Chapter 4 Plane problems in linear isotropic elasticity

4.1 Introduction

The basic problem in the theory of elasticity is to solve a system of differential equations with appropriate boundary conditions (Chou & Pagano, 1967; Slaughter, 2002). For a static isothermal case this system is reduced to fifteen equations with the unknowns being the stresses, strains and displacement (see Appendix I). The general solution of these equations is often too difficult. Thus one seeks for simplifications of the problem with regard to the distribution of stresses or strains. In several cases of engineering interest, simplifying assumptions can be adopted, without loss of accuracy in the obtained solution. These approximations make the solution of the problem much simpler.

Below we define two important cases of plane problems of elasticity. These are the case of plane strain and that of plane stress. The solution of these problems is discussed in terms of Airy's stress functions only. The short presentation of the plane elasticity serves as an introduction to the analysis stress distribution around a wedge and the s of fracture which is presented at the end of this chapter.

4.2 Conditions of plane strain

Consider a long prismatic bar under the action of lateral forces and/or (Figure 4.1). Assume that the component of the body force along X3 is zero while the body forces along X1 and X2 are only functions of x1 and x2.

)x,P(x 21 )x,Q(x 21

Due to the large dimension along the X3 axis, it may be assumed that at some distance from the ends of the bar, the displacements u1 and u2 are functions of x1 and x2 only. Moreover, if the bar is very long (i.e., of infinite length) or its ends are fixed, we assume that u3 = 0 at every cross section of the bar. In such cases the strain components are,

1

111

ux

∂ε =

∂, 2

222

ux

∂ε =

∂, 1 2

122 1

u u12 x x

⎛ ⎞∂ ∂ε = +⎜ ∂ ∂⎝ ⎠

⎟ (4.1)

and

3

333

u 0x

∂ε = =

∂, 31

133 1

uu1 02 x x

⎛ ⎞∂∂ε = + =⎜ ⎟∂ ∂⎝ ⎠

, 3223

3 2

uu1 02 x x

⎛ ⎞∂∂ε = +⎜ ∂ ∂⎝ ⎠

=⎟ (4.2)

The state of deformation so defined is called plane strain. With the strain components defined by (4.1) and (4.2), the stress components

11σ ,

22σ ,

33σ and 12

σ are obtained from Hooke's law (eqs I.3, Appendix I). These are,

( )( ) ( )11 11 22E 1- ν ν

1 ν 1- 2νσ = ε + ε⎡⎣+

⎤⎦ (4.3a)

( )( ) ( )22 22 11E 1- ν ν

1 ν 1- 2νσ = ε + ε⎡⎣+

⎤⎦ (4.3b)

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( )( ) ( )33 11 22νE

1 ν 1- 2νσ = ε + ε

+ (4.3c)

12 21E

1 νσ = ε

+ ,

(4.3d) 23 31 0σ = σ =

Figure 4.1 State of plane strain in a prismatic body.

The strain components are obtained by solving eqs (4.3),

( )11 11 221 ν 1 ν ν

E+

ε = − σ − σ⎡⎣ ⎤⎦ (4.3a')

( )22 22 111 ν 1 ν ν

E+

ε = − σ − σ⎡⎣ ⎤⎦ (4.3b')

12 121 ν

E+

ε = σ (4.3c')

Equations (4.1,2) and (4.3a’) show that the stress components are functions of x1 and x2 only. Accordingly, the equations of equilibrium become,

11 121

1 2

f 0x x

∂ σ ∂ σ+ + =

∂ ∂ , 12 22

21 2

fx x

∂ σ ∂ σ 0+ + =∂ ∂

(4.4)

We impose the same restriction on the surface forces. Normally, the forces and should be functions of x1 and x2 only, with = 0, for plane strain to prevail. Thus, the boundary conditions, when forces are prescribed are,

1t 2t

3t

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Figure 4.2 Surface forces and boundary conditions.

, (4.5) 1 11 1 12t n= σ + σ 2n 2n2 12 1 22t n= σ + σ

where n1 and n2 are the components of the unit vector n along x1 and x2 ( Figure 4.2).

When the stresses are the unknowns, the compatibility equations must be used. Under the assumption of plane strain stated above the compatibility equation that is not identically satisfied is,

2 2 2

11 22 122 2

2 1 1

2x x x

∂ ε ∂ ε ∂ ε+ =

∂ ∂ ∂ ∂ 2x. (4.6)

Therefore, in the case of plane strain problems, eight quantities must be determined by solving eqs (4.1), (4.3) and (4.4)

satisfying the boundary conditions (4.5). However, these eight equations can be reduced to three in the following manner.

11 22 12 11 22 12 1 2( , , , , , , u , u )ε ε ε σ σ σ

First : introduce eqs (4.3a',b',c') in (4.6),

( ) ( )22 2

1211 22 22 112 2

2 1

1 ν ν 1 ν ν 2x x

∂ σ∂ ∂− σ − σ + − σ − σ =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦∂ ∂ 1 2x x∂ ∂

(4.7)

Second : differentiate the equations of equilibrium (4.4), the first with respect to x1 and the second with respect to x2 and add these two equations,

2 2 2

12 11 22 1 22 2

1 2 1 2 1 2

f f2x x x x x x

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

⎞⎟⎠

(4.8)

Finally substituting the last equation into (4.7) we have,

( ) ( )2 21

11 222 21 2 1 2

f f1x x 1

2

ν x x⎛ ⎞∂ ∂ ∂ ∂

+ σ + σ = − +⎜ ⎟∂ ∂ − ∂ ∂⎝ ⎠ (4.9a)

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which in the absence of body forces reduces to,

( )

2 2

11 222 21 2

0x x

⎛ ⎞∂ ∂+ σ + σ⎜ ⎟∂ ∂⎝ ⎠

= (4.9b)

According to the foregoing analysis, we have now, a set of three equations. Two equations of equilibrium (4.4) and one equation derived above (4.9). These three equations have as unknowns three quantities, i.e,

11σ ,

22σ and

12σ . This set of equations, with the boundary

conditions equations (4.5) can be used when one seeks a solution for a plane strain problem. After the determination of the stress components, the strains are computed from equation (4.3) and the displacements from (4.1).

