Mathematical Modeling of Deep Drawing Force with Double Reduction

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Mathematical Modeling of Deep Drawing Force with Double Reduction Powered By Docstoc
					                                                                  61

                                 ISSN 1392 - 1207. MECHANIKA. 2008. Nr.2(70)

Mathematical modeling of deep drawing force with double reduction of
wall thickness

I. Karabegović*, E. Husak**
*University of Bihać, Dr. Irfana Ljubijankića bb, 77 000 Bihać, Bosnia and Herzegovina, E-mail: tfb@bih.net.ba
**University of Bihać, Dr. Irfana Ljubijankića bb, 77 000 Bihać, Bosnia and Herzegovina,
E-mail: erminhusak@yahoo.com

1. Introduction                                                        tion, revitalization and controlling of processes and sys-
                                                                       tems.
          Experimental research is used for inspection, cor-                     Regarding that, the main aim of process and sys-
rection and verification of numerical results and at stochas-          tem modeling is the construction of mathematical models.
tic modeling which describes the most realistically machin-            The main aim of experimental research is to get exact, ap-
ing processes and systems. Technological construction and              proximately correct data which will serve as relationships,
projecting of the modern machining processes demand to                 necessary for mathematical model. Mathematical model is
analyze all technical and technological parameters of the              necessary to start optimization of process. Regarding that,
process and to apply scientific methods for modeling and               the main aim of modeling and optimization of machining
defining of optimal conditions of machining processes and              process and systems is cheaper and higher quality produc-
systems.                                                               tion [1].
          The main goal of modeling and optimization of                          Mathematical model of deep drawing force with
machining process is to increase productivity, economy,                double reduction of wall thickness is presented in this pa-
total quality of the product or partial segments (machined             per.
surfaces, tool durability, etc.) also to decrease material
costs, energy, machining time, and machining costs per                 2. Election of significant factors
one piece of the product.
          Using theoretical analytical models it is hard to                  For significant factors electing it is used criteria
define precisely parameters of machining processes like:           that elected factors are not related and not depends from
wasting tool, optimal geometric shape, deformation ap-             outside factors. Outside factors are unelected factors which
pearance at tool or press die, limitation level of deforma-        also belong to the process of deep drawing but are not sig-
tion, tribologic (friction) processes, tool loading. In each of    nificant for it. Significant factors values have to be appli-
the mentioned machining processes a lot of significant             cable for the measuring process. For the process of deep
factors and theirs interactions were applied. Therefore the        drawing are elected three significant factors deformation φ,
application of experiments and analysis of their results is        diameter ratio d1/d2 and friction coefficient μ.
unchangeable in developing the new and improve existing                      Variation of the factors limits are shown in Ta-
machining processes and systems.                                   ble 1 [1 - 4]. The experiment was conducted with the varia-
          The main goal of modeling is to define mathe-            tions factors of two levels. The experiment was repeated
matical model which is necessary in optimization, simula-          three times for each sample.
                                                                                                                         Table 1
                                                Significant factors of machining
   EXPERIMENT LEVEL                                                          Significant factors
                                              Deformation φ                 Diameter ratio d1/d2         Friction coefficient μ
   Maximal                                        1.22                            1.055                           0.20
   Minimal                                        0.95                            1.017                           0.10

3. Equipment for researching                                           d0=26.4 mm (Fig. 3). Force of this process is measured on
                                                                       the same hydraulic testing machine.
          In experiment planning it was decided to use three                     Force measuring equipment in this machining
independent changeable input values with two levels which              process are presented in Fig. 4 [6]: 1 - hidraulic testing
make eight samples. For each sample measuring was re-                  machine, 2 - signal acquisition unit, 3 - monitor for presen-
peated three times which makes twenty four measuring                   tation of data and 4 - unit for data processing.
times.
          This experiment of deep drawing process with                 4. Experiment results
double reduction of wall thickness is executed on hydraulic
testing machine Amsler 300 kN (Fig. 1). Sensors for meas-                        After decision which significant factors would be
uring friction coefficient (Fig. 2) were installed [5, 6]. The         used in this process, the number of variation levels and
material of raw specimen is CuZn28 which before deep                   decision about the number of repeating for each sample,
drawing process was prepared by heating, washing and                   start phase of the experiment process which are presented
lubricating.                                                           Figs. 1 - 4.
          Outer diameter of the raw specimen is
                                                           62




                   a                                                                b
                 Fig. 1 Testing equipment: a - hydraulic testing machine, b - specimen grips




