[CIG-SHORT] step14 friction negative

[Birendra jha]

Hi Brad

Can you please suggest how you would calculate the friction coeff? Why is the evolution of theta not captured in the calculation of mu?

The equation for the friction coefficient mu is explicit in terms of the state parameter theta. So I use theta at a time step to calculate mu at that time step. I use Eq (6.70) from the manual, which linearizes the rate term during zero slip velocities. For this simulation, linear_slip_rate = 1e-9 dimensionless = 3.17e-14 m/s. 

Also, what about the shear traction being higher than the coulomb friction stress? 

Thanks
Birendra

--------------------------------------------
On Thu, 9/17/15, Brad Aagaard <baagaard at usgs.gov> wrote:

 Subject: Re: [CIG-SHORT] step14 friction negative
 To: cig-short at geodynamics.org
 Date: Thursday, September 17, 2015, 10:28 AM
 
 On 9/16/15 9:27 PM,
 Birendra jha wrote:
 > Hi
 >
 > Step 14 example: This
 is an example with a rate and state fault in a
 > quasi-static simulation.
 >
 > Q1. I want to check
 the evolution of the friction coefficient. When I
 > calculate the friction coeff in paraview
 calculator as follows
 >
 > mu =
 >
 0.4+0.002*ln(mag(slip_rate)/2e-11)+0.08*ln(2e-11*state_variable/0.05)
 >
 >  where 0.4 is the
 reference friction coeff. Above equation shows that
 > mu becomes negative when rupture happens.
 What am I doing wrong
 > here?
 
 
 
 > Q2. When I plot shear traction vs normal
 traction at a point in the
 > middle of
 the fault, I see that the shear tractions increase above
 > 0.4 x normal traction, where mu_0= 0.4.
 Why is so?
 
 Rate and
 state friction is much more complex than linear slip
 weakening 
 because the state variable is
 sensitive to the initial conditions. You 
 are probably not accounting for how it has
 evolved from the initial 
 conditions.
 
 Regards,
 Brad
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