4.2.1 The Airy stress function in plane strain problems

The problem of plane strain can further be reduced to one equation that contains a single parameter, called Airy’s stress function. It is not difficult to verify that the equations of equilibrium are satisfied if the stress components are given by the derivatives of a function ψ , i.e.,

2

11 22x

∂ ψσ =

∂,

2

22 21x

∂ ψσ =

∂,

2

12 1 2 2 11 2

f x f xx x∂ ψ

σ = − − −∂ ∂

(4.10)

Neglecting the body forces we have,

2

11 22x

∂ ψσ =

∂,

2

22 21x

∂ ψσ =

∂,

2

121 2x x

∂ ψσ = −

∂ ∂. (4.11)

Substituting (4.11) into (4.9b) we obtain,

4 4 4

4 2 2 41 1 2 2

2x x x x

∂ ψ ∂ ψ ∂ ψ+ +

∂ ∂ ∂ ∂0= , (4.12)

which is the governing equation for the Airy's stress function ψ . Realizing that

( )24 4 4 2 2 22

4 2 2 4 2 21 1 2 2 1 2

2x x x x x x

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

∇ (4.13)

equation (4.12) can be written as,

(4.14) 4 0∇ ψ =

which is a standard way to present a biharmonic equation.

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The particular problem of elasticity is thus reduced to identifying a function , that satisfies the boundary conditions of the problem at hand. Knowledge of the function allows for the determination of the stresses from (4.11), strains from (4.3a’-c’) and displacements from (4.1).

ψ

4.3 Conditions of plane stress

Let us consider now Figure 4.3 where is some how the other extreme of the long cylinder in Figure 4.1. Suppose we have a body whose dimension along x3 is very small compared to the dimensions along x1 and x2 (Figure 4.3). We also assume that loading is on the boundary and parallel to the plane of the plate. We further consider that the body force along x3 is zero and those along x1 and x2 are functions of x1 and x2 only. From the considerations regarding the geometry of the body and the loading we see that at x3 = ±h/2 there are no external forces. Consequently the stress components

33σ ,

13σ and

23σ are zero there. If the plate is sufficiently

thin, we can assume without substantial error that these components are zero across the thickness and that the other stress components

11σ ,

22σ and

12σ remain practically constant

across the thickness. Such a state of stress is called plane stress. That is when 33

σ =13

σ =23

σ = 0

and 11

σ , 22

σ and 12

σ are functions of x1 and x2 only. In such cases, the equilibrium equations are the same as equations (4.4). The boundary conditions are also the same as equation (4.5).

To obtain the stress-strain relations we use Hook's law (eqs I.6), with 33

σ =13

σ =23

σ , to obtain,

(11 11 221 ν E

ε = σ − σ ) (4.15a)

(22 22 111 νE

ε = σ − σ ) (4.15b)

12 121 ν

E+

ε = σ (4.15c)

and

, 13 23 0ε = ε = (33 11 22νE

)ε = − σ + σ (4.15d)

The stress components are obtained by solving eqs (4.15),

[11 11 222

E ν1 ν

σ = ε + ε−

] (4.16a)

[22 22 112

E ]ν1 ν

σ = ε + ε−

(4.16b)

12 21E

1 νσ = ε

+ (4.16c)

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By solving for ( and inserting the result into (4.15b))11 22σ + σ 2 we obtain,

(33 11 22ν

1)

νε = − ε + ε

− (4.17)

The last equation gives the "out of plane" normal strain in terms of the "in plane" normal strains. Note that

33ε is not a part of the expressions for plane stress. However, it can be

obtained independently using (4.17). Naturally, then can be obtained by integrating 3

u

33 3 3u xε = ∂ ∂ . The displacements and are independent of x3 and the strain displacement relations are given by (4.1).

1u

2u

Figure 4.3 Conditions of plane stress.

As in the case of plane strain, the equations of plane stress can be reduced to three equations involving the three components of stresses

11σ ,

22σ and

12σ . This is not difficult to see when we

realize that eqs (4.1) and (4.6) apply also to plane stress. Substituting the strain components (4.15) into (4.6) and using the equilibrium eqs (4.4), we obtain,

( ) ( )2 2

1 211 222 2

1 2 1 2

f f 1 νx x x x

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

⎞⎟⎠

. (4.18)

This last equation together with the equilibrium equations is a system of three equations with three unknowns.

4.3.1 The Airy’s stress function in plane stress problems

A function that satisfies the equilibrium equations can be defined in a similar manner as in plane strain discussed earlier. By defining the stress components with eqs (4.10) and introducing them in (4.18) and setting the body forces zero we obtain,

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(4.19) 4 0∇ ψ =

In summary, it has been shown above that for the plane problems of elasticity, plane strain and plane stress, the stress function is governed by the same differential equation provided that the body forces are zero. The difference among these cases is that after the stresses are calculated, the strains must be calculated from eq. (4.3) for plane strain and (4.16) for plane stress. In a similar manner we could arrive at formulating the plane problem in terms of a single function when the body forces are conservative, or when they are derived by a known potential [2].

Note: By using appropriate combinations with the elastic constants, the equations of plane strain can be converted to those of plane stress (Chou & Pagano, 1967). Namely,

a. From plane strain to plane stress

The constitutive equations for plane strain (eqs 4.3) can be converted to those for plane stress (eqs 4.15 and 4.16) if in the place of Young modulus, E, in eqs 4.3 we use E(1+2ν)/(1+ν)2 and in the place of Poisson's ratio ν, we use ν/(1+ν),

b. From plane stress to plane strain

Similarly, the constitutive equations for plane stress (eqs 4.15 and 4.16) can be converted to those for plane strain (eqs 4.3) if in the place of Young modulus E, in (4.15) and (4.16) we use E/(1-ν2) and in the place of Poisson's ratio ν, we use ν/(1-ν). Thus, the solution for a plane stress problem can be determined from the solution for the corresponding plane strain problem, and vice versa.

