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Viscoplastic Models for Polymeric Composite

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									Viscoplastic Models for
 Polymeric Composite
              Mentee
           Chris Rogan
        Department of Physics
         Princeton University
          Princeton, NJ 08544

             Mentors
Marwan Al-Haik & M.Y Hussaini
    School of Computational Science
        Florida State University
        Tallahassee, FL 32306
Part-1 Explicit Model

Micromechanical
Viscoplastic Model
                               Explicit Model

    Viscoplastic Model Proposed by Gates and Sun


                             t    =       e      +       p

    e      = /E                                       p = A()n
The elastic portion of the strain is          The plastic portion of the strain is represented by
determined by Hook’s Law, where E is          this non-linear equation, where A and n are
Young’s Modulus                               material constants found from experimental data


 Gates, T.S., Sun, C.T., 1991. An elastic/viscoplastic constitutive model for fiber reinforced
 thermoplastic composites. AIAA Journal 29 (3), 457–463.
                              Explicit Model
   The total strain rate is composed of elastic and plastic components



      dt/dt = de/dt + dp/dt
        de/dt = (d/dt)/E
dvp /dt = dvp’/dt + dvp’’/dt
The elastic portion of the strain rate is the elastic component of the strain differentiated
with respect to time. The component of the strain rate is further divided into two viscoplastic
terms,
                       Explicit Model
    The first component of the plastic strain rate is the plastic strain
    differentiated with respect to time




   dvp’/dt = A(n)()n-1(d/dt)
The second component utilizes the concept of ‘overstress’, or - *, where * is
the quasistatic stress and and  is the dynamic stress. K and m are material
constants found from experimental data.



   dvp’’/dt = (( - *)/K)1/m
                            Tensile Tests
                                     Tensile Tests at Different Temperatures

               400


               350       T= 25 C

                         T= 35 C

               300       T= 45 C

                         T= 50 C

                         T= 55 C
               250
Stress (MPa)




                         T= 60 C

                         T= 65 C
               200
                         T= 75 C

                         Poly. (T=
               150       75 C)
                         Poly. (T=
                         35 C)
                         Poly. (T=
               100       25 C)
                         Poly. (T=
                         45 C)
                         Poly. (T=
                50       55 C)
                         Poly. (T=
                         60 C)
                         Poly. (T=
                 0       65 C)
                         Poly. (T=
                     0      0.001
                         50 C)              0.002      0.003            0.004   0.005   0.006
                                                       Strain (mm/mm)



                                          Figure 1
                           Methodology
Firstly, the tensile test data (above) was used to determine the material constants A,
n and E for each temperature. E was calculated first,




                                     = /E +                  A()n

fitting the linear portion of the tensile test curve to reflect the elastic component of
the equation as shown in Figure 2. Next, the constants A and n were calculated by

plotting Log() vs. Log( -   /E) and extracting n and A from a linear fit as the
slope and y-intercept respectively. Figure 3 displays the resulting model’s fit to the
experimental data.
                         Log(s) vs. Log(e - s/E) @ 45

            0

           -1

           -2
                2.3   2.35        2.4           2.45         2.5

                                          y = 7.1791x - 21.826
                                                                           2.55
                                                                                                                            A-n
                                               R2 = 0.925
  Log(s)




           -3



                                                                                        Log( - /E) =
           -4

           -5

           -6
                                    Log(e - s/E)                                          nLog() + LogA
                                        Figure 2
                                                                                                                    Stress vs. Strain @ 45
                             Stress vs. Strain @ 45

                                                                                                 350
           140
                                                                                                 300
           120
                                                                                                 250
           100




                                                                                        Stress
                                                                                                 200                                                        Exper.
Stress




           80
                                                                 y = 65922x                      150                                                        Modeled
           60
                                                                 R2 = 0.9989                     100
           40
                                                                                                 50
           20
                                                                                                  0
            0
                                                                                                       0   0.001   0.002   0.003    0.004   0.005   0.006
                  0   0.0005      0.001            0.0015        0.002         0.0025
                                                                                                                           Strain
                                          Strain



                               Figure 3                                                                                    Figure 4
                                 Table 1




