Mechanics of Solids and Fracture

HO SUNG KIM MECHANICS OF SOLIDS AND FRACTURE

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CONTENTS

Preface 7

1 Stress and strain 9 1.1 Stress at a point 9 1.2 Relation of principal stress with other stress components 10 1.3 Stresses on oblique plane 11 1.4 3D Mohr’s circle representation 17 1.5 Strain at a point 19

2 Linear elastic stress-strain relations 26 2.1 The Hooke’s law 26 2.2 Calculation of stresses from elastic strains 29 2.3 Plane stress and plane strain 31 2.4 Strain energy 34 2.5 Generalised Hooke’s law 36 2.6 Elastic properties dependant on orientation 40

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3 Thin circular plates 44 3.1 Stress and strain 44 3.2 Bending moment 47 3.3 Slope and deflection without boundary conditions 49 3.4 A general axi-symmetric case where a circular plate is subjected to combined uniformly distributed load (p) and central concentrated load (F) 50 3.5 A case where a circular plate with edges clamped is subjected to a pressure 52 3.6 A case where a circular plate with edges clamped is subjected to a centrally concentrated load 54 3.7 A case where a circular plate with edges freely supported is subjected to a pressure 56 3.8 A case where a circular plate with edges freely supported is subjected to a central concentrated load 59 3.9 A case where a circular plate with edges freely supported is subjected to a load round a circle 62 3.10 A case where an annular ring with edges freely supported is subjected to a load round a circle 67

4 Fundamentals for theory of elasticity 72 4.1 Equilibrium and compatibility equations 73 4.2 Airy’s stress function 76 4.3 Application of equilibrium equations in photo-elastic stress analysis 80 4.4 Stress distribution in polar coordinates 85

5 S-N Fatigue 95 5.1 Various types of uniaxial loading 97 5.2 S-N curve models 100 5.3 Compatibility concept of fatigue damage and axioms 104 5.4 Fatigue damage domains 106 5.5 A damage function and validation 114 5.6 Numerical determination of exponent n 116 5.7 Examples for predicting of the remaining fatigue life 117 5.8 Effect of mean stress on fatigue 119

6 Linear elastic stress field in cracked bodies 121 6.1 Complex stress function 123 6.2 The stress around a crack tip 125 6.3 Stress intensity factor determination 127 6.4 Stress intensity factor with crazing 132

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6.5 Semi-elliptical crack 137 6.6 ‘Leak-before-burst’ criterion 137 6.7 Relation between energy release rate G and KI 138 6.8 Fracture criteria for mixed mode loading 139

7 Plastic deformation around a crack tip 147 7.1 One-dimensional plastic zone size estimation 147 7.2 Two dimensional shape of plastic zone 150 7.3 Three dimensional shape of plastic zone 155 7.4 Plastic constraint factor 159 7.5 The thickness effect 163 7.6 Thickness of adhesive layer 165 7.7 Experimental determination of KIc 168

8 Crack growth based on the energy balance 174 8.1 Energy conservation during crack growth 174 8.2 Griffith’s approach (1921) 176 8.3 Graphical representation of the energy release rate 178 8.4 Analytical Approach 182 8.5 Non-linear elastic behaviour 186 8.6 Crack growth resistance curve (R-curve) 189 8.7 R-Curve and stability 191 8.8 Geometric stability factors in elastic fracture 192 8.9 Testing machine stiffness 195 8.10 Essential work of energy 197 8.11 Impact fracture toughness 203

9 Single crack approach for fatigue 207 9.1 Temperature and frequency effects on fatigue crack growth 211 9.2 Fatigue crack life calculations 214 9.3 Overload retardation and crack closure 215 9.4 Variable amplitude loading 220 9.5 Fatigue near threshold and measurement methods 222 9.6 Interpretation of fatigue crack growth in P − u and R – A Diagrams 229 9.7 Short crack behaviour in near-threshold fatigue 232

References 239

Index 247

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PREFACE

The level of knowledge content given in this book is designed for the students who have completed elementary mechanics of solids for stresses and strains associated with various geometries including simple trusses, beams, shafts, columns, etc. At the successful completion of understanding the content provided, the students would be able to reach a stage where they can do self-directed learning at any further advanced level in the area of mechanics of solids. The emphasis is given on the fundamental concepts for students to quickly follow through for an advanced level if required in the future. Fracture mechanics is included in this book with necessary preliminary steps for those who might have had difficulties with the subject in the past.

The essence of mechanics of solids is lies in stress-strain analysis ultimately for the fail-safe design of structural components. It is important to keep in mind that such analysis would be useless without various criteria for yielding, failure, fracture, and fatigue. Materials behaviour is more complex than some students might think. Materials fail sometimes at higher or lower stress than the stress calculated. Some materials are more sensitive in failure to stress concentration than some other materials. They fail sometimes in a ductile manner and some other times in a brittle manner. The ductility of a material is not only a material property but also is affected by its geometry and loading condition. One approach would be applicable to some particular cases while the other approach is more appropriate for some other cases.

Engineering practitioners will be able to find this book useful as well for the fail-safe design, and for a way of thinking in making engineering decisions.

The 3rd edition of this book may be the first one that deals with a new concept of S-N fatigue damage. The fatigue damage has been a research topic since 1924 but without much progress. It is only recently that the author and Ms Hoda Eskandari have made a small step forward for the fatigue damage theory. Thank you Hoda! She has inspired me in various ways.

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Part of the book owes to the Lecture Notes developed for many years in the past. I would like to thank Ms Carol Walkins of the University of Newcastle, Callaghan, for typing in earlier years. I am grateful to Mr Kam Choong Lee of PSB Academy in Singapore for the feedback on Lecture Notes before I transformed that into this book, and for further proofreading of the manuscript. Also, thanks go to Ms Haleh Allameh Haery of the University of Newcastle, Callaghan, for assisting with some graphic material, invaluable feedback and proofreading. I thank my friend Mr Ron Williams, who is a man of ingenuity, for general discussions and encouragement.

Ho Sung Kim

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1 STRESS AND STRAIN

z z

zy zx yz

xz σy xy ∆z yx y x ∆x ∆y x (a) (b)

Figure 1.1 (a) A body subjected to uniform stress; and (b) one of cubes in ‘(a)’ subjected to uniform stress distribution.

1.1 STRESS AT A POINT The stress components on a cubical element may be useful for describing fundamental relations with reference to the coordinate system. The cubical element is one of building blocks constituting the elastic body. Figure 1.1b shows a body subjected to normal uniform stress distribution. The body is assumed to consist of infinite number of cubical elements. Figure 1.1b shows one of cubes, representing a point in the body, in which nine stress components are used to describe a stress state in terms of location, magnitude and direction:

C S C S D x xy xz T D xx xy xz T

D yx y yz T or D yx yy yz T . D T D T E zx zy z U E zx zy zz U

The first subscript of each stress component indicates plane and the second direction. The nine stress components can be reduced to six components because some of stress components are equal. This can be found by taking the summation of the moments (SM) about z-axis, y-axis and x-axis:

SMz=0 for z- axis,

xy ( y z) x yx ( x z) y (1.1)

and therefore xy yx.

Similarly for x and y axes, we find that yz zy and zx xz. Consequently, the state of stress at a point can be now described by six components: x , y , z , xy, xz and yz .

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1.2 RELATION OF PRINCIPAL STRESS WITH OTHER STRESS COMPONENTS

K (Principal stress)

Ө3 x xy yx y 1 2 xz yz y 0 zx zy L

z J x

Figure 1.2 Stress components on a tetrahedron.

When a body subjected to external forces, a range of different planes may considered for stress analysis. The planes where no shearing stresses but normal stresses exist are called the principal planes. The normal stress on each principal plane is referred to as the principal stress. Figure 1.2 shows the principal stress on area JKL as a result of choosing the coordinate system in a particular orientation and for a particular position.

Let cosq1 = l, cosq2 = m and cosq3 = n. The areas on the tetrahedron are found in relation with A or area JKL:

Area K0L = Al (or = A cosq1) Area K0J = Am Area J0L = An.

The principal stress (s) in Figure 1.2 may be related with the stress components by taking the summation of the forces in x, y and z directions:

SFx=0,

( x )Al yx Am zx An 0 (1.2a)

SFy=0,

xy Al ( y )Am zy An 0 (1.2b)

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SFz=0,

xz Al yz Am ( z )An 0 . (1.2c)

These three equations are compacted for relations between the principal stresses and other stress components:

I( x ) yx zx YI l Y L LL L (1.2d) J xy ( y ) zy ZJmZ 0 L LL L K xz yz ( z )[Kn [ The direction cosines l, m and n can be eliminated from the three equations to find an expression for s:

3 2 I1 I 2 I 3 0 (1.2e) where

I1 x y z 2 2 2 I 2 x y y z x z xy xz yz (1.2f) 2 2 2 I 3 x y z 2 xy yz xz x yz y xz z xy or

I1 1 2 3

I 2 1 2 2 3 1 3 (1.2g)

I 3 1 2 3 .

As seen in Equation (1.2e), I1, I2 and I3 are not functions of direction cosines. They are independent of the coordinate system location and therefore they are called the invariants.

1.3 STRESSES ON OBLIQUE PLANE Any other planes than the principal planes may be called the oblique planes in which always shear stress exists when subjected to external forces (Figure 1.3a). The total stress (S ) on the oblique plane can be resolved into three components (Sx, Sy, and Sz) (Figure 1.3b) and

2 2 2 2 S Sx S y Sz (1.3a)

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Taking the summation of the forces in the x, y and z directions yields:

S x x l yxm zx n (1.3b) S y xyl y m zy n

S z xzl yzm z n

The normal stress s( n) may be found in terms of Sx, Sy and Sz (Figure 1.3c) by projecting the total stress components (Sx, Sy, and Sz) onto the normal stress direction:

n S xl S y m Sz n (1.3c)

2 2 2 Also, S n . (1.3d)

Therefore, the shear stress is found as a function of principle (normal) stresses ( 1 , 2 , and 3 ):

2 2 2 2 2 2 2 2 2 2 ( 1 2 ) l m ( 1 3 ) l n ( 2 3 ) m n (1.4)

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z z n Sz S (Total stress) Principle plane S (Total stress)

Sy Sx x yx τ τxy Oblique plane y y y τxz τ τzx τzy τyz

z x (a) x (b) z

Sz σn θ3

θ1 θ2 Sy Sx y

Figure 1.3 Stress components on oblique plane: (a) total stress, S, consisting of normal stress (sn) and

shear stress (τ); (b) total stress components (Sx, Sy, and Sz) in x, y, and z directions; and (c) normal stress components of the total stress can be obtained by projecting total stress components onto the normal stress direction.

The principal (maximum) shear stresses (t1, t2, and t3) occur at an angle of 45˚ with the three principal axes as shown in Figure 1.4 and found to be 2 3 1 2

1 3 2 (1.5) 2 1 2 . 3 2

2 3

1 1

3 1 1

max=2=( 1 - 3)/2 3=( 1 - 2)/2 2 2

1 > 2> 3 3

1=( 2 - 3)/2 2

Figure 1.4 The maximum shear planes at an angle of 45˚ with the three principal axes.

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The maximum shear stress criterion (or Tresca yield criterion) assumes that yielding occurs when the maximum shear stress [ max ( 1 3 )/2] reaches its yielding point. In the case of uni-axial loading, the maximum principle stress ( 1 ) reaches its yielding point ( ys ) so that 1 ys , 2 3 0, and the maximum shear stress becomes:

0 1 3 1 ys . (1.6) max 2 2 2

Figure 1.5 Rubber modified epoxy showing cavities on fracture surface. The arrow indicates fracture propagation direction and the bar represents 10 μm. [After Kim and Ma 1996]

In general, the deformation of an element consists of volume and shape changes. The volume change is a result of proportional change in element edge lengths. In contrast, the shape change is a result of disproportional change in element edge lengths as well as element corner angle change. The former is associated with volumetric modulus (K ) and hydrostatic stress (or mean stress) while the latter is associated with shear modulus (G ) and shear stress. For example, the hydrostatic stress creates cavities during deformation as shown in Figure 1.5 or increases the brittleness while shear stress contributes to the material flow. The total stress for deformation consists of hydrostatic stress and stress deviator i.e.

Total stress = Hydrostatic (or mean) stress (sm) + Stress deviator.

The hydrostatic stress or mean stress is defined as

I 1 x y z 1 2 3 (1.7) m 3 3 3

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and the stress deviator ( ij) can be found by subtracting the hydrostatic stress from the total stress:

F x xy xz V F m 0 0 V G W G W yx y yz 0 m 0 G W G W G W G W H zx zy z X H 0 0 m X

F2 x y z V G xy xz W G 3 W 2 G y x z W (1.8) G yx 3 yz W G 2 z x y W G zx zy W H 3 X

This relation is graphically shown inFigure 1.6.

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( x + y + z)/3 z

zy zx yz

xz y ( x + y + z)/3 xy yx - x ( x + y + z)/3

( x - y -2 z)/3

zy zx yz xz = ( x -2 y - z)/3 xy yx

(2 x - y - z)/3

Figure 1.6 Superposition of stress components.

It can easily be shown that the stress deviator involves the principal shear stresses. For example,

2 x y z . (1.9) x 3

If we choose principal stresses in the equation, the stress deviator becomes a function of principal shear stresses,

2s -s -s 2 (s -s ) + (s -s ) 2 s ¢ = 1 2 3 = 1 2 1 3 = (t +t ) (1.10) 1 3 3 2 3 3 2 Similarly,

2 2 ( ) ( ) 2 2 1 3 2 1 2 3 ( ) (1.11) 2 3 3 2 3 3 1

2 2 ( ) ( ) 2 (1.12) 3 1 2 3 1 3 2 ( ) 3 3 3 2 3 2 1

where τ1 τ2 and τ3 are the principal shear stresses.

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2=0 1= 2=0 3=25MPa

1 1=50 MPa

3=0 2= 3=0

(a) 2=50 MPa

2=0 1 3=25MPa 3 1 1=50 MPa 2

3= - 50 MPa

(b)

2=50 MPa 1=2=3=0

1=50 MPa 1= 2= 3=50MPa

3=50 MPa

(c)

Figure 1.7 Various states of stress on elemental cube and Mohr’s circles; (a) uni-axial tension; (b) tension and compression without hydrostatic stress; and (c) hydrostatic stress without shear stress.

1.4 3D MOHR’S CIRCLE REPRESENTATION

The three principal stresses (s1, s2, s3) with the maximum shear stresses (t1, t2, t3) can graphically be represented as shown in Figure 1.7. The radius of each circle represents the maximum (or principle) shear stress. Accordingly, Equation (1.5) can be found from the Mohr’s circle.

Figure 1.7a shows a case of uni-axial tensile loading (s1) with s2=s3=0. If s3 varies from zero to -50MPa (compressive stress), the principal shear stresses (t1 and t2) increase and becomes a state of stress where hydrostatic stress is zero as given in Figure 1.7b, resulting in more chances for material flow or high ductility than that of the case inFigure 1.7a because of the increase in shear stress. Figure 1.7c is the limiting case where the three circles reduces to a point where s1=s2=s3=50MPa and the three principle shear stresses are zero.

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In this case, no material ductility is possible in the absence of stress deviator. Therefore, it is theoretically possible to have a stress state where the hydrostatic stress component exists without the stress deviator. Examples for the state of stress where relatively large principal shear stresses are involved are found in various processes in metal forming (e.g. wire drawing through a die) involving lateral compressive stresses and a longitudinal tensile stress. Also, some localized deformation of reinforcing particles on fracture surfaces of advanced materials is caused by such a state of stress involving large shear stresses as shown in Figure 1.8.

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Figure 1.8 Fracture surface of hollow microsphere reinforced epoxy under plane strain in the vicinity of initial crack tip. Each hollow-microsphere experienced a tensile stress in the direction perpendicular to the fracture surface and simultaneously lateral compressive stresses. The crack propagation direction is from top to bottom. The scale bar represents 100 μm. [After Kim 2007]

1.5 STRAIN AT A POINT As previously discussed, the deformation is due to the volume and shape change. We need to define the displacement components. When a point P moves to P’ in coordinates x, y, z as shown in Figure 1.9, respective u, v and w are called displacement components. To find a general form of strain, let us consider first the length change using an element subjected to a load in the x-direction as shown in Figure 1.10. When the load is applied, the solid line becomes the dashed line. Accordingly, A moves to A’ and B moves to B’ and the x direction normal strain of the infinitesimal segment is given by

u dx dx dx AB AB u (1.13a) e e x . x xx AB dx x

Similarly, the displacement derivatives for y and z directions can be found:

v w e , e . (1.13b) yy y zz z

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P P'

y x y+v z x z+w

x+u z

Figure 1.9 Displacement of a point P.

A A' B B' x dx

u+(du/dx)dx

u dx+(du/dx)dx

Figure 1.10 Deformation in the x-direction.

y du C' D' D C

exy dy

eyx dv x B dx

Figure 1.11 Angular distortion of an element.

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For further displacement derivatives, let us consider a shape change using an element in the xy plane, which is subjected to shear deformation as shown in Figure 1.11. The element has undergone an angular distortion and thus the angular displacement derivatives along the x and y axes are given by

BB v e (1.13c) yx AB x and DD u e xy DA y (1.13d) respectively. Similarly, the rest of components can be found:

Fu u u V G W x y z Fexx exy exz V G W G W v v v e e e e G W (1.13e) ij G yx yy yz W G x y z W G W Hezx ezy ezz X Gw w wW G W H x y z X

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In general, components such as eexy ,,yx etc., other than those for length change produce both shear strain and rigid-body rotation. For example, those given in Figure 1.12a represent 1 C u v S a pure rotation with an average of D T and Figure 1.12b a pure rotation with an 2 E y x U 1 C v u S average of D T . Thus, the rigid body rotation components ( ij ) can be identified as: 2 E x y U

F 1 C u v S 1 C u w SV G 0 D T D TW G 2 E y x U 2 E z x UW Fxx xy xz V G W G C v u S C v w SW 1 D T 1 D T . (1.14) ij G yx yy yz W G D T 0 D TW 2 E x y U 2 E z y U G W G W H zx zy zz X G1 C w u S 1 C w v S W G D T D T 0 W H2 E x z U 2 E y z U X

Accordingly, the strain components ( ij ) can be found by subtracting rigid body rotation components ( ij ) from the displacement derivatives:

F u 1 C u v S 1 C u w SV G D T D TW G x 2 E y x U 2 E z x UW F xx xy xz V G W G C u v S v C v w SW 1 D T 1 D T . (1.15) ij G yx yy yz W G D T D TW 2 E y x U y 2 E z y U G W G W H zx zy zz X G1 C u w S 1 C v w S w W G D T D T W H2 E z x U 2 E z y U z X

C u S C u u S 1D ui j T 1D i j T For short, , ij and are called the strain tensor and the rotation ij 2D T 2D T E x j xi U E x j xi U tensor respectively. Also,

eij ij ij (1.16)

exy= - eyx y -exy= eyx

x x

Figure 1.12 Pure rotation without shear.

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Referring to Figure 1.13, engineering shear strain components ( ij) are defined as

xy exy e yx xy yx 2 xy , xz 2 xz , yz 2 yz (1.17)

In summary, strain components are

u v w , , , x x y y z z u v u w v w , , (1.18) xy y x xz z x yz z y

y du y

exy = exy + eyx dy

eyx dv x x B dx

Figure 1.13 Engineering shear strain g.

The volume strain (D) (see Figure 1.14) is defined as

Final volume - Original volume Original volume or

1 1 1 dxdydz dxdydz x y z dxdydz (1.19) 1 x 1 y 1 z 1

x y z for small deformation.

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dy dx

Figure 1.14 An elemental cube.

Themean strain or the hydrostatic component of strain, which contributes to volume change, is also defined as

x y z kk . (1.20) m 3 3 3

E @

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' Then, the strain deviator ( ij) which contributes to shape change, can be obtained by subtracting m from each of the normal strain components:

F x m xy xz V G W ' ij G yx y m yz W G W H zx zy z m X

F2 x y z V (1.21) G xy xz W G 3 W 2 G y z x yz W G yx 3 W G 2 z x y W G zx W H zy 3 X

In complete analogy between stress and strain equations, the principal strains are the roots of the cubic equation:

3 2 I1 II23 0 (1.22) where

I1 x y z 1 I ( 2 2 2 ) (1.23) 2 x y y z z x 4 xy zx yz 1 1 I ( 2 2 2 ). 3 x y z 4 xy zx yz 4 x xy y zx z yz

As already discussed for stress, I1, I2 and I3 are not functions of direction cosines. As such, they are independent of the coordinate system location and therefore they are called ‘invariants’.

Also, the principal (engineering) shear strains are

1 2 3,

2 1 3 ( max ) (1.24)

3 1 2.

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2 LINEAR ELASTIC STRESS- STRAIN RELATIONS

2.1 THE HOOKE’S LAW The elastic stress ( ) is linearly related to elastic strain ( ) by means of the modulus of elasticity (E ) for the isotropic materials:

E . (2.1)

This relation is known as the Hooke’s law.

σx σx x

Figure 2.1 Deformation of an element subjected to a tensile force.

A tensile force in the x direction causes an extension of the element in the same direction. Simultaneously it also causes a contraction in the y and z directions (see Figure 2.1). The ratio of the transverse strain to the strain in the longitudinal direction is known to be constant and called the Poisson’s ratio, denoted by the symbol v.

v x y z v x (2.2) E

The principle of superposition is then can be applied to determine the strain produced by more than one stress component. For example, the stress x produces a normal strain x and two transverse strains y v x and z v x. Similarly, other strain components can be found as listed in Table 2.1.

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Strain in the Strain in the Strain in the Stress x direction y direction z direction

v v x x x x x y z E vE v x x x x vE x v y y z x vEy x E x x Ex z v y y E y y E x v y EE z v vE y Ey y x z y v y x vEy v y z z y E E E x v y y z z z E z vEEz z E x z z vE y E z vEz z z z x y E E E z E

Table 2.1 Strain components for superposition.

x, y, Accordingly, the components of strain in the and z directions are found: 1 v x E x y z 1 v (2.3) y E y z x 1 v z E z x y

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The shear stresses acting on the unit cube produce shear strains independent of normal stresses:

xy G xy , yz G yz , xz G xz (2.4)

The proportional constant G is the modulus of elasticity in shear, or the modulus of rigidity. Values of G are usually determined from a torsion test.

Another elastic constant is the bulk modulus or the volumetric modulus of elasticity (K). The bulk modulus is the ratio of the hydrostatic stress or the hydrostatic pressure to the volume strain that it produces m p 1 K (2.5) where p is the hydrostatic pressure and b is the compressibility. It is applicable to both fluid and solid.

Some useful relationships between the elastic constants (E, G, v, and K ) may be derived. Adding up the three equations in Equation (2.3),

1 2v (2.6) x y z E x y z

It is noted that the terms on the left of Equation (2.6) is the volume strain (D), and the terms ( x y z ) on the right is 3 m . Accordingly,

1 2v 3 (2.7a) E m or

m E K (2.7b) 3 1 2v

The following equation is often introduced in an elementary course of mechanics of solids for a relationship between E, G, and n:

E G . (2.8) 2 1 v

Using Equations (2.7) and (2.8), other useful relationships can be found:

9K E , (2.9a) 1 3 K G

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1 2G 3K , (2.9b) 2 2G 3K

3 1 2v K G , and (2.9c) 2 1 v

E K . (2.9d) 9 3E G

2.2 CALCULATION OF STRESSES FROM ELASTIC STRAINS The strains are measurable while the stresses can be calculated. It may be useful to have stresses as functions of strains. From Equation (2.6),

E . (2.10) x y z 1 2v x y z

We eliminate ey and ez in Equation (2.10) using Equation (2.3):

1 v v ( ) . (2.11) x E x E x y z

Substitution of Equation (2.10) into Equation (2.11) gives:

E vE (2.12a) x 1 v x 1 v 1 2 x y z where vE (2.12b) 1 v 1 2v

and is known as the Lamé’s constant. Further, using the volume strain x y z , and shear modulus (G):

x 2G x . (2.13)

In this way, more relations can be found for other stress components ( y , x , and so on). It may be timely to introduce the tensorial notation to deal with a large number of equations and a specified system of components. All the stress components can now be expressed as

ij 2G ij kk ij (2.14)

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where i and j are free indexes, k is a dummy index, and dij is the Kronecker delta i.e.

dij = 1 if i = j j dij = 0 if i j.

The free index assumes a specified integer that determines all dummy index values. The dummy index takes on all the values of its range. Upon expansion, Equation (2.14) gives three equations for normal stress and six equations for shear stress using indexes for a range of x, y and z. Equation (2.14) may be expanded in a matrix form:

F x V F 2G 0 0 0 VF x V G W G W G W G y W G 2G 0 0 0 WG y W G W G 2G 0 0 0 WG z W G z W G WG W (2.15) G xy W G 0 0 0 G 0 0 WG xy W G W G 0 0 0 0 G 0 WG W G yz W G WG yz W HG zx XW HG 0 0 0 0 0 GXWHG zx XW

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F x V F 1 0 0 0 VF x V G W G WG W G y W G 1 0 0 0 WG y W G W z 1 G 1 0 0 0 WG W G W G WG z W (2.16) G xy W E G 0 0 0 2(1 ) 0 0 WG xy W G W G 0 0 0 0 2(1 ) 0 WG W G yz W G WG yz W HG zx XW HG 0 0 0 0 0 2(1 )XWHG zx XW

As previously discussed, the stresses and the strains can be broken into deviator and hydrostatic components. The distortion is associated with the stress/strain deviator and its stress-strain relation is given by

' E ' ' ij 2G ij . (2.17) ij 1 v

Also, the stress-strain relationship between hydrostatic stress and mean strain components in tensorial notation is given by

E K . (2.18) ii 3 1 2v kk kk

2.3 PLANE STRESS AND PLANE STRAIN Plane stress or plane strain is a state of stress/strain (Figure 2.2). An example is given for plane stress in Figure 2.2a, in which two of the faces of the cubic element are free of any stress. Another example is given for plane strain in Figure 2.2b, which occurs to the situations where the deformations take place within parallel planes. In practice, the plane strain often occurs internally within a structural component in which stress distribution is non-uniform when stress raisers such as rivet hole and notch are present whereas the plane stress occurs on its surfaces.

For a case of plane stress (s3 = 0 or sz = 0), Equation (2.3) becomes 1 v 1 E 1 2 (2.19a) 1 v (2.19b) 2 E 2 1 . (2.19c) 3 E 1 2

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Then, two stress-strain relations can be obtained by solving simultaneously two of the equations: E E v 1 2 1 2 (2.19d) 1 1 v 2 1 v 2 1 v

E 2 E 2 2 v 1 . (2.19e) 21 v 2 v1 . 1 v2

y (or 2) y (or 2)

Fixed support x (or 1) x (or 1) Fixed support z (or 3) z (or 3) (a) (b)

Figure 2.2 Example for: (a) plane stress; and (b) plane strain (no displacement in the z-direction).

For a case of plane strain 3 0 ,

1 v 0 (2.20a) 3 E 3 1 2 so that,

3 v 1 2 (2.20b)

Therefore, a stress exists even though the strain is zero in the z (or 1) direction. Substituting this value into Equation (2.3), we get

1 v 2 C v S D T (2.20c) 1 E E 1 1 v 2 U

1 v2 C v S 2 D 2 1 T (2.20d) E E 1 v U

3 0 (2.20e)

v 1 v 2 1 Note that when and are replaced with v and respectively plane stress 1 v E E equations are obtained.

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σx σx x

dx du

du z

(a) (b)

Figure 2.3 (a) An elemental cube subjected to a tensile stress. (b) Force-displacement (P-du) curve and strain energy.

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2.4 STRAIN ENERGY In general, the strain energy is graphically an area under a force-displacement (P-du) diagram (Figure 2.3). When an elemental cube is subjected to a tensile stress in the x-direction, its elastic strain energy (L) is given by 1 1 d Pdu A dx 2 2 x x (2.21) 1 Adx . 2 x x Equation (2.21) describes the elastic energy absorbed by the element volume (A dx). If we define the strain energy density (L0) as the energy per unit volume, it is given by

1 1 2 1 x 2 E (2.22) 0 2 x x 2 E 2 x

Similarly, the strain energy per unit volume of an element subjected to pure shear ( xy ) is given by

2 1 1 xy 1 2 G . (2.23) 0 2 xy xy 2 G 2 xy For a general three-dimensional stress distribution, it may be obtained by superimposing the six components:

1 0 x x y y z z xy xy xz xz yz yz (2.24) 2 or in tensorial notation,

1 (2.25) 0 2 ij ij

To identify volume- and shape- dependent quantitative characteristics, we first find an expression for strain energy per unit volume (L0) as a function of the stress and the elastic constants. Substituting the equations of Hooke’s law [Equations (2.3) and (2.4)] into Equation (2.24),

1 v 1 2 2 2 2 2 2 (2.26a) 0 2E x y z E x y y z x z 2G xy xz yz

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9K 1 2G 3K where E and .. The strain energy density L( ) may be rewritten for 1 3K G 2 2G 3K 0 separate volume and shape dependent parts:

2 I1 1 2 0 (I1 3I 2 ) (2.26b) 18K 6G where I1 1 2 3 (first invariant) and I 2 1 2 2 3 1 3 (second invariant).

The strain energy density ( 0) can be found for incompressible materials i.e. K = ∞:

1 ( )2 ( )2 ( )2 . (2.27) 0 12G 1 2 2 3 3 1 1 A uni-axial yield stress ( ) can be related for the distortion energy ( ) to be 2 ys 0 0 6G ys for 1 ys 2 0 , and 3 0 when subjected to a uni-axial loading. Accordingly, Equation (2.27) becomes

2 2 2 2 2 ys ( 1 2 ) ( 2 3 ) ( 3 1) (2.28)

This equation is known as the distortion energy criterion or von Mises’ yield criterion.

To find stress-strain relations involving the strain energy density, the following equation is first found by substituting Equation (2.15) into Equation (2.24):

1 1 2 G 2 2 2 G 2 2 2 (2.29) 0 2 x y z 2 xy xz yz and then we find that the derivative ofL O with respect to any strain component gives the corresponding stress component and vice versa. For example, 0 2G x x (2.30a) x

0 2G y y (2.30b) y

0 2G z z (2.30c) z

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0 (2.30e) y y

0 z . (2.30f) z This mathematical concept is applicable to a large structural component for force-deflection relation as described in the Castigliano’s theorem.

