Introduction to deformable solid mechanics: educational manual 9786010439290

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UDC LBC Iskakbaev
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Introduction

AL-FARABI KAZAKH NATIONAL UNIVERSITY

Alibay Iskakbayev

INTRODUCTION TO DEFORMABLE SOLID MECHANICS Educational manual

Almaty «Qazaq University» 2019

1

Introduction to Deformable Solid Mechanics

UDC 531 (075) LBC 222я73 I-85 Recommended for publication by the Scientific Council of the Faculty of Mechanical Mathematics and RISO of Al-Farabi Kazakh National University (Protocol №3 dated 06.02.2019) Reviewers: Doctor of technical sciences, professor B.B. Teltaev Doctor of physical and mathematical sciences, professor G.B. Sheryazdanov Translated from Russian language edition: «Введение в механику деформируемого твердого тела» (published by Palmarium Academic Publishing, German in 2014) and edited by: A.A. Iskakbayeva, A.Zh. Bibossinov

I-85

Iskakbayev Alibay Introduction to deformable solid mechanics: educational manual / Alibay Iskakbayev. – Almaty: Qazaq University, 2019. – 256 p. ISBN 978-601-04-3929-0 This manual, which is distinguished by its extremely complete terminology on the mechanics of deformable solid body with explanations that are accessible to perception, will help students to become familiar with the logic of constructing models of the mechanics of deformable solid body. The manual contains more than 150 figures and 10 tables, which favorably affects the students' mastering of the course material of the mechanics of deformable solid body, regardless of the form of education. The book meets all modern requirements for the provision of complete and accessible educational information to students with any form of education. The educational manual is intended for university students and can be used in preparation and promotion knowledge in the field.

UDC 531 (075) LBC 222я73 © Iskakbayev Alibay, 2019 © Al-Farabi KazNU, 2019

ISBN 978-601-04-3929-0

2

Introduction

INTRODUCTION

The main task of the mechanics of a deformable solid (MDTT) is the modeling of the processes of deformation and fracture. Destruction refers to one of the types of durability. Violation of the strength of materials and structural elements may occur as a result of: 1) excessive (elastic or plastic) deformation; 2) loss of stability; 3) fracture. Fracture may be partial or complete. In case of partial destruction in the body, material damage occurs in the form of individual cracks or in the form of material defects distributed over the volume, resulting in a change in the mechanical properties of the material. With complete destruction, the body is divided into parts. Fracture can also be divided into the following types: 1. Plastic fracture. Occurs after a significant plastic deformation that occurs throughout (or almost all of) the volume of the body. A type of plastic failure is a rupture after 100% neck contraction under tension resulting from the loss of the ability of a material to resist plastic deformation. This case can also be attributed to one of the types of durability. 2. Brittle fracture. Occurs as a result of propagation of the main crack after plastic deformation, concentrated in the area of the mechanism of destruction. In case of ideally brittle fracture, there is no plastic deformation, therefore, after fracture, it is possible to re-compose the body of the previous size from fragments without gaps between them. 3

Introduction to Deformable Solid Mechanics

In the case of a quasi-brittle fracture, there is a plastic zone in front of the edge of the crack and riveted material at the crack surface. The other, and a much larger, volume of the body is in an elastic state. At present, a fracture, in which the breaking stress in the net section is higher than the yield strength, but lower than the tensile strength, is called quasi-brittle. 3. Fatigue damage. Occurs with cyclic (repeated) loading as a result of accumulation of irreversible damage. The fracture is macroscopically brittle, however, the material at the surface of the fracture is substantially riveted. There are fatigue and low-cycle fatigue. Fatigue is characterized by nominal stresses, lower yield strength, repeated loading is macroscopically observed in the elastic region, the number of cycles before failure is large. Low-cycle fatigue is characterized by nominal stresses, high yield strength; macroscopic plastic deformation occurs at each cycle; the number of cycles before failure is relatively small. 4. Deformation and creep rupture. The sample under load at elevated temperatures, even under conditions of a uniform stress state, begins to break down. It forms a system of microscopic cracks along the grain boundaries. In fact, at the grain boundaries, pores are first formed, then these pores merge into cracks with rounded ends, extending within one grain. Over time, the number of microscopic cracks increases, they merge into one or several through cracks and destruction occurs. The question of the movement or equilibrium of a macroscopic crack in such a formulation is not significant, the main role is played by the kinetics of accumulation of microcracks. When the density reaches a critical value, destruction occurs. The position and shape of the fracture macrocrack are more or less random. 5. Corrosion destruction. There are three main mechanisms of influence of corrosive environments on the crack resistance of structural materials: adsorption reduction of strength, hydrogen embrittlement and corrosion dissolution. The adsorption of surfactants on the surface of a highly stressed material in the tip of a crack causes a decrease in the surface energy and facilitates destruction (the Rehbinder effect). Adsorption effects can be successfully used to increase the efficiency of metalworking. 4

Introduction

Note that of the three main mechanisms the adsorption effect is dominant at high values of the stress intensity factor when due to highspeed subcritical crack growth, other mechanisms do not have time to occur. It is well known that the effect of moisture on metals leads to corrosion and destruction. Corrosion often does not change the mechanical properties of the material, but leads to a gradual uniform decrease in the size of the loaded part, for example, due to gradual dissolution. As a result, the stresses acting in the dangerous section increase, and when they exceed the permissible level, destruction will occur. According to modern concepts, the main process that accelerates subcritical crack growth, leading to accidents, is hydrogen embrittlement of a small region near the tips of the cracks. In most cases of corrosion growth of cracks, the processes of adsorption, hydrogen embrittlement and corrosion dissolution are mutually interconnected, and one process causes the manifestation of others. The interrelationship of these processes is complicated by the influence of the metal structure, the type of stress state, and external loading conditions. The study of this linkage is the subject of corrosion fracture mechanics. This tutorial covers modeling of the processes of deformation and fracture, which are the main tasks of the mechanics of a deformable solid body. It contains the basic concepts and models of the theory of elasticity and viscoelasticity, the theory of strength and cracks, the theory of stability of deformable bodies, the theory of plasticity, the mechanics of composite materials and damage mechanics. Students reveal the mechanisms of violation of the strength of solids.

5

Introduction to Deformable Solid Mechanics

1 SOME INFORMATION FROM CONTINUUM MECHANICS

1.1. Stress condition Stress condition is characterized by a symmetric stress tensorat in any point of a continuous medium   x  xy  xz    ˆ    xy  y  yz ,    xz  yz  z 

(1.1)

where  x ,  y ,  z are normals,  xy , yz , xz are tangential stresses on the areas vertical to x, y, z coordinate axis.  The stress vector pï on an arbitrary orientated area with the unit  normal n (Figure 1.1) is determined by Cauchy formula:  p x   x nx   xy n y   xz nz ,   p y   xy nx   y n y   yz nz ,  p   n  n   n , xz x yz y z z  z

 where n x , n y , n z are components of the unit normal vector n . 6

(1.2)

1. Some information from continuum mechanics

Figure 1.1. Components of stress vector



Calculating vector pï on the direction of the normal, we obtain the normal stress  n on this area

 n   x n x2   y n y2   z n z2   2 xy n x n y  2 yz n y n z  2 zx n z n x .

(1.3)

The tangential stress rate  n is equal to:



 n  p 2x  p 2y  p 2z   2n



1/ 2

.

(1.4)

There are three mutually perpendicular planes at any point of a continuum, on which the tangential stresses are zero. The directions of the normals to these areas form the main directions of the stress tensor and do not depend on the initial coordinate system x, y, z. This means that any stress state at the considered point can be caused by the extension of the point area in three mutually perpendicular directions. The corresponding stresses are called the principal normal stresses; let us denote them by  1 ,  2 ,  3 and let us enumerate the principal axes so that

1   2   3 . 7

(1.5)

Introduction to Deformable Solid Mechanics

Using equations (1.2)-(1.4) it is easy to find that in the cross-sections dividing in half angles between the general planes and passing, respectfully, through the general axes 1, 2, 3, the absolute values of tangential stresses are equal to 1  2   3 , 2 1 1  3   1 ,  1   2  . 2 2

Figure 1.2. Principal tangential stresses

The tangential stresses in these cross-sections reach extremal values and are called the principal tangential stresses:

1 

2 3 2

,

2 

 3  1 2

, 3 

1   2 2

.

(1.6)

The value of actual tangential stress  n changes with changes in the area orientation. The biggest value  n at a certain point is called the maximal tangential stress  max . If condition (1.5) is satisfied, then from (1.6) we find

 max   2  8

1   3 2

.

1. Some information from continuum mechanics

The normal stress  n at the area does not depend on the choice of the coordinate system and changes only when the area rotates. The principal invariants of the stress state as the simplest symmetric functions of arguments in the principal stresses are written as:

 1 ˆ    1   2   3  3 ,   2 ˆ    1 2   2 3   3 1 ,   ˆ      , 1 2 3  3

Here



1  x   y   z , 3

(1.7)

(1.8)

is called as average (or average hydrostatic) pressure (it is also called as average normal stress) at the point. As a rule, materials have different mechanical resistance to the shear and to the uniform integrated compression, it is convenient to write the stress tensor as (1.9)-(1.11):

ˆ  ˆ1  ˆ 2 ,

(1.9)

 0 0    ˆ1   0  0 , 0 0   

(1.10)

 x   xy  xz    ˆ2    xy  y   yz ,    yz  z    xz

(1.11)

here ˆ 1 is a spherical tensor (1.10) corresponding to the average

pressure at the point, ˆ 2 is a tensor (1.11) which characterizes 9

Introduction to Deformable Solid Mechanics

tangential stresses at a given point and is called a stress deviator. The main directions of the stress deviator ˆ 2 and stress tensor ˆ coincide. The deviator's invariants are easily obtained from Eq. (1.7) if 1, 2 , 3 are replaced by  1   ,  2   ,  3   :

 1 ˆ 2   0,  1 2 2 2   2 ˆ 2    1   2    2   3    3   1  , 6    3 ˆ 2    1    2    3   ,





(1.12)

The nonnegative quantity (1.13)

i   2 ˆ2   

          6 6

1

2

x

y

2

y

z

2

z

x

2 xy

 yz2  xz2



1/ 2

(1.13)

is called intensity of tangential stresses. Stress intensity is a quantity:

i 



1 1   2 2   2   3 2   3  1 2 2



1/ 2

.

(1.14)

The stress intensity in Eq. (1.13)-(1.14) vanishes to zero only when the stress state is the hydrostatic pressure. For a pure shear

1   ,  2  0,  3   , where  is a shear stress. So,  i   . In the case of extension (compression) in the x axis direction

 x   0 ;  y   z   xy   yz   zx  0; 10

1. Some information from continuum mechanics

then

i  0 , i 

0 3

.

(1.15)

The inequality established by A.A. Ilyushin:

1

i 2  .  max 3

(1.16)

Octahedral plane. Let's consider an area inclined at the same angle to the main planes (see Figure 1.3) and calculate normal and tangential stresses on the area called octahedral. The coordinate axes are directed along the normals to the main segments. The direction cosines for the octahedral plane with respect to coordinates are 1 . n1  n2  n3  3 From Eq. (1.2) for the stress vector on the octahedral plane we get Eq. (1.17)

p n2 





1 2  1   22   32 , 3

(1.17)

thus, the square of the total stress on the octahedral plane is equal to the average sum of the squares of principal stresses.

Figure 1.3. Octahedral plane

11

Introduction to Deformable Solid Mechanics

Using Eqs. (1.3)-(1.4) and Eq. (1.17) we find:

 oct 

1  1   2   3    , 3

(1.18)

so the normal stress on the octahedral plane is equal to the average stress for a given point;



1  1   2 2   2   3 2   3   1 2 3 2 2  i  i, 3 3

 oct 



1/ 2



(1.19)

in other words, the octahedral tangential stress is directly proportional to the calculated stress. The above normal and tangential stresses on the octahedral plane, also called the octahedral stresses, are the same for all eight segments which can be drawn in all octants. Mohr diagram. Let's find a plane on which normal and tangential stresses have predetermined values  n and  n . The problem is to find out three unknowns n12 , n 22 , n32 from Eqs. (1.2)-(1.4) to which we add Eq. n12  n 22  n 32  1 . The desired solution is written as n12 

f1n , n  , 1 2 1 3  n32 

where

n22 

f2  n , n  , 2 1 2 3 

(1.20)

f3 n , n  3 1 3 2  2

2

2

2

2

2

 3  2 3   f1  n , n    n2    n  2  ,   2   2     1    3  1   f 2  n , n    n2    n  3  ,   2   2      2   1   2   f 3  n , n    n2    n  1  .   2   2   12

(1.21)

1. Some information from continuum mechanics

As Eq. (1.5) is fulfilled and the left parts of these equations (1.20) are nonnegative, we get

f1  0,

f 2  0,

f 3  0,

(1.22)

In other words, stresses n ,  n are inside the area limited by semicircles and crosshatched regions shown in Figure 1.4.

Figure 1.4. The Mohr diagram for stresses

Deviator plane. Let's choose a Cartesian coordinate system  1 ,  2 ,  3 (Fig. 1.5). The stressed state in some points is characterized by a vector with the components  1 ,  2 ,  3 .

 1   2   3  0,

(1.23)

The plane, equation (1.23), goes through the origin of coordinates and is equally inclined to the principal axes. It is obvious that any vector lying in this plane characterizes the stress deviator of a stress state, that is why the plane (Eq. 1.23) is called the deviator plane.

1   2   3 ,

(1.24)

The line (Eq.1.24) declined at the same angles to all three axes of coordinates  n1  n2  n3  1  , determines the position of the points 3   1   2   3 corresponding to the hydrostatic state. 13

Introduction to Deformable Solid Mechanics

The diagrams of the distribution of normal (continuous lines) and tangent (dashed lines) stresses during extension of a plated sample depending on cross-section position (Fig. 1.6) (numbers indicate the angles between the normal to the area and to the tensional force).

Fig. 1.5. Deviator plane

Fig. 1.6. The diagrams of the distribution of normal and tangent stresses during extension of a plated sample

1.2. Deformation Deformation of a continuum is characterized by a symmetrical deformation tensor (Eq. 1.25): 14

1. Some information from continuum mechanics

  x  1 ˆ     xy 2 1   xz 2

1  xy 2

y 1  yz 2

1   xz  2  1   yz , 2   z  

(1.25)

here  x ,  y ,  z – relative elongations,  xy ,  yz ,  zx – relative shears. The differences (Eq. 1.26) are called the principle shears.

1 ,  2 ,  3 are called the principal elongations (Fig. 1.7).  1   2   3 ,  2   3  1,  3  1   2 ,

(1.26)

The maximum shear at a given point is called the maximum shift

 max .

The deformation tensor, as any symmetric tensor, is reduced to the principal axes. This means that any deformation can be fulfilled by a simple extension in three mutually perpendicular directions (main directions).

Fig. 1.7. The Mohr diagram for deformation

The invariants of the strain tensor in the principal axes have the form of Eq. (1.27): 15

Introduction to Deformable Solid Mechanics

1 ˆ   1   2   3 ,  2 ˆ   1 2   2 3   31 ,

(1.27)

 3 ˆ   1 2 3 .

