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