2.5 GENERALISED HOOKE’S LAW The generalized Hooke’s law is not only for three-dimensional loading but also for all the possible linear elastic material properties. It may be expressed as

ij Cijkl kl (2.31)

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ij Sijkl kl (2.32)

where Cijkl is the compliance tensor and Sijkl is the stiffness tensor (physically elastic constants). Equation (2.32) represents:

11 = S111111 + S111212 + S111313 + S112121 + S112222 + S112323 + S113131 + S113232 + S113333 12 = S121111 + S121212 + S121313 + S122121 + S122222 + S122323 + S123131 + S123232 + S123333 13 = S131111 + S131212 + S131313 + S132121 + S132222 + S132323 + S133131 + S133232 + S133333 21 = S211111 + S211212 + S211313 + S212121 + S212222 + S212323 + S213131 + S213232 + S213333 = + + + + + + + 22 S221111 S221212 S221313 S222121 S222222 S222323 S223131 (2.33) S223232 + S223333 23 = S231111 + S231212 + S231313 + S232121 + S232222 + S232323 + S233131 + S233232 + S233333 31 = S3111 11 + S3112 12 + S3113 13 + S3121 21 + S3122 22 + S3123 23 + S3131 31 + S3132 32 + S3133 33 32 = S3211 11 + S3212 12 + S3213 13 + S3221 21 + S3222 22 + S3223 23 + S3231 31 + S3232 32 + S3233 33 33 = S3311 11 + S3312 12 + S3313 13 + S3321 21 + S3322 22 + S3323 23 + S3331 31 + S3332 32 + S3333 33

We know that sij and eij are symmetric tensors (eij = eji, Cijlkeij = Cijkleji, sij = sji, sij = sji). This leads to simplification of Equation (2.33):

11 = S111111 + S112222 + S113333 + S1123(223) + S1113(213) + S1112(212) 12 = S121111 + S122222 + S123333 + S1223 (223) + S1213(213) + S1212(212) 13 = S131111 + S132222 + S133333 + S1323 (223) + S1313(213) + S1312(212) 21 = S211111 + S212222 + S213333 + S2123 (223) + S2113(213) + S2112(212) 22 = S221111 + S222222 + S223333 + S2223 (223) + S2213(213) + S2212(212) (2.34) 23 = S231111 + S232222 + S233333 + S2323 (223) + S2313(213) + S2312(212) 31 = S3111 11 + S3122 22 + S3133 33 + S3123 (2 23) + S3113(2 13) + S3112(2 12) 32 = S3211 11 + S3222 22 + S3233 33 + S3223 (2 23) + S3213(2 13) + S3212(2 12) 33 = S3311 11 + S3322 22 + S3333 33 + S3323 (2 23) + S3313(2 13) + S3312(2 12) or knowing engineering shear strain g (=2e),

11 = S111111 + S112222 + S113333 + S1123 23 + S1113 13 + S1112 12 ………………………………………………………………………………… ………………………………………………………………………………… (2.35) 23 = S231111 + S232222 + S2333 33 + S2323 23 + S2313 13 + S2312 12 ………………………………………………………………………………… …………………………………………………………………………………

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Similarly, Equation (2.31) is expanded:

11 = C111111 + C112222 + C113333 + 2C1123 23 + 2C1113 13 + 2C1112 12 ………………………………………………………………………………… ………………………………………………………………………………… (2.36) 23 = 2 23 = 2C231111 + 2C232222 + 2C2333 33 + 4C2323 23 + 4C2313 13 + 4C2312 12 ………………………………………………………………………………… …………………………………………………………………………………

If contracted notation is used to follow the usual convention, only two subscripts instead of four for compliance and stiffness tensors are sufficient and Equations (2.35) and (2.36) are expressed as

11 = S1111 + S1222 + S1333 + S14 23 + S15 13 + S16 12 22= S2111 + S2222 + S2333 + S24 23 + S25 13 + S26 12 33 = S3111 + S3222 + S33 33 + S34 23 + S35 13 + S36 12 23 = S4111 + S4222 + S43 33 + S44 23 + S45 13 + S46 12 = + + + + + 13 S51 11 S52 22 S53 33 S54 23 S55 13 S56 12 (2.37) 12 = S6111 + S6222 + S6333 + S64 23 + S65 13 + S66 12 ( 21 = S6111 + S62 22 + S63 33 + S64 23 + S65 13 + S66 12) (31 = S31 11 + S32 22 + S33 33 + S34 23 + S35 13 + S36 12) = + + + + + ( 32 S31 11 S32 22 S33 33 S34 23 S35 13 S36 12) and

11 = C1111 + C1222 + C1333 + C14 23 + C15 13 + C16 12 22= C2111 + C2222 + C2333 + C24 23 + C25 13 + C26 12 33= C31 11 + C32 22 + C33 33 + C34 23 + C35 13 + C36 12 23 = C41 11 + C42 22 + C43 33 + C44 23 + C45 13 + C46 12 (2.38) 13 = C51 11 + C52 22 + C53 33 + C54 23 + C55 13 + C56 12 12= C61 11 + C62 22 + C63 33 + C64 23 + C65 13 + C66 12.

It can be noted that the subscripts of coefficients have been rearranged systematically: 11→1,

22→2, 33→3, 23→4, 13→5, 12→6, 21→6, etc and S2322= S42, S1122= S12, C1122= C12, 2C2311=

C41, 4C2323= C44, etc.

The elastic stiffness and compliance constants are defined as

11 23 S11= , S44= , etc 11 23 and

11 23 C11= , C44= , etc. 11 23

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In general, Sij= Sji and Cij= Cji for linear elastic materials. This can be easily shown as follows.

For

0 = 11 = S11 11 + S12 22 + S13 33 + S14 23 + S15 13 + S16 12, 11 the second derivative is given by

2 0 = S12. (2.39a) 11 22

For

0 = 22 = S21 11 + S22 22 + S23 33 + S24 23 + S25 13 + S26 12, 22 another second derivative is given by

2 0 = S21. (2.39b) 22 11

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Therefore,

2 2 0 0 (2.39c) = = S12 = S21. 11 22 22 11

Now, we started with 36 elastic constants as given in Equation (2.41), but as a result of analysis, these can be reduced to 21 independent elastic constants.

2.6 ELASTIC PROPERTIES DEPENDANT ON ORIENTATION The elastic properties such as elastic modulus and Poisson’s ratio may be characterised by a set of planes of symmetry in a particular orientation. Each plane of symmetry is defined as a plane to which elastic properties are symmetric. A material having an infinite number of sets of such planes in any orientation is called an isotropic material and otherwise is called an anisotropic material.

3 or z

1 or x

2 or y Figure 2.4 Orthotropic material: it has three mutually perpendicular orientations for respective three sets of planes of symmetry.

One of the important classes of engineering materials is one that has three mutually perpendicular orientations for respective three sets of planes of symmetry. Materials in such a class are called the orthotropic materials (see Figure 2.4). Examples for orthotropic materials include unidirectional fibre reinforced laminates and highly textured cold rolled metal sheets. For orthotropic materials, constants Sij in Equation (2.41) and Cij in Equation (2.41) reduces to

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FS11 S12 S13 0 0 0 V G W S S S 0 0 0 G 12 22 23 W GS S S 0 0 0 W 13 23 33 (2.40a) Sij G W G 0 0 0 S44 0 0 W G 0 0 0 0 S 0 W G 55 W HG 0 0 0 0 0 S66XW and

FC11 C12 C13 0 0 0 V G W C C C 0 0 0 G 12 22 23 W G W C13 C 23 C33 0 0 0 (2.40b) Cij G W G 0 0 0 C 44 0 0 W G 0 0 0 0 C 0 W G 55 W HG 0 0 0 0 0 C66 XW respectively. Thus, the stress-strain relations for an orthotropic material are given by

11 = C1111 + C1222 + C1333 22= C1211 + C2222 + C2333 33= C1311 + C2322 + C3333 (2.41a) 23 = C44 23 13 = C55 13 12= C66 12 or

x = C11x + C12y + C13z y= C12x + C22y + C23z z= C13x + C23y + C33z (2.41b) yz = C44 yz xz = C55 xz xy = C66 xy

The constants in Equation (2.41) can be related to elastic moduli and Poisson’s ratios or directly determined by conducting the tests. For example, stress components for a uni-axial tensile test in the x-direction are given by: sx 0, sy = 0, and sz = 0. From Equation (2.41),

1 x = C11x = x, y= C12x, z= C13x. (2. 42) E x 1 Accordingly, C11 is now determined to be and further E x y y yx zx C12 = , and C13 = (2. 43) x x E x E x E x

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y z where and zx . Similar relationships can be obtained by applying stresses in yx x x different directions. Therefore,

1 xy xz C11= C12 = ,, C13 = E x Ey Ez

xy 1 yz C12= , C22= ,, C23= Ey E y Ez yz zy 1 C31= , C32= , C33= Ez Ey E x 1 C44 = (2. 44) Gyz 1 C55 = Gxz 1 C66 = . Gxy

Another class of materials are transversely isotropic. When two of three sets of symmetry planes for the orthotropic properties become an infinite number of sets, the properties are called transversely isotropic. Figure 2.5 illustrates an example for a transversely isotropic material using a unidirectional fibre reinforcedcomposite. Therefore,E x=Ez, Gyz=Gxy, and nyz =nxy.

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3 or z

1 or x

2 or y Figure 2.5 Transversely isotropic material.

For isotropic materials, Ex = Ey = Ez = E, Gyz = Gxz = Gxy= G and nxy= nyx = nzy =nzx =n . Accordingly, the following equations can be recovered for isotropic materials:

1 v x E x y z 1 v (bis 2.3) y E y z x 1 v z E z x y

xy G xy, yz G yz , xz G xz (bis 2.4).

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3 THIN CIRCULAR PLATES

The plates are meant to be subjected to the bending loads. Some examples for the use of plates include pressure vessel end caps and piton heads. An analysis can be conducted for the axi-symmetric loading with the benefit of the circular geometry. The analysis is based on the linear stress distribution across the thickness. For a circular plate (Figure 3.1), x in the coordinate system may be exchangeably used with r to indicate the radial direction.

y Radial x or r

x z Tangential

(a) (b) Figure 3.1 (a) A circular plate. (b) An infinitesimal element for directions in the axi-symmetric analysis.

3.1 STRESS AND STRAIN The general relations between stresses and strains previously discussed for isotropic materials are applicable:

1 v x E x y z 1 v (bis 2.3) y E y z x 1 v z E z x y

The state of plane stress is also applicable to the circular plate. The stresses in the radial (x) and tangential directions (z) in the current coordinate system (Figure 3.1) are given by:

E v x 1 v2 x z (bis 2.19) E v z 1 v2 z x

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dθ θ R

dx dθ dy θ u x

Figure 3.2 Cross section of circular plate.

The following procedure is given for finding strainsε ( x and εz) in above equations as functions of slope ( ).

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In general, the following relation is applicable to pure bending of an infinitesimal element having a liner strain distribution:

M 1 (3.1) EI R where M is the bending moment, R is the radius of curvature, E is the elastic modulus and I is the second moment of area.

The strain in thex -direction due to pure bending in x-y plane (Figure 3.2) is u x (3.2) Rxy where u is the distance of any point from the neutral axis.

In general, the curvature (1/R) for small deflection (Figure 3.2) is given by

1 d 2 y , (3.3) R dx 2 and for a small angle,

dy tan . (3.4) dx Therefore, the curvature in thex -y plane is given by

1 d 2 y d 2 (3.5) Rxy dx dx

dθ θ Assumed circle

M M u

a x uθ

Figure 3.3 An exaggerated cross section of circular plate forming assumed spherical deflection for small deformation when a couple (M) consisting of two equal and opposite forces is applied.

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and bending strain in radial direction in the circular plate (εx) is given by

d u . (3.6) x dx For a plate subjected to a couple (M ) consisting two equal and opposite forces is applied

(Figure 3.3), the circumferential strain (εz) at a (= εz), to which the distance from the neutral axis is u, due to the pure bending:

x u x u . (3.7) z x x Thus, using the plane stress equation, the stresses in the radial (x) and tangential directions (z) are found as functions of slope (q ):

E Eu C d S v D v T (3.8) x 1 v 2 x z 1 v 2 E dx x U and

Eu Eu C d S z z v x D v T . (3.9) 1 v 2 1 v 2 E x dx U Equations (3.8) and (3.9) will be useful for finding stresses when the slope θ( ) and its derivative are known.

σr or σz u h

Unit length

Figure 3.4 A section of the circular plate.

3.2 BENDING MOMENT Let us consider the small section of the circular plate with a unit length (Figure 3.4). From the simple bending theory,

M I h3 M (3.10) I u u 12u

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where M is the bending moment per unit length, the bending moment (Mr ) due to the stress in the radial direction (σx),

h3 h3 E C d S C d S M x D v T DD v T (3.11) r 12u 12 1 v 2 E dx x U E dx x U

h3 E where D . 12 1 v2

Similarly, the bending moment (Mz) due to the stress in the tangential direction (σz),

C d S M DD v T (3.12) z E x dx U

h3 E where D . 12 1 v2

It is useful to know that, unlike beams, the bending moments (Mr and Mz) here will be eliminated rather than calculated.

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3.3 SLOPE AND DEFLECTION WITHOUT BOUNDARY CONDITIONS Consider an infinitesimal element in Figure 3.5 to relate deformation with the shear force (Q ) and then external forces such as concentrated force and pressure.

Mz Q+dQ

Mr +dMr Mr

dø/2 dø Q Unit thickness Mz

x dx

Figure 3.5 An infinitesimal element of circular plate.

The moments in the radial and tangential directions per unit length are Mr and Mz respectively and Q is the shear force per unit thickness.

Taking the moments about the outside edge under the equilibrium:

1 (M dM )(x dx)d M xd 2M dxsin( d) Qxddx 0 . (3.13) r r r z 2 Neglecting small quantities, this reduces to

Mrdx dM r x M zdx Qxdx 0 . (3.14) and rearranging,

dM M x r M Qx. (3.15) r dx z

To eliminate moments, substituting

C d S C d S M r DD v T and M DD v T (bis 3.11 & 3.12) E dx x U z E x dx U into Equation (3.15) yields,

d 2 1 d Q (3.16) dx 2 x dx x 2 D

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d C 1 d(x ) S Q D T . (3.17) dx E x dx U D For a circular plate, it is convenient to replace x with r so that

d C 1 d dy S Q D (r )T . (3.18) dr E r dr dr U D

dy Note that and y in equation above are the slope (θ) and the deflection respectively. The dr following section will show how to relate Equation (3.18) with external forces.

3.4 A GENERAL AXI-SYMMETRIC CASE WHERE A CIRCULAR PLATE IS SUBJECTED TO COMBINED UNIFORMLY DISTRIBUTED LOAD (P) AND CENTRAL CONCENTRATED LOAD (F) p

dr Mr r

Figure 3.6 A circular plate subjected to pressure (p) with reactions (Mr and F0 ).

A general axi-symmetric case where the pressure (p) is applied uniformly on the circular plate is given in Figure 3.6 without boundary conditions yet. Theshear force per unit length (Q ) may be found from the equilibrium at any radius (r):

pr Q 2r p r 2 Q (3.19) 2 so that Equation (3.18) is related with an external load, pressure (p),

d C 1 d dy S Q pr D (r )T . (3.20) dr E r dr dr U D 2D

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F p

dr Mr r

Figure 3.7 A circular plate subjected to pressure p and point force F with

reactions (Mr and F0 ).

Similarly, for the case (Figure 3.7) where both pressure (p) and central concentrated load (F ) on the circular plate are applied, the shear force per unit length (Q ):

pr F Q 2 r p r 2 F Q (3.21) 2 2 r so that

d C 1 d dy S F pr F V 1 D (r )T . (3.22) dr E r dr dr U GH 2 2rXW D

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dy To find the slope (( ) and the deflection (y), this equation may be integrated, dr

d dy 1 F pr F V 1 F pr3 Fr V (r ) O G Wdr G ln rW C1r (3.23) dr dr D H 2 2rX D H 4 2 X

where C1 is an integration constant. Integrating again,

dy pr3 Fr C r C 2ln r 1 1 2 . (3.24) dr 16D 8D 2 r Then,

pr 4 Fr 2 C r 2 y ln r 1 1 C ln r C . (3.25) 64D 8D 4 2 3 The integration constants can be determined according to the boundary conditions as will be introduced for different cases.

3.5 A CASE WHERE A CIRCULAR PLATE WITH EDGES CLAMPED IS SUBJECTED TO A PRESSURE

2R0

(a) (b)

Figure 3.8 Circular plate with edges clamped: (a) cross sectional view; and (b) perspective view of plate.

To determine the integration constants in Equations (3.24) and (3.25) for the case where a circular plate with edges clamped is subjected to a pressure (p) with F=0 (Figure 3.8):

pr3 C r C 1 2 . (3.26a) 16D 2 r and

pr 4 C r 2 y 1 C ln r C . (3.26b) 64D 4 2 3

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The slope θ( ) is zero at r =0, then, C2 should be zero if the slope θ is not to approach infinity near the centre of the plate. If the centre of the circular plate is taken as the origin, deflection y = 0 at r = 0, and then C3=0. At the clamped edge where r = R0, θ = dy/dr=0, from Equation (3.26a), pr 3 C r C pR 3 C R 1 2 0 1 0 0 16D 2 r 16D 2 (3.27a) pR 2 C 0 . 1 8D Therefore, the three integration constants are determined.

The maximumdeflection (y ) of the plate occurs at r = R : max 0 4 2 2 4 pR0 pR0 R0 pR0 y (3.27b) max 64D 8D 4 64D To determine stresses ( x and r ) using Equations (3.8) and (3.9), the slope (θ ) is first determined,

3 pr C1r C2 pr 2 2 (r R0 ) (3.27c) 16D 2 r 16D and then,

d p 2 2 (3r R ) . (3.27d) dr 16D 0 Therefore,

Eu C d S Eu C pr 2 pR 2 S D 0 T (3.28a) r x 2 D v T 2 D (3 ) (1 )T 1 v E dx x U 1 v E 16D 16D U and

Eu C d S Eu C pr 2 pR 2 S D 0 T . (3.28b) z 2 D v T 2 D (1 3v) (1 v)T 1 v E r dr U 1 v E 16D 16D U

The maximum stress σ( rmax) will occur at the edge at which r = R0 and at the surface where u = h/2,

3 pR 2 0 (3.28c) r max 4 h2 and σzmax takes place at r=0 so that 3pR2 0 z max 2 (1 v) . 8h (3.28d)

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3.6 A CASE WHERE A CIRCULAR PLATE WITH EDGES CLAMPED IS SUBJECTED TO A CENTRALLY CONCENTRATED LOAD F F

2R0

(a) (b)

Figure 3.9 Circular plate with a radius of R0, subjected to a centrally concentrated load (F ), where edges of the plate are clamped: (a) cross sectional view; and (b) perspective view.

To determine the integration constants in Equations (3.24) and (3.25) for a case where a circular plate with edges clamped is subjected to a centrally concentrated load (F ) as given in Figure 3.9, the slope (θ) at the centre is zero at r=0 so that C2=0. If the origin of the coordinate system is taken as the centre of the plate, y = 0 at r = 0, therefore, C3=0. Also, to determine C , a boundary condition is θ = 0 at r=R , therefore, from Equation (3.24), 1 0 Fr C r F C ln R 1 S 2ln r 1 1 →C D 0 T (3.29) 8D 2 1 D E 2 4 U

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The maximum deflection (ymax) will occur at r =R0, according to the current coordinate system. From Equation (3.25),

FR 2 y 0 (3.30a) max 16 D or

2 3FR0 2 y (1 v ) . (3.30b) max 4Eh 3

To determine σr at r = R0, the slope (θ) and its derivative need to be determined first:

dy pr3 Fr C r C 2ln r 1 1 2 (bis 3.24) dr 16D 8D 2 r

C Fr Fr S Fr D ln r T 2ln R 1 , (3.31a) E 4D 8D U 8D 0

d F 2F r (2ln r 2 2ln R0 ) (ln 1) (3.31b) dr 8D 8D R0 and C F S F F D ln r T ln R (ln R ln r) . r E 4D U 4D 0 4D 0 (3.31c)

Therefore, from Equations (3.8) and (3.9),

Eu C d S F Eu C r S D T (3.32a) r 2 D v T 2 D (ln 1) (ln R0 ln r)T 1 v E dr r U 4D 1 v E R0 U and Eu C d S F Eu C r S D T z 2 D v T 2 D(ln R0 ln r) v(ln 1)T . (3.32b) 1 v E r dr U 4D 1 v E R0 U

The stress distribution according to Equation (3.32a) is shown inFigure 3.10.

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Figure 3.10 Stress distribution in the radial direction.

h h3 E Accordingly, the stresses are found at r R and u with D : 0 2 12 1 v2 Eu C d S 3F rR D v T (3.33a) 0 1 v2 E dr r U 2 h2 and

Eu C d S 3F D v T . (3.33b) z R0 1 v2 E dr r U 2 h2

3.7 A CASE WHERE A CIRCULAR PLATE WITH EDGES FREELY SUPPORTED IS SUBJECTED TO A PRESSURE

2R0

(a) (b)

Figure 3.11 Circular plate with edges freely supported: (a) cross sectional view; and (b) perspective view.

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To determine the integration constants in Equations (3.24) and (3.25) for a case where a circular plate with edges freely supported is subjected to a pressure (p) as given in Figure 3.11, the boundary conditions may be considered with: dy pr 3 Fr C r C 2ln r 1 1 2 , dr 16D 8D 2 r (bis 3.24)

pr 4 Fr 2 C r 2 y ln r 1 1 C ln r C . (bis 3.25) 64D 8D 4 2 3

At r=0, the slope (θ) at the centre is zero and therefore C2 = 0 if the slope (θ ) is not to approach infinity near the centre of the plate.

Again, if the centre of the circular plate is taken as the origin of the coordinate system, deflection y = 0 at r = 0 and therefore C3=0.

To determine C1, we consider that the bending moment (Mr ) is zero at any free support

(r = R0 ). From

C d S M DD v T 0, (bis 3.11) r E dr r U

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d v . (3.34) dr r

Using Equations (3.34) and (3.24), we find,

2 d 3pr C1 V W 2 2 dr 16D 2 3pr C1 C pr C1 S W → D T 2 D T (3.35a) pr C1 W 16D 2 E 16D 2 U W r 16D 2 X and

C pR 2 S (3 ) D 0 T C1 D T . (3.35b) E 8D U (1 )

The maximumdeflection (y ) occurs at r =R if we use the current coordinate system: max 0 C pR 4 S (5 ) D 0 T ymax D T (3.35c) E 64D U (1 )

h3 E with D , or 12 1 v2

4 3pR0 ymax (5 )(1 ) . (3.35d) 16Eh3

To determine stresses ( r and z) using Equations (3.8) and (3.9), we need to determine d the slope (θ ) and first: dr

3 3 2 pr Fr C1r C2 pr r C pR0 S (3 ) 2ln r 1 D T , (3.36a) 16D 8D 2 r 16D 1E16D U (1 )

2 2 pr C pR0 S (3 ) D T , (3.36b) r 16D E16D U (1 ) and

2 2 d 3pr C pR0 S (3 ) D T . (3.36c) dr 16D E16D U (1 )

Therefore, the radial stress ( r ) is found,

Eu C d S Eu C pr 2 C pR 2 S S D v T D (1 ) D 0 T(3 )T . (3.37a) r 2 2 D D T T 1 v E dr r U 1 v E 16D E 16D U U

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The maximum stress σrmax occurs at r = 0, Eu C pr 2 C pR2 S S Eh / 2 C pR2 S D (3 ) D 0 T(3 )T D 0 T(3 ) r max 2 D D T T 2 D T (3.37b) 1 v E 16D E16D U U 1 v E16D U

h3 E with D , or 12 1 v2

C 3pR 2 S D 0 T (3.38a) r max D 2 T(3 ) E 8h U Similarly, the maximum stress in the tangential direction occurs at r = 0 and

C 3pR2 S D 0 T z max r max D 2 T(3 ) . (3.38b) E 8h U

3.8 A CASE WHERE A CIRCULAR PLATE WITH EDGES FREELY SUPPORTED IS SUBJECTED TO A CENTRAL CONCENTRATED LOAD F F

2R0

(a) (b)

Figure 3.12 Cross section of a circular plate with a radius of R0, subjected to a point force (F), where edges are freely supported: (a) cross sectional view; and (b) perspective view.

To determine the integration constants in Equations (3.24) and (3.25) for a case where a circular plate with edges freely supported is subjected to a central concentrated load (F ) as given in Figure 3.12, the boundary conditions may be considered with:

dy pr 3 Fr C r C 2ln r 1 1 2 , (bis 3.24) dr 16D 8D 2 r

pr 4 Fr 2 C r 2 y ln r 1 1 C ln r C . (bis 3.25) 64D 8D 4 2 3

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At r=0, the slope (θ) at the centre is zero and therefore C2 = 0 if the slope (θ ) is not to approach infinity near the centre of the plate.

Again, if the centre of the circular plate is taken as the origin of the coordinate system, deflection y = 0 at r = 0 and therefore C3=0.

To determine C1, we consider that the bending moment (Mr ) is zero at any free support

(r = R0 ). As before, from,

C d S M DD v T 0, (bis 3.11) r E dr r U we find,

d v . (bis 3.34) dr r Using Equations (3.34) and (3.24) again,

d F C 1 S C 2F C 1 S C D2ln r 2r 1T 1 Dln r T 1 (3.39a) dr 8D E r U 2 8D E 2 U 2

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F C 2ln r 1 1 . (3.39b) r 8D 2

Equating these two equations with r = R0

F C 1 S C D2ln R T . (3.39c) 1 4D E 0 1 U

The maximum deflection (ymax) occurs at the supports (r =R0), if we use the current coordinate system. From Equation (3.25), (p = 0, C2= C3=0)

FR 2 3 y 0 (3.39d) max 16D 1

h3 E with D or 12 1 v2

2 2 2 2 FR 3 FR 3 12(1 v ) 3FR y 0 0 0 (3 )(1 v) . (3.39e) max 16D 1 16 1 Eh 3 4Eh 3

This maximum deflection is approximately 2.5 times that of the plate with the clamped edge for Poisson’s ratio ν = 0.3.

To determine stresses ( r and z) using Equations (3.8) and (3.9), we need to determine d the slope (θ ) and first. From Equation (3.24), dr

F C1r 2r ln r r (C2 = 0, p = 0), (3.40a) 8D 2

d F C R S (3.40b) Dln 0 T , dr 4D E r 1 U and

F C 1 R S D ln 0 T . (3.40c) r 4D E1 r U

Accordingly,

Eu C d S Eu F F C R SV 0 (3.41a) r 2 D v T 2 G D(1 )ln TW 1 v E dr r U 1 v H4D E r UX

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h3 E with D or 12 1 v2 3F(1Q ) R V ln 0 . (3.41b) r 2Sh2 r

Note that σr is zero at the edge and infinite at the centre. In practice, the concentrated load is on a finite area.

The stress in the z-direction,

Eu §T dT · V ¨ v ¸ (3.41c) z 1 v 2 © r dr ¹

Eu ª F § R0 ·º « ¨(1 v)ln (1Q )¸» 1 v 2 ¬4SD © r ¹¼

h3 E with D oror 12 1 v2 3F ª§ R ·º V 0 Q z 2 «¨(1 v)ln (1 )¸» . (3.41d) 2Sh ¬© r ¹¼

3.9 A CASE WHERE A CIRCULAR PLATE WITH EDGES FREELY SUPPORTED IS SUBJECTED TO A LOAD ROUND A CIRCLE

(a) (b)

Figure 3.13 Circular plate with a radius of R0, subjected to a load round a circle (F ), where edges are freely supported: (a) cross-section (b) perspective view.

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To determine the integration constants in Equations (3.24) and (3.25) for a case where a circular plate with edges freely supported is subjected to a load (F ) round a circle as given in Figure 3.13, the boundary conditions may be considered with:

dy pr 3 Fr C r C T 2ln r 1 1 2 , (bis 3.24) dr 16D 8SD 2 r 4 2 2 pr Fr C1r y ln r 1 C2 ln r C3 . (bis 3.25) 64D 8SD 4

There will be two sets of integration constants because of the discontinuity between two parts. For the inner part of the circular plate, r < R1, p = F = 0. From Equation (3.25), the deflection (y) is found to be

C r 2 y 1 C ln r C (3.42a) 4 2 3 and the slope (θ) is found from Equation (3.24),

C r C T 1 2 . (3.42b) 2 r

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If the centre of the circular plate is taken as the origin of the coordinate system, deflection

y = 0 at r = 0 and therefore C3=0. For non-infinite slope at the centre,C 2=0.

Thus, thedeflection (y) for the inner part of the circular plate (r < R1),

2 C1r y (3.43a) 4 and

C r T 1 . (3.43b) 2

The constantC 1 is to be determined later.

For the outer part of the circular plate, r ≥ R1, another set of integration constants (C1c , C2c , and C3c) may be introduced and Equations (3.24) and (3.25) reduce respectively to

dy Fr Ccr Cc T 2ln r 1 1 2 (3.44a) dr 8SD 2 r and

Fr 2 Ccr 2 y ln r 1 1 Cc ln r Cc . (3.44b) 8SD 4 2 3 c Now, four constants (C1 , C1c , C2c , and C3) need to be determined and we need to find four simultaneous equations. Inner and outer plate parts are common at r = R1 for deflection (y) and slope (θ). Accordingly, equating Equation (3.43a) to Equation (3.43b), and Equation (3.43b) to Equation (3.44b), two of the four simultaneous equations are found:

C1R1 FR1 C1cR1 C2c 2ln R1 1 (3.45a) 2 8SD 2 R1 and

C R 2 FR 2 CcR 2 1 1 1 ln R 1 1 1 Cc ln R Cc . (3.45b) 4 8SD 1 4 2 1 3 Also, using

§ dT T · M D¨ v ¸ , (bis 3.11) r © dr r ¹ with the common Mr at r = R1, two more equations can be found.

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For the inner part (F=0, r

C r T 1 (3.46a) 2 then,

dT C 1 (3.46b) dr 2 and

T C 1 . (3.46c) r 2

For outer part (F≠0, r≥R2), using Equation (3.44a),

dT F § 1 · Cc Cc ¨2ln r 2r 1¸ 1 2 (3.47a) dr 8SD © r ¹ 2 r 2 and, for r = R1,

§ dT · F Cc Cc ¨ ¸ 2ln R 1 1 2 . (3.47b) dr 8SD 1 2 2 © ¹ r R1 R1 Also,

T F Cc Cc 2ln r 1 1 2 (3.47c) r 8SD 2 r 2

and for r = R1, §T · F Cc Cc ¨ ¸ 2ln R 1 1 2 . (3.47d) r 8SD 1 2 2 © ¹r R1 R1 Substituting Equations (3.46) and (3.47) into Equation (3.11) and then equating resulting two equations:

C F Cc Cc 1 (1Q ) 2ln R (1Q ) (1Q ) 1 (1Q ) 2 (1Q ) . (3.48) S 1 2 2 8 D 2 R1 This is the third equation of the four simultaneous equations.

