= (2-114)

Using the above expression we may obtain a quantity IK ∗ using the nodal point displacement iu∗ at some point ( , )r θ close to the crack tip.


45

[ ]2 ( , )I i iK G f urπ θ ν∗ ∗=

(2-115)

The stress intensity factor can be inferred by plotting the quantity IK ∗ against distance from the crack tip, and extrapolating to r = 0. This technique was the predominant one prior to the discovery of the quarter point singular element. In early finite element studies of LEFM, it was recognized that unless singular elements could be used, it would be necessary to have a very fine mesh at the crack tip to approximate the stress singularity with non-singular elements. Singular Elements Certain element/node configurations produce strain singularities. Forcing the elements at the crack tip to exhibit a strain singularity greatly improves accuracy and reduces the need for a high degree of mesh refinement at the crack tip. It has been shown that ordinary quadratic isoparametric elements can be degenerated to give desired singularity by moving the mid-side nodes to the l/4 points.

Figure 31 Energy Approach, Elemental Crack Advance We have already shown that the energy release rate can be obtained from the rate of change in total potential energy with crack growth. Assume a two-dimensional body with unit thickness. If two separate numerical analyses are performed, one with the crack length a, and the other with the crack length a + ∆a, the energy release rate is given by,

Ga

ΔΠ⎛ ⎞= −⎜ ⎟Δ⎝ ⎠

(2-116)


46

This technique is easy to implement, since the total strain energy is a natural output by many commercial analysis codes. This technique is also more efficient than the point matching methods, since global energy estimates do not require refined meshes. One disadvantage of this method is that it requires multiple solutions, while other methods infer the desired crack tip parameter from a single analysis. Assignment 4: Consider a center-cracked plate of the AISI 4340 steel (E = 210 GPa, ν = 0.3), which has dimensions, as follows: width 2W = 76 mm; length-to-width ratio 2H/2W = 5; and thickness B = 6 mm. The plate contains an initial crack of length (half) a = 1 mm. It is subjected to a tension loading of P = 240 kN. Compute the SIF using the available closed form solutions and FEM. Compare the results.

Energy Method, Virtual Crack Extension, Stiffness Derivative Formulation In this section we will briefly explain the method developed by Parks and Hellen. Consider a two-dimensional cracked body with unit thickness, subject to Mode I loading. The potential energy of the body, in terms of the finite element solution, is given by

[ ] [ ][ ] [ ] [ ]12

T TKΠ = −u u u F (2-117)

Accordingly, it can be shown that the energy release rate is proportional to the derivative of the stiffness matrix with respect to crack length.

[ ] [ ] [ ]12

T KdGda a

∂Π= − = −

∂u u

(2-118)

The implementation of the above expression does not require changing all of the elements in the mesh. Instead, we may accommodate the crack growth by moving only the elements near the crack tip as illustrated in Fig. 32.


47

Figure 32 Each of the elements between Γ0 and Γ1 is distorted, such that its stiffness changes. The energy release rate is related to this change in element stiffness as follows:

[ ] [ ] [ ]1

12

CNT i

i

KG

a=

⎧ ⎫∂= − ⎨ ⎬∂⎩ ⎭

∑u u (2-119)

where [ ]iK are the elemental stiffness matrices and NC is the number of elements between Γ0 and Γ1. It should be noted that in this analysis, it is sufficient merely to calculate the change in elemental stiffness matrices corresponding to shifts in the nodal coordinates and there is no need to generate a second mesh with a longer crack. One problem with the stiffness derivative approach is that it involves cumbersome numerical differentiation. A more recent formulation of the virtual crack extension method overcomes these difficulties, as discussed below. Energy Method, Virtual Crack Extension, Continuum Approach In this section we will briefly explain the method developed by deLorenzi. Figure 33 illustrates a virtual crack advance in a two-dimensional continuum. Material points inside Γ0 experience rigid body translation a distance ∆a in the x1 direction, while points outside of Γ1 remain fixed. In the region between contours, virtual crack extension causes material points to translate by ∆x1. For an elastic material, deLorenzi showed that energy release rate is given by,

11

1

1 jij i

iA

u xG w dAa x x

σ δ∂⎛ ⎞ ∂Δ

= −⎜ ⎟Δ ∂ ∂⎝ ⎠∫ (2-120)

where w is the strain energy density and the crack growth is assumed in the x1 direction. The integration need only be performed over the annular region between Γ0 and Γ1.


