## Velocity solutions in a perfectly plastic medium

We now use the stress solutions, Equations (10.18), to obtain solutions for the velocities from Equations (10.5) through (10.7). From Equations (10.5) and (10.6) we obtain:

and from Equation (10.7):

Let us first examine the applicable boundary conditions. The stress solutions are valid only for the thickness h = k/pgx. Therefore, we seek a velocity solution that will maintain this thickness. Because there is accumulation, bn (or ablation, -bn) at the surface, we know that:

We will now show that dw/dx = 0. The stresses are independent of x, and the material is the limiting case of a purely viscous material in which stresses determine strain rates. Therefore, the strain rates must be independent of x. In particular, dw/dz is independent of x, so:

Therefore, w/ x = const, independent ofz. Then, because w/ x = 0 on the upper and lower boundaries from the boundary conditions, dw/dx = 0 everywhere.

Equation (10.21) now becomes:

u kz

Combining Equations (10.20) and (10.22) to eliminate X yields:

Differentiating with respect to x and again making use of the fact that w is continuous inx and z gives:

where the last equality results from the fact that dw/dx = 0. Thus from Equation (10.20), differentiating with respect to z, we obtain:

d 2u d2w

dzdx dz2

which has the solution:

Using the first boundary condition, w = bn on z = 0, yields c2 = bn, whereupon the second boundary condition, w = 0 on z = h, yields 0 = cih + bn. The solution for the velocity in the z direction thus becomes:

Note that w varies linearly with depth. We discussed this in Chapter 5 (pp. 87-90) and will analyze it in greater detail later.

Using this solution for w in Equation (10.20) we obtain:

In the coordinate system we have chosen, azx is negative for positive z. Therefore ezx, andhence in Equation (10.27), du/dz, must be negative so that a negative stress produces a negative strain rate. (In other words, the horizontal velocity must decrease with depth.) Thus, when bn is positive in Equation (10.27) we use the upper sign, and conversely. Equation (10.27) thus becomes:

Integrating Equation (10.26) yields:

(a) Accumulation area; bn positive

(b) Ablalation area; bn negative

Figure 10.4. Velocity solutions for a deforming slab of material with a perfectly plastic rheology.

where, as in Equation (10.15), f (z) represents some function of z alone. Taking the derivative of this with respect to z, substituting the result into Equation (10.28), and integrating gives:

Thus, the solution for the velocity in the x-direction is:

These velocity solutions are illustrated in Figure 10.4. On z = 0, u = (bnx/h) + 2 |bn| + c and w = bn, while onz = h, u = (bnx/h) + c and w = 0. Thus evidently (bnx/h) + c is the "sliding" speed. Note that Equation (10.29) implies that the ice must be free to slide on the bed at a speed determined by bn, and independent of azx and bed roughness.

If, on the contrary, the sliding speed were presumed to be a function of azx and bed roughness, the distribution of stress and hence of w could not be independent of x, and the ice mass would not remain a uniform slab. A reasonable presumption is that the sliding speed would not increase sufficiently rapidly with x, and that conservation of mass would then require that the ice thickness increase upglacier, leading to a convex surface profile. Depending on the degree of convexity and the consequent change in ice thickness, such a profile would offer the potential for increasing azx downglacier. This would provide the required increase in mass flux.

If the boundary conditions were selected such that u = 0 at x = 0, z = h, then c would be 0, but the original differential equations are z z not necessarily valid at the ends of the slab. Note also that between any two vertical sections, x1 and x1 + Ax, u increases by bnAx/h, which would be the increase in mean velocity required to transmit the additional flux, bnAx, downglacier - the balance velocity that we discussed in Chapter 5.

If there is no accumulation or ablation, bn = 0, and thus u = c and w = 0 throughout the block. In this case there is no internal deformation. All movement is confined to sliding. The yield criterion is then satisfied only at the bed and axx can take any value between the limits shown in Figure 10.2.

## Post a comment