                                               Plug


                                               d0=Ø 26.4

                                   12º            Ø 22



                                                  d1
                        Die

                                         12º


             α
                                                                               Sensor for measuring
                                          Piece                                normal contact
                                                                               tension pn
        Sensor for measuring                          d2
        tangential contact
        tension τn

        Fig. 2 Process of deep drawing with the system for measuring of contact tensions pn and τn

                   d0=Ø26.4                                                                             1
                     Ø22
                                                                                                        2


                                                                                                        3
h0=40




                                                                                                        4




           Fig. 3 Raw specimen
                                                                     Fig. 4 Measuring system of force
                                                                                           63

                                                                                                                                                                   Table 2
                                                                               Experiment results
                          Significant factors data              Results after measuring of deep           Arith-                    Square          Degree of   Varia-
     No of sample

                                                                         drawing force                    metic                      sum            freedom      tion
                                                                                                          value
                                                                                                                             ∑ ( y ji − y j )
                                                                                                                              n                 2
                                                                 Fj1 , kN   F j 2 , kN     F j 3 , kN     yj = F                                        fj       S2
                                φ        d1/d2            μ                                                                  i =1
                                                                                                                                                                  j


          1.              0.95           1.017           0.10       34        33            34.2           33.7                      0.83              2        0.415
          2.              1.22           1.017           0.10       46        43            45.2           44.7                      4.83              2        2.415
          3.              0.95           1.055           0.10      28.1      27.4           27.9           27.8                      0.26              2         0.13
          4.              1.22           1.055           0.10      42.6      43.1           42.3           42.7                      0.33              2        0.165
          5.              0.95           1.017           0.20      34.8      34.8           36.1           35.3                      1.14              2         0.57
          6.              1.22           1.017           0.20      47.1      47.2           46.1           46.8                      0.74              2         0.37
          7.              0.95           1.055           0.20      29.5      28.1           29.8           29.2                      1.66              2         0.83
          8.              1.22           1.055           0.20       48        48            47.5           47.8                      0.17              2        0.085

          Changing the value of all or just some input pa-                                      5. Processing of experimental results
rameters and repeating the experiment the matrix of output
result values is completed. Repeating of experiment in
                                                                                                          After inspection of the homogeneity of experi-
same sample with the same values of significant factors
                                                                                                mental results, the next step would be the calculation of
was founded three different results. Final experimental
                                                                                                regression coefficients. Regression coefficients are calcu-
results are given in Table 2. For further procedure it is im-
                                                                                                lated using the following formulas (models) [1, 7]
portant to find arithmetical mean value of three repeated
experiments. On these bases it is possible to finish inspec-
                                                                                                               1    N
tion of homogeneity of result dispersion.                                                               bi =
                                                                                                               N
                                                                                                                   ∑ X ij y j , i = 1, 2, ...., k                        (4)
          Inspection of homogeneity of dispersion for                                                               j =1

sorted level of reliability (P=0.95) has been done by
Cohran’s criteria [1, 7]                                                                        and
                                                                                                                1       N

                        max S 2                                                                         bim =        ∑ X ij X mj y j                                     (5)
        Kh =             N
                              j
                                          (
                                    ≤ Kt f j , N     )                               (1)                        N     j =1


                         ∑ S 2j                                                                 where X ij is the value of X i in j-th experiment presented
                         j =1

                                                                                                in Table 3, y j is measured value in j-th experiment and
where K t is value by Cohran’s criteria.
                                                                                                 N is the number of experiments (samples).
        Dispersion (variation) is calculated by the next                                                  For correct structuring of mathematical model of
model (2) (Table 2)                                                                             any process in machining it is very important to choose a
                                                                                                correct approximate mathematical model. If the first is not
        S2 =
         j
                           1 3
                        n j − 1 i =1
                                     (
                                ∑ y ji − y j     )
                                                 2
                                                         j = 1, 2 ,...,8             (2)        correct it is necessary to repeat all the steps for new ap-
                                                                                                proximate model construction.
                                                                                                          In this paper the function of deep drawing force
          f j = n j −1                                                               (3)        correlates with the mathematical model presented in for-
                                                                                                mula (6) where is k=3
where f j is degree of freedom, n j = 3 is the number of
                                                                                                        Y = b0 + b1 X 1 + b2 X 2 + b3 X 3 + b12 X 1 X 2 +
repeating for one sample
          By Cohran’s criteria:                                                                              + b23 X 2 X 3 + b13 X 1 X 3 + b123 X 1 X 2 X 3              (6)
- after calculation process with the results in Table 2
                                                                                                         After calculation of regression coefficients results
        K h = 0.4849                                                                            are presented below