48

Figure 33 *Calculation of SIF using the concept of Energy-Domain Integral will be discussed in Lecture Note 3.

Crack Tip Plasticity As it was shown before, under LEFM assumptions, the magnitude of stress at the crack tip is theoretically infinite. However, it is obvious that every material has a finite strength and, as a result, there will always be a small damaged zone around the crack tip. For metals, this damaged zone is referred to as the crack tip plastic zone. For materials like ceramics and concrete it is often called fracture process zone. The question is how the use of LEFM parameters like G and K can be justified under these circumstances. The answer relies on the spread of the damage zone compared to the size of what we may call the K-dominant region. By the latter we mean the small region around the crack tip for which the states of stress, strain, and displacement are determined by the singular term (first term) of Eq. (2-69). If the size of the damaged zone is small enough that it is contained within the K-dominant region, we may conclude that the similitude condition exists. Hence, different cracks with equal stress intensity factors will have equal damaged zones and will behave similar to each other. On the other hand, if this zone is larger than the K-dominant region, then our linear elastic assumptions are not correct, i.e. LEFM is not applicable, and a nonlinear model must be used.


49

Size of the plastic zone Based on the above arguments, it seems necessary to have a reliable estimate of the size of the crack-tip plastic-zone. We start by considering the plane stress condition and assume that the boundary between plastic and elastic regions is the locus of the stresses, given by Eq. (2-63), which satisfy a yield criterion. However, as a first approximation, we assume that yielding occurs when 22σ equals the uniaxial yield strength of the material ( ysσ ). On the crack plane we have θ = 0 and the normal stress 22σ in a linear elastic material is given by Eq. (2-63).

σys

σ22

rry

Figure 34 Substituting the yield strength into the left side of Eq. (2-63) and solving for r gives a first order estimate of the plastic zone size:

22

2 2

2 2

2

2

2 2

I

Iys

y

Iy

ys ys

Kr

Kr

K ar

σπ

σπ

σπσ σ

=

=

⇒ = =

(2-121)

However, as depicted in Fig. 34, we have actually ignored the load represented by the hatched area in our derivation. It can be expected that due to this extra load the actual plastic zone size should be larger than yr .


50

Irwin Plastic zone correction In order to give a better estimation of the plastic-zone size, Irwin argued that consideration of a larger plastic zone may be taken equivalent to the assumption of a larger crack. This can be readily seen from the last expression of Eqs (2-121). Hence, we may define an effective crack length whose length is equal to the size of the actual crack plus a correction ρ. The next step is to repeat the previous procedure for plastic zone size estimation for the effective crack. However, we consider the extra length ρ large enough to carry the extra load ignored by truncating the asymptotic stress distribution.

σys

σ22

rΔρaeff

a

Figure 35

Hence, we may write:

2

2

22

( )2

Iys

yys

K a

a r

ρσ σπ

σ ρσ

+= ≡

ΔΔ

+⇒ Δ = ≈

(2-122)

Equating the two hatched areas, we have:

0 2ys ys

a drrρρσ σ σ

Δ⎛ ⎞+= −Δ⎜ ⎟⎜ ⎟⎝ ⎠∫ (2-123)

Since ρ is small, it can be neglected compared to the crack length in the above integration and we may write using (2-122):

( )

( )22 22

2

2 4

y ys y

y y yys

y

r ar

ar r r

r

ρ σ σ

σρσ

ρ

+ =

⇒ + = ≡

⇒ =

(2-124)


