56
Geotechnical News December 2010
GEO-INTEREST
in academia knew any better. So slow-
ly, as work allowed, I set about trying
to figure it out for myself – from first
principles.
Over the past two decades I’ve spent
much of my spare time thinking about
the real basics of pore water pressure
in both saturated and unsaturated soils,
doing so in the hope that I might even-
tually come up with a rational explana-
tion for each of these soil conditions.
In this series of articles I want to con-
centrate on pore pressure generation in
saturated granular soils, leaving unsat-
urated soils and cohesive materials for
another occasion.
Later in this series I will propose the
following equation for the generation
of excess pore water pressure at any
point within a saturated granular soil
experiencing deformation:
K ( 24 µ + ρ D v ) L v / 2 D
I won’t complicate things right now
by explaining each of the terms, other
than to say that the only unfamiliar
terms are “K” and “L”, and that these
two will be fully developed in subse-
quent articles. Incidentally, nothing
more than the early bits of Physics 101
will be needed to follow my line of ar-
gument.
Square One
In making a fresh start I had the luxury
of deciding where to begin. And the
easiest place for me to get going was
liquefaction. Apart from being an
attention grabber, I see liquefaction as
a physical activity where it is easiest to
grasp what’s happening in the motion
between the two phases (solid and
liquid).
I do hope readers don’t get too hung
up on the term “liquefaction”: This
simple concept has been much abused
over the years. So I suggest for those
folk who believe dense sand, or a well
graded granular fill can liquefy, you
read instead something like “total col-
lapse of a saturated soil-structure”. In
any event what I have in mind here is
what happens when, in a fully satu-
rated environment, a very loose mass
of similarly sized sand grains falls
into a denser arrangement due to some
change in the stress system which had
been keeping it in a precarious struc-
ture of mutual support.
The reason I think liquefaction is
a good point of departure is this. As
a consequence of collapse the soil-
structure can no longer act as a rigid
formwork for the discrete grains, and
for some time thereafter they no longer
interact or support each other. It is dur-
ing this momentary separation of the
two phases, as the two soil components
merge into a composite fluid, that para-
doxically, an opportunity is afforded to
view the particles as acting indepen-
dently and be apprehended in isolation
as separate individuals.
To focus my attention on this par-
ticular phenomenon I dreamed up a
cartoon of liquefaction in the sim-
plest form I could imagine. I call this
“thought-experiment” the three beaker
question, and I will now described how
it goes.
The Weight of Failure
Figure 1 shows three identical beakers
containing particles submerged in
water. In fact what I really have in mind
is the same beaker at three different
times. The number and size of the solid
particles and the amount of water is
exactly the same in each beaker. The
beakers sit on weighing scales. The
particle packing in the “before” beaker
is as loose as can be and consequently
is at the point of structural collapse.
The “during” beaker has been subjected
to a jolt which causes failure of the
structure, so what is represented here
is a soil being weighed during failure
of the soil-structure. The “after” beaker
is the situation prevailing shortly after
failure when the new soil-structure
has settled into a denser, more stable,
packing arrangement.
The question is: Is there a difference
in the weight between the three bea-
kers? More specifically, is the weight
of the middle beaker different from that
of the ones on either side.
I believe there are several “right”
answers to that question, depending on
particular details of particle size and
the time at which the weight of the ac-
tive beaker is recorded. My answer is
that in most cases, most of the time,
the weight of the active beaker is less
than the other two – the other two (in-
active) beakers being exactly the same
in weight. My thinking goes along the
following lines.
Since the two outside beakers (be-
fore and after failure) have the same
mass of solids and water, and since
they are static, just sitting there with
nothing moving, there is no reason to
justify a difference in weight. Using
“g” for gravitational attraction, the
weight of the first and third beakers is
simply equal to m times g.
But things are different in the “dur-
ing” beaker: There is movement, and
that movement is downwards into the
gravitational field. More than that, be-
cause the solids/particles were initially
Figure 1. The three beaker question.