                                 Stress*
T \ %Strain      30%      40%        50%      60%      70%      80%
         25   124.054   160.19     200.82   242.14   285.19   322.12
         35    101.93   140.73      177.8   229.92   276.68   316.46
         45    85.878   135.66     173.91   201.93   224.98   277.78
         50    81.821   117.35     154.97   181.73   196.57   245.75
         55    69.657   110.49      132.5   173.12   189.18   235.28
         60     68.61   103.32     138.24    145.5   187.69   215.91
         65     63.11    94.13     102.65   135.77   150.44   178.86
         75    58.604   76.479     101.95   125.25   136.44   167.97
                           Load Relaxation Tests
                                               Stress Relaxation @ 45 C

                350


                300


                250

                                                                                               30% Strength
Stress [ MPa]




                200
                                                                                               40% Strength
                                                                                               50% Strength
                150                                                                            60% Strength
                                                                                               70% Strength
                100                                                                            80% Strength


                50


                 0
                      0   1000   2000   3000        4000      5000        6000   7000   8000

                                  tests
The data from the load relaxationTime [s] was used to determine the temperature-
dependent material constants K and m. For each temperature, the load relaxation test
was conducted at 6 different stress levels, as shown in Figure 4.
  Curve Fitting of Load Relaxation




                                   Figure 5
Firstly, the data from each different strain level at each temperature was isolated. The noise
in the data was eliminated to ensure that the stress is monotonically decreasing, as dictated
by the physical model (Figure 5). The data was then fit into two different trends; exponential
and polynomial of order 9 functions (Figures 6 and 7).
      Figure 6                   Figure 7

   0 = d/dt = (d/dt)/E + (( - *)/K)1/m =>
Log(-(d/dt)/E) = (1/m)(Log( - *) – (1/m)Log K
   From the exponential fits the constants K and m were calculated by plotting Log(-
   (d/dt)/E) vs. Log( - *), and calculating the linear fit, as shown in Figures 8 and 9. The
   tabulated material constants for each temperature are pictured below.



                                         Table 2
Temp (Deg)         25         35         45         50         55       60         65         75
A (Mpa)    10^-12.479 10^-6.8025 10^-12.156 10^-41.478 10^-19.4 10^-19.257 10^-28.563 10^-9.1163
n             3.6026       1.296    3.2313     15.305     6.3367   6.6161     10.165     2.5115
K          1.44E+07 9.21E+06 1.80E+13 3.50E+11 2.24E+07 4.39E+07 3.61E+06 4.00E+07
m            0.64654 0.74623        1.1965     1.0403 0.58915 0.71173 0.54771            0.8108
E(Mpa)         81081      72514      65922      65224      62014    60527      58331      46611
                                                                Figure 9
                        Figure 8

                    = */E + A(*)n
For each temperature and strain level, the quasistatic stress was found by solving the above non-
linear equation using Newton’s method. The quasistaitc stress values are displayed in Table 1.
      Simulation of Explicit Model
 -(d/dt)/E = (( -                    *)/K)1/m

The total strain rate is zero
during the load relaxation test,
leading to the differential equation
above. The explicit model solution
was generated by solving this
differential equation using the
fourth order Runge-Kutta method.
Different step-sizes were
experimented with, and an
example solution is shown in
Figure 10.




                                         Figure 10
   Part 2: Implicit Model

Generalizing an Implicit
 Stress Function Using
   Neural Networks
Neural Networks (NN)
                      The Implicit Model consists of
          creating an implicit, generalized stress function,
          dependent on vectors of temperature, strain
          level and time data. A generalized neural
          network and one specific to this model are
          shown in Figure 11. A neural network consists
          of nodes connected by links. Each node is a
          processing element which takes weighted
          inputs from other nodes, sums them, and then
          maps this sum with an activation function, the
          result of which becomes the neuron’s output.
          This output is then propagated along all the
          links exiting the neuron to subsequent neurons.
          Each link has a weight value to which traveling
          outputs are multiplied.
            Procedures for NN
Based on the three phases of neural networks functionality
(training, validation and testing),the data sets from the load
relaxation tests were split into three parts. The data sets for
three temperatures were set aside for testing. The other five
temperatures were used for training, excluding five specific
combinations of temperature and strain levels used for
validation.
           Pre-processing

Before training, the data vectors were put into
random order and were normalized by the equation
                         Training NN
Training a feed-forward backpropagating neural
network consists of giving the network a
vectorized training data set each epoch. Each
individual vector’s inputs (temperature, strain
level, time) are propagated through the
network, and the output is incorporated with
the vector’s experimental output in the error
equation above. Training the network consists
of minimizing this error function in weight’s
space, adjusting the network’s weights using
unconstrained local optimization methods. An
example of a training session’s graph is shown
in Figure 12, in this case using a gradient
descent method with variable learning rate and
momentum terms to minimize the error
function.




