[cig-commits] r11339 - short/3D/PyLith/trunk/examples/twocells/twoquad4

[brad at geodynamics.org]

Author: brad
Date: 2008-03-06 07:33:23 -0800 (Thu, 06 Mar 2008)
New Revision: 11339

Removed:
   short/3D/PyLith/trunk/examples/twocells/twoquad4/README.tractions
Modified:
   short/3D/PyLith/trunk/examples/twocells/twoquad4/axialtract.cfg
Log:
Moved info in README.tractions to axialtract.cfg.

Deleted: short/3D/PyLith/trunk/examples/twocells/twoquad4/README.tractions
===================================================================
--- short/3D/PyLith/trunk/examples/twocells/twoquad4/README.tractions	2008-03-06 04:23:55 UTC (rev 11338)
+++ short/3D/PyLith/trunk/examples/twocells/twoquad4/README.tractions	2008-03-06 15:33:23 UTC (rev 11339)
@@ -1,37 +0,0 @@
-The traction example is a simple problem to demonstrate the usage of
-Neumann (traction) boundary conditions. A constant traction of 4.0e8 Pa
-is applied to the right edge of the mesh, the left edge is pinned in x,
-and the bottom edge is pinned in y. The constant normal traction yields
-a constant stress field within the domain, where the xx-component is equal
-to the applied traction. The yy and xy components are both zero. The
-corresponding strain field is:
-
-exx = (1 - nu^2) * N/E
-eyy = -nu * (1 + nu) * N/E
-exy = 0
-
-where nu is Poisson's ratio (0.25 for this problem), E is Young's modulus
-(75 GPa for this problem), and N is the applied normal traction (400 MPa
-for this problem). Integrating the strain-displacement relations and using
-the applied displacement BC, the corresponding displacement field is:
-
-u = (1 - nu^2) * N * (x - x0)/E
-v = nu * (1 + nu) * N * (y0 - y)/E
-
-where x0 and y0 are the minimum x and y-values respectively, along which
-displacements are fixed. Using the mesh, properties, and boundary conditions
-for this problem, the solution should be:
-
-exx = 5.0e-3
-eyy = -1.66667e-3
-exy = 0.0
-
-(-2,-1): u = 0.0,  v = -0.003333
-(-2, 1): u = 0.0,  v =  0.0
-( 0,-1): u = 0.01, v = -0.003333
-( 0, 1): u = 0.01, v =  0.0
-( 2,-1): u = 0.02, v = -0.003333
-( 2, 1): u = 0.02, v =  0.0
-
-Users should compare the computed solution to the analytical solution to
-insure that everything is computed correctly.

Modified: short/3D/PyLith/trunk/examples/twocells/twoquad4/axialtract.cfg
===================================================================
--- short/3D/PyLith/trunk/examples/twocells/twoquad4/axialtract.cfg	2008-03-06 04:23:55 UTC (rev 11338)
+++ short/3D/PyLith/trunk/examples/twocells/twoquad4/axialtract.cfg	2008-03-06 15:33:23 UTC (rev 11339)
@@ -1,10 +1,8 @@
 # -*- Python -*-
 
-# To run this problem, type "pylith axialtract.cfg". The settings in
-# pylithapp.cfg will be read by default. See the README for how to run
-# other problems in this directory.
+# This is a simple problem to demonstrate the usage of Neumann
+# (traction) boundary conditions. 
 #
-#
 # >----------------------- ->
 #  |                     |
 #  |                     | ->
@@ -14,12 +12,48 @@
 # >----------------------- ->
 #  ^          ^          ^
 #
-# Left boundary is pinned in the horizontal direction.
-# Bottom boundary is pinned in the vertical direction.
-# Right side has a uniform horizontal traction.
+# Left boundary is pinned in the horizontal direction (x direction).
+# Bottom boundary is pinned in the vertical direction (y direction).
+# Right side has a uniform horizontal (x) traction of 4.0e+08 Pa
+#
+# The uniform normal traction generates a uniform stress field within
+# the domain, where the xx-component is equal to the applied
+# traction. The yy and xy components are both zero. The corresponding
+# strain field is:
+#
+# exx = (1 - nu^2) * N/E
+# eyy = -nu * (1 + nu) * N/E
+# exy = 0
+#
+# where nu is Poisson's ratio (0.25 for this problem), E is Young's modulus
+# (75 GPa for this problem), and N is the applied normal traction (400 MPa
+# for this problem). Integrating the strain-displacement relations and using
+# the applied displacement BC, the corresponding displacement field is:
+#
+# u = (1 - nu^2) * N * (x - x0)/E
+# v = nu * (1 + nu) * N * (y0 - y)/E
+#
+# where x0 and y0 are the minimum x and y-values respectively, along which
+# displacements are fixed. Using the mesh, properties, and boundary conditions
+# for this problem, the solution should be:
+#
+# exx = 5.0e-3
+# eyy = -1.66667e-3
+# exy = 0.0
+#
+# (-2,-1): u = 0.0,  v = -0.003333
+# (-2, 1): u = 0.0,  v =  0.0
+# ( 0,-1): u = 0.01, v = -0.003333
+# ( 0, 1): u = 0.01, v =  0.0
+# ( 2,-1): u = 0.02, v = -0.003333
+# ( 2, 1): u = 0.02, v =  0.0
+#
+# To run this problem, type "pylith axialtract.cfg". The settings in
+# pylithapp.cfg will be read by default. See the README for how to run
+# other problems in this directory.
+#
 
 [pylithapp]
-
 # ----------------------------------------------------------------------
 # problem
 # ----------------------------------------------------------------------