It is convenient to use the strain tensor in the form of Eqs. (1.28)(1.31): (1.28) ˆ  ˆ1  ˆ2 ,   0 0  3    (1.29)  ˆ 1  0 0 ,   3   0 0  3  (1.30)   x   y  z,

1   x   3  1 ˆ2    xy  2  1   xz  2

1  xy 2 1 y   3 1  yz 2

1   xz  2  1  yz .  2 1  z    3 

(1.31)

Here ˆ1 is the spherical tensor corresponding to the volumetric

expansion and ˆ2 is the deviator of deformation (Eq. 1.31), which characterizes the change in the shape of the element of a continuous medium caused by the shifts. In the case of small deformation, the relative change in the volume is determined by the relation (1.30). The invariants of the deformation deviator are equal to: 1 ˆ2   0,





1 1   2 2   2   3 2   3  1 2 , 6 1  1  1    3 ˆ3     1     2     3   . 3  3  3    2 ˆ2  

16

(1.32)

1. Some information from continuum mechanics

A non-negative quantity

i 2 2ˆ2 





1/ 2

2 x y2 y z 2 z x2 32 xy2 yz2 zx2  3 

(1.33)

is called the shear strain rate. In the case of a pure shift,

 x   y   z   yz   xz  0,  xy  

(1.34)

substituting the values from Eq. (1.34) into Eq. (1.33), we find:

i   . The dependence of the strain intensity on the principal linear deformations has the form

i 

1 2 i  ( 1   2 ) 2  ( 2   3 ) 2  ( 3  1 ) 2 . 3 3

(1.35)

For the particular case of uniaxial stretching of an isotropic material

1   0 ;  2   3  

0   2

according to the equation (1.35), we find:

i  0   .

(1.36)

In solid mechanics we usually use the Lagrangian method to describe the continuous medium movements. The vector of the    displacement of a particle is denoted by u  u(x, t) , here x is the material coordinates of a particle of a continuous medium. We assume 17

Introduction to Deformable Solid Mechanics

that a certain coordinate system xi (i  1,2,3) is introduced with local





basis vectors ei and a mutual basis e i . We represent the gradient of the displacement vector in x i coordinates in the chosen basis as follows:   (1.37) Grad u  wij ei e j . The tensor (1.37) in the literature is known as the tensor of distortion. We assume that the condition given below is fulfilled:

 Grad u  1.

(1.38)

If condition (1.38) is satisfied for all points of the medium and for any moment of time, then the deformations are considered as small ones. In this case the spatial coordinates of particles are different from material coordinates by infinitesimal quantities. Let's divide the distortion tensor into symmetric and antisymmetric parts: (1.39) ui , j   ij   ij , where

 ij 

1 ui, j  u j ,i , 2

(1.40)

ij 

1 ui, j  u j ,i . 2

(1.41)



A symmetric tensor ˆ   ij ei e j is called a small deformation tensor and its calculation using displacement vector (1.40) is called the  Cauchy relation. An antisymmetric tensor ˆ   ij ei e j is called a rotation tensor. The axial rotation vector can be related to it: 

1 2



  rotu . 18

(1.42)

1. Some information from continuum mechanics

In Eqs (1.39)-(1.41) the comma means the partial differentiation corresponding to the coordinate. Cylindrical coordinates ds2  dr 2  r 2d 2  dz 2  . Equations for the physical components of the strain tensor:  ur  r  r ;   1 u ur  ;    r  r   u  z  z ; z 

1  ur u z  r 2  z

 ,  1  u 1 u z  z     , 2  z r   1  1 ur u u     r   , r 2  r  r 

 rz  

(1.43)

here ur , u , u z are the physical components of the displacement vector. Spherical coordinates ( ds2  dr2  r 2d 2  r 2 sin2 d2 ). Physical components of deformation are:

 u 1 u u r  r  r ;    ; r  r r   u u 1 u ;     r  ctg   r r sin   r   1  u u 1 u r   ,   r     r r r    2 sin     1 u u 1 u r   r        ; 2 r r r     u ctg u 1 u     1  1       r r sin   2  r 

(1.44)

  . 

ur , u , u are components of the displacement vector. 19

Introduction to Deformable Solid Mechanics

The problem is set to determine the displacement vector using a given linear tensor of deformation ˆ . In other words, we have to integrate a system of six differential equations (1.40), the given right sides of which are continuous together with their first and secondorder derivatives. The number of equations (six) exceeds the number of unknowns (three), so the problem can have a solution only under the condition  (1.45) rot ( rot  )*  0 ,

 where  is the deformation tensor, and (*) is the sign of transposition, the condition of Eqn. (1.45) is the integrability condition for the system (1.40); it is also called – the condition of continuity and the compatibility condition of deformations. The great importance of such conditions in the mechanics of continuum was pointed out by Saint-Venant, therefore the term "Saint-Venant's dependencies" is also used. 1.3. Equilibrium equation of the continuum Differential Equilibrium Equations (1.46)

 ij , j  Fi  0,

(1.46)

where  ij are components of the stress tensor (Fig. 1.8), Fi is the body force. Cylindrical coordinates ( r ,  , z ). Differential Equilibrium Equations:  r 1  r  rz  r        Fr  0, r r  z r  r 1    z 2 (1.47)     r   F  0, r  r r z  rz 1  z  z 1     rz  Fz  0. r r  z r 20

1. Some information from continuum mechanics

Spherical coordinates ( r, , ). Differential Equilibrium Equations: 1  r 1  r 2 r     r ctg  r     Fr  0, r r r sin  r   r 1  1  3 r  2ctg   F  0,   r r r sin  r  1  1  3 r    ctg  r    F  0.  r r sin  r  r

(1.48)

Fig. 1.8. Components of the stress tensor

1.4. Linear model of elastic body The law describing the state of a linearly elastic body in an isothermal deformation process will be written in the form:

ˆ   Eˆ   2  ˆ ,

(1.49)

here Eˆ is the unit tensor,  is the volume expansion (1.30), and  ,  are the constant elastic modules called the Lame coefficients. The law of elastic change of the volume and the law of elastic change of the shape are written in the form:

  K , 21

(1.50)

Introduction to Deformable Solid Mechanics

ˆ 2  2 ˆ 2 ,

(1.51)

where K is the bulk modulus of elasticity 2 K    . 3

(1.52)

The volume module K determines the resistance of the material to the volume change (Eq. 1.50), which is not accompanied by a change in shape (the case of hydrostatic pressure). The law for the shape change (Eq. 1.51): the stress and strain components corresponding to the shape change are proportional to each other. The shear modulus μ determines the resistance of the material when its shape changes which is not accompanied by a change in volume. It is necessary to add relations (1.53) or (1.54) to equations (1.50)(1.51):

 i  3  i ,

(1.53)

 i   i .

(1.54)

In equations (1.49), the stress components are expressed in terms of the deformation components. It is often necessary to have the inverse relations, i.e. deformations expressed in terms of stresses:













1  x   y   z  , E 1  y   y   z   x  , E 1  z   z   x   y  , E

x 

 xy 

 xy G

,  yz  22

 yz G

,  zx 

(1.55)

 zx G

,

1. Some information from continuum mechanics

where E is the elasticity modulus, G is the shear modulus, and  is the Poisson coefficient. The elasticity law in this form is usually used for an isotropic body. For an isotropic homogeneous elastic medium:

E  2(1  )G.

(1.56)

Comparing the laws (1.49) and (1.55) - (1.56), we find:

  G,  

E 2 . , K 3(1  2 ) 1  2

(1.57)

Now taking into account the relations (1.56)-(1.57), (1.49) and (1.55), we have:    (1.58) ˆ  2G  Eˆ  ˆ ,  1  2 

ˆ 

1    3Eˆ .  ˆ  2G  1  

(1.59)

1.5. Specific potential energy. Elastic potential Specific potential energy is the energy accumulated upon deformation in a unit volume of a material near a given point. The expression for determining the potential deformation energy of an elastic body is written in the form:

1 U  ˆ  ˆ, 2

(1.60)

where ˆ is the stress tensor, ˆ is the deformation tensor, and (∙∙) is the sign of scalar multiplication. Taking into account the law of elasticity (1.49), from Eq. (1.60) we find: 23

Introduction to Deformable Solid Mechanics

1 U   2    ij2 , 2

(1.61)

here  is the volume deformation (1.30),  ij is the strain tensor. From the energy concepts it follows that

U  0.

(1.62)

The statement (1.62) is a property attributed to an elastic body, there are no zones with U  0 . In a linearly elastic isotropic body it must be ensured by the requirements imposed on the elastic modules (1.63)-(1.64):

  0,   0,

(1.63)

1 E  0, G  0,  1    . 2

(1.64)

Differentiating (1.61) with respect to

 ij 

U  ij   ij

 ij , we obtain:

  ij  2  ij ,

(1.65)

 ij is the Kronecker tensor. Eq. (1.65) is the condition that the elastic energy U is a stress potential. The elastic potential U has a direct mechanical meaning. It is the potential energy of elastic deformation accumulated in the body. As isotropic materials can withstand very large hydrostatic pressures without the appearance of fluidity, the energy calculated in (1.60) is divided into two parts to better match the theory with the experiment: the energy from a change in the body volume (1.66): 24

1. Some information from continuum mechanics

1 1 U1  ˆ1  ˆ1  K 2 , 2 2

(1.66)

where K  0 (1.52); and the other part of energy is from the distortion of the body shape (1.67): 2

1 1   U 2  ˆ 2  ˆ2     ij     2 3   1  x   2   y   2   z   2  2  xy2   yz2   xz2   4G 1  x   y 2   y   z 2   z   x 2  6  xy2   yz2   xz2 .  12G











(1.67)



So we obtain: U  U1  U 2 ,

where U1 is the energy consumed by the change in the volume at a given point, U 2 is the energy consumed by the shape change. If we compare (1.12) and (1.19) with (1.67), we see that 2 9 оct  6 I 2 (ˆ 2 )  12 GU 2 ,

the square of the tangential octahedral stress, the second invariant of the stress deviator and the specific energy of shaping are proportional to each other. The elastic work of the generalized stress at a given point of the generalized deformation is equal to the amount of specific energy of the shape change at the same point (1.68):

1 U2   i i , 2

(1.68)

Here the law of elasticity is determined by the relation (1.53). 25

Introduction to Deformable Solid Mechanics

1.6. Classification of stressed states Table 1 Main types

Stress state diagram

Main stresses

Mohr circles

Linear or Uniaxial stress state τ

z Stretching

1   z  0 2  y  0 y

σ

3   x  0

x Compression

z

τ

1   x  0 2   y  0 y

σ

3   z  0

x

Two-axle compression

Two-axle stretching

Plane or biaxial stress state

z

1   z  0 2  y  0 y

τ

3  x  0

σ

x z

y

1   x  0 2  z  0 3   y  0

x

26

τ

σ

1. Some information from continuum mechanics

Main stresses of different signs

z

y x

1   z  0 2   y  0

τ

3   x  0

Pure shear

z

σ y x

Three-axis stretching

Volumetric or triaxial stress state

z

1   z  0 2   y  0 y

τ

σ

3   x  0

x

Three-axis compression

z

τ

1   z  0 2   y  0 y

3   x  0

σ

Main stresses of different signs

x z

1   z  0 2   y  0 y

3   x  0

x

27

τ

σ

Introduction to Deformable Solid Mechanics

z

1   z  0 2   y  0 y

3   x  0

τ

σ

x

1.7. Classical theory of strength 1.7.1. Theory of maximum normal stresses This theory is based on the following hypothesis: the strength will be violated if the maximum normal stresses exceed certain limiting values that do not depend on the type of stress state (Galileo's hypothesis). The condition of the onset of a dangerous state

 1   dan

 or  3   dan ,

(1.69)

where  dan  and ( dan ) are the limiting (dangerous) stresses under stretching and compression. The criterion (1.69) is qualitatively confirmed in the experiments on tension and torsion of samples from a very brittle material.

1.7.2. Theory of the maximum linear deformations The theory is based on the Mariotte hypothesis: the strength will be violated if the absolute value of the greatest relative linear strain exceeds a certain limiting value which is independent of the type of the stressed state. 28

1. Some information from continuum mechanics

The condition of a dangerous state is written in the form (1.70):

 maxE  1  ( 2   3 )   b ,  min E   3  (1   2 )   b or

  ( 2   3 )   b ,  eqv   1  3  ( 1   2 )   b  ,

(1.70)

where  eqv is the equivalent stress,  b is the strength limit of the sample. It is possible to explain the destruction of brittle materials not only during tension and torsion but also in compression with the help of the Mariotta criterion. The diagram of the development of the concepts of destruction is given in Fig. 1.9 (the time of final destruction is shown by a circle): 1 – to the XVII century; 2 – XVII and XVIII centuries; 3 – XIX century; 4 – the beginning of the XX century; 5 and 6 – 30-60s of the XX century.

Fig. 1.9. Scheme of the destruction development

The first measured mechanical characteristic was the resistance force to fracture without taking into account the elastic deformation of 29

Introduction to Deformable Solid Mechanics

the body (curve 1 in Figure 1.9), such a body is a hard destructive one; when elastic strain was taken into account a deformation diagram of a fragile body (curve 2) was obtained which was subsequently supplemented by a nonlinear deformation (curve 3). For a long time, the stress at the highest point max of the diagram (curve 4) was considered a characteristic of destruction. It has been preserved to this day in the designation  b where the index b is the initial letter of the German word bruch – destruction. Much later it was assumed that the maximum force on curve 4 was associated not with the destruction but with the loss of plastic stability or the formation of a neck on the specimen. Then it was assumed that the end point ê of the diagram (curve 5) and not the maximal point was associated with destruction, and destruction resistance was expressed in actual stresses. However, now it is clear that this point is not the beginning and not the end of the destruction but the critical state of destruction, corresponding to the loss of the body stability with a crack. Destruction ends when the crack completely crosses the body or specimen at an effort decreasing to zero (curve 6). 1.7.3. Theory of the maximum tangential stresses The theory is based on the Coulomb-Tresk hypothesis: the strength will be violated if the largest tangential stresses exceed a certain limiting value that does not depend on the type of the stress state. This criterion was proposed by Treska on the basis of experiments on stamping and pressing metals in plasticity condition. The condition when a dangerous state may arise:

 max   dan    , here  T is the plasticity limit of the sample (Figure 1.10). Taking into account condition (1.5) and formulas (1.6), we finally obtain:

 eqv   1   3   dan . 30

(1.71)

1. Some information from continuum mechanics

Fig. 1.10. The relation between stress intensity and strain

This criterion is sufficiently well confirmed in the plastic state of the material. This theory of strength is not very acceptable in the fragile state of the material. 1.7.4. The theory of full potential energy This theory is based on the following hypothesis: strength will be violated if the specific potential energy of deformation exceeds a certain limiting value which is independent of the form of the stressed state (Beltrami's hypothesis). Beltrami's criterion is practically not used. Numerous experiments show that with uniform regular compression, the destruction (or fluidity) of materials does not occur even at very high stresses. At the same time, the energy accumulated in the test sample is significantly higher than the dangerous value obtained in the experiments on uniaxial tension or compression. 1.7.5. Distortion energy theory This theory is based on the following hypothesis: the strength will be violated if the specific potential energy of shaping exceeds a certain limiting value which is independent from the stressed state form (Huber's conjecture). 31

Introduction to Deformable Solid Mechanics

The condition when a dangerous state may arise:

 eqv  12   22   32  (1 2   2 3   31 )   dan .

