FEAP Example Manual - Civil and Environmental Engineering
Contents
1. Plasticity Problem In this example we consider the solution of a plane strain two dimensional continuum problem with an elasto plastic J gt model The domain for the problem is a square with side lengths of 200 mm which has a central circular hole with radius 10 mm The top and bottom boundary surfaces are subjected to a uniform normal loading of 450 MPa Due to the symmetry of the problem only one quadrant is modeled and symmetry boundary conditions are imposed as uy 0 on x7 0 Ug 0 on zo 0 The data for 992 nodes and 900 elements is generated using the commands FEAP 000 2 2 4 PARAmeters r 10 100 0 4 0 35 15 2xm BBQRp tou ow A SNODes 100 sind 45 rx cosd 45 0 Y O 01H YN hMoryroaok kK rPyrPyrowK Ao 51 CHAPTER 9 FINITE DEFORMATION PLASTICITY 90 f h 10 d h d h SIDE polar 2 3 1 polar 3 4 1 cart 25 8 cart 3 6 10 cart 479 BLENd surf nmOOl1 2563 BLENd surf nmOOl1 3674 The boundary conditions and loading are specified by the commands EBOUn 1010 2001 CSUR line 1 h h 450 2 O h 450 A plot of the mesh for the region modeled is shown in Fig 9 1 The material properties for the Jp elasto plastic model are bulk modulus K 52 16 4206 x 10 MPa Shear modulus 0 801938 MPa and uniaxial yield stress o 450 MPa The material has no hardening The material is nearly incompressible even in the elastic range consequently a mixed formulation is used with a Q1P0 for2. Table 2 2 Data for Patch Test CHAPTER 2 PATCH TESTS 5 MATErial COORdinates ELEMents BOUNdary restraints FORCes END BATCh END STOP FEAP interprets only the first four characters of each command These are shown as upper case letters to indicate the minimum amount which may be used to identify each command Either upper case or lower case letters may be used to identify each command Thus MATE or mate may be used to identify the material property data sets After a FEAP command and the control record the commands before the first END may be in any order and define the mesh for the problem The commands after the BATCH describe the solution algorithm for the problem and terminate with the second END command Finally the STOP command informs FEAP that no more data exists Only one problem may appear in the file which contains the problem initiation command FEAP See example in Chapter 5 for a way to run multiple problems The control record defines the size of finite element problem to be solved The first field defines the number of nodes NUMNP the next field is the number of elements NUMEL followed by the number of material sets NUMMAT the spatial dimension for the mesh NDM the number of degrees of freedom for each node NDF and the maximum number of nodes on any element NEN The patch test has a mesh with 9 nodes 4 elements and 1 material set The problem is 2 dimensional has 2 degrees of freedom at each node
3. 0 00E 00 Time 1 00E 01 Time 1 00E 00 Figure 6 4 Temperature contours at t 0 1 and t 1 0 Chapter 7 Coupled Thermo mechanical Problem In this example we consider the solution of a thermo mechanical problem in a state of plane strain The domain for the solution is a square with side lengths of 5 units A vertical roller support is specified at the upper right hand corner and a pin support at the lower right hand corner This is just sufficient to prevent rigid body motions for a static problem Loading is provided by a transient thermal analysis in which the left side has a specified unit temperature T 1 suddenly applied at time zero and held constant The material parameters are set as follows 1 Solid E 100 v 0 4995a a 0 25 To 0 2 Thermal The problem is solved using a 10 x 10 uniform mesh of 9 node quadrilateral elements The mixed formulation is used With the above properties and element the material data is specified as MATErial 1 SOLID ELASTIC ISOTROPIC 100 0 4995 THERMAL ISOTROPIC 0 25 0 0 FOURIER ISOTROPIC 10 0 1 0 DENSITY MASS 0 10 MIXED The mesh is generated using the block command with the data 42 CHAPTER 7 COUPLED THERMO MECHANICAL FEAP o 0 0239 BLOCK CARTESIAN A 0 0 0 0 EDIS 100 0 1 MATE 1 material properties as above END A plot of the mesh is shown in Fig 7 1 ir Figure 7 1 Mesh of 9 node
4. 9 node radial displacement contour test requirements for stability and consistency Chapter 3 Truss Problem As a next example for the use of FEAP consider a simple truss problem The mesh nodal and element numbers loading restraints and material properties are shown in Figure 3 1 2 Figure 3 1 Mesh for Truss Example The complete data file for the problem is shown in Table 3 1 and a description for each part of this data follows The control record indicates the mesh has 5 nodes 7 elements 2 material sets It is a 2 dimensional problem with 2 degrees of freedom at each node and 2 nodes for each truss element The MAT Eeral property sets for all members are identical except for the cross sectional area of the members the first numerical field on the CROSs SECTion record Two types are indicated Set 1 has an area of 10 units while set 2 has an area of 5 units The coordinates are input as for the patch example described above Element properties 15 CHAPTER 3 TRUSS PROBLEM FEAP 2 D Truss Problem 57222 2 MATErial 1i TRUSS ELAStic ISOTropic 1000 0 CROSs SECTion 10 0 Blank termination MATErial 2 TRUSS ELAStic ISOTropic 1000 0 CROSs SECTion 5 0 Blank termination COORdinates 100 0 0 2 0 200 0 3 0 400 0 4 0 100 0 5 0 300 0 RRR 0O00O0O0O oooo oo Blank termination ELEMents 0 Be NOOO FP WD ooooo RRENNRRR BWNHONFNE oode RUN Blank termination BOUNdary restraints 101
5. and each element has 4 nodes The first data set is identified by the MATErial command This record also must contain a material set number ranging from 1 to the maximum number of sets needed The next records consist of commands which describe the type of element see Chapter 6 of the User Manual for the types of elements included with FEAP and the material parameters associated with the set The data shown in Table 2 1 indicates a SOLId continuum element is used the problem is plain strain and the material parameters are associated with a linear elastic isotropic material Except for the element type record other data may be in any order and terminates with a blank record comments are permitted on records and begin with the exclamation point The values of the nodal coordinates for the patch are specified using the COORdinate command Each record defines a node number a generation parameter and the x and y coordinate values Nodes may be in any order but are shown in increasing order in Table 2 1 Input terminates with a blank record The manner in which nodes are connected to form individual finite elements and their CHAPTER 2 PATCH TESTS 6 association to the type of element and material parameters is described by the data following the ELEMent command Each record defines the element number which must be in increasing numerical order a generation parameter to be described later the material data set associated wit
