store them (the displacement ux uy uz) as state variables (SVARS array in your fuel... so if you have 3 disp.. 4 nodes for 1 element.. svars should be 12 by 1.

thanks Ditho Ardiansyah Pulungan for your response.

According to your comment, I stored updated "u" in SVARS arrays, but it is not solved yet. I'm not very familiar with UEL subroutine, so maybe some problem in my code. could you explain more?

it's kind of you if look into my subroutine and attached pic.

highly appriciated.

SUBROUTINE UEL(RHS,AMATRX,SVARS,ENERGY,NDOFEL,NRHS,NSVARS,

1 PROPS,NPROPS,COORDS,MCRD,NNODE,U,DU,V,A,JTYPE,TIME,

2 DTIME,KSTEP,KINC,JELEM,PARAMS,NDLOAD,JDLTYP,ADLMAG,

3 PREDEF,NPREDF,LFLAGS,MLVARX,DDLMAG,MDLOAD,PNEWDT,

4 JPROPS,NJPROP,PERIOD)

INCLUDE 'ABA_PARAM.INC'

DIMENSION RHS(MLVARX,*),AMATRX(NDOFEL,NDOFEL),

1 SVARS(NSVARS),ENERGY(8),PROPS(*),COORDS(MCRD,NNODE),

2 U(NDOFEL),DU(MLVARX,*),V(NDOFEL),A(NDOFEL),TIME(2),

3 PARAMS(3),JDLTYP(MDLOAD,*),ADLMAG(MDLOAD,*),

4 DDLMAG(MDLOAD,*),PREDEF(2,NPREDF,NNODE),LFLAGS(*),

5 JPROPS(*)

REAL kx,KY,kz

INTEGER I,J,jj,N1,N2,N3,N4,m4,m5,ii

REAL SRESID(NDOFEL),Force(451),z3(451),u1(451)

Data u1 / 451*0 /

Data force / 451*0 /

DO I = 1, NDOFEL

DO J = 1, NDOFEL

AMATRX(I,J) = 0.D0

END DO

END DO

IF ((kstep.EQ.1).OR.(kstep.EQ.2)) THEN

kx=0

KY=100000

force(jelem)=ky*u(2)

ELSE IF (kstep.EQ.3) THEN

if (u(2).gt.0) THEN

kx=0

ky=0

force(jelem)=ky*u(2)

else

kx=0

ky=5000

force(jelem)=ky*u(2)

end if

u1(jelem)=u(2)

coords(2,1)=u1(jelem)

ELSE IF (kstep.EQ.4) THEN

if (u(2).gt.0) THEN

kx=0

ky=0

force(jelem)=0

else

ky=100000

force(jelem)=ky*u(2)

end if

END IF

AMATRX(1,1:3)=(/KX, 0.,0./)

AMATRX(2,1:3)=(/0., KY,0./)

AMATRX(3,1:3)=(/0., 0., KZ/)

DO I = 1, NDOFEL

RHS (I,1) = 0.D0

SRESID(I) = 0.

END DO

SRESID(2)=FORCE(jelem)

SRESID(1)=abs(kx*u(1))

RHS(1,1)=RHS(1,1)-SRESID(1)

RHS(2,1)=RHS(2,1)-SRESID(2)

RHS(3,1)=RHS(3,1)-SRESID(3)

SVARS(1)=u(1)

SVARS(2)=u(2)

SVARS(3)=COORDS(1,1)

SVARS(4)=COORDS(2,1)

SVARS(5)=RHS(1,1)

SVARS(6)=RHS(2,1)

RETURN

END