CFVR0853 - Linear combination test
Example to check the linear combinations utility
This example checks the functionality of the combinations module.
The model is a shell platform supported on four columns. The columns are european IPE 180 steel sections and the shell is a 3cm thickness steel shell The spans are 10m, 16m and 10m long and are meshed with two elements each.
Three load cases have been defined. Each of them applies a pressure of 100kN/m2 on a third of the shell.
The combination rules defined are:
- Combination rule 1: Addition with variable coefficients [1.0 or 2.0 * H1 + 1.0 or 2.0 * H2]
- Combination rule 2: Addition [Combination 1 + H3]
Element types used in the model: BEAM4, SHELL63 Needed CivilFEM Modules: |
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| Model Statistics | |
| Number of elements | 58 |
| Number of nodes | 74 |
| Number of civil materials | 1 |
| Number of cross sections | 1 |
| Number of shell vertices | 1 |
Log file: CFVR0853.DAT
FINISH ~CFCLEAR,,1 AnsLic='ansys' NomFile='CFVR0853' /TITLE, %NomFile%, Linear combination test ! --------------------------------------------------------------------------------- ! Model definition and solve ! --------------------------------------------------------------------------------- ~UNITS,,LENG,M ~UNITS,,TIME,S ~UNITS,,FORC,KN ~CODESEL,EC3-92,EC2-91,EC2-91,,EC8-94 /PREP7 ! Materials ~CFMP,1,LIB,STEEL,EA,A42 ! Element types ET,1,Beam4 ET,2,63 ! ELASTIC SHELL63 ! Sections ~SSECLIB,1,1,1,6 ! IPE 180 ~SHLSTL,1,0.03,1 ! SHELL VERTEX ~SHLMDF,1,NAME,,,Shell Vertex 1 ! Beam & Shell Property ~BMSHPRO,1,BEAM,1,1,,,4,1,0,,Beam 1 ~BMSHPRO,2,SHELL,1,1,1,1,63,,,0,Shell 2 ! Nodes & Elements ! Beams TYPE,1 MAT,1 REAL,1 K, 1,0 K, 2,3 K, 3,3,,2 K, 4,0,0,2 K,5,0,1 K,6,3,1 K,7,3,1,2 K,8,0,1,2 L,1,5 L,2,6 L,3,7 L,4,8 LESIZE,ALL,1000 LMESH,ALL ! Shell K,9,1,1 K,10,2,1 K,11,1,1,2 K,12,2,1,2 A,5,9,11,8 A,9,10,12,11 A,10,6,7,12 TYPE,2 MAT,1 REAL,2 AMESH,ALL ! Boundary conditions NSEL,S,LOC,Y,0 D,ALL,ALL ALLSEL,ALL /VIEW,1,0.4,0.3,0.9 EPLOT /IMAGE,SAVE,%NomFile%,BMP ! Initial hypothesis solve /SOLU ! Hypothesis 1: LS1 /TITLE, H1: LS1 SFA,1,2,PRES,-100 SOLVE SFADELE,ALL,2,PRES SFEDELE,ALL,ALL,ALL ! Hypothesis 2: LS2 /TITLE, H2: LS2 SFA,2,2,PRES,-100 SOLVE SFADELE,ALL,2,PRES SFEDELE,ALL,ALL,ALL ! Hypothesis 3: LS3 /TITLE, H3: LS3 SFA,3,2,PRES,-100 SOLVE SFADELE,ALL,2,PRES SFEDELE,ALL,ALL,ALL /POST1 ! --------------------------------------------------------------------------------- ! Definition and solving of combinations ! --------------------------------------------------------------------------------- ~CMBCLR ! Combinations definition ! Combination 1: ADDITION [1.0 or 2.0 * H1 + 1.0 or 2.0 * H2] ~CMBDEF,1,ADDVC,2 ~STSTDEF,1,1,LSTEP,1 ! H1: LS1 ~STSTDEF,1,2,LSTEP,2 ! H2: LS2 ~STSTCFT,1,1,1.00,2.00 ~STSTCFT,1,2,2.00,1.00 ! Combination 2: ADDITION CMB1+H3 ~CMBDEF,2,ADD,2 ~STSTDEF,2,1,CMB,1 ! Cmb1 ~STSTDEF,2,2,LSTEP,3 ! H3: LS3 ~STSTCFT,2,1,1.00, ~STSTCFT,2,2,1.00, ~LINCMB ! --------------------------------------------------------------------------------- ! DATA CHECK ! --------------------------------------------------------------------------------- ! Data comparison number NComp =4 NComp_ch = 0 ! Matrix dim. *DIM,LABEL,CHAR,Ncomp,1 *DIM,LABEL_CH,CHAR,Ncomp_ch,1 *DIM,VALUE,,Ncomp,3 *DIM,VALUE_CH,CHAR,Ncomp_ch,3 *DIM,TOLER,,Ncomp,2 ! Correct values ! --------------------------------------------------------------------------------- ! My Element 5 End J VALUE( 1,1) = 15.277+9.668 ! LS1+LS2 MY=24.945 Minimum value VALUE( 2,1) = 2*15.277+2*9.668+3.167 ! 2LS1+2LS2+SL3 MY=53.057 Maximum Value ! Concomitant SX on element 4 (beam) VALUE( 3,1) = 601.241E3+365.87E3 ! LS1+LS2 SX=967.111E3 VALUE( 4,1) =2*601.241E3+2*365.87E3+123.45E3 ! 2LS1+2LS2+LS3 SX=1691.80E3 ! Obtained values ! --------------------------------------------------------------------------------- ~CFSET,,4 ~CFGET,ValAux, ELEMENT,5, FORCE, MY, J MaxLS = 4 MinLS = 4 MinVal = ValAux MaxVal = ValAux *DO,II,5,11 ~CFSET,,II ~CFGET,ValAux, ELEMENT,5, FORCE, MY, J *IF,MinVal,GT,ValAux,THEN MinVal = ValAux MinLS = II *ENDIF *IF,MaxVal,LT,ValAux,THEN MaxVal = ValAux MaxLS = II *ENDIF *ENDDO ! Test 1 and 3: ~CFSET,,MinLS ~CFGET,VALUE(1,2), ELEMENT,5, FORCE, MY, J ~PLLSSTR,SX,1 *GET,VALUE(3,2),ELEM,4,ETAB,CFETAB_J ! Test 2 and 4: ~CFSET,,MaxLS ~CFGET,VALUE(2,2), ELEMENT,5, FORCE, MY, J ~PLLSSTR,SX,1 *GET,VALUE(4,2),ELEM,4,ETAB,CFETAB_J ! Labels ! --------------------------------------------------------------------------------- LABEL( 1) ='Min_MY' LABEL( 2) ='Max_MY' LABEL( 3) ='Conc_SX' LABEL( 4) ='Conc_SX' ! Warning and error tolerances TOLER( 1, 1)= 1E-3 $ TOLER( 1, 2)= 1E-2 TOLER( 2, 1)= 1E-3 $ TOLER( 2, 2)= 1E-2 TOLER( 3, 1)= 1E1 $ TOLER( 3, 2)= 1E2 TOLER( 4, 1)= 1E1 $ TOLER( 4, 2)= 1E2 ! --------------------------------------------------------------------------------- ! Results comparison ! --------------------------------------------------------------------------------- COMPARA.MAC |
Results
| Label | Target | CivilFEM | Ratio | Tolerance |
| Min_MY | 24.945 | 24.945 | 1.000 | 0.01 |
| Max_MY | 53.057 | 53.058 | 1.000 | 0.01 |
| Conc_SX | 9.6711e+005 | 9.6711e+005 | 1.000 | 100 |
| Conc_SX | 2.0577e+006 | 2.0577e+006 | 1.000 | 100 |
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