(1.72)

Huber's criterion has been satisfactorily confirmed experimentally for various structural materials that are in a plastic state and are equally resistant to tension and compression. Various experimental data make it possible to form a more or less complete picture of the nature of the destruction of materials, depending on the type of stress state. Table 2 shows a graphic diagram of brittle and plastic fractures for typical tests (Ya.B. Fridman). Table 2 Type of load

Type of fracture Separation

Stretching

Compression

Torsion

Bend

32

Cut

1. Some information from continuum mechanics

1.8. Elastic parameters of isotropic materials at normal temperature

Density, g/cm3

Coefficient of linear expansion α,10-6grad-1

Shear modulus G, kgs/mm2

Poisson's ratio ν

Elastic modulus Е, kgs/mm2

Material

Table 3

Aluminum

70008000

0,32-0,34

26002700

20-24

2,72,9

Brass

1000011000

0,33-0,36

37004200

19,820,9

8,3

Copper

1180012000

0,33-0,36

40504700

16,016,7

8,9

Cast iron

910014700

0,21-0,30

36504050

10,5

6,97,5

Steel

1960022400

0,26-0,29

77008400

9,9-12,7

7,8

Stainless steel

1960021000

0,30

7420

14,816,9

7,88,0

Titanium

1080011600

0,34

4200

8,8

4,43

Glass

50008000

0,21-0,27

26503280

6,0-9,5

2,43,8

33

Introduction to Deformable Solid Mechanics

1.9. Speed of pressure wave propagation in various materials Table 4 Е



Е

Lead

m/с 1320

Cadmium

 m/с 2310

Brass Copper

3500 3666

Constantan Rubber (vulcanized)

4300 43

Magnesium Nickel

4600 4970

Cork Ebonite

500 1570

Steel Iron

4982 5000

Ice Granite

3160 5100

Aluminum

5104

Glass

5500

Material

Material

The shear modulus  has the same value in adiabatic and iso/ thermal processes. Table 5 gives the ratios K for some metals

K

(at 20 °C); here K / is called adiabatic, K – isothermal modulus of volume compression. According to the data given in Table 5 it is necessary to make a distinction between the adiabatic and isothermal modules K/  K. The data of Table 4 show that the rate of elastic deformation is very high, in any case, much higher than the practical rates of loads. Basic properties of a linear elastic medium: 1) independence of the elastic properties of the strain rate (Table 4); 2) independence from the loading history (unloading law: Figure 1.11). Table 5 Element 1 Aluminum Molybdenum

Element

K/ K 2 1.043 1.007

3 Iron Copper

34

K/ K 4 1.016 1.028

1. Some information from continuum mechanics 1 Tungsten Silver Manganese Lead

2 1.006 1.004 1.044 1.067

3 Cobalt Nickel Platinum Gold

Fig. 1.11. Regularities for a hyperelastic medium

35

4 1.020 1.021 1.020 1.038

Introduction to Deformable Solid Mechanics

2 BASIC CONCEPTS AND PROBLEMS OF THE LINEAR THEORY OF ELASTICITY

The classical theory of elasticity has its own special place in the science of the behavior of deformable solid. Its initial definitions are common to all sections of this science, the methods of setting and solving problems are representative.

2.1. Basic relations of the linear theory of elasticity 2.1.1. Basic equations of the linear theory of elasticity These equations are determined by three groups of relations. The first group is represented by the equations of statics in the volume.  (2.1) divˆ  F  0 ,



where ˆ is a stress tensor, F is a volume force vector. The second group of equations contains the definition of the  linear deformation tensor ˆ through the displacement vector u :

ˆ 





* 1  u  u  , 2 36

(2.2)

2. Basic concepts and problems of the linear theory of elasticity

 is the nabla operator, and (*) is the transposition sign. There are six equations that determine the components of the strain tensor according to the first derivatives of the three components of the displacement vector. In the third group, the physical state of a linearly elastic body is described by six equations. The generalized Hooke's law, for an isotropic body and in isothermal or adiabatic processes, is written as

   Eˆ  ˆ , 1  2   

ˆ  2  

(2.3)

or in the form of an inverse relation

ˆ 

3 ˆ  1  E ,  ˆ  1  2  

(2.4)

here   G is the shear modulus,  – the Poisson ratio,  – the volume expansion (Eqn. 1.30),  is the hydrostatic pressure (1.8), Eˆ – the unit tensor. Fifteen equations of three groups contain the same number of unknowns: twelve components of two symmetric tensors of the second  rank ˆ , ˆ and three components of the u vector.

2.1.2. Boundary conditions

The conditions on the boundary surface (S) are added to the system of equations (2.1) - (2.3) determining the behavior of a linearly elastic body at points of its volume (V). There are three types of problems. In the first problem, the kinematic boundary condition is set:

  u S  u* ( x1 , x2 , x3 ), 37

(2.5)

Introduction to Deformable Solid Mechanics

S is a surface of V volume. The coordinates x1 , x 2 , x3 are related by the surface equation,  The second boundary problem is static. The distribution of q surface forces is given and the equation of equilibrium on the surface is the boundary condition

  n  ˆ S  q ,

(2.6)

 where n is the unit vector of the outer normal to the body surface. The third boundary problem is mixed. The kinematic boundary condition is set on the S1 part and the static boundary condition is given on the S2 part:   u S  u* ( x1 , x2 , x3 ), 1 (2.7)   n  ˆ S  q. 2

Two methods for solving problems in the theory of elasticity are  known. The first one is aimed at finding a displacement u vector, using which it is not difficult to calculate the strain tensor ˆ , and using the last one it is easy to find the stress tensor. The second method is the method of solving the problem in stresses. In this method it is necessary to find such a statically possible stress tensor ˆ that helps to find a strain tensor ˆ which satisfies the continuity condition (1.45).  The displacement u vector is found by Cesaro's formula. 2.1.3. Differential equations of the theory of elasticity in displacements

Substituting in the static equation (2.1) the expression for the  stress tensor (2.3) through the displacement u vector (2.2), we find the equality      2  div   Eˆ   u   u *   F  0, 1  2  38

2. Basic concepts and problems of the linear theory of elasticity

which after substituting is     div u , div  Eˆ  Eˆ  grad   graddiv u ,    div  u     u   2 u ,

 2  Laplace operator , ,

 *  *  div  u      u   e S

  2u S    u k  e k et  et  x S xt xt x S

  graddiv u ,

leads to the desired differential equation   1  1 graddiv u   2 u  F  0 .  1  2

(2.8)

Projecting it on the axes of the Cartesian system, we obtain three equations, which are called the differential equations of the theory of elasticity in displacements

1 1  2 1 1  2 1 1  2

 1   2 u  Fx  0,  x  1   2 v  Fy  0,  y  1   2 w Fz  0,  z

(2.9)



 u v w    divu . x y z

(2.10)

so

They were first obtained by Navier and Cauchy. These equations (2.9) are the differential equation for the volumetric expansion  1 1  2 (2.11)  2  divF  0. 2  1  39

Introduction to Deformable Solid Mechanics

Using the relation (2.12)

   rotrot u  graddiv u   2u ,

(2.12)

it is possible to give equation (2.8) another form   1  2(1  ) graddiv u  rotrot u  F  0.  1  2

In the absence of volume forces, the volume expansion harmonic function according to Eq. (2.11)

2  0,

(2.13)



is a

(2.14)

 u is a biharmonic vector:  4 4 4  4 u  0,  u  0,  v  0,  w  0 .

(2.15)

The boundary condition (2.6) is written in terms of the displacement vector in the form

       q  n  ˆ  2   n  n  ˆ    1  2     1     2  n  n   u  n  rot u . 2  1  2 

(2.16)

2.1.4. Influence of the temperature field

At  temperature and in the absence of external forces (then

ˆ = 0), the strain tensor in the test cube is spherical and is determined by the equality

ˆ  ˆ1   Eˆ , I1 ˆ     3 , 40

(2.17)

2. Basic concepts and problems of the linear theory of elasticity

here α is the coefficient of linear expansion. On the basis of the superposition principle, from Eqs (2.17) and (2.4) we find:

ˆ  So we have

3 1   Eˆ   Eˆ ,  ˆ  1  2  

(2.18)

1     Eˆ  ˆ   Eˆ . 1  2  1  2 

ˆ  2  

(2.19)





  It is further assumed that there are no external forces F  0, q  0 . Assuming that the heat conduction problem can be considered independently of the elasticity problem, repeating the derivation of the equations in the displacements (2.8) taking into consideration the temperature term in Eq. (2.19), we obtain:

  1 1  graddiv u   2 u  2 grad  0. 1  2 1  2

(2.20)

The boundary condition (2.16) can be written in the form



    1  1   ndiv u  n   u  n  rot u  n  0. 1  2 2 1  2

(2.21)

2.1.5. Solution in stresses. The Beltrami-Michell dependences

In the absence of volume forces, the first invariant of the stress tensor  will be a harmonic function:

2  0.

(2.22)

The stress tensor ˆ that satisfies the equations of statics in the volume must be chosen so that the strain tensor, calculated from it, satisfies the consistency conditions (1.45): 41

Introduction to Deformable Solid Mechanics

Inkˆ  rot (rot )* 

1 3 ˆ   E   0, Ink ˆ  2 1   

(2.23)

where Ink is the first letters of the word "inkompatibilitat" – incompatibility. Using the statics equation, you can convert this relation to an easily observable and memorizable form:  2ˆ 

3   0. 1 

(2.24)

In components of the Cartesian coordinate system:

 2 x 

3  2  0, 1  x 2

 2 xy 

3  2  0, 1  xy

 2 y 

3  2  0, 1  y 2

 2 yz 

3  2  0, 1  yz

 2 z 

3  2  0, 1  z 2

 2 zx 

3  2  0. 1  zx

(2.25)

2.2. The Clapeyron equation. The Kirchhoff theorem

The potential energy of deformation of an elastic body is determined by the integral over the volume of the specific potential energy:

~ U   Udv.

(2.26)

V

This quantity is equal to half of the work of external forces on the sequence of equilibrium states in the process of transition of a linearly elastic body from its natural state. The proof is based on the equality

     ˆ u  ( div   F )dv   (q  n  ˆ )  uds  0. 

V

S

42

(2.27)

2. Basic concepts and problems of the linear theory of elasticity

 The formula for the vector divergence ˆ  u is:

  div (u  ˆ )  u  div ˆ  ˆ  ˆ.

(2.28)

Now taking into account the determination of the specific potential energy (1.60) and relation (2.28), we obtain:





 u  divˆdv   div u  ˆ dv   ˆ  ˆdv 

V

V

    n  ˆ  uds   2Udv. S

V

(2.29)

V

The substitution of Eq. (2.29) into Eq. (2.27) leads to the desired relation    ~ 1   (2.30) U    F  u dv   q  u ds , 2 V S 

~ here U is determined by the relation (2.26). Equation (2.30) is the Clapeyron equation. The potential energy of deformation of an elastic body is equal to half of the work of external forces on the sequence of equilibrium states of a linearly elastic body from its natural state. From these energy concepts it follows that ~ U 0, such a statement is equivalent to the local property (1.62). The Kirchhoff theorem. The following assumptions are introduced: 1) the initial state of the body is natural; 2) the specific potential energy satisfies the condition (1.62) at each point of the body; 3) the neglect in changing the body shape is generally accepted in the linear theory of elasticity when the boundary conditions are formulated – bounding the elastic body, the surface S in the equilibrium state as it is in the natural state 43

Introduction to Deformable Solid Mechanics

 ij  1, k  1 .

(2.31)

The boundary problems with the given conditions have the only one solution (Kirchhoff's theorem). Indeed, assuming the existence of   two different solutions u , ˆ  and u , ˆ  with the same volume forces V , surface forces on S 2 and the displacement vector on obtain that the differences

   u  u   u ,

S1 , we would

ˆ  ˆ   ˆ 

(2.32)

are the solution of the homogeneous boundary problem divˆ  0, ˆ  ˆ * ,    Eˆ  ˆ ,  1  2    1 ˆ  u  u * , 2

(2.33)

ˆ  2 





 u S  0,

 n  ˆ

1

S2

 0.

(2.34)

From these equations it follows (see Eq. 2.28)









 u  divˆdv   divu ˆ dv   ˆ  ˆdv   n ˆ  uds   ˆ  ˆdv  0.

V

V

V

S

V

But the surface integral is zero according to (2.34), so

 ˆ  ˆdv  2 Udv 0 ,

V

(2.35)

V

and because of the positive sign of the potential strain energy U  0,

ˆ  0, ˆ  0. 44

(2.36)

2. Basic concepts and problems of the linear theory of elasticity

From Eq. (2.32) we obtain

  u   u , ˆ   ˆ , which contradicts the assumption that there are two solutions different from each other.

2.3. The plane problem of the elasticity theory 2.3.1. Flat deformation

In the problem of plane (flat) deformation, we consider a particular solution of equations of the elasticity theory in which the displacements are

u  u( x, y), v  v( x, y), w  const.

(2.37)

An obvious consequence of these assumptions is the absence of stresses  zx , zy :

 x   x (x, y),  y   y (x, y),  z   z (x, y),  xy  xy (x, y),  zx  0,  zy  0.

(2.38)

The generalized Hooke's law under these conditions is written as:

 v v  u   v 1  ,  y  2  1  , y  x   1  2v  1  2v

 x  2 

 u

v 

 xy     ,  y x  here

1 

 z  2

v 1 , 1  2v

u v 1  2   ( x   y ). x y 2 45

(2.39)

(2.40)

Introduction to Deformable Solid Mechanics

On the basis of Eqs. (2.39), (2.40), we transform the expression for  z to the form (2.41)  z   ( x   y ), after that the problem of  z determination is not urgent, so we start to calculate the plane stress field  x ,  y , xy .

Plane deformation is realized in a prismatic body, theoretically of infinite length loaded by surface and volume forces perpendicular to the z  x3 axis, the intensity of which does not depend on x3 (Figure 2.1: b). Then all cross sections of the body are in the same conditions, which justifies the expression of displacements in the form (2.37). An approximately flat deformation occurs at a distance from the ends of the body in its middle part on a finite extension along the z  x3 axis.

Fig. 2.1. a is a plane stress state; b is a plane deformation

The static equations of a plane problem in a volume are written as  x  xy   Fx  0, x y  yx  y   Fy  0, x y

(2.42)

but on the surface, in other words, on the Г contour of the cross section of the body, 46

2. Basic concepts and problems of the linear theory of elasticity

 :  x n x   xy n y  q x ,

(2.43)

 yx n x   y n y  q y . 2.3.2. Airy stress function

Consider homogeneous statics equations:  x  xy   0, x y

 yx x



 y y

 0.

