[CIG-SHORT] Fault_edge problem

[Brad Aagaard]

Li Teng,

I have annotated your Model_2.jpg figure (attached) to show the nodesets 
you should be using for buried edges. The main fault has two vertices on 
"buried edges", one at each end shown in dark blue (assuming the fault 
ends before the domain boundary). Each secondary fault has two vertices 
on buried edges (one at each end) shown in red and light blue.

When you create the faults, you MUST specify the main fault first.
interfaces = [main, secondary1, secondary2]

I changed the names just so that the order is clear.

Brad


On 08/29/2017 10:20 PM, Li, Teng wrote:
> Hi,
> 
> I am now creating a rectangular model with one main fault and two 
> secondary faults. Please see the attached picture for model details. 
> Because there are no offsets between the main fault and secondary 
> faults, so I create 3 groups in trellis model which contains the nodes 
> that are the fault edge:
> groups: inter, inter1, inter2
> Group inter1 has one node, 15, which is the node linking main fault and 
> the lower secondary fault.
> Group inter2 has one node, 17, which is the node linking main fault and 
> the higher secondary fault.
> Group inter has 2 nodes which are nodes 15, 17, two linking nodes.
> 
> Then I specify fault edge like this:
> 
> Main Fault:
> [pylithapp.timedependent.interfaces.fault]
> open_free_surface = False
> 
> # Material id
> id = 100
> 
> # Name of nodeset from CUBIT
> label = fault
> edge = inter
> 
> Secondary Fault 1:
> [pylithapp.timedependent.interfaces.fault1]
> open_free_surface = False
> 
> 
> # Material id
> id = 101
> 
> # Name of nodeset from CUBIT
> label = fault1
> edge = inter1
> 
> 
> Secondary Fault 2:
> [pylithapp.timedependent.interfaces.fault2]
> open_free_surface = False
> 
> 
> # Material id
> id = 102
> 
> # Name of nodeset from CUBIT
> label = fault2
> edge = inter2
> 
> After the program running successfully, I use Matlab to post process the 
> data on the main fault. However, I find the initial shear traction is 
> strange on the main fault. The initial shear traction are zero at the 
> position of two secondary faults. Actually, I add some initial shear 
> traction on the two secondary faults, however, the direction is also 
> left lateral. So if this is because the link of the two faults, then the 
> shear traction should be bigger. Or is it because there is something 
> wrong in fault edge setting?
> 
> Please see the picture of the initial shear traction on the main fault. 
> The two red lines are reference lines which indicate the secondary 
> faults positions.
> 
> Best,
> Teng
> 
> 
> Teng Li
> 
> Master Candidate in Structures
> 
> Department of Civil and Environmental Engineering
> 
> University of Illinois at Urbana-Champaign
> 
> 205 North Mathews Ave, Urbana, IL. 61801
> 
> Phone:(217)8196210, Email: tengli2 at illinois.edu
> 
> 
> 

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