Creep of floating ice shelves

Ice shelves around Antarctica play an important environmental role, as they act as dams, restraining the flow of ice from the interior of the continent. Were they to break up, ice levels in the interior would decrease rapidly over a period of a few centuries, and sea level would rise accordingly. Break up of ice shelves in northeastern North America may have contributed to the collapse of the Laurentide Ice Sheet at the end of the Wisconsinan. Thus, understanding the flow of ice shelves is a problem of both academic and environmental significance.

The problem of ice shelf flow is unique because tb is likely to be quite low where the shelf is grounded, and goes to zero in the limiting case when the shelf is afloat. Herein, we consider only the latter case. Weertman (1957b) was the first modern glaciologist to study this problem, and our approach follows his initially, but then incorporates some important modifications introduced by Thomas (1973a).

The coordinate system to be used is shown in Figure 12.11. The origin is at sea level, but is within the ice mass. The z-axis is vertical and positive upward. H is the thickness of the shelf, and h is the height of the surface above sea level. Inland, the surface rises gradually and the base drops further below sea level, so Hand h both increase. As long as the ice shelf does not become grounded, however, we assume that hydrostatic equilibrium is maintained; therefore, assuming a constant density and thus ignoring the low density snow and firn at the surface, (H - h)pw = Hpi, where pw and pi are the densities of water and ice, respectively.

At the risk of being repetitive, it is convenient, once again, to write out the momentum balance equation in the z-direction:

dTx7 dTv7 dT77

dx dy dz

Our objective is to obtain an expression for tXx, and then to use the flow law to solve for exx.

Because shear stresses are zero at the bed and surface, it is reasonable to assume that axz = azx = ayz = azy = 0. This means that velocities and strain rates are independent of z. Equation (12.31)can thus be integrated:

Thus, azz is simply the hydrostatic pressure at depth (h - z). Suppose that field measurements at some point give:

Then, from the incompressibility condition:

Both n and v are functions of horizontal position, but because strain rates are independent of z, n and v are also independent of z. Then because we assume that ice is isotropic, we have ¿¡j = kaj where, as before, X = an-1 /Bn. Thus, a'yy = naX x' aXy = va'x x' and a'zz = -(1 + V)aXx

From the last of these expressions, converting to total stresses, we obtain axx - a'zz = a'xx + (1 + nKx = (axx - P) - (azz - P)

When n = 0, this expression reduces to one that often appears in analyses in plane strain. It can be derived, for example, from Equations (10.18) and (10.19).

It is interesting to consider the implications of this relation: azz varies linearly with depth (Equation (12.32)) but exx is independent of depth. However, because the temperature of an ice shelf is normally well below 0 °C at the surface and close to the pressure melting point at the base, B, and hence a'xx, also vary strongly with depth (Figure 12.5). Thus, axx varies with depth in a way that is not intuitively obvious. We will avoid this problem by seeking an expression for exx in terms of the depth-integrated values of B and axx.

Let us proceed by determining the total force per unit width. To do this, integrate Equation (12.35) from the base, b, to the surface, s:

[ ( h2 (h — H)2 a xxdz — pig I — —--2-+ h(h — H)

or defining:

F is the total force opposing movement of a vertical section, of unit width normal to the flow direction, of the ice shelf.

We now need to use the flow law to express the left-hand side of Equation (12.36) in terms of strain rates. First, the effective stress is

2 aijaij

2(1 + n2 + 1 + 2n + n2 + 2v2)a = (1 + n + n2 + v 2)1/2|ax xl Thus, from the flow law:

If n = 3, we can drop the absolute value signs, which we now do. Thus, rearranging:


As strain rates are assumed to be independent of z, Equation (12.36) now becomes s f a' dz =-


Because B varies with depth, we define a depth-averaged B by

We also define 0 by

Solving for x now yields


To proceed further, we need to evaluate F, the force per unit width opposing motion. We do this for two special situations. In the first, the ice shelf is free to expand in both the x- and y-directions, and movement is restrained by seawater pressure only. Then, n = 1 and

Making use of the condition of hydrostatic equilibrium, pw(H - h) = piH, yields

Fw = - Pw g — H2 = - pi g\ — ]H 2 Pw 2 Pw and Equation (12.37) becomes:

2 Pw

The term in the inner brackets on the right-hand side is simply h so this becomes

As this expression is always positive, strain rates will always be extending. Note that the surface slope does not appear in this solution; thus, even an iceberg with a horizontal surface will deform. This solution does not apply very near a calving face where bending moments are present.

It is instructive to compare this expression with that developed in Chapter 5 (Equation (5.3) with (5.2c)) for ezx at the bed of a land-based xx s n

xx n

glacier in the absence of significant longitudinal strain:

Assuming that n = 3 and noting that h = (1 - pi/pw)H « 0.1 H, and that 0 = 1/9 when n = 1, v = 0, Equation (12.38) becomes:

0.024pi gH B

Thus, the driving stress (Pigh) in an ice shelf is comparable to that in a land-based glacier of the same thickness with a surface slope of « 0.024. However, because B increases with decreasing temperature, strain rates in ice shelves are normally less than those in land-based glaciers of comparable thickness.

The second example is that of an ice shelf between approximately parallel valley walls. In this case, F = Fw + Fs, where Fs is the shear force on the valley sides. Utilizing the expression for Fw just obtained, s,* becomes

Here, exx can be negative, or compressive, if Fs is sufficiently large.

Fs merits further comment. Suppose that a is the distance from the centerline of an ice shelf to the valley wall. Suppose further that the depth-averaged drag on a valley wall is Ts. Then TsH is the force on the valley wall per unit length along the direction of flow. This force must balance forces acting in the direction of flow over the half-width of the ice shelf. In the absence of basal drag, it is reasonable to assume that any vertical slice of unit width extending through the ice shelf and parallel to the direction of flow will be restrained equally by this side drag. Thus, any such slice will experience a drag of Ts H/a per unit length along the direction of flow. Noting that Ts is a negative quantity, as it is directed in the upflow direction (Figure 12.11), the resisting force per unit width is

Here, x is the coordinate position where the calculation is being made, and L is the x-coordinate of the edge of the shelf. Note that, consistent with being a force per unit width, Fs has the dimensions N m-1.

Equation (12.41) says that Fs increases as the distance to the edge of the shelf increases. Thus, from Equation (12.40), exx may change from extending nearer the shelf edge to compressive farther inland. This is the reverse of the normal situation in a grounded glacier, in which compressive flow is the rule in the ablation area and extending flow xx

Figure 12.12. Effect of longitudinal extension on an inclined borehole.

in the accumulation area. The implications of this are fascinating. With extending flow nearer the shelf edge, a positive emergence velocity would occur only if the product of the velocity times the surface slope were high enough to offset any downward vertical velocity resulting from the extension. In the absence of such conditions, a steady state can exist only if the mass balance is positive, as, in fact, is typically the case. This means that ice shelves with ablation (= melt) zones near the shelf edge should be uncommon. Furthermore, if the mass balance near the shelf edge is positive, it must also be positive at higher elevations further inland. Thus, if Fs ever became large enough to make è■xx compressive, the ice shelf would increase in thickness unstably until it became grounded.

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