For one more simultaneous equation, we may use Mr = 0 at the outside edge (r = R0 ) with Equation (3.11),

F Cc Cc 2ln R (1Q ) (1Q ) 1 (1Q ) 2 (1Q ) 0 . (3.49) S 0 2 8 D 2 R1

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Thus, four simultaneous Equations (3.45a), (3.45b), (3.48), and (3.49) have been found for four unknowns. The solution yields

F R (1Q )(R 2 R 2 ) 0 0 1 C1 (1 2ln 2 R1 ) , (3.50a) 4SD R1 (1Q )R0

F § (1Q )(R2 R2 ) · c ¨ 0 1 ¸ C1 ¨2ln R0 2 ¸ , (3.50b) 4SD © (1Q )R0 ¹

FR 2 Cc 1 , (3.50c) 2 8SD and

FR 2 Cc 1 (ln R 1) . (3.50d) 3 8SD 1

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The deflection (y) for the outer part of the circular plate, r ≥ R1 with the origin of the current coordinate system at the centre of the plate, Equation (3.44b) yields:

Fr 2 Ccr 2 y ln r 1 1 Cc ln r Cc 8SD 4 2 3

Fr 2 r 2 F § (1Q )(R2 R2 ) · FR 2 FR 2 ¨ 0 1 ¸ 1 1 (3.51) ln r 1 ¨2ln R0 2 ¸ ln r (ln R1 1). 8SD 4 4SD © (1Q )R0 ¹ 8SD 8SD

The maximumdeflection (ymax) occurs at the supports (r =R0 ) and is given by

F ª§ 3Q · 2 2 2 R0 º y y ¨ ¸(R R ) R ln . max r R0 «¨ ¸ 0 1 1 » (3.52) 8SD ¬© 2(1Q ) ¹ R1 ¼

The stress σrmax for this case occurs at r = R1. Using Equation (3.8),

3F R R 2 R 2 ½ V Q 0 Q 0 1 r max 2 ®2(1 )ln (1 ) 2 ¾ . (3.53) 4Sh ¯ R1 R0 ¿

3.10 A CASE WHERE AN ANNULAR RING WITH EDGES FREELY SUPPORTED IS SUBJECTED TO A LOAD ROUND A CIRCLE

(a) (b)

Figure 3.14 Annular ring with a radius of R0, subjected to a load round a circle (F ), where edges are freely supported: (a) cross-section; and (b) perspective view.

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The plate subjected to a load round a circle (p = 0) shown in Figure 3.14 is an annular ring with edges freely supported. The following equations derived previously for the outer part of the circular plate in Figure 3.13 is directly applicable for the annular ring:

dy Fr Ccr Cc T 2ln r 1 1 2 (bis 3.44a) dr 8SD 2 r and

Fr 2 Ccr 2 y ln r 1 1 Cc ln r Cc . (bis 3.44b) 8SD 4 2 3

To determine constants in equation above, we may use Mr = 0 at both r = R1 and r = R0 with § dT T · M r D¨ v ¸ . (bis 3.11) © dr r ¹

Differentiating Equation (3.44a),

dT F § 1 · Cc Cc ¨2ln r 2r 1¸ 1 2 , (3.54a) dr 8SD © r ¹ 2 r 2 then,

§ dT · F Cc Cc ¨ ¸ 2ln R 1 1 2 (3.54b) dr 8SD 1 2 2 © ¹r R1 R1 and

§ dT · F Cc Cc ¨ ¸ 2ln R 1 1 2 . (3.54c) dr 8SD 0 2 2 © ¹r R0 R0 Using Equation (3.44a),

T F Cc Cc 2ln r 1 1 2 , (3.54d) r 8SD 2 r 2

§ T · QF CcQ CcQ ¨Q ¸ 2ln R 1 1 2 (3.54e) r 8SD 1 2 2 © ¹r R1 R1 and

§ T · QF C1cQ C2cQ ¨Q ¸ 2ln R0 1 . (3.54f) r 8SD 2 2 © ¹r R0 R0

Setting Equation (3.11) to zero for r = R1,

F Cc Cc ^`2(1Q )ln R (1Q ) 1 (1Q ) 2 (1Q ) 0 (3.55) S 1 2 8 D 2 R1

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and for r = R0,

F Cc Cc ^`2(1Q )ln R (1Q ) 1 (1Q ) 2 (1Q ) 0 . (3.56) S 0 2 8 D 2 R0 From the two simultaneous Equations (3.55) and (3.56), two of the three integration constants are found to be

F (1Q ) R 2R 2 R c 0 1 0 C2 2 2 ln (3.57a) 4SD (1Q ) R1 R0 R1 and

F (1Q ) 2(R 2 R 2 ) R ½ c 0 1 0 (3.57b) C1 ® 2 2 ln ¾ . 4SD ¯(1Q ) R0 R1 R1 ¿

If we use the same coordinate system, y = 0 at r = R1, FR 2 1 (1Q ) R 2 R ½ c 1 0 0 C3 ®1 2 2 ln ¾ . (3.57c) 8SD ¯ 2 (1Q ) R0 R1 R1 ¿

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More cases with different boundary conditions are shown in Figure 3.15. The maximum stress (V max) for all those cases can be in a generalised form given by

k pR 2 V 1 0 (3.58a) max h 2 or

k F V 1 (3.58b) max h 2

R0 where k1 is a factor dependent on the boundary condition, Poisson’s ratio, and . Likewise, R1 the maximum deflection (ymax) for the same cases is given by k pR 2 y 2 0 (3.58c) max Eh 2 or

k FR 2 y 2 0 (3.58d) max Eh 2

R0 where k 2 is a factor dependent on the boundary condition, Poisson’s ratio, and . R1

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R0 R1

p h

F p p

F p

F F p

Figure 3.15 Cross sections of various cases for different boundary conditions.

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4 FUNDAMENTALS FOR THEORY OF ELASTICITY

In the elementary mechanics of solids, assumptions are used for simplification before arriving at solutions. For example, a linear stress distribution is assumed for a beam or a shaft. In the theory of elasticity, however, the stress distribution is to be found by satisfying the equilibrium equations, compatibility equations, and boundary conditions without assumptions. It is a mathematical process.

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4.1 EQUILIBRIUM AND COMPATIBILITY EQUATIONS

Vz+ wV z dz wz

wW zy Wzy+ dz Wzx+ wW zx dz wz wz V wW x wW Wxz+ xz dx Wyz+ yz dy wx Wxy wy Wyx wV σ W Vy+ y dy y xz wy wW y wV wW Wyx+ yx dy dz Vx+ x dx W xy xy+ dx wy wx Wzxx W wx yz Wzy dx dy x Vz

(a) (b)

Figure 4.1 (a) A body subjected to external forces. (b) One of the elements where stress varies from one face to another.

For the stress variation within a elastic body (Figure 4.1a), let us consider one of the stress elements given in Figure 4.1b. To find the equilibrium equations, we need to consider the forces acting on the element. Theforces are found by multiplying the stress on any face by the surface area. Also, we need to consider a body force though the centroid of the element and having components X, Y, Z per unit volume. Taking the summation of forces in the x, y and z directions results in the following differential equations of equilibrium: for x-direction,

wV wW yx wW x zx X 0 ; (4.1) wx wy wz for y-direction,

wV y wW xy wW zy Y 0 ; (4.2) wy wx wz for z-direction,

wV wW wW z yz xz Z 0 . (4.3) wz wy wx

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Also, the body forces are given by

w: X , (4.4) wx w: Y , (4.5) wy w: Z (4.6) wz where W is called the potential function. Thebody forces are to deal with gravitational forces, magnetic forces and/or inertia forces. The force of one body acting on another by a direct contact is the surface force. It may be noted that the equilibrium Equations (4.1–4.3) do not provide a relationship between the stresses and the external loads, although they give the rate of change of the stresses at any point in the body. One of the requirements to establish such a relationship is that the deformation continuity of each element must be preserved. This means that the displacement in components must be continuous and single-valued functions. Certain relationships between the strain components must be satisfied to meet the requirement. These relationships are called the equations of compatibility. The relationship between the stresses and the external loads is required to also satisfy the boundary conditions.

To derive the equations of compatibility, let us consider the strain-displacement relations previously given in Equation (1.18):

wu H , (a) x wx wv H y , (b) wy ww H , (c) z wz wu wv J xy , (d) wy wx wu ww J , (f) xz wz wx wv ww J yz . (g) wz wy

To eliminate the displacements from the above equations, we differentiate Equation (a) twice with respect to y and Equation (b) twice with respect to x and Equation (d) once with respect to x and then once with respect to y results in the following compatibility equation,

w 2H w 2H w 2J x y xy . (4.7) wy 2 wx 2 wxwy

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Thus, the compatibility equations are to establish relationships between different strain components as well.

Two additional compatibility equations may be obtained in a similar way:

2 2 2 w H y w H w J yz z , (4.8) wz 2 wy 2 wywz w2H w2H w2J z x xz . (4.9) wx2 wz 2 wxwz The following equations may be found from Equation (1.18) for more compatibility equations:

wH w 3u x , (4.10) wywz wxwywz

2 3 3 w J xy w u w v , (4.11) wxwz wxwywz w 2 xwz w2J w3u w3w xz , (4.12) wxwy wxwywz w2 xwy

2 3 3 w J yz w v w w . (4.13) wx2 wx2wz w2 xwy

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We may add Equation (4.11) and Equation (4.12) together and, then, subtract Equation (4.11) to get

2 2 2 3 w J xy w J w J yz w u xz 2 . (4.14) wxwz wxwy wx2 wxwywz

From Equation (4.10) and Equation (4.14), we find anothercompatibility equation,

w 2H w § wJ wJ wJ · x ¨ yz xz xy ¸ 2 ¨ ¸ . (4.15) wywz wx © wx wy wz ¹

Similarly, two more compatibility equations can be found: w2H wJ wJ y w § wJ xz xy yz · 2 ¨ ¸ (4.16) wxwz wy © wy wz wx ¹ 2 wJ wJ w H z w § xy yz wJ xz · 2 ¨ ¸ . (4.17) wxwy wz © wz wx wy ¹

4.2 AIRY’S STRESS FUNCTION As discussed, to find equations for stress distribution on a solid body, any candidate equations are required to satisfy the boundary conditions, equilibrium equations and compatibility equations. This procedure can be simplified using the Airy’s stress function (F) which is defined by the following three equations: w 2) V x : (4.18) wy 2 w 2) V : (4.19) y wx 2 w 2) W xy . (4.20) wxwy

The three equations above containing the Airy stress function (F) satisfy the equilibrium equations for two dimensional cases. Thus, the procedure for finding equations for stress distribution involves finding the Airy stress function (F) and satisfying the compatibility equations. The compatibility equations and Airy stress function (F) will be further discussed in relation with plane stress and plane strain.

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4.2.1 PLANE STRESS

Equations (4.18)–(4.20) are substituted into the following equations for plane stress

(V z W xz W yz 0),

1 H V QV (bis 2.19a) x E x y 1 H V QV (bis 2.19b) y E y x Q H V V (bis 2.19c) z E x y 1 2(1Q ) J xy W xy W xy (bis 2.4 & 2.8) G E and then into the compatibility equation (4.7) to obtain w 4) w 4) w 4) § w 2: w 2: · Q ¨ ¸ . 4 2 2 2 4 (1 )¨ 2 2 ¸ (4.21) wx wx wy wy © wx wy ¹

The symbol is called del operator and 4 is called the biharmonic operator defined as

w 4 w 4 w 4 4 2 (4.22) wx 4 wx 2wy 2 wy 4 and 2 is called the Laplacian operator defined as

2 2 2 w w . (4.23) wx 2 wy 2

Thus, Equation (4.21) becomes

4) = - (1-Q)2 : (4.24a)

Note that the remaining five compatibility equations (4.8), (4.9), (4.15–4.17) have not been used. Two of these compatibility equations vanish because of the stress field here is independent of z but the other three will not be satisfied. However, the stresses above are known to be good approximations. For the case of zero body forces Equation (4.24a) reduces to the so called the biharmonic equation:

4) = 0. (4.24b)

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4.2.2 PLANE STRAIN

We employ the same Airy stress function F(x, y) [see Equations (4.18)–(4.20)] as for the plane stress case in conjunction with the following relations for plane strain (ez = 0):

V z v V x V y (bis 2.20b) 1 v2 § v · H ¨V V ¸ (bis 2.20c) x E © x 1 v y ¹ 1 v2 § v · H ¨V V ¸ . (bis 2.20d) y E © y 1 v x ¹

For the plane strain case, five of the compatibility equations are satisfied, leaving only 2 2 2 2 2 Equation (4.7), w H x /wy w H y /wx w J xy /wxwy , to be considered. If we consider the compatibility Equations (4.15), (4.16) and (4.17), we obtain

1 4 2: . (4.24c) 1Q

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When there are no body forces, the same biharmonic equation as that for plane stress is obtained as

4) = 0. (4.24d)

Therefore, to find the equation for a stress distribution, we need to find an Airy’s stress function with the biharmonic equation satisfied.

The following is an example for using the Airy stress function to find equations for stress distribution. A case is given in Figure 4.2, in which the pressure p varies along the bar. The stress function Φ = By3 may be considered for it (B is a constant). It is found that The stress function satisfies the biharmonic equation (4) =0) and therefore produces the following equations for stress distribution:

w 2 ) V x 6 (4.25a) wy 2 w 2) V 0, and (4.25b) y wx 2 w 2) W 0 . (4.25c) xy wxwy

To determine the constant (B ), boundary conditions are used:

σx = pA = 0 at y=0 and

σx = 6Bl = pB at y=l.

p Accordingly, B B and therefore the stress distribution is described by 6l w 2) y V x 6By pB . (4.25d) wy 2 l

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y pB B

p l

A x

Figure 4.2 A bar subjected to a linear stress distribution.

For many problems, the stress function may be possible to be in the form of the following polynomial expression:

) Ax 2 Bxy Cy2 Dx3 Ex 2 y Fxy 2 Gy3 (4.26) Hx4 Jx 3 y Kx2 y2 Lxy3 My4 Nx5 Px 4 y Qx3 y2 Rx 2 y3 Sxy 4 Ty5 ...

Terms containing x or y up to the third power satisfy the biharmonic equation. However, terms containing higher powers remain in the biharmonic equation. Those terms can be sometimes vanished by relating associated coefficients.

4.3 APPLICATION OF EQUILIBRIUM EQUATIONS IN PHOTO-ELASTIC STRESS ANALYSIS The equilibrium equations may be useful for photo-elastic stress analysis.

For example,

wV wW wW x yx zx X 0 (bis 4.1) wx wy wz can be rewritten for plane stress as

wV wW yx x 0 (4.27a) wx wy

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wW yx dV x dx . (4.27b) wy

Therefore, the normal stress in thex -direction may be obtained by integration

wW yx V dx c (4.27c) x ³ wy where c is an integration constant.

'V Accordingly, a stress difference ( x) between any two points (x0 and x) is found to be,

x wW yx 'V x dx . (4.27d) ³ wy x0 For numerical calculation, if a stress at the point x , V is known, then the stress at 0 x x x0 the point x (V x) can be translated into

n 'W V V yx 'x . (4.27e) x x x x0 ¦ 1 'y

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Equation (4.27e) may be practically useful in conjunction with photo-elastic stress analysis for finding stress values.

Photoelasticity has been used as an experimental method for finding stress distributions on various geometries. It is based on a material property called birefringence (double refraction) which reacts to stresses. In practice, there are two types of patterns on the model for stress analysis may be used – isoclinic and isochromatic. As shown in Figure 4.3a, an isochromatic is the locus of the points, along which the difference between major and minor principal stresses is constant while an isoclinic is a locus of points at which the principle stresses are all in the same direction.The loci appear on the photoelastic model in a form of lines called fringes – isoclinics appear in black and isochromatics in other colours as shown in Figure 4.3b.

y σ2 σ2

σ2 σ1 σ1 σ2 σ2 σ1 σ1 σ2 σ2 σ1 σ1

σ1 σ1 σ2 σ1 σ2 σ1 σ2

σ1 σ1 σ2 σ2

(a) (b)

Figure 4.3 (a) Isoclinics with the principle stress directions – the principle stress directions of each point are inclined at a constant angle to x and y axes. (b) Fringes on a plate with a hole.

When a ray of plane polarized light pass through a photo-elastic material model, it resolves along the two principal stress directions and each of these components experiences different refractive indices as they travel at different velocities within the model. When the light comes through the analyzer (Figure 4.4), the phase difference or relative retardation (R ) in wave lengths between the two resolved rays is given by (Frocht 1941)

R Ct(V1 V 2 ) 2CtW max (4.28a) where C is a constant known as the stress optic coefficient, t is the thickness of the model plate, and σ1 and σ2 are major and minor the principal stresses.

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Light source

Polariser

Model

Analyser

Eye

Figure 4.4 A photoelasticity arrangement to view a fringe pattern on the model. The fringe pattern viewed is due to the plane polarized light. Each fringe is a locus of isochromatic points at which the difference between two principal stresses or the maximum shear stress is constant.

When a material of birefringence is stressed, fringes are created. Fringes for stress may be similar to contour lines on a map where a close spacing between contour lines indicates a high slope and vice versa. Each fringe line is a locus of points of constant difference between the major (σ1) and minor (σ2 ) principal stresses and can be used for stress calculation using nf V V (4.28b) 1 2 t where n is the fringe number or fringe order, f is the model material fringe value, and t is the model thickness.

Equation (4.28b) can be used for finding shear stress W( yx) with the stress relation in Figure 4.5.

σ2

σ1

y τxy

σy x

Figure 4.5 Principal stresses with other stress components.

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A relation according to the equilibrium in the x-direction (Figure 4.5) is

V1 sinT cosT W xy V 2 cosT sinT 0 (4.28c) so that

V1 V 2 W xy sin 2T . (4.28d) 2

In practice, the variables in Equation (4.28d) can be measured using the photo-elastic experiment and, then, stress at any point can be calculated according to Equation (4.27e).

Further, the stress distribution for a three-dimensional model or real component can be obtained using the same theory if reflective surface and photo-elastic coating are used as shown in Figure 4.6.

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Component Reflective or model surface

Light source

Eye

Photo- elastic coating

Figure 4.6 A photoelasticity arrangement for a three dimensional component or model.

4.4 STRESS DISTRIBUTION IN POLAR COORDINATES The polar coordinate system is useful for some particular geometries such as cylinder and circular plates. Stress components on an infinitesimal element in polar coordinates are given in Figure 4.7.

wW y wVT rT V dT W rT dr T wT wr dr wW wW W rT dT W rT dT rT wT rT wT

τθr τrθ σ r dθ/2 σθ r

dθ θ

Figure 4.7 Stress components on an infinitesimal element in the polar coordinate system.

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The equilibrium equations in radial and tangential directions are given by

wV 1 wW V V r rT r T R 0 (4.29) wr r wT r

1 wV T wW T W T r 2 r T 0 (4.30) r wT wr r where R is a radial body force and T is a tangential body force.

The normal stress distributions in radial and tangential directions and shear stress are given by

1 w) 1 w2) V r , (4.31) r wr r 2 wT 2 w2) VT , (4.32) wr 2 1 w) 1 w2) w § 1 w) · W T ¨ ¸ (4.33) r r 2 wT r wrwT wr © r wT ¹ where Φ is the Airy stress function.

The biharmonic equation without the body force is

§ w 2 1 w 1 w 2 ·§ w 2) 1 w) 1 w 2) · 2 2 ) ¨ ¸¨ ¸ (4.34) ( ) ¨ 2 2 2 ¸¨ 2 2 2 ¸ 0 © wr r wr r wT ¹© wr r wr r wT ¹

4.4.1 THICK WALLED CYLINDER

The Airy stress function for the general continuous axi-symmetric stress distributions independent of θ can be found by solving the biharmonic equation (4.34). The biharmonic equation independent of θ is given by

§ w 2 1 w ·§ w 2) 1 w) · 2 2 ) ¨ ¸¨ ¸ (4.35) ( ) ¨ 2 ¸¨ 2 ¸ 0 © wr r wr ¹© wr r wr ¹ or

§ w 4) 2 w3) 1 w 2) 1 w) · 2 2 ) ¨ ¸ ( ) ¨ 4 3 2 2 3 ¸ 0 . (4.36) © wr r wr r wr r wr ¹ Solving the differential equation,

) A Bln r Cr 2 D(ln r)r 2 . (4.37)

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Consequently, the normal stress distributions in radial and tangential directions and shear stress are given by

B V 2C , (4.38) r r 2 B VT 2C (4.39) r 2

W rT 0 . (4.40)

The stress distributions for a thick walled cylinder shown inFigure 4.8 may be determined using the following boundary conditions when pressures exist both internally (pi) and externally (po):

V r pi at r =ri

V r po at r =r0

W rT 0 at both r =ri and r =r0.

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pi ri

Figure 4.8 Thick walled cylinder cross section with

an internal pressure (pi ) and an external pressure (po ).

Therefore,

B ( p p )r 2r 2 p r 2 p r 2 V i o o i o o i i r 2 2C 2 2 2 2 2 (4.41) r r (ri ro ) (ri ro ) and B ( p p )r 2r 2 p r 2 p r 2 V i o o i o o i i . T 2 2C 2 2 2 2 2 (4.42) r r (ri ro ) (ri ro )

4.4.2 STRESS DISTRIBUTION FOR AN INFINITELY LARGE THIN PLATE WITH A SMALL CIRCULAR HOLE

σ ө x

Figure 4.9 An infinitely large thin plate with a small circular hole.

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The Airy stress function Φ for the stress distribution for an infinitely large thin plate with a small circular hole (Figure 4.9) is found to be

V (r 2 A2 )2 ½ ) 2 2 T ®r 2A ln r 2 cos 2 ¾ (4.43) 4 ¯ r ¿ where σ is a uni-axial tensile stress applied in a remote place. Accordingly, the stress distributions are given by

1 w) 1 w 2) V A2 4A2 3A4 ½ V T r 2 2 ®1 2 (1 2 4 )cos 2 ¾ (4.44a) r wr r wT 2 ¯ r r r ¿

w2) w2 V (r 2 A2 )2 ½ V 2 2 T T 2 2 [ ®r 2A ln r 2 cos 2 ¾] (4.44b) wr wr 4 ¯ r ¿

w § 1 w) · V § 2A2 3A4 · W ¨ ¸ T (4.44c) rT ¨ ¸ ¨1 2 4 ¸sin 2 wr © r wT ¹ 2 © r r ¹ where A is the radius of the hole.

4.4.3 STRESS DISTRIBUTION ACTING ON A STRAIGHT BOUNDARY OF A SEMI-INFINITE PLATE SUBJECTED TO A NORMAL LINE FORCE

r dθ

σr

Figure 4.10 A semi-infinite plate subjected to a normal line force P( ).

The Airy stress function Φ for the stress distribution due to a concentrated normal force (P ) acting on a straight boundary of a semi-infinite plate (seeFigure 4.10) is given by

Φ = Ar θ sin θ (4.45a)

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1 w) 1 w 2) cosT V 2A (4.45b) r r wr r 2 wT 2 r w 2) w 2 (ArT sinT) V T 0 (4.45c) wr 2 wr 2 w § 1 w) · W T ¨ ¸ 0 (4.45d) r wr © r wT ¹ where A is a constant. The radial stress (σr ) appears to be a principal stress in the absence of shear stress (W rT 0). The constant A is determined according to the equilibrium of forces acting on any cylindrical surfaces of radius r so that

2P cosT V . (4.46a) r S r

A locus for a constant radial principle stress (σr ) for a given load (P ) may be found by eliminating θ, given that d cosT r in the circle with a diameter (d ) shown in Figure 4.10, which is given by 2P V r . (4.46b) Sd

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In other words, the principle stress direction constantly varies along the circle although the radial stress (σr ) is constant.

r σθ=0 θ

σr= -2P/πd

d: circle diameter

Figure 4.11 A locus of radial stress (σr ) for a given load (P ).

4.4.4 STRESS DISTRIBUTION IN A CIRCULAR DISK

In order to obtain the stress distribution for a finite circular disk subjected to a force (P ), the stress distribution in the semi-infinite plate may be used. If two equal tensile radial stresses 2P (V r ) with another force P are added on the circle circumference in Figure 4.11, then, Sd the radial compressive stresses are offset and, as a result, no stress exists on the circumferential surface and the circle is equivalent to a finite disk as shown in Figure 4.12. Therefore, the stress distribution within the disk can be calculated by superimposing the two equal tensile radial stresses on the previous stress distribution in the circle.

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σr=2P/πd

θ r

M σr=2P/πd

x r1

θ1

d: Disk diameter P Figure 4.12 A circle is isolated from the semi-infinite plate subjected to the force

P and then two equal tensile radial stresses (V r 2P/Sd) with another force P are added on the circle circumference to offset the compressive stresses. As a result, no stress exists on the circumferential surface and the circle is equivalent to a finite disk.

At point M’ (for solid arrow) σy

ΣFy=0 σx 2 σy= σrsinθsinθ = σrsin θ

σr ΣFx=0 σr 2 M’ σx= σrcosθcosθ = σrcos θ θ At point M’ (for dashed arrow)

σr r ΣFy=0 σr 2 x σy = σrcosθcosθ = σrcos θ σx

σy ΣFx=0 2 σx=σrsinθsinθ = σrsin θ

At point M’ (for both arrows) d: Disk diameter

ΣFy=0 2 2 σy=σr(sin θ+ cos θ)= σr

ΣFx=0 2 2 σx= σr(sin θ+ cos θ)= σr

Figure 4.13 The stress distribution caused by the tensile stresses added along the circumference is uniform

within the disk i.e. σx=σy=σr without force P.

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P 2P

σθ=0 θ r = + σr= -2P/πd

(a) (b) (c)

Figure 4.14 A finite disk with a diameter ofd subjected to line force P.

Before we obtain the stress distribution for a finite circular disk, we need to know that the two equal tensile radial stresses (V r 2P/Sd ) added along the circumference causes a uniform stress distribution where V x V y 2P /Sd within the disk as shown in Figure 4.13. Accordingly, we use the superposition as shown in Figure 4.14 to find the stress distribution on the horizontal diametral section of the circular disk subjected to a force P [Figure 4.14(a)]. The

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stress in the y-direction (σy ) on the horizontal diametral section where the stress directions are symmetric [Figure 4.14(c)] is given by

4P cos 3 T V 2V cos 2 T (4.47a) y r S r and the stress in the y-direction (σy ) due to the two equal tensile radial stresses at any point in Figure 4.13(b) is given by 2P V y . (4.47b) Sd

Superimposing these together,

4P cos 3 T 2P V y . (4.47c) S r Sd

The maximum compressive stress (= minor principal stress) along the horizontal diameter occurs at the center of the disk (q=0) and is found to be

6P V . (4.47d) y max Sd

Similarly, the stress in the x-direction along the horizontal diameter (σx ) is given by 4P cosT 2P V ( )sin 2 T (4.47e) x S r Sd and the stress at the center of the disk (= major principal stress) is found to be

2P V x max . (4.47f) Sd

This equation is useful for calculating the splitting tensile strength of brittle materials (ASTM D3967-16). Also, it follows that

8P V x V y V1 V 2 , (4.47g) Sd which may be useful for the photo-elasticity calibration.

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5 S-N FATIGUE

∆σ (stress range) σa (stress amplitude)

) Specimen σ

σmax Stress ( σmean σmin

Time Loading

Figure 5.1 Cyclic loading and definitions of terms for applied stress.

Fatigue may be the most common cause of service failure in mechanical components and structures. It is a type of damage caused by cracking and deformation. Fatigue failure takes place when a material is subjected to cyclic loading lower than the static strength, involving various stresses including stress range (∆σ), stress amplitude (σa), mean stress (σmean), valley

(minimum) stress (σmin), and peak (maximum) stress (σmax) (Figure 5.1).

There have been two different ways to approach the fatigue damage. One of the ways is to deal with cracks individually using fracture mechanics parameters such as the stress intensity factor. The other way is to deal with relatively “small” cracks (Kitagawa and Takashima 1976) for fatigue life in a collective manner including crack initiation at a relatively large scale of un-notched geometry with respect to the initial flaws, which requires an S-N curve model for characterisation. (S and N in the acronym stand for stress and number of fatigue loading cycles at failure respectively.) Some S-N curve models have been derived from fatigue damage formulations while some others are intuitively obtained empirically or statistically using the Weibull distribution (Weibull 1951). The S-N curve models have been used for both monolithic and composite materials. A significant difference between the two types of materials in S-N curve modelling has not been found although the fatigue damage growth pattern varies dependent on material structure. Small cracks, for example, grow in polymeric matrix in the case of fiber reinforced composites but a follow-up crack propagation stage leading to the final failure (or complete separation into two pieces) is often not clearly defined (Wang and Chim 1983, Kawai and Hachinohe 2002) whereas the initiation site(s) and crack propagation traces in the case of monolithic materials are usually identifiable on the fracture surfaces (Rhines 1979).

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An S-N curve is essential not only for characterising fatigue test results from un-notched specimens but also for the remaining fatigue life predictions involving fatigue damage quantification and stress ratio effect. Thus, the S-N curve behavior has been a backbone of fatigue life studies since the 19th century. A historian refers to the fatigue test results published in 1837 by Albert (Schütz 1996) and to the term “fatigue” appeared in 1854 (Braithwaite) as the first ones. Some early fatigue studies, in fact, had begun without an S-N curve model (Wöhler 1870). Also, the S-N fatigue study development in history has been somewhat disorderly in sequence of events, slowing down the knowledge advancement. For example, the research of stress ratio effect on fatigue life (e.g. Gerber 1874, Weyrauch 1891, Goodman 1899, Soderberg 1930) mostly appears earlier than S-N curve modelling (e.g. Basquin 1910, Weibull 1952, Sendeckyj 1981, and Hwang and Han 1986a) although S-N curve modelling should be a prior task to the stress ratio effect for fatigue life prediction. One would be surprised to find the fact that the stress ratio effect has still been a contemporary research topic (e.g. Kawai and Koizumi 2007, and Kawai and Itoh 2014) since the 19th century. On the other hand, the study on fatigue damage (e.g. Palmgren 1924, Hwang and Han 1986b, Poursatip, Ashby, and Beaumont 1986, Poursatip and Beaumont 1986, Eskandari and Kim 2017) for fatigue life prediction seems to be concurrent with modelling of S-N curve (Hwang and Han 1986a, Kawai and Koizumi 2007).

