Dear Sir/Madam,
I need to work with the CREEP subroutine. I have written a CREEP subroutine
which is rather complicated. My subroutine works well for simple examples
like uniaxial condition (I have verified its correctness through comparison
with outcome of some matlab routines).
But when I apply it to my more complicated geometry, its convergence rate
is too slow. I think the problem should be in definition of DECRA(5) (or
DECRA(2)) in my subroutine and I would appreciate your support.
I read the ABAQUS theory manual and got confused with the way I should
define my parameters for explicit and/or implicit LEXIMP = 0 or 1. (LEND?)
Could you please help me and inform me if there is anything wrong in the
algorithm I used in the below subroutine?
Is it fine to use EC(1)?
Is the definition of DECRA(5) correct?
Do I need to define DECRA(2)? how?
Anymore points I should take care?
Thank you in advance for your support.
Kind regards
Ehsan Hosseini
Notes:
1My creep model defines the absolute value of creep strain based on a
stress, (factious) time, and temperature. (it is not creep strain rate
based)
2I cannot analytically derive an expression for DECRA(5) and I need to
calculate it numerically.
3I have the idea that I do not need to define DECRA(2) (because my
creep model defines the absolute value of creep strain base on a stress).
But should I define DECRA(2) ? how?
My subroutine looks like:
------------------------------------------------------------
SUBROUTINE CREEP(DECRA,DESWA,STATEV,SERD,EC,ESW,P,QTILD,
1 TEMP,DTEMP,PREDEF,DPRED,TIME,DTIME,CMNAME,LEXIMP,LEND,
2 COORDS,NSTATV,NOEL,NPT,LAYER,KSPT,KSTEP,KINC)
.
------------------------S = QTILD
CALL
EHCreep(EC(1),S,TEMP,DTIMEH ,de)
DECRA(1) = de
------------------------dQTILD = 1.D-10
S = QTILD + dQTILD
CALL
EHCreep(EC(1),S,TEMP,DTIMEH,de)
DECRA(5) = (de- DECRA(1))/ dQTILD
RETURN
END
-----------------------------------------------------------subroutine EHCreep(EC(1),S,TEMP,DTIMEH,de)
------------------------Determine a fictues time (i.e. t_f) which gives creep strain EC(1)
(i.e. EC(1)= f_creep (S,TEMP,t_f))
t_f =
...
------------------------de = f_creep (S,TEMP,t_f + DTIMEH) - EC(1)
RETURN
END
-----------------------------------------------------------My .sta file looks like:
6
466
1
0
5
5
4.27E+05
1.50E+04
23.12
6
467
1
0
5
5
4.27E+05
1.50E+04
23.12
6
468
1U
0
4
4
4.27E+05
1.50E+04
23.12
6
468
2
0
2
2
4.27E+05
1.50E+04
5.78
6
469
1U
0
4
4
4.27E+05
1.50E+04
5.78
6
469
2U
0
4
4
4.27E+05
1.50E+04
1.445
6
469
3
0
2
2
4.27E+05
1.50E+04
0.3612
6
470
1
0
1
1
4.27E+05
1.50E+04
0.5418
6
471
1U
0
4
4
4.27E+05
1.50E+04
1.084
6
471
2
0
1
1
4.27E+05
1.50E+04
0.2709
and my .msg file:
CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION
AVERAGE FORCE
6.612E+04
1
1.001E+05
TIME AVG. FORCE
LARGEST RESIDUAL FORCE
2.747E+03
AT NODE
18900
DOF
1
LARGEST INCREMENT OF DISP.
3.618E-02
AT NODE
83
DOF
2
LARGEST CORRECTION TO DISP.
9.228E-05
AT NODE
2219
DOF
2
FORCE
EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.
NUMBER OF EQUATIONS =
2.24E+09
254268
USING THE DIRECT SOLVER WITH
CHECK POINT
CHECK POINT
NUMBER OF FLOATING PT. OPERATIONS =
2 PROCESSORS
START OF SOLVER
END OF SOLVER
ELAPSED USER TIME (SEC)
=
1.3000
ELAPSED SYSTEM TIME (SEC)
=
0.10000
ELAPSED TOTAL CPU TIME (SEC) =
1.4000
ELAPSED WALLCLOCK TIME (SEC) =
2
CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION
AVERAGE FORCE
6.612E+04
2
1.001E+05
TIME AVG. FORCE
LARGEST RESIDUAL FORCE
-477.
AT NODE
109077
DOF
1
LARGEST INCREMENT OF DISP.
3.618E-02
AT NODE
83
DOF
2
LARGEST CORRECTION TO DISP.
1.016E-05
AT NODE
5407
DOF
1
FORCE
EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.
NUMBER OF EQUATIONS =
2.24E+09
254268
USING THE DIRECT SOLVER WITH
CHECK POINT
CHECK POINT
NUMBER OF FLOATING PT. OPERATIONS =
2 PROCESSORS
START OF SOLVER
END OF SOLVER
ELAPSED USER TIME (SEC)
=
2.1000
ELAPSED SYSTEM TIME (SEC)
=
0.0000
ELAPSED TOTAL CPU TIME (SEC) =
2.1000
ELAPSED WALLCLOCK TIME (SEC) =
2
CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION
3
AVERAGE FORCE
6.612E+04
1.001E+05
TIME AVG. FORCE
LARGEST RESIDUAL FORCE
-364.
AT NODE
51249
DOF
2
LARGEST INCREMENT OF DISP.
3.618E-02
AT NODE
83
DOF
2
LARGEST CORRECTION TO DISP.
4.382E-07
AT NODE
12787
DOF
1
FORCE
EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.
NUMBER OF EQUATIONS =
2.24E+09
254268
USING THE DIRECT SOLVER WITH
CHECK POINT
CHECK POINT
NUMBER OF FLOATING PT. OPERATIONS =
2 PROCESSORS
START OF SOLVER
END OF SOLVER
ELAPSED USER TIME (SEC)
=
1.3000
ELAPSED SYSTEM TIME (SEC)
=
0.0000
ELAPSED TOTAL CPU TIME (SEC) =
1.3000
ELAPSED WALLCLOCK TIME (SEC) =
1
CONVERGENCE CHECKS FOR EQUILIBRIUM ITERATION
AVERAGE FORCE
6.612E+04
1.001E+05
LARGEST RESIDUAL FORCE
367.
4
TIME AVG. FORCE
AT NODE
51249
DOF
2
LARGEST INCREMENT OF DISP.
3.618E-02
AT NODE
83
DOF
2
LARGEST CORRECTION TO DISP.
-3.349E-07
AT NODE
12787
DOF
1
FORCE
EQUILIBRIUM NOT ACHIEVED WITHIN TOLERANCE.
NUMBER OF EQUATIONS =
2.24E+09
254268
USING THE DIRECT SOLVER WITH
CHECK POINT
CHECK POINT
NUMBER OF FLOATING PT. OPERATIONS =
2 PROCESSORS
START OF SOLVER
END OF SOLVER
ELAPSED USER TIME (SEC)
=
2.0000
ELAPSED SYSTEM TIME (SEC)
=
0.0000
ELAPSED TOTAL CPU TIME (SEC) =
2.0000
ELAPSED WALLCLOCK TIME (SEC) =
1