4.4 Stress field in a plate with a sharp notch

It has been demonstrated by several experiments that most engineering materials fail well below their theoretical strength. This "strength loss" has been attributed to the presence of defects in a body. Although their geometry is not that of a hole, an ellipse or of a crack, quite often they can be approximated as such for the purpose of analytical treatment of the problem. In this section we look at the stress field around a sharp wedge (or a sharp notch) on a plate in order to obtain an insight on the problems of stress distribution around the tip of a crack. The branch of mechanics that addresses such problems is called fracture mechanics (Anderson, 1995; Broek, 1986; Hellan, 1984)

A plate in a state of plane stress or plane strain is bounded by two straight edges with an angle of 2(π-α) and remote boundaries (see Figure 4.4) at infinity. We are interested in the stress fields around the tip of the wedge. The boundary conditions of the problem are,

(4.20) θθ rθ 0 for α , r 0σ = σ = θ = ± >

Although there are methods to solve for the stresses and strains around cracks or notches (Hellan, 1984), we are following here an elegant method of solution which is based on the concept of self-similarity of the stress field (see Appendix II) around the tip of the wedge (or crack) and the Airy’s stress function,. The concept of self-similarity states that the distribution of the field under question remains similar to itself when a change in the intensity of the field is imposed. On the analytical level, self-similarity reduces the number of unknowns by using concepts of symmetry.

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Figure 4.4 Stress and displacement components in the vicinity of a sharp notch.

To solve the problem under consideration we assume that the stress distribution is self-similar. Thus, employing cylindrical coordinates the following stress function is assumed,

( ) ( )λr,θ r θ−ψ = φ (4.21)

Function must satisfy the biharmonic eq. (4.14) or (4.19) in cylindrical coordinates, (r,θψ )

( )2 2 2

222 2 2 2

d1 1 r 0r r r r d

− λ+⎡ ⎤⎡ ⎤ ⎛ ⎞∂ ∂ φ∂+ + λ φ + =⎢ ⎥⎜ ⎟⎢ ⎥∂ ∂ ∂θ θ⎣ ⎦ ⎝ ⎠⎣ ⎦

This equation can be written as,

( ) ( )2 4

2 22 22 4

d d2 2d d

φ φ⎡ ⎤ ⎡ ⎤λ λ + φ + λ + + λ + =⎣ ⎦ ⎣ ⎦ θ θ0 (4.22)

which is an ordinary differential equation for the function φ .

We assume a form of the solution, φ = eaθ. Taking the derivatives of , inserting them in (4.22) and eliminating eaθ we have,

φ

( ) ( )2 22 2λ λ 2 λ 2 λ a a⎡ ⎤+ + + + + =⎣ ⎦2 4 0

2

)

Defining a2 = m, we obtain the following quadratic equation,

( ) ( )2 22 2λ λ 2 λ 2 λ m m 0⎡ ⎤+ + + + + =⎣ ⎦

whose solutions are,

and 21m = −λ 2

2m ( 2= − λ +

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or 21,2a i= ± −λ = ± λ , ( ) ( )2

3,4a 2 i 2= ± − λ + = ± λ +

Thus, the solution of the differential eq. (4.22) is,

( ) ( ) ( )2 i 2 ii i1 2 3 4c e c e c e c eλ+ θ − λ+ θλθ −λθφ θ = + + + (4.23)

where ci (i = 1, ..., 4) are constants to be determined. Using Euler's formula,

ie cos i sin± λθ = λ θ ± λ θ

eq. (4.23) can be written as,

( ) ( ) ( )A cos Bsin Ccos 2 Dsin 2φ θ = λθ + λθ + λ + θ + λ + θ (4.24)

To proceed further with the solution we need to look at the symmetry properties of the solution with respect to the type of loading. In fracture mechanics, we consider three types of loads as shown in Figure 4.5.

In the most general case, a solid with a crack in it can be subjected to all three modes. For the problem we are solving (plane stress or plane strain), Mode III does not apply since the loads are normal to the plane of the plate. Thus, we look at Modes I and II only in our solution. Going one step further, with regard to the symmetries of the loading and geometry, we note that Mode I is symmetric in that the stresses at a point P are the same as these at point P'. With respect to Mode II we note that it is antisymmetric in that the stress state at P is antisymmetric of that at P' (Figure 4.5). These two symmetries are reflected in the solution (4.24) if we realize that,

(4.25a) (A cos A cosλ θ = − λ θ)

⎤⎦

)

(4.25b) ( ) ( )C cos 2 C cos 2λ + θ = − λ + θ⎡⎣

(4.26a) B sin B sin( )λ θ = − −λ θ

(4.26b) ( ) ( )(D sin 2 D sin 2λ + θ = − − λ + θ

Accordingly, the terms (4.25) correspond to Mode I and the terms (4.26) to Mode II. In the following sections we study Mode I only. The same methodology can be applied for Mode II.

4.4.1 Mode I-loading

The stress function for Mode I is deduced from (4.24) by accounting for (4.25) and (4.26),

(4.27) ( ) ( )s A cos C cos 2φ θ = λ θ + λ + θ

where the subscript s indicates the symmetric part of (4.24). The stress components in cylindrical coordinates are (Botsis & Deville, 2006),

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2

r r 2 2

1 1r r r

∂ψ ∂ ψσ = +

∂ ∂θ,

2

2rθθ

∂ ψσ =

∂,

2

r 2

1 1r rθ r

∂ψ ∂ ψσ = −

∂θ ∂θ ∂ (4.28,a,b,c)

where, . ( )sr−λψ = φ θ

Figure 4.5 Modes of fracture: Mode I - opening; Mode II – shearing; Mode III – tearing.