51

It turns out that the new plastic zone size is twice as large as our first estimate:

2

12 Ip y

ys

Kr rρπ σ⎛ ⎞

= Δ + = = ⎜ ⎟⎜ ⎟⎝ ⎠

(2-125)

The main effect of plasticity is that the stresses close to the crack tip are relaxed by yielding at the expense of increased stresses outside the plastic zone. The effect of the plastic zone on the stress intensity factor might then be approximated by adding half the plastic zone length to the real crack before calculating K as follows,

( )I yK a rσ π= + (2-126) Of course this procedure should be limited to yr a . Thus, the calculation procedure for plastic zone size adjustment consists of calculating the elastic stress intensity based on the real crack length, calculating the plastic zone dimension through Eq. (2-121), adding it to the real crack length, and recalculating the stress intensity through Eq. (2-126). Crack Tip Opening Displacement One key advantage of plastic-zone correction is that it gives a meaning to the crack-tip-opening displacement (CTOD). It is clear that CTOD, obtained from a linear elastic analysis, is zero.

2 2

2 2

2v

4 0x a

a xE

a xE

σ

σδ δ=

= −

= − ⇒ =

(2-127)

However, if we introduce the effects of crack tip plasticity, we will have:

2 2 2

2

4 2

40 2

4

eff y t y y

y t y

It

ys

a a r a ar r aE

r arE

KCTODE

σδ

σδ

δπ σ

= + ⇒ = + + −

→ ⇒ =

≡ =

(2-128)

We will use the concept of CTOD in relation with the crack tip plastic deformation in our discussions on elasto-plastic fracture mechanics.


52

Dugdale Approach An alternative approach for estimation of the plastic-zone size was proposed by Dugdale and Barenblatt. In this approach it is assumed that the plastic deformation is confined to a localized strip in front of the crack. The formulation of the model starts by considering a virtual crack which is larger than the real crack by the amount 2∆. It is assumed that the virtual crack is closed by the amount ∆ at every tip by the use of compressive stresses equal to the yield stress.

2aσys σys

2a + 2Δ

Figure 36 The next step is to find the stress function required for the solution of this problem by superimposing two stress functions. The first is for an infinite plate, containing a central crack with the length 2a + 2∆, under the far-field stress σ, for which we have:

( )

1 22

z

z a

σΦ =

− + Δ (2-129)

The second stress function is required for the same crack under a distributed load ysσ over the distance ∆. This stress function can be obtained by integration of the solution for the same crack under a point load located at a distance b from the centerline, for which we have:

( )( ) ( )

( )( ) ( )

( )( )

( )

2 2

22 2 2

2 2

2 22 2 2

221 1

2 2 222

2

for this problem

2

2cos cot

ysb

a ys

a

ys

z a b

z a z b

b a

z a adz

z a z a

z az a aa z a az a

σ

π

σ

π

σπ

+Δ

− −

+ Δ −Φ =

− + Δ −

=

+ Δ −⇒Φ =

− + Δ −

⎡ ⎤⎛ ⎞− + Δ⎛ ⎞⎢ ⎥⎜ ⎟⇒ Φ = −⎜ ⎟⎢ ⎥⎜ ⎟+ Δ⎝ ⎠ + Δ −− + Δ ⎝ ⎠⎣ ⎦

∫

(2-130)


53

Now we may obtain the required stress function for the Dugdale model by superposition of the two functions as follows: 3 1 2Φ = Φ −Φ (2-131)

However, we do not need to go through the complete solution. All we need is to set the sum of the singular terms of the two functions equal to zero, because we know that in reality the stresses at the location of the tip of the virtual crack are finite. Accordingly, the plastic-zone size ∆ can be determined as:

( ) ( )1

2 22 2

2 2

2

2

2cos 0

cos 12 8

8

ys

ys ys

I

ys

zz aaz a z a

a aa a

K

σσ

π

πσ π σσ σ

πσ

− ⎛ ⎞− =⎜ ⎟+ Δ⎝ ⎠− + Δ − + Δ

⎛ ⎞ − Δ⇒ = ⇒ − =⎜ ⎟⎜ ⎟ + Δ⎝ ⎠

⎛ ⎞⇒ Δ = ⎜ ⎟⎜ ⎟

⎝ ⎠

(2-132)

which is quite similar to Eq. (2-125), except for a slight difference in the coefficient. Crack-Tip 3D State of Stress Both Irwin and Dugdale estimations of the crack-tip-plastic-zone size assume uniaxial plastic deformation. Nevertheless, the actual state of stress at the crack tip region is triaxial. In order to see how this 3D state of stress can develop we assume a roll of material in the interior of the specimen at the crack tip, as depicted in Fig. 37.

Elastic

Plastic Zone σ11

σ33

σ33

σ22

σ22

Figure 37


54

Due to very high stresses in this region, the material undergoes a large extension in the x2 direction and also tends to contract in the x1 and x3 directions to maintain the condition of constant volume required by plastic deformation (ε11 + ε22 + ε33 = 0). However, the material in this zone is continuously attached to a larger mass of surrounding material, which is at a low stress and has no tendency to contract. As a result, tensile stresses develop in other two directions, as shown in Fig. 37. Thus, in the interior of the specimen, the material in the crack tip region experiences a state of plane strain due to the constraints imposed by the surrounding material. However, at the surfaces of the specimen there can be no stress in the x3 direction and a state of plane stress exists. Due to the above arguments, it is clear that a very precise determination of the shape and size of the plastic zone is not an easy task. Nevertheless, we may go one step further by first ignoring the stress redistribution due to plastic deformation. Next we may combine our knowledge of plane strain/plane stress transition, the elastic solution for the crack tip stresses, and the von Mises yield criterion to obtain a better estimation of the shape and size of the plastic zone. The von Mises yield condition written in terms of principal stresses is: ( ) ( ) ( )2 2 2 2

1 2 2 3 3 1 2 ysσ σ σ σ σ σ σ− + − + − = (2-133)

Using Eqns (2-63) we may write the crack-tip principal stresses as:

( )

11/ 2

2

31 2

1 sin( / 2)cos( / 2)

1 sin( / 2)(2 )

0

IKr

plane stressplane strain

σ θθ

σ θπ

σν σ σ

+⎧ ⎫ ⎧ ⎫=⎨ ⎬ ⎨ ⎬−⎩ ⎭⎩ ⎭⎧

= ⎨ +⎩

(2-134)

Substituting the above stresses into Eq. (2-133) and solving for rp(θ) results in:

( )

( ) ( ) ( )

2

2

2 22

1 1 3 1sin cos2 2 2 2

1 21 3 sin 1 cos2 4 2

Ip

ys

Ip

ys

Kr plane stress

Kr plane strain

θ θ θπ σ

νθ θ θ

π σ

⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ −= + +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(2-135)

Figure 38 shows the plastic zone shapes, which are the locus of the above expressions

normalized through a division by 2

12

Iy

ys

Krπ σ⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

, which was obtained from Eq. (2-121).


55

|

||

|

|

|

|

|

|

|

|

1.0

1.0

Plane Stress

Plane Strain

Figure 38 Plane Stress/Plane Strain Transition Figure 39 shows how the plastic zone size gradually decreases from the plane stress size at the surface to the plane strain size in the interior of the component. This phenomenon has very important implications for toughness evaluation, i.e., different toughness values may be measured depending on the dimensions of the specimens.

Plane Stress

Plane Strain

Figure 39 It has long been observed that thicker components have a greater tendency to fracture. This effect can be attributed to the size of the crack tip plastic zone relative to the thickness. In thin components, the plastic zone is large compared to the thickness, whereas in thick components it is very small. In general, fractures of thick specimens are more brittle in appearance, being flat with no evidence of ductility, while the fractures of thin specimen often show 45° shear lips over parts of the fracture surface.


