Chapter II: Continuum Theory of Dislocations
2.2 Review of Classical Linear Elasticity Theory
Linear elasticity is a specialization of continuum mechanics to a subclass of materials in which the stress tensor in any element is determined exclusively by the strain tensor in the same element and, furthermore, the relationship between these two tensors is a linear one [44]. In other words, in classical linear elasticity, the stress tensor at a point inside the body is a linear function of the strain tensor at the same point. This is known as(generalized) Hooke’s law. There are several references on the subject of linear elasticity. Classic ones include references [45–48]. However, [49–51] were also found to be excellent references. As a result of this vast literature, we do not attempt to give here an extensive survey of the topic. Interested readers should consult the references just mentioned for that purpose. Instead, we only revisit those equations as given in [44] and [26] that will be relevant to subsequent sections of this chapter. Finally, here and throughout this thesis, we will use index notation and Einstein’s summation convention on repeated indices.
In classical linear elasticity theory, the displacement field is the variable of interest.
This is denoted by u or, using index notation, ui wherei= 1,2,3. The position of a material point inside the body is denoted by xor simplyxi. In linear elasticity theory, deformations are small enough that the components of u and x can be taken with respect to the undeformed or deformed configurations. Furthermore, the theory also assumes that the distortions ∂ui/∂xj are small, leading to the following expression for the strain tensor ε at a point inside the body:
εij = 1 2
∂ui
∂xj + ∂uj
∂xi
or
εij = 1
2(ui,j+uj,i) , (2.2)
where we have used the comma to represent differentiation with respect to the corresponding component of x.
Notice that
εij =εji (2.3)
follows immediately from the above definition. Hence, the strain tensor issymmetric.
The stress tensor σ on the other hand is such that σij gives the ith component of the force per unit area on a plane whose outward normal is parallel to the positive xj direction—see Figure 2.2.
Figure 2.2: Stress distribution on an infinitesimal volume element.
Given a body B of densityρ with body force per unit volumef, the stress state at every point inside B must satisfy
σij,j +fi =ρu¨i, (2.4)
where u¨i =∂2ui/∂t2 is the acceleration of the material point. As given, Equation (2.4) is the momentum conservation equation.
For problems in elastostatics, i.e. whereu¨i = 0, Equation (2.4) reduces to
σij,j +fi = 0, (2.5)
which is the equilibrium equation for classical elasticity.
If no internal torques are present, conservation of angular momentum requires that
σij =σji, (2.6)
which implies that the stress tensor is also symmetric.
Furthermore, at an external surface of the body, we either have
σijnj = ¯ti, (2.7)
where n is the outward normal to the surface and t¯the applied traction per unit area, or
ui = ¯ui, (2.8)
where u¯ is the prescribed displacement at the surface of the body.
The union of all traction boundary surfaces is usually denoted by ∂B2 and that of all displacement boundary surfaces by ∂B1 so that ∂B = ∂B1 ∪∂B2 with
∂B1∩∂B2 =∅—see Figure 2.3.
Figure 2.3: Body under displacement boundary conditions u¯i on its boundary ∂B1
and traction boundary conditions t¯i on its boundary ∂B2. Note that if no traction or displacement is prescribed at a boundary, this is equivalent to having a traction
of ¯ti = 0 at said boundary.
Next, recall that linear elasticity is characterized by a linear relationship between the stress and strain tensors. Specifically, we write
σij =Cijklεkl, (2.9)
where Cijkl are the elastic constants of the material.
It immediately follows from Equations (2.3) and (2.6) that Cijkl=Cjikl =Cijlk =Cjilk,
and we say that the elastic modulus tensor has minor symmetry. Though we don’t prove it here, the elastic modulus tensor also hasmajor symmetry, viz
Cijkl =Cklij.
As a result of these symmetries, the number of independent elastic constants decreases from 81to 21.