Thus, ( ) ( ) ( )2s1 r− λ+

θθσ = λ λ + φ θ , ( ) ( ) ( )2 sr

d1 r

d− λ+

θ

φ θσ = λ +

θ

Using the boundary conditions (4.20) we obtain,

( ) ( ) ( ) ( )2sr, 0 1 r 0− λ+

θθ θ=±α θ=ασ θ = ⇒ λ λ + φ θ = (4.29a)

( ) ( ) ( )2 sr

dr, 0 1 r 0d

− λ+θ θ=±α

θ=α

φσ θ = ⇒ λ + =

θ (4.29b)

Note here that r > 0 along θ =± α. Thus, to have eqs (4.29) identically zero, function should satisfy the conditions,

sφ

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, ( )s 0φ α = ( )sd 0dφ

α =θ

Using these two conditions in eq. (4.27) we find,

(4.30a) ( )A cos C cos 2 0λ α + λ + α =

(4.30b) ( ) ( )A sin C 2 sin 2 0− λ λ α − λ + λ + α =

For a non trivial solution for A and C we need,

( )

( ) ( )cos cos 2

0sin 2 sin 2

λα λ + α=

λ λ λ + λ + α

The last condition leads to the following equation for λ,

( ) ( ) ( )2 cos sin 2 cos 2 sin 0λ + λ α λ + α − λ λ + α λ α =

Using trigonometry, the last equation results in,

2

sin 2tan 22 cos

λ αα = −

⎡ ⎤λ + λ α⎣ ⎦

which is a transcendental equation for the unknown λ.

To simulate a "mathematical crack" we let α approach π. Therefore, tan 2π = 0, and

sin 2λπ = 0

which has as roots,

λ = n2

; n being integer.

To each value of λ, or n, (an infinite number of them) there corresponds a relationship between the coefficients A and C (eqs 4.30). This is found by introducing the value of λ in (4.30b), i.e.,

n n4 nA C

n+

= − or n nnC A

4 n= −

+ (4.31a)

Consequently, the symmetric part of the stress function (4.21) takes the form,

( )λs(r, ) r θ−ψ θ = φ (4.31b)

takes the form,

n 2n

n

nθ n n(r,θ) A r cos cos 2 θ2 n 4 2

− ⎡ ⎤⎛ ⎞ψ = − +⎜ ⎟⎢ + ⎝ ⎠⎣ ⎦∑ ⎥ (4.31c)

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The stress components are obtained using (4.31c) and (4.28) in a series form,

2

(2 n 2)rr n

n

n n nθ n n n nA r 1 cos 2 cos 2 θ2 2 2 n 4 2 2 2

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

∑ (4.32a)

(2 n 2)θθ n

n

n n nθ n n1 A r cos cos 2 θ2 2 2 n 4 2

− + ⎧⎡ ⎤ ⎛ ⎞σ = + − +⎨ ⎜ ⎟⎢ ⎥ +⎣ ⎦ ⎝ ⎠⎩ ⎭∑ ⎫

⎬ (4.32b)

(2 n 2)r n

n

n n nθ n n n nA r 1 sin 2 1 sin 2 θ2 2 2 n 4 2 2 2

− +θ

⎧ ⎫⎡ ⎤ ⎛ ⎞⎛ ⎞ ⎛ ⎞σ = − + + + + +⎨ ⎬⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥ +⎣ ⎦ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎩ ⎭∑ (4.32c)

However, we can reject a part of the values of n (or λ) due to the following physical reasoning,

(a) for n22

⎛ ⎞+ = λ + =⎜ ⎟⎝ ⎠

2 0 , the stresses do not depend on r. This is physically unrealistic.

(b) for n22

⎛ ⎞+ = λ + <⎜ ⎟⎝ ⎠

2 0 , the stresses go to infinity as r goes to infinity. This is also

unrealistic.

(c) for n22

⎛ ⎞+ = λ + >⎜ ⎟⎝ ⎠

2 0 , we obtain the only realistic constraint for the values of λ.

Accordingly, λ can take only the values,

1 3 50, , 1, , 2, , ...2 2 2

λ = ± ± ± (4.33)

Furthermore, several of the remaining terms in the expressions (4.32) decay very fast as r are increased. Therefore their contribution to the stress state in the vicinity of the crack is not important and can be neglected. Other terms, give unlimited energy within a volume surrounding the crack tip and as such must not be part of the series solutions in (4.32). The only term with the largest contribution to the stresses in vicinity of the crack tip, is the term that corresponds to the value of n = -3, or λ = - 3/2. For this case, we obtain from (4.31a) that

and, 3A C=

- from (4.32a) rr3A 35cos cos

2 24 rθ θ⎡σ = −⎢⎣ ⎦

⎤⎥ (4.34a)

- from (4.32b) 3A 33cos cos2 24 rθθ

θ θ⎡σ = +⎢⎣ ⎦⎤⎥ (4.34b)

- from (4.32c) r3A 3sin sin

2 24 rθ

θ θ⎡σ = +⎢⎣ ⎦⎤⎥ (4.34c)

The constant A is determined by the boundary conditions. It is usually renamed and for Mode I is expresses as,

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I3A K 2π= (4.35)

The new parameter , is called stress intensity factor (SIF). Thus, the stress components (4.34) take the form,

IK

Irr

K 5 1 3cos cos4 2 4 22 r

θ⎡σ = −⎢π ⎣ ⎦θ⎤

⎥ (4.36a)

IK 3 1 3cos cos4 2 4 22 rθθ

θ⎡σ = +⎢π ⎣ ⎦θ⎤

⎥ (4.36b)

Ir

K 1 1 3sin sin4 2 4 22 rθ

θ⎡σ = +⎢π ⎣ ⎦θ⎤

⎥ (4.36c)

It is worth making the following important observations:

1. The stresses tend to infinity when r approaches the crack tip. This singularity is characteristic of solutions of crack problems using linear elasticity, and is not valid at r=0. In reality the material yields, or develops some kind of damage. However, the solution (4.36) accurately provide the stress state within an area bounded by the frontier of the damaged zone and a circle of radius r, that is small in comparison to the crack size.

