Here is the way I related the Paterson ice flow power law formulation for ice rheology to that used by the ADINA-F finite element model system - with help from Nadine and EdW.

Problem:

Relate the flow law of Paterson to the “Power Law” material model of ADINA-F.

From Paterson:

(1) e_ij=B t_ij^m where e_ij is meant to mean strain rate not strain – I can’t make the “dot”

m is meant to be what Paterson calls “n” – generally we use 3

B is meant to be what Paterson calls “A” – e.g. 4.9x10²⁵ s^-1 pa^-3

From ADINA: (eqn 2.1-5 in tmg-tf)

(2) t_ij=2 m e_ij where e_ij is strain rate and m is viscosity. The power law relationship between stress and strain rate is expressed through the dependence of m on strain rate.

From ADINA:

(3) m = A e_ijⁿ

For the ADINA FEM, we must specify A, n and a m_o value to be used depending on whether n>=0 or n<0.

In the former case m_o is used if it is greater and m_o is used if it is less than A e_ijⁿ

So, first we find an expression for m in terms of Paterson's “m” and “B” and then write the FEM parameters A & n in terms of those “m” and “B”.

Starting with Paterson’s (1)

(4) e_ij=B t_ij^m

Then substitute from (2) for t_ij into (4)

(5) e_ij=B(2 m e_ij)ⁿ

Solve for m

B^-1/m/2 e_ij^(1/m-1)= m

And from ADINA (3)

m= A e_ijⁿ

We find that in terms of Paterson's "B" and "m", the FEM values for "A" and "n" are:

A= (B^-1/m)/2

and

n=(1/m - 1)

From Paterson we assume m=3 which implies that the FEM value n:

n=-2/3

From Paterson, "B" for -10C ice = 4.9x10²⁵ s^-1 pa^-3

A=6.34x10⁷ s^1/3 pa

Not sure what to use for a m_o

In the model then, we specify the following for now - neglecting the heat stuff. Note that the 2-D problems are done in the Y-Z plane with Z positive upward so we specify -9.8m/s/s acceleration due to gravity. This will be the only "loading" prescribed in the model - though it is not considered "loading"

Here are some comments from Ed that got me on back on track for describing what to use for m_o

FROM EDW:

Hi Tony -

This sounds like the right direction to go to convert Glen into Adina. I think we need to be careful to use the invariants like "D" in the second box in the derivation.

Otherwise you may get a result applicable only to a case where there is only one nonzero stress component, like simple shear.

The equation (3) from ADINA says

mu = A eps_{ij}^n

but eps_{ij}^n is a tensor and mu is a scalar, so (3) should more properly

be

mu = A D^n

like (2.3.1). And Paterson's Glen law should be

eps_{ij} = B S^{m-1} tau_{ij}

where S^2 = tau_{ij} tau_{ij} is a scalar, and the invariants are related

by

D = B S^m

eps_{ij} = B S^{m-1} tau_{ij} from Glen

= B S^{m-1} ( 2 mu eps_{ij} ) using (2)

so

B S^{m-1} 2 mu = 1

and

mu = (1/2B) S^(1-m)

= A D^n from ADINA definition

= A ( B S^m )^n

= A B^n S^{nm} from Glen law

Equating coefficients and powers on left and right

(1/2B) = A B^n and nm = 1-m

A = (1/2) B^-(n+1) n = 1/m - 1

And using the result for n,

A = (1/2) B^(1/m)

TGADES NOTE: slight error on this step exponent of B should be negative 1/m

so my answer is the same as yours after all.

The goal of the ADINA formulation appears to be to prevent the

viscosity from going to infinity as stress goes to zero when m>1

or going to zero when m<1.

So mu_0 should be the viscosity at the "crossover stress" S_0 at which Glen's law goes linear. Based on ice cube squashing tests, this stress must be below 0.3 bar, and probably around 0.1 bar, based on Dave's thesis (tests for Taylor Dome flow model), the poster about the Siple Dome vertical strain experiment outside ATG 717, and some models for Siple Dome that Erin has done.

mu_0 = (1/2B) S_0^(1-m)

Charlie's model currently has both a linear and cubic term in it all the

time, rather than an abrupt transition like ADINA apparently has.A

It will be a very interesting test whether you can actually build your own viscosity function to replace the ADINA power law, because that is what we will need to do to include anisotropy and other rheological variants.

cheers, Ed

Ed Waddington

Geophysics Program Box 351650

University of Washington

Seattle WA 98195 USA

So now we take Paterson Table 5.2 for N=3, A values are in columns 2,3 and convert them to the language of ADINA:

TEMP A (S^-1 kpa^-3)

0.0000000e+000 6.8000000e-024

-2.0000000e+000 2.4000000e-024

-5.0000000e+000 1.6000000e-024

-1.0000000e+001 4.9000000e-025

-1.5000000e+001 2.9000000e-025

-2.0000000e+001 1.7000000e-025

-2.5000000e+001 9.4000000e-026

-3.0000000e+001 5.1000000e-026

-3.5000000e+001 2.7000000e-026

-4.0000000e+001 1.4000000e-026

-4.5000000e+001 7.3000000e-027

-5.0000000e+001 3.6000000e-027

So here are the values to use for -10C ice (A) and for -25C ice (B):

(A) Calculating values for the FEM.

At -10C (A_paterson = 4.9e-25 s^-1 pa^-3

n_model = 1/n_paterson - 1

= -2/3

A_model = 1/2 A_paterson^(-1/n_paterson)

=1/2 (4.9E-25 s^-1 pa^-3)^(-1/3)

= 6.3422e7

mu_0 = 0.5(1/A_paterson) S_0^(1-m) (assume transition occurs at 0.1bar = 1e4 pascals)

= 1.0204e16

Menu option: MODEL->MATERIALS->POWER LAW->ADD

(B) At -25C (A_paterson = 9.4e-26 s^-1 pa^-3)

n_model = 1/n_paterson - 1

= -2/3

A_model = 1/2 A_paterson^(-1/n_paterson)

= 1.1e8

mu_0 = 0.5(1/A_paterson) S_0^(1-m) (assume transition occurs at 0.1bar = 1e4 pascals)

= 5.319e16