6. mulation The analysis is performed as a finite deformation problem The data for the element type and material properties is given as PARAmeters mu 80193 8 k 1642060 nu 3 k 2 mu 2 mu 6 k e 2 1 nu mu y 450 MATErial 1 solid CHAPTER 9 FINITE DEFORMATION PLASTICITY 53 Figure 9 1 Mesh of elasto plastic tension strip elastic isotropic e nu plastic mises y mixed finite Note that the element requires elastic properties to be given as Young s modulus and Poisson ratio which are computed from the bulk and shear modulus Since the mesh consists of two blend regions it is necessary to TIE them together prior to performing the analysis The problem is subjected to the cyclic loading shown in Fig 9 2 and performed using a sequence of time steps of At 0 1 CHAPTER 9 FINITE DEFORMATION PLASTICITY 54 p t Load 1 5 2 2 5 t Time Figure 9 2 Proportional loading The command language statements for the analysis are given as TIE BATCh PROP DT 0 1 END 25 0 0 11 20 3 1 40 BATCh NOPRint PLOT RANGe 0 440 LOOP 40 TIME LOOP 30 TANG 1 NEXT PLOT PSTRe 6 1 NEXT END CHAPTER 9 FINITE DEFORMATION PLASTICITY 59 The solution summary for the finite deformation case is shown below times are for a PIII 650MHz processor SOLUTION SUMMARY Total Iter Forms wW Y aw O 0 Y O0000dgnm0aNn ana aaa YOU Yau Y NAONA WWW OOOO ORO OR OR ORO ORO ORO OR ORO OR ORONO OR ORO
7. requests the SOLId element type See Chapters 6 and 7 of the User Manual for ad ditional information on types of elements and material models permitted The third record defines the material constitution as isotropic linear elastic and sets the parame ters as E 1000 y 0 25 The fourth record sets the material density as p 0 1 The fifth record specifies a 2 x 2 Gauss quadrature to compute arrays and output element results The input of material parameters terminates with a blank record The data following the SOLId specification may be in any order Also any data not needed may be omitted Since the disk analysis is quasi static the material density is unnecessary and could be omitted Also default quadrature values will be used if this record is omitted For the current analysis the minimum material data is MATErial 1 SOLId ELAStic ISOTropic 10000 0 25 E and nu Blank termination record The END command signals FEAP that the specification of the mesh and initial loading conditions is complete An mesh may be modified during the solution phase by re entering MESH generation or by manipulating the mesh to merge parts using the TIE command or setting boundary conditions to satisfy LINK or ELINk conditions We next describe some of the other options which are available to generate and or manipulate a mesh Again we consider the example problem to make the discussion specific The preceding format to generate the data for a finite elem
8. 2 O 3 800E 00 39 5 O 3 900E 00 40 5 O 4 000E 00 eerrrrerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrerr Residual Initial 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 FPRrRFWWWWKRPPOFPBORFPNNFPOOWWRFWWWKRKBTTDPWBAWREROWNNN Norm Final ADMAADAWAIAMWAIAIAIAWAIMWMAAAIAAAWiMAiWAAIMAWAAAAAAWMMWMAAAARAAARNA Energy Norm Initial 78E 01 78E 01 T8E 01 T8E 01 18E 01 18E 01 18E 01 18E 01 78E 01 78E 01 18E 01 T8E 01 78E 01 78E 01 18E 01 18E 01 18E 01 18E 01 18E 01 18E 01 T8E 01 78E 01 78E 01 78E 01 18E 01 18E 01 18E 01 18E 01 18E 01 718E 01 78E 01 78E 01 78E 01 T8E 01 78E 01 18E 01 18E 01 18E 01 718E 01 18E 01 000000 RrRrRROdmAaROO0oORAS RASO No00 erre Rgn0aA WOOP CO ONN 60E 26 82E 26 18E 26 33E 26 47E 24 89E 18 28E 26 93E 16 OOE 21 O5E 17 44E 25 30E 25 19E 25 10E 25 O3E 25 D6E 26 13E 26 60E 26 68E 19 22E 22 33E 19 ODE 27 81E 16 86E 16 33E 27 26E 23 OOE 19 T5E 15 83E 23 17E 24 43E 25 18E 25 20E 25 O3E 25 84E 26 45E 26
9. OOE 26 64E 26 41E 19 14E 24 Note that linear elasticity requires only 2 iterations during elastic behavior CPU Time Final Seconds
10. OR ORO ORONO ORO OR ORO OR ORCS BWWWWWWWWWWNNNNNNNNNNRPRPRFPRFPRPRFPRFP RFP RPP OONOOBRWNEF Solution Time OO0E 01 OO0E 01 OO0E 01 OOOE 01 OOOE 01 OO0E 01 OOOE 01 OO0E 01 OOOE 01 OOOE 00 100E 00 200E 00 300E 00 400E 00 500E 00 600E 00 7O0E 00 800E 00 900E 00 OOOE 00 100E 00 200E 00 300E 00 400E 00 500E 00 600E 00 7O0E 00 800E 00 900E 00 OOOE 00 100E 00 200E 00 300E 00 400E 00 500E 00 600E 00 T00E 00 800E 00 900E 00 OOOE 00 eerrrrerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrerr Residual Initial 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 14E 03 0Nn0I00000O0o0No0swv0nainaaoaanaarr0odsSan0os00o0ooro0o0o o 0om0umr 00000 Norm Final O 0 0 01010 00 0 070 000 0 0 0 0000700 000000 000000 000 00 Energy Norm Initial 78E 01 78E 01 18E 01 19E 01 79E 01 79E 01 79E 01 S0E 01 S0E 01 S0E 01 81E 01 S0E 01 S0E 01 S0E 01 S0E 01 79E 01 79E 01 79E 01 78E 01 T8E 01 78E 01 TTE 01 TTE 01 TTE 01 TTE 01 T6E 01 T6E 01 T6E 01 T5E
11. a z translation That is a displacement field which causes no strain is restricted to 1 3 8572E 04 9 8218E 03 5 2832E 03 1 5889E 03 4 9187E 02 where A is any constant Eigenvalue 2 3 0844E 04 9 3116E 03 3 6854E 03 1 5347E 03 5 0762E 12 3 1 6764E 04 7 8663E 03 3 0857E 03 1 1869E 03 uy 0 and uz A 4 1 5316E 04 5 5895E 03 2 1230E 03 5 3488E 02 CHAPTER 2 PATCH TESTS 12 2 2 2 Consistency test The consistency test considers behavior under specified linear displacement fields that in general will have non zero strains and stresses as well as non zero body forces For a part of the consistency test we consider the test applied to a displacement field expressed as u A Br and u 0 From this we obtain strains B and for an isotropic material with constitution Tij AE Ou 2 U Eij where is the volume change in small deformation the stresses A 28 24B 1 A 28 Bi r A A A 28 2 8 r r 0 Inserting these stresses into the equilibrium equation requires a non zero radial body force to be A This body force distribution must be coded into each element to permit the testing of a patch performance The isotropic elastic properties in feap are given in terms of Young s modulus and Poisson s ratio v From these we obtain E 2v d A E ta which are used to compute the boundary tractions and body force b For the example we let A B 1 0 The input data is given
12. by CHAPTER 2 PATCH TESTS 13 FEAP Axisymmetric patch test 000229 PARAM mu 1000 2 5 la 0 5 mu t1 2 2xla 2 mu 5 t2 2 1 la 2 mu 5 br la 2 mu MATErial SOLID AXISymmetric ELAStic ISOTropic 1000 0 0 25 BODY PATCh br 0 BLOCK CARTesian 4 4 QUADrilateral 1 5 0 2 10 0 3 10 0 4 9 Ooo0oooo0o 5 0 7 75 NaO O 25 CSURf Tractions on radial boundary LINEAR Note t1 t2 multiplied by r 1 5 5 ti 2 5 0 ti LINEAR 1 10 0 t2 2 10 5 t2 EBOUn Set u_z 0 2 0 0 O 1 2 5 0 0 1 END and solution commands by BATCh TANG 1 PLOT CONT 1 END CHAPTER 2 PATCH TESTS 14 Additional commands may be used to output displacements and stresses 21 22 23 24 25 b g b 2 bo s S Figure 2 3 a 4 node mesh b 9 node mesh A rectangular region between 5 lt r lt 10 and 0 lt z lt 5 is used Note that it is important not to use a patch which includes a zero radius as no radial displacement is allowed there The meshes shown in Fig 2 3 are use for the patch tests and results for the 9 node mesh are shown in Fig 2 4 results for a 4 node element are very similar Based on the above assessment we find that the element as coded passes the patch DISPLACEMENT 1 6 00E 00 6 71E 00 7 43E 00 8 14E 00 8 86E 00 9 57E 00 1 03E 01 1 10E 01 Current View X 5 00E 00 Y 0 00E 00 Max 1 10E 01 X 1 00E 01 Y 0 00E 00 Time 0 00E 00 Figure 2 4