For an isotropic material, this is further reduced to only two elastic constants called Lamé constants. In this case, we have
Cijkl=λδijδkl+µ(δikδjl+δilδjk) , (2.10) whereδij is the Kronecker delta, being 1wheni=j and zero otherwise. The usual elastic constants can be expressed in terms of λ and µas follows:
Young’s modulus: E =µ(3λ+ 2µ)/(λ+µ), Poisson’s ratio: ν=λ/[2 (λ+µ)] ,
Shear modulus: µ ,
Bulk modulus (or inverse compressibility): κ=λ+ 23µ .
It should be noted that each of the above moduli can be similarly expressed in terms of any other two. The resulting table can be found on pages xxvii and 74 of references [51] and [50] respectively.
Using Equation (2.10), Hooke’s law for isotropic solids becomes
σij =λεkkδij + 2µεij. (2.11) Alternatively, strains can be expressed in terms of stresses as
εij = 1 +ν
E σij − ν
Eσkkδij. (2.12)
The strain-energy density function, which represents the energy per unit volume associated with the deformation of the solid, is given by
W = 1
2σijεij = 1
2Cijklεijεkl. (2.13) To summarize, the classical problem of static linear elasticity consists of solving the field Equations (2.5), (2.2), and (2.9) for the displacement field u, the stress and strain tensors σ and ε subject to the boundary conditions (2.8) and (2.7) on ∂B1 and ∂B2 respectively—see Figure 2.3. Next, we look at two special cases arising from these boundary-value problems.
Plane strain
In this section, we consider so-called plane strain problems in an isotropic medium.
These are problems for which it is known a priori—or assumed—that the displace- ment field depends only on two components of position, with the displacement along the third component being identically zero. For instance, for a plane strain problem along the x3 axis, the displacement field is of the form
u1 =u1(x1, x2), u2 =u2(x1, x2), u3 ≡0. (2.14) Consequently,
ε13= 0, ε23 = 0, ε33= 0. (2.15) When combined with Equation (2.11), (2.15) implies that
σ33=λ(ε11+ε22) , σ13 = 0, σ23= 0. (2.16) Therefore, for a body in static equilibrium with no body forces, (2.5) becomes
σ11,1+σ12,2 = 0, (2.17a)
σ12,1+σ22,2 = 0, (2.17b)
where we have used symmetry of the stress tensor to writeσ12,1 instead of σ21,1
in Equation (2.17b).
Equations (2.17) are satisfied automatically if there exists a function Ψ such that σ11= Ψ,22, σ22 = Ψ,11, σ12=−Ψ,12. (2.18) Such a function Ψis commonly referred to as the Airy stress potential.
On the other hand, one can use Equation (2.2) to easily show that
ε11,22+ε22,11 = 2ε12,12. (2.19) Inserting (2.12) into (2.19) and making use of relations (2.18) to simplify while keeping in mind that λ=νE/[(1 +ν) (1−2ν)], one arrives at
∂4Ψ
∂x41 + 2 ∂4Ψ
∂x21∂x22 + ∂4Ψ
∂x42 = 0, or
∂2
∂x21 + ∂2
∂x22 2
Ψ = 0. (2.20)
Solving the above biharmonic equation yields the stressesσ11, σ22,andσ12, which in turn yield the corresponding strains through Equation (2.12). The displacements u1 and u2 follow from direct integration of the strains.
Antiplane strain
We now consider an isotropic solid in a state of antiplane strain. In this case, two components of the displacement field are identically zero, with the nonzero component depending only on the position components along which the displace- ments are zero. For instance, for a state of antiplane strain along the x3 axis, the displacement field is of the form
u1 ≡0, u2 ≡0, u3 =u3(x1, x2). (2.21) Consequently, all strain components are zero except for ε13 and ε23 which in this case are given respectively by
ε13= 1
2u3,1 and ε23= 1
2u3,2. (2.22)
Accordingly, the nonzero components of stress are
σ13 = 2µε13=µu3,1 and σ23= 2µε23 =µu3,2. (2.23) Hence, assuming no body forces in the solid, equilibrium equation (2.5) simplifies to
∇2u3 = 0. (2.24)
Thus, the antiplane strain problem reduces to solving Laplace’s equation for u3 together with the relevant boundary conditions.