2.2  Fundamental equations

In elasticity theory, an element of volume experiences forces via stresses applied to its surfaces by the surrounding material. Stress is the force per unit area of surface. A complete description of the acting stresses therefore requires not only specification of the magnitude and direction of the force but also of the orientation of the surface, for as the orientation changes so, in general, does the force. Consequently, nine components must be defined to specify the state of stress. They are shown with reference to an elemental cube aligned with the x,y,z axes in Figure 2.2. The component σij, where i and j can be x, y or z, is defined as the force per unit area exerted in the +i direction on a face with outward normal in the +j direction by the material outside upon the material inside. For a face with outward normal in the -j direction, σij is the force per unit area exerted in the -i direction. For example, σyz acts in the positive y direction on the top face and the negative y direction on the bottom face.

Each point in a strained body is displaced from its original position in the unstrained state. The point displacement is represented by the vector

u = (ux,uy,uz ).
(2.1)

Let us consider the body in the condition of mechanical equilibrium. In such situation, no net torques taken about x,y and z axes placed through the center of the cube are present in the element, so that

σij = σji.
(2.2)

Thus, the order in which subscripts i and j are written is immaterial. The three components σxx, σyy, σzz are the normal components. From the definition given above, a positive normal stress results in tension and a negative one in compression. The effective pressure acting on a volume element is therefore

p = -1-
3(σxx + σyy + σzz) . (2.3)
The six components with ij are the shear stresses.

As a consequence of equilibrium, no net force can act on the element, so that

∂σxi-+ ∂σyi-+ ∂-σzi=  0,i = x, y,z,
∂x      ∂y     ∂z
(2.4)

equations (2.4) are the so-called equilibrium equations of classic elasticity.

When acted upon by stresses, the body deforms. The displacement u has the components ux, uy, uz representing its projections on the x, y, z axes, as shown in Figure 2.3.


PIC

Figure 2.3: Displacement of P to P’ by the displacement vector u.

Let us consider now the response to stress. In linear elasticity, the strains are defined in terms of the first derivatives of the displacement components

εij = 1-
2(           )
  ∂ui-+  ∂uj-
   ∂j    ∂i. (2.5)
The nine components of strain are therefore
εxx = ∂ux
----
 ∂x,εyz = εzy = 1
--
2(           )
 ∂uy    ∂uz
 ---- + ----
  ∂z     ∂y,
εyy = ∂uy-
 ∂y,εzx = εxz = 1-
2(            )
  ∂uz-   ∂ux-
   ∂x  +  ∂z,
εzz = ∂uz-
∂z,εxy = εyx = 1-
2(            )
  ∂ux- + ∂uy-
   ∂y     ∂x. (2.6)
Partial differentials are used because in general each displacement component is a function of position (x,y,z).

The strains εxx, εyy, εzz defined in (2.6) are the normal strains. They represent the fractional change in length of elements parallel to the x, y and z axes, respectively, e.g. the length lx of an element in the x direction is changed to lx(1 + εxx).

The volume V of a small volume element is changed by strain to (V + ΔV ) = V (1 + εxx)(1 + εyy)(1 + εzz). In the linear approximation, the fractional change in volume Δ, known as the dilatation, is therefore

Δ =  ΔV ∕V  = (εxx + εyy + εzz).
(2.7)

Δ is independent of the orientation of the axes x,y,z.

Each component for ij is half of the shear strain ηij, which is usually defined in engineering as

ηij = 2εij,ij. (2.8)
The six components εyz,εzy,εxz,εzx,εxy, and εyx have a simple physical meaning. This is demonstrated by εxy in Figure 2.4, in which a small area element ABCD in the xy plane has been strained to the shape ABCD without change of area. The angle between the sides AB and AD that was initially parallel to x and y, respectively, has decreased by 2εxy. By rotating, but not deforming the element as in Figure 2.4, it is seen that the element has undergone a simple shear.

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Figure 2.4: (a) Pure shear and (b) simple shear of an area element in the xy plane.

The shear strain ηxy is related to the angles of shear by the relation

ηxy = ζx + ζy = ∂u
--x-
∂y + ∂u
---y
 ∂x. (2.9)
The above expressions contain only first-order terms in derivatives of stress and displacement and are exact only in the limit that the stress and displacement approach zero, i.e, the magnitude of these components is 1. This procedure results in the linear theory of elasticity, which is a good approximation for small displacements. The inclusion of second-order terms makes the theory much more complicated, thus such second-order corrections are not considered here.

For small distortions (∂ui∕∂xj 1), each stress component is linearly proportional to each strain. The relationship between stresses and strains in linear approximation is called Hooke’s law. Thus, the stresses depend only on the strains

σ   = c   ε .
  ij    ijkl kl
(2.10)

The coefficients cijkl are the elastic constants, also called stiffness constants. From (2.2) and the relation εkl = εlk, it follows directly that

cijkl = cjikl = cjilk = cjilk.
(2.11)

The combination of the last two relations yields the expression

σij = cijkl∂uk-.
         ∂xl
(2.12)

The further combination of the last equation with (2.4) gives

      2
    -∂-uk--
cijkl∂xj∂xl =  0.
(2.13)

When a unit volume element deforms reversibly by differential strain increments dεij, the stresses work on the element by an amount

dw =  σijdεij = cijklεijdεij.
(2.14)

If the deformation is reversible and isothermal, and if the work is restricted to that of elastic deformation, the differential work dw is equal to the differential change in Helmholtz free energy dF of the element

dF  = dw  = cijklεkld εij.
(2.15)

Thus

   2
--∂-F--- = c   .
∂εij∂εkl    ijkl
(2.16)

The free energy is a state function and dF is an exact differential, so that the order of differentiation in the previous equation is immaterial. As a consequence it follows

cijkl = cklij.
(2.17)

The strain-energy-density function (the strain energy per unit of volume) is determined by the integration of (2.15)

     1
w =  -cijklεijεkl.
     2
(2.18)

A typical solution to an elasticity problem involves the determination of the displacements uk from (2.13), given a set of boundary conditions. Stresses and strains then follow from (2.5) and (2.12), and the strain energy follows from (2.18).