13. figure Similarly the upper beam is designated as contact surface 2 and facets are numbered from left to right to give an outward normal pointing down The two surfaces are specified to interact using the PAIR command with surface 1 being CHAPTER 8 CONTACT PROBLEM 49 the slave and surface 2 the master A node to surface contact with a penalty solution scheme is adopted command NTOS with an augmented Lagrange correction applied AUGM The data for a contact follows the mesh description and for the present example is given by CONTact SURFace 1 LINE 2 FACEt 1 1 11 10 10 0 2 1 SURFace 2 LINE 2 FACEt 1 1 12 13 11 0 22 23 PAIR 1 NTOS 1 2 SOLM PENAlty 2e3 AUGMent TOLE 1e 5 1e 5 1e 5 END A solution to the problem is obtained using the commands BATCh PROP DT 0 1 END BATCh LOOP time 50 TIME LOOP augment 4 LOOP newton 30 TANG 1 NEXT AUGMent NEXT NEXT END Note especially the extra loop required to perform the augmentation Results for the deformed shape and contours of the final gap achieved constructed using the plot command PLOT CVAR 9 9 gap are shown in Fig 8 2 CHAPTER 8 CONTACT PROBLEM CONTACT VAR 9 1 00E 06 2 00E 07 6 00E 07 1 40E 06 2 20E 06 3 00E 06 3 80E 06 4 60E 06 Current View Min 1 23E 06 X 1 00E 01 Y 1 68E 00 Max 9 03E 01 X 0 00E 00 Y 0 00E 00 Time 5 00E 00 Figure 8 2 Contact deformed shape and gap Time 5 50 Chapter 9
14. file or in a file which is referenced by an include file Finally no STOP command should be referenced by the ISTRIP file FEAP will perform all three problems and place the output sequentially in the file named OSLIT The same methodology also may be used to run a sequence of different problems In CHAPTER 5 STRIP WITH HOLE AND SLIT PARAm c cosd 45 0 s sind 45 0 a cosd 22 5 b sind 22 5 Termination BLOCk CART n n d e m 1 1 0 ooN RAUN NPY NNU QQ Termination Figure 5 5 Tension Strip with Slit Block Generation of Quadrant EBOUndary 2 8010 2 8010 Termination CBOUndary node 6 0 1 0 Termination EDISplacements 2 80 0 5 2 8 0 0 5 Termination MATErial 1 SOLId ELAStic ISOTopic 10000 0 25 Termination Figure 5 6 Tension Strip with Slit Boundary and Material Parameters 34 CHAPTER 5 STRIP WITH HOLE AND SLIT 35 STRESS 2 Min 1 17E 02 Max 2 32E 03 2 31E 02 5 79E 02 9 26E 02 1 27E 03 1 62E 03 1 97E 03 Current View Min 1 17E 02 X 5 94E 00 Y 7 10E 00 Max 2 32E 03 X 1 01E 00 Y 4 26E 18 Figure 5 7 Tension Strip with Slit Contours of oy PARAmeter n 4 INCLude ISTRIP PARAmeter n 8 INCLude ISTRIP PARAmeter n 16 INCLude ISTRIP STOP Figure 5 8 Include data for tension strip refinements CHAPTER 5 STRIP WITH HOLE AND SLIT 36 this case a reference by the include statement would name each of the problem files to be
15. implies the value of the corresponding displacement component will have specified values default values are 0 0 whereas a zero code the default restraint code value indicates the component is unknown and must be computed The inputs use generation similar to the COOR data Thus the If the field is omitted a default value of 1 is assumed CHAPTER 4 CIRCULAR DISK 23 first pair of records will also generate restraints for nodes 2 3 and 4 A negative or zero code will propagate to the generated node whereas a positive code will become a zero values on the generated nodes Thus the first pair will generate a set of restraints such that the first displacement component u1 will be an unknown and the second component uz will be set to a specified value The last pair of records also generates restraints for the first degree of freedom of nodes 11 and 16 For degree of freedoms with zero restraint codes FEAP will add a force value to each node default is 0 0 whereas for non zero restraint codes FEAP will impose a displace ment value default is 0 0 Non zero forces may be specified using a FORCe command other options also exist as described later Non zero displacements may be specified using a DISPlacement command again other options exist Thus for the example problem a concentrated load F gt 5 is applied on node 19 by the FORCe command set shown above The MATErial record specifies the set number as 1 The next record
16. restraints 10 010 20 001 Blank termination record CFORce Coordinate specified forces NODE 0 0 1 0 0 0 5 0 CHAPTER 4 CIRCULAR DISK 27 Blank termination record MATErial 1 SOLId ELAStic ISOTropic 10000 0 25 DENSity data 0 10 QUADrature data 2 2 Blank termination record END TIE Tie nodes with same coordinates accomplish these steps and complete the description for the mesh and its material properties Note particularly the TIE command following the generation of the mesh This command will merge the three blocks to form a mesh which is equivalent to the original mesh however this mesh has different numbering for nodes and elements than the first mesh generated The numbers for the active nodes after the merge and the element numbers produced by the block commands are shown in Figure 4 5 The advantage of this form however is the ease by which the mesh can be refined Indeed by assigning the parameter n and m a value of 5 one obtains a mesh with 75 elements as shown in Figure 4 6 Figure 4 6 Mesh for Circular Disk 75 Elements Since the boundary conditions are associated with an edge all nodes which lie on the edge will be identified and restrained Similarly the node with the specified coordinates will be identified for the applied F 5 0 force Additional details on options for the CFORce command are given in the mesh user manual pages in Appendix A Once the mesh is described the problem ma
17. 01 T5E 01 T4E 01 T5E 01 75E 01 T5E 01 T6E 01 T6E 01 T6E 01 TTE 01 TTE 01 TTE 01 XANPRERERRRRRERJONJPOoORDm AaSNOoO gt RRRRRRRR0000000J R R SCS Rp 46E 19 47E 19 46E 19 47E 19 TTE 22 33E 18 04E 22 59E 22 41E 22 14E 17 49E 19 48E 19 48E 19 48E 19 48E 19 47E 19 47E 19 46E 19 22E 20 01E 22 14E 18 26E 18 13E 16 22E 15 91E 22 T4E 22 35E 18 14E 22 13E 22 92E 22 42E 19 43E 19 43E 19 43E 19 44E 19 44E 19 44E 19 45E 19 S0E 18 80E 22 CPU Time Final Seconds CHAPTER 9 FINITE DEFORMATION PLASTICITY 56 The solution summary for a small deformation case remove the FINIte material record is given below times are for a PIII 650MHz processor SOLUTION SUMMARY Load Total Solution Step Iter Forms Time 1 2 O 1 000E 01 2 2 O 2 000E 01 3 2 O 3 000E 01 4 2 O 4 000E 01 5 5 O 5 000E 01 6 5 O 6 000E 01 7 6 O 7 000E 01 8 6 O 8 000E 01 9 7 O 9 000E 01 10 8 O 1 000E 00 11 2 O 1 100E 00 12 2 O 1 200E 00 13 2 O 1 300E 00 14 2 O 1 400E 00 15 2 O 1 500E 00 16 2 O 1 600E 00 17 2 O 1 700E 00 18 2 O 1 800E 00 19 5 O 1 900E 00 20 5 O 2 000E 00 21 5 O 2 100E 00 22 6 O 2 200E 00 23 5 O 2 300E 00 24 5 O 2 400E 00 25 6 O 2 500E 00 26 6 O 2 600E 00 27 6 O 2 700E 00 28 7 O 2 800E 00 29 8 O 2 900E 00 30 8 O 3 000E 00 31 2 O 3 100E 00 32 2 O 3 200E 00 33 2 O 3 300E 00 34 2 O 3 400E 00 35 2 O 3 500E 00 36 2 O 3 600E 00 37 2 O 3 700E 00 38
