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Geophysical Mass Flow Modeling
Modeling Geophysical Mass Flows
E. B. Pitman
Overview of issues:
1. Geometry in 3D means how to incorporate elevation data into PDEs.
Tedious but straightforward
2. Flow modeling. What is the ‘right’ level of detail?
 In FLOW3D, for example, individual ‘particles’ flow ‘downhill’.
What about flowing up and over obstacles? How to include fluid into
a mixture?
Here we take a stab at continuum plasticity modeling. This approach opens
possibilities for us later on as we look at computing and modeling concerns,
and provides tie-ins with more classical fluids and solids mechanics.
We stay with dry material for now, in 2D. The extension to 3D is relatively
straightforward. The incorporation of fluid is harder.
1 STRESSES
Let Fi is the force in the ith direction. Consider an area element , and let i
be the force in the ith direction that  exerts on the continuum. Then
d i ds = Fi
Cauchy showed linear dependence of i on direction. That is,
i= ij j
Then ij is the stress acting in the jth direction across an element whose
normal is parallel to the ith axis.
FACT: ij = ji
Otherwise one would generate an infinite angular moment.
Order the eigenvalues of  = (ij) as 1  2
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If the coordinate directions were aligned with the eigendirections, then the
traction across a surface element with normal  is T = (1 1 , 2 2 ). The
normal stress  = T   and the tangential stress  = [T  T - 2]1/2.
Consider now a normal making an angle  with the x1 axis, so  =
(cos(), sin()). Then  = 1 cos2 () + 2 sin2 () = (1 + 2 )/2 + (1
- 2 )/2 cos(2) and 2 = (1 )2 cos2 () + (2 )2 sin2 () - 2 = [(1 - 2
)/2 sin(2)]2.
{Truth to tell, there is an ambiguity in choosing a sign that arises here; we’ll
resolve later.}
The Coulomb Yield condition says that material will deform when
 =  tan() + c
where c is a cohesion (we’ll set c=0). This condition is akin to static friction
– sliding occurs when the tangential force is equal to a friction coefficient
times the normal force. Putting the conditions together, we get yield on a
plane disposed at an angle  = ½ (/2 - ) from the direction of major
stress (the eigendirection associated with the largest eigenstress). Here we
must choose sign correctly. One can show the Coulomb condition is
equivalent to
1 /2 = (1+ sin())/(1- sin())
We ask, then, how to relate the principal stresses to the components of ?
11 12
=
(
)
21 22
(11 - 22 )/2
1 0
= (11 + 22 )/2
(
12
)+ (
0 1
)
21
-(11 - 22 )/2
The eigenvalues are  = (11 + 22 )/2  {((11 - 22 )/2)2 + (12) 2}.
Unraveling all this results in a cone in stress space, with the cone angle =
tan-1 (sin()):
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Geophysical Mass Flow Modeling
(11 - 22 )2 + 4 (12) 2 = sin2 () (11 + 22 )2
This relation is the Mohr-Coulomb Condition. Stress inside the cone is
insufficient for yield (elastic behavior if we were considering elastic-plastic
deformation), and stresses outside the cone are prohibited.
2 EQUATIONS OF MOTION
To describe motion of a mass, we must specify six variables (in 2D), density
, stresses ij, i,j, = 1,2, and the velocities ui, i, = 1,2,. We have 4
governing equations, continuity of mass, momentum balance, and MohrCoulomb. We need to determine additional equations. First assume the
material is incompressible, removing  as a variable. The equations of
motion are
 t ui +  uj j ui + xj ij = gi
 i ui = 0
To complete the model, we combine the Mohr-Coulomb yield condition
with an relation between the stress and the strain-rate tensors
Let  = ½ tr(), and define the strain-rate Vij = - ½ (j ui + i uj).
We can write the yield condition as
|dev()|2 = (1 -  )2 + (2 -  )2 = k2  2 where k = 2 sin().
The so-called Flow Rule tells of an alignment of the directions of the stress
deviator and the strain-rate
Vij =  dev(), >0
Using these assumptions, one can write  = I + dev() = I + -1 V. But
-1 = | dev()| / |V| = k / |V|. Combining all these assumptions, we write
 =  (I + k V/|V|).
Note: If  < 60, I + k A is positive definite, where A=V/|V|. To see this,
observe a1 + a2 = 0, and = a1 2 + a2 2 = 1, so a1,2 =  2-1/2 .
Fact: The governing system is ill-posed.