56

z zx

x

y

y

Plane Stress Plane Strain

thickness

toughness

τ τ

σ1 σ1

σ2

σ2 σ3

σ3

Figure 40 Moreover, as depicted in Fig. 40, in the case of plane stress, the planes of maximum shear stress are located at angles 45° from the directions of σ1 and σ3. Although in the plane strain condition σ1 and σ2 have the same magnitude as in plane stress, but the third principal stress equals ν(σ1 + σ2). As the requirement for constant volume dictates ν = ½ for plastic deformation, the maximum shear stress will be much lower and occurs on planes rotated 45° from the directions of σ1 and σ2. The effect of Elastic T Stress We have already shown that the Williams solution for the crack tip stress field in an isotropic elastic material can be expressed as an infinite power series, whose leading term exhibits a 1/ r singularity. The third and higher terms in the Williams solution have positive exponents on r and vanish at the crack tip. However, the second term remains finite. The two-term expansion is:

0 0

( ) 0 0 02 0 0

Iij ij

TK f

r Tσ θ

π ν

⎡ ⎤⎢ ⎥= + +⎢ ⎥⎢ ⎥⎣ ⎦

(2-136)

where T is a uniform stress in the x direction, which in turn, results in the stress Tν in the z direction in plane strain. It has been shown that this second term can affect the plastic zone shape and the stresses deep inside the plastic zone. The influence of the T stress can be investigated by constructing a circular model that contains a crack, as depicted below.


57

On the boundary of this model, we may apply in-plane tractions that correspond to a two-term expansion of the Williams series. To ensure the validity of the boundary conditions, the size of the plastic zone which develops at the crack tip must be small relative to the size of the model. This configuration simulates the near-tip conditions in an arbitrary geometry and is often referred to as the modified boundary layer analysis.

1 1( )2

Iij ij i j

K f Tr

σ θ δ δπ

= +

Figure 41 FRACTURE TOUGHNESS TESTING Fracture toughness tests are carried out to determine the resistance of materials to crack growth. A variety of specimens have been proposed for fracture toughness testing. The single-edge-notch bend specimen, SENB, and the compact-tension, C(T), are the most common designs and have been standardized by ASTM.

a

W1.25W

S=4W

0.6W

W

B

a

B

SENB

C(T)

Figure 42

The stress intensity calibration for the SENB specimen is:

( ) ( )( ) ( ) ( ){ }( ) ( )

3/ 2

1/ 2 2

3/ 2

( / ) (2-137)

3 / 1.99 / 1 / 2.15 3.93 / 2.7 /( / )

2 1 2 / 1 /

IPSK f a W

BW

a W a W a W a W a Wf a W

a W a W

=

⎡ ⎤− − − +⎣ ⎦=+ −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦


58

in which P is the applied load and the dimensional parameters are shown in Fig. 42. For the C(T) specimen the K calibration is:

( )( )

( ) ( ) ( ) ( )

1/ 2

2 3 43/ 2

( / ) (2-138)

2 /( / ) 0.886 4.64 / 13.32 / 14.72 / 5.6 /

1 /

IPK f a W

BWa W

f a W a W a W a W a Wa W

=

+ ⎡ ⎤= + − + −⎣ ⎦−

At the end of this notes, we will show how a simple analytic K-expression can be derived for the C(T) specimen using an energy approach. The obtained expression is:

( )( )

1/ 2

3/ 21/ 2

6.75 / 8.251 /I

a WPKBW a W

+=

− (2-139)

Figure 43 shows the comparison between the above expressions.

Figure 43 Assignment 5: Using FRANC2D, obtain a K-calibration for the standard C(T) specimen and compare it with the above expressions.