(2.44)

Each of the two homogeneous static equations (2.44) is identically satisfied by the function Ф(х,у), the Airy stress function:

x 

 2Ф , y 2

y 

 2Ф , x 2

 xy  

 2Ф . xy

(2.45)

Certainly, they are easily verified by substitution in Eq. (2.44). Using  x0 ,  y0 , xy0 particular solutions (not depending on z) of the equations of statics (2.42), determined by the action of volume forces, we have

 x   x0 

 2Ф , y 2

 y   y0 

 z   x0   y0  2Ф,

 2Ф , x 2

0  xy   xy 

 2Ф , xy

(2.46)

 2 is Laplacian with respect to two variables x, y. 2.3.3. The differential equation for stress function

In the conditions of plane deformation (2.41): 

1  2  . 3 47

(2.47)

Introduction to Deformable Solid Mechanics

It remains to require that the Airy function satisfy the homogeneous Beltrami-Michell relations (2.25): 2

 2  2 2  2  Ф  0, у 2 x

 2

2

 2  2 2  2    0, x 2 y

 2Ф  2 2    0,  2  2   0,  xу xу

which leads to the equation  2 2   0 .

(2.48)

Thus, the Airy function satisfies the biharmonic equation (2.48). 2.3.4. The plane stress condition

In this case the volume and surface forces are also perpendicular to the z  x3 Fz  0, qz  0 axis (Figure 2.1: a). It is assumed that a particular solution, corresponding to the action of body forces in an elastic body. is known. Therefore, in the subsequent discussion, the volume forces are not taken into account. We consider a stress state in which there are no stresses on areas perpendicular to the z axis:

 z  0,  xz  0,  yz  0.

(2.49)

It is called a plane stress state. Of course, homogeneous static equations in the volume are written in the form (2.44), and they can be satisfied by introducing the Airy stress function

x 

 2 , y 2

 xy  

 2 , x  y 48

y 

 2 . x 2

(2.50)

2. Basic concepts and problems of the linear theory of elasticity

However, it does not mean that this function, as well as the stresses, does not depend on z. Indeed, the Beltrami-Michell dependences should be written in the form 2

 2  4 1 2 2      0, у 2 z 2 у 2 1   x 2

2

 4 1 2 2  2     0,  x 2 z 2 x 2 1   y 2

 2

 4  2 1 2   2   0,  2 xy z xy 1   xy

1 2  2   0, 1   yz

(2.51)

1 2  2   0, 1   zx

1 2 2    0. 1   z 2

Combining the first and second equations and taking into account the sixth, it is easy to see that the stress function is biharmonic in the variables x and y in the plane stress state:

 . 

(2.52)

 2    c,  2   cz  a( x, y), z

(2.53)

 2 2  0

 2 2 2    2  2 x y 

The three last equations (2.51) give

where c is a constant, and a(x,y) – a harmonic function. Taking into account Eq. (2.49), from (2.51)-(2.53) we obtain the following expression for the stress function: ( x, y, z )  1 ( x, y) 

 z2 2  1 ( x, y)  z 2 ( x, y)  2(1  )

1  cz( x 2  y 2 ), 4 49

(2.54)

Introduction to Deformable Solid Mechanics

where 1 is a biharmonic function,  2 is a harmonic function:

 2 2  1 ( x , y )  0,

 2 2 ( x, y )  0.

In this case the stresses turn out to be quadratically dependent on z. Therefore, a plane stress state is realized in an elastic body only under the condition that the forces on its lateral surface are distributed according to the same law. 2.3.5. Generalized plane stress condition

Defining the body size along the z axis as 2h and assuming that the stress state in the body is symmetric with respect to the median plane of the body z = 0, so we can rewrite (2.54) in the form 



 h2    z 2   2  , 2(1   )  3 

(2.55)

here 1   



h2 2  , 2(1   ) 3

 2 1   2 ,

Ψ is a biharmonic function. The stresses determined by formulas (2.50) are equal to

х 

2 2 2   h2 2       z  y 2 , y 2 2(1   )  3 

у 

2 2 2   h2 2       z  x 2 , x 2 2(1   )  3 

 xy  

(2.56)

  2 2  2   h2   z 2   . xy 2(1   )  3  xy

Then it is assumed that the body represents a plate of thickness 2h, small in comparison with its dimensions in the plan. This allows 50

2. Basic concepts and problems of the linear theory of elasticity

us to assume that its stress state can be described with sufficient accuracy by specifying the average thickness values of the stresses h

h

1 1  x    x dz,  xy   xy dz, 2h h 2h h

h

1  y    y dz. 2h h

(2.57)

 h2 2   h 3  z  dz  0 we obtain h

Then using the average values of

x 

 2 , y 2

 xy  

 2 , xy

y 

 2 . x 2

(2.58)

These equations (2.58) determine the generalized plane stress state. The surface forces should also be averaged over the thickness of the plate: h h 1 1 (2.59) qx  q ( z ) dz , q  q y ( z )dz. y x 2h h 2h h The generalized Hooke's law in the absence of  z , zx , zy is written in the form u v 1   xy ,  y x  v w  yz    0, z y w u  zx    0. x z

u 1  ( x  y ), x E v 1  y   ( y  x ), y E w  z    ( x   y ), z E

x 

 xy 

(2.60)

To determine the average values of the u, v displacement values, we use equations u 1 v 1  ( y  x ),  ( x  y ), y E x E u v 1    xy . y x  51

(2.61)

Introduction to Deformable Solid Mechanics

Then according to the third relation (2.60), we find: w( x , y , h )  w( x , y ,  h )  

2 h 1 

 u  v   .   x y 

(2.62)

The remaining equations (2.60) determine the differences between u, v values on the sides of the plate through x, y derivatives of the average value w ( x , y ) but the latter remains unknown. 2.3.6. The plane problem

Further in the notations of averaged by thickness values describing the generalized stress state, the dashes are not used and the notation Φ is used instead of  for the stress function. Then the equations in § 2.3.5 can be rewritten as:  2  2  2    ,  ,  , xy y xy y 2 x 2 u 1   0 ),  ( x  x 2 1  v 1 u v 1      xy . ( y   0 ), y 2 y x  1 

x 

Here

0   x   y.

(2.63)

(2.64)

(2.65)

In the case of plane deformation the equations (2.63) are preserved, but the generalized Hooke's law according to Eq. (2.39) and Eq. (2.40) can be written in the form v 1 u v 1 u 1  ( x 0 ),  ( y 0 ),    xy, x 2 y 2 y x 

 0 is determined by Eq. (2.65). 52

(2.66)

2. Basic concepts and problems of the linear theory of elasticity

From the comparison of these formulas it follows that having the solution for the plane deformation problem and using the substitution



 1

(2.67)

it is possible to obtain a solution for the corresponding problem for a generalized plane stress state. In both cases, we consider the same biharmonic boundary-value problem, called the plane problem of the elasticity theory, or the theory of plane elasticity. 2.3.7. Displacements in the plane problem

The definition of u and v displacements reduces to integrating the system of equations (2.66), in which the stresses are replaced by their relations (2.63) in terms of ( x, y) biharmonic stress function. This system of three equations, containing two unknown functions, is integrable, since the Beltrami-Michell relation (2.48) is equivalent to the continuity conditions. 2.3.8. Polarity coordinates on the plane

In these coordinates, denoted by r ,  , the equilibrium equations take the form (1.47):

 r 1  r  r      0,  r r r   r 1   2    r  0. r r  r

(2.68)

The deformation coordinates are given by equations

r   r

ur 1 u ur  , ,   r  r r u u 1 ur      . r r  r 53

(2.69)

Introduction to Deformable Solid Mechanics

Finding out the stress equations through the stress function is identical to the homogeneous equilibrium equations (2.68) which are represented as  

 2 , r 2

 r

1  2  1   r 2  2 r r   1     . r  r  

r 

(2.70)

Here (r, ) satisfies the biharmonic equation (2.48), where the Laplace operator is: 2 

2 1  1 2   . r 2 r r r 2  2

(2.71)

2.4. Concentration of stresses around holes 2.4.1. All-sided plane stretching with a circular hole

Let's consider an elastic unbounded flat plate which is stretched in all directions by constant stresses  0 at infinity. In this case, the plate has a generalized plane stress state with a uniform stress distribution in the xy plane which coincides with the middle plane of the plate:  x   y   0 ,  xy  0 ,  z  0 ; or in a cylindrical coordinate system (r, θ, z)

 r      0 ,  r  0 ,  z  0 .

(2.72)

Draw a circle with а radius which is centered at the origin and mentally delete the interior of the circle. The action of the mentally discarded part is replaced by surface forces applied along the contour of the circle: 54

2. Basic concepts and problems of the linear theory of elasticity

 r      0 , r  0 when r  a . Suppose that the external stresses on the circumference contour are mentally (quasistatically) decreasing to zero. In this case a redistribution of stresses occurs in the plate. Let us determine the distribution of stresses in a flat, uniformly stretched  0  const , unlimited at infinity, plate with a circular hole with а radius. There are no external forces on the boundary of the hole:

r r

r a r 

 0;   0,

(2.73)

here  0 is a constant stress. It is obvious that the stress state of a uniformly stretched plate with a circular hole is independent of the  angle. The (r ) general integral of the biharmonic equation (2.48, 2.71) has the form

 ( r )  A ln r  Br 2 ln r  Cr 2  D ,

(2.74)

where A, B, C, D are arbitrary constants. The equations (2.70) expressing the components of the stress tensor through the Airy function (r) (2.74) are

A  2B ln r  B  2C , r2 A     2  2 B ln r  3B  2C , r  r  0.

r 

Using the boundary conditions (2.73), we find

A   0 a 2 , B  0, 2C   0 . 55

Introduction to Deformable Solid Mechanics

Therefore, the solution of the problem in stresses is represented by 

 r   0 1  



    0 1  

a2  , r 2  a2  , r 2 

(2.75)

 r  0,  z  0. The stress component

  increases when it approaches the boun-

dary of the circular hole and the maximum value   is reached at the boundary of the cut circle (Figure 2.2):

 r a  ( )max  2 0  2( )

(2.76)

In this case, it is said that the concentration coefficient is equal to 2. In the general case, it is calculated using

 max , H where  max is the maximum local stress caused by the stress concentrator, and  H is the "nominal" stress that would have occurred in the absence of the concentrator.

Fig. 2.2. Uniform stretching of the plane with a circular hole

56

2. Basic concepts and problems of the linear theory of elasticity

2.4.2. Variable strengthening of a plane with a circular hole (the Kirsh problem) (figure 2.3)

If we remove some of the loads and leave only the tension along one axis (Figure 2.3), as Kirsch showed, the stress concentration does not decrease, it conversely increases significantly. Kirsch's solution can be written as: 1 2

 

 a 2   3a 4 4a 2    1  4  2  cos 2  , 2  r   r r  

1



a2  

1



 r   0 1 



3a 4 

    0 1  2   1  4  cos 2  , r  2  r    3a 4

(2.77)

2a 2 

 r   0 1  4  2  sin 2 . r r  2 

From the equation (2.77) it follows that on the circumference of the hole when r  a :  r  0,  r  0 , then

    0 (1  2 cos 2 ). The   value reaches a maximum at the point   (0,  ) (Figure 2.3); at these points (A) is equal to

 ra  ( )max  30 , so, the concentration coefficient is equal to 3 in this problem. y

Fig. 2.3. Uniaxial stretching of a plane with a circular hole

57

Introduction to Deformable Solid Mechanics

2.4.3. Variable strengthening of a plane with an elliptical hole (Kolosov-Inglis problem) (figure 2.4)

The obtained equations are very complicated, therefore they are not given. We only note an extremely important result: the maximum stresses are observed at the vertices A of the ellipse, where

 

a b

 y   0  1  2 .

(2.78)

Fig. 2.4. Uniaxial stretching of the plane with an elliptical hole

According to this equation the stresses at the vertices of a narrow ellipse ( a b – large) can become very large. 2.4.4. Variable strengthening of a plane with any shape holes (figure 2.5) b2 (called the radius of curvature a at the hole vertex) to the equation (2.78), we obtain

If we introduce a quantity  



 y   0 1  2 

a .   

(2.79)

It turned out that in this form the expression for concentrations is applicable not only to elliptical holes, but also to holes of any shape, 58

2. Basic concepts and problems of the linear theory of elasticity

on the contour of which there is a point with a small radius of curvature (Figure 2.5). In any case, the concentration of stresses is determined by the depth of the cut and the radius of curvature of the contour at its vertex. It does not depend on the shape of the contour.

Fig. 2.5. Uniaxial stretching of the plane with any shape hole

2.5. Stability of elastic bodies 2.5.1. Problems of stability of deformable bodies

To study the stability of continuous media let's consider the stability of equilibrium forms and call this state unperturbed. We designate the corresponding components of the stress and strain tensors as the basic ones and denote all basic quantities by the index "0": (2.80)  ij0 ,  ij0 , ui0 ,... Along with the basic forms (the unperturbed equilibrium form) we can consider the perturbed equilibrium forms. All the quantities related to the perturbed equilibrium form are denoted with a prime:

ij , ij , ui ,... 59

(2.81)

Introduction to Deformable Solid Mechanics

There are two approaches to the study of the stability of elastic bodies: 1) a dynamic approach; 2) a static approach. The study of stability under a dynamic approach is aimed at investigation of the behavior of perturbations in time. Let's start from the static stability and assume that the external loads are proportional to some parameter. So it is necessary to find such a parameter value, for which there exists an unperturbed equilibrium form and the perturbed form at the same time. This problem is to find the bifurcation point of equilibrium forms. Following Poincaré, we use the linearized equations (equations in variations) to study the equilibrium bifurcation. The parameter value is called critical when it corresponds to the bifurcation point. Following Poincaré, we write:

ij  ij0 ij ,

ij  ij0  ij , ui  ui0  ui ,...,

(2.82)

here  ij ,  ij , u i are the perturbation components and they are considered small values in comparison with the values with index "0". So, the critical value of the parameter is found from the solution of the linearized problems. Some information about the methods of solving the elastic stability problem. Not long ago it was believed that the elastic stability problem of the bodies' equilibrium can be solved only by the methods of the applied theory of elasticity or resistance of materials without the equations of the mathematical elasticity theory. As A.Yu. Ishlinsky noticed, "Kirchhoff's well-known theorem about the uniqueness of the solution of the elasticity theory equations played a negative role here as the problem of elastic bodies' stability in the framework of the mathematical theory of elasticity is unsolvable. However, it is necessary to remember that the proof of Kirchhoff's theorem is essentially based on the usage of the boundary conditions of the problem about the equilibrium of an elastic body not taking into account the deformation of the boundary surface, so, it cannot be considered sufficiently substantiated». 60

2. Basic concepts and problems of the linear theory of elasticity

V. Novozhilov considered the general problem of equilibrium stability of the elastic body in terms of the nonlinear elasticity theory which takes into account the rotation components both in the equilibrium equations and in the boundary conditions. There were some significant difficulties in solving specific problems. Only when A.Yu. Ishlinsky proved that the loss of stability in elastic systems occurs mainly due to the deformation of the surface boundary, the account of which allows us to find successful solutions to the problems of stability of elastic systems by the methods of the mathematical elasticity theory, it became possible to use the solutions of the linear elasticity theory to find the critical force. 2.5.2. Stripe strength

Let's solve this problem using L. Leibenson – A. Ishlinsky method. Consider a rectangle stripe with 2a length and 2 b thickness with a >> b condition and this stripe is compressed by a longitudinal force p uniformly distributed along its thickness (Figure 2.6). A uniformly distributed pressure q is applied along the long sides of the stripe. The own weight of the stripe is not taken into account.