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3 ‘SHALLOW GRAIN THEORY’ (Savage and Hutter, JFM 199 1989)
Consider a mass of material sliding down an incline at an angle . Let
y=h(x,t) be the equation describing the (top) free surface of the mass, and
y=b(x) be the basal surface. The equations of motion for the mass are
 t ui +  uj j ui + xj ij = gi
 i ui = 0
Assuming the free surface is stress-free, so a function FS(x,t) is zero there,
the boundary conditions are
t FS + i FS ui = 0, and    = 0
At the base, we have that u   = 0 (flow is tangent to the surface), and a
solid friction law relating shear and normal stresses and a bed friction angle,
   -  (    ) = - (uS / |uS|) (    ) tan() where uS is the
basal slip velocity and  is a basal friction angle [this is a restatement of the
Coulomb law  =  tan() at the base, with the internal friction angle 
replaced by a basal friction angle ].
Scale the downslope direction by L and the normal direction by H, time by
{L/g}, the downslope velocity by {gL} and the normal velocity by
(H/L{gL}). Scale the xx- and yy-stresses by gH cos(), and the xystresses by g H sin(). Let  = H/L.
The governing equations take the form
t u + u x u + v y u = sin()[1- y XY ] -  cos()X XX
[t v + u x v + v y v] = - cos()[1+ y YY ] -  sin()X XY
X u + Yv = 0
In the limit as   0 and incorporating the stress-free surface condition, the
second momentum equation reduces to hydrostatic equilibrium YY (x,y,t) =
h(x,t) – y. Setting =0 in the first momentum equation simplifies the
system too much to be of use in this modeling. The underlined term,
although formally small, is O(1) because the derivative is large. We shall
depth-average this first momentum equation and the equation of
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incompressibility. Checking back after averaging will show this term to be
O(1).
To proceed, then, integrate the x-momentum equation from y=b to y=h to
find
t u dy + x u2 dy – [u (t h + u x h – v)]y=h - [u(u x b – v)]y=b
= sin()[(h-b) - XY (h) + XY (b)]
-  cos() [x XXdy - XX(h) x h + XX(b) x b]
Several terms here vanish due to boundary conditions.
First, we have t h + u x h – v = 0 at y=h(x,t), and u x b – v = 0 at y=b(x).
Second, the stress-free condition implies - cos() XX x h + sin()XY =
0 and - sin() XY xh + cos()YY = 0 at y=h(x,t).
With the shallowness assumption, frictional sliding at the base implies
XY = -sgn(u) YY cot() tan() +O(2) at y=b(x).
The general constitutive relation to be used relies on an assumption from soil
mechanics, that the pile is sufficiently shallow that, throughout, the stress
normal to and parallel to the bed are in constant proportion, XX = kap YY
Where the value of the so-called earth pressure coefficient depends on
whether the pile is elongating (active) or compressing (passive) along the
bed direction. Thus
kap = 2[1 -/+ {1-(1+ tan2()) cos2 ()}1/2 ] / ( cos2 () –1)
with kap = ka and we choose the – sign if x u > 0, and kap = kp and we choose
the + sign if x u < 0.
From these conditions, we have that, at the free surface, XX = kap YY . This
further implies that, at y=h(x,t), XY = 0 = YY = XX .
With these, the momentum equation simplifies to read
t u dy + x u2 dy = sin()[h-b] + sin()XY (b)
-  cos() [x XXdy + XX(b) x b]
Define the average values as, for example, uh = u dy, where h = h-b.
There is one term requiring special consideration, namely
u2h =  u2 dy =  (u)2 h
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The value of  depends on the velocity profile. For example, for a
parabolic velocity field and vanishing basal velocity,  = 6/5, while for a
uniform velocity,  = 1. Because experiments suggest significant sliding is
present and confined to a very thin layer, we make simplify our model by
setting  = 1.
From the condition of hydrostatics we have
YY (x,b,t) = h(x,t) – b(x) = h(x,t)
and
h YY (x,b,t) = ½ (h(x,t) – b(x))2 = ½ h2 (x,t).
Combining all assumption finally yields
t (hu) + x h(u)2 h = sin() h - cos() sgn(u) tan() h
-  cos()kap [x (½h2) + h x b] .
Depth averaging the continuity equation yields
t h + x hu = 0.
Given initial profiles for h and an initial velocity for this pile, given internal
and basal friction angles, and the topography (the angle  and the profile
b(x)), we can solve for the evolving pile height and velocity.
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