JOHNSON-COOK MODEL.SIMPLE TENSION.

 material constants set number ....     1

           material model .............    12
           equation-of-state model ....     1
           hourglass control model ....     1
           bulk viscosity model .......     1

     den .............................. =  7.8900E+00
     hourglass coefficient ............ =  1.0000E-01
     quadratic bulk viscosity ......... =  1.5000E+00
     linear bulk viscosity #1 ......... =  6.0000E-02
     stress rate option ............... =    1
        eq.1:  jaumann
        eq.2:  green-nagdi
        eq.3:  hyperelastic
     g ................................ =  7.6000E+01
     a ................................ =  1.7512E-01
     b ................................ =  8.0000E-01
     n ................................ =  3.2000E-01
     c ................................ =  6.0000E-02
     m ................................ =  5.5000E-01
     melt temperature ................. =  1.8110E+03
     room temperature ................. =  2.9800E+02
     eps0 ............................. =  1.0000E-06
     specific heat .................... =  4.5200E-04
     pressure cutoff .................. = -1.0000E+30
     spall type ....................... =  1.0000E+00
     plastic strain iteration flag .... =  0.0000E+00
     eff plastic strain at failure .... =  0.0000E+00
     failure parameter option ......... =  0.0000E+00
        eq.1 delete element when D > 1
     d1 ............................... =  5.5000E+00
     d2 ............................... =  0.0000E+00
     d3 ............................... =  0.0000E+00
     d4 ............................... =  0.0000E+00
     d5 ............................... =  0.0000E+00


    EQUATION OF STATE (volumetric behavior):  This is linear polynomial

     c0 ............................... =  0.0000E+00
     c1 ............................... =  1.4000E+02
     c2 ............................... =  0.0000E+00
     c3 ............................... =  0.0000E+00
     c4 ............................... =  0.0000E+00
     c5 ............................... =  0.0000E+00
     c6 ............................... =  0.0000E+00
     e0 ............................... =  0.0000E+00
     initial relative volume .......... =  1.0000E+00