Fig. 2.6. Stripe loading scheme

The main stress condition for the stripe of the mentioned system of external forces is determined by:

 x0   p,  y0  q,  xy0  0, here p  q . 61

(2.83)

Introduction to Deformable Solid Mechanics

Under some values of parameters of the external p force alomg with the main equilibrium condition when the boundaries (long sides of the stripe) are straight there may occur another condition with not straight boundaries. The elements of the body boundary make rotation with respect to its coordinate axes.  Let  and m be directing cosines of n normal to the curvative boundary of the stripe (Figure 2.7).  The directing cosine is   cos( n 0 , x )  sin  . Considering  is infinitely small, we can write   ,   tg   

v , so x

v , x



(2.84)

where u 2  v , u1  u (2.81). Moreover,

 m  cos( n 0 , y )  cos   1 .

(2.85)

Fig. 2.7. Directing cosines of the normal to the curvative boundary of the stripe

The boundary conditions on the curvative boundary are:

 

0 x 0 xy

    

 m   qm .

  x    xy0   xy m   q  ,   xy

0 y

62

 y

(2.86)

2. Basic concepts and problems of the linear theory of elasticity

Substituting the values from Eq. (2.83) into Eqs. (2.86) taking into account Eqs. (2.84)-(2.85) and neglecting small values, we obtain

 y  0,  xy  ( ð  q )

(2.87)

v  0. x

They are linearized boundary conditions of the elastic stable stripe. According to Leibenson – Ishlinsky method, the stress components  x ,  y ,  xy can be determined by the Airy function (Eqn. 2.63). The solution of biharmonic equation (2.48) for the stripe is  ( x , y )   Ach   Bsh   C  ch   D  sh  cos

here  

 

 

x,

(2.88)

y,  is the half-wave length of the stripe boundary distur-

bance; A,B,C,D are arbitrary constants. Using boundary conditions of elastic stability (Eq. 2.87) considering relations (2.63), (2.66) and (2.88), we find out critical stress:

 p  q 0  2

chsh  

1  2v chsh  

,

(2.89)



b. When q  0 , we obtain the Ishlinsky equation. He  showed that there is no loss of the stripe stability under all-around compression in the condition of the plane problem.

where  

2.5.3. Stability of circular cavity

The stability of the equilibrium state around a circular cavity under all-around compression is studied according to the LeibenzonIshlinsky approach. The case of plane deformation in the xoy plane is studied (Fig. 2.2:  0   p ). 63

Introduction to Deformable Solid Mechanics

The problem of the compression stability of a plane weakened by a circular notch under the conditions of plane deformation is considered. The plane at infinity is compressed by uniformly distributed p forces. Then the subcritical stress state will be written as: 

 r0   p1  

a2  , r 2 



 0   p1  

a2  , r 2 

(2.90)

 r0  0, here а is the radius of a cavity. It is necessary to find out if along with the basic equilibrium states (2.90) of the circular cavity there is another equilibrium state, infinitely close to the basic one, in which the cavity boundaries are also free of stress. The stresses  r , and  r have to satisfy the boundary conditions at r = a:  r  0, (2.91) 1 ur  0,  r   0 r  here ur and u are infinitely small additional displacements of the plate points from their position in u r and u initially deformed state (2.82). The displacements due to locality of the perturbation zone satisfy the following condition: 0

0

ur r   0, u r   0.

(2.92)

According to the Leibenzon-Ishlinsky approach, the components

 r ,   ,  r  can be determined through the Airy function (2.70). For a circular cavity, the Airy function is chosen in the form:





 ( r ,  )  Ar m  Br m  2  Cr  m  Dr  m  2 cos  , (2.93) here m  2,...,, A, B, C, D are arbitrary constants. 64

2. Basic concepts and problems of the linear theory of elasticity

The exact solution of problem (2.90) - (2.93) will be written as (A.V. Navoyan):  m2  1 (2.94) , p mm  2  2 m  1 where  is the shear modulus, v is the Poisson ratio. The right side of Eq. (2.94) is a monotonous function with respect to m , the critical load: 3 (2.95) p0  . 42  3v 

2.6. Solving the problem of elastic semi-plane by means of fourier transformation

Integral transformations allow us to reduce the problem of integrating partial differential equations to the integration of the ordinary differential equation system to represent the unknown functions. Let's remind the definition and the basic properties of the Fourier transformation. If f(x) is an integrable function, x    ,   then the Fourier image is a function of the real variable p f ( p) 

1 2



 f ( x)e

ipx

dx,



Conversely, if the Fourier image of certain functions f (p) is known, then the function itself is reconstructed by f ( x) 

1 2



 f ( p )e

 ipx

dp .



Let's consider the following problem. The elastic medium fills the half-plane    y  0 , the x-axis is the boundary of the half-plane. Suppose, it is given on the boundary: 65

Introduction to Deformable Solid Mechanics

 y ( x , 0 )   q ( x ),

 xy ( x , 0 )  0 .

(2.96)

Write out the equilibrium equations:

 x  xy   0, x y

 xy x



 y y

 0.

(2.97)

Multiply equations (2.97) by e  ipx and integrate with respect to x in the range from   to   . Using the properties of the Fourier transformation, we obtain the equilibrium equations

 ip x   xy  0,  ip xy   y  0,

(2.98)

where the primes denote differentiation with respect to the variable y. Using the Fourier transformation for the boundary conditions (2.96), we obtain (2.99)  y ( p , 0 )   q ( p ),  xy ( p , 0 )  0 . It follows from Eq. (2.22) and Eq. (2.24) that

 2 2ˆ  0,

(2.100)

thus, each of the stress tensor components is a biharmonic function, so

 4 y x 4

2

 4 y x 2 y 2



 4 y y 4

 0.

Applying the Fourier transformation to this equation, we obtain for the Fourier image  y the following ordinary differential equation:

 yIV  2 p 2 yII  p 4 y  0 . The equation solution is written as  y   A Bye p y . 66

(2.101)

2. Basic concepts and problems of the linear theory of elasticity

It can be called gluing of two solutions,  p and  p are the double roots of the characteristic equation, the solution is considered in the lower half-plane  y  p ,    0 , so if p  0 then we have to

take a solution corresponding to  p root, and if p  0 , we take the solution corresponding to  p root. Differentiating Eq. (2.101) with respect to y, we obtain

 y  A p  B p y  B e p y . Substituting into the boundary condition (2.99) considering Eq. (2.98), we obtain

A  q( p), B p q( p). Now, Eq. (2.101) can be rewritten as

 y  q ( p)(1  p y)e p y .

(2.102)

Using Eq. (2.98), we find

 xy  iq ( p) pye p y ,

 x  q ( p )(1  p y )e p y .

(2.103)

Using the equation, we find stresses 

x  

q ( ) y ( x   ) 2 d ,   r4

y  

q ( ) y 3 d ,   r 4

 xy  

q ( ) y 2 ( x   ) d ,   r4

2

2

2







here r 2  (x  )2  y2. 67

(2.104)

Introduction to Deformable Solid Mechanics

2.7. The concentrated force on the boundary of the half-plane. Contact problem

Consider

 P / 2 ,       . q ( )       ,    0, When   0 , we find

x  

2P x 2 y 2P y 3 ,   ,  y  r4  r4

 xy  

2P xy2 .  r4

(2.105)

These equations give a solution to the problem of a concentrated force applied to the boundary of an elastic half-plane. The obtained solution, as any other solution to the problem of action of a concentrated force, should not be understood word-for-word according to the paragraph title. Indeed, when x  y  0 the stresses are infinitely big. The equations (2.105) can be used in two cases: 1. The load of high intensity is distributed in a small area, for example, in a section    x   , where 

 qdx  P,



then the Eqs. (2.105) are valid in the region x , y  . 2. From the Eqs. (2.105) we can obtain the inverse transition (Eq. 2.104). To do this in Eq. (2.105), we must replace x coordinate by x   , that is, obtain a solution for the concentrated force, applied at x   point. Then P force is assumed to be equal to q( )d and the integration with respect to  is carried out. The obtained equations (2.104) contain convergent integrals and the stresses are finite if q() function is bounded. 68

2. Basic concepts and problems of the linear theory of elasticity

The stress state described by simple equations (2.105) are even simpler if we transform the tensor components   to polar coordinates, taking the point of applied force as the origin. Instead, we immediately derive the corresponding equations in the polar coordinates. Let us assume that

r  

2P cos ,   0,  r  0. r

(2.106)

Fig. 2.8. Concentrated force on the boundary of the half-plane

Here the  angle is measured from the line of force action as it is shown in Fig. 2.8. By a direct verification, we see that the equilibrium equations (2.68) will be fulfilled identically. The BeltramiMichell compatibility equations (2.24) lead to a unique condition  2 r  0 , which is also fulfilled identically. After verification, we see that equations (2.105) and (2.106) coincide. Comparison of the experimental a and theoretical b characters of the planes confirms the coincidence of the results of the elasticity theory with the experiment (Figure 2.9). We calculate the displacement  using Hooke's law

v 1   y   y   x . y E 69

Introduction to Deformable Solid Mechanics

Fig. 2.9. Comparison of the experimental a and theoretical b characters of the planes confirming the coincidence of the results of the elasticity theory with the experiment

It was accepted for certainty that a plane stressed state was realized. The values of the constants change for plane deformation. Integrating this relation, we obtain v( x,0) 

1 E



  y  x dy, 0

here   is determined by the Eqs. (2.105). Let's calculate the first integral 

0

y 3dy 1 2P  y dy    E  x 2  y 2 E 0





2



y2 z  x2

0 0  1 1  2P P   y 2  ln1  2   1.    ln1  z    2  E E   x     2   

(2.107)

When we substitute the lower limit to the right part in Eq. (2.107), the right part becomes infinite at any x value. The concentrated force in the plane problem causes infinite displacements not only at the point of its application, which would be natural, but everywhere. 70

2. Basic concepts and problems of the linear theory of elasticity

This circumstance seems paradoxical but it is an inevitable consequence of the plane problem formulation. The integral  x in the same range is a finite value which is independent of x and the value of this integral is not written out. Differentiating the obtained relation (2.107) with respect to x , we have 0

P   y2  2P 1 v ( x ,0 )  . ln 1  2    E  x  x    E x x Thus, although the displacement is infinite everywhere on y = 0 line, the derivative of the displacement or the angular tangent coefficient to the deformed boundary is finite everywhere except x  0 point. Let now the boundary of the half-plane be normally loaded q(), the angle of the tangent inclination to the curved boundary  (x) is determined by superposition

 ( x) 

2 E



q ( ) d  . x  



(2.108)

The contact problem for a half-plane is a mixed problem of the elasticity theory. When one part of the boundary is force-free or there is a force acting on it, on the other side of the boundary there is a contact with an elastic or rigid body pressed into the half-plane. Here we consider the simplest contact problem: a rigid stamp without friction is pressed into the half-plane on x   a , a  area; thus, on the contact site v x,0   g  x . Everywhere  xy  0 ,  y is equal to zero outside the contact area, on the contact area  y  q( x) . Supposing (x)  g(x) and substituting into Eq. (2.108), we obtain 1 q ( ) d     E g ( x ). 2   x a a



71

(2.109)

Introduction to Deformable Solid Mechanics

Let's consider the problem of action of a loaded plane rigid stamp on the half-plane, so that g( x)  const, g  0. Solution of Eq. (2.109) is written as:

q ( x) 

1 a  x2 2

const.

The constant coefficient is easily determined from a

 q( x)dx  P,

a

where P is the force acting on the stamp (Figure 2.10). Hence we obtain the Chaplygin-Sadovsky equation:

q(x) 

P

 a2  x2

.

(2.110)

It should be noted that peculiarity at the stamp edge is the same as in the fracture problem considered in Section 3.

Fig. 2.10. Contact problem for the half-plane

72

2. Basic concepts and problems of the linear theory of elasticity

2.8. Lame problem for a thick-walled tube (flat deformation) Lame's problem. A thick-walled cylindrical tube, the dimensions of which are shown in Fig. 2.11, is under the influence of internal pressure p, the tube is closed from the ends. Using Huber's criterion (the theory of distortion energy), find p value at which the yield stress is firstly reached.

Fig. 2.11. The loading scheme of a thick-walled tube

Solution. We introduce the cylindrical coordinates with z axis along the tube axis (Figure 2.12).  rr

 zz

 

Fig. 2.12. Components of the stress tensor

It is assumed that there is no longitudinal deformation of the tube. The deformation is a purely radial displacement u r  u  r  when the  pressure is uniform along the tube (Figure 2.11). Therefore rotu  0 and the Navier equation is reduced to: 73

Introduction to Deformable Solid Mechanics

 divu  0.

(2.111)

So

 1 d  ru  divu   rr      zz   rr      const  2c, r dr d u  cr  , d  const. r

Components of the strain tensor:

du d c 2 , dr r u d  c 2. r r

 rr  

Components of the stress tensor:

 rr  2 c  2   c  

d , r2 

 

d , r2 

   2 c  2   c   zz    rr     ,

where  ,  are the Lame coefficients, v is the Poisson ratio. Unknowns c and d can be found from the boundary conditions:

 rr   p when r  a, when r  b,  rr  0 c

pa 2 , 2   b 2  a 2 

d

pa 2 b 2 . 2  b 2  a 2  74

2. Basic concepts and problems of the linear theory of elasticity

The components of the stress tensor of a thick-walled tube in the elastic state are determined by the relations:

 rr 

pa 2  b2  1   , b2  a 2  r 2 

(2.112a)

 

pa 2  b2  1   , b2  a 2  r 2 

(2.112b)

 zz  2

pa 2 pa 2  , b2  a 2 b2  a 2

(2.112c)

here   0.5 . In this case, Hubert's fluidity criterion is:

 rr        zz    zz   rr  2

2

or

2

 6 s2 ,

i   s,

(2.113а) (2.113b)

where  i is the intensity of tangential stresses,  s is the yield strength of the material under pure shear. From Eqs. (2.112) and Eqs. (2.113) we find: 1 pa 2 b2     rr   2 2  2 , 2 b a r a  r  b,

i 

 i  max 

pb2   s. b2  a 2

Whence we have:

 a2  ps   s  1  2  .  b  75

(2.114)

Introduction to Deformable Solid Mechanics

When p  ps Eq. (2.114), two areas appear in the tube (Fig. 2.13)

k Fig. 2.13. Two areas in the thick-walled tube

2.9. Lame problem for a thick-walled spherical vessel (shell) Lame problem. A thick-walled spherical elastic shell, the dimensions of which are given in Fig. 2.14, is under the influence of increasing pressure p. Using Hubert fluidity criterion (theory of distortion energy), it is necessary to find a pressure, at which the plastic state first appeared.  Solution. We introduce spherical coordinates  r,,  . u vector

is directed everywhere along the radius and is a function only of r.  Therefore rotu  0 , Navier equation also takes the form of Eq. (2.111). From Eq. (2.111) we find:

 1 d r u divu  2  co n s t  3c, r dr d u  cr  2 . r 2

Components of the strain tensor:

 rr  c 

2d , r3

    c  76

d . r3

2. Basic concepts and problems of the linear theory of elasticity

a

r

p

b

Fig. 2.14. The loading scheme of a thick-walled spherical shell

Radial stress:

 rr 

E 2E d , c 1  2 1  r 3

where E is Young's modulus, v is Poisson's ratio, 1    1 . 2 Unknown c and d are determined from the boundary conditions

 rr   p

when

r  a,

 rr  0

when

r  b.

