帮忙开下这个子程序哪里出问题了，谢谢

nuaa10526

      subroutine usdfld(field,statev,pnewdt,direct,t,celent,
     1 time,dtime,cmname,orname,nfield,nstatv,noel,npt,layer,
     2 kspt,kstep,kinc,ndi,nshr,coord,jmac,jmtyp,matlayo,laccfla)
c redefine field variables at a material point.
      include 'aba_param.inc'
c
      character*80 cmname,orname
      character*3  flgray(15)
      dimension field(nfield),statev(nstatv),direct(3,3),
     1 t(3,3),time(2)
      dimension array(15),jarray(15),jmac(*),coord(*)         !jmatyp(*),
c this subroutine must call utility routine GETVRM to access material point data.
c Get temperatures from previous increment
      call getvrm('TEMP',array,jarray,flgray,jrcd,
     $     jmac, jmtyp, matlayo, laccflg)
      temp = array(1)
c field(1) : la valeur du degré de cuisson dans chaque point d'integration de l'incrément. 0.0001 est la valeur initiale définie.
c kinc: Increment number
      if(kinc.eq.1)then
        statev(1)=1E-4   
        else  
        field(1)=statev(1)
        endif
c
      return
      end
      SUBROUTINE HETVAL(CMNAME,TEMP,TIME,DTIME,STATEV,FLUX,
     &PREDEF, DPRED)
C DEFINE THE HEAT FLUX DUE TO INTERNAL HEAT GENERATION (THE CHEMICAL REACTION)

      INCLUDE'ABA_PARAM.INC'
C
c      real*8 STATEV
      CHARACTER*80 CMNAME
      DIMENSION TEMP(2),STATEV(3),PREDEF(1),TIME(2),FLUX(2),DPRED(1)


C TEMP(1): CURRENT TEMPERATURE
      IF(TEMP(1).LT.358.)THEN
        STATEV(2) = 0.0
      ELSE


      STATEV(2)=1.5E10*EXP(-11396.4/TEMP(1))*((1-STATEV(1))**1.877)
     &*(STATEV(1)**0.45)


C STATEV(1): THE VALUE OF DEGRE OF CURE PASSED FROM THE SUBROUTINE USDFLD,
C STATEV(2): da/dt  (a=degre of cure)
      STATEV(1) = STATEV(1)+ STATEV(2)* DTIME
      END IF
C FLUX(1) HEAT FLUX J/TIME/VOLUME, AT THIS MATERIAL CALCULATION POINT.
      FLUX(1) = 242971740* STATEV(2)
C
      RETURN
      END
程序如上所示，计算结果STATEV(2)一直为0，若去掉 &*(STATEV(1)**0.45)则可以计算，麻烦纠正一下，谢谢！