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A primary reason for the slow progress in S-N fatigue life prediction studies despite the long history may be that the studies on other topics such as stress ratio effect and fatigue damage for fatigue life predictions cannot be efficiently conducted without knowing an adequate deterministic S-N curve model. Other reasons may be as follows: (a) researchers have been persistent in adhering to particular S-N curve model(s) or producing their own model(s) regardless of the benefits of existing models, and (b) researchers tend to choose particular damage accumulation rules which are not necessarily valid. Palmgren (1924) and Miner (1945) model may be one of those chosen by many researchers even though the Palmgren and Miner damage model is not valid as Schütz (1996) in his historical review pointed out. Also, evidence for the slow progress has been found in the literature. Philippidis and Passipoularidi (2007) evaluated nine selected damage models involving residual strength degradation in relation to fatigue life. They found that no model is consistent in accurately predicting the residual strength degradation behavior of different composite laminates for fatigue life. Post et al (2008) also selectively evaluated twelve fatigue life prediction models for spectrum loading applications; they summarized that most of the investigators who proposed models simply fit them to their experimental results with their predictive ability remaining uncertain. It is only recently that Eskandari and Kim (2017) returned to the fundamentals and provided mathematical reasoning for how fatigue damage models in the past went invalid for predicting the remaining fatigue life. In this chapter, the fundamentals of S-N fatigue (Eskandari and Kim 2017) will be discussed.

5.1 VARIOUS TYPES OF UNIAXIAL LOADING A logical sequence of uni-axial fatigue loading types to be dealt with for developing various predictive models may be dependent on the complexity of cyclic loading or number of independent variables. Various types of fatigue loading are shown in Figure 5.2. The simplest cyclic loading type may be the one given in Figure 5.2a. It is a constant amplitude loading at stress ratio R(=σmin/σmax)=0 (or alternatively at R≠0). An S-N curve model can be developed for this case to characterise the fatigue behaviour prior to dealing with complex loading cases or more independent variables. Then, two different avenues may be considered. One avenue is for dealing with the case in Figure 5.2b for predicting an S-N curve at a different stress ratio. The other avenue may be for dealing with so called ‘two stage block’ loading at R=0 as given in Figure 5.2c, which may be either ‘high to low’ or ‘low to high’ loading. The former (Figure 5.2b) involves understanding of stress ratio effect onS-N curve as illustrated in Figure 5.3 while the latter (Figure 5.2c) involves understanding of fatigue damage to find iso-damage curves or an equivalent damage point (see Figure 5.4). Then, the cyclic loading illustrated in Figure 2d may be dealt with by combining the prior two approaches used for fatigue damage (Figure 2c) and stress ratio (Figure 2b). If theoretical modelling for stress ratio effect is not resolved, the loading illustrated in Figure 2e may

Download free eBooks at bookboon.com 97 MECHANICS OF SOLIDS AND FRACTURE S-N Fatigue still be dealt with by omitting the stress ratio effect. When the loading amplitude varies, the load sequence effect due to the crack closure (Elber 1971) may be considered. The load sequence effect in the cases given in Figure 2c and d may occur but only once. However, the random loading at R=0 in Figure 2e involves the sequence effect multiple times. The random loading at R≠0 illustrated in Figure 2f may be the most complex loading involving all the effects mentioned in the other loading cases.

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Remaining fatigue S-N curve can be predicted using R-ratio Fatigue life life can be predicted effect model expended using S-N curve V model V

Number of cycles (N) Number of cycles (N)

(a) (b)

High load (Fatigue life High load Low load (Fatigue life expended) Low load (Fatigue life to be expended) (Remaining predicted) V fatigue life to V be predicted)

Number of cycles (N) Number of cycles (N)

Fatigue life Remaining Fatigue life Remaining fatigue expended fatigue life to expended life to be predicted be predicted

V V

Number of cycles (N) Number of cycles (N) (e) (f)

Figure 5.2 Uni-axial cyclic loading types and expected sequence of stages for various model developments: (a) constant amplitude loading at R=0 for S-N curve; (b) constant amplitude loading at R≠0 for S-N curve (c) two stage block loading at R=0; (d) two stage block loading at R≠0 (e) random amplitude loading at R=0; and (f) random amplitude loading at R≠0.

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σmeam=0 R = -1 0

σa

-∞

R = -∞ 0 0 -1

0 -∞

Figure 5.3 Stress ratio ranges on - diagram for the loading types: T-T, T-C, σa σmean

C-T, and C-C. Figure 5.4 Iso-damage line and equivalent damage points A and B

(DA=DB) on S-N plane for two stage block loading

5.2 S-N CURVE MODELS An S-N curve model is a line shape function for representing the phenomenon of fatigue dependent on fatigue damage. Numerous S-N curve models have been produced since 1910 (Basquin). When an S–N curve model is to be selected for applications, the following criteria may be useful (Burhan and Kim 2018): (a) capability of curve fitting, (b) efficiency

Download free eBooks at bookboon.com 100 MECHANICS OF SOLIDS AND FRACTURE S-N Fatigue of fitting parameter determination, (c) adaptability at various stress ratios, (d) representability of fatigue damage at failure, and (e) satisfaction of the initial boundary condition of S-N curve at the first loadcycle. An S-N curve model developed by Kim and Zhang (2001) meets the criteria as will be discussed in this chapter. Some other capable S-N curve models include those proposed by Weibull (1952), and Sendeckyj (1981) but they do not meet the criterion ‘(d)’.

Kim and Zhang (2001) quantified the fatigue damage at tensile fatigue failure (DfT ) for S-N curve derivation: V max D fT 1 (5.1) V uT and they found that a fatigue damage rate follows

wDfT E D(V max ) . (5.2) wN f

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The number of cycles at failure (Nf ) in this equation can be obtained by integration, yielding an S-N curve as a function of applied peak stress (σmax ):

1E V E ª§V · º N uT «¨ max ¸ 1» N (5.3a) f D(E 1) ¨ V ¸ 0 ¬«© uT ¹ ¼» or 1 §D(E 1)(N N ) ·1E V V ¨ f 0 ¸ (5.3b) max uT ¨ E 1¸ . © V uT ¹

The term N0 in equation is to adjust the initial number of cycles for the first cycle failure point according to the quantised analysis. The two parameters α( and β) in Equation (5.2) can be determined for a set of fatigue data, given that Equation (5.2) is numerically,

σmax(i) = (σmax(i) – σmax(i-1))/2. Two different ways may be considered for numerically calculating

ΔDf(i)/ ΔNf(i) and corresponding σmax(i). One of the ways is to directly use experimental data points on an S-N plane by choosing a pair of close data points for each value. The other way is to draw the best fit line manually through S-N data points for digitisation (http://arohatgi.info/WebPlotDigitizer/app/) and then collect a set of digitised points. Then, α log(ΔDf(i)/ ΔNf(i)) is plotted as a function of log σmax(i) such that β is a slope and 10 is an intercept. It is practically efficient with a minimal iteration for the best fit. Examples are given in Figure 5.5 for determining fitting parametersα and β. Figure 5.5a and b show experimental data fitted to Equations (5.2) and (5.3) respectively.

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(a) (b)

Figure 5.5 Experimental data obtained at R=0.2 for 24S-T Alclad aluminium alloy (Miner 1945): (a) least square line for fitting parametersα = 10-13.84 and β=3.25; and (b) fittedS-N curve.

The similar procedure based on the same principle applies to compressive fatigue failure cases such that fatigue damage at compressive failure (Dfc ) is given by:

V min DfC 1 . (5.5) V uC

If fatigue damage fits the following equation:

wDfT E D V min , (5.6) wN f the number of cycles at failure (Nf ) becomes a function of applied valley stress (|σmin|):

E 1E V ª V º uC min . N f «1 » N0 (5.7) D(E 1) V ¬« uC ¼»

For finding two parametersα and β, the similar procedure described above can be implemented using

'D E fi . | D V min i (5.8) 'N fi

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5.3 COMPATIBILITY CONCEPT OF FATIGUE DAMAGE AND AXIOMS Quantification of fatigue damage may be the first step for fatigue life prediction. It may now be considered from the possible different fatigue systems in a broad sense as shown in Figure 5.6, depending on how we select system elements (i.e. variables and constants). An open system consists of elements for input, output, and system constants, in which two different types of variables and one set of system constants may be identified. Dependent variables (e.g., residual strength, residual modulus, and number of cycles at failure Nf ) form one of the types and are useful for description of the consequential property change. Independent variables form another type and are useful for description of the loading conditions [e.g. applied peak stress level (σmax) and number of applied loading cycles (N)]. The system constants are also useful not only for loading conditions (e.g., stress ratio and frequency) but also can be selected as independent variables depending on the purpose of a system set-up. Some dependent variables may also be selected as system constants for a simple analysis. In a system, input may consist of independent variables for loading condition while the output is damage quantity from damage function (D) for fatigue life prediction. Thus, the output is dependent on the independent variables and system constants we select and on what damage function is used. Discovering a valid damage function may not necessarily be straightforward as learned from the fatigue history, and the difficulty increases with an increasing number of independent variables. If we reduce the number of

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Input (Independent Consequential properties variables for loading (dependent variables e.g. residual condition such as N strength, modulus, etc) and σmax, etc)

System constants (e.g. stress ratio, frequency, etc) Output (fatigue damage quantity for fatigue life prediction)

Figure 5.6 A fatigue system structure consisting of input and output with system constants.

For conceptualizing the damage compatibility, we may consider fatigue damage (D)

N D (5.9) N f which is a function of a single independent variable (N ) with the number of cycles to failure (Nf ) as the system constant or boundary condition. It may be valid for damage quantification for fatigue life prediction but only at a givenσ max and system constants. If we choose N and σmax as two independent input variables, we may consider the Palmgren (1924) and Miner (1945) model

k N N N N ¦ i 1 or 1 2 k 1 (5.10) i 1 N fi N f 1 N f 2 N fk where subscript i indicates different peak stress levels. This equation may be seen as being capable of predicting the remaining fatigue life for certain sets of σmax and N values, given that the equation indicates that, for example, damage (D) calculated from N1=500 cycles and Nf1=1,000 cycles at an applied peak stress level is equal to that from different values

N2=50 cycles and Nf2=100 cycles at another peak stress level. However, one problem here is that we do not know if two different damage conditions indicated byD value (50/100=0.5) provide the same capability of resisting fatigue because the damage compatibility between

N and σmax for damage (D) is not known. Another problem is that Nf (which is a system constant in the previous equation, D =N/Nf ) turned into a variable dependent on applied peak stress (σmax) as σmax is selected as an additional independent variable. As a result, damage compatibility conditions between N and σmax must be set up. The compatibility is to ensure that no overlaps in damage quantities develop when damage varies as a function of multiple variables such as N, Nf , and σmax such that a damage quantity should be a single value for a given damage severity. If a damage prediction equation lacks the compatibility,

Download free eBooks at bookboon.com 105 MECHANICS OF SOLIDS AND FRACTURE S-N Fatigue for example, a single damage quantity is produced for different damage severities or vice versa. Also, it is mandatory that the compatibility conditions comply with the principles given in the following damage axioms at any point on the S-N plane (see Figure 5.7):

a) Fatigue damage increases monotonically with an increasing number of loading cycles (N ). b) Fatigue damage increases monotonically with an increasing applied peak stress level

(σmax ). c) Residual strength decreases monotonically with increasing fatigue damage accumulated within a material.

5.4 FATIGUE DAMAGE DOMAINS It may be convenient to divide the damage into three domains as illustrated in Figure 5.7: Domain 1, Domain 2, and Domain 3.

5.4.1 DAMAGE IN DOMAIN 1 AND BOUNDARY CONDITIONS

The damage at failure (Df ) appearing on an S-N curve or at (Nf -0.5) to be exact may be used as the reference damage for Domain 1 if quantifiable because it is physically known damage such that damage at all other points on S-N plane can be determined with reference to Df . A candidate reference damage equation in Domain 1 (Df ) is required to satisfy the following equation:

wD f A (5.11) wV max where A is a constant. This requirement is useful for finding an optimum damage quantity

(Df ) for Domain 1 (Figure 5.7). For example, a candidate damage quantity (Df ) with A=0 violates the axiom ‘(c)’ above. If A ≠ constant, a candidate damage quantity Df would be convoluted and unnecessarily complex. Therefore, A should be a constant. It is noted that V max D fT 1 (bis 5.1) V uT or

V min DfC 1 . (bis 5.5) V uC satisfies Equation (5.11).

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Domain 2

Domain 1 on S-N curve max σ Domain 3 Log

Log Nf

Figure 5.7 Three fatigue damage Domains on S-N plane.

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a Va Da=Df N=1.3

Db> Da Vmax b Vb c (a) Dc> Db Vc

0 0.5 1.5 2.5 N

N=0.5 cycle V = V V V V V max a N=1.3 max < b max = b a N=1.5 V max b

(b) (c) (d)

Figure 5.8 Schematic S-N curve and cyclic loading: a) failure points (a, b and c) on S-N curve on linear-linear scales; (b) point a on cyclic loading curve corresponding to that on S-N curve; (c) failure stress at N=1.3, which is lower than at N=1.5and also lower than the expected stress at N=1.3 on S-N curve; and (d) point b

on cyclic loading curve corresponding to N=1.5 at max = b.

5.4.2 DAMAGE IN DOMAIN 2 AND BOUNDARY CONDITIONS

For the damage in Domain 2 and boundary conditions, we need to know: (a) at what loading cycle (N ) the S-N curve should begin with to include a data point for static strength; and (b) what damage quantity between zero loading and failure point at the initial N should be. To this end, it may be conceptually important to differentiate between damage accumulated before failure (DABF) and failure caused by damage (FCD) for understanding the nature of the S-N plane and for finding boundary conditions with quantizedN . Also, it may be equally important to find a common property between DABF and FCD. The failure in fatigue occurs after damage takes place although the damage can occur without failure. Thus, DABF and FCD are different from each other and occur sequentially at different stages. A common property between the two, however, is residual strength because it indicates part of a condition for either DABF or FCD such that it can be used as a parametric quantity between DABF and FCD. Accordingly, the residual strength itself is not necessarily the compatible damage quantity, but damage (Df ) at failure can be a function of the residual strength.

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As the number of cycles to failure (Nf ) increases, damage and failure may more and more practically behave in a continuous manner on an S-N curve. However, when Nf is small, DABF and FCD behaviours tend to be dependent on the quantized variable N. As an S-N curve is illustrated in Figure 5.8a for stress ratio R=0 on a linear coordinate, when N=0, no damage is created by loading, and its initial material property (e.g., strength) is neither known nor recordable on an S-N curve that is practically a locus of FCD points. The FCD at the highest peak stress level may occur at N=0.5 (to be precise) as indicated with point a on the S-N curve in Figure 5.8, given that the failure is not expected to occur during unloading between N=0.5 and 1, although it may be nominally regarded at N=1. Also, it should be noted that DABF exists for N<0.5 but does not manifest itself as failure at any peak stress level above the fatigue limit. The next higher number of cycles for FCD corresponding to a peak stress level would occur at N=1.5 (point b in Figure 5.8a and b), and then N=2.5 (point c in Figure 5.8) and so on. Thus, the highest applied stress level a damaged specimen can resist prior to failure, which is a residual strength or stress at

FCD, is equal to the applied peak stress level (σmax) if we use a number of cycles N-0.5. According to the damage axiom ‘(a)’, the relativity in damage quantity on the S-N curve can be found. For example, the damage accumulated (Db) up to N=1.5 is higher than Da at N=0.5, Dc is higher than Db, and so on. Therefore, its damage behaves as a function of discrete variable N-0.5 where N is an integer N≥1.

Incidentally, however, the failure stress levels at a relatively small value of N-0.5 other than integers (N ) but greater than N=1 (e.g., 1.3, 1.4, etc.) would theoretically deviate from the continuous S-N curve depending on the initially set applied peak stress because the failure stress is not equal to the peak stress at N-0. For example, σmax at N=1.3 as given in

Figure 5.8 is lower than σb at N=1.5 and also lower than the expected stress at N=1.3 on S-N curve because of probabilities that the expected stress at the first cycle is not equal to that at the failure (Figure 5.8c).

No practical difference between discrete numbers (e.g., 0.5, 1.5, etc.) and other continuous numbers may be found for a relatively large Nf . Accordingly, most of the S-N curve may be treated as a continuous curve for analytical purposes. At the same time, such an analytical description for a small and quantized N is useful for finding boundary conditions. Now, the two damage values at N=0 and at N=0.5 may be considered at certain stress levels for boundary conditions on the S-N plane for damage at FCD and DABF. The first one (N=0) has been used in the literature for zero damage but it should not be used as a boundary condition because zero damage at N=0 is neither part of a continuous S-N curve nor part of a continuous S-N plane at any stress level except a stress level of zero. However, the damage values at Nf =0.5 is legitimate for boundary conditions if the S-N curve is assumed to be a continuous line at R=0.

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The following equation for damage at failure (Df ) may be generally considered as a candidate reference damage or a damage boundary line: V max Df B C for N ≥ 0.5 (5.12) V a where B and C are constants, and V a (see Figure 5.8) is the breaking stress at N=0.5 cycle V which is similar to the static ultimate stress ( u) i.e. σa≈ σu (ultimate strength). The damage at failure (Df) depends on constants B and C, and should not be zero at a peak stress level for N=0.5 as discussed above. Accordingly, it is required that B>Cσmax/σa or B>C. If we choose C =1, any value greater than 1 for B, Equation (5.12) satisfies Equation (5.11) and therefore valid. If we use relatively large values for B for a boundary condition, however, there will be a difficulty in determining DABF values at different peak stress levels because we do not know yet how DABF varies at N=0.5 for different peak stress values. Therefore, if we adopt a value approaching unity for B, Df at N=0.5 may be used as a value approaching zero, then damage D at N=0.5 at all different stress levels approaches zero as well [e.g. Equation

(5.1) or (5.5)]. The value Df on S-N curve can, thus, be found with B=1. (Note the way of taking the value for B does not violate the relativity of damage quantities.) Therefore, the value of zero for damage at all different peak stress levels at N=0.5 may be used as the quasi-boundary conditions for damage at other points. A reason why it is termed ‘quasi’ is

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Download free eBooks at bookboon.com Click on the ad to read more 110 MECHANICS OF SOLIDS AND FRACTURE S-N Fatigue that (a) the damage may still vary as a function of N but smaller than 0.5cycle so that it is a function of a sub-scale variable different from multiples of 0.5cycle; and (b) damage values at different peak stress levels are not theoretically known but zero is practically acceptable. Equation (5.1) or (5.5) is the one with B=C=1.

5.4.3 DAMAGE IN DOMAIN 3 AND COMPATIBILITY CONDITIONS

The damagecompatibility in Domain 3 (Figure 5.7) is to ensure that no overlaps in damage quantities develop when damage varies as a function of multiple variables such as N and

σmax so that a damage quantity should be a single value for a given damage severity. If a damage function lacks the compatibility, for example, a single damage quantity for different damage severities is produced. The compatibility conditions are essential for finding a valid iso-damage line or an equal damage point that can be practically used for predicting the remaining fatigue life at a changed stress level.

For understanding the characteristics of relative damage quantities, let us consider damage at any point between N=0.5cycle and Nf on the S-N plane, which may be represented by point C as shown in Figure 5.9. The point C is located on the horizontal line DA and intersected by a vertical line BE . Other points such as points A, B, D and E are accordingly dependent on the location of point C. In other words, if point C moves towards point A, points A and B become closer to each other on S-N curve, and simultaneously point E moves along the axis of log N towards S-N curve. The damage at point C (denoted by

DC ) needs to be quantified relative to the reference damage at points on S-N curve for for predicting the remaining fatigue life at a changed stress level. According to the damage axioms, we find that DC is lower than damage at point B (DfB ) because it is subjected to a lower applied peak stress at a given number of cycles. It is also lower than damage at point

A (DfA ) because it is subjected to a lower number of cycles (N ) at a given applied peak stress level. Accordingly, the following damage order is preserved:

DDBC

where DDBC = DfB – DC and DDAC = DfA – DC . Accordingly, we find that 'D BC 1 (5.15) 'DAC

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for point C. Also, damage at any point in hatched area ABC (DABC) and damage at any point in hatched rectangular area CDEO (DCDE0) (Figure 5.9) are found to be

DCDEO ≤ DC ≤ DABC (5.16) and

DD < DC < DB. (5.17)

Also, for iso-damage lines in relation with S-N curve, if damage at point b is equal to that at point B (Figure 5.9), the angle (θb) of line connecting between point B and point b (see Inset in Figure 5.9 is also required to be

1 'V max S tan Tb . (5.18) 'log N f 2

B B D’

T b Vmax A C V D A b maxA V C b H maxH 'dc InsetInse

O E N=0.5 (or log N=-0.3) Log N

Figure 5.9 Schematic representation of S-N plane for fatigue damage characteristics.

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∆ dfBb

DfA A

∆DAC DfB B b Damage (D) DC C ∆DBC 0 0.5 1

df of point C

Figure 5.10 Damage for points b, B and C as a function of general relative location factor

(df ) of a point on line DA on S-N plan (see Figure 5.9). The upper curve represents damage

(D) of point B at df of point C and the lower curve represents damage (D) for points b and C.

To quantify a location at a given σmax on the S-N plane, a general relative location factor

(df ) of point C in Figure 5.9 is defined as, (log N 0.3) C d fC (5.19) (log N fC 0.3)

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where NC is the number of cycles at point C and NfC is the number of cycles at failure at point A (e.g. dfC =0 at point D and dfC =1 at point A). Then, valid relative relationships between the three different points (i.e. B, b, and C) at a given applied peak stress level (σmax) above the fatigue endurance limit may be discussed using a schematic in Figure 5.10. The upper curve in Figure 5.10 represents damage (D) of point B at df of point C in Figure 5.9 and the lower curve in Figure 5.10 represents damage (D) for points b and C at different locations. The upper curve can be found from an S-N curve and Equation (5.3) or (5.7) but the lower curve is not known until a damage function (D) is found. It is known, however, that point b should be on the right side of point B at a given damage in Figure 5.9 and that Point C should be lower than point B at a given df as shown in Figure 5.10. The upper curve would be either concave or convex depending on an S-N curve shape. If S-N curve is linear, it becomes linear. If part of S-N curve is convex, its corresponding part becomes concave, and vice versa. It is noted, if two curves vary smoothly (Figure 5.10), that both

∆DBC and ∆dfBb (=dfb – dfB ) should increase initially and then decrease as point C moves towards point A on S-N plane (Figure 5.10) for any non-linear monotonic variation of damage while ∆DAC always decreases at a given peak stress level. The compatibility between log N and σmax for damage on S-N plane can be achieved when a valid set of points of b for iso-damage lines is found. One of the necessary conditions for the compatibility is that df of point b should be higher than point C (Figure 5.10) or the location of point b on S-N plane is required to be between points C and A (Figure 5.9) according to the axioms. Hence, the general relative location factors at points b and C are required to be

d fb ! d fC for Domain 3 (i.e. 1! d fC ! 0). (5.20)

Also, it may be summarized:

d fb | d fC |1 for Domain 1 (i.e. at failures on S-N curve) (5.21) and

d fb | d fC | 0 at N=0.5 cycle for Domain 2. (5.22)

5.5 A DAMAGE FUNCTION AND VALIDATION A damage function (D) is required to be monotonic over an interval between Domain 1 and Domain 2 at a given peak stress level. Accordingly, the following damage function (D) may be considered:

n D Df d f (5.23a)

Download free eBooks at bookboon.com 114 MECHANICS OF SOLIDS AND FRACTURE S-N Fatigue where exponent n is an unknown constant. Equation (5.23a) satisfies the boundary conditions

(i.e. D=0 at dfC=0 or logN=-0.3, and D= Df at dfC=1 or at logN = logNf). Also, it is noted n1 that n governs the change rate of damage [wD / wN nD f d f ] at a given applied peak stress level such that, unless a transition of damage mechanism occurs for another mechanism, n should be a single value at various applied peak stress levels. The value ofn is greater than unity and depends on non-linearity of S-N curve as described in Figure 5.10. However, it is not known yet whether it satisfies Equation (5.20), which is a necessary condition, for the compatibility in Domain 3, given neither n value nor an S-N curve is known. If a single critical value of n exists for a given S-N plane satisfying Equation (5.20), it would be sufficiently valid because it represents a unique characteristic value for a given S-N curve.

For two stage block loading, let us consider damage on two horizontal lines DA and DcB on S-N plane (Figure 5.9) for finding an equivalent damage point b (Figure 5.4). The upper curve (Figure 5.10) is to represent a damage function (DA) for the points on line DA (Figure 5.9) given by

n DA DfAd fA (5.23b)

and the lower curve (Figure 5.10) is to represent a damage function (DB) for the points on line DcB (Figure 5.9) by

n DB D fB d fB . (5.23c)

The general relative location factor for point b (dfb) whose damage is equal to damage at point B (Figure 5.9) is found to be:

1 § D · n log N 0.3 d ¨ fB ¸ b fb ¨ ¸ (5.24) © D fA ¹ log N fA 0.3 where DfA and DfB are known damage values at two points A and B respectively according to Equation (5.1) or (5.5) (with constants B = C =1).

To find an exponent n in Equation (5.23) for a given S-N curve, typically an n value of 1 may be incremented until it reaches the critical value. If multiple critical values of n exist in an acceptable small range, a highest critical value of n may be taken as an approximately valid one.

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5.6 NUMERICAL DETERMINATION OF EXPONENT N To determine numerically the exponent n value in Equation (5.23) for a given S-N curve, two different stress levels ofσ max may be firstly chosen within the highest and lowest σmax bounds of an S-N curve, and then dfb and dfC are calculated until a positive value of (dfb – dfC) is obtained. A sequence given in Figure 5.11 for calculating (dfb – dfC) may be useful.

The twoσ max bounds should be between tensile (or compressive) strength and assumed fatigue limit (or substantially low σmax). Two different numerical schemes may be possible for calculating (dfb – dfC): (1) the highest σmax (σHmax) is fixed and then lowestσ max (σLmax) is incremented – this process is repeated with the next highest (σHmax(i)) until σHmax(i)= σLmax; and (2) both two adjacent σHmax and σLmax (= σLmax < σHmax) are decremented; or both two adjacent σHmax and σLmax are incremented between the two σmax bounds. The first scheme may be comprehensive but requires many iterations whereas the second scheme is relatively simple and efficient for execution. Thus, the second scheme may be sufficient for most of cases because the angle θb [see Equation (5.18)] tends to be small for a small interval

(=σHmax – σLmax).

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Determine σHmax and σLmaxwith a small interval (=σHmax-σLmax) for a chosen n value

Calculate DfB and DfA for σHmax and σLmax Calculate LogNf for point B at σHmax and point C E V E ª 1 º respectively using max V §V · D fT 1 at σLmax using uT « ¨ max ¸ » V N f 1 ¨ ¸ N0 uT D(E 1) V ¬« © uT ¹ ¼»

Calculate dfC at σLmax using 1 § · n log N fB 0.3 D fB ¨ ¸ d fC where NfB and NfC are Calculate dfb using d fb with ¨ ¸ log N fC 0.3 © D fA ¹ Nf at σHmax and σLmax respectively obtained DfB and DfA

Calculate dfb – dfC

Figure 5.11 A sequence for calculating (dfb – dfC).

5.7 EXAMPLES FOR PREDICTING OF THE REMAINING FATIGUE LIFE Experimental results obtained by Broutman and Sahu (1972) were employed as an example for predicting the remaining fatigue life at a changed stress. Figure 5.12 shows an S-N curve represented by Equation (5.3) fitted to the experimental data for cross-ply glass fibre reinforced plastic laminate with an ultimate strength of 448 MPa and a stress ratio (R) of 0.05 (which is close to zero). Fitting parameters in Equation (5.3) were found to be α=2.8×10-41 and β=14.2. The dashed lines on both sides of the solid line represent experimental failure probabilities 10% and 90%. The experiment for predicting the remaining fatigue life was conducted with two sequential blocks of loading. Figure 1.12 shows an example where a specimen was cycled at a peak stress level 290 MPa for the first block of loading and then changed to 241MPa for the second block of loading. It also shows an equivalent damage point at σmax=241MPa predicted using Equation (5.23) and then a predicted failure point that is close to the S-N curve.

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Figure 5.12 An S-N curve fitted to data of Broutman and Sahu (1972) for cross-plyglass fibre reinforced plastic laminate with an ultimate strength of 448 MPa and a stress ratio (R ) of 0.05. The dotted lines on both sides of the solid line represent failure probabilities 10% and 90%.

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5.8 EFFECT OF MEAN STRESS ON FATIGUE

Stress (σ) Constant stress ratio line σmax Stress (σ)

σmean σmax R= -1

σu Stress (σ) A Constant fatigue life lines σmax C σmax R=0

σmean D σa R=0.5

(σa )o R≈1

B 0 σu σmean

Figure 5.13 Constant fatigue life (CFL) diagram for various stress ratios.

The fatigue life (N ) generally increases as the mean stress (σmean) or stress ratio (R ) increases at a constant stress amplitude (σa). Also, we know that a material breaks when the maximum stress (σmax) reaches the ultimate strength (σu). It is useful to use a σa – σmean plane for a relation between the variables. Let us assume σu is equal to yield stress. On the σa – σmean

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σu. If we connect the two points, a straight line (which is a locus of breaking points at the 1st cycle) is found for a constant life at the same maximum stress but at different stress ratios. The same principle for a constant life may be applied to different stress amplitudes at R=-1. Then, another point C at a lower stress may be found from an experiment for another constant fatigue life (CFL) line CB but for a longer fatigue life. The point B is used for all other stress levels, given that it is a known condition for any stress ratios. If a fatigue limit is available from an S-N curve for R=-1, the fatigue limit line DB may be found. As a result, it is possible from a single S-N curve to predict a series of different values of σmean and σa for different stress ratios for each constant fatigue life. On the other hand, the dash-dot line in Figure 5.13 represents a constant stress ratio for different fatigue lives for a given stress ratio (R), and its slope is

V 1 R a (5.25) V mean 1 R

where R V min /V max (V mean V a ) /(V mean V a ) .

In this way, series of CFL lines and constant stress ratio lines can be found. Note that different CFL types are possible dependent on ultimate tensile strength σ( uT ), ultimate compressive strength (σuC ), and yield strength.

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6 LINEAR ELASTIC STRESS FIELD IN CRACKED BODIES

Most engineering materials contain small cracks or defects produced during service or manufacturing. When an engineering component is fractured, new surfaces are created. They are caused by the rupture of atomic bonds due to high local stresses. The phenomenon of fracture may be approached at different scales. As the crack size decreases, smaller scale analyses would be required. At a small scale for some cases, the phenomena of interest may be considered within distances of the order of 10-7 cm so that the problem is studied using the concepts of uncertainty. However, as the crack size increases, the material behaviour based on continuum mechanics may be more appropriate. The complex nature of cracking behaviour prohibits a unified approach of the problem, and the existing theories deal with the subject from either the microscopic or the macroscopic point of view. In this chapter, the linear elastic stress analysis for cracked bodies will be introduced as part of the continuum mechanics.

Mode I Mode II Mode III

Figure 6.1 The three modes of cracking.

When we consider a two-dimensional crack extending through the thickness of a flat plate, three different cracking modes need to be defined by the loading position and direction. These three basic modes are illustrated in Figure 6.1, which presents three types of relative displacements of the crack upper and lower surfaces. Mode I, mode II and mode III are also called opening mode, shearing mode and tearing mode respectively. Some practical loading examples of testing for such modes are given in Figure 6.2.