- using data from Table 1 founded in reference [1] it is                                                b0 = 38.5, b1 = 7 , b2 = −1.625, b3 = 1.275
obtained                                                                                                b12 = 1.375, b13 = 0.525, b23 = 0.35, b123 = 0.4

                    (        )
        K t f j , N = K t (2,8)                                                                           The next step is testing of significances of regres-
                                                                                                sion coefficients. Each coefficient should be tested for im-
        K t = 0.516                                                                             portance in our mathematical model of deep drawing force
                                                                                                (function) (6). In this calculation were use coded values.
          Regarding that K h = 0.4849 < K t =0.516, process                                               Evaluation of significances of regression coeffi-
of getting mathematical model to be continued. Cohran’s                                         cients is going to be done separately one by one. Regres-
criteria confirms that homogeneity of experimental results                                      sion coefficients which are not significant should be re-
dispersion satisfy.                                                                             moved from the mathematical model. It is not necessary to
                                                                                                do value correction for significant regression coefficients.
                                                                                                    64

                                                                                                                                                                                             Table 3
                                                                                         Code value of process
     No of               Variables of physical process                              Code variable of process
                                                                                                                                                                     (y              )
                                                                                                                                                                                         2
     sample                                                                                                                 y jE                yR
                                                                                                                                                 j
                                                                                                                                                                          E
                                                                                                                                                                          j   − yR
                                                                                                                                                                                 j
                          φ                     d1/d2                    μ          X1        X2           X3
        1.               0.95                   1.017                  0.10         -1        -1           -1             33.7               33.7                             0
        2.               1.22                   1.017                  0.10         +1        -1           -1             44.7               44.7                             0
        3.               0.95                   1.055                  0.10         -1        +1           -1             27.8               27.8                             0
        4.               1.22                   1.055                  0.10         +1        +1           -1             42.7               42.7                             0
        5.               0.95                   1.017                  0.20         -1        -1           +1             35.3               35.3                             0
        6.               1.22                   1.017                  0.20         +1        -1           +1             46.8               46.8                             0
        7.               0.95                   1.055                  0.20         -1        +1           +1             29.2               29.2                             0
        8.               1.22                   1.055                  0.20         +1        +1           +1             47.8               47.8                             0

          Two known criteria for the evaluation of signifi-                                                     b3 = 1.275 > 0.28183                                 significant
cances of regression coefficients can be used: t – Student’s
criteria or Fisher’s criteria and correlation [1, 7] existing                                                   b12 = 1.375 > 0.28183                                significant
between them                                                                                                    b13 = 0.525 > 0.28183                                significant
                                                                                                                b23 = 0.35 > 0.28183                                 significant
      F (1, f ) = t 2 ( f )                                                                   (7)
                                                                                                                b123 = 0.4 > 0.28183                                  significant
         For the evaluation of significances of regression
coefficients of model bi , it would be used t’s criteria or                                              where
Student’s test.                                                                                                 t ( f 1 ,ε ) = t ( f E ,α ) = t (16,0.05) = 1.75                                (13)
         The formula for significance testing of regres-
sions coefficients bi with t - criteria is                                                                        After significance testing of regression coeffi-
                                                                                                         cients in the mathematic model, the conclusion is that all
                                                                                                         regression coefficients are significant. The next step is to
              bi             bi        N n                                                               return original values for all regression coefficients in
      tri =              =                       ≥ tt                                         (8)
              Sbi                  Sy                   ( f y ,α )                                       model (6). After this model will have the structure pre-
                                                                                                         sented in formula
for i = 0, 1, 2, ...., k or
                                                                                                                Y = 38.5 + 7 X 1 − 1.625 X 2 + 1.275 X 3 +
                                                                       Sy
        bi ≥ Δbi = ±tt                         Sbi = tt                                       (9)                    + 1.375 X 1 X 2 + 0.525 X 1 X 3 + 0.35 X 2 X 3 +
                                  ( f y ,α )              ( f y ,α )   N n
                                                                                                                     + 0.4 X 1 X 2 X 3                                                          (14)
for i = 0, 1, 2, ..., k
          The variation of experimental error can be de-                                                          Model Eq. (14) presents deep drawing force as the
scribed by the model                                                                                     function of significant parameters but in the code value.
                                                                                                         Used those coded values the values of deep drawing force
                                                                                                         are calculated. They are presented in Table 3.
               ∑ ∑ (y ji − y j )
                N        n
                                                  2