2. The solution for the stresses does not contain the elastic constants of the material.

3. They are applicable for both plane stress and plane strain.

4. The exact expression for KI depends on the loading and boundary conditions. When these conditions are known, the coefficient A and thus, KI can be determined. However, proceeding with the present method to determine KI is not realistic due to the series nature of the solution for the stresses. On the other hand, direct analytical methods can be used for the expressions of KI in terms of the applied load and geometry (Anderson, 1995; Hellan, 1984). When closed form solutions are not available, numerical methods are used to obtain expressions for the SIF (Tada, Paris & Irwin, 2000). Selected cases for SIF, using numerical means, are shown in Appendix III.

Figure 4.6 shows a plate with a central crack under remote stress ∞

σ . For such a case, the SIF is related to the applied stress and crack size in the following way,

IK ∞= σ πa (4.37)

With the stresses known, we can calculate the strains using Hook’s law. For example, when plane stress prevails, we have,

( )rr rr1E θθε = σ − νσ , r r

1Eθ θ

+ νε = σ , ( rr

1Eθθ θθ )ε = σ − νσ (4.38)

Further, integrating the strain-displacement relations (Chou & Pagano, 1967),

March 2006 4-13

rrr

ur

∂ε =

∂, r

ru uu1

2 r r r rθ θ

θ

⎛ ⎞∂ ∂ε = + −⎜ ⎟∂ ∂⎝ ⎠

, ru ur r

θθθ

⎛ ⎞∂ε = +⎜ ∂θ⎝ ⎠

⎟ (4.39)

we obtain the displacement components and for Mode I are (Anderson, 1995),

Figure 4.6 Specimen loading configuration in Mode I.

( ) ( )Ir

K 1 ν r θ 3θu 2κ 1 cos cos2E 2π 2 2

+ ⎡= −⎢⎣ ⎦⎤− ⎥ (4.40a)

( ) ( )Iθ

K 1 ν r θ 3θu 2κ 1 sin sin2E 2π 2 2

+ ⎡= − + +⎢⎣ ⎦⎤⎥ (4.40b)

where, 3 νκ1 ν

−=

+

The displacement components are very important in the analysis of plates with cracks. In particular, they offer the means to determine the SIFs experimentally by using surface displacement measurements. Such an example for Mode I is shown in Appendix IV. When plane strain prevails, we use the Hooke's law for plane strain and proceed in exactly the same manner to arrive at the same expressions as (4.40) for the displacement components. The only difference being in the constant κ which for plane strain is,

κ 3 4ν= −

Equations (4.40) can be transformed on a Cartesian coordinate system by the usual orthogonal transformation of a vector,

( ) ( )Ix

K 1 ν r θ 3θu 2κ 1 cos cos2E 2π 2 2

+ ⎡= −⎢⎣ ⎦⎤− ⎥ (4.40c)

( ) ( )Iy

K 1 ν r θ 3θu 2κ 1 sin sin2E 2π 2 2

+ ⎡= +⎢⎣ ⎦⎤− ⎥ (4.40d)

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4.4.2 Mode II-loading

Taking the antisymmetric part of the solution (eqs 4.26) and following the same procedure as that for Mode I, we can obtain the expressions for Mode II fracture. These are (Anderson, 1995),

a. for the stress components,

IIrr

K 5 3 3sin sin4 2 4 22 r

θ θ⎡σ = − +⎢π ⎣ ⎦⎤⎥ (4.41a)

IIK 3 3 3sin sin4 2 4 22 rθθ

θ θ⎡σ = − −⎢π ⎣ ⎦⎤⎥ (4.41b)

IIr

K 1 3 3cos cos4 2 4 22 rθ

θ⎡σ = +⎢π ⎣ ⎦θ⎤

⎥ (4.41c)

b. for the displacement components,

( ) ( )IIr

K 1 ν r θ 3θu 2κ 1 sin 3sin2E 2π 2 2

+ ⎡= − − +⎢⎣ ⎦⎤⎥ (4.42a)

( ) ( )IIθ

K 1 ν r θ 3θu 2κ 1 cos 3cos2E 2π 2 2

+ ⎡= − + +⎢⎣ ⎦⎤⎥ (4.42b)

Figure 4.7 Specimen loading configuration in Mode II.

Here KII is the Mode II stress intensity factor and κ has similar meaning as in (4.39). It should be added that all four observations stated earlier for Mode I fracture apply also for Mode II.

For a through crack in an infinite plate, the SIF for Mode II is given by the following expression (Anderson, 1995),

IIK τ πa∞= (4.43)

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4.4.3 Mode III-loading

Using a similar method we can obtain the stress and displacement fields in the vicinity of a Mode III crack. The details of the analysis can be found in Hellan (1984). Under Mode III, there are only two non-zero stress components and one non-zero displacement component. These are,

IIIrz

K θsin22π r

σ = (4.44a)

IIIθz

K θcos22π r

σ = (4.44b)

( )IIIθ

4K r θu 1 ν sinE 2π 2

⎡ ⎤= +⎢ ⎥⎣ ⎦ (4.45)

The stress intensity factor KIII, is also expressed in terms of the boundary conditions and the applied load. A Mode III loading is shown in Figure 4.8. The stress

∞τ , is applied at the

remote boundaries.