18. 076E 01 9953E 01 3438E 01 5 2171E 01 O666E 02 3559E 01 O666E 02 3559E 01 1 0 0 0 0 10 O 1 1 1 ROPNDNEPNER O PON 0 0 No factoring performed EIGE VECTor Eigenpair for last element Eigenvalue 2 3 OOOOE 04 6 1457E 03 6205E 03 9 2513E 02 Eigenvectors 2 3500E 01 2700E 01 5400E 01 2700E 01 5400E 01 2700E 01 3500E 01 2700E 01 3 1106E 01 2639E 01 4839E 01 2639E 01 4839E 01 2639E 01 1106E 01 2639E 01 Eigenvectors 6 0466E 02 6701E 01 4 4533E 01 7990E 02 2 9436E 01 1870E 01 7990E 02 2 9436E 01 7 1 4 7 2928E 01 4533E 01 4 de 6 5 lt 1 5 4 4533E 03 8724E 12 4 5699E 02 4460E 01 4896E 01 2482E 01 4896E 01 2482E 01 5699E 02 4460E 01 8 2360E 17 0000E 01 1113E 16 0000E 01 4079E 17 10 CHAPTER 2 PATCH TESTS The test is repeated for the mesh shown in Fig 2 2 b using the mesh data 2 0666E 02 6 2171E 01 2 0666E 02 6 6701E 01 4 4533E 01 2 1870E 01 4 4533E 01 5 0000E 01 7 0466E 02 1 2928E 01 1 7654E 17 FEAP Axisymmetric patch test 0 00229 MATErial SOLID ELAStic ISOTropic 1000 0 0 25 AXISymmetric BLOCK CARTesian 2 2 QUADrilateral 9 END 1 0 0 2 10 0 3 10 0 4 0 0 0 0 1 0 0 10 0 5 0000E 01 The resulting eigenvalues are vectors are not included here In both cases there is only one rigid body modes namely
19. 1 3001 Blank termination FORCe 5 0 0 0 10 0 Blank termination END INTEractive STOP Table 3 1 Data for Truss Analysis Problem record record record record record record 16 CHAPTER 3 TRUSS PROBLEM 17 are also input in a similar way however note that the material property set in the third field now is set to either 1 or 2 depending on whether the cross section has 5 or 10 units of area Boundary restraint conditions are imposed so that node 1 is restrained to have zero u and v displacements while node 3 has only a restrained v displacement Finally a single load in the vertical direction with magnitude 10 0 is applied to node 5 This problem requests an INTEractive mode of solution In the interactive mode users must give the solution commands needed for each solution step For example when the command FORM is given FEAP will compute the residual and then prompt for another command input Similarly giving the commands for output will display the request on the screen and also place the information in the output file Chapter 4 Circular Disk Subjected to Point Loading The next problem considered is a circular disk loaded by two concentrated forces of 10 units each directed along a diagonal see Figure 4 1 The material of the disk is assumed to be linearly elastic Furthermore for simplicity we consider the loads to be slowly applied so that inertial effects may be ignored Thus the model to be solved
20. 1 2 180E 01 1 293E 00 9 207E 01 28 3 50 2 496E 01 1 651E 01 1 304E 00 8 316E 01 29 3 25 2 614E 01 1 143E 01 1 313E 00 7 428E 01 30 3 00 2 695E 01 6 483E 02 1 319E 00 6 543E 01 31 2 75 2 740E 01 1 640E 02 1 322E 00 5 658E 01 32 2 50 2 756E 01 3 160E 02 1 324E 00 4 774E 01 33 2 25 2 740E 01 7 961E 02 1 322E 00 3 891E 01 34 2 00 2 695E 01 1 280E 01 1 319E 00 3 006E 01 39 1 75 2 614E 01 1 775E 01 1 313E4 00 2 120E 01 36 1 50 2 496E 01 2 283E 01 1 304E4 00 1 233E 01 37 1 25 2 328E 01 2 812E 01 1 293E 00 3 424E 02 38 1 00 2 101E 01 3 363E 01 1 278E 00 5 505E 02 39 0 75 1 788E 01 3 947E 01 1 261E 00 1 447E 01 40 0 50 1 358E 01 4 553E 01 1 240E 00 2 346E 01 41 0 25 7 526E 02 5 208E 01 1 216E 00 3 248E 01 42 0 00 1 527E 02 5 886E 01 1 188E 00 4 151E 01 Table 7 1 Solution for displacements at time step 1 and 20 46 Chapter 8 Contact Problem In this example we consider the solution of a two dimensional contact problem Two beams are placed a short distance apart with the top beam loaded by a uniformly distributed vertical load Each beam has a length of 20 units and the spacing between the beams is 0 5 units The upper beam is divided into 11 equal length elements and the lower beam into 10 equal length elements Each beams is clamped at both ends Beams are modeled using the finite deformation FRAMe element The material model i
21. 2 This solution was computed using the commands BATCh TANGent 1 Solve problem PLOT CONTour 1 Contour solution END The transient solution is computed using a transient solution This may be accom plished by inserting an ORDEr command after the mesh END command and before the first solution execution The order command is given as CHAPTER 6 THERMAL PROBLEM 39 DISPLACEMENT 1 0 00E 00 1 43E 01 2 86E 01 4 29E 01 5 71E 01 7 14E 01 8 57E 01 1 00E 00 Current View Min 0 00E 00 X 5 00E 00 Y 0 00E 00 Max 1 00E 00 X 0 00E 00 Y 0 00E 00 Time 0 00E 00 Figure 6 2 Steady state temperature solution Patch test ORDER 1 where the 1 restricts transients for the heat equation to a first order differential equa tion Any time integrator could be tried however in the results given below the back ward Euler scheme was used i e TRANsient BACKward The solution was performed for 20 time steps of At 0 005 with the command language program BATCH PARAmeter DT dt Sets to parameter value TRANs BACK LOOP 20 TIME TANG 1 Problem linear no iterations PLOT RANG O 4 0 9 PLOT WIPE PLOT CONT 3 Temperature contours NEXT END dt 0 0001 End of parameter input A solution at later times is continued using larger time steps Generally the solution can be computed by increasing the time increment by factors of 10 every logarithmic decade of time using a FEAP function program The fu
22. FEAP A Finite Element Analysis Program Version 8 4 Example Manual Robert L Taylor Department of Civil and Environmental Engineering University of California at Berkeley Berkeley California 94720 1710 E Mail rltGce berkeley edu May 2013 Contents 1 Introduction 2 Patch tests 2 1 Plain Pateh Leste te te era ete heed See Gok gd A a RE E 2 2 Axisymmetric patch test er day Se cae hho wee Howe OE A BR A ee ee EE ee BEE es 2 2 2 Consistency test Sa RA BA AE eS la ee Be ee 3 Truss problem 4 Circular disk 5 Strip with hole and slit 6 Thermal Problem 7 Coupled Thermo mechanical 8 Contact Problem 9 Finite Deformation Plasticity O nN bw N 15 18 30 37 42 47 51 Chapter 1 Introduction In this manual we provide some examples of problems which can be set up and solved using the FEAP program We begin by describing some of the methods which may be used to define an input data file for some simple finite element analyses The manual is organized to start with very basic methods for inputs and precedes to more general methods to describe input data and problem solutions FEAP is controlled using a set of commands Each command performs a basic step in either describing a problem or solving a problem Commands are divided into three basic groups 1 Mesh description commands 2 Problem solution commands and 3 Graphical display commands The appendices of the User Manual contain the optional f
