Page 261 - Intro to Tensor Calculus

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Basic Equations of Linear Elasticity
• Conservation of linear momentum.

ij j j
σ + %b = %f j =1, 2, 3. (2.4.27(a))
,i
j
j
where σ ij is the stress tensor, b is the body force per unit mass and f is
j
the acceleration. If there is no motion, then f = 0 and these equations
reduce to the equilibrium equations
j
σ ij + %b =0 j =1, 2, 3. (2.4.27(b))
,i
• Conservation of angular momentum. σ ij = σ ji
• Strain tensor.
1
e ij = (u i,j + u j,i ) (2.4.28)
2
where u i denotes the displacement field.
• Constitutive equation. For a linear elastic isotropic material we have
E i E k i
i
σ = e + e δ i, j =1, 2, 3 (2.4.29(a))
j
j
k j
1+ ν (1 + ν)(1 − 2ν)
or its equivalent form
r i
i
i
σ =2µe + λe δ i, j =1, 2, 3, (2.4.29(b))
j j r j
r
where e is the dilatation. This produces 15 equations for the 15 unknowns
r
u 1,u 2 ,u 3 ,σ 11 ,σ 12 ,σ 13 ,σ 22 ,σ 23 ,σ 33 ,e 11 ,e 12 ,e 13 ,e 22,e 23 ,e 33 ,

which represents 3 displacements, 6 strains and 6 stresses. In the above
equations it is assumed that the body forces are known.

Navier’s Equations

The equations (2.4.27) through (2.4.29) can be combined and written as one set of equations. The
resulting equations are known as Navier’s equations for the displacements u i over the range i =1, 2, 3. To
derive the Navier’s equations in Cartesian coordinates, we write the equations (2.4.27),(2.4.28) and (2.4.29)
in Cartesian coordinates. We then calculate σ ij,j in terms of the displacements u i and substitute the results
into the momentum equation (2.4.27(a)). Differentiation of the constitutive equations (2.4.29(b)) produces

σ ij,j =2µe ij,j + λe kk,j δ ij . (2.4.30)