      Sy =
       2       j =1 i =1
                                                                                             (10)        6. Structuring of final mathematical model
                                  fy
or                                                                                                                 After calculating and writing down the results of
                     2                                                                                   deep drawing force, the next step is to test this mathemati-
                S    y
      Sbi =
       2
                             , i = 0, 1, 2, ..., k                                           (11)        cal model (14) for adequacy. For this purpose, Fisher’s
               N n                                                                                       criteria presented below are used
where                                                                                                           Fa < Ft                                                                         (15)

                     (             )
              N
        f y = ∑ n j − 1 = N ( n − 1)                                                         (12)        where Ft is value founded in Table 3 in reference [1]
              j =1
                                                                                                         where is determinate by level of significance
and f y is total number of degree of freedom, n j is the                                                  p (Fa > Ft ) = α = 0.05, or (1 − α ) = 0.95 , it is 95% reli-
number of experiment repeating in j-th line of matrix,                                                   ability. Where is
when is the same repeating number n = n j .
                                                                                                                        2
S b ⋅ t = 0.28183 , in the condition for being significant. The                                                 Fa =
                                                                                                                       Sa
                                                                                                                        2
                                                                                                                                                     (
                                                                                                                          ≤ Ft ( f 1 , f 2 ) = Ft f a , f y      )                              (16)
                                                                                                                       Sy
coefficients which satisfy this condition bi > S b ⋅ t , are:

       b0 = 38.5 > 0.28183                                                    significant                for Sa > S y
                                                                                                              2     2



       b1 = 7 > 0.28183                                                       significant                               2
                                                                                                                       Sy
       b2 = − 1.625 > 0.28183                                               significant
                                                                                                                Fa =    2
                                                                                                                                                     (
                                                                                                                             ≤ Ft ( f 1 , f 2 ) = Ft f a , f y   )                              (17)
                                                                                                                       Sa
                                                                         65

for S y > Sa .
      2    2
                                                                              F, kN
         The value of dispersion of adequacy is determined
by the following model                                                            55           d1/d2=1.017
                                                                                  50           µ = 0.20
               ∑ n(y E − y R )
                  N
                                      2
                     j     j                                                      45        d1/d2=1.017
               j =1
        Sa =
         2
                                                                  (18)            40        µ = 0.10
                                 fa
                                                                                  35                                         d1/d2=1.055
                                                                                  30                                         µ = 0.20
where f a = N − k − 1 is the number of degree of freedom
                                                                                  25
which is related to the dispersion of adequacy.                                                                              d1/d2=1.055
         The corresponding mathematical relation is ob-                           20                                         µ = 0.10
tained using the model (16) when S y > Sa
                                      2   2                                       15
                                                                                  10
                2                                                                 5
               Sa
        Fa =    2
                  =0                                                              0
               Sy                                                                     0.7   0.8 0.9        1.0 1.1 1.2 1.3 φ
                                                                              Fig. 5 Theoretical correlation between deep drawings force
        From this it can be concluded that the mathemati-                            F, kN and deformation degree φ
cal model 100 % describes finished experiment. That is
                                                                              F, kN
        Fa = 0 < Ft = 3
                                                                                                     φ=1.22       φ=1.22

∑ n(y                 )
 N                                                                                                                µ = 0.10
                  R 2                                                                                µ = 0.20
         E
         j   −y   j           is a part of model (19). Model (19) pre-            50
 j =1

sents coefficient of multiple regression. Result of                               45

∑ n(y E − y R )
 N
                      2                                                           40
      j     j             equals zero (Table 3). That shows coeffi-
j =1                                                                              35
cient of multiple regression will equals 1.                                       30                              φ=0.95
                                                                                                   φ=0.95
         When regression model correctly describes the                            25               µ = 0.20       µ = 0.10
process R → 1 . Coefficient of multiple regression is de-
                                                                                  20
scribed by the model (19)
                                                                                  15
                                                                                  10
                          ∑ (y E − y R )
                          N
                                          2

                          j =1
                               j     j                                            5
        R = 1−                                                    (19)            0
                          ∑ (y E − y E )
                          N
                                          2
                               j                                                           1      1.01 1.02 1.03         1.04 1.05 1.06 d1/d2
                          j =1

