[CIG-SHORT] step14 friction negative

[Brad Aagaard]

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