                      Figure 12
                     2 Hidden Layers NN

      The architecture of the neural network is difficult to
      decide. Research by Hornik et al. (1989) suggests that
      a network with two hidden layers can approximate any
      function, although there is no indication as to how many
      neurons to put in each of the hidden layers. Too many
      neurons causes ‘overfitting’; the network essentially
      memorizes the training data and becomes a look-up-
      table, causing it to perform poorly with the validation
      and training data that it has not seen before. Too few
      neurons leads to poor performance for all of the data.
Hornik, K., Stinchocombe, M., White, H., 1989. Multilayer feedforward networks are universal
approximators. Neural Networks, 359–366.
                              Error Surface
Figure 13 shows the resulting mean
squared error performance values for
neural networks with different
numbers of neurons in each hidden
layer after 1000 epochs of training.




                                        Figure 13
                  Figure 14                                    Figure 15
Figures 14 and 15 display similar data, except that only random data points are used in
the neuron space and a cubic interpolation is employed in order to distinguish trends in
the neuron space. As figure 15 shows, there appears to be a minimum in the area of
about 10 neurons in the first hidden layer and 30 in the second. A minimum did in fact
occur with a [10 31 1] network.
  Genetic Algorithm (GA) Pruning
A genetic algorithm was used to try to
determine an optimal network
architecture. Based on the results of
earlier exhaustive methods, a domain
from 1 to 15 and 1 to 35 was used for
the number of neurons in the first and
second hidden layers respectively.
A population of random networks in this
domain was generated, each network
encoded as a binary chromosome. The
probability of a particular network’s
survival is a linear function of its rank in
the population.
Stochastic remainder selection without
replacement was used in population
selection. For crossovers, a two-point
crossover of chromosomes’ reduced
surrogates was used as shown in Figure
                                               Figure 16
16.
           GA-Pruning
This method allows pruning of not only neurons
   but links, as each layers of neurons is not
   necessarily completely connected to the
   next, and connections between non-               Figure 17
   adjacent layers is permitted. The genetic
   algorithm was run with varying parameter
   values and two different objective functions;
   one seeking to minimize only the training
   performance error of the networks and
   another minimizing both the performance
   error and the number of neurons and links.
   Figure 17 displays an optimal network when
   only the performance error is considered.
   Figure 18 shows and optimal network when
   the number of neurons and links was taken
   into account.                                   Figure 18
                 GA-Performance
Figure 19 shows the results of an exhaustive architecture search in a smaller
domain than earlier, the first arrow pointing to a minimum that coincides with
the network architecture displayed in Figure 17.




                                                        Figure 19
             Results of NN Implicit Model
            A network architecture of [10 31 1] was used for the training and testing of the
            neural networks. Several different minimization algorithms were tested and compared
            for the training of the network and are listed in Figures 20 and 21. These two figures
            display the training performance error and gradient over 1000 epochs.




                                                       Figure 21
Figure 20
  Training- Validation & Testing of Final
              NN Structure
Figure 22 shows the testing, validation and training performance for the
Gradient Descent algorithm while Figure 23 shows the plot of a linear least
squares regression between the experimental data and network outputs for
the Polack Ribiere Conjugate Gradient method.




                                            Figure 23


     Figure 22                                          Figure 23
            Figure 24 displays the final performance of both models
            compared to the experimental data. The Quasi-Newton
Comparing   BFGS algorithm was used for the Implicit model, as it
            performed the best. The Implicit model ultimately
Explicit    outperformed the Explicit, and required only the load
            relaxation data to generate the solution.
and
Implicit
Models


Figure 24
            Conclusion

The Implicit model(NN+GA) ultimately
outperformed the Explicit( Gates),
and required only the load relaxation
data to generate the solution.

								
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