The stress intensity factor for Mode III is expressed as,

IIIK πa∞= τ (4.46)

Note that for all modes, the stress components for all three modes can be expressed as,

( )I,II,III I,II,IIIij ij

Kf

2 rσ = θ

π (4.47a)

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( )I,II,IIII,II,III I,II,IIIi

K ru g2E 2

=π ij θ (4.47b)

a product of the stress intensity factor K (for any of the three modes), and a function of r and θ . While K represents the intensity of the stress field and is a function of the geometry and loading conditions, the other part of the expressions relate to the distribution of the local stress field.

Figure 4.8 Loading specimen configuration in Mode III.

4.4.4 Principle of superposition for the stress intensity factors

Owing to the linear elastic model, the principle of superposition applies for each of fracture. Namely, if for n loading types in Mode I the resulting stress intensity factors are, KI

(1), KI(2), ...

KI(k), ... KI

(n), the total stress intensity factor when all loads are applied simultaneously is,

(4.48) n

total (i)I

i 1K

=

= ∑ IK

Similar relations hold true for the other modes of fracture. This principle is of great importance in obtaining stress intensity factors of various complicated specimen loading configurations. On the other hand, the SIF from different modes can not be added.

4.5 Relationship between KI and GI

Consider a specimen with a through crack of initial length equals to a + ∆a in Mode I. We apply a pressure over the increment ∆a so that the crack closes over that interval and becomes a crack of length a.

March 2006 4-17

Figure 4.9 Crack increment and stress distribution along its axis.

Clearly this closing pressure is given by the stress obtained from the asymptotic analysis shown before in this chapter (eq. 4.36b). Thus,

( ) ( )Iyy

K ax

2 xσ =

π (4.49)

Under load controlled conditions, G is given by (see chapter 3),

I a 0

UG lima∆ →

∆=

∆ (4.50)

where ∆U is the work done by the closing stresses (4.49) over the crack opening displacement and is given by,

(4.51) ( )a

0

U dU x∆

∆ = ∫

To calculate the work we proceed as follows. The increment of work is,

( ) ( ) ( )y y1dU x 2 F x u x2

= (4.52a)

with (4.52b) ( ) ( )y yyF x x dx= σ

Using (4.40d) for and setting θ = π E / 2( 1)µ = ν + we obtain,

( ) ( ) ( ) ( ) ( )I Iy

1 K a a 1 K a aru x2 2 2

κ + + ∆ κ + + ∆ a x2

∆ −= =

µ π µ π (4.52c)

Combining (4.49) and (4.52a,b,c) we obtain,

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( ) ( ) ( ) ( )I I1 K a K a a a xdU x dx2 2 x

κ + + ∆ ∆ −=

µ ⋅ π

and

( ) ( ) ( ) aI I

0

1 K a K a a a xU d4 x

∆κ + + ∆ ∆ −∆ =

µπ ∫ x

Using the definition (4.50) we obtain,

( ) ( ) ( ) aI I

I a 00

1 K a K a a a xG lim d4 a x

∆

∆ →

κ + + ∆ ∆ −=

µπ∆ ∫ x

where ( )aa

1

0 0

a x x adx a x x a sinx a

∆∆−⎡ ⎤∆ − π∆

= ∆ − + ∆ =⎢ ⎥∆⎣ ⎦∫ 2

Thus, ( ) ( ) ( )I I

I a 0

1 K a K a a aG lim4 a 2∆ →

κ + + ∆ π∆=

µπ∆

as ∆a tends to zero, → K(IK a a+ ∆ ) I (a)

and ( ) 2I

I

1 KG

8κ +

=µ

(4.53a)

It should be noted that the values of G are not additive for the same Mode. However, they can be added for the different Modes. Assuming a self similar crack growth where all three Modes apply, the total energy release rate is,

2 2 2I II III' '

K K KGE E 2

= + +µ

(4.53b)

where for plane stress 'E E=

( )' 2E E / 1= − ν for plane strain

and ( )E / 2 1µ = + ν

4.6 Mixed mode fracture

4.6.1 Uniaxial remote loading

When a single Mode is present in a structural component, crack initiation is imminent when the corresponding stress intensity factor is equal to the fracture toughness. For example, in Mode I we have,

March 2006 4-19

or equivalently IK K= IC

2IC

I IC '

KG GE

= = (4.54)

If more than one Mode operate, a condition for crack initiation is not so obvious. A typical case of mixed loading (Modes I & II) is shown in Figure 4.10.

The energy release rate for a crack subjected to all three s is given by (4.53b). This relation has been derived under the assumption of self-similar crack growth (crack growing along its plane or on planes tangent to the original plane). Due to this assumption, (4.53b) can not be used to determine values of at crack initiation, if crack growth does not satisfy the self-similar growth pattern.

IG

Figure 4.10 Case of mixed loading under uniaxial remote stress.

For several materials under mixed loading, crack growth does not follow the self similar assumption. This is true when a Mode II is present. Thus, eq. (4.53b) can not be used1. However, if the loading is mixed, relation (4.53b) serves as a useful guide for extrapolation. Namely, if a specimen is subjected to all three Modes, but 'dominated' by Mode I, one can use the following relation to obtain critical values of stress or crack length at crack growth,

2

2 2 IIII II

KK K K1

+ + =− ν

2IC (4.55)

If another Mode dominates, the right hand side is replaced by the appropriate critical value of the SIF (if Mode III dominates eq. 4.55 is set equal to 2

IIICK /(1 v)− ).

When none of the Modes dominates the fracture process, relation (4.55) is not realistic. We may then search for a function to describe the onset of crack growth, 1 Indeed, it has been observed experimentally that if Mode II is present, it tends to change crack growth from the self-similar pattern.

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(4.56) ( I II IC IIC iK ,K ,K ,K , ,...... 0Ω β ) =

Here K have the usual meaning and (i = 1, ... n) are parameters. iβ

If this function is satisfied, crack growth is imminent. If it is less than zero, crack growth does not occur. Values of this function larger that zero are not admissible. The explicit from of (4.56) is established experimentally and for each type of material and conditions of loads.