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(a) (b) (c)

Pin F

Specimen D E F DF/(D+E) + + Bond line - (E-D)F/(D+E)

Shear force diagram

(e) (d) (f) F

D E F DF/(D+E) + +

- (E-D)F/(D+E)

Shear force diagram

(g)

Figure 6.2 Practical loading examples of testing for different modes: (a) mode I, (b) mode I, (c) mode I, (d) mode II, (e) mode II [after Kim and Ma 1998], (f) mode II, and (g) mode III.

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6.1 COMPLEX STRESS FUNCTION The stress field around a crack tip can be found mathematically using the equations with the Airy’s stress function (F) as discussed previously: w 2) V : (bis 4.18) x wy 2 w 2) V : (bis 4.19) y wx 2 w 2) W . (bis 4.20) xy wxwy

The complexfunction is defined as

Z(z) Re Z i Im Z (6.1) where z = x + iy.

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The Cauchy-Riemann conditions for the complexfunction are given by dZ w Re Z w ImZ Re (6.2) dz wx wy

dZ w Im Z w Re Z § dZ w Im Z w Re Z · Im ¨or Im ¸ w w ¨ w w ¸ (6.3) dz x y © dz x y ¹ . The Airy’s stress function (Westergaard 1939) is given by

) = Re Z y ImZ (6.4)

dZ dZ dZ where Z , Z and Zc . dz dz dz Therefore, the stresses are found:

V x Re Z yImZc (6.5)

V y Re Z y ImZc (6.6)

W xy y Re Zc . (6.7)

y τxy

Vx Vx r τ kV 2v xy θ Vy kV x 2a

V (a) τ

y τxy

Vx Vx

2v r τxy θ Vy x 2a

τ (b)

Figure 6.3 Crack in an infinite plate: (a) mode I and II; and (b) mode III.

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6.2 THE STRESS AROUND A CRACK TIP

Let us consider mode I crack problem when k =1 in Figure 6.3 with a crack length of 2a in an infinite plane under biaxial stresses. The boundary conditions of the problem at infinity and on the crack surface may be stated as:

2 2 Vx = Vy = V and Wxy = 0 for z x iy x y o f

and along the crack face andfor V y 0 andW xy 0 for y=0, -a

The stressfunction for symmetric crack problems satisfying the boundary conditions is Vz Z . (6.8) z 2 a2

The equation is analytic except for –a ≤ x ≤ a at y = 0.

To move the origin of the coordinate system to the crack tip (z = a) from the middle of the crack, z is replaced by z +a:

2 3 Vz V (z a) ° 1 z 1 3 § z · 1 3 5 § z · °½ Z ®1 ¨ ¸ ¨ ¸ ....¾ . (6.9) z 2 a 2 2az ¯° 2 2a 2 4 © 2a ¹ 2 4 6 © 2a ¹ ¿°

For small z , K I V S Z I where K I a . (6.10) 2Sz

Using the polar coordinates with z = r(cosT +i sinT)=reiθ, the stresses near the crack tip are obtained for mode I:

KI T § T 3T · V x cos ¨1sin sin ¸ (1 k)V (6.11a) 2Sr 2 © 2 2 ¹

KI T § T 3T · V y cos ¨1 sin sin ¸ (6.11b) 2Sr 2 © 2 2 ¹

KI T T 3T W xy cos sin cos (6.11c) 2Sr 2 2 2

V z Q V x V y for plain strain (6.11d)

where KI V Sa .

The last term( 1 k)V in the equation for V x is obtained separately for k z1 by the superposition principle.

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Also, the vertical displacement (v) along the crack:

2V 2 22 ½ v 1Q axfor plane strain E ° ¾ (6.12) 2V 22 ° v= ax for plane stress ° E ¿ .

The stresses around the crack tip for Mode II are given by:

K T ª T 3T º V II x sin «2 cos cos » (6.13a) 2Sr 2 ¬ 2 2 ¼

KII T T 3T V y sin cos cos (6.13b) 2Sr 2 2 2

K T ª T 3T º W II xy cos «1 sin sin » (6.13c) 2Sr 2 ¬ 2 2 ¼

V z Q V x V y for plain strain (6.13d)

where KII W Sa .

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The stresses around the crack tip for Mode III are given by:

W T ½ ª xz º K III ªsin 2º « » « »° ¬W yz ¼ 2Sr ¬cosT 2¼¾ (6.14) VVW ° x y xy 0 ¿

where KaIII WS.

The stress intensity factors (KI , KII and KIII ) given above are for an infinity body. Obviously, the finite size of the cracked body is expected to have an influence upon crack tip stress field. Accordingly, the expressions for the stress intensity factor have to be modified to account for this effect. A more general expression for the stress intensityfactor may take the form:

KY VS a (6.15) where Y is1 a factor which accounts for geometric effect andY 1 for an infinite plate. Some authors do not incorporate S in the expression. Mode I is the usual one for fracture toughness tests and a critical value of stress intensity factor (K Ic) determined for this mode would be KYIc VSc a.

6.3 STRESS INTENSITY FACTOR DETERMINATION The stress intensity factors may be determined for various loading cases using the stress intensity factor for a case given in Figure 6.4 with the superposition principle.

1/ 2 P § a2 b2 · ¨ ¸ Z ¨ 2 2 ¸ (6.16) S (z b) © z a ¹ and, accordingly, the stress intensity factors for A and B sides are found (Sih et al 1962) to be

b P B x A

P a a

Figure 6.4 Crack subjected to a point forces P.

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Figure 6.4 shows a plate containing a crack subjected to point forces (P) at crack surfaces, which may resemble a practical case where a crack originates at a bolt or rivet hole under loading. The stressfunction satisfying the boundary conditions is given by

P ab K IA (6.17a) B Sa ab and

P ab K IB (6.17b) B Sa ab where B is the thickness, and (K IA and K IB) denote the stress intensity factors for A and B sides respectively. When b = 0 for a centrally located point force (P), the equations reduce to P K IA K IB . (6.18) B Sa

Equation (6.18) describes that the stress intensity factor decreases for increasing crack size at a constant P. It is therefore possible that a crack can be arrested after some growth when its stress intensity factor falls below a critical value (K Ic).

The superposition principle can be used to calculate the stress intensity factor if the same stress field equations are applicable for mode I cases or mode II cases or mode III cases. However, it is not permitted for a combination of different fracture modes because of different stress fields.

As an example for the calculation of a stress intensity factor, let us consider the case of a crack with an internal pressure. Figure 6.5a shows a plate without a crack under uni-axial tension and hence the stress intensity factor KIa 0 . The stress distribution in Figure 6.5a may be equivalent to a case given in Figure 6.5b where a crack with a length of 2a is made at the centre of the plate and an external stresses (σ) are applied to the crack edges.

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σ σ

σ σ (a) (b)

= +

σ (c) (d)

Figure 6.5 Superposition.

Case (b) in Figure 6.5 is a case where a plate given in Case (c) with a central crack under uni-axial tensile stress (s) is superimposed with a plate given in Case (d) with a crack having uniformly distributed stress (s) along its edges. Accordingly, the stress intensity factor for Case (c) is found

KIc KId KIb 0 orKIc KId V Sa (6.19)

A case where a crack is subjected to an internal pressure P is equivalent to the case in Figure 6.5d except the pressure acting in an opposite direction to s. If the sign of K in Equation (6.19) is reversed, the stress intensity factor for a crack with internal pressure is found to be

KI paS . (6.20)

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Further examples for the determination of stress intensity factor will follow. For cracks emanating from a loaded rivet hole (Figure 6.6), it can now be derived using the superposition principle. The hole is assumed to be small with respect to the crack. The case given in Figure 6.6a is broken up into components (b), (d) and (e). The components(b) and (d) can be obtained first with satisfied equilibrium conditions, and then component(e) is found to take away the stress (σ) and force (P ) used for the equilibrium in (b) and (d) respectively. Accordingly, the stress intensity factor (K Ia) is given by:

KIa KIb KId KIe (6.21)

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Rivet and crack

P P

σ σ W

P=σWB P

= 2a + -

P P

σ σ

(a) (b) (d) (e)

Figure 6.6 Cracks emanating from loaded rivet hole and superposition.

and, given that KKIa Ie : 1 1 VW KIa () KK Ib Id VS a . (6.22) 2 22Sa

The internal pressure (p) [Equation (6.20)] is equivalent to a series of evenly distributed point forces. This allows us to use Equation (6.17) for determining stress intensity factors by integration for various cases. For example, Equation (6.19) can be found by integration:

p a ax ax ½ K I ³ ® ¾dx Sa o ¯ ax ax ¿

a a dx 2p ³ S o a 2 x2 a ª a x º «2p cos 1 » p Sa . S a (6.23) ¬« ¼»0

Thus, the two methods validate each other.

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6.4 STRESS INTENSITY FACTOR WITH CRAZING Crazing is a phenomenon which occurs in polymers when crack-like discontinuities are formed, in which fibrils connect the two faces of the crack. The restraining of the faces may be described by a uniform stress V c over the crack faces (Figure 6.7a) and from Equation (6.19) we have:

KaIcc VS (6.24)

where 2a is the length of the craze and K Ic is the stress intensity factor due to the crazing.

The applied stressσ at infinity also gives rise to a stress intensityfactor (K Icc),

KIcc V Sa (6.25)

so that the net stress intensity factor (K I) is given by

KI KIccKIc VV c Sa . (6.26)

This is illustrated inFigure 6.7 using the superposition principle.

σ σ

σc

= +

σc

σ σ (a) (b) (c)

Figure 6.7 Superposition with crazing: (a) fully crazed to resist applied stress;

(b) sc represents resisting stress by crazing; and (c) a crack subjected to stress, s.

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σ σ σ

σc σc

= + a1 a

σ σ σ (a) (b) (c) (d)

Figure 6.8 Craze development and superposition of stress intensity factors for crazing: (a) fully crazed

to resist applied stress; (b) partial craze breakage; c) sc represents resisting stress due to crazing at the tips; and (d) a crack subjected to stress, s.

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σc σc

Figure 6.9 Stresses in a craze.

As the applied stress (σ) increases, the crack length increases and becomes partially crazed as illustrated in Figure 6.8. The stress intensity factor due to the craze (Figure 6.8c) may be derived:

2 a a V c dx K Ic (6.27) S ³ 2 2 a1 a x i.e.

2 aa1§ · Kc V cos ¨ 1 ¸ . (6.28) IcS © a ¹

The stress intensityfactor (K Icc) due to the applied stress (σ) is the same as before and hence the net value of the stress intensity factor (K I) for the case given in Figure 6.8b or Figure 6.9 is given by: ª º 2 V c 1 a1 K I K IccK Ic V Sa «1 cos » . (6.29) ¬« S V a ¼»

To calculate the craze zone size (rp) or a plastic zone size in the case of metals using the model given in Figure 6.9, we need to find a condition for it. It can be supposed that the crazes grow as the load increases but cease to grow at a stage where no further craze

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stress increases i.e. the craze stress and length remain constant at V c . Therefore, the critical condition is found to be K I 0 . Accordingly, setting Equation (6.29) to zero leads to

V 2 1 a1 cos . (6.30) VSc a

If we consider small values of stress, i.e. V V c and rp (a a1 ) a for approximation, we find,

a SV 1 S 2V 2 1 | cos 1 2 . (6.31) a 2V c 2 4V c

Therefore, the craze size r( p) becomes

VS22a

rp 2 . (5.32) (5.32) 8V c

If we let KaI VS, then,

2 S § KI · rp ¨ ¸ . (6.32) 8 © V c ¹

Figure 6.10 Comparison between true and approximated approaches.

The relationship between applied stress and craze length at the critical condition is shown in Figure 6.10 for both approximated and true values. There is very rapid change in a for s greater than about 0.8sc. Dugdale (1960) conducted an experiment and found that there is a good agreement between experimental data and theory for a steel.

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B a 2c

σ a/c=0.02

a/c=0.1 3.0

2.5

a/c=0.2

2.0

0.5 Me/Q a/c=0.4

1.5

a/c=0.8

1.0

0.5 a/c=1

0 0.2 0.4 0.6 0.8 1.0 σ a/B

(a) (b)

Figure 6.11 (a) A surface crack subject to uniaxial tension and (b) the associated elastic magnification factors on stress intensity. [After Atkins and Mai 1988]

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6.5 SEMI-ELLIPTICAL CRACK The semi-elliptical crack resembles a surface crack occurring in practice (Figure 6.11). A three-dimensional geometry is useful involving crack depth (a), crack length (2c) and thickness of plate (B ) to model a semi-elliptical part-through crack. An empirical stress intensity factor (Newman and Raju 1981) subjected to a remote stress (σ) is given by

K I V Sa M e Q (6.33)

where Me, is called an elastic magnification factor on stress intensity, and Q is an elastic shape factor for an elliptical crack. Me and Q are functions of a B and a c;; plots of the factor Me Q are given in Figure 6.11b. It is found that long, shallow cracks have high

Me Q values increasing with a B whereas short, deep cracks have essentially constant low

Me Q values. Equation (6.33) may be useful in the design of pressure vessels.

6.6 ‘LEAK-BEFORE-BURST’ CRITERION The safety may be one of the most important factors for consideration in the pressure vessel design. There are two different possibilities in pressure vessel failure process. When the fracture toughness of a material chosen is sufficiently high and the growing surface crack reaches the other external surface, the pressure vessel starts to leak before it bursts (Figure 6.12a). However, if the fracture toughness is low and the growing crack reaches its critical value (KIC ) before it leaks, the pressure vessel would burst.

We may consider crack geometry and fracture toughness for ‘leak before burst’ design. One of the conditions to be satisfied for design is

a > B as given in Figure 6.12b. Another condition is that fracture toughness should be sufficiently high. We can find a ‘leak before burst’ criterion from Equation (6.33) satisfying the conditions:

K IC !V SB M e Q . (6.34)

Note that the crack depth (a) is replaced with the thickness (B).

a B Crack

2c Pressure a ≈ B (a) (b)

Figure 6.12 Cross section of pressure vessel: (a) different stages of semi-elliptical crack growth; and (b) assumed crack geometry for ‘leak before burst’.

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6.7 RELATION BETWEEN ENERGY RELEASE RATE G AND KI Consider an infinite plate (for plane stress) with fixed ends containing a crack size a as shown in Figure 6.13. Two different stages are shown – before and after crack length increment over a distance 'a . If we want to close the crack over an infinitesimal distance 'a , the strain energy for the closure ('/) is calculated as:

'a 2 V y v BK '/ 2B ³ dr I 'a (6.35) 0 2 E where B is the thickness. When the crack length (a) increases, the strain energy ('/) will be released and its release rate, GI (strain energy release rate), is defined as, '/ G . (6.36) I B'a

V (applied stress)

y 2V v a2 x2 (plane stress) E

KI T T 3T V y cos (1 sin sin ) 2Sr 2 2 2

x=a-'a+r x r dr

a 'a

Figure 6.13 Before and after crack length increment over a distance ('a).

Combining Equations (6.36) and (6.37) yields:

K 2 G I (plane stress) (6.37) I E and

K 2 G I (1Q 2 ) (plane strain). (6.38) I E

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6.8 FRACTURE CRITERIA FOR MIXED MODE LOADING In mechanics of solids, various criteria are used for yielding, failure, and fracture. The fracture criteria to be introduced here are those involving the stress intensity factor for mode I and mode II.

KIIc = KIc KIIc

Circle 2 2 2 K I + K II = K Ic KIIc

KII Ellipse 2 2 (KI/KIc) +(KII/KIIc) = 1

KIc KI Figure 6.14 Fracture criterion based on the energy balance for combined loading.

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The stress fields under mode I and mode II loading can be characterised by stress intensity factors K I V Sa and KaII WS respectively. When the stress intensity factors increase under mixed mode loading, fracture must be assumed to occur when a certain combination of the two stress intensity factors reaches a critical value.

One of the fracture criteria is based on an energy balance principle. According to the energy conservation, the total energy release rate (Gt) is the sum of individual contributions for I–II mixed mode loading and assumed to be a constant:

Gt GI GII = constant (6.39)

2 2 2 K I 22 where GI 1v KI E (for plane stress), G (for plane stress), GII 1 vKII E (for I E K 2 plane strain) and G II . Alternatively, the fracture condition would be: II E

2 2 KI KII = constant. (6.40)

2 2 According to Equation (6.40), when KII = 0 for mode I cracking, KI KIc = constant, and, 22 when KI = 0 for mode II cracking, KKII IIc . Consequently,

2 2 2 2 KI KII KIc KIIc . (6.41)

This is depicted in Figure 6.14. However, Equation (6.41) is problematic if KIc ≠ KIIc. In practice, unfortunately KIc ≠ KIIc is the case, indicating the energy consumption for creating fracture surfaces under mode I loading is different from that under mode II loading. Also, it is usually observed that crack extension under mode II loading takes place at an angle with respect to the original crack direction. The fracture condition is then empirically modified to satisfy the condition KIc ≠ KIIc:

22 § K · § K · ¨ I ¸ ¨ II ¸ 1. (6.42) © K Ic ¹ © K IIc ¹ As shown in Figure 6.14, it is elliptical.

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y σθ Vr

dr τθr τrθ

τθr τrθ

σr

σθ r

dθ

Crac θ x

Figure 6.15 Stress components in the polar coordinate system.

Another criterion proposed by Erdogan and Sih (1963) is based on the postulation that crack growth occurs in a direction perpendicular to the maximum principal stress to derive the fracture condition under mixed loading.

It is convenient to use the polar coordinate system for analysis (Figure 6.15). The stresses in the polar coordinate system are given by

K TT22K TT V I cos (1 sin )II sin (13 sin ) (6.43a) r 22Sr 22Sr 2 2

K I 3 T K II T 2 T V T cos 3sin cos 2Sr 2 2Sr 2 2 1 T ª T 3 º (6.43b) 2 T cos «K I cos K II sin » 2Sr 2 ¬ 2 2 ¼

K I T 2 T K II T 2 T W rT sin cos cos (1 3sin ) 2Sr 2 2 2Sr 2 2 . 1 T (6.43c) cos >@K I sinT K II 3cosT 1 2 2Sr 2

Note that the stress fields around the crack tip are obtained by superimposing the stress fields from mode I and mode II.

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The cracking angle T( m) with respect to the x-direction can be found if the major principal stress direction is known. The stressV T will be the principal stress (V 1 ) if W rT 0 as we readily find that a mode I crack extends along q = 0. Accordingly, setting Equation (6.43b) to zero:

KKIsinTT m II 3 cos m 10 . (6.44) T T T Equation (6.44) can be rewritten using cos 2 m sin 2 m 1 and cosT 1 2sin 2 m as 2 2 m 2 T T § T T · § T T · 2K sin m cos m 3K ¨cos 2 m sin 2 m ¸ K ¨sin 2 m cos 2 m ¸ 0 (6.45) I 2 2 II © 2 2 ¹ II © 2 2 ¹ which yields a quadratic equation, TT 2Ktan2 m KK tan m 0 . (6.46) II 22I II

Solving this equation, the cracking angle (T m) is found to be 2 § T m · 1 K I 1 § K I · ¨tan ¸ ¨ ¸ 8 (6.47) © 2 ¹1,2 4 K II 4 © K II ¹

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2 § T m · 1 K I 1 § K I · ¨tan ¸ ¨ ¸ 8 . (6.48) © 2 ¹1,2 4 K II 4 © K II ¹ +σ

+τ

KI/KII 1 KII/KI +T

-T 0.5 Pure mode II Pure mode I

+τ 0 -80 -60 -40 -20 0

T m (degrees) +σ Figure 6.16 Sign convention and theoretical crack extension angle according to Equation (6.48)

Accordingly, the principal stress (V 1) is also found as

K I 3 Tm K II Tm 2 Tm V1 V T T Tm cos 3sin cos . (6.49) 2Sr 2 2Sr 2 2

A sign convention and theoretical crack extension angle as a function of KII /KI according to Equation (6.48) are given in Figure 6.16. According to the sign convention, the crack propagation angle under mode II loading is negative.

Figure 6.17 Comparison between theory [Equation (6.51)] and experimental data for PMMA. (1963) (1963) (1963) (1963)

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To find the fracture condition under mixed loading, we postulate that the crack extension takes place if V 1 under mixed loading has the same value as V 1 at fracture under mode I loading. Theprincipal stress for pure mode I at fracture is given by

KIc V 1 ()T 0 (6.50) 2Sr from

KI T § T 3T · V y cos ¨1 sin sin ¸ . (bis 6.11b) 2Sr 2 © 2 2 ¹

The fracture condition under mixed loading, then, is found by equating Equations (6.49) and (6.50): T T T K K cos 3 m 3K sin m cos 2 m . (6.51) Ic I 2 II 2 2

A comparison between theory [Equation (6.51)] and experimental data for PMMA is shown in Figure 6.17. The theoretical prediction appears to be conservative compared to experimental data.

σ'x y x' τ'xy β β τ'xy σ'y y' x cσ 2a cσ

σ'y

τ'xy

σ σ'x (a) (b)

τ'xy β σy'

1 cσ m

σ (c)

Figure 6.18 (a) An infinite plate subjected to remote stressess and cs biaxially. (b) Applied stresses at a different angle to find separate mode I and mode II. (c) Stress components in equilibrium.

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Let us consider an infinite plate containing a crack of length 2a at an angle (E ) to the y direction as shown in Figure 6.18a. It is subjected to stresses V and cV in the y - and x - directions respectively at infinity. We need to find stresses for mode I and mode II to use the fracture criterion under mixed mode loading. To this end, we need to find the stress components defined in Figure 6.18b in relation with those in Figure 6.18a using the equilibrium condition as shown in Figure 6.18c. Consequently, the stress intensity factors for this case are obtained as

KI (1 2)>@c1 c1 cos2E V Sa (6.52) and

c 1 K sin 2EV Sa (6.53) II 2 for mode I and mode II respectively.

Example) A thin walled cylindrical pressure vessel with a large radius of R and a wall thickness of B contains a through-the-thickness crack oriented at an angle b with the circumferential direction as shown in Figure 6.19. Determine the stress intensity factors of the crack when the vessel is subjected to an internal pressure, p. Assume the geometry factor is 1.

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Solution) The hoops h and longitudinal sl stresses for the cylindrical vessel are

pR pR V l and V h respectively. The ratio c=1/2 in Equations (6.52) and (6.53). Thus, the 2B B stress intensity factors due to hoop and longitudinal stresses are given by

pR 2 pR K Is 0.5>@0.51 0.5 1 cos2E Sa = 0.5 >@1sin E Sa (6.54a) B B and

0.51 pR pR KIIs sin2E Sa = 0.5 sin2E Sa . (6.54b) 2 B 2B

The stress intensity factors due to pressures over the cracked surfaces are

KIp p Sa and KIIp 0 . Therefore, the total stress intensity factors by superposition are

ª R º pR 2 E S and E S . K I p« 1 sin 1» a KII sin2 a 6.54c) ¬ 2B ¼ 4B

E E

Vl 2a Vl + p

Figure 6.19 A cylindrical pressure vessel with an inclined ‘through the thickness’ crack and superposition.

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7 PLASTIC DEFORMATION AROUND A CRACK TIP

7.1 ONE-DIMENSIONAL PLASTIC ZONE SIZE ESTIMATION The elastic stress field around the crack tip is very high so that a cracked body is usually accompanied by plastic deformation and non-linear effects. There are, however, cases where the extent of plastic deformation and the non-linear effects are very small compared to the crack size. In such cases, the linear elastic theory is still validly used to address the problem of stress distribution in the cracked body. The elastic stress field solutions discussed in the previous chapter show a stress singularity exists at the tip of a crack i.e. the stress approaches infinity. However, the stress in the vicinity of a crack tip, in reality, is limited to a yield stress when subjected to loading, and deform plastically. A simplistic estimate of the size of the plastic zone can be made, whether in plane strain or in plane stress. Let us consider first a plane stress case for a one-dimensional horizontal extent of plastic zone, which occurs on the surface of a plate.

Vys

θ = 0 x

a rp

σ Figure 7.1 One-dimensional plastic zone with stress

distribution of sy and yield stress (sys).

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The stress distribution of sy on a plate with a yield stress of sys for θ = 0 when subjected to an applied stress (s) is shown in Figure 7.1 according to Equation (6.11b),

KI T § T 3T · V y cos ¨1 sin sin ¸ . (bis 6.11b) 2Sr 2 © 2 2 ¹

One can realise that the stress (sy) cannot increases in a real material beyond the yield stress

(sys). Accordingly, the corresponding distance from the crack tip (rp) to the yield stress (sys) may be used as a simplistic estimate for the plastic zone size. Substituting θ = 0, sy =sys and r rp into Equation (6.11b),

2 2 K I V a rp 2 2 . (7.1a) 2SV ys 2V ys

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σy

A σys B

* r a rp G O

aeff

Figure 7.2 Modified one-dimensional plastic zone size

In this calculation, though, the hatched area in Figure 7.1 is ignored. If we compensate for the loss of the hatched area, the actual plastic zone size must be larger than rp [see Equation (7.1a)]. Such shortcomings may be reduced if the material immediately ahead of the plastic zone (rp) is allowed to carry some more stress by introducing an effective crack size (aeff ) (Irwin 1960) which is longer than the physical crack length (a). To this end, the crack tip position can be shifted for calculation. Then, the effective crack size becomes aeff = a+d where d is the length contributed by the hatched area as shown in Figure 7.2. Accordingly, the plastic zone is calculated by adding l and d together. The distance l is found by replacing a with a+d in calculation using Equation (6.11b):

KI V S (a G ) V y (when T 0) o V ys (7.1b) 2Sr 2Sr or

V a V ys | (7.1c) 2O for d <

V 2a O 2 rp . (7.1d) 2V ys

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The distanced is obtained by equating area A to area B in Figure 7.2:

O O O V Sa K I eff V y dr OV ys dr OV ys dr OV ys ³ ³ 2Sr ³ 2Sr 0 0 0 (7.1e) V (a G ) O 1 OV GV ³ dr ys ys 2 0 r so that, for d <

2 2 V a K I G | 2 2 . (7.1f) 2V ys 2SV ys

Accordingly, it is found that

G rp (7.2)

* and that the modified plastic zone size (rp ) is given by

* rp OG 2rp . (7.3)

* The size of the plastic zone (rp ) calculated according to the second model (Figure 7.2) appears twice as large as the one calculated according to the first model (Figure 7.1).

7.2 TWO DIMENSIONAL SHAPE OF PLASTIC ZONE The two dimensional shape can be obtained by examining the yield condition around the crack tip. Either the Tresca criterion or the Von Mises criterion may be adopted. The Von Mises yield criterion, in terms of the principal stresses, is given by

2 2 2 2 2V ys V1V 2 V 2 V 3 V 3V1 (bis 2.28)

where sys in the uniaxial yield stress.

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The crack tip stress field equations in terms of principal stresses can be found by substituting the following equations [for the case where k =1],

K T § T 3T · V I V x cos ¨1 sin sin ¸ (1 k) (bis 6.11a) 2Sr 2 © 2 2 ¹

K T § T 3T · V I y cos ¨1 sin sin ¸ (bis 6.11b) 2Sr 2 © 2 2 ¹

KI T T 3T W xy cos sin cos 2Sr 2 2 2 (bis 6.11c)

V z Q V x V y for plain strain (bis 6.11d)

into Equation (7.4) for two dimensional principal stresses (s1 and s2),

2 1/ 2 V V ª§V V · º V V x y r ¨ x y ¸ W 2 1 or 2 «¨ ¸ xy » (7.4) 2 ¬«© 2 ¹ ¼»

E @

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KI T § T · V1 cos ¨1sin ¸ (7.5a) 2Sr 2 © 2 ¹ K T § T · V I 2 cos ¨1 sin ¸ (7.5b) 2Sr 2 © 2 ¹

KI T V 3 v V1V 2 2v cos (plane strain) (7.5c) 2Sr 2 or

V 3 0 (plane stress). (7.5d)

The two-dimensional plastic zone (rp) as a function of q can be obtained by substituting Equation (7.5) into distortion energy criterion (or Von Mises criterion) Equation (2.28):

K 2 I 3 2 T 2 T rp 2 >@2 sin 1 2v 1 cos for plane strain (7.6) 4SV ys and

K 2 I 3 2 T T rp 2 >@1 2sin cos for plane stress. (7.7) 4SV ys

Substituting q = 0 in Equation (7.7) for plane stress, we recover the one-dimensional estimate:

2 K I rp 2 . (bis 7.1a) 2SV ys

1 1 Plane stress 2 Rp/(KI/πσ) 0.5

Plane stress Plane strain Plane strain

(a) (b)

Figure 7.3 Plastic zone shapes calculated according to Von Mises and Tresca yield criteria: (a) Von Mises criterion; (b) Tresca criterion.

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The two-dimensional shape can be shown using Equations (7.6) and (7.7) [see Figure 7.3(a)]. It is seen that the plastic zone in plane strain is smaller than that in plane stress.

Similarly, the Tresca yield (or maximum shear stress) criterion may be employed for the two- dimensional plastic zone. As already discussed, the Tresca yield criterion assumes that yielding occurs when the maximum shear stress W max (V1 V 3 )/2 or W max (V1 V 2 )/2 , whichever is the largest, reaches its yielding point. In the case of uni-axial loading, when the maximum principle stress (V 1) reaches its yielding point (V ys ), V 1 V ys , V 2 V 3 0 . Accordingly, the maximum shear stress is found,

V V V 0 V ys W 1 3 1 . (bis 1.6) max 2 2 2

Substituting Equation (7.5) into the maximum shear stress (W max) or the maximum shear stress criterion (or Tresca yield criterion),

2 K 2 ª TT§ ·º r I cos¨1 sin ¸ for plane stress (7.8) p 2 « © ¹» 22SV ys ¬ 2 ¼ and the larger of

2 K 2 T ª T º K 2 I 2 I 2 T for plane strain. (7.9) rp 2 cos «1 2v sin » and rp 2 sin 2SV ys 2 ¬ 2¼ 2SV ys The two-dimensional shape then can be shown by plotting Equations (7.8) and (7.9) as shown non-dimensionally in Figure 7.3b. The difference is seen between Von Mises and Tresca plastic zone shapes and sizes. The Tresca plastic zone size appears slightly larger than Von Mises plastic zone size.

Similar calculations are made for modes II and III and plastic zone shapes based on the Von Mises yield criterion are shown in Figure 7.4.

2 Plane rp /(KI /SV ys ) stress Plane strain Crack Crack Mode III

Mode II

Figure 7.4 Plastic zone shapes based on Von Mises for modes II and III.

Figure 7.5 shows a plastic deformation zone obtained experimentally on a steel using an etching technique. We can find some similarities in plastic zone profile. The etching response

Download free eBooks at bookboon.com 153 MECHANICS OF SOLIDS AND FRACTURE Plastic deformation around a crack tip is sensitive to grain orientations. Nonetheless, it offers a good guide for understanding the simple theoretical calculations. It is noted that the plastic zone size has been shown to be 22 proportional to KI/ V ys regardless of the different calculation methods, which may be a basis for developing a valid practical testing method of fracture toughness.