59

Plane Strain Fracture Toughness The details of the procedure for determining plane strain fracture toughness of materials are described by the ASTM Standard E 399. As mentioned before, fracture toughness varies with the specimen thickness. Thus, the specimen should be thick enough to ensure the minimum toughness corresponding to the plateau shown in Fig. 40. The measured critical SIF is then called the plane-strain fracture toughness for mode I (designated as KIc) and is a true material property. The specimens should contain the sharpest possible crack to represent the most sever situation. The usual practice is to machine a notch with certain geometry and size in the specimen and then produce a fatigue crack in the root of the notch by cycle-loading the specimen. It should be noted that the successful usage of the experimentally determined KIc values in real applications requires that the test materials be identical to those used in real situations in terms of metallurgy and manufacturing procedure like heat treatment, etc. The direction of the initiating crack relative to any metallurgical orientation is also important. The test environment should simulate the real situation particularly with respect to temperature, loading rate, and corrosive media. The experiments are usually performed using servo-hydraulic mechanical testing machines equipped with autographic instrumentation to record load and displacement. Displacement is usually measured using a clip-gauge which is mounted at the mouth of the machined notch. If the specimen fails in a linear brittle manner, as shown in Fig. 44, the failure load Pcrit is used to calculate KIc using the appropriate K calibration expression.

Figure 44

In practice, however, the load/displacement behavior shows some degree of non-linearity as depicted in Fig. 45. This effect can be attributed to limited stable crack extension or the plastic zone growth prior to instability. In this situation, failure is defined as 2% crack growth; which in normal specimen types would result in a deviation from a linear


60

trace of about 5%. Thus, the intersection of a line with a 5% smaller slope than the elastic line with the load-displacement curve is considered as the critical load.

Figure 45

The obtained critical K, designated as KIq, should be further checked to meet a criterion with regard to size and constraint. The plastic zone size is calculated using the KIq and compared with the crack length and thickness. If the following criterion is met:

2

or 2.5 Iq

ys

Ka B

σ⎛ ⎞

≥ ⎜ ⎟⎜ ⎟⎝ ⎠

(2-140)

then KIq is considered as KIc.

Load-COD Analysis of the C(T) Specimen In this section we present the derivation and experimental verification of a compliance expression for the standard C(T) specimen. Accordingly, we will derive a simple and accurate analytic expression for the stress intensity calibration for the C(T) specimen using an energy approach. More detailed information about the procedure can be found in: Mirzaei, M., "Analytic Compliance and Stress Intensity Factor Expressions for C(T) Specimen," Modarres Technical and Engineering Journal, No. 2, 1996, 44-49. Based on the theory of elasticity, the analytical solution for such problems starts with the definition of a stress function, which simultaneously satisfies the equilibrium equations and the specified boundary conditions. However, the simplifying assumptions required to find the solution for a finite geometry, such as a C(T) specimen, influences the accuracy of the final expression. Accordingly, the most accurate expressions found in the literature have been derived using numerical techniques. The expression proposed by Newman (also reported in the ASTM proposed recommended practice for R-Curve determination)


61

derived from a combination of the complex variable method of Muskhelishvili with an improved boundary-collocation method, is in the original form of:

( ) ( ) ( ) ( )2 3 4v 120.7 1065.3 / 4098 / 6688 / 4450.5 /EB a W a W a W a WP

= − + − + (2-141)

where E is the modulus of elasticity, B is the specimen thickness, v is the CMOD (crack mouth opening displacement), P is the applied load, a is the crack length, and W is the width of the specimen. The above expression can be solved for either compliance or stiffness of the specimen and has an accuracy of ±0.4 percent within the range of 0.35 < a/W < 0.6. However, the expression dramatically deviates from experimental data outside the specified range, particularly at low a/W ratios. Based on the Newman's results and by the use of Wilson's deep crack analysis, Saxena and Hudak proposed elastic compliance expressions for a wider range of crack lengths, i.e., 0.2 < a/W < 0.975. They used Newman's results to calculate the location of the axis of rotation of the crack surfaces at various crack lengths. This, accompanied by an extrapolation technique, resulted in the initial form of the compliance expression. The obtained expression was then transformed into a polynomial form using curve fitting to give:

( ) ( )