So we have:

1  2 pa 3  3 ,  b  a3 1   pa 3b3 d  .  b3  a 3 c

The stress tensor components of a spherical shell in the elastic state are determined by the expressions:

 rr  p

a 3  b3  1  , b  a 3  r 3  3

77

(2.115a)

Introduction to Deformable Solid Mechanics

      p

a3  b3   . 1  b3  a 3  2r 3 

(2.115b)

Considering Eq. (2.115), we obtain

i 

1 3 pa 3b 3 , a  r  b,     rr   3 2 b  a 3  r 3 3

 i  max 

3 pb 3  3 . 2 b  a 3 

Using Hubert plasticity criterion

i s, we find

ps 

a3  2   s 1  3 . 3  b 

(2.116)

when p  p s (2.116) two areas appear in the spherical shell (Figure 2.15).

Fig. 2.15. Two spheres of a spherical shell

78

2. Basic concepts and problems of the linear theory of elasticity

2.10. Lame problem for a thick-walled flat disk Lame problem. A thick-walled disk of constant thickness, the dimensions of which are shown in Fig. 2.16, is under the influence of an internal pressure p. Using Huber's criterion (the theory of deformation energy), find the value at which the yield stress is first reached. Solution. Consider the elastic equilibrium of a thick-walled disk (Figure 2.16), with pressure p inside, and no pressure outside. If the thickness of the disk is small in comparison with the radius, we can assume that the stresses are uniformly distributed over the thickness (Fig. 2.16) and, therefore, are independent of z coordinate. The equation of equilibrium has the form

d rr  rr      0. dr r

(2.117)

The solution of Eq. (2.117) is found in the form

 rr 

1 dF  F '. F ,    r dr

The F  r  function is found from the equation of compatibility of deformations:

1 1 F ''  F '  2 F  0. r r The general integral of this equation will be

B F (r)  Ar  . r Then

 rr  A 

B B ,    A  2 . 2 r r 79

(2.118)

Introduction to Deformable Solid Mechanics

The unknown A and B are found from the boundary conditions:

 rr   p

when

 rr  0

when

A p

r  a, r b,

a2 a 2b2 , B   p . b2  a 2 b2  a 2

Fig. 2.16. Loading scheme of thick-walled thin disk

Substituting the obtained values in Eq. (2.118), we have:

 b2   1  2  , r   2  a b2   ,  p 2  1 b  a 2  r 2   zz  0.

 rr  p  

a2 b2  a2

In this case, the shear stress intensity is

i 

1 1  rr2  2   rr  2 , 3

considering Eq. (2.119) we find: 80

(2.119a) (2.119b) (2.119c)

2. Basic concepts and problems of the linear theory of elasticity 1

 a2 b4  2  i  p 2 1 3 ,   b  a 2  3  r 4  a  r  b, 1

 i  max 

 pa 2 b4  2 1 3 .   a 4  3 b2  a 2  

Using Hubert fluidity criterion (theory of the deformation energy)

i  s, we find  b2  b4  ps  3 s  2  1   1  3 4  a  a 

1

2

.

(2.120)

There are two areas in the disk (Figure 2.17) when p  p s (2.120).

k Fig. 2.17. Two areas in a thick-walled thin disk

81

Introduction to Deformable Solid Mechanics

3 LINEAR FRACTURE MECHANICS

When we talk about destruction of solids, it means their destruction into pieces which is connected with formation of new surfaces. The process of the new surfaces formation can be described as fracture grows. The term "crack" is used here in general sense and it refers only to geometrical change. In fact, the process of fracture growth can proceed differently depending on the material and the conditions of its loading. This process is described by different models. Linear fracture mechanics, which originated in well-known Griffith’s research, is one of the fields of fracture mechanics. In analyzing and evaluating this theory it is necessary to distinguish whether we observe submicroscopic cracks in the crystalline grain where the lattice near the crack can be considered correct, or we see the fracture in a big body where the stress concentration region covers large volumes of polycrystalline material. In the first case, V.V. Novozhilov gave an original interpretation of crack formation as instability of a nonlinear mechanical system, which is an atomic lattice. The analysis of the plastic stressed and deformed state near the end of the crack is crucial to judge about the possibility of further propagation of the appeared macrocrack in a polycrystalline material. In low-plastic materials it is not necessary to give a detailed description of the state near the end of the crack but it is sufficient to know the value of the coefficient at the singular term. In solving the problem 82

3. Linear fracture mechanics

of the elasticity theory, this coefficient is compared with a certain integral constant of the material. To find this constant, the experimental methods included in the standards are used. Linear fracture mechanics is based on the following general idea. Material properties – elastic or plastic – are assumed to be unchanged and predetermined. Spreading of a crack is the result of local processes occurring at its end and is described by means of some methods (Rabotnov Yu.N.).

3.1. Microstructural mechanisms of deformation and fracture Consider the ideal crystal structure in the form of an ordered series of molecules (atoms) (Fig. 3.1) interacting in accordance with the Van der-Waals law

  Ar   Br   ,   P   . S 

(3.1)

Fig. 3.1. The ideal crystal structure in the form of ordered rows of molecules (atoms) interacting in accordance with the Van der-Waals law

Here  is the specific force, P is the tensile force, S  nb2 , n is the number of molecules on S area when x2  0 , b and a are the 83

Introduction to Deformable Solid Mechanics

intermolecular distances in x1 and x 2 directions. The variable r is the current distance between the neighbor horizontal rows of molecules (Figure 3.2), A, B, ,  are constants, characterizing the type of interaction. A characteristic diagram of the dependence (3.1) is shown in Fig. 3.2. The a point corresponds to the distance between the horizontal rows in the initial unstressed state and on the basis of Eq. (3.1) is given by 1

 B    a  .  A

(3.2)

The angular slope of the curve at r  a point, the value r  r0 corresponding to  maximum, and this maximum  th considering Eq. (3.2) are determined by 1    d        1  B    , r0  a   , a    dr r  a    1      B    a   ,  th       

where  th is the theoretical strength.

 th a

r0

Fig. 3.2. The dependence of Eq. (3.1)

84

r

(3.3)

3. Linear fracture mechanics

The specific force  in the macro experiment corresponds to the normal stress in x2 direction. It corresponds to axial deformation

r a

  1.

(3.4)

Therefore, the first of the equations (3.3) determines the  ~  coupling coefficient at the initial segment, which in the macro experiment is perceived as the Young's modulus d d

E

 B    a   ,

(3.5)

 0

so, from Eq. (3.3) we find

E   th    



    , 

(3.6)

here E is the modulus of elasticity in the macro experiment. Thus, the theoretical strength  th is determined only by the Young's modulus and  and  exponents. For a metallic type of connection, we can use

  7,   14,

(3.7)

thus, we found the result obtained by V.D. Klyushnikov:

 th 

E . 28

(3.8)

The actual strength b for different materials is  b  (0,01  0,1)  th , and it means that the given calculation does not reflect the real behavior of the atomic-molecular structure. 85

Introduction to Deformable Solid Mechanics

There is an even greater discrepancy between the theoretical and real value of the limiting force in the case of action on the ordered molecular structure of the tangential force Q in x1 direction (Figure 3.1). If x1 = x is the displacement of atoms from equilibrium positions (Fig. 3.3), then  should grow when x increases, so  x . This dependence can be approximately written as

Fig. 3.3. Displacement of atoms from equilibrium positions

   th sin

2 x Q ,   , b S

(3.9)

thus, it is assumed that there is a sinusoidal relief (Fig. 3.4) for the forces of interatomic interaction, here  th is the theoretical shear resistance. Similar to the previous one, introducing the macrodeformation of the γ shift and the corresponding shear modulus G x a

  , G

d d

,

(3.10)

 0

on the basis of Eq. (3.9) we obtain

 th 

G b . 2 a 86

(3.11)

3. Linear fracture mechanics

This estimation (3.11) was first obtained by Ya.I. Frenkel.

 th

Fig. 3.4. Dependence     x 

Many theorists refined this estimation after Frenkel but their refinements only slightly changed the main result. When external forces are applied to an ideal crystal, the atoms are displaced from the equilibrium positions coinciding with the nodes of the crystal lattice.

Fig. 3.5. a, c – elastic deformation of the crystal lattice, b, d – plastic deformation of the crystal lattice

There is an elastic deformation of the crystal, which is characterized by the fact that the displacement of the atoms is not accompanied by a change in their nearest neighbors (i.e., there is no reconnection of the interatomic bonds), while stresses appear in the 87

Introduction to Deformable Solid Mechanics

body (Figure 3.5: a, c) (the interatomic bonds are strained). If you b unload a crystal, which is under elastic deformation  x   , it retakes  4 its original shape and dimensions. The maximum force is required to b achieve the displacement (Figure 3.4), as at this position of the node 4 the bonds are reconnected (plastic shear starts) (Fig. 3.5: b, d). An irreversible change in the shape and size of the crystal occurs in the result of plastic deformation. Note that macroscopically the same shape change can occur due to elastic or plastic deformation (Figure 3.5: c, d), in the first case (elastic deformation) the identical environment is not broken (stresses occur), in the second case there are no stresses (plastic deformation). The stress necessary for the passage of plastic deformation is the theoretical fluidity  th , in which the interatomic bonds are reconnected simultaneously over the whole plane (3.11). When the value of the limiting shear stress is b ~ a, it gives an excess over the actually observed value, which reaches three or four orders of magnitude. Based on the concept of an ideal crystal it is impossible to explain the huge difference between th and T (the critical shear stress), during which translational slip practically proceeds. The deformation pattern will be different if we assume that plastic slip is realized by moving special defects of the crystal lattice (dislocations) (Figure 3.6) (a is an ideal crystal, b is an edge dislocation and c is a spiral dislocation).

Fig. 3.6. a is an ideal crystal; b is an edge dislocation; c is a spiral dislocation)

88

3. Linear fracture mechanics

Note that something that was called above as theoretical strength, has a different meaning for stretching and shear. If in the first case the consequence of its exhaustion is the destruction in the form of division of the body into parts, then the post-critical behavior under shear is a transition to a new, irreversible type of deformation, as a rule, without loss of continuity, which brings the appearance of plastic deformations. 3.2. The variety of cracks. The Griffith experiment. The Griffith idea

A new direction in fracture mechanics is associated with the name of the English engineer, A.A. Griffith. Griffith attempted to achieve theoretical strength in experiments on breaking of freshly-drawn fine glass fibers and found that with decreasing fiber diameter their strength sharply increases (Figure 3.7) and it becomes comparable with theoretical estimates. Griffiths explained significant differences between the strength of the vast majority of real solids and the theoretical values by the cracks in them, perhaps invisible but considerably exceeding in size intermolecular distances. What is the mathematical model of the crack? Considering a real crack (Figure 3.8) in a deformable solid, it is possible to find on its boundary a line 1 – the front of the crack, on which the cavities meet, 2 – the edges of the crack. Obviously, the greatest concentration of stresses will be observed in the vicinity of the front, so in this area the material will be destructed. Griffith linked the factors of crack development in the body with the processes of accumulation and release of deformation energy in it. He realized that for the crack growth, it is necessary to do work on the formation of new material surfaces proportionally to the crack length (Figure 3.9). When the crack is spreading, the material is discharged in the adjacent areas (shaded areas), the released energy flows to the tops of the crack and is used there to destroy the material (Figure 3.9). Griffith formulated two conditions, each of which is necessary for the propagation of a crack. Simultaneous fulfillment of these two conditions is sufficient for the crack growth. 89

Introduction to Deformable Solid Mechanics

Fig. 3.7. Dependence between the  â strength of glass fibers and their thickness discovered by Griffith

y 1 2 0

x

z Fig. 3.8. The front of a real crack in a deformable solid

The first condition is that the process of crack growth should be energetically favorable, and the second condition refers to the existence of a micro mechanism capable of transforming stored energy (the presence of such a mechanism of energy conversion distinguishes fragile materials from viscous ones). In 1920 Griffith proposed the idea of determining the strength of an elastic material, in which the reason for the discrepancy  b and

 th was the cracks present in the material, which at the microstructure level means the presence of group vacancies in the form of elongated voids (cavities) that are big in comparison with molecular distances. 90

3. Linear fracture mechanics

Fig. 3.9. The material is unloaded in the adjacent areas (shaded areas) during the crack spreading, the released energy flows to the top of the crack and is used for the material destruction

Let us consider an example where a crack represents a cavity with 2c width going through an elastic body with a typical orthogonal cross section, shown in Fig. 3.10. Let such body be loaded at infinity by uniformly distributed forces, which are normal to the middle plane of the crack. So if there is no crack in the body, the uniform stress state with a normal component 22  would be realized. The crack presence leads to the disappearance of a part of the elastic energy. This energy is equal to the energy in the cylinder surrounding the crack with a square cross section 2 c  2 c . Calculated per unit length of the body in the direction orthogonal to the x1 x 2 plane, as in all subsequent calculations, it will be

4c2

2 2E

,

(3.12)

where λ is an undefined coefficient. As the stress  increases, there may come a moment when the width of the cavity strats to expand. This moment of cracks initiation corresponds to the beginning of fracture and can be considered as the ultimate. In an infinitely close moment to the crack initiation, the crack width increases to 2 dc and in the result of it there are new free surfaces with a total area equal to 4dc. 91

Introduction to Deformable Solid Mechanics

Fig. 3.10. Group vacancies in the form of a crack

It is normal to bind some specific energy  S with a newly appeared unit of a free surface, as the energy is required to create these new surfaces

4 S dc ,

(3.13)

here  S is the specific surface energy of the substance. The source of this energy is the elastic energy stored in the body, which should lose a part, additional to Eq. (3.12) and equal to Eq. (3.13), when the crack moves. Therefore, the energy balance must be fulfilled at the moment of the crack initiation

2  d   2 c 2   4 S , dc   

(3.14)

whence for the limiting (critical) value    0 it means

0 

E S . c

(3.15)

The material constant  S in this equation can be estimated from Eq. (3.1) as the work on separation of two molecular series 92

3. Linear fracture mechanics 



a

a





2 S    (r )dr   Ar   Br   dr, whence

S  On the other hand, 

S

Ea . 2(  1)(  1)

(3.16)

could be considered as the same macro-

characteristics as E module and  S could be determined from the experiment on the maximum load for one fixed crack length if there was a definite constant λ. But it is problematic to obtain an exact  0 value using Eq. (3.15) if the exact shape of the crack (cavity) and λ are unknown. As it is known (Figure 2.5), the stress concentration near the hole is determined by the hole depth and the radius of the curve contour at its top but it does not depend on the contour shape. Therefore, the main geometric parameters in the considered problem are c and  . This leads to the hypothesis of replacing the actual shape of the cavity with a simpler one. For example, according to the observed case, when we assume that the middle surface of the cavity is close to the plane, and instead of the true shape, we take a thin ellipse with c and b semiaxes along x1 and x2 to make the minimum radius of its curve



b2 c

(3.17)

which corresponds to the true radius of the curve at the tip of the real cavity, we can use the well-known solution of the classical linear theory of elasticity. The risk will be justified, firstly, if the value  is much greater than the intermolecular distance (otherwise the methods of continuum mechanics are not applicable), and secondly, if the obtained maximum deformations are small in comparison with unity (otherwise the equations of linear theory of elasticity are not applicable). 93

Introduction to Deformable Solid Mechanics

Fig. 3.11. The Griffith Problem

From Eq. (2.79) it follows (when x1   c ):



 22  1  2 

c c   2  .   