23. count nodes e 0 To make feap count elements Terminator REGIon 2 Assign 2nd and 3rd quadrant TRANsform Reverse x axis for second quadrant 1 0 0 0 1 0 0 0 1 0 0 0 INCLude IHQUAD TRANsform Reverse x y axis for third quadrant 1 0 0 0 1 0 0 0 1 0 0 0 INCLude IHQUAD REGIon 3 Assign 4th quadrant to region 3 TRANsform Reverse y axis for fourth quadrant 1 0 0 0 1 0 0 0 1 0 0 0 INCLude IHQUAD END Figure 5 2 Region Transformation and Include Structure CHAPTER 5 STRIP WITH HOLE AND SLIT 33 Figure 5 3 Tension Strip Block Structure Before Merges Pe Figure 5 4 Tension Strip After Merge of Each Region with Itself Finally the displacements on the top and bottom edges are specified using the com mands given in Figure 5 6 An analysis using this data produced the results for the oyy stress shown in Figure 5 Since no explicit use of node element numbers appears in any of the data it is possible to run a number of problems using the same data but different values for the mesh gen eration parameter n Indeed all problems may be performed during the same execution of the program by defining a master input file say ISLIT which has the values for the parameters The file ISLIT can contain the data shown in Fig 5 8 It is necessary to remove the definition of the parameter n from the ISTRIP file It is further assumed that the solution commands are also contained in the ISTRIP
24. ction the nodal forces on the boundary are 2 5 at the corners and 5 0 at the midside The left side of the mesh is restrained so that no x direction displacements occur and the lower left corner is also restrained in the y direction These restraints prevent rigid body motions as well as ensure that a correct solution to the constant stress problem may be obtained For a linear elastic problem with the isotropic properties Young s modulus 1000 and Poisson ratio 0 25 the plane strain solution has displacements u 0 009375 2 v 0 003125 y CHAPTER 2 PATCH TESTS 3 Figure 2 1 Patch Test Mesh and stresses Org 1 0 Oy 0 0 Ory 0 0 Ozz 0 25 The input data for each problem to be solved by FEAP is prepared and placed in a file on disk It is recommended that the filename for the input data have a first character of I The filename should not exceed 14 characters 12 in PC mode of which the last 4 may be used for an extender xxx When using the program on a PC it is recommended that no extender be used for the main data input file Feap uses this file for generating other files which do contain extenders and thus an error could occur The filename for the input data of the patch test will be called Ipatch for the discussion below When FEAP is run the names for other files will be assigned by replacing the first character i e the I by one indicating the type of file For example the file containing the desc
25. e element records defines the increment to apply to nodes in order to generate missing elements Thus from the first two pairs of records it is evident that elements 2 3 and 4 are missing The generation CHAPTER 4 CIRCULAR DISK 22 increment to apply to the nodes on element 2 is specified as 1 on the element 1 record Accordingly the generated elements will have the following sequence of nodes 2 2 3 8 7 3 3 4 938 4 4 510 9 The material identifier number will be the values specified in the third field and for the data above will be 1 for all generated elements the material identifier for generated elements is taken from the first record of each generation pair Multiple ELEMent sets may be used or ELEMent may be combined with other options e g BLOCk type generations To complete the problem specification it is necessary to impose constraints on the nodal displacements along the symmetry axes specify the applied concentrated load and define the material set properties A set of commands which accomplish this is given by BOUNdary restraint codes 11 1 1 50 0 1 6 5 1 0 190 1 0 Blank termination record FORCes on nodes 1900 5 0 Blank termination record MATErial 1 SOLId ELAStic ISOTropic 10000 0 25 E and nu DENSity data 0 1 rho QUADrature data 2 2 Blank termination record END The set of records following the BOUNdary code command imposes the necessary restraints on nodes A non zero restraint code
26. e is constant along the line 11 0 25 and equals 0 70261 at time step 1 and 0 96363 at step 20 CHAPTER 7 COUPLED THERMO MECHANICAL 45 DISPLACEMENT 3 STRESS1 1 70E 03 5 28E 00 4 00E 01 1 00E 00 5 00E 01 7 60E 01 6 00E 01 5 20E 01 7 00E 01 2 80E 01 8 00E 01 4 00E 02 9 00E 01 2 00E 01 1 00E 00 1 54E 00 Current View Current View Min 1 70E 03 Min 5 28E 00 X 4 99E 00 X 1 59E 00 Y 2 50E 01 Y 5 13E 00 Max 1 00E 00 Max 1 54E 00 X 6 70E 02 X 1 89E 00 Y 5 64E 00 Y 2 47E 00 Time 5 00E 03 Time 5 00E 03 Figure 7 2 Thermal stress problem Time 0 005 DISPLACEMENT 3 STRESS1 5 37E 01 9 98E 01 4 00E 01 1 00E 00 5 00E 01 7 60E 01 6 00E 01 5 20E 01 7 00E 01 2 80E 01 8 00E 01 4 00E 02 9 00E 01 2 00E 01 1 00E 00 3 49E 01 Current View Current View Min 5 37E 01 Min 9 98E 01 X 4 93E 00 X 2 59E 00 Y 3 87E 00 Y 6 07E 00 Max 1 00E 00 Max 3 49E 01 X 1 28E 00 X 1 86E 00 Y 6 40E 00 Y 2 98E 00 Time 1 05E 01 Time 1 05E 01 Figure 7 3 Thermal stress problem Time 0 100 CHAPTER 7 COUPLED THERMO MECHANICAL Time 0 005 Time 0 10 Node 29 Uy TA Uy TA 22 5 00 1 527E 02 5 254E 01 1 188E 00 1 370E 00 23 14 75 7 526E 02 4 576E 01 1 216E 00 1 279E 00 24 4 50 1 358E 01 3 921E 01 1 240E 00 1 189E 00 25 14 25 1 788E 01 3 314E 01 1 261E 00 1 099E 00 26 4 00 2 101E 01 2 731E 01 1 278E 00 1 010E 00 27 13 75 2 328E 0
27. elements 43 The solution of the solid mechanics problem is assumed to be at a time scale for which inertial effects may be ignored The heat equation however is to be solved using a CHAPTER 7 COUPLED THERMO MECHANICAL 44 transient solution This may be accomplished by inserting an ORDEr command after the mesh END command and before the first solution execution The order command is given as ORDER O O 1 where the 0 denotes a zero order i e static differential equation form and the 1 restricts transients for the heat equation to a first order differential equation Any time integrator could be tried however in the results given below the backward Euler scheme was used i e TRANsient BACKward The solution was performed for 20 time steps of At 0 005 with the command language program BATCH DT 0 005 TRANs BACK PLOT DEFOrm PLOT FACT 0 7 to keep plot in window LOOP 20 TIME LOOP 5 Note that at least 3 iterations TANG 1 are needed to converge since no NEXT compling tangent is included PLOT WIPE PLOT RANG O 4 0 9 PLOT CONT 3 Temperature contours PLOT RANG 1 0 0 2 PLOT STRE 1 11 Stress contours NEXT END Results for the contours after the first and last step are shown in Fig 7 2 and 7 3 Numerical results for displacements and temperatures along the row of nodes at x 0 25 center of left row of elements are included in Table 7 1 The temperature for the row of nodes shown in the tabl