Crack

Figure 7.5 Plastic zone around a crack tip, boundaries of which were traced out from an experimental plastic deformation image obtained by an etching technique. [Hahn Mincer and Rosenfield 1971]

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7.3 THREE DIMENSIONAL SHAPE OF PLASTIC ZONE The three dimensional plastic zone around a crack tip may be theoretically estimated when the plane strain condition exists for a thick plate. The plane stress condition is, also, applicable depending on how far the location of interest is away from the plate surface. The plane stress exists at the surface of the plate if the surface is free from stresses (V z V 3 0).

In contrast, plane strain prevails in the interior of the plate because the stress s3 gradually increases from zero at the surface towards the middle of the plate. The three dimensional plastic zone is illustrated schematically in Figure 7.6 using the previous calculations based on the Von Mises yield criterion.

Figure 7.6 Three-dimensional plastic zone shape based on the Von-Mises yield criterion.

Thickness

Free yielding Constrained yielding

Figure 7.7 Yielding at different states of stress.

Figure 7.7 shows a cross section schematically of the three dimensional plastic zone shown in Figure 7.6. The plastic deformation at the surface takes place more freely than that in the interior because the plane stress condition is dominant on the surface. Concurrently, the plastic deformation in the interior is much more constrained than that at the surface because the plane strain condition (ez = 0) is dominant in the interior. Therefore, more hydrostatic component than deviatoric stress component prevails internally, resulting in the smaller plastic zone and more brittleness. Such different stress conditions may be shown using the Mohr’s circles and stress elements under mode I loading in Figure 7.8. For q ≈

0, the stresses sy, sx and sz near the crack tip correspond to the principal stresses s1, s2 and s3 respectively according to Equation (7.5). In the case of plane stress, the maximum W o shear stress ( max) occurs at planes inclined at angles of 45 to the directions of s2, and s3 as shown on the stress element in the figure. In the case of plane strain, s1 and s2 have

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the same magnitude as that in plane stress but V 3 v V1 V 2 is acting in the z-direction. The Mohr’s circles represent such a difference between plane stress and plane strain for ν =

0.5. Accordingly, the hydrostatic stress (V m)

I V x V y V z V V V V 1 1 2 3 (bis 1.7) m 3 3 3 is not only higher in plane strain than in plane stress but also the maximum shear stresses W ( max) occurs at planes inclined at angles of 45˚ to s1 and s2 directions (Figure 7.8).

y V1 Plane stress

x V2 V 1 V3=0 z

V2 V2 V 1 V3 Plane strain V2 V3 θ Crack V2 V3=0 Plane stress V1

(a)

Wmax=W2

W3 W1 V V3 V1 V2 Plane stress

Wmax

V V1 2 V3

Plane strain (b)

Figure 7.8 (a) Planes of maximum shear stresses near the crack tip for θ ≈ 0; and (b) Mohr’s circle representation for plane stress and plane strain.

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Such maximum shear stress planes manifest themselves in a form of slip bands as depicted graphically in Figure 7.9. The 45˚ slip bands appear internally at the cross section perpendicular to the specimen surface in the case of plane stress while they appear also, in the case of plane strain, internally on the cross section but parallel with the specimen surface. In the plane strain case, the 45˚ slip bands constantly varies as a function of θ because the principal stress directions for s1 and s2 varies although the principal stress direction for s3 is always in the z direction. Figure 7.10 shows sketches of experimental slip bands with plastic deformation on the specimen surface in plane stress.

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Slip bd

(a) (b)

Figure 7.9 Deformation patterns around the crack tip: (a) 45o shear planes in plane stress; and (b) hinge type deformation in plane strain in the middle section.

Crack

Figure 7.10 Experimental plastic zones in plane stress (Hahn, Mincer and Rosenfield 1971): (a) front surface section, (b) cross section normal to the front and back sections, and (c) back surface section. [Sketches were provided by Haleh Allameh Haery.]

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(a)

(b)

Figure 7.11 Compact tension specimen after fracture: (a) depression along the crack; and (b) flat fracture surface in the middle section and slant fracture along the edges.

Also, the plane stress and plane strain deformations affect the failure mode as shown in Figure 7.11. The slant regions so called ‘shear lips’ are formed on the specimen surfaces along the edges and the flat fracture surfaces are created in the middle section [Figure 7.11(b)]. The shear lips coincide with the 45o shear planes indicating that their formation is associated with the ductile failure mode. However, the flat fracture surfaces do not coincide with the shear planes but appear to be caused directly by the maximum principal stress involving the brittle failure mode.

7.4 PLASTIC CONSTRAINT FACTOR The yielding behaviour in the vicinity of a crack tip is affected by plane thickness. For instance, the plane strain plastic zone is significantly smaller than the plane stress plastic zone. Such a difference is caused by different constraints. A plastic constraint factor (p.c.f.) may be introduced for quantification defined as

V 1 p.c.f. (7.10) V ys

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where V1 is the maximum principal stress. To relate V1 with other principal stress components, let

VV21 n and VV31 m .

From the Von Mises yield criterion,

2 2 2 2 2V ys (V1 V 2 ) (V 2 V 3 ) (V 3 V1 ) (bis 2.28) the following relation is found:

2 22 VV22 >@ 1n nm 12 m 1 ys . (7.11)

Therefore,

1 V max V 1 22 p.c.f. 1 n m n m mn 2 . (7.12) V ysV ys

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From the stress field equations,

KI T § T · V1 cos ¨1sin ¸ (bis 7.5a) 2Sr 2 © 2 ¹

K I T § T · V 2 cos ¨1sin ¸ (bis 7.5b) 2Sr 2 © 2 ¹

KI T V 3 v V1V 2 2v cos (plane strain) (bis 7.5c) 2Sr 2 or

V 3 0 (plane strain) (bis 7.5d) we found,

n 1sinTT 21 sin 2 (7.13) and

m 21QT/( sin / 2 ) (for plane strain) (7.14a)

m 0 (for plane stress). (7.14b)

Accordingly, we found p.c.f = 1 for plane stress when q = 0, and p.c.f. = 3 (and n = 1, m = 2n=0.67) for plane strain when q = 0 and n = 1/3. The maximum stress in plane strain appears as high as three times the uni-axial yield stress.

A comparison of approximate stress distribution between plane stress and plane strain based on the calculations is shown in Figure 7.12. In the case of plane strain, the stress continues to rise beyond V ys until it becomes 3V ys around the crack tip.

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Vys

θ = 0 x

σ (a)

3Vy

2Vy

Vy θ = 0 x

(b)

Figure 7.12 Comparison of approximate stress distribution between plane stress

and plane strain in relation with yield stress (sys): (a) plane stress; and (b) plane strain.

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7.5 THE THICKNESS EFFECT As discussed, the failure mode (e.g. ductile mode) is affected by plastic deformation and hence by the thickness of specimen. Therefore, the fracture behaviour is ultimately affected by the specimen thickness until it reaches a point where the plane stress condition is negligibly small. The transition from plane stress dominant deformation to plane strain dominant deformation is graphically illustrated in Figure 7.13. Figure 7.13a shows a thin specimen with plastic zone shape and size according to the Von Mises yield criterion. As the thickness (B ) increases, the proportion of plastic deformation governed by plane stress is maintained until it reaches a stage where the plastic deformation occurs with plane stress slip planes as shown in Figure 7.13b. Eventually, the thickness reaches another stage (Figure 7.13c) where the plastic deformation occurs with slip planes generated by both plane stress and plane strain.

h o The maximum depth r( p ) of the zone can be shown to be at about 80 and to have a value:

rh 2.59 r p p T 0 (7.15a)

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h rp Bcrit (7.15b)

at the transitional stage [Figure 7.13(b)]. Therefore, the critical thickness (Bcrit) is found: 2 2 2.59 § K · § K · B ¨ I ¸ 0.41¨ I ¸ (7.16) crit S ¨V ¸ ¨V ¸ 2 © ys ¹ © ys ¹ According to the ASTM standard, the minimum specimen thickness requirement for plane strain fracture toughness test is given by

2 § · KIC B t 25. ¨ ¸ . (7.17) © V ys ¹ A higher stress intensity and a lower yield stress give rise to a larger plastic zone. As a result, a larger thickness is required for the plane strain fracture toughness test.

r h Crack ө r h rp (a)

h rp (b

Bcrit

(c)

Figure 7.13 Graphical representation of plastic deformation transition from plane stress to plane strain at the back face of the specimen as the specimen thickness increases: (a) plane stress; (b) at the transition; and (c) plain strain deformation in addition to that of plane stress. The plastic zone shape in ‘(a)’ is based on the Von Mises yield criterion. [After Hahn, Mukherjee and Rosenfield 1971]

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The dependence of K1c on thickness is illustrated given in Figure 7.14. (The critical stress intensity for cracking is usually denoted by Kc, but the notation K1c will be adopted here to indicate mode I cracking for both plane stress and plane strain.) The figure shows also cross sections for shear lips and flat fracture surface regions corresponding to K1c. The curve suggests that, beyond a certain thickness (Bs), a state of plane strain prevails and toughness reaches the plane strain toughness value (KIc) practically independent of thickness for B>

Bs. It also suggests there is an optimum thickness B0 where the toughness reaches its highest level. In the transitional region between B0 and Bs, the toughness has intermediate values.

For thicknesses below B0, it is possible that there is not much material available for the plastic flow before the fracture, resulting in low values ofK 1c as the thickness decreases.

Shear lip

K1c

KIc

B0 Bs

Thickness, B

Figure 7.14 Toughness as a function of thickness and cross sections of specimens with different thicknesses.

7.6 THICKNESS OF ADHESIVE LAYER The adhesion between different components is important for the integrity of engineering structure made of composites. The thickness of adhesive layer significantly affects the fracture toughness (GIc) of adhesively jointed section although, for highly brittle adhesives, this parameter may be not as much significant. Figure 7.15 shows the fracture toughness as a function of adhesive layer thickness for both toughened and un-toughened epoxies.

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GIc (max) 3 Toughened epoxy 2 GIc (kJ/m2) 1 Un-toughened epoxy

1 2 ham Adhesive layer thickness (ha)

Figure 7.15 Adhesive fracture energy (GIc) as a function of thickness (ha) of the adhesive layer for joints consisting of steel bonded with a rubber-toughened or un-toughened epoxy. [After Kinloch and Shaw 1981]

A relatively complex behaviour with toughened adhesives arises from the plastic deformation in the vicinity of the crack tip, which is highly constrained with high modulus and high yield strength substrates such as steel or aluminium alloy. The constraint of adhesive joint may be higher than that of an adhesive without substrates. It may restrict the full volume development of the plastic zone in the adhesive layer ahead of the crack tip (Figure 6.16).

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Since the toughness is largely derived from the energy required for forming the plastic zone, the adhesive fracture energy (GIc) steadily increases as the adhesive layer thickness (ha) increases up to a certain value. The maximum toughness, GIc(max) occurs when the adhesive h layer thickness and the plastic-zone height (rp ), are similar to each other (Figure 7.15). Accordingly, the following equation based on plastic zone size calculation [see Equation

(7.6)] would provide a good guidance for the adhesive thickness (ham) at GIc(max): § · h ¨ 1 EG Ic control ¸ ham| rp | (7.18) ¨ S V 2 ¸ © ys ¹ where E is the elastic modulus and V ys is the yield stress. Table 7.1 lists some experimental results.

Substrate

h Crack r p ham Adhesive layerr

Substrate

Figure 7.16 Elastic-plastic model for plastic deformation zone at a crack tip in the adhesive layer with high yield stress (elastic) substrates.

h Temperature Log (rate GIc GIc(max) ham r p of test) (Control) (Joint) (oC) (m/s) (kJ/m2) (kJ/m2) (mm) (mm)

20 -6.08 2.10 3.90 1.0 0.85

20 -4.78 1.85 3.65 0.8 0.70

20 -3.78 1.55 3.55 0.55 0.49

20 -3.08 1.50 3.15 0.4 0.43

50 -4.66 4.70 2.95 1.1 1.6

37 -4.66 3.75 2.85 0.9 1.16

25 -4.66 2.70 3.85 0.6 0.57

0 -4.66 1.65 3.00 0.5 0.39

-20 -4.66 1.00 3.15 0.25 0.15

-40 -4.66 0.75 2.50 0.1 0.05

Table 7.1 Comparison of measured adhesive layer thickness (ham) at maximum adhesive fracture energy (GIcm) and h calculated plastic zone diameter (rp ). [After Kinloch and Shaw, 1981]

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7.7 EXPERIMENTAL DETERMINATION OF KIC

The experimental determination of plane strain fracture toughness (KIc) is based on the theories discussed up to now to obtain reproducible values of KIc under the conditions of maximum constraint. The plastic zone size in the vicinity of a crack tip must be very small relative to the specimen dimensions. The procedure for measuring KIc has been standardised by the American Society for Testing and Materials (ASTM) to meet the requirements. In this section, the salient points of the ASTM standard test method will be introduced.

7.7.1 TEST SPECIMEN DIMENSIONS

The dimensions of specimens are specified for the minimum thickness (B) for a valid plane strain fracture toughness (KIc) is given by 2 § K · B t25. ¨ Ic ¸ (bis 7.17) ©V ys ¹ and the crack length (a) is given by

2 § K · a t25. ¨ Ic ¸ . (7.19) © V ys ¹ The ASTM E 399 describes many pre-cracked test specimens such asthree-point bend specimen, compact tension specimen, arc-shaped specimen, and disk-shaped compact specimen. The three-point bend specimen and compact tension specimen are shown in Figures 7.17 and 7.18. The stress intensity factor expressions (Strawley 1976) for the standard specimens are:

W a

B S=4W±0.2W

Figure 7.17 Three-point bend specimen.

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.25W±.005W DIA 2 HOLES

.275W .6W±.005W ±.005W

.275W ±.005W .6W±.005W .6W±.005W

a W±.005W

1.25W±.010W B= W ±.010W 2

Figure 7.18 Compact tension specimen.

a 12ª a a § a a2 ·º § · § · 3¨ ¸ « 1. 99¨ 1¸¨ 2 .. 15 3 93 2 . 7 2 ¸» PS © W ¹ ¬ W © W ¹© W W ¹¼ KI 32/ (7.20) BW 32 § a ·§ a · 21¨ 2¸¨ 1 ¸ © W ¹© W ¹

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2 34 § a · ª a § a · § a · § a · º ¨2 ¸ « 0.. 886 4 64 13 . 32¨ ¸ 14 . 72¨ ¸ 5 . 6¨ ¸ » P © W ¹ ¬ W © W ¹ © W ¹ © W ¹ ¼ (7.21) K I 12 32 BW § a · ¨1 ¸ © W ¹ for compact tension specimen. The dimensions a, W and B are shown in Figures 7.17 and 7.18. The P in equations is a measure of the load, and S is the distance between the points of support of the beam in Figure 7.17. Equation (7.20) is accurate within ±0.25 per cent, over the entire range of a/W (a/W < 1). Equation (7.21) is also accurate within ±0.25 per cent for 0.2< a/W < 1.

Fatigue crack W

Chevron notch 1 3 a 2 a a

Figure 7.19 Chevron Notch: (a) Cracked surface with different crack lengths; and (b) side view. (see Figure 7.11)

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100 1/2 MPa m 1c K 50

0 1.5 0.5 1 ρ mm1/2

Figure 7.20 Effect of notch radius (r) on the critical stress intensity factor K1c. [After Irwin 1964]

7.7.2 PRE-CRACK

The pre-crack of test specimen is made of mechanical and fatigue cracks as shown in Figure 7.19. The mechanical crack is first machined for a chevron starter notch and then a fatigue crack follows. There is an advantage of using the chevron notch in that it forces crack initiation into the centre so that a reasonable symmetric crack front is obtained before testing. If the initial machined notch front is straight, the subsequent fatigue crack tends to initiate from a corner. The prepared crack front may be not straight so that an average of three crack lengths is used. One of the crack lengths is measured in the centre of the crack front, and the other two lengths are measured in the midway between the centre a a a and the end of the crack front, giving a 1 2 3 . The reason for using the fatigue crack 3 is that the crack tip radius should be sufficiently small. The effect of the notch radius (r) on the stress intensity factor (K1c) is shown in Figure 7.20. The stress intensity factor (K1c) decreases with decreasing notch radius (r) until a transitional point is reached, and then

Download free eBooks at bookboon.com 171 MECHANICS OF SOLIDS AND FRACTURE Plastic deformation around a crack tip a plateau value is found. The fatigue loading should satisfy some requirements to achieve such a small radius of crack tip and consistent results. The maximum stress intensity factor during fatigue cycling should not exceed 60% of KIc.

Pmax Pmax Pmax = PQ PQ

P5 P5 = PQ P5 P5

Type I P Type II Type III

0 0 0

u (Displacement)

Figure 7.21 Determination of PQ for three types of load-displacement response according to ASTM standards.

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7.7.3 INTERPRETATION OF TEST RECORD AND CALCULATION OF KIC The procedure for conducting the test is straightforward. A typical instrumentation for measurement requires a clip gauge to produce a load-displacement curve. A typical record of load-displacement for metallic materials would look like one of the three curves shown in Figure 7.21. Type I represents nonlinear behaviour involving a large plastic deformation, type III dominantly linear response and type II reflects the phenomenon of ‘pop-in’. From the output record, three values PQ, P5 and Pmax are extracted – PQ is the load for calculation of the fracture toughness [see Equations (7.20) and (7.21)], P5 is the limit of allowable plastic or non-linear deformation, and Pmax is the maximum load. To identify the three different loads, a secant line 0P5 is drawn through the origin with a slope equal to 0.95 of the slope of the tangent to the initial linear part of the record. The load P5 corresponds to the intersection of the secant with the test record. The loadP Q is then determined as follows. If the load at every point on the record between the initial tangent line and a secant line

0P5 is lower than PQ as in Type I, then PQ = P5. If, however, there is a load higher than P5 between the initial tangent line and a secant line 0P5, then PQ is equal to this higher load as in Types II and III. Furthermore, the test is not valid if Pmax/PQ is greater than 1.10, where Pmax is the maximum load the specimen was able to resist. When a test is invalid, it is necessary to use a larger specimen to determine KIc.

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8 CRACK GROWTH BASED ON THE ENERGY BALANCE

The theories of crack growth based on the energy conservation for fracture will be introduced in this chapter. They not only complement various methods based on the linear stress analysis for the fracture toughness determination but also are capable of dealing with non- linear materials behaviour. Also, the equivalence of the energy conservation approach to that based on the linear stress analysis will be found. The energy principles, thus, provides further benefits for understanding the fracture behaviour of materials.

8.1 ENERGY CONSERVATION DURING CRACK GROWTH The fracture process is associated with the energy conservation. The energy is supplied to the structural system by the externally applied load, and is simultaneously consumed when the rupture of atomic bonds of a material takes place for a new crack surface formation, for elastic and plastic deformations, and for kinetic behaviour. Let us consider a cracking body creating a cracked area A [= thickness (B ) × crack length (a)]. According to the law of conservation of energy, we have,

. . .. WK /* (8.1)

. . . where is the work W performed per unit time by the applied load, / and K are the rates of . change for the strain energy and kinetic energy of the body respectively, and * is the energy per unit time for increasing the crack area. (A dot over a letter denotes differentiation with respect to time.)

The strain energy L can be broken up into two parts i.e. one for elastic work and the other for plastic work,

ep // / (8.2) where /e is the elastic strain energy and /p the plastic strain energy.

If the crack grows slowly in a stable manner, the kinetic term K is negligible and can be omitted. Since all the changes with respect to time are caused by change in crack size, we find that

w wA w w A A t 0 (8.3) wt wt wA wA

Download free eBooks at bookboon.com 174 MECHANICS OF SOLIDS AND FRACTURE Crack growth based on the energy balance and Equation (8.1) becomes

Equation (8.4) describes the energy balance during the crack growth. In other words, the work rate supplied to the cracking body by the applied load is balanced with the rate of the elastic strain work, the rate of plastic strain work, and the energy consumption rate for crack surface creation. From Equation (8.4), the potential energy (3) in the system may be defined as

w3/w p w * (8.5) wAAAw w where

3 /e W . (8.6)

Equation (8.5) describes that the rate of potential energy reduction during the crack growth is balanced with the rate of energy consumed for plastic deformation and crack surface creation.

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8.2 GRIFFITH’S APPROACH (1921) The energy consumed for plastic deformation in an ideally brittle material is negligibly small and can be omitted from Equation (8.4). Then, Equation (8.4) is rewritten as

wW w/e w* G . (8.7) wA wA wA

The symbol G is introduced in the equation and represents the crack driving force involving W and /e . Equation (8.7) describes that the crack driving force is balanced with the resistance of the material having a characteristic value of * .

Two limiting loading cases may be considered in practice – one is the constant displacement with varying load and the other is the constant loading with varying displacement. In the wW case of constant displacement, 0 in Equation (8.7). Therefore, we find, wA w/e G . (8.8) wA

Equation (8.8) describes the energy release rate when the energy stored in material is released for crack growth. Hence, the symbol G is usually referred to as the elastic strain energy release rate. In the case of constant loading, the work performed by the constant load is approximately twice the increase of elastic strain energy wW wA 2w/e wA Consequently, Equation (8.7) becomes,

w/e G . (8.9) wA

In this case, the energy required for crack surface creation is supplied by the external load.

Thus, G is found to be independent of the loading method and Equations (8.8) and (8.9) can be put in the form for ideally brittle materials:

w3 G (8.10) wA where the potential energy 3 is defined in Equation (8.6). Also, the energy balance in general may be written as w 3* 0 . (8.11) wA

The elastic strain energy (/e) can be calculated using the stress field and displacement (ν) around a crack tip. Let us consider a line crack of length 2a in an infinite plate subjected to a uniform stress (s), perpendicular to the crack (Figure 8.1). The change in elastic strain energy ('/) due to the crack length increment ('a) is found:

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σ y

∆a a

σ Figure 8.1 An infinite plate with a thickness B( ) and a crack length (a) subjected to a remote stress (s).

'a 2 V y v BSaV '/ 2B ³ dr = 'a for plane stress. (bis 6.36) 0 2 E

According to the following definition for a critical energy release rate (Gc ),

'/ G (8.12) c B'a the critical stress (V c) required for crack growth is given by

EG V c (8.13) c Sa for plane stress, and

EG V c (8.14) c Sa 1 v 2 for plane strain. One of conclusions drawn by Griffith (1921) is as follows:

“The breaking load of thin plate of glass having in it a sufficiently long straight crack normal to the applied stress, is inversely proportional to the square root of the length of the crack. The maximum tensile stress in the corners of the crack is more than ten times as great as the tensile strength of the material, as measured in an ordinary test.”

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8.3 GRAPHICAL REPRESENTATION OF THE ENERGY RELEASE RATE The graphical representation of the energy balance for crack growth is useful for interpretation of experimental results for finding the energy release rate. The load-displacement response of cracked plate as can be obtained from a testing machine will be discussed for three different cases: (a) constant displacement, (b) constant load, and (c) generalised case of changing both the load and the displacement.

a1 A

a2 a2 P

C 0 u

Figure 8.2 Load-displacement response of a cracked plate to a crack length change from

length a1 to a2 under constant displacement.

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8.3.1 CONSTANT DISPLACEMENT CASE The load-displacement response of a cracked plate is represented in Figure 8.2. Two different crack lengths (a1 and a2) are shown for a2 > a1. The straight line OA is a linear response of the cracked plate with a crack length of and the other straight line OB is that with a crack length of a2. It is noted that the cracked plate with a shorter crack is stiffer than that with a longer crack. The magnitudes of strain energy stored at point A and point B are represented by area OAC and area OBC respectively. If the crack length changes from a1 to a2, the load drops from point A to point B and hence the strain energy in the cracked plate is reduced by a magnitude represented by area OAB. Therefore, the elastic energy release rate (G ) equivalent to Equation (8.8) is graphically obtained as:

area 0AB G (8.15) B'a

a1 a2 A a2 B

P E

0 C D u

P, u

Figure 8.3 Load-displacement response of a cracked plate to a crack length change

from a1 to a2 under constant load.

8.3.2 CONSTANT LOAD CASE The load-displacement response of a cracked plate is represented in Figure 8.3. Two different crack lengths (a1 and a2) are again shown for a2 > a1. The straight line 0A is a linear response of the cracked plate with a crack length of a1 and the other straight line 0B is that with a crack length of a2 as before. The strain energy stored in the cracked plate at point A is represented by area 0AC. If the crack length changes from a1 to a2, the displacement (u) increases from point A to point B. At this point, the total energy supplied by the load (P ) is represented by area 0ABD and the strain energy stored in the cracked plate is represented by area 0BD. Thus, the strain energy released from the cracked plate due to the crack length change is represented by

area 0BD – area 0AC

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Also, the total energy lost from the total energy supplied due to the crack length change is represented by area 0AB. However,

area 0AB ≈ area 0BD – area 0AC because area ABE diminishes as the crack length change Da approaches zero. Therefore, the elastic energy release rate (G ) equivalent to Equation (8.9) is graphically obtained as:

area 0AB G . (8.16) B'a

8.3.3 GENERALIZED CASE

The previous two cases are the limiting ones and the crack growths cannot be produced directly by the load (P ) without assistance. The different crack lengths (ai) in the generalised case shown in Figure 8.4 are, however, directly produced by the applied load. The crack length a1 is the initial crack length and a1

area 0Ai1 Ai G (8.17) B ai1 ai with i =1, 2, 3, etc.

a2 A1

a3 A2

A3 a4 P

A4 a5 A5

0 u

Figure 8.4 Load-displacement response of a cracked plate to crack propagation

with crack lengths a1< a2< a3< a4< a5

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In an experimental determination of G, the locations of different crack lengths are recorded on the P-u output and the corresponding radial stiffness lines 0Ai are drawn for finding areas. The linear elastic behaviour of the cracked plate is verified by unloading if P-u follows the radial stiffness lines.

8.3.4 G - a REPRESENTATION

The elastic energy release rate (G ) may be represented as a function of crack length (a) as shown in Figure 8.5. It is given by [see Equation (8.13) and (8.14)]

SaV 2 G k (8.18) E where k =1-n2 for plane strain and k =1 for plane stress. Figure 8.5 shows three different stresses s3>s2>s1 for various crack lengths. According to Equation (8.18) at a given stress s1, G linearly increases with increasing crack length and reaches a critical point for fracture

(G= Gc). At a lower stresses (s2 and s3), though, longer crack lengths are required to reach the same critical point for fracture (G= Gc ).

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V1>V2>V3 V1 V2 V3

a1 a2 a3

Figure 8.5 The elastic energy release rate (G) versus crack length (a) for a crack of length 2a in an infinite plate subjected to a uniform stresss , perpendicular to the crack axis.

8.4 ANALYTICAL APPROACH The analytical approach is based on the energy balance principles. From Equation (8.7) wW w/e w* G (bis 8.7) wA wA wA we find that

Pdu d/ GdA. (8.19)

The Λ without the subscript e denotes the elastic strain energy unless otherwise stated. The Pdu in Equation (8.19) represents dW, the infinitesimal amount of external work done, during the crack area growth (dA), and d/ GdA the internal work done for strain energy and crack growth. Equation (8.19) is the basis for the following derivations.

Figure 8.6 Stain energy in a linear elastic system.

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In a linear elastic system, the strain energy (L) is the triangular area under a given stiffness line (Figure 8.6) so that

1 d/ d 2 Pu . (8.20a)

Thus, from Equation (8.19) for the applied force P and associated displacement u, we find,

1 Pdu 2 Pdu udP GdA (8.20b) or

Pdu udP 2 GdA . (8.20c)

Dividing both sides by P2, we have

d u P 2G (8.21a) dA P 2 or

Equation (8.21) is practically useful for the fracture toughness determination. The derivative d u / P /dA in the equation is the rate of change of (u/P ) with respect to crack area. In other words it represents the slopes of the curve in Figure 8.7 (b). It can be found experimentally using multiple specimens for a series of different crack lengths as shown for sequence in

Figure 8.7. The specific work of fracture (R) is equal to Gc for linear elastic fracture and given by

where Pc is the fracture load at a crack length a6 (Figure 8.7).

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a1 a6

a2 a3 P u/P a5 a4 a4 a3

a5 Fracture point a2 a1

u a

(a) (b)

Fracture point

d(u / P) da

(c)

Figure 8.7 Analysis for experimental results from multiple specimens for different crack lengths: (a) a series of stiffness lines for different crack lengths; (b) compliance (u/P) versus crack length; and then § u · (c) d¨ ¸ / dA versus crack length with a critical value at fracture. © P ¹

More analytical expressions can be derived from Equation (8.19) for quasi static linear elastic cracking. They are

2 2R uc ªd P º « u » (8.22a) « dA » ¬ ¼ P Pc

Pc ªwu º R (8.22b) 2 «wA» ¬ ¼ P Pc

u ªwPº R c (8.22c) 2 «wA» ¬ ¼u uc

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P B

a h

Figure 8.8 DCB Specimen

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If the compliance (u/P) is theoretically known, a single specimen instead of multiple specimens may be sufficient for the fracture toughness determination. For example, the compliance of a double cantilever beam (DCB) specimen shown in Figure 8.8 is determined using the deflection formula for a cantilever beam. The theoretical compliance is given by

u a 3 u 2 (8.23a) P 3EI so that

d u P 2a 2 . (8.23b) dA EIB

Using Equation (8.21c), we find the critical fracture load (Pc ) as

2 REIB Pc . (8.23c) a 2

8.5 NON-LINEAR ELASTIC BEHAVIOUR The non-linear elastic behaviour may be analyzed using the same energy principles. The symbol J is commonly used for non-linear rate of change of potential energy with respect to crack area to be distinguished from G (strain energy release rate) for linear behaviour. When quasi-static fracture occurs, J or G has the critical value JC or GC which exactly matches the specific work of fracture given byR . Accordingly,

P /

Figure 8.9 The strain energy (L) and complementary strain energy (W).

Pdu d/ GdA (bis 8.19)

Download free eBooks at bookboon.com 186 MECHANICS OF SOLIDS AND FRACTURE Crack growth based on the energy balance is rewritten as

Pdu d/ JdA (8.24) for non-linear elastic cracked bodies. The strain energy L( ) for a non-linear elastic cracked body (Figure 8.9) is given by

/ ³ Pdu (8.25) and the complementary strain energy (W) is given by

: ³ udP . (8.26)

From : Pu / and Equation (8.24), we have

udP d: JdA (8.27)

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a1< …< a6 a1

a2 a3

P a5

u1 u2 u3 u4 u5 u (a)

u3 u4 u2 Fracture point a5 u5 Jc u1 / a4 J a3 a 2 a a1 6 a5 a4 a3 a2

u1 u u3 a1 a2 a3 a4 a5 a6 2 u4 u5 a u

(b) (c)

Figure 8.10 Analysis for experimental results from multiple specimens for different crack lengths: (a) a series of curves for strain energy from different crack lengths; (b) strain energy versus crack

length; and then (c) J versus displacement (u) with a critical value at fracture (Jc).