( ) ( ) ( )

22

3 4 5

0.25 1 /1 1.61369 12.6778 / 14.2311 // 1 /

16.6102 / 35.0499 / 14.4943 /

EB a W a W a WP a W a W

a W a W a W

ν +⎛ ⎞⎛ ⎞ ⎡= + + −⎜ ⎟⎜ ⎟ ⎣−⎝ ⎠⎝ ⎠⎤− + − ⎦

(2-142)

Notwithstanding the above approaches, the elastic deformation of a C(T) specimen can be quantified through a simple analysis of the deformation components. Figure 46 is a schematic presentation of one-half of the specimen in both its undeformed and deformed configurations. Because of its symmetry only half of the specimen is considered.

Figure 46


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The analysis starts by assuming that each half of the specimen consists of two major segments separated by an imaginary boundary, CD. The next assumption is that the line OO' suffers no deformation so that the vertical displacements of the points O and O' are identical. According to Fig. 40 the application of the load P forces the block AA'CD to deform in shear and the block DCEF to deform in bending. Thus, the overall vertical displacement of the point O' can be considered to consist of three different components, i.e., the vertical displacement of the point B due to the deformation of the block AA'CD, the vertical displacement of the point C due to the deformation of the block DCEF, and the rotation induced by the deformation of the block DCEF. Starting with the first component, for the element of length dx the strain energy due to shear, Ush , may be written as:

2

2shU B dx dyGτ

= ∫ (2-143) Here, τ can be defined as:

Q A yIB

τ = (2-144) in which, A is the area of the shaded region of the cross section, y is the distance of its centroid from the neutral axis, Q is the cross-section shear force, and I is the second moment of area. Substituting for τ in Eq. (2-143) results in:

22/ 2 2

/ 22 2 4H

sh H

Bdx Q HU y dyG I−

⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦∫ (2-145)

where H equals the height of the block AA'CD. The integration results in:

2 5

2240shBQ HU dx

GI= (2-146)

To obtain the total strain energy due to shear, t

shU , the above expression must be integrated along the length of the block which is equal to a. In this case Q is considered to be constant and equal to P so we may write:

2 5

20 240at

shBP HU dx

GI= ∫ (2-147)

which upon integration gives:


63

23

5tsh

P aUHBG

= (2-148) Here, using Castigiliano's second theorem, the vertical displacement of point A' can be calculated as:

65

tsh

BU PaP HBG

δ ∂= =

∂ (2-149)

For a standard C(T) specimen H is equal to 0.6W. Thus, the above expression can be written as:

2B

PaWBG

δ = (2-150) Based on a similar approach the vertical displacement of point C due to the deformation of block DCEF can be obtained by integration of the resultant strain along the distance h and may be written as:

( )2

6 / 2C

Ph a rEBr

δ+

= (2-151) in which r is the unbroken part or the remainder of the specimen and h is the distance of the location of M, the applied moment, from the line of symmetry. The above expression upon substitution of r by (W-a) and h by 0.4W can be rearranged into the following form:

( )( )2

1.2C

PW a WEB W a

δ+

=−

(2-152)

The third component of the vertical displacement of the point O' is caused by the rotation angel θ. The obtained expression, when rearranged for a standard C(T) specimen, is in the following form:

( )( )( )2

1.5 1.5R

P a W a WEB W a

δ+ +

=−

(2-153)

The total vertical displacement of the point O' is the sum of the three individual components and can be written as:

( )( )( )2

1.5 2.3 2T

a W a WP aB WGE W a

δ⎡ ⎤+ +

= +⎢ ⎥−⎢ ⎥⎣ ⎦

(2-154)


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The above expression was derived for half of the specimen. In practice, the CMOD for the whole specimen, v, is two times Tδ . In order to simplify the obtained expression, G can be substituted by E/2(1 +ν). Accordingly, the final expression for compliance can be written as:

( )( )( )

