A square patch made of one quadrilateral plate elements

Basic implementation test with applied loads.

Testing the tangent stiffness computation.

free   free
 ^     ^
 |     |
 3-----2 -> free
 |     | >
 |  a  | > (w = 1.0)
 |     | >
 0-----1 -> free

width:  10.
height: 10.

Material parameters: St. Venant-Kirchhoff, plane stress
    E  = 10.0
    nu =  0.30
    t  =  1.0

Element loads:
    node 0: [ 0.0, 0.0]
    node 1: [ 5.0, 0.0]
    node 2: [ 5.0, 0.0]
    node 3: [ 0.0, 0.0]

Author: Peter Mackenzie-Helnwein

import numpy as np

from femedu.examples import Example

from femedu.domain import System, Node
from femedu.solver import NewtonRaphsonSolver
from femedu.elements.linear import Quad
from femedu.materials import PlaneStress


class ExamplePlate07(Example):

    def problem(self):

        params = dict(
            E  = 10., # Young's modulus
            nu = 0.3,   # Poisson's ratio
            t  = 1.0,   # thickness of the plate
            fy = 1.e30  # yield stress
        )

        a = 10.     # length of the plate in the x-direction
        b = 10.     # length of the plate in the y-direction

        model = System()
        model.setSolver(NewtonRaphsonSolver())

        nd0 = Node( 0.0, 0.0)
        nd1 = Node(   a, 0.0)
        nd2 = Node(   a,   b)
        nd3 = Node( 0.0,   b)

        # nd0.fixDOF('ux', 'uy')
        # nd1.fixDOF('uy')
        # nd3.fixDOF('ux')

        model.addNode(nd0, nd1, nd2, nd3)

        elemA = Quad(nd0, nd1, nd2, nd3, PlaneStress(params))

        model.addElement(elemA)

        elemA.setSurfaceLoad(face=1, pn=1.0)

        model.plot(factor=0.0, title="Undeformed system", filename="plate07_undeformed.png", show_bc=1)

        # %%
        # We can have a quick look at the stiffness mode shapes using the
        # buckling-mode plotter.  These are simply eigenvalues and eigenvectors of Kt
        # at the current load level (0.0)
        #
        model.setLoadFactor(0.0)
        model.solve()

        np.save('../../../Kplate.npy', model.solver.Kt)

        for k in range(8):
            name = f"plate07_mode{k:2d}.png"
            model.plotBucklingMode(mode=k,filename=name,factor=1.0)

        # %%
        # Note the three rigid body modes (lam=0.0). It can be shown that all three
        # are limear combinations of translations in x and y-directions and a
        # rigid body rotation.
        #

        # %%
        # Now it is time to add boundary conditions, apply loads
        # and check the convergence behavior.
        #

        nd0.fixDOF('ux', 'uy')
        nd1.fixDOF('uy')
        nd3.fixDOF('ux')

        model.setLoadFactor(1.0)
        model.solve()

        # %%
        # The output shows that we do have a quadratic rate of convergence.

        # %%
        # Let's finish off with a nice plot of the deformed system.

        model.plot(factor=1.0, filename="plate07_deformed.png")

        model.report()

Run the example by creating an instance of the problem and executing it by calling Example.run()

if __name__ == "__main__":
    ex = ExamplePlate07()
    ex.run()

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System Analysis Report
=======================

Nodes:
---------------------
  Node_1388:
      x:    [0. 0.]
      fix:  ['ux', 'uy']
      u:    [0. 0.]
  Node_1389:
      x:    [10.  0.]
      fix:  ['uy']
      u:    [1. 0.]
  Node_1390:
      x:    [10. 10.]
      u:    [ 1.  -0.3]
  Node_1391:
      x:    [ 0. 10.]
      fix:  ['ux']
      u:    [ 0.  -0.3]

Elements:
---------------------
  Quad_1803: nodes ( Node_1388 Node_1389 Node_1390 Node_1391 )
      material: list
      strain (0): xx=1.000e-01 yy=-3.000e-02 xy=-4.760e-18 zz=-2.100e-02
      stress (0): xx=1.000e+00 yy=0.000e+00 xy=-1.831e-17 zz=0.000e+00
      strain (1): xx=1.000e-01 yy=-3.000e-02 xy=8.302e-17 zz=-2.100e-02
      stress (1): xx=1.000e+00 yy=0.000e+00 xy=3.193e-16 zz=0.000e+00
      strain (2): xx=1.000e-01 yy=-3.000e-02 xy=-7.511e-17 zz=-2.100e-02
      stress (2): xx=1.000e+00 yy=0.000e+00 xy=-2.889e-16 zz=0.000e+00
      strain (3): xx=1.000e-01 yy=-3.000e-02 xy=1.857e-17 zz=-2.100e-02
      stress (3): xx=1.000e+00 yy=0.000e+00 xy=7.142e-17 zz=0.000e+00