Jökull - 01.12.1966, Blaðsíða 15
The formula indicates that even a small force
K gives an uplift u0, but this may be too
insignificant, say 1 cm, to have any interest.
We ask, therefore, what h must be if u() is 1
m, and we find h = 47.5 m. In this case L
turns out to be 351 m, as found from K =
2pL/3, i. e. 2L/h = 14.8.
M o
Fig. 3. Lifting of a verv long beam by force
2K in its centre.
3. mynd. Lyfting mjög langs bita með krafti
i miðju hans.
Tlius even for such a very thin ice sheet the
uplift is just sufficient to give room for a
little outflow of lava. But after the space had
been filled with quenched lava, no more lift
would occur in the purely elastic case we are
considering. A thickness of 500 m is a realistic
figure for a Pleistocene ice sheet in Iceland,
but the theoretical lift u0 is here 10~4 cml Even
for the thickness h = 100 m the lift is only 1
cm, which must in reality mean blocking of
the lava outflow.
d. We have sofar not taken the limited
strength of the ice into consideration. In Hand-
book of Applied Hvdrology, Ed. Ven Te Chow,
1964, the following values are given as typical
in laboratory tests with conventional machines:
Crushing strength 37 kg/cm2; tensile strength
9 kg/cm2; bending strength 13 kg/crn2; shear
strength 7 kg/cm2.
We use the bending strength Mn/W =
6M0/h2 = 14 10-2 . K2/h3 (K in kg, h in m).
For a thickness of 100 m we find 25 kg/cm2,
i.e. double the bending strength. For li = 130 m
we find 11 kg/cm2, or less than the bending
strength.
4. We have hitherto considered the magma
dyke to be 2 m broad. If this dyke communi-
cates with a lava pond at the bottom of the
glacier we have by the principle of the hyd-
raulic press a force K corresponding to the
diameter of the pond. If this pond has the
length of the dyke and a diameter of, say 200
m, then we need an ice sheet of 2800 xn thick-
ness to withstand the bending moment. But
such a pond is quite unrealistic in the presence
of water. We could visualize a “lake” of say
10 m diameter in average along the dvke and
this would be withstood by an ice sheet of
380 m thickness. But by the presence of water it
is really very questionable whether any “pond”
at all can exist along the dyke and in com-
munication with it, and an escape af the lava
at the base of the glacier would be a much
more likely yield than a wholesale uplift of a
thick glacier. Here the vapour also enters the
picture. If critical vapour permeated the ridge
and there were no cracks for its escape, then
lifting of a very thick ice would be possible.
But if by the presence of much meltwater the
vapour temperature is not higher than 200° C
then lifting would not be possible for thicker
ice than about 150 m.
5. Central volcano. If the magma rises not
in a fissure but in a circular funnel of radius
r0, the lava pressure pi corresponds to a central
force P = apir02. The central uplift is now
found by the formula u0 = 147 P2/Eh4, see
Appendix.
Putting r0 = 5 m, P = 500 • a • 25 • 104 kg =
3.92 • 108 kg, and E = 105 kg/cm2, we get:
h u0
100 m 2.2 • 10-2 cm
50 m 0.35 cm
25 m 5.6 cm
We can only conclude that blocking by a
50—100 m thick ice sheet must be practically
complete.
6. General considerations. We can now dis-
cuss a sub-glacial eruption in a more general
xvay. The first contact of lava and ice causes
some melting and the produced water in turn
causes quenching. A cooling by 100—200° C
suffices to make the lava immobile and if it
remained compact, further loss of heat would
be slow. But observations in Surtsey (see for
instance Einarsson 1966) make it likely that at
least partly the lava would crumble. The frag-
ments would then cool rather rapidly, not by
melting more ice with which the fragments
would not be in contact, but by heating the
soaking water. As likely as not this water would
JÖKULL 169