( )2

3 2.3 8 1v a W a W aEBP WW a

ν+ + += +

− (2-155)

which can be rearranged in terms of the a/W ratio to give:

( )( )( )

( )( )2

3 / 1 / 2.3v 8 1 // 1

a W a WEB a a WP a W

ν+ +

= + +−

(2-156)

Figure 47 is a comparison between Eq.(2-156), with Eq.(2-141) proposed by Newman, and Eq.(2-142) proposed by Saxena and Hudak. In order to present the entire a/W range, a logarithmic scale is used for the normalized compliance. It is evident that there is an excellent agreement between the three expressions within the middle range of a/W. Above a/W = 0.6 and below a/W = 0.35 Newman's expression deviates from the other two equations.

Figure 47

As shown before, the stress intensity factor can be related to compliance through the following expression:


65

2

2 G2EP CK E

B a∂

= ≡∂

(2-157) in which G is the energy release rate and C is the compliance. In this equation, however, compliance should be defined for the point of application of load. For this purpose Eq.(2-156) can be modified for the load line to give:

( )( )( )

LL2

3 / 1 / 1.5v/ 1

a W a WEB EBCP a W

+ += =

− (2-158)

In the above vLL is the displacement at the load line. Finally, the required expression for K can be obtained from Eqs (2-157) and (2-158) as:

( )( )

1/ 2

3/ 21/ 2

6.75 / 8.251 /I

a WPKBW a W

+=

− (2-159)

Plane Stress Fracture Toughness Design based on KIc can be considered quite conservative for relatively thin sections in the structures because materials usually show much higher fracture toughness in the state of plane stress (see Fig. 40). In our previous discussion on K-G relationship we showed that:

( )2

2

2

1 ,

,

I

I

KG plane strainE

KG plane stressE

ν= −

=

Hence, the critical stress intensity factor, Kc, can be used to calculate the critical energy release rate, R, and vice versa. Also in our previous discussions on energy approach we saw that some materials, particularly in the state of plane stress, exhibit a rising R curve. In these situations, it is possible to define the critical stress intensity at the point where G is tangent to the R curve, as depicted in Fig. 48 (Note that in general G can be a non-linear function). However, this instability point does not represent a material property, because it depends on the shape of the R curve, which in turn, is governed by the geometry of the cracked body. Thus the usage of such Kc values in design should be carried out with some precaution, regarding the similitude between the size and geometry of laboratory specimens and real components. For instance, the thickness of the laboratory specimen is normally fixed to match the thickness of the real component.


66

a0

RK /Ec

2

Figure 48 One problem with thin sheet fracture toughness testing is that the specimens are subject to out-of-plane buckling. Consequently, an anti-buckling device should be fitted to the specimen, which makes the test more difficult. The ASTM Standard E 561 describes alternative methods for computing both the stress intensity and the crack extension in an R curve test. In general as the crack grows, the load-displacement curve deviates from its initial linear shape because of the continuous change in compliance. The test can be carried out under displacement-control condition to ensure stable crack growth. Instantaneous crack length can be obtained by partial unloading, measuring the compliance (see Fig. 49), and using the compliance expressions like (2-142) and (2-156).

1/C

Displacement

Partial unloadingsLoad

Figure 49


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The instantaneous stress intensity is related to the current values of load and crack length. Usually, the stress intensity should be corrected for plasticity effects by determining an effective crack length. Some authors use “1” as the subscript to designate the plane-stress fracture-toughness for mode I as K1c. These notes have been prepared as a student aid and should not be considered as a book. Little originality is claimed for these notes other than selection, organization, and presentation of the material.

The following references have been used for preparation of the lecture notes and are recommended for further study in this course.

References: Anderson, “Fracture Mechanics Fundamentals and Applications.” Meguid, “Engineering Fracture mechanics” Kanninen, "Advanced Fracture Mechanics" Broek, “Elementary Engineering Fracture Mechanics” Dowling, "Mechanical Behavior of Materials" “Structural Integrity Lecture Notes”, by Prof. Gray

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