28. ent mesh for the example problem is quite restrictive If it is desired to increase the number of nodes and elements it is necessary to restart the process from the very beginning Thus we now consider options for describing the mesh which permit the problem to be described more easily as well as permit the number of nodes and elements to be increased or decreased CHAPTER 4 CIRCULAR DISK 24 FEAP has some powerful options which permit the generation of problems in a form amenable to refinement For example the control record pair may be specified with 0 nodes 0 elements and or 0 materials indicated FEAP will use the subsequent data to compute the number of nodes elements and material sets in the mesh Accordingly it is possible to use FEAP Circular Disk Block inputs 000224 Without additional features this is of little merit However nodes and elements may be generated using a BLOCk command as PARAmeter m 2 n 2 End of parameters BLOCk 1 CARTesian m n 1 1 1 s 0 0 0 0 op OOM OCOooos n 0 0 4 5 PUNE Blank termination record A BLOCk command uses a regular subdivision of the parent master element shown in Figure 4 2 to describe the mesh within its perimeter In Figure 4 2 the coordinate directions 1 and 2 are local axes for the BLOCk generation i e the natural coordinates for an isoparametric master element Using this convention BLOCk 1 is described by a 4 node quadrilateral master element
29. ents methods to input non zero values are described in Section 5 7 of the User Manual however data is given to impose non zero forces using the FORCe command Each force record defines a node number a generation parameter and the force values for each degree of freedom associated with the node Thus for the data given in Table 2 1 nodes 3 and 9 have x force values of 2 5 units and node 6 has an x force value of 5 0 units Input terminates with a blank record The final command after the force values is the END command which terminates input of the data describing the finite element mesh The set of commands shown in Table 2 2 define the solution algorithm to be used in solv ing the problem The execution is initiated by the BATCh command alternatively it is possible to perform an interactive execution where users enter each command as needed see next example and Chapter 11 of the User Manual The FORM command instructs FEAP to form the residual for the equilibrium equations written as R u F P u where F is the vector of applied nodal forces u is the vector of nodal displacements and for a static linear elastic problem P is defined as P Ku in which K is the stiffness matrix A solution to the problem is defined by requiring the residual to be zero In FEAP the solution may be computed using Newton s method CHAPTER 2 PATCH TESTS T which solves a sequence of linear problems Thus the TANGent command requests the tan
30. gent matrix to the residual about the current solution state u at start of execution the value is zero where the tangent matrix is defined as Kiang a E Thus for a linear elastic static problem the tangent matrix is just K The SOLVe command instructs FEAP to solve the incremental Newton equations Kiang Au R and to update the solution as u u Au For a linear problem this solution sequence should converge in one iteration thus the residual would be zero if the command FORM is given again At this point FEAP has computed the solution however it is necessary to issue additional commands to output the values for each type of solution quantity The DISPlacement command instructs FEAP to output values for the nodal generalized displacement parameters associated with each nodal degree of freedom for linear elas ticity using solid elements these are the values of the u and v displacements at a node The option ALL requests the displacement values for all active nodes see manual pages in Appendix A for other options Similarly the command STREss ALL re quests output values for stresses within all the active elements see Appendix A for other options All output is placed in the output file 2 2 Axisymmetric patch test A three dimensional solid is described by a body of revolution and is to be analyzed for the case of axisymmetric behavior Only small deformations are considered For this case the displacement field is g
31. h the element and the list of nodes connected to the element For the elements shown in Figure 2 1 the node sequence must start with a node at one vertex and then proceed with the nodes on vertices traversed counter clockwise around the element Input terminates with a blank record The degree of freedoms for each node may have known applied loads nodal forces or may be restrained to satisfy specified nodal displacements In FEAP all degree of freedoms are assumed to have specified loading applied unless a restraint code is set The BOUNdary restraint command may be used to assign restraints to degree of freedoms which are to have specified displacements Each record defines a node number a generation parameter and the restraint codes for each degree of freedom associated with the node A non zero value for the restraint code indicates that the associated degree of freedom must satisfy a specified nodal displacement value default is zero whereas a zero restraint code indicates the associated degree of freedom has a specified nodal force also zero by default Thus for the data shown in Table 2 1 node 1 has both the u and v displacements restrained nodes 4 and 7 have the u displacement restrained and the y force specified All other nodes have both the x and the y forces specified since no restraints are specified Input terminates with a blank record It is evident from the remaining data that no data is provided to impose non zero displacem
32. hird quadrant 7 Assign the fourth quadrants to REGIon 3 8 Set the TRANsformation to reflect y axes Use an INCLude IHQUAD to generate the 2 blocks for the fourth quadrant The commands to perform the transformations and read the include files are summa rized in Figure 5 2 The outlines for all the blocks formed after this step are shown in Figure 5 3 It is necessary now to merge these blocks to form the final mesh while retaining the slit which if not properly treated will be merged also during the use of a TIE command It is at this time that the utility of the REGIon descriptions is used A summary of the merge order is as follows 1 Merge each of the regions with itself The result of this step is shown in Figure 5 4 2 Merge region 1 with region 2 also merge region 2 with region 3 This will produce the final mesh whose outline was shown in Figure 5 1 The TIE commands to achieve these two steps are TIE REGlon 1 1 TIE REGIon 2 2 TIE REGIon 3 3 TIE REGIon 1 2 TIE REGIon 2 3 The generation of the two blocks forming each quadrant is achieved using the commands shown in Figure 5 5 CHAPTER 5 STRIP WITH HOLE AND SLIT 32 FEAP Tension Strip With Hole and Slit 0 0 0 2 2 4 PARAmeters d 1 First node number e 1 First element number 1 Material set number n 8 Size of blocks Terminator REGIon 1 Assigns 1st quadrant to region 1 INCLude IHQUAD Input first quadrant PARAmeters d 0 To make feap