Experiments have shown that for certain materials in structural components, subjected to Modes I and II, the following equation describes the data very well,

2

I II

IC IIC

K K 1K K

⎛ ⎞+ ⎜ ⎟

⎝ ⎠= (4.57a)

For materials other than those described by (4.57a), different forms of equations result. These different forms come from fitting the data. One may generalize (4.57a) in the following way,

m n

I II0

IC IIC

K KC 1K K

⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (4.57b)

where m , n and C0 are parameters determined experimentally.

A criterion for crack growth under mixed based on the maximum stress has also been proposed1. This criterion postulates that crack growth occurs on directions normal to the maximum principal stress or parallel to planes where the shear stress is zero (recall that on the planes of principal stresses the shear stress is zero).

Suppose that a specimen is subjected to Modes I and II. The normal stress , and the shear stress components , at a point ahead of the crack tip are (eqs 4.36b,c and 4.41b,c),

θθσ

rθσ

IK 3 1 3cos cos4 2 4 22 rθθ

θ θ⎡ ⎤σ = +⎢ ⎥π ⎣ ⎦

Ir

K 1 3 1sin sin4 2 4 22 rθ

θ⎡σ = +⎢π ⎣ ⎦θ⎤

⎥ Mode I

IIK 3 3 3sin sin4 2 4 22 rθθ

θ θ⎡ ⎤σ = − −⎢ ⎥π ⎣ ⎦

IIr

K 1 3 3cos cos4 2 4 22 rθ

θ⎡σ = +⎢π ⎣ ⎦θ⎤

⎥

Mode II

The crack growth direction results from the following condition,

1 F. Erdogan and G. C. Shih (1963), Journal of Basic Engineering, vol. 85, pp. 519-527.

March 2006 4-21

I II3K sin sin K cos 3cos 02 2 2 2θ θ θ θ⎡ ⎤ ⎡+ + + =⎢ ⎥ ⎢⎣ ⎦ ⎣

3 ⎤⎥⎦

(4.58a)

It is further assumed that crack growth is imminent when the principal stress as a result of the combined loading equals to the critical stress in an equivalent Mode I. That is when,

( ) IC0r K /θθ

θ=σ = 2π

Using the stress components , for Modes I & II, there results, θθσ

I II3K cos 3cos K 3sin 3sin 4K2 2 2 2

θ θ θ⎡ ⎤ ⎡ ⎤θ + + − − =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦IC

3 (4.58b)

If KI = KIC and KII = 0

one can easily see that,

θ = θc = 0 and KI = KIC

If KII = KIIC and KI = 0

the solution of the equation (see 4.58a)

3cos 3cos 0

2 2θ θ

+ =

leads to 0c

1arccos 70.63

−θ = θ = =

and II IIC IC3K K K4

= = (4.59)

The foregoing analysis has worked well in a variety of materials. If the material follows (4.57a) and (4.59) the stress at the onset of crack growth is (Figure 4.10).

2

IC2

161 tanK3 38 sina∞

1+ α −σ =

απ (4.60)

Equation (4.60) can be obtained following the steps,

1. The SIF for the case in Figure 4.10 are,

2IK cos∞= σ α πa (4.61a)

IIK cos sin∞= σ α α πa (4.61b)

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2. Introduce (4.61) in (4.57a) and use (4.59) to obtain a second order equation for ∞σ and solve this equation for . ∞σ

4.6.2 Biaxial loading

Consider a plate loaded along two directions as shown in Figure 4.11. Assuming that R = / the stress components and 2σ 1σ 1σ 2σ contribute to SIF as follows,

on Mode I loading :

- stress : 1σ (1) 2 2I 1 I(0)K cos a K cos= σ β π = β

- stress 2σ : (2) 2 2 2I 2 1 I(0)K cos a R sin a RK sin

2π⎛ ⎞= σ + β π = σ β π = β⎜ ⎟

⎝ ⎠

Adding these two contributions, the total SIF is,

(4.62) ( 2I I(0)K K cos R sin= β + )2 β

Figure 4.11 Case of biaxial loading.

on Mode II loading :

- stress 1

σ : (1)II 1 I(0)K cos sin a K cos sin= σ β β π = β β

- stress 2

σ : (2)II 1 1 I(0)

π πK cos sin πa Rσ cosβ sinβ πa RK cosβ sinβ2 2

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

March 2006 4-23

By adding these two contributions, the result for the total SIF is,

(4.63) (II I(0)K K cosβ sinβ 1 R= )−

4.7 Fracture toughness Testing

4.7.1 Single fracture -plane strain fracture toughness test

The goal of a laboratory test is to recreate the conditions occurring in an actual structure in a smaller test specimen. One cannot necessarily test the entire structure, however we only need to create identical conditions in both the specimen and structure in the region of interest (i.e. near the crack tip), as shown in Figure 4.12. For this task, standard test specimens have been designed (ISO or ASTM standards). The specimen design standards ensure that there are identical stress and strain conditions near the crack tip in both the structure and test specimen (Anderson, 1995; Broek, 1986).

Figure 4.12 Test specimen and real structure conditions. Detail shows singularity dominated zone.

One important standard test is the measurement of the fracture toughness of a material in Mode I, KIC. This is a measure of the critical value of KI at which the crack extends. In order to ensure that the KIC measured is truly a material property and not a function of the specimen geometry, we must enforce the following conditions on the singularity dominated zone (if the results were a function of the specimen geometry, we could not transfer them from the test specimen to the structure):

(1) The size of the plastic zone is small compared to the length of the crack.

(2) A state of plane strain exists at crack tip.

The first condition allows one to apply linear elastic fracture mechanics near the crack tip. The importance of the second condition is shown in Figure 4.13.

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Figure 4.13 Dependence of measured KI on specimen thickness.