(3.18)

When the phenomenological and microstructural approaches are combined in the limiting state:

 22   th .

(3.19)

Now, considering Eqs. (3.18), (3.6), (3.15), and (3.16) on the basis of Eq. (3.19) we find

 2 2 a      1  1  

2

    0. 

(3.20)

This equation determines the relationship between the curve at the tip of the cavity and the intermolecular distance а in the limiting state. The Eq. (3.20), obtained by V.D. Klyushnikov, can be interpreted as follows. In a real body, which has a molecular structure, there can exist only cavities with a minimum radius of sharp edge curve, bigger than 94

3. Linear fracture mechanics

in Eq. (3.20); in the supercritical mode, when  max   th such cracks propagate at an accelerated rate. Cracks with the minimum curve radius at the edge when they are smaller than  0 self-collapse. For the ,  values used above in Eq. (3.7), Eq. (3.20) gives 20





a.

(3.21)

4 so it can be recognized that 3 0  a and therefore the risk of using continuity properties is justified, the same conclusion can be made about the maximum deformation:

As it will be seen further,  

 max 

 th

 0,035.

E

(3.22)

The solution of the problem about stretching of an elliptical cavity leads to the following expression for  22 on the x1 axis:

 22  

1 m  1 m  R2 ( R2  1)  2 1 2 1    R   , R2  m   R2  m  2 R2  m 





(3.23)

where

R

 1 1  m t  t 2  4m 2 2 1 m 





 ,  

(3.24)

x c b m , t  1. cb c Asymptotic analysis for small in comparison with unity

α 1 m 

ξ 2b ,  t  1, cb c 95

(3.25)

Introduction to Deformable Solid Mechanics

 is counted from the top of the cavity x1  c

where

(Figure 3.11)

and b   c , gives 2

c  ρ  1  . ρ  2ξ  ρ  2ξ 

σ 22  σ

(3.26)

Hence Eq. (3.26) gives two results:

 22 | 0  2  22 |  0  

c



, (3.27)

c . 2

The first has already been used above and refers to a narrow elliptical cavity (3.18), the second represents the limiting case of such a cavity and unlike the first, it does not contain an almost immeasurable parameter  . The second result is the solution of a simpler boundary problem, that indicates the perspective direction in which the second of the asymptotes (3.27) takes place.

3.3. Linear fracture mechanics

The presented arguments urge us to accept a mathematical cut as a model form for the cavity. In this case, the fracture condition is 

c  const. 2

This approach to the strength problem was the basis for the fracture theory proposed by Irwin, which is usually called the linear fracture mechanics (LFM) in the framework of the linear theory of elasticity. 96

3. Linear fracture mechanics

Let's consider on the basis of LFM the above problem of the orthogonally loaded crack – normal fracture crack (Figure 3.12).

Fig. 3.12. The Irwin problem

We use the Kolosov-Muskhelishvilii's method to solve this problem.

3.3.1. Kolosov-Muskhelishvilii equation

Consider

z  x1  ix 2 ,

z  x1  ix 2 .

If f  z , z  is a given function, then

f f f   , x1 z z

f  f f   i  . x2  z z 

(3.28)

The Laplace operator of f function is transformed into: 2 f  4

2 f . zz

(3.29)

We now write out Lame's equations for plane deformation, when: u3  0 :  2u1       ,1 , 2u2     ,2 ,   u , ;   1,2 . 97

Introduction to Deformable Solid Mechanics

We assume that u1  u1  z , z  and u 2  u 2  z , z  , we use Eq.(3.29) and write two Lame equations as one equation:

4

2w     ,1  i,2 , w  u1  iu2. zz

(3.30)

In the absence of volume forces,  is a harmonic function, hence it can be regarded as a real part of the analytic function   i  f (z). From the Cauchy-Riemann relation

 ,1   , 2  , 2   ,1 it follows

 ,1  i  , 2   ,1  i  ,1  f ( z ). Here f (z ) is a function that is a complex conjugate to f (z) . Equation (3.30) takes the form 4

2 w      f ( z ). zz

We integrate both sides of the z variable, and obtain 4

w       f ( z )  g ( z ), z

(3.31)

where g(z) is an arbitrary function of a complex variable z . We again integrate Eq. (3.31) with respect to the z variable and get the following result:

4 w      z f ( z )  g ( z )  h ( z ) ,

(3.32)

here h (z ) is an arbitrary function of a complex variable z . So from Eq. (3.28) we can write: 98

3. Linear fracture mechanics

u1,1  iu2,1 

w w  w w   , u1, 2  iu2, 2  i  i , z  z z  z

it gives

u

1,1

 u2, 2   i u2,1  iu1, 2   2

w . z

w . Thus, the real parts of forz mula (3.31) and the 2f  z  function are equal

Consequently,   u1,1  u2, 2  2 Re





Re      f  z   g  z   2 Re f ( z ). But the functions f  z  and f z  differ only in the sign for the imaginary part, the real parts of them are the same; so it gives      f z   g  z   2f ( z ).

So f ( z) 

g ( z ) ,   3

and Eq. (3.32) can be written as:

4w  

 zg ( z )  g ( z )  h ( z ).   3

Usually the general solution of the plane deformation problem is written in the following form: 2  u1  iu 2   æ  z   z  z    z , g h   3   ,   , æ  3  4 . 2æ 2 

(3.33)

Here   z  and   z  are two arbitrary analytic functions of the complex variable. The functions  and  differ from g and h only by 99

Introduction to Deformable Solid Mechanics

constant factors. This equation (3.33) was obtained by Kolosov and Muskhelishvilii in a different way. Here is given Eq. (3.33) obtained by the method of Rabotnov. This representation is convenient as the equations for stresses do not contain elastic constants:

 11   22  4 Re  z ,  22   11  2i 12  2z   z     z .

(3.34)

The obtained equations (3.33)-(3.34) can be used not only for the plane deformation state, but also for the plane stress state if

æ

3  . 1 

3.3.2. Stress condition near crack

The elastic space is cut into parts of the plane  c  x1  c, x2  0 at infinity  22   , 11  12  0 . In x1 x2 plane of the crack corresponds to the one shown in Fig. 3.12, the cut is between x1   c zones. From the symmetry conditions it follows that there should be  12  0 on x1 axis. This condition can be satisfied if we take

  z    z  z     z .

(3.35)

Substituting Eq. (3.35) into formula (3.33), we get 2  u1  iu 2   æ   z   z  z   z     z  .

Separating the real and imaginary parts, we find the following expressions for displacements: 100

3. Linear fracture mechanics

u1  1  2  Re  z   x2 Im z , u2  21   Im  z   x2 Re  z .

(3.36)

Using Eq. (3.34)

 22   11  2i 12  2 z  z   z . From this relation and Eq. (3.34) we find

11  2Re  x2 Im, 12  2x2 Re,  22  2Re  x2 Im.

(3.37)

Eqs. (3.36) and Eqs. (3.37) determine the field of displacements and stresses by means of  z  function. When x2  0  12  0,

 11  2 Re  /   22 ,

u1  1  2  Re   z ,

u2  21    Im  z .

We should select   z  function in a such way, that:

1 Re  0, x2  0, x1  c, ,  z    2 . z  Consider:

 z   D z 2  c 2 ,  z    z  

D z 2  c2



z

Dz 2 2

 c2

Dz z 2  c2



3

,

.

Substituting Eqs. (3.38) into Eqs. (3.37), we find 101

(3.38)

Introduction to Deformable Solid Mechanics

11   22  2D, 12  0. If we consider 2 D   , , the function   z   D z 2  c 2 is used to solve the problem of all-round stretching of the body with a crack. We are primarily interested in the distribution of  22 stress near the

end of the crack, where it is big. Let us find the stresses  22 and

displacements u 2 for the points of the plane x2  0 . From Eqs. (3.36) – (3.38) we find

 22  0,  22 

u2 

x1

 1   2 2 c  x1 , 

, u2  0,

x12  c 2

x1  c,

(3.39) x1  c.

Near the end of the crack  x1  c  r  c  (Figure 3.12), after expanding equations for u2 and  22 in power series of r, we obtain the following formulas for the main members of the expansion: u2 

K I (1  ) 2r





,

 22 

KI 2r

,

K I   c ,

(3.40)

here K I is the stress intensity factor. If     x1  , then KI 

1

c

 x  c  1

c

c  x1 dx1. c  x1

(3.41)

Using the principle of superposition, any kind of crack loading can be represented as a sum of three elementary types of loading, shown in Figure 3.13. Using the same method as for type I, it is possible to determine SSS (stress-strain state) near the end of the crack for the II and III types: for the II type crack 102

3. Linear fracture mechanics

 12  u1 

 II 2r

,  11   22  0 ( x 2  0, x1  c ),

K II (1  ) 2r





, u 2  0 ( x2  0, x1  c),

(3.42)

 II   c,

for III type crack

 23  u3 

KIII 11   22  ...  13  0 x2  0, x1  c, 2r

KIII



2r



(3.43)

, u1  u2  0 x2  0, x1  c , KIII  q c.

Fig. 3.13. a – type I – crack of normal separation, b – type II – crack of a flat shear, c – type III – crack of transverse (antiplane) shear (force q is directed along the х3 axis).

As we see, the K  coefficients are the same for both plane deformation and for a plane stress state.

3.3.3. Fracture force criterion

In 1957 Irwin showed that for each type of cracks there is a critical value of stress intensity coefficient (SIC), reaching which the crack begins to grow and the body fractures: 103

Introduction to Deformable Solid Mechanics

K  Kc , (  I , II , III.)

(3.44)

here K Ic , K IIc , K IIIc are the material constants (dimension – kgs/cm3/2), determined from the experiment for the corresponding type of loading. Note that the most dangerous are cracks of type I, so K Ic is given a special name – fracture toughness (crack resistance). For cracks of a mixed type, the strength criterion should be represented by

F ( KI , KII , K III , KIc , KIIc , KIIIc )  0.

(3.45)

Obviously, if at least one of the conditions (3.44) is satisfied, then the crack must develop. This circumstance, within the framework of the force criterion, is the only condition imposed on the form of F function.

3.4. Energy criterion of fracture

Consider a body that in the initial state occupies volume V with S  S 2  S1 surface, where S 2 is a part of the surface with the given forces, S1 is a part of surface with the given displacements. The body contains a cavity with v volume and with  surface, which is free from stresses. The initial state of the body with the cavity is characterized by  ij stresses,  ij deformations and ui displacements. Let the cavity volume at fixed external parameters

qi S  0, 2

ui S  0,

(3.46)

1

change by v , its surface change by   . This causes a change in the stress-strain state of the body by  ij ,  ij , ui . If such a transition to a new state is quasistatic, then the energy balance equation must be satisfied 104

3. Linear fracture mechanics ~

A   U  Ï ,

(3.47)

~

where  U is the change in the elastic potential, A is the change in the work of external forces, Ï is the change in the surface energy of the body, which is characterized by  s parameter,

~ U  W ij dV, A   qiui ds, Ï   s  .

(3.48)

~ A U  Q

(3.49)

S2

V

Denote

and taking into account the change in the volume of the body,

Q   qi ui dS   W  ij dV  S2

V

 W 

ij

  ij dV .

(3.50)

V  v

Considering Eqs. (3.46) and the fact that there are no forces on the newly formed surface, the first integral (3.50) can be extended to the whole surface of the body in the form:

 q u dS   q i

S2

i

i S   

 qi u i dS 

 

ij

  ij  ij dV .

V  v

Now, dividing the second integral (3.50) into areas V   and

 , we represent the Eqs. (3.50) in the form Q 

  W  ij dv  v

 

ij



  ij  ij W  ij   ij   W  ij  dV.

(3.51)

V v

For a linearly elastic body, taking into account Eqs. (3.46) and the absence of forces on the boundary of the initial cavity, we represent the second integral in Eqn (3.51) in this form: 105

Introduction to Deformable Solid Mechanics

 

ij



   ij   ij  W  ij    ij   W  ij  dV 

V  v



1 1   ij   ij dV   q i  u i dS   2 V  v 2 S    



1 1  qi  u i d     qi  u i d  .  2  2 

Substituting this expression into Eq. (3.51), we finally have

Q   W  ij dv  v

1 qi ui d  . 2 

For v  0, crack-cut, we get

 Q 

1 1 qi ui d    qi ui d  ,  2   2   



and as on the boundaries of the cut qi   qi , we have

 Q 

1 qi ui d  , 2 

(3.52)

where ui  is the displacement jump on the cut. Considering that the crack is located along x1 axis, taking into account the sign of the normal to of the cut boundaries

q1  12,





q2  22,

and the symmetry in separation

q3  32,

u1  2u1 , u2   2u2, 106

3. Linear fracture mechanics

using (3.52) we obtain c

Q 

 



u   22u2   32u3 dx1 ,

 12 1

(3.53)

0

where the end-point is taken as the origin of coordinates and consequently, the integration extends over a small distance from the end of the cut, so that the Eqs. (3.40), (3.42), (3.43) are sufficient to calculate the integral. In this case, the stresses should be considered as they were in their initial state, and displacements should correspond to the opening of the cut boundaries after the cut moves by c . Therefore the displacements in Eq. (3.53) must be calculated by asymptotic formulas with a shift of the origin by c , i.e. with a replacement of 2r  2x1 by 2 c  x1  . Thus, from Eq. (3.53) and Eq. (3.40) for the limiting state of the I type crack, we have c

K2 c  x1 K2 QI  Ic æ  1  dx1  Ic æ  1c. 4 x1 8 0

(3.54)

Similarly for II and III type crack, we get QII  QIII 

2 K IIc æ  1c, 8

(3.55)

2 IIIc

K c. 2

Returning to the energy balance equation (3.47), we represent  for the crack-cut in the form

  Gc     Gc c. 107

(3.56)

Introduction to Deformable Solid Mechanics

For plane deformation (PD) and for a plane stress state (PSS), from Eqs. (3.47), (3.49) and (3.54) in the limiting state, we have

GI c  (1  2 ) K Ic2 / E (PD),

(3.57)

GI c  K / E (PSS). 2 Ic

It is possible to extend these equations to the pre-limit state GI  (1  2 ) K I2 / E,

GI  K I2 / E ,

(3.58)

attributing to the GI parameter the meaning of the energy stream at the crack end under subcritical loading, we obtain G1 





dQ1 d ~ A U .  dc dc

or the meaning of resistance to the crack propagation. In this case, the limiting condition is the equality GI  GI c ,

(3.59)

where GI c is the material constant. If relations (3.57) are satisfied, then the proportionally force and energy approaches are equivalent separately for the all mentioned above types of problems. However, in qualitative terms, the energy approach has greater heuristicity. Thus, it removes the uncertainty in the form of a limiting condition for mixed type cracks, similar to Eq. (3.45). Indeed, according to the principle of superposition, the total energy stream to the crack end (or the total resistance to the crack propagation) is: G  GI  GII  GIII 

1  2  2 K2  K I  K II2  III 1  E  108

 , 

(3.60)

3. Linear fracture mechanics

where GI is defined by Eq. (3.58), GII and GIII are determined by the relations (3.55) – (3.56). Now the strength of the material can be determined by the condition

G  Gc ,

(3.61)

here G is determined by the relation (3.60) and Gc is the material constant. In the particular case with the cracks of the normal fracture, the energy criterion gives a possibility to determine the value of the indefinite factor  in Eq. (3.15). As Gc  2 s , then in the conditions of plane deformation, the first of the formulas (3.58) with account for Eq. (3.40) in the limiting state, gives 2 E s . 1  2 c

0 





Comparing this result with Eq. (3.15), we obtain



 1  2  2

,

1  3

    1,39,

1 2

    1,18.