33. is a simple linear elasto statics problem Since the loading is symmetric and we assume the material to be isotropic and thus also symmetric it is only necessary to construct a mesh for one quadrant of the circular disk This problem has curved boundaries and requires a general mesh to define the finite element solution thus more details are described for the input data options available in FEAP Figure 4 1 Circular Disk with Point Loading In the two dimensional capabilities included in FEAP the solid elements permit an analyst to use elements with between 3 and 9 attached nodes A 3 node element is a plane triangle with nodes located at each vertex A 4 node element is a quadrilateral 18 CHAPTER 4 CIRCULAR DISK 19 with nodes at each vertex A 9 node element is quadrilateral in shape but may have curved sides defined by nodes located in the mid part of each edge as well as one additional node in the interior Omitting some midside nodes and or the center node produces quadrilateral elements with between 5 and 8 nodes Let us assume that a mesh will be constructed using 4 node quadrilateral elements The nodes are defined by a sequence starting with Node 1 and concluding with the maximum number Node NUMNP The elements also are defined by a sequence starting with Element 1 and concluding with Element NUMEL A 4 node quadrilateral element is defined by the node numbers associated with each vertex A simple mesh for one quadrant
34. isplacement relations become ON Or 0 Na r ON Oz 0 ON yo Oz ue 0 ug ON Or The variational problem generates the residual equation again without the factor 2 7 Re Fa f Blordrdz Q CHAPTER 2 PATCH TESTS 9 in which nodal forces are given by F p Nbr dr dz Nebr ds Q TD Linearizing generates the tangent stiffness used in solution of problems by a Newton method Accordingly after the linearization step we obtain Koa f B7D7 Bgr dr dz Q as the stiffness matrix in which Dry is the matrix of tangent moduli for each material The patch test satisfaction requires two parts A stability test and a consistency test Here the stability is assessed by computing the eigenvalues for a single element f d 2 8 Figure 2 2 a 4 node element b 9 node element 2 2 1 Stability test An eigenpair assessment is performed for the 4 node element shown in Fig 2 2 a The geometry and material properties are given by the commands FEAP Axisymmetric patch test 000224 MATErial SOLID ELAStic ISOTropic 1000 0 0 25 AXISymmetric BLOCK CHAPTER 2 PATCH TESTS CARTesian 1 QUADrilateral 4 END To compute the eigenvalues and eigenvectors the command language statements needed are BATCh 1 0 0 2 10 0 3 10 0 4 0 0 TANG 1 END and results obtained are 1 0593E 04 3 5958E 03 1 9953E 01 3438E 01 8127E 01 0076E 01 8127E 01 0
35. iven by u e a 2 1 u r 2 That is the displacements do not depend on the 9 coordinate however in FEAP all axisymmetric problems are formulated for a one radian sector in the 0 direction and the factor 27 is omitted The stress and strain for an axisymmetric case may be ordered as a o Opr Ozz O66 pz 2 2 and ds c Err Ezz 00 Ane 2 3 CHAPTER 2 PATCH TESTS 8 where 7 2 Using the displacements given by Eq 2 1 the strain displacement relations are given by Reference 1 S Sokolnikoff Mathematical Theory of Elasticity McGraw Hill New York 1956 pp 182 184 T u Or Ou OU u Our T 2 4 Or Oz r Oz Similarly for the above displacement field the equations of equilibrium are given by Or OO rz Orr 090 b 0 Or Oz r Oa Oo Oo rz rz 22 b i 2 5 Or r Oz 23 in which body forces also are axisymmetric and thus are given by b b r z For a displacement formulation a variational problem for linear elasticity may be written as H u p Wie raraz f W brdrdo f ut ras min Q Q Ti where W is the stored energy function and ds is a differential length on the two di mensional r z boundary Note also that the factor 27 is again not included for implementation into FEAP 2 6 Introducing an isoparametric interpolation for the finite element model as r NAL z NAS ur N uz Uz Nal uz where are natural coordinates the strain d
36. mmands STREss ALL REACtion ALL produce outputs for all the element variables stresses and any other values included in the element output section and reactions for all the nodes The order of output is controlled by the order of issuing the commands If the solution is converged then nodal reactions should be zero numerically for all degree of freedoms which are not restrained Restrained degree of freedoms report the reaction force necessary to impose the specified displacement constraint Alternatively graphical outputs may be requested For example the commands CHAPTER 4 CIRCULAR DISK 29 PLOT MESH PLOT LOAD 1 produce the results shown in Figure 4 6 Contour plots are also possible Shaded contours for the vertical displacements are shown in Figure 4 7 and are obtained using the command PLOT CONT 2 Additional information on solution and plot commands is included in Chapters 11 and 12 of the User Manual respectively DISPLACEMENT 2 2 24E 03 1 92E 03 1 60E 03 1 28E 03 9 59E 04 6 39E 04 3 20E 04 3 76E 08 Current View Min 2 24E 03 X 0 00E 00 1 00E 00 Y M X Y Time 0 00E 00 Figure 4 7 Contours of Vertical Displacement for Circular Disk Chapter 5 Strip with Hole and Slit As a next example we consider the analysis of a tension strip which contains a hole but has a slit between the hole and the right boundary as shown in Figure 5 1 The strip is to be loaded by applying
37. nction program is contained in a file with the extender fcn For example a routine for the time increment can be given by CHAPTER 6 THERMAL PROBLEM 40 dt 10 dt stored in the file tinc fcn The solution commands are then performed using the set of commands BATCh DT dt LOOP 4 FUNC tinc LOOP 9 TIME TANG 1 NEXT NEXT END This will continue the solution to time 10 Results for the temperature contours at selected times are shown in Figs 6 3 to 6 4 Note that this problem produces one dimensional results and could be solved using a much simpler meshing DISPLACEMENT 1 DISPLACEMENT 1 3 60E 14 2 06E 03 4 00E 01 4 00E 01 5 00E 01 5 00E 01 6 00E 01 6 00E 01 7 00E 01 7 00E 01 8 00E 01 8 00E 01 9 00E 01 9 00E 01 1 00E 00 1 00E 00 Current View Current View Min 3 60E 14 Min 2 06E 03 X 4 00E 00 X 5 00E 00 Y 5 00E 00 Y 5 00E 01 Max 1 00E 00 Max 1 00E 00 X 0 00E 00 X 0 00E 00 Y 0 00E 00 Y 0 00E 00 Time 1 00E 03 Time 1 00E 02 Figure 6 3 Temperature contours at t 0 001 and t 0 01 CHAPTER 6 THERMAL PROBLEM Al DISPLACEMENT 1 DISPLACEMENT 1 5 06E 01 9 99E 01 4 00E 01 4 00E 01 5 00E 01 5 00E 01 6 00E 01 6 00E 01 7 00E 01 7 00E 01 8 00E 01 8 00E 01 9 00E 01 9 00E 01 1 00E 00 1 00E 00 Current View Current View Min 5 06E 01 Min 9 99E 01 X 5 00E 00 X 5 00E 00 Y 5 00E 00 Y 4 50E 00 Max 1 00E 00 Max 1 00E 00 X 0 00E 00 X 0 00E 00 Y 0 00E 00 Y