Equation (8.28) form the basis for the experimental determination of J and Jc. The quantity of w/ /wA in Equation (8.28b) is the rate of change of strain energy with respect to crack area (A=Ba). Accordingly, the strain energy (Λ) is obtained from P-u curves (Figure 8.10a) for a constant displacement (u) to construct a strain energy versus crack length (a) diagram shown in Figure 8.10b. Then, the values for w/ /wa is plotted as shown in Figure 8.10c.

The value ofJ c corresponds to the point of fracture.

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If the non-linear elastic behaviour can theoretically be characterised by a power relation

un Cn (8.29a) P where n is a constant, the strain energy (Λ) is given by U n / ³³Pdu du (8.29b) Cn

Also, from Equation (8.24), we find,

1 ª du dPº J «nP u » (8.30) 1 n ¬ dA dA¼ and we have J in terms of Cn, 1 dC 1 u1n dC 1n n 1 1n n n J P Cn 2 . (8.31) 1 n dA 1 n Cn dA

For n = 1, we recover the linear elastic parameter (G),

8.6 CRACK GROWTH RESISTANCE CURVE (R-CURVE) It is possible that, as the crack grows under loading, much more energy is consumed in the plastic deformation under plane stress than under plain strain condition. Also, such a plastic deformation takes place as the crack grows. The crack resistance (R ) curve is useful for describing the crack growth involving a relatively large plastic deformation. The theoretical basis is found from Equation (8.5) and R is defined as w*/w p R . (8.32) wAAw

The crack resistance (R ) consists of the energy consumption rate for crack surface creation and the rate of plastic strain work as previously discussed.

In the case of plane strain or small plastic deformation,

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and Gc or R is a constant as shown in Figure 8.11a. Otherwise, R increases non-linearly as shown in Figure 8.11b. The R-curve is known as a unique property independent of the initial crack size and the geometry of the specimen.

Gc or R

G R

a a

(b) (b)

Figure 8.11 Typical load (P)-crack length ( a) curves for: (a) plane strain and (b) plane stress.

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8.7 R-CURVE AND STABILITY

The rate at which strain energy may be released depends on the geometry of a cracked body and the conditions of loading. Figure 8.12 shows an R-curve and a set of radial lines of 2 2 § wG · 1§SV · SV a slope ¨ ¸ = tan ¨ ¸ for the geometry of a small crack in a large sheet for which G © wa ¹V © E ¹ E is applicable. (The radial lines for some cracked body geometries would not necessarily be straight if a geometry factor is considered.) The slope increases as the applied stress σ( ) increases. As the stress increases from zero, the available G increases from point A to point B without the crack growth. The point B represents the minimum fracture toughness of the material before any subsequent increase in R along the R-curve. Cracking can thus commence at point B stably. We may compare the set of radial lines of slopes represented § wG · by ¨ ¸ with the other set of slopes represented by dR which is independent of the initial © wa ¹V da crack length (a0). At point B, and we see that dR § wG · !¨ ¸ . (8.34a) da wa © ¹V V B

§ wG · As the R increases, the slope ¨ ¸ also increases. As the crack further grows quasi-statically © wa ¹V under increasing load, point C is reached. At point C, we find that

The crack growth between point B and point C is stable because of the balance between energy consumption rate and energy supply rate. However, beyond point C along the R-curve, we find that

dR da ww G a V (8.34c) representing an instability condition because of the higher energy supply rate than the energy consumption rate. The part of the R curve beyond C at which σ = σc and a = ac is not observable with the given geometry unless a longer initial crack (a0) or a stable geometry is used.

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at σc

C R curve G orJ

a0 ac a

Figure 8.12 G versus a curve for constant applied stress σ ; R versus a curve is also superimposed.

8.8 GEOMETRIC STABILITY FACTORS IN ELASTIC FRACTURE The cracking stability is also dependent on the testing machine type because the testing machine itself stores the strain energy and/or the energy balance is not exactly maintained unless a special control circuitry is used for controlling a crosshead. In practice, there are two typical types of testing machines viz displacement controlled machines and load controlled machines (Figure 8.13). The increment of the crosshead (du) in a displacement controlled machine is always positive because it does not reverse the loading direction during testing. In a load controlled machine, on the other hand, the increment of load (dP ) is always positive.

Crosshead

(a) (b)

Figure 8.13 Testing machines: (a) Displacement controlled; and (b) load controlled.

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We may consider the following equation for a linear elastic cracked plate,

Since R may vary during the crack propagation, the variation of R with respect to incremental input P and output A (thickness × crack length a) is obtained by differentiating Equation (8.21c) to have

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For stability in a load controlled testing machine, the condition dP > 0 applies so that

d 2 u P d 2 P u d P u 1 dR 2 2 t dA dA 2 dA . (8.36) R dA d u P d P u P u dA dA

Similarly, from Equation (8.22a)

d 2 P u d 2 u P d u P 1 dR dA2 dA2 dA t 2 . (8.37) R dA d P u d u P u P dA dA

In the equation, d P u is negative if the stiffness decreases with increasing crack length. dA 1 dR In the right hand sides of Inequalities (8.36) and (8.37), R dA is called the geometry stability factor (GSF ) of a test specimen. It can be calculated for the stability. For example, P u 32 EI a 3 for a DCB specimen is calculated to have d dA P u 9EI 2a 4 B and d 2 dA2 P u 36EI 2a5B2 to give

14dR ! R dA A for stability in a displacement controlled testing machine. If R is constant in this condition, GSF becomes zero and accordingly the stability condition becomes 0 > -4/A. Therefore, the DCB specimen satisfies the stability condition. If a test specimen does not satisfy the stability condition, instability of cracking is expected. Sometimes it may be possible that test specimens with satisfied stability conditions have instabilities if the crack front is blunt.

2 Also, the GSF can be calculated using a stress intensity factor relation with G (GI KI E ) for plane stress. From Equation (8.21b), we obtain § EP 2 · d u P 2 ¨ ¸ 2V 2S K I EG I ¨ ¸ Y a (8.38) © 2B ¹ da

Download free eBooks at bookboon.com 194 MECHANICS OF SOLIDS AND FRACTURE Crack growth based on the energy balance where Y is the geometry factor. Accordingly, for displacement controlled machine

8.9 TESTING MACHINE STIFFNESS The total deflection (u *) of the system including test specimen and testing machine is given by

u* uCP (8.40) where C is the compliance of testing machine. It is found that § u *· § u · d¨ ¸ d¨ ¸ © P ¹ © P ¹ (8.41a) dA dA and § u *· § u · d 2 ¨ ¸ d 2 ¨ ¸ © P ¹ © P ¹ (8.41b) . dA2 dA2

Therefore, it follows from Equation (8.36) that the stability of test specimen in the load controlled machine is unaffected by the flexibility of the testing machine. However, the situation is different under du ! 0 for displacement controlled machine, as Puz Pu*. Now, we have for du ! 0 , Pu Pu* (8.41c) 1 CPu so that d P u d P u * dA (8.41d) dA >@1C P u 2 and

2 d 2 P u ªd P u º >@1C P u 2C d 2 P u * dA2 ¬« dA ¼» (8.41e) . dA2 >@1C P u 3

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Therefore,

d 2 P u d P u 2C 1 dR 2 ! dA dA . (8.42) R dA d P u 1C P u dA

Inequality (8.42) may be compared with Inequality (8.37) for testing machine stiffness effect on the stability. For increasing crack length, d P u is negative and hence the value dA of right hand side of the Inequality (8.42) increases due to the additional term. Therefore, the stability is decreased by the flexibility of the testing machine.

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G-locus for infinitely stiff testing machine G-locus for displacement controlled testing machine with finite compliance P A Minimum crack displacement for B stability

Figure 8.14 Effect of machine stiffness on load -deflection diagram and onR -locus.

The testing machine stiffness effect is illustrated in load (P )-displacement (u) diagram given in Figure 8.14. Two G-loci are shown: one for infinitely stiff testing and the other for displacement controlled testing machine with a finite compliance. Points A and B indicate the transition between crack lengths for stability in infinitely stiff testing machine and displacement controlled testing machine respectively. They also indicate the minimum crack displacements. Point B lies on a longer crack length with a lower load for transition than point A. It is noted that, for a constant load, the energy stored in a testing machine with a finite compliance is larger than that with infinitely stiff testing machine.

8.10 ESSENTIAL WORK OF ENERGY Resistance to tear is one of important mechanical properties of flexible materials such as thin polymer sheets, rubbers, etc. The trousers test under mode III loading has drawn attention for material evaluation since Rivlin and Thomas (1953) considered trouser tear criterion for rubbers. Joe and Kim (1990) analysed the load-displacement records to determine the critical J-integral (or just J) value and crack resistance (R ). The determination of the critical J value, however, requires the detection of the crack initiation which is not an easy task for highly deformable materials. Alternatively, the resistance to tear may be evaluated using the essential work of fracture (EWF) approach which was first developed for mode I fracture of ductile metals (Cotterell and Reddel 1977). The EWF approach was further developed by Mai and Cotterell (1984) for elasto-plastic fracture of thin metal sheets under mode III loading using trouser specimens with various widths, taking into account the work done in plastic

Download free eBooks at bookboon.com 197 MECHANICS OF SOLIDS AND FRACTURE Crack growth based on the energy balance bending and unbending of the trousers. Muscat-Fenech and Atkins (1994) expanded this Mai and Cotterell’s work for a wide range of specimen dimensions and geometric change. For tearing of polymer sheets, however, the work for plastic bending and un-bending of the trousers is negligible due to their low stiffness, and deformation reflected in the model for metals cannot be translated into that for the tearing of thin polymer sheets.

Zone I

Zone V Zone S Zone I

Zone V h hs

Ligament li length Whitened zone Li lv ls Lv Ls Zone S l

40mm

(a) (b)

1mm

(c)

Figure 8.15 (a) Configuration of trousers test. (b) Plastic zone model consisting of zones I, V and S for tear specimens with sufficiently long ligaments. (c) Polarised light microscopic image for plastic deformation in PET (0.25 mm thick) near the crack tip showing zones I and V. [After Kim and Karger-Kocsis 2004]

Wong et al (2003) proposed a two-zone model for deformation and tearing behaviour of thin polymer sheets under mode III loading. In the first zone, which is called zone A in their paper and is adjoining the initial crack tip, the outer plastic zone height lineally increases with the torn ligament and thus the zone is of triangular shape. At the end of zone A, the deformation enters zone B. The height of the zone B remains constant with further increase of torn ligament length. The zone A, though, did not consider plastic deformation caused by loading prior to tearing (which is referred to as initial plastic zone in the paper). The two-zone model hence would lead to overestimation of EWF if tearing is the case where increasing height of the initial plastic zone does not coincide with that of

Download free eBooks at bookboon.com 198 MECHANICS OF SOLIDS AND FRACTURE Crack growth based on the energy balance subsequently following zone B. In the light of the deficiency, Kim and Karger-Kocsis (2004) developed a three-zone model for tear fracture under mode III loading to include the initial plastic deformation and analysis based on the EWF approach for prediction of overall tear resistance. In tearing of thin ductile polymeric sheets, mode III fracture mode is expected at the beginning of loading but the mode tends to be mode I later so that tearing becomes virtually mixed mode. In this section, the three-zone model for tear fracture under mode III loading will be introduced.

As schematically indicated for a specimen with sufficiently long ligament in Figure 8.15 (if the ligament is not long enough, only partial deformation occurs), generally two different types of deformation appear along the torn ligament after tearing. One is plastic deformation and the other characterised by whitening. The whitening, accompanied by a weak change of transparency, is found in some polymers and is of non-plastic deformation as seen in a polarized image along the torn ligament – note photo-elastic fringe pattern does not appear in plastic deformation. Also it did not cause any visible change in the surface texture of the specimens. This indicates that its deformation energy would be small compared to the plastic deformation which requires relatively large deformation energy as expected.

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Therefore, the whitened zone is not included in the model formulation despite its considerable size. The plastic zone is found to have three distinctive zones as detailed in Figure 8.15. The three zones will be referred to as: zone I (initial), zone V (v-shape) and zone S (saturation). Zone I is initially formed as loading increases prior to the crack propagation. After zone I fully developed, the following events took place sequentially to form zone V: (a) change of specimen configuration as a result of further lining up of the trouser legs with increasing load; (b) change in the fracture mode from mode III to mode I to include more mode I component due to rotation of the area around the crack tip; and (c) gradual increase in duration of straining applied to the plastic zone around the crack tip as a result of increase in plastic deformation size as the tear progresses at an almost constant speed (e.g. material at l = Lv is subjected longer straining time than that at l = Li because material at l = Lv is strained and plastically deformed more ahead in time before the crack tip arrives at it than material at l = Li). Thus, the zone V is the result of an evolutionary process before it saturates. The height (h) of zone V continues to extend as the crack propagates until it reaches zone S where the plastic deformation is stabilized and thus the height (h) becomes constant. Based on the deformation described above, the following analysis is given. The total work of fracture (Wf ) for a pre-cracked test specimen can be generally divided into two components:

Wf = We + Wp (8.43)

where We is the energy for yielding and tearing of the inner fracture process zone, which is referred to as the essential work, and Wp is the work for the outer plastic deformation zone which is geometry dependant, non-essential work.

For 0 < li < Li,

the total work of fracture (Wif ) for a specimen with a ligament length of li is given for zone I as

Wf = Wif = Wie + Wip = wielit + wipAit (8.44a)

where subscript ‘i’ indicates zone I, wie is the specific EWF, wip is the specific non-EWF, t is the thickness and Ai is the outer plastic deformation zone area given by

2 Ai = lih/2 = li . (8.44b)

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This equation considers the fact that the profile of the initial plastic zone is inclined approximately 45° to the tear path. Although different materials might have different angles, it appears reasonable for approximation. Thus,

2 Wif = wielit+ wip li t (8.44c) or

Wif wf = wif = wie wipli (8.44d) lit

where wif is the specific total work of fracture. In practice, it is difficult to use this equation for determination of wie because li is often too small to be varied in test specimens. The total work of fracture (Wvf ) for a specimen with a ligament length of (Li + lv) can similarly be written for zone V as

Wf = Wvf = Wve + Wvp = wve(Li + lv)t + wvpAvt (8.45a)

for 0 < lv < Lv where subscript ‘v’ indicates zone V and the outer plastic zone area (Av) is obtained as

2 2 Av Li 2Lilv Dlv (8.45b) where a is the taper angle given by

hs 2Li D (8.45c) Lv where hs and Li would be directly measured from specimens after testing or Lv can be estimated from a plot of specific total work of fracture versus ligament length of specimen if it is difficult to be identified under a microscope. The specific total work of fracture (wvf ) in this case becomes

Wvf w f wvf (Li lv )t 2 2 (8.45d) wvp (Li 2Lilv Dlv ) wve Li lv

where wvp and wve can be found by the linear regression analysis using a plot of wvf versus 2 2 ( Li 2Lilv Dlv )/( Li lv ) . For l = Li or lv = 0, this equation becomes

wvf wve wvp Li . (8.45e)

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(a)

(b) (c)

Figure 8.16 PET with a thickness of 2.5 mm for essential work of fracture: (a) load-displacement curves; (b) specific total work of fracture as a function of torn ligament length obtained from both experiments and predictions based on Equations (8.44d), (8.45e) and (8.46b); and (c) linear plots 2 2 with the least square lines for data shown in ‘(a)’ using L* which is (Li +2Lilv+αlv )/(Li+lv) for V 2 2 zone data or (Li +2LiLv+αLv +hsls)/(Li+Lv+ls) for S zone data. [After Kim and Karger-Kocsis 2004]

The total work of fracture (Wvf ) for a specimen with a ligament length of (Li + Lv + ls) can similarly be given for zone S as

2 2 As= Li + 2Li Lv + aLv + hs ls (8.46a) so that

W 2 2 sf Li 2Li Lv DLv hsls wf =wsf = = wse + wsp (8.46b) ( Li Lv ls )t Li Lv ls

where wsp and wve can be found by linear regression analysis using experimental data from 2 2 Li 2Li Lv DLv hsls specimens with ligament length longer than (Li + Lv) for a plot of wsf versus Li Lv ls but preferably the data can be combined with those for 0 < lv < Lv if a full range of data is used. The intercepts and slopes can be used for we (=wse = wve) and wp (=wsp = wvp) respectively in a linear plot (see Figure 8.16).

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8.11 IMPACT FRACTURE TOUGHNESS Structural materials used for functioning components are very likely subjected to impact loading. Accordingly, impact strength is often the deciding factor in materials selection for such an application. The impact test methods in general may fall into two categories according to the relative amounts of energy between striker and specimen viz: (a) limiting energy methods, in which the striker energy is adjusted until a set damage to specimen is found; and (b) excess energy methods, in which the kinetic energy of the striker is always greater than the energy required to break the specimens. The falling weight test falls into the first category, and the Charpy, Izod and tensile impact tests typically fall into the second.

The conventional test methods have the advantages of being easily and rapidly performed. However, their results are dependent on the notch size. The problem of specimen geometry dependence can be approached in different ways based on the fracture mechanics. One of the ways is to obtain force (P ) – displacement (u) curves from a single specimen test with instrumented striker. The other way is to make variations in notch depth with a sharp radius for multiple specimens.

In the Charpy test (Figure 8.17), a bar specimen is placed on horizontal supports attached to up right pillars for central striking. The impact energy is an amount lost from the kinetic

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Download free eBooks at bookboon.com Click on the ad to read more 203 MECHANICS OF SOLIDS AND FRACTURE Crack growth based on the energy balance energy of striker for breaking a specimen. The energy measurement in the Izod test is based on the same principle as for the Charpy test. The difference between two tests is that the lower half of a specimen in the Izod test is clamped for cantilever loading. The clamping, though, generates the complex stress field around a notch tip, making it difficult for analysis. In this section, a theory and method based on fracture mechanics for the Charpy impact test will be introduced.

Pendulum

(a) (b)

Figure 8.17 Impact test type: (a) Charpy test; and (b) Izod test.

The impact specimen breaks and flies away after being struck by the striker, involving kinetic energy of specimen. The impact energy (LE) measured is hence the sum of elastic strain e energy (L ) and kinetic energy of specimen (Ks), i.e.

e LE =L + Ks. (8.47)

P/2

a S P

P/2 B W

Figure 8.18 Impact specimen with a crack length a.

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The elastic strain energy L( e) is given by

Pu P2 u /e . (8.48) 2 2 P

On the other hand, from the elementary beam theory with the section modulus (Z ) and bending moment (M ),

P S M V 2 2 2 (8.49a) Z BW 6

We find that

2 VBW 2 P . (8.49b) 3 S

We know

K 2 Y 2V 2Sa 9S P 2 S 2 a G I (1Q 2 ) (1Q 2 ) Y 2 (1Q 2 ) (8.49c) E E 4 B 2W 4 E where Y is a geometry factor. For the three point bend specimen with S=4W:

In instrumented tests, the peak force Pc is measured and GIc is found directly from Equation (8.49c). Otherwise, for non-instrumented tests with varying notch depth (Figure 8.18), we obtain from

§ u · d¨ ¸ P 2 © P ¹ (bis 8.21b) G , 2 dA that § u · d¨ ¸ P 2G 9 S 2Y 2Sa © ¹ Ic Q 2 (8.49d) 2 2 4 (1 ) dA Pc 2 B W E and

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To determine the integration constant, we use the deflection formula of a simply supported beam for a = 0,

u 1 S 3 C . (8.49f) P 4 EBW 3

Substituting Equations (8.49b) and (8.49e) into Equation (8.48), we find a practical formula for an impact test:

/ E BWIGIC K s (8.50a) where the factor f can be obtained experimentally or can be calculated from

SW S ³Y 2 ada I 18 . (8.50b) Y 2 aW

The specific energy release rate (GIc) is obtained from the slope for a linear regression line as shown in Figure 8.19.

0.20

0.15

0.10

Impact energy (Joule) 9.5 9.5 45 E × × / specimen 0.05

20 60 100 140 BWI (m2) ×10-7

Figure 8.19 Measured energy versus BWf. [After Marshall et al 1973]

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9 SINGLE CRACK APPROACH FOR FATIGUE

The S-N curve approach in fatigue does not account for the details of a crack although it is useful to deal with a case where the failure is caused by the multiple cracks. A single crack approach provides another aspect of fundamental understanding of the fatigue phenomenon by modelling the fatigue crack initiation and propagation processes. The fatigue initiation may be analysed at a smaller scale while the fatigue crack propagation at a larger scale. When a component is subjected to cyclic loading, energy is consumed in the neighbourhood of inherent small defects, which grow and coalesce, for forming a crack to be large enough to be analysed by the principles of continuum mechanics. The crack propagation leading to the catastrophic failure is more predictable than the initiation of a fatigue crack.

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Fatigue life

af ) a

Useful life

au Crack length ( length Crack

ai al Limit of non-destructive detection

N (number of cycles)

Figure 9.1 Typical form of crack size versus number of cycles curve for constant amplitude loading

∆K Crack K

Kmax Kmean Kmin

Time Loading Figure 9.2 A sinusoidal load with a constant amplitude and frequency for stress intensity factor (K).

Figure 9.1 illustrates some characteristic crack lengths (a) dependant on number of loading cycles (N ). There are relative four different crack lengths. The smallest crack length (ai) represents the one that is big enough for fracture mechanics to apply but too small to be detected by the non-destructive inspection technology until it grows to al. The crack length further grows to reach the limit of useful life (au) before the catastrophic failure takes place (af ).

Fatigue crack propagation data at a stress ratio (R= Kmin/Kmax = σmin/σmax) are obtained from experiments on pre-cracked specimens subjected to cyclic loading, and the change in crack length (a) is recorded as a function of loading cycles (N). The crack growth rates (da/dN ) are then numerically calculated for corresponding stress intensity factor ranges (∆K ) from

Download free eBooks at bookboon.com 208 MECHANICS OF SOLIDS AND FRACTURE Single crack approach for fatigue the raw data. The experimental results are usually plotted in a log (DK ) versus log (da/dN ) diagram. The load is usually sinusoidal with constant amplitude and frequency (Figure 9.2).

A typical plot of a log (DK ) – log (da/dN ) curve is shown in Figure 9.3. Three characteristic Stages may be identified. In Stage I, da/dN diminishes rapidly to a vanishingly small level, and for some materials there might be a threshold of the stress intensity factor range (DKth), below which no crack propagation takes place. In Stage II, a linear log (DK ) – log (da/dN ) relation is usually found. As da/dN further increases, it reaches Stage III in which the crack growth rate (da/dN ) curve rapidly rises and the maximum stress intensity factor (Kmax) in the fatigue cycle becomes equal to the critical stress intensity factor (Kc) leading to catastrophic failure. Experimental results indicate that the fatigue crack growth rate curve depends on the stress ratio (R ), and is shifted towards higher da/dN values as R increases. The Stage I has been known to be sensitive to variations of mean stress, microstructure and environment as expected at low stress intensity factor values and extremely slow crack growths. Stage II is not as sensitive as Stage III to mean stress and specimen thickness because of relatively small plastic zone sizes at low stress intensity factors compared to those in Stage III. Also, it represents a wide range of DK.

Stage I Stage II Stage III Large influence of: Little influence of: (a) Mean stress Large influence of: (a) Mean stress (b) Thickness (a) Mean stress (b) Thickness (b) Microstructure (c) Environment Catastrophic Failure da/dN m/cycle) Log ( μ Threshold

Log ∆K

Figure 9.3 Typical form of the fatigue crack growth rate curve.

One of the most widely used fatigue crack propagation empirical models for Stage II is proposed by Paris and Erdogan (1963), which is represented by

da m CK ' (9.1) dN

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where 'K Kmax Kmin , and C and m are constants for materials. Equation (9.1) indicates a linear relationship between log (DK ) and log (da/dN ). The constants C and m may be experimentally determined for the effects of mean stress, frequency, and temperature variation. Equation (9.1) does not, however, describe the crack growth rates in Stages I and III. At high DK values in Stage III, as Kmax approaches the critical level Kc, the crack growth rate approaches infinity. Stages II and III can be represented by a modification of the Paris and Erdogan Equation, i.e.

da C('K)n C('K)n

n n dN 1 (K max / Kc ) 'K ½ (9.2) 1 ® ¾ ¯Kc 1 R ¿

where RK min K max and C and n are material constants. The fatigue crack growth rate

(da/dN) in Equation (9.2) approaches infinity if Kmax Kc , satisfying the requirement of the curve.

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9.1 TEMPERATURE AND FREQUENCY EFFECTS ON FATIGUE CRACK GROWTH Materials such as polymers are readily influenced by temperature variation. The fatigue crack growth rate (da/dN ) generally increases with increasing temperature although some materials display a different response. Arrhenius (Glasstone et al 1941) proposed an expression to account for the influence of temperature on the rate (k ) of inversion of sucrose: § 'H · k A1 exp ¨ ¸ (9.3a) © RT ¹ where A1 is a quantity independent of, or varies relatively little, with temperature, ∆H is the activation energy (kJ/mol), R(= 8.31J/mol K) is the universal gas constant and T is the absolute temperature (K).

Krausz and Krausz (1984) related the rate constant (k ) to a crack velocity based on an atomistic model as

da da dt da 1 (9.3d) dN dt dN dt f where f is the cyclic load frequency. The fatigue crack process is affected not only by temperature but also by the stress intensity in the vicinity of a crack. We see that the higher activation energy (∆H ) the slower crack growth – ∆H is an energy barrier – but the higher stress intensity factor the faster crack growth is expected. Accordingly, an apparent activation energy (∆Ha) may be used to account for this and we find a termJ log 'K satisfying the Paris equation for the energy barrier reduction:

'Ha 'H J log 'K (9.3e) where γ is a constant and ∆K is the stress intensity factor range. Then, the fatigue crack growth rate (da/dN ) takes the final form for the temperature effect (Kim and Mai 1993),

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The time t( ) dependence for polymers may be expressed as

k E Et0 (9.5a)

dlnE where E is the tensile modulus, E is the unit time modulus (at time t 1), and k . 0 dln t Although Marshall et al (1974) indicated that k decreases at extremes of rate or temperature, the constant (k ) is assumed to be approximately constant in a certain range for any visco- elastic process. Williams (1977) related da/dN to frequency (f ) based on the line zone model by the following relationship

da v f km (9.5b) dN where m is the Paris equation exponent which is insensitive to temperature and frequency for many polymers so that k m may be an approximately constant.

To accommodate both temperature and frequency in one equation, the following procedure is conducted (Kim and Wang 1994). Taking log in Equation (9.5b), we have

Accordingly, a series of straight lines with a slope of -k m for a given DK, one line for each temperature, can be obtained in a plot of log (da/dN ) against log f.

Since the fatigue crack growth rate as influenced by temperature at a given frequency can be described by Equation (9.4), it allows us to relate frequency to temperature by

log aT km (9.5d) log a f where aT

f and a f . The subscript, r, denotes an arbitrarily chosen reference point in the coordinate fr system. Therefore, we obtain fatigue crack growth rate (da/dN ) as

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Since Equation (9.6) has been developed for the Stage II governed by the Paris equation it can be equated to the Paris equation. Taking logs on both equations, the following relations are obtained J m r (9.7a) 2. 303 RT and

km 'H log A log f logC (9.7b) 2.303RT

km where C is Bfrr . It should be noted that the constant B is dependent on frequency. However, the constant C here is independent of frequency and temperature. Also, Equation

(9.7a) indicates J r , is independent of frequency and temperature. Hence, Equation (9.6) can be simplified to

da km ª 'HJ log 'K º f Cexp « » (9.8) dN ¬ RT ¼

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This equation expresses the combined effects of frequency and temperature on the fatigue crack growth rate. Equations (9.5d) and (9.7) can be used to plot experimental data and determine the constants in Equation (9.8). Equation (9.4) is recovered from Equation (9.8) for a constant frequency, Bf km C.

9.2 FATIGUE CRACK LIFE CALCULATIONS The fatigue crack life or a number of load cycles (N ) required for a crack to grow one- dimensionally from a certain initial crack size ao to the maximum permissible crack length ac is easily calculated using the Paris equation.

Consider a fatigue crack of length (a0) in a plate subjected to a uniform stress s perpendicular to the plane of the crack (Figure 9.4). The stress intensityfactor (KI) is given by

KI YV Sa (6.15 bis) where Y is a geometry factor and a function of a/W.

Integrating dN of the Paris equation, we find

a a c da c da

N m m . (9.9) ³ C 'K ³ ao ao C>@Y'V Sa Usually Y varies with the crack length a and the integration cannot be performed directly but by a numerical method. However, we may assume for estimation that Y is an approximately constant if the initial and final crack lengths are very small compared to the width (W ).

The crack lengtha c is calculated from KIc.

Crack length a

Figure 9.4 Fatigue specimen geometry.

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9.3 OVERLOAD RETARDATION AND CRACK CLOSURE The fatigue crack propagation discussed so far has been concerned with constant amplitude loads. It is one of types. Another type is of variable amplitude loads. In the case of constant amplitude loads, the crack growth is more predictable. In other words, a higher fatigue crack growth rate is expected when subjected to higher amplitude of stress intensity factor. However, when a single overload is applied as shown in Figure 9.5, the crack length does not increase as same rate as expected. Surprisingly, its rate is, in fact, lower than it would have been under constant amplitude loading. This effect is shown schematically in Figure 9.5. The crack retardation takes place when a tensile overload follows a constant amplitude cyclic load. An explanation of the crack retardation phenomenon may be obtained by examining the stress distribution in the wake of the plastic zone formation ahead of the crack tip. The plastic deformation creates a compressive residual stress field reducing mode I stress intensity factor for any subsequent lower load. The compressive residual stress tends to close the crack. The overload leaves a larger plastic zone size than the subsequent regular constant amplitude load. The reduction of mode I stress intensityfactor depends on the difference between the overload and the regular constant amplitude load. Accordingly, the crack propagates after overloading at a decreased rate into the zone of residual compressive stresses. Once it passes through the plastic zone created by the overload, its expected growth rate is recovered as the residual stress diminishes.

Overload

Number of cycles Crack length Crack Constant amplitude loading growth

Retarded growth due to overload

Number of cycles Figure 9.5 Typical form of crack length versus number of cycles curve for constant amplitude loading and constant amplitude plus overloading.

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Plastic zone starts influence linearity C Plastic zone ) P Force (

B Crack fully open

Closed part A Closed crack starts to open

Displacement (u)

Figure 9.6 Force-displacement diagram showing the non-linearity caused by configuration change and plastic zone formation.