(3.62)

The simplest way to determine G c is to use the so called compliance method. Now using Eq. (3.62), the Griffiths formula can be written in the form of Eq. (3.63):

0  0 

2E s (PD) . 1  2 c





(3.63)

2 E s (PSS) . c

Here, PD is a plane deformation, PSS is a plane stress state. 109

Introduction to Deformable Solid Mechanics

3.5. Some applications of the crack theory

The materials of this section are presented according to V.Z. Parton's book. 3.5.1. In an aluminum panel with b = 2 m width and h = 100 mm thickness, a flat full depth crack in the welded seam was found. The panel is loaded with F = 400 ton force, the crack length  = 20 mm is located perpendicular to the direction of extension in the central part of the panel. Material – aluminum alloy 5083 – 0 with a fracture 1 toughness of 25 MPa·m 2 . Is it safe to use this panel? Solution. The length of the crack is small in comparison with the width of the panel. The stress intensity factor is calculated by the Eq.  KI    . The criterion of brittle fracture K I  K Ic determines the 2 critical size of the crack:

с 

2K12c

 2

.

(3.64)

Before substituting concrete numerical values in Eq. (3.64), we translate all the data into the SI system. The force F = 1400 tf = 1.37*107N, the stress 

F 1,37 107 S   69МРа . bh 2  0,1m2

Eq. (3.64) gives

 c  0 ,085 m , thus, the critical crack length is

 с = 85 mm, which means that the

existing crack length  = 20 mm is not critical. However, before confirming the safety of the panel usage, the engineer must find out a reason for the crack appearance and growth up to 20 mm, also the engineer has to understand whether its growth will continue due 110

3. Linear fracture mechanics

to fatigue or corrosion, and how soon it can reach a critical length of 85 mm. 3.5.2. A transverse crack with the length  = 30mm is found in the lower shelf of a steel crane beam, the width of which is b = 254 mm. The beam is operated at a maximum tensile stress  = 172 MPa. Is the operation safe if the steel fracture toughness is

K 1c  165 MPa · m

1

2

. Solution. The test can be carried out in two ways: 1

2 1. Calculate K I =    = 37.3 MPa·m . Let's compare the 2 found value with the fracture toughness, since K I is much smaller

than K Ic . The beam is assumed absolutely safe considering the conditions mentioned in example 3.5.1. 2. We calculate the critical length of the defect by the Eq. (3.64):  c  568 мм ,

as it is seen, the length of the existing crack is almost 20 times less than the critical length. Let's see now how to take into account the final dimensions of the item and introduce a correction for the plastic zone near the top of the crack. In order to take into account the real geometry, we introduce a calibration factor Y corresponding to the stress intensity factor K, for example, for a strip with b width and edge notch with



length;

KI   () ,

(3.65)

 is a b relative depth of the notch. The Eq. (3.65) gives a good accuracy up to  = 0.7. Irvine's correction to the material plasticity is a fictitious where   1,99  0 , 41  18 ,70  2  38 , 48  3  53 ,85  4 ,  

111

Introduction to Deformable Solid Mechanics

extension of the crack length by a small amount, which is approximately equal to the radius of the plastic zone near the crack top:

rT 

K I2 , 2 T2

(3.66)

where  T is the yield point. Let's return to the analyzed example 3.5.2 about the crane beam and now suppose that: 1) the transverse crack in the lower shelf is located at the edge of the beam and not in the middle part; 2) the crane should work in the open air and according to the weather forecast the temperature can fall to zero degrees at night. In the handbook it is written that at this temperature the fracture toughness drops to 60 MPa·m 12 , and the yield strength decreases to 480 MPa.

 = 0,12, for such b a depth K, the calibration factor Y according to Eq. (3.65) is equal to In this example, the relative crack depth is  

 (0,12)  2,15

We calculate the Irvine correction for plasticity by the Eq. (3.66) rT 

1 60 2 ( )  2,5mm 2 480

Again, we check the safety of the beam in two ways: 1 1) K I     rT  66,7 MPa  m 2 (remember that the fracture 1

toughness of steel can drop to 60 MPa·m 2 at night); K 2)  с  ( 1c ) 2  26,3 мм (remember that there is a crack in the  beam length  = 30 mm!). Both calculations indicate that the crane 112

3. Linear fracture mechanics

beam will be on the verge of a catastrophe at night. Of course, this example is conditional in the sense that the fracture toughness of the real crane beams must be sufficiently high at low temperatures. It is more difficult to predict the fatigue cycle life but not much more difficult. Paris' law is most often used to describe the growth of fatigue cracks: d  A(K ) n . dN

(3.67)

Here A and n are the empirical coefficients, K  K max  K min is the change in the stress intensity factor in one loading cycle, and N is the number of cycles. Numerous experimental studies well confirm this equation, the exponent n ranges from 2 to 7 for different materials (most often n=4). The more brittle the material is in the test, the higher is the exponent n. Paris wrote about Eq. (3.67): "It is curious that such a simple law can describe the data for the materials with an extremely different microstructure! Obviously, the mechanism of crack growth for all of them is the same regardless of the material microstructure characteristics." Using Eq. (3.65), we express the change in the intensity coefficient over the loading cycle

K    ,

(3.68)

where    max   min is the value of the stress change during the loading cycle. To find the dependence of  on N, it is necessary to substitute Eq. (3.68) into Eq. (3.67):

d  A(  ) n dN and integrate l

d

  nn 2

 A(   ) n N .

l0

113

(3.69)

Introduction to Deformable Solid Mechanics

From the criterion of the brittle fracture K  K c we find the critical length  c , and the durability N c is obtained by substituting

 c instead of  in Eq. (3.69). In the general case Y depends on  ,

and it is difficult to integrate Eq. (3.69). We simplify the problem by taking the calibration correction Y = const. Then from Eq. (3.69): (1 n2 )

o



(1 n2 )

n  (  1) A( ) n N . 2

(3.70)

Let's consider a concrete example. 3.5.3. The examination by the nondestructive testing methods made it possible to detect a surface crack with  = 4,5 mm in depth in the anchor rod of the extrusion press. The diameter of the steel rod is 300 mm, the force of 1850 tons, developed in each loading, is equally distributed on four such rods. Let's assume that the number of loads per month is approximately 9600, the critical depth of the crack is approximately 60 mm. The law of fatigue crack growth is established by the experiment.

mm d ,  1,7 1011(K)3,5 dN cycle 3

if K is measured in H  mm 2 . The plant needs a working press. How much time do they have before the rod will be out of work? The factory engineer is facing a choice: 1) in 24 months the rod will be removed as nonoperable; 2) in 14 months the press will be stopped as the scheduled repair period comes; 3) the shortest period is 2.5 months in which an identical rod may be delivered; 4) the shortest replacement period of the damaged rod by a temporary one is about 2 weeks; 114

3. Linear fracture mechanics

5) press operation is unsafe, it must be stopped until the rod is replaced. Solution. Let's use the SI system. Maximum load: 1850 tf = 18.1 (0,3) 2 2 MN, the total cross-sectional area of the rod: 4 м  0,283м 2 , 2 the maximum stress in the rods: 18.1 / 0.283 MPa  64 MPa, the lowest voltage: 0, the stress range;  = 64 MPa, initial crack depth:  o = 0.0045 m. According to the table of stress intensity factors (Parton V.Z.), we find that for a surface annular crack Y = 1.88 for a depth of  o = 0.0045 m and Y = 1.01 for a depth of  o = 0.060 m. We take the greatest value Y = 1.88 to obtain the lower estimate of durability. If the crack does not completely encircle the cross section, then replacing its annular part leads to a conservative estimation of the rod durability. Thus, the equation of the crack growth (3.70) takes the form

( 0,0045 )  0 , 75   0 , 75  0,75  1,7  10 11  (1,88  64 ) 3 , 5  N or

  (57,56  2,44 104  N ) 1,33 m. In 24 months, i.е. when N = 24·9600 = 23.04·104 loading cycles, the depth of the crack will be equal to  = 0.675 m = 675 mm, which exceeds more than twice the diameter of the rod! In 14 months, i.е. when N = 14·9600 = 13.44·104 cycles, the depth of the crack is  = 14 mm. Hence, the press can be safely operated until the planned repair, because the critical depth is

 с =60 mm.

3.5.4. A zone with one boundary crack is subjected to cyclic stretching. In this case, the range of the stress intensity factor (Parton V.Z.) is equal to  K  1,12     . The zone material is 3

maraging steel A514 (  T  700H / мм2 , K1c  5300H / мм 2 ). The 115

Introduction to Deformable Solid Mechanics

initial crack length is

 max = 320 N/mm , 2

 o = 7.6 mm, the loading cycle parameters are:  min = 175 H/mm2,  =145 H/mm2. Suppose

that the processing of the fatigue testing results obtained for the samples of the given steel in accordance with the Paris formula yields

d mm ,  3,5531013 (K )2,95 dN cycle 3

if K is measured in H·mm 2 . The critical crack length is determined in accordance with the Irwin criterion Kmax  KIc K1c c  ( )2  70mm 1,12 max 

Integrating the Paris equation, we find that 82000 cycles are needed for the crack to propagate from  o = 7.6 mm to  c = 70 mm. If it is required that the construction withstand, for example, 100000 cycles, then an engineer has the following instructions to ensure this durability.

 c by applying a material with a higher K Ic value or reducing the calculated stress  max . 1. Increase the critical crack length

2. Reduce the stresses range  for reduction of  K and, consequently, for reduction of the rate of crack growth. This causes a corresponding increase in the number of cycles when the crack grows from  o to

 c . The speed d / dN

is related to  nonlinearly, and

a small change in  causes a fairly large change in d / dN . 3. Change the technology and control the construction structure in order to reduce the initial length of the crack

 o It can be seen from

the   N diagram that a region of small crack lengths influences 116

3. Linear fracture mechanics

durability. Therefore, a small reduction in the initial crack length should give a significant increase in durability. In the considered example, a reduction of the initial crack length to

 o = 4.7 mm gives an increase in durability up to 20.700 cycles,

during which the crack grows from 4.7 to 7.6 mm. In this case the total durability is equal to 102700 cycles.

3.6. Fracture mechanics and non-destructive testing of structures

The materials of this section are presented according to the book of V.Z. Parton. Modern non-destructive testing makes it possible to determine material defects, including the internal ones, and to determine their type, size, and location. However, non-destructive testing also has a weak point. A defect is found, and what's next: leave it unattended, continue the working process, but periodically repeat the control, stop the process and start repairing the construction, or maybe remove it? Nondestructive testing specialists have difficulties in accepting the decision as they are extremely concerned with the integrity of the construction. Let's analyze the situation using the simplest examples of interaction between nondestructive testing and fracture mechanics. Analysis of the structural element integrity includes calculation of its stress state and determination of a connection between the external loads and the maximum internal stresses. The classical strength calculation recognizes the construction as acceptable if these maximum stresses do not exceed the ultimate strength of the material or the yield strength taking into account the appropriate safety factor. In the simplest constructions, like Galileo considered, the beam fixed firmly at one end into the wall holds P load hung at the free end. According to the technical theory of bending: 117

Introduction to Deformable Solid Mechanics

 max 

6 PL , bh 2

(3.71)

if L, b and h are the length, width and height of a beam with a rectangular section, so it is believed that the beam is safe at a given P load if

 max 

T n

,

(3.72)

where  T is the yield strength of the material, n is the safety factor, which takes into account inaccuracy of manufacturing and the spread of material properties. Thus, the classical theory limits the beam strength: P  T

bh 2 . 6 Ln

(3.73)

Suppose that a crack-defect of  length appeared in the most dangerous place where the beam surface is embedded in the wall, i.e. where the tensile stresses are maximum. For a crack of a shallow depth, according to the table, the stress intensity factor is: (3.74) K I  1,12 max  . The operation of the construction will be safe by Irwin's criterion if

KI 

1 K Ic , m

(3.75)

where K Ic is the fracture toughness, and m is another safety factor. Substituting Eq. (3.71) and Eq. (3.74) into formula (3.75), we obtain: P

K Ic bh2 .  6,72mL 118

(3.76)

3. Linear fracture mechanics

Note that the geometric dimensions of the beam are present in the formulas (3.73) and (3.76) in exactly the same way, the same can be said about the strength characteristics of the material  T and KIc . However, Eq. (3.76) contains a new physical parameter – the size of a crack-like defect, which is just the link that connects fracture mechanics and non-destructive testing. Pressure vessels can be mentioned as another example of the fracture mechanics application for analyzing the results of non-destructive testing. Consider a thin-walled cylindrical vessel with hemispherical bottoms filled by gas or liquid under p pressure. The length of its cylindrical part is L, the radius of the cylinder and the bottoms are R, the wall thickness of the vessel is h (h Let us consider it connected. Let's denote the deformations and stresses of the springs as  i and  i i  1,...,M  , and deformations and stresses

of the pistons as  j and  j  j  1,..., N  . Suppose there are L kine-

matic N

M

 A    A    A  , k  1,..., L i 1

ki i

j 1

kj

j

k

(4.77)

and S-static M

N

 B    B   B  . k  1,..., S  i 1

ki

i

j 1

kj

j

k

(4.78)

structural relations. We leave the defining relations unchanged, i.e. in the form (4.1), (4.2), M  N . Thus, for 2M  N   2 quantities  i,  i,  j,  j,  ,  there are only defining and structural relations

M  N  L  S . Due to the assumed coupling of the system, there must be a single relation between the total deformation  and the total stress  , i.e. it is necessary that

M  N  L  S  2M  N   1 139

(4.79)

Introduction to Deformable Solid Mechanics

or

L  S  M  N  1.

Substituting the defining relations (4.1) and (4.2) into (4.77) and multiplying them by d operator, we obtain M

 A i 1

ki

  d i N  Akj j  Ak d , k  1,..., L. . Ei j 1  j

(4.80)

The relations (4.78), (4.80) are regarded as a system of L  S linear algebraic equations with respect to L  S (in virtue of equation (4.79)) unknowns  i,  j ,  . This system is consistent in physical meaning, i.e. the rank of the system matrix, shown schematically in Fig. 4.17, is equal to the rank of the extended (shaded) matrix.

Fig. 4.17. A generalized model of a viscoelastic body

The shaded part of the matrix contains d operator as a factor. The determinant of the M  N  1 order of this system is a differential operator, therefore it is convenient to represent the solution of the algebraic system (4.78), (4.80) in the form: 







  Q  / P or P   Q  , 140

(4.81)

4. Behavior characteristics of renom models

here 

m



i 0 n

P   ai d i ,

(4.82)

Q   bjd j ,

(4.83)

j 0





P and Q are differential operators, m = min {L, M}, n = min {L, M + 1} (Figure 4.17). So n=m or

n = m + 1.

If the number of kinematic relations L is equal to or less than the number of springs M, then the orders of the differential operators are equal. If M+1