If the thickness of the specimen is sufficiently large (B > Bmin) a state of plane strain exists at the crack tip and. the measured KI is independent of B. However if B < Bmin the KI increases with decreasing B until the state of plane stress is reached. Since the lowest value of KI measured is that at plane strain, we can use the plane strain KI as a conservative estimate of KIC for any stress/strain condition.

4.7.2 A standard specimen to measure KIC (based on the ASTM standard2)

The Compact Tension specimen is shown in Figure 4.14. The specimen is loaded by a pin load P. Once the arbitrary dimension W has been fixed, the other dimensions are given as in Figure 4.14. Thus this specimen can be arbitrarily scaled. The reasons for the dimensions specified include:

(1) The choice of a = 0.45-0.55W ensures that crack length is large compared to the size of the plastic zone.

(2) The choice of B > 0.25W sufficiently large ensures that the plane stress region at the surface is relatively small, thus one can assume plane strain at the crack tip.

2 ASTM Designation : D 5045-99

March 2006 4-25

Figure 4.14 Dimensions of standard compact tension specimen.

The stress intensity factor for a crack of length a in the compact tension specimen due to an applied load P is given by the following formula, empirically determined from curve fits of compact tension test data

1 3 5 72 2 2 2

I 1 2

P a a a a aK 2.96 185.5 655.7 1017 639BW W W W W W

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥= − + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎣ ⎦

92

(4.64)

4.8 Precracking by fatigue

The introduction of a sharp crack of length a, into a test specimen without significantly damaging the material near the crack tip is not trivial. Consequently a standard has been developed for the growth of the crack in fatigue, to ensure ideal crack tip conditions (i.e. similar to that observed in real structures). First, a Chevron notch, as shown in Figure 4.15, is cut into the specimen at the desired location of the crack. This notch shape has the following advantages:

(1) The notch forces initiation of the crack at the center of the specimen.

(2) A crack starts almost immediately upon cyclic loading.

(3) A straight, sharp crack is produced.

Figure 4.15 Chevron notch.

In order to initiate and then propagate the crack, the specimen is then loaded in fatigue according to a prescribed loading history as shown in Figure 4.16(a). The load typically cycles between a minimum value of approximately 40%PC and a maximum value of approximately 60%PC, where PC is the critical load. For each series of specimens, PC is determined from the load-displacement curve as shown in Figure 4.16(b).

March 2006 4-26

(a) (b)

Figure 4.16 (a) Load history for growth of crack in fatigue: (b) load-displacement curve to obtain PC.

4.9 Mixed fracture -Brazilian disk test

Since the loading of an actual structure is often not limited to uniaxial tension, it is important to develop a criterion for mixed- fracture. While we measured the critical fracture toughness in Mode I, KIC, in the previous section, we could also perform a similar test to measure KIIC. However, the values of KIC and KIIC for a given material are not sufficient to describe a criterion of mixed- fracture in which both KI and KII are present. Thus, we need to develop a more general mixed- fracture criteria, i.e. an equation of the form,

f (KI,KII) = 0

which describes the failure envelope (in terms of the variables KI and KII) at which a crack will propagate. Naturally, this fracture criterion should satisfy the boundary conditions,

f (KIC, 0) = 0 and f (0, KIIC) = 0

Several authors have proposed the following form for f (see 4.57b),

( )m n

I III II 0

IC IIC

K Kf K , K C 1K K

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠− (4.65)

where m, n, and C0 are material parameters determined experimentally.

A standard method to study KI-KII mixed fracture in brittle materials is the Brazilian disk test for which the specimen geometry is in shown in Figure 4.17.

March 2006 4-27

Figure 4.17 Geometry of Brazilian disk specimen.

The loading is applied as a point load P in compression. By simply varying the angle θ between the applied load and the crack axis, one can vary the ratio of loading in Modes I and II. For a given crack length l, the SIFS KI and KII are given by3,

( )( )

( )( )

22 2 2I

22II

PK 1 4sin 4sin 1 4cos aaPK 2 5 8cos a sin 2a

l l

l l

⎡ ⎤= − θ + θ − θ⎣ ⎦π

⎡ ⎤= + − + θ θ⎣ ⎦π

(4.66)

Thus, by fabricating a series of Brazilian disk specimens of the same material, and loading each at a different angle θ, one can determine empirically the constants m, n, and C0 which define the failure envelope. To obtain loading in pure Mode II (i.e. to obtain KIIC), the angle θ, necessary can be calculated from eq. (4.66), for a given length l (l «a). An example collection of data taken from soda-lime glass at various θ is shown in Figure 4.18. The data are plotted in terms of KI vs. KII to show the failure envelope.

( )( )22 2 2I

PK 1 4sin 4sin 1 4cos a 0a

l l⎡= − θ + θ − θ⎣π⎤ =⎦

(4.67)

4.9.1 Pre-Cracking

As for the previous specimen, it is important to grow the crack in a controlled manner so as to produce a straight, sharp crack. Thus, as before, the crack tips of the central notch are cut in the form of a chevron notch. In addition, for the Brazilian disk, the crack must be initialized in Mode I. Thus, the disk is placed in the loading frame such that θ = 0 and loaded until a short, initial crack (often called the pre-crack) is formed. At this point the specimen is unloaded,

3 C. Atkinson, et al. (1982), International Journal of Fracture, Vol. 18, pp. 279-291.

March 2006 4-28

rotated until the desired θ, and then the fracture toughness test is performed. Figure 4.19 shows an actual specimen, after failure, in which one can see the original notch with pre-crack, and the crack at failure, parting at an angle dependent upon θ, the angle of loading.

Figure 4.18 Results of mixed fracture tests for soda-lime glass specimen4

Figure 4.19 Fracture pattern in a Brazilian disk soda-lime glass after mixed fracture test4.

4 D. K. Shetty, et al. (1987), Engineering Fracture Mechanics, Vol. 26, pp. 825-840.

March 2006 4-29