Extension to three dimensions and introduction of deviatoric stresses

It has been found empirically that, to a first approximation, deformation of ice subjected to a normal stress is independent of the hydrostatic pressure or mean stress, P (see discussion of Equation (4.9)). This might well be anticipated from the observation that ice is (nearly) incompressible. In three dimensions, the mean stress is given by:

Because deformation is independent of P, we define a new set of stresses, denoted by primes, by a 'xx = axx — P, a'yy = ayy — P, and a'zz = azz — P. These stresses are variously known as deviatoric stresses, stress devia-tors, or non-hydrostatic stresses. "Deviator" refers to the fact that they are deviations from the mean stress.

A more compact relation for the deviatoric stresses is:

Here, we have introduced the Kronecker 8; 8ij takes the values:

We have also introduced the summation convention. Whenever two subscripts are repeated in the same term, as in akk, that term is summed over all possible combinations of the subscripts. Equation (9.7), therefore, represents nine equations, of which three are identical owing to the symmetry of the tensor. Two of the nine are:

As you see, deviatoric shear stresses are identical to their non-deviatoric (or total) counterparts. Only the normal stresses are different. In general, deformation depends only on these non-hydrostatic components of the stress field.

If we were to go through a derivation similar to that above (Equations (9.1)-(9.5)) in three dimensions (Johnson and Mellor, 1962, pp. 23-25), we would find that there were three invariants having the form:

J2 = < + < + < — « — « — « (9.8a)

J3 = Kxa'yya'zz + 2a'xya'yza'zx — v'xxtfz — ^yy^z — ^z^y

If total stresses were used instead of deviatoric stresses, the right-hand sides of Equations (9.8a) would be the same, except for the primes, but on the left, by convention, we would use I rather than J, thus:

12 = &Hy + &^z + &lx - &xx&yy - &yy&zz - &zz&xx (9.8b)

13 = &xx&yy&zz + 2&xy&yz&zx - &xx&^z - &yy&Hz - &zz&ly

It is easy to show that J is 0. Just use Equation (9.7) to express the deviatoric stresses in terms of their total counterparts, and simplify. Note also that:

Let us now derive an alternative expression for J. To do this, square the first of Equations (9.8a), thus:

j2 = ++&2+2(&x x&yy+&y,y&L+<&Xx) =0 This expression equals zero because J = 0, so we have:

&'2 + &a + &'2 = —2(&' &' + &' &' + &' &' )

Substituting this into the expression for J yields:

2 J2 = &x2x + &* + &zz + + 2&^2 + 2&zx (9.10)

or, using the summation convention:

The reader will recognize the right-hand side of Equation (9.10) as 2&e2 (Equation (2.10)). Thus, the effective shear stress that we have mentioned several times previously is, in fact, the square root of the second invariant of the stress tensor: &e = VJ2 = [2 &ij&ij]1/2.

Using the summation convention, the effective strain rate Equation (2.11) can also be written more compactly as ee = [|¿y£;j]1/2.

0 0