2.2.1  Matrix notation

Hooke’s law (2.10) written as an equation between matrices becomes

{σij} = {cijkl}{εkl}.
(2.19)

The matrix cijkl is relating the nine elements σij to the nine elements εkl. According to (2.17), the matrix {cijkl} is symmetric. The elastic coefficients are often written in a contracted matrix notation as cmn where m and n are each indices corresponding to a pair of indices ij or kl, according to the following reduction:

ijorkl112233233112321321,
morkl123456789. (2.20)
Thus, by definition,
c11 = c1111c12 = c1122
c44 = c2323c46 = c2312
...... (2.21)
(2.11) and (2.17) indicate that due to symmetry there are only 21 independent elastic constants among the 81 in cmn. Thus (2.19) reduces to
(    )                                                (   )
  σxx     ( c   c    c    c   c    c   c    c   c  )    εxx
|| σyy||    |  11  12   13   14   15   16   14   15  16|  || εyy||
| σzz|    | c12 c22  c23  c24  c25  c26  c24  c25 c26|  | εzz|
|| σ  ||    || c13 c23  c33  c34  c35  c36  c34  c35 c36||  || ε ||
||  yz||    || c14 c24  c34  c44  c45  c46  c44  c45 c46||  ||  yz||
| σzx| =  | c15 c25  c35  c45  c55  c56  c45  c55 c56|  | εzx|  .
|| σxy||    || c   c    c    c   c    c   c    c   c  ||  || εxy||
|| σzy||    |(  16  26   36   46   56   66   46   56  66|)  || εzy||
( σxz)      c14 c24  c34  c44  c45  c46  c44  c45 c46   ( εxz)
  σ         c15 c25  c35  c45  c55  c56  c45  c55 c56     ε
   yx                                                    yx
(2.22)

Because of the symmetry the previous relation is reduced to the following 6×6 representation

(    )    (                           ) (     )
  σxx       c11  c12 c13  c14  c15  c16     ϵxx
|| σyy||    || c12  c22 c23  c24  c25  c26|| ||  ϵyy||
| σzz|    | c13  c23 c33  c34  c35  c36| |  ϵzz|
||    ||  = ||                           || ||     || .
| σyz|    | c14  c24 c34  c44  c45  c46| | 2ϵyz|
( σzx)    ( c15  c25 c35  c45  c55  c56) ( 2ϵxz)
  σxy       c16  c26 c36  c46  c56  c66    2ϵxy
(2.23)

The cmn in the 6×6 matrix are the same as the cmn occurring in the 9×9 matrix. The 6×6 matrix is also symmetric. When the axes of reference are rotated, stresses, strains, and elastic constants must be transformed accordingly. Transformations are performed most conveniently in the complete 9×9 scheme. This procedure is described in Section 2.3. For most crystals, the number of independent elastic constants is reduced further from 21 because of crystal symmetry. As an example, only five constants are independent for hexagonal crystals:

(                           )
  c11  c12 c12   0   0    0
| c12  c11 c12   0   0    0 |
|| c    c   c     0   0    0 ||
||  12   12  11              || .
|  0    0   0   c44  0    0 |
(  0    0   0    0   c55   0 )
   0    0   0    0   0   c66
(2.24)

For cubic crystals, the independent constants are three:

(                           )
  c11  c12 c12   0   0    0
|| c12  c11 c12   0   0    0 ||
| c12  c12 c11   0   0    0 |
||                           || .
|  0    0   0   c44  0    0 |
(  0    0   0    0   c44   0 )
   0    0   0    0   0   c44
(2.25)

For isotropic solids only two material properties – λ and μ, called Lamé constants – are required:

σxx = 2μεxx + λ(εxx + εyy + εzz) ,
σyy = 2μεyy + λ(εxx + εyy + εzz) ,
σzz = 2μεzz + λ(εxx + εyy + εzz) ,
σxy = 2μεxy,σyz = 2μεyz,σzx = 2μεzx. (2.26)
The shear modulus μ is calculated as  [57]:
μ = -----1------
2(s  -  s  )
   11    12. (2.27)
Other elastic constants are frequently used, the most useful being Young’s modulus Y, Poisson’s ratio ν, and the bulk modulus H. Under uniaxial, normal loading in the longitudinal direction, Y is the ratio of longitudinal stress to longitudinal strain  [57], and ν is minus the ratio of lateral strain to longitudinal strain  [57], H is a function of the fractional change in volume Δ and the effective pressure p (see equations (2.3) and (2.7)):
Y = 1∕s11,
ν = -s11∕s12,
H = 2p∕Δ. (2.28)
Since only two material parameters are required in Hooke’s law, these constants are interrelated  [57]. For example,
Y = 2μ(1 + ν)ν = λ∕2(λ + μ)H =   Y
1 --2-ν. (2.29)