                                                                              Fig. 6 Theoretical correlation between deep drawing force
         As the model (14) adequately describes mean                                 F, kN and diameter ratio d1/d2
value of deep drawing force F in relation with significant
parameters: φ, d1/d2 and μ, then the next proceeding is the
                                                                                 F, kN              d1/d2=1.017       d1/d2=1.055
conception of final mathematical model. In order to get
                                                                                                    φ = 0.20          φ = 0.20
mathematical model with real coefficients it is necessary to
finish decoding of the model using transformation equation                             50
[1].                                                                                   45
         Mathematical model for deep drawing the force                                 40
where force is related with significant parameters is de-
scribed in the next model                                                              35
                                                                                       30
     F = 213.4 − 30.46ϕ − 213.58 d1 d 2 + 3049.14 μ +                                  25
                                                                                                        d1/d2=1.017          d1/d2=1.055
        + 68ϕ ( d1 d 2 ) − 3149.88ϕ μ − 2999.9 μ ( d1 d 2 ) +                          20               φ = 0.10             φ = 0.10
        + 3115.69ϕ μ ( d1 / d 2 )                                 (20)                 15
                                                                                       10
         The correlation between the force and parameters                              5
can be shown graphically Fig. 5, 6 and 7. Diagrams (Fig. 5,                            0
6 and 7) present the correlation of deep drawing force with                               0      0.05     0.10    0.15   0.20    0.25 μ
double reduction of wall thickness with: deformation de-
                                                                              Fig. 7 Theoretical correlation between deep drawings force
gree φ, diameter ratio of press die d1/d2 and friction coeffi-
                                                                                     F, kN and friction coefficient μ
cient μ.
                                                                  66

7. Conclusions                                                         tinis modeliavimas įgalina teisingai analitiškai aprašyti
                                                                       deformacijos procesą. Turint analitiškai aprašytą deforma-
         Experiments of deep drawing process with double               cijos procesą, galima skaičiuoti giliojo ištempimo jėgą ir
reduction in press die of wall thickness were performed.               optimizuoti visą deformacijos procesą bei ištempimo jėgą.
         The following conclusions were made:                          Straipsnyje aprašoma planuojamo eksperimento eiga, pra-
     - final mathematical model is adequate for describ-               dedant svarbiausių parametrų įvertinimu, sienelės storio
       ing the process of deep drawing what is confirmed               tolygumo kitimo, eksperimente naudojamos įrangos, gautų
       by Fisher’s test;                                               rezultatų palyginimu su žinomais analiziniais modeliais bei
     - final mathematical model is absolutely correct for              galutinio matematinio modelio giliojo ištempimo jėgai
       describing the process of deep drawing what is con-             nustatyti sudarymu.
       firmed by multiple regression coefficient. Which is
       equal 1;
     - deep drawing force is in line function of independ-             I. Karabegović, E. Husak
       ent input parameters: deformation degree φ, diame-
       ter ratio of press die d1/d2 and friction coefficient μ;        MATHEMATICAL MODELING OF DEEP DRAWING
     - final mathematical model should be used for opti-               FORCE WITH DOUBLE REDUCTION OF WALL
       mization of deep drawing process.                               THICKNESS
         The main goal of the optimization of this mathe-
                                                                       Summary
matical model is to get minimal deep drawing force with
the reduction of wall thickness with optimal input parame-                      The main goal of this paper is to get a mathema-
ters (factors). The importance of this force minimizing                tical model for the process of deep drawing force with the
would be multiple: from minimizing energy consumption                  reduction of wall thickness. Mathematical model should
to decreasing of intensity of wasting on press die guides              have a correct analytic description of this deformation
and other pieces of press die.                                         process and along with that a possibility to calculative deep
                                                                       drawing force and after that a possibility of optimization of
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                                                                       ход планируемого эксперимента, начиная с оценки
I. Karabegović, E. Husak                                               основных параметров, изменения равномерности ши-
                                                                       рины стенки, в эксперименте используемого оборудо-
GILIOJO IŠTEMPIMO JĖGOS, PERPUS                                        вания. Полученные результаты сравнены с результата-
SUMAŽINANČIOS SIENOS STORĮ, MATEMATINIS                                ми, полученными при использовании известных ана-
MODELIAVIMAS                                                           литических моделей и в конечном счете создана мате-
                                                                       матическая модель для определения силы глубокой
Reziumė                                                                вытяжки.
         Straipsnis skirtas plastinio deformavimo jėgos,
                                                                                                        Received January 21, 2008
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čios sienelės storį, matematiniam modeliavimui. Matema-