The crack closure and plastic zone formation can be detected on the force-displacement (P-u) diagram as Elber (1971) suggested. Since the non-linearity of a linear elastic material

E @

Download free eBooks at bookboon.com Click on the ad to read more 216 MECHANICS OF SOLIDS AND FRACTURE Single crack approach for fatigue on a force-displacement diagram can only be caused by two reasons – change of geometric configuration, and material plasticity, as illustrated in Figure 9.6. When an elastic body with a closed crack is under loading, it displays a linear behaviour until the closed-crack starts to open at point A. As the crack opens, the crack length increases, causing the change in geometrical configuration. Accordingly, non-linearity continues from point A until point B at which the crack is fully open. The linearity remains from point B to point C at which the plastic deformation sufficiently large to change the linear behaviour again.

A retardation factor may be defined using plastic zones in the wake of overloading (Wheeler

1972). Let us consider a crack-tip plastic zone of length (rpo) (Figure 9.7) at a crack length

(a0) by an overload of stress (σo) given by K 2 V 2a I 0 0 rp0 2 C 2 (9.10a) 2SV ys V ys

rpo (Plastic zone due to overload)

λ rpi (Plastic zone due to constant load after overload)

Figure 9.7 Plastic zones: small one produced by constant amplitude and large one by overload.

and another plastic zone size (rpi) when the crack has propagated to a length ai at a stress

(σi) is calculated as V 2 a i i rpi C 2 (9.10b) V ys where C is a constant. The plastic zone (rpi) due to the stress (σi) is within the overload plastic zone (rpo). The retardation is due to the difference (=λ – rpi) and a retardation factor f is given by

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where O ar0 po a i and m is an empirical parameter. Then, the retarded crack growth rate § da · ¨ ¸ for ari pi a0 rpo for is given by © dN ¹R § da · § da · ¨ ¸ I¨ ¸ (9.10d) © dN ¹R © dN ¹ where da/dN is the constant amplitude crack growth rate unaffected by the overload. We see that, when the crack has propagated through the overload plastic zone, the crack length ai rpi becomes greater than a0 rpo and the retardation factor also becomes f = 1.

Elber (1971) introduced a model based on the crack closure for stress ratio (R ) effect on fatigue crack growth.

R>0 R=0

Kmax K

Keff

Kmin Kop

Kmin Number of cycles

Figure 9.8 Stress intensity factor at crack opening at different stress ratio (R) values.

It is based on the fact that the faces of a fatigue crack subjected to zero-tension loading close during unloading, and compressive residual stresses act on the crack faces at zero load at R=0. An effective stress intensityfactor range is defined by

'KKKeff max op (9.11a)

where Kop corresponds to the point at which the crack is fully open (Figure 9.8). Using the Paris equation we can find for Stage II,

da m C U'K (9.11b) dN where

K max Kop U (9.11c) K max K min

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It was experimentally found that

UR 05. 0 .4 (9.11d) where

K R min for 0.. 1 ddR 0 7 . (9.11e) Kmax

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10-6

-5 10-5 10

10-6 10-6 (m/cycle) (m/cycle) (m/cycle)

da/dN 10-7 R 10-7 da/dN 0 0.33 0.5 10-8 10-8 0.7

0 10 20 30 40 5 10 15 3/2 ∆K (MN/m3/2) ∆K(MN/m ) eff (a) (b)

Figure 9.9 Crack growth rate and stress intensity factor range for different stress ratios, R= 0, 0.33,

0.5, and 0.7: (a) crack closure included [After Hudson 1969]; and (b) crack closure excluded for ∆Keff. [After Elber 1971]

Figure 9.9a shows crack growth rate (da/dN ) as a function of stress intensity factor range (∆K ) for different stress ratios, R= 0, 0.33, 0.5, and 0.7, displaying the stress ratio effect. The crack growth rate (da/dN ) is re-plotted as a function of effective stress intensityfactor range (∆Keff ) in Figure 9.9b according to Equation (9.11b). It appears that a single curve fits the data from a wide range of stress ratios.

9.4 VARIABLE AMPLITUDE LOADING The prediction of the fatigue crack growth under a variable amplitude loading by simply summing up the individual fatigue lives from respective constant amplitude loads in the loading history may lead to conservative values due to the overload effect. However, the

Paris equation may be applicable if we find an appropriate distribution function of ∆K for a small block of loads. Barsom (1973) demonstrated that the root-mean-square value of the stress intensity factor ∆Krms is useful, which is given by

n f ' 2 6 Ki i 1 (9.12a) 'K rms n f

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where nf is the number of loading amplitudes for each block or random cycles with a stress intensity factor range of DKi for various variable loading types as given in Figure 9.10. Accordingly, the Paris equation becomes

da m C 'K rms . (9.12b) dN

It has been found that the average fatigue crack growth rate (da/dN) under random sequence or ordered-sequence loading fluctuation spectra is approximately equal to the rate of fatigue crack growth obtained under constant amplitude cyclic loading.

Load

(a)

(b)

(c)

(d)

2 blocks (1000cycles)

Figure 9.10 Variable amplitude loading: (a) random sequence, (b) descending sequence, (c) ascending sequence, and (d) combined ascending-descending sequence.

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9.5 FATIGUE NEAR THRESHOLD AND MEASUREMENT METHODS

The fatigue threshold stress intensity factor range (∆Kth) is the one that corresponds to zero crack growth although it can be defined by an arbitrary crack growth rate for practicality.

Most fatigue data do not show a clear ∆Kth. In designing structural components subjected to cyclic loading it is important to determine the fatigue threshold stress intensity factor (∆Kth), below which a crack does not grow. However, important as it is, a ‘true’ ∆Kth is difficult to measure since this requires very long testing times. Usually, near-threshold fatigue crack -10 growth rates of less than 10 m/cycle are determined and then used to estimate ∆Kth Even so, obtaining the near-threshold crack, growth data is a tedious time-consuming procedure. Also, the threshold data vary depending on the experimental technique so that a thorough understanding of various techniques is important for all users of threshold data.

The method desired should be able to reduceD K as quickly as possible without load history effects to reach a very low crack growth rate (da/dN ), but if DK is reduced abruptly it causes the retardation in fatigue crack growth resulting in a higher value than true near-∆Kth. Load shedding can be conducted either manually or continuously by computerised automated control. The automated technique is preferred to avoid the intensive manual labour for processing raw data from measurements of crack length positions with corresponding loads.

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Also, a load-shedding schedule is required to efficiently minimise the load retardation effects. In this section, various methods employing load-shedding schedule will be introduced.

9.5.1 CONTINUOUS K-DECREASING METHODS

A continuous K-decreasing method was proposed by Saxena et al (1978) and recommended by ASTM E24 Committee with

'K 'Ki exp >@C aai (9.13a)

Here and ('Kaii, ) are ( 'Ka, ) initial and instantaneous values respectively of applied stress intensities and crack lengths. The constantC has a physical dimension of length given by

1 d'K C d0.08mm1 (9.13b) 'K da A limit on C assumes that there is a gradual decrease in DK so that the rate of the fractional change of the estimated plastic zone size (rp) remains constant with increase in a and that there is no overload effect on crack growth if the decrease is sufficiently gradual. The acceptable values of C depend on test conditions. If K-increasing and K-decreasing fatigue data agree with each other, then the chosen value of C is permitted. This means that C can only be selected from separate experiments if it is not already established for the particular material to be tested. Accordingly, this method requires very long testing times.

9.5.2 LOAD SHEDDING USING A DAMPING COEFFICIENT

A load shedding method proposed by Bailon et al (1981) employs

'P 'Pi exp QN (9.14a)

where DP and DPi are the instantaneous and initial load, N is the number of elapsed cycles and Q is a damping coefficient given by

1 drp Q (8.32b) (9.14b) rp dN

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) 1/2 4 (MPam K

∆ 2

∆Kth

17 18 19 20 21 22 Crack length (mm)

Figure 9.11 Load shedding using a damping coefficient. [After Bailon et al 1981]

The basic principle of this method is to approach ∆Kth by steps of load shedding according to Equation (9.14a) until crack arrests at ∆Ka (subscript a denotes arrest) which is larger than

∆Kth due to overloading effects (Figure 9.11). The test is resumed with a new set of values for Q (half the previous magnitude) and ∆Pi (and hence ∆Ki, which is half the sum of the original ∆Ki and the associate ∆Ka ). Values of ∆K a, which are similar for the last two or three iterative steps indicate that ∆Kth has been reached. As opposed to the ASTM method for which dK/da is maintained constant, this method uses decreasing dK/da gradients in the load shedding program. It was claimed that this method provides 50% better efficiency than the ASTM method. However, it also requires some preliminary tests to determine the best damping coefficient Q.

9.5.3 CONDITIONAL LOADING BY ITERATION

The principle of the method, proposed by Kim et al (1988) is to searchD K corresponding to a given da/dN by an iteration scheme conditionally. The condition is imposed on the crack growth rate. If the current crack growth rate (da/dN )c is higher or lower than the initially set (da/dN )i, Kmax is either decreased according to

Kmax (i) K K (9.15a) max n max n 1 2n

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A n=1 B

D E n=3 F C n=4

H ∆ K n=2 G n=5 I n=4 J n=5 n=5 n=5 n=6 n=6 n=6

No of cycles

Figure 9.12 Illustration of conditional loading.

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Kmax ( i) K K (9.15b) max n max n 1 2n where n is the number of iterations and Kmax(i) is an initially set relatively high stress intensity factor range. The procedure follows as illustrated in Figure 9.12. Point A is at the first allocated Kmax (1) and hence at the largest plastic zone size at a stress ratio (R) so that (da/ dN )c is higher than (da/dN )i. Accordingly, Equation (9.15a) applies to get to point B with n=2 at which overloading effect is high due to the large drop of Kmax (1) to Kmax (2) and as a result, the condition (da/dN )c<(da/dN )i is found at point C and Equation (9.15b) with n=3 applies to get to point D for Kmax (3). If (da/dN )c>(da/dN )i at point E and crack grows out of calculated plastic zone due to overload, Equation (9.15a) with n=4 applies to decrease the loading. Further, decrease in loading follows since (da/dN )c>(da/dN )I with the same n=4 to reach point G. The same conditional loading continues for the subsequent points H, I, J and so on. The iteration is terminated when the following convergence criterion is satisfied,

K K Toleranced max n 1 max n . (9.15c) Kmax n1 At the end of the iterative procedure the current and previous plastic zone sizes become essentially the same, being defined by a low tolerance (say 2%) set in the computer program. Also, to avoid overloading effects, the crack growth may be allowed to advance twice as long as the plastic zone created by the previous DK only when (da/dN )c>(da/dN )i. The crack tip plastic zone size is calculated according to the Dugdale’s plastic zone model i.e. 2 rp S 8 Kmax V ys . The procedure is summarised in Figure 9.13 and comparisons of the efficiency of ASTM and the present methods for near-threshold crack growth measurement at R=0.1 and 5Hz are given in Table 9.1.

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Input parameters:

da/dN, Kmax(i), Hz, Stress ratio, σys

Initialisation: V=0, n=0

Cycling No Yes Kmax(n-1) < Tolerance Calculate ∆K Stop/(׀ (Kmax(n-1) - K max(n׀)

Yes 2 Yes ∆N

Current crack length?

Yes n ∆a/∆N < given ∆a/∆N V=V+1 n=n+1 Kmax (n)=Kmax (n-1)+Kmax(i)/2

No n=n+1 Yes V=0

No n=n-1

n Kmax (n)=Kmax (n-1) - Kmax(i)/2

Figure 9.13 Flowchart of the conditional loading by iteration: Resolution is for a travelling microscope for

crack length measurement; R is the stress ratio; and σys is yield stress. [After Kim and Mai 1988]

The threshold fatigue data points at low da/dN values measured with the present method for a uPVC pipe material is shown in Figure 9.14.

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a Duration of test da/dN ∆Ki Tolerance 1/2 (m/cycle) (MPa m ) (%) ASTM Current method

4.3×10-9 0.7 6.67 53.10 17.34

1.49×10-9 0.7 2.17 160.80 22.45

10×10-9 0.2 1.8 -b 225.66

Table 9.1 Comparison of the efficiency of ASTM and the present methods for near-threshold crack growth measurement at R=0.1 and 5Hz. [After Kim and Mai 1988] a Tolerance limit set for the present test method only but ASTM method with C= 0.08 mm-1 in Equation (9.13a). b Not available because of unnecessary long times.

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-5

-6

-7

-8 da/dN (m/cycle)

Log -9

-10

-11 -1.0 -0.5 0.0 0.5 1.0 Log 'K (MPam1/2 )

Figure 9.14 Fatigue crack growth of uPVC pipe material. [After Kim and Mai 1988]

A variation of the method can be made if we keep Kmax constant for measuring a ∆K corresponding to a da/dN. The same algorithm can be used by replacing Kmax with ∆K in

Equation (9.15a) and (9.15b). Since there is no overloading effect when Kmax is constant, near- ∆Kth can be obtained more quickly than any other method. However, it is difficult to obtain near- ∆Kth for low stress ratios because it is not possible to obtain near- ∆Kth at a particular stress ratio nominated.

9.6 INTERPRETATION OF FATIGUE CRACK GROWTH IN P − U AND R – A DIAGRAMS

The incorporation of the threshold∆ Kth on a force (P)-displacement (u) diagram may be useful for understanding from a different perspective. As shown in Figure 9.15 (a) and (b) for stress ratios, R =0 and R > 0 respectively, three different Stages are indicated. Stage I is the area of near-threshold fatigue growth. The crack growth in Stage II is governed under the Paris equation, and Stage III includes near- and catastrophic failure, as also described in Figure 9.3. On loading from point 0 to B, fatigue crack starts to grow, which is well below the static fracture point C. As the fatigue crack further grows, stiffness decreases and reaches point D at which catastrophic fracture takes place. When the loading is not high enough,

Download free eBooks at bookboon.com 229 MECHANICS OF SOLIDS AND FRACTURE Single crack approach for fatigue however, for crack growth, the threshold at point A and along the line AE can be identified for specimens with different crack lengths. Threshold-Kmax can be converted into G (energy 2 2 § K · § Kmax (th) · ¨ max (th) )¸ ¨ Q 2 ¸ release rate) and a Gth locus of magnitude ¨ ¸ for plane stress or ¨ (1 )¸ plane © E ¹ © E ¹ strain is found, the shape of which of course depends the geometry of the test specimen.

Stress ratio=0 P Static cracking begins P for fatigue max C crack growth Catastrophic failure D after fatigue Pmax B Stage III

G locus th Gc- locus A Stage II

Stage I E Pmin=0 0 u (Displacement)

(a)

Stress ratio>0 P Static cracking begins Pmax for fatigue crack growth C Catastrophic failure after fatigue B D Pmax A Stage III

E Stage II Gc - Pmin Gth locus Stage I

0 u (Displacement)

(b)

Figure 9.15 Interpretation of constant load range fatigue crack growth in P-u diagram

when there is a fatigue threshold Gth.

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Stress ratio=0

Pmax=cons t Fatigue crack Static cracking begins starts to grow

R B 2 or Stage III VSa

G G Stage I A E

Stage II 'Gth

0 a (Crack length)

(a)

Stress ratio>0

Pmax=cons t Fatigue crack Static cracking begins

starts to grow Pmin=const GC

R B Stage III VS2 or a A G G E

'Gth Stage II

Stage I

0 a (Crack length)

(b)

Figure 9.16 Interpretation of constant load range fatigue crack growth in G-a diagram

when there is a fatigue threshold DGth.

The same information (with the same symbols) can be shown on a G-a (or R-a) diagram (Figure 9.16). For both stress ratios, R = 0 and R > 0, radial lines are drawn for constant V 2S loads, Pmax and Pmin,, and varying crack length (a) according to G a / E . The fatigue crack growth start at point A and grows until point B at which G becomes the critical value Gc (or R) and catastrophic failure takes place. As the crack length decreases for a given loading condition, Stage I area is found, at which near-threshold fatigue crack growth takes place. Again, the crack growth in Stage II is governed under the Paris equation, and Stage III includes near- and catastrophic failure, as also described in Figure 9.3.

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9.7 SHORT CRACK BEHAVIOUR IN NEAR-THRESHOLD FATIGUE The short crack is referred to as the one that is comparable with microscopic features such as grain size and small defects. An example of short surface crack behaviour is given in Figure 9.17. Its crack growth rate does not vary monotonically but fluctuate, displaying peaks and valleys, which is sensitive to grain boundaries. The short cracks are also sensitive to the orientation of the grain. Thus, the crack growth would smoothly increase if all grains are favourably oriented, or zigzag otherwise, with partial or complete arrest in some cases. Such behaviour is illustrated in comparison with a long crack in Figure 9.18. Also, it is obvious that the crack growth rates of short cracks are higher than that of the long crack. The anomalous behaviour of the small cracks does not obey the same propagation laws which we apply to the long cracks. The stress intensity factor range (DK) is not as much useful for the fail-safe design if the crack is smaller than some critical length, typically 1 mm in metallic or polymeric materials.

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10-9

10- 10

10- 11

10- 12

Crack growth rate rate growth Crack (m/cycle) 10- 13 2 60 100 140 180 0 Half crack length (Pm)

Figure 9.17 Growth behaviour of short cracks in Al 2024-T3 during cyclic loading at R=-1 and 20 kHz. [Blom et al 1986]

Another aspect of short crack is associated with the crack closure. According to one- dimensional plastic zone size (rp) equation

2 2 K I V a rp 2 2 , (bis 7.1a) 2SV ys 2V ys rp is proportional to the crack length (a). In other words, as the crack length decreases, the compressive residual stress created by the plastic deformation decreases and hence the crack closure diminishes. Some supporting evidence is given in Figure 9.19. Near-threshold stress intensity for small and large cracks are shown in the figure for difference between effective

∆Kth and apparent ∆Kth due to diminished crack closure effect.

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10-7

Long crack 10-8 Micro cracks (m/cycle)

-9 da/dN 10

10-10 1 2 4 6 8 10 ∆K (MPam1/2) 20 40 80 160 320 μm Figure 9.18 Crack growth behaviour of short and long cracks in aluminium alloy. [After Chan and Lankford 1983]

10 ) 1/2 (MPam ∆Kth for long Crack th

∆ K Crack closure Effective ∆K th contribution

10 100 100 10000 0 Crack length (μm) Figure 9.19 Data for threshold stress intensity for small and large cracks for difference

between effective ∆Kth and apparent ∆Kth due to diminished crack closure effect. [After Blom et al 1986]

Scatter is an essential feature of short-crack data because individual small cracks behave differently. The analysis based on continuum mechanics for long cracks is hardly applicable to the anomalous behaviour of the short cracks. Then, an important question arises as to how we conduct the fail-safe design against small cracks. The important steps may be to

a) define the difference between short and long cracks, b) find common variables between short and long cracks, and c) apply the relevant theories for short and/or long cracks.

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We know that the behaviour of small cracks is reflected in a S-N curve with scattered data points and the long crack is still validly treated within the framework of continuum mechanics. We also know that the common variables are applied stress and crack length (a), allowing us to display two different equations based on the two different approaches together on a σ-a plane. The stress intensityfactor for threshold (∆Kth) with a corresponding applied stress range (∆σth) is given by

'Kth 'V th Sa (9.16a) so that

'K 1 log 'V log th log a . (9.16b) th S 2

This equation is plotted on logarithmic scales in Figure 9.20 with the fatigue limit (∆σ0). The fatigue limit is independent of the crack length.

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Fatigue limit (∆σ0)

∆σ0

(log scale) (log scale) Equation (9.16b)

th ∆σ

Crack length, a (log scale)

Figure 9.20 Kitagawa and Takashima (1976) plot of threshold stress range ∆σ0 as a function of

crack length and fatigue limit (∆σ0).

The limitations of the stress intensity factor approach are clear in the figure. As the crack length approaches zero, ∆σth approaches ∞ with a slope of 0.5 according to Equation (9.16b). However, we know that if the crack length is zero, for a perfectly polished specimen, the threshold stress for fatigue is not infinity, but is equal to the fatigue limit (∆σ0). Therefore, the crack length at which two lines intersect with each other becomes the demarcation point between short and long cracks. The representation shown in Figure 9.20 is often called a ‘Kitagawa’ plot after one of its originators. The plot also explains that the small cracks grow at applied DK values lower than DKth measured for long cracks.

In practice, the experimental data for near-threshold takes the form shown in Figure 9.21. The measured threshold data points deviate from Equation (9.16) for the long crack and eventually merge with the fatigue limit. The curved region on the figure is lower than the fatigue limit or the long crack threshold stress. It may be useful to describe the characteristics of two different crack lengths in addition to the demarcation point (a0) although those are not explicitly definitive due to the smooth transition:

a) a1 is the length at which the fatigue behaviour deviates from the fatigue limit. As such, it is the longest crack length at which the fatigue limit is still a material

property. Therefore, if inherent crack lengths of a material are longer than a1, its fatigue limit may be lowered and hence is no longer the material property. Accordingly, it is possible that some materials would not have a fatigue limit if they contain relatively long cracks produced during manufacturing.

b) a2 is the length at which its behaviour deviates from that of long-crack for the transitional behaviour.

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Such a transitional behaviour (El Haddad et al 1979) can be described by adding a constant length (a0) to the crack length (a), i.e.

''Kth VS th >@ aa 0 . (9.17)

As the crack length decreases, according to this equation, the constant length (a0) constitutes an increasing fraction of (a0 +a) until at very short lengths. An example is given in Figure

9.22(a) for a0 =2 and ∆Kth = 20. The curve and fatigue limit are dependent on the choice of a0 value. If we choose a shorter length a0 =1, the curve displays a higher fatigue limit as shown in Figure 9.22(b). Equation (9.17) may be useful for an initial estimation using only near threshold-DKth. The value ofa 0, however, does not a physical basis for understanding the transitional behaviour.

∆σ0 500 Fatigue limit (∆σ0) (log scale) (log scale)

th 10 ), ∆σ 5 ( MPa a2

Near-threshold stress range a1 a0 0.05 0.1 0.5 1 5

Crack length (mm), a (log scale)

Figure 9.21 Typical experimental behaviour of short cracks, plotted on the Kitagawa diagram. [After Kitakawa and Takahashi 1976]

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Equation (9.16b)

Equation (9.17)

a0=2

(a)

Equation (9.16)

Equation (9.17)

a0=1

(b)

Figure 9.22 Comparison of Equations (9.16) and (9.17) for different crack lengths (a0) = 1

and 2 with ∆σth =31.5 (or log (∆σth) =1.5).

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REFERENCES

Atkins, A.G. and Mai, Y.-W., Elastic and Plastic fracture, Ellis Horwood Limited, 1988.

Bailon, J.P., Chappuis, P. and Masourrave, J., “A rapid experimental method for measuring the threshold stress intensity factor”, In: Fatigue Thresholds (Ed. J. Backlund, A.F. Blom and C.J. Beevers), EMAS, Warly, UK, Vol 1, p. 113, 1981.

Barsom, J.M., “Fatigue-crack growth under variable-amplitude loading in ASTM A514–B steel”, In: Progress in Flaw Growth and Fracture Toughness Testing, ASTM STP 536, American Society for Testing and Materials, pp. 147–167, 1973.

Basquin, O.H., “The exponential law of endurance test”, ASTM STP, Vol 10, pp. 625–630, 1910.

Blom, A.F., Hedlund, A., Zhao, W., Fathulla, A., Weiss, B. and Stickler, R., “Short fatigue crack growth behaviour in AL 2024 and Al 7475, In: The Behaviour of Short Fatigue Cracks, EGF Pub. 1 (Edited by K.J. Miller and E.R. De los Rios) Mechanical Engineering Publication, London, pp. 37–66, 1986.

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Braithwaite, F., “On the fatigue and consequent fracture of metals”, In: Minutes of the Proceedings of Civil Engineers; Institution of Civil Engineers: London, UK, 1854; Vol 13, pp. 463–467.

Broutman, L.J. and Sahu, S., “A new theory to predict cumulative fatigue damage in fiberglass reinforced plastics”, Composite materials: Testing and design (Second Conference), Vol 497, pp. 170–188, 1972.

Burhan, I. and Kim, H.S., “S-N curve models for composite materials characterisation: an evaluative review”, Journal of Composites Science, Vol 2, pp. 1–29, 2018.

Chan, K.S. and Lankford, J., “A crack-tip strain model for the growth of small fatigue cracks”, Scripta Metallurgica, Vol 17, pp. 529–532, 1983.

Cotterell, B. and Reddel, J.K., “Theessential work of plane stress ductile fracture”, International Journal of Fracture, Vol 13, pp. 267–277, 1977.

Dugdale, D.S., “Yielding of Steel Sheets Containing Slits”, Journal of Mechanics and Physics of Solids, Vol 8, pp. 100–101, 1960.

El Haddad, M.H., Smith, K.N. and Topper, T.H., “Fatigue crack propagation of short cracks”, Journal of Engineering Materials and Technology”, Transactions of the ASME, Vol 101, pp. 42–46, 1979.

Elber, W., “The significance of fatigue crack closure”, Damage Tolerance in Aircraft Structures, ASTM STP486, pp. 230–242, 1971.

Erdogan, F. and Sih, G.C., “On the crack extension in plates under plane loading and transverse shear”, Journal of Basic Engineering, pp. 519–727, December 1963.

Eskandari, H. and Kim, H. S., “A theory for mathematical framework and fatigue damage function for S-N plane”, ASTM STP1598 on Symposium on Fatigue and Fracture Test Planning, Test Data, Acquisitions and Analysis, pp.299-336, 2017. doi:10.1520/STP159820150099.

Frocht, M.M., Photoelasticity, New York: John Wiley & Sons, Inc., 1941.

Gerber, W.Z., “Bestimmung der zulassigen Spannungen in Eisen-Constructionen”, [Calculation of the allowable stresses in iron structures]. Z Bayer Arch. Ing. Ver. 6, pp. 101–110, 1874.

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Glasstone, S., Laidler, K.J. and Eyring, H., The theory of rate processes, McGraw-Hill, New York, 1941.

Goodman, J., Mechanics Applied to Engineering, 1st ed.; Longmans, Green and Co., London, UK, 1899.

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INDEX

A energy 139, 174, 175, 176, 243 adhesive 243 equilibrium 50 Airy’s stress function 76, 79, 123, 124 essential 96, 111, 197, 200, 202, 234, 240, 243, 246 ASTM 172, 228, 239, 240, 245 experiment 120 average 171 F axi-symmetric 44 factor 70, 95, 113, 115, 127, 128, 129, 130, 132, B 134, 137, 139, 145, 159, 168, 171, 172, 191, bending 44, 245 194, 195, 203, 205, 206, 208, 209, 211, 214, biharmonic equation 77, 79, 80, 86 215, 217, 218, 220, 221, 222, 226, 232, 235, boundary condition 54, 70, 101, 105, 109, 110 236, 239, 244, 245 bulk modulus 28 failure 208, 209 fatigue 119, 171, 209, 227, 239, 240, 241, 242, C 243, 244, 246 circular hole 88, 89 fatigue life 120 circular plate 68 fatigue limit 120 compatibility 105 fracture 140, 173, 174, 208, 239, 241, 242, 243, 245 compatibility equation 74, 76, 77 fracture surface 14, 19, 159, 165 composite 42, 95, 97, 240, 241, 244, 245 function 12, 16, 34, 74, 76, 78, 79, 80, 86, 89, 100, crack 171, 174, 175, 176, 193, 194, 208, 209, 227, 102, 103, 104, 105, 108, 109, 111, 113, 114, 228, 238, 239, 240, 241, 242, 243, 244, 246 115, 123, 124, 125, 128, 143, 152, 157, 165, curve 33, 95, 96, 97, 99, 100, 101, 102, 103, 106, 166, 181, 202, 208, 214, 220, 236 108, 109, 110, 111, 112, 113, 114, 115, 116, G 117, 118, 120, 165, 173, 183, 189, 190, 191, G 43, 176, 239, 240, 241, 243, 244, 245, 246 192, 207, 208, 209, 210, 215, 220, 235, 237, glass 117, 118, 177, 242 240, 241 cycle 101, 102, 108, 109, 110, 114, 120, 209, H 222, 228 Hooke’s law 26 hydrostatic stress 14, 15, 17, 18, 28, 31, 156 D damage 105, 106, 240, 242, 243 I deflection 36, 46, 49, 50, 52, 53, 55, 57, 58, 60, impact 243 61, 63, 64, 67, 70, 186, 195, 197, 206 instability 194 deviatoric stress 155 isochromatic 82, 83 displacement controlled 194 isoclinic 82 Domain 106, 108, 111, 114, 115 Iso-damage line 100

E J elasticity 26 J-integral 197

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K R

KIc 172, 173 R-curve 189, 190, 191, 241 Kitagawa 95, 236, 237, 243 reference 9, 106, 110, 111, 212 retardation 82, 215, 217, 218, 222, 223 L load 67, 68, 172, 173, 174, 175, 176, 192, 194, S 208, 209 shear 13, 14, 25, 240 load controlled 192, 194 Small cracks 95 S-N 7, 95, 96, 97, 99, 100, 101, 102, 103, 106, M 107, 108, 109, 110, 111, 112, 113, 114, 115, measure 170, 222 116, 117, 118, 120, 207, 235, 240 mode I 140 S-N curve 95, 96, 97, 99, 100, 101, 102, 103, 106, mode II 140 108, 109, 110, 111, 112, 114, 115, 116, 117, mode III 121, 122, 124, 128, 197, 198, 199, 200 118, 120, 207, 235, 240 moment 46, 47, 48, 57, 60, 205 stability 194 N stiffness 194 STRAIN 26 normal 25, 26, 158 Strain energy 34 O Stress 27 Overload 215 stress intensity factor 171, 208, 209, 239, 244, 245 superposition 26, 27 P peak stress 102, 104, 105, 106, 109, 110, 111, 114, T 115, 117 tear 197, 198, 199, 200, 201, 241, 243 Photoelasticity 240 test 171, 173, 194, 228 planes of symmetry 40 testing machine 194 plane strain 19, 31, 32, 76, 78, 138, 140, 147, 152, tetrahedron 10 153, 155, 156, 157, 158, 159, 161, 162, 163, three-point bend specimen 168 164, 165, 168, 177, 181, 189, 190, 230 threshold 209, 227, 228, 239, 243 plane stress 31, 32, 44, 47, 76, 77, 78, 79, 80, 138, time 174 140, 147, 152, 153, 155, 156, 157, 158, 159, toughness 173, 174, 241, 242, 243, 245 161, 162, 163, 164, 165, 177, 181, 189, 190, V 194, 230, 240 valid 97, 104, 105, 110, 111, 114, 115, 154, 168, plastic deformation 173, 175, 176 173 plastic zone 209 variable 105, 109, 111, 215, 220, 221, 239, 244 plate 44, 68, 193, 245 pressure 44 W principal strains 25 work 174, 175, 176 principal stress 10, 82, 90, 94, 141, 142, 143, 144, 157, 159, 160 Z zone 241 Q quasi static 180, 184

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