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April 2, 2024 at 7:35 pm
Shahrukh Ahmed
SubscriberHi everyone,
I am trying to perform non linear buckling analysis in ANSYS Mechanical APDL 2021 R1.
But the bending stress capacity i am getting (product of reduction factor and yield stress) theoretically is nowhere matching with the numerical results.
fbd(reduction factor x yield stress(250MPa) = 201.2 MPa
fbd(as per model) = 21MPa
Here is my codeÂblen=10.0depth=0.55fwidth=0.210Âfthk=17.2/1000wthk=11.1/1000ÂÂÂTAREA,0,0,0,depth/2.0,0,blen,1TAREA,0,0,depth/2.0,depth,0,blen,1TAREA,-fwidth/2.0,0.0,0,0,0,blen,2TAREA,-fwidth/2.,0,depth,depth,0,blen,2TAREA,0,fwidth/2.0,0,0,0,blen,2TAREA,0,fwidth/2.0,depth,depth,0,blen,2!TACUT,0,0,0,depth,0,blen,0.3Âpoi=0.3Âdamp1=0.02!Material Number 1ex, 1,   2e5*1000ey, 1,   2e5*1000nuxy,1,   poigxy, 1,   2e5*1000/(2*(1+poi))dens,1,   2.5alpx,1,   9.50E-06alpy,1,   9.50E-06damp,1,   damp1ÂET, 11, shell181 !WallÂSECTYPE,111,SHELLSECDATA,fthk,ÂSECTYPE,112,SHELLSECDATA,wthk,ÂTASEL,0,0,0,depth,0,blen,1AATT,1,,11,,112ÂÂTASEL,-fwidth/2.0,fwidth/2.0,0,0,0,blen,2AATT,1,,11,,111ÂTASEL,-fwidth/2.,fwidth/2.0,depth,depth,0,blen,2AATT,1,,11,,111Âallsel,allÂaclear,allesize,0.1amesh,all,allÂÂddele,all,alltnsel,0,0,depth/2,depth/2,0,0,d,all,uy,0d,all,rotz,0Âtnsel,0,0,depth/2,depth/2,blen,blen,d,all,uy,0d,all,uz,0d,all,rotz,0ÂÂtnsel,0,0,0,depth,0,0,0,d,all,ux,0Âtnsel,0,0,0,depth,blen,blen,d,all,ux,0Â!LeftSidesÂfdele,all,alllogforce=1/(0.5328*2)Âtnsel,-fwidth/2,-fwidth/2,0,0,0,0,0,f,all,fz,-logforceÂtnsel,-fwidth/2,-fwidth/2,depth,depth,0,0,0,f,all,fz,logforceÂtnsel,fwidth/2,fwidth/2,0,0,0,0,0,f,all,fz,-logforceÂtnsel,fwidth/2,fwidth/2,depth,depth,0,0,0,f,all,fz,logforceÂÂtnsel,-fwidth/2,-fwidth/2,0,0,blen,blenf,all,fz,logforceÂtnsel,-fwidth/2,-fwidth/2,depth,depth,blen,blenf,all,fz,-logforceÂtnsel,fwidth/2,fwidth/2,0,0,blen,blenf,all,fz,logforceÂtnsel,fwidth/2,fwidth/2,depth,depth,blen,blenf,all,fz,-logforceÂÂallsel,all Âeplot ÂÂ!*change materialÂ/VIEW,1,1,1,1 Â/ANG,1 Â/REP,FAST Â/DIST, 1 ,1.082226,1/REP,FAST Â/DIST, 1 ,1.082226,1/REP,FAST Â/DIST, 1 ,1.082226,1/REP,FAST Â/DIST, 1 ,1.082226,1/REP,FAST Â/DIST, 1 ,1.082226,1/REP,FAST ÂFINISH Â/SOLFINISH Â/PREP7 Â!* ÂTB,BISO,1,1,2, ÂTBTEMP,0TBDATA,,250000,0,,,,ÂÂÂÂ!*static anlaysisÂÂÂ/VIEW,1,1,1,1 Â/ANG,1 Â/REP,FAST ÂFINISH Â/SOLPSTRES,1!* ÂANTYPE,0solve Â!* Â!* Â!* Â!* Â!* Â!* ÂFINISH ÂÂ!* eigenvalue buckling analysis/SOLUTION ÂANTYPE,1!* ÂFINISH Â/POST1 ÂFINISH Â/PREP7 ÂFINISH Â/SOL!* ÂBUCOPT,LANB,10,0,0,CENTER Âsolve ÂFINISH Â/POST1 Â!* Â/EFACET,1 ÂPLNSOL, U,SUM, 0,1.0SET,LIST,999)/GOP  ÂÂ!*********************************************UPGEOMallsel,alleplot/PREP7 ÂUPGEOM,10/350,1,6,'test','rst',' ' ÂFINISH Â/SOLÂeplotÂ!*change forcefdele,all,alllogforce=1000/(0.5328*2)Âtnsel,-fwidth/2,-fwidth/2,0,0,0,0,0,f,all,fz,-logforceÂtnsel,-fwidth/2,-fwidth/2,depth,depth,0,0,0,f,all,fz,logforceÂtnsel,fwidth/2,fwidth/2,0,0,0,0,0,f,all,fz,-logforceÂtnsel,fwidth/2,fwidth/2,depth,depth,0,0,0,f,all,fz,logforceÂÂtnsel,-fwidth/2,-fwidth/2,0,0,blen,blenf,all,fz,logforceÂtnsel,-fwidth/2,-fwidth/2,depth,depth,blen,blenf,all,fz,-logforceÂtnsel,fwidth/2,fwidth/2,0,0,blen,blenf,all,fz,logforceÂtnsel,fwidth/2,fwidth/2,depth,depth,blen,blenf,all,fz,-logforceÂallsel,alleplotÂ!*NLGEOMÂ!* ÂANTYPE,0ANTYPE,0NLGEOM,1NSUBST,100,100,1OUTRES,ERASEOUTRES,ALL,ALL ÂRESCONTRL,DEFINE,ALL,ALL,-1ÂTIME,1 ÂFINISH Â/POST1 ÂFINISH Â/SOL!* Â!* ÂNLGEOM,1NROPT,AUTO, ,OFFLUMPM,0ÂEQSLV, , ,0, ,DELE ÂMSAVE,0ÂPCGOPT,0, ,AUTO, , ,AUTOPIVCHECK,0 ÂPSTRESS,0 ÂTOFFST,0, Â!* Âsolve ÂÂeplot
What is wrong with my analysis ?? -
April 3, 2024 at 5:29 pm
dlooman
Ansys EmployeeThe APDL Materials Reference gives the example input below for temp-dependent bilinear isotropic hardening. You appear to have only specified the yield stress. Perhaps a shell element isn't the best choice for a plasticity analysis. Perhaps a thin solid with several elements through the thickness would be more accurate. Â
/prep7 MPTEMP,1,0,500 ! Define temperatures for Young's modulus MPDATA,EX,1,,14E6,12e6 MPDATA,PRXY,1,,0.3,0.3 TB,PLAS,1,2,,BISO ! Activate a data table TBTEMP,0.0 ! Temperature = 0.0 TBDATA,1,44E3,1.2E6 ! Yield = 44,000; Plastic Tangent modulus = 1.2E6 TBTEMP,500 ! Temperature = 500 TBDATA,1,29.33E3,0.8E6 ! Yield = 29,330; Plastic Tangent modulus = 0.8E6
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