38. of the disk is shown in Figure 3 1 The figure shows the numbers associated with each node and element This mesh has 19 nodes NUMNP 19 and 12 elements NUMEL 12 Figure 4 2 Finite Element Mesh for Circular Disk with Point Loading The control data for the mesh shown in Figure 4 2 is FEAP Circular Disk Basic inputs 19121224 The remainder of the finite element mesh is described using the data set concepts introduced in the previous examples There are options to specify the number and coordinates for each nodal point The one used to this point is the COORdinate option Since only the first 4 characters of each command are interpreted by FEAP the use of COOR or COORDINATE produces identical results After the COOR command individual records defining each nodal point and its coordinates in the present case the x and x2 coordinates are specified as CHAPTER 4 CIRCULAR DISK 20 N NG X 1 X 2 where N Number of nodal point NG Generation increment to next node X 1 value of x 1 coordinate X 2 value of x 2 coordinate Thus for the mesh shown in Figure 3 1 the coordinate data may be specified as COORdinates 1 1 0 0000 0 0000 5 0 1 0000 0 0000 6 1 0 0000 0 2500 8 1 0 4500 0 2000 10 0 0 9239 0 3827 11 1 0 0000 0 5000 13 1 0 4000 0 4000 15 0 0 7010 0 7010 16 0 0 0000 0 7500 17 0 0 2913 0 6869 18 0 0 3827 0 9239 19 0 0 0000 1 0000 Blank termination record The missing node numbers and their co
39. ordinate values are generated using linear in terpolation on the NG generation sequence given Thus the first pair of records also generates nodes 2 to 4 with coordinates 2 0 2500 0 0000 3 0 5000 0 0000 4 0 7500 0 0000 Comments may be appended after the second character of any line by using an excla mation followed by the text Other options to define coordinates are discussed later There are also different options which may be used to generate the element connection data One is the ELEMent command which is given as N NG MA N 1 N 2 N 3 N 4 CHAPTER 4 CIRCULAR DISK 21 where E N Number of element NG Generation increment for node numbers MA Material identifier associated with element N 1 Node number for first vertex N 2 Node number for second vertex N 3 Node number for third vertex N 4 Node number for fourth vertex Any vertex of the quadrilateral may be used to define N 1 The remainder however should be specified using a counter clockwise sequencing around the nodes as shown in Figure 4 3 Figure 4 3 Node Sequencing for 4 node Quadrilateral The element connection data for the example problem is given by ELEMents 1 12 7 6 6 7 12 11 11 12 17 16 13 14 17 14 15 18 17 16 17 18 19 Blank termination record fae o OrrrRrRR PA BPRRPRR E N Elements within a data set must be in order however gaps may occur and missing elements will be generated The second field in each of th
40. orms which each input com mand may have It is suggested that new users of FEAP carefully read this manual in its entirety before starting to generate their own input data The later examples provide ways to manage the data for problems in separate files using an include option Also the form of the data input records may be constructed using parameters to which numerical values may be assigned either in numeric or expression form see also Chapter 4 of the User Manual Chapter 2 Patch tests The first problems considered are simple patch tests in which all the data is speci fied explicitly Later examples will illustrate how FEAP can add missing data or use repeating similar parts to generate a mesh The patch test is a simple test which should be performed when first using any finite element program see Chapter 11 Zienkiewicz and Taylor 4th ed Vol 1 for description of patch test The patch test both ensures that the theory for the finite element formulation has been correctly implemented and that installation of the analysis system is also correct 2 1 Plain Patch Test For a plain strain or plane stress problem in linear elasticity one patch test is uniform stress in the x direction This condition may be imposed on a square region divided into 4 elements and loaded by a constant x traction on the right side as shown in Figure 2 1 node 6 is located at y 5 units For a square with side lengths 10 and a 10 unit per length tra
41. r disk then may be given as BLOCK 2 CARTesian 10 5 Blank termination record BLOCk 3 CARTesian n n 10 4 0 4 CHAPTER 4 CIRCULAR DISK 26 01 01 OP WD nooo oo aoro anon Blank termination record The PARAmeter command also permits arithmetic calculations to be performed see Chapter 4 of the User Manual for further information In the above the parameter p is set to 0 257 i e 45 degrees and s and c are the sin and cosine of 22 5 degrees The second and third blocks are also 2 x 2 meshes of quadrilaterals since the value of n has not been changed however each of these blocks are described on an element with one edge curved along the circular boundary of the disk Note also that after defining the initial generation sequence for the node and element numbers on the first block all others have zero numbers FEAP will be able to compute the number for the first node and element of each block so that the final mesh has 12 elements and 27 nodes it is necessary to merge the blocks to produce a mesh with only 19 active nodes needed to define the problem Figure 4 5 Finite Element Mesh for Circular Disk using Block Commands and Merging The boundary conditions and applied nodal load may also be defined such that it is not necessary to know node numbers by using EBOUndary for edge boundary specification and CFORced commands for coordinate point force loading respectively Accordingly EBOUndary Edge boundary
42. ription of the mesh and solution results is called the output file and for the above choice for the name of the input data file will be named Opatch The complete input data file for the patch test problem is shown in Table 2 1 and a description for each part of this data follows Input records for FEAP are free format Each data item is separated by a comma and or blank characters If blank characters are used without commas each data item must be included That is multiple blank fields are not considered to be a zero Each data item is restricted to 14 characters 15 including the blank or comma Comments may be appended to any data record after the character e g see Table 2 1 The input file Ipatch consists of several data sets For the patch test mesh given in Table 2 1 the data sets are given by the commands shown without indentation in the table FEAP x 4 Element Patch Test CHAPTER 2 PATCH TESTS FEAP 4 Element Patch Test 9 4 1 2 2 4 MATErial 1i SOLId PLANe STRAin ELAStic ISOTropic 1000 0 0 25 Blank termination record COORdinates 1 O 4 10 000000000 a ONO0O0dIndnoooo oo0oodcdad ooo Ooo0ooIandnooo OONDOAEWHN 0 1 4 1 10 1 Blank termination record ELEMents 1111254 2112365 3114587 4115698 Blank termination record Blank termination record Blank termination record Table 2 1 Data for Patch Test BATCh FORM residual TANGent SOLVe DISPlacement ALL STREss ALL END STOP
43. s elastic E 20 000 with a cross sectional area A 0 1 and moment of inertia J 1 The upper beam is loaded by a uniform load which produces a vertical load of 20 x t units at each node where t is time During the analysis t increases from zero to 5 in equal increments of 0 1 units The mesh is generated using the commands FEAP O 0 O 2 3 2 PARAmeters pr 20 Nodal loading a 20 Lower beam length b 20 Upper beam length h 0 5 Spacing between beams BLOCK CARTesian 101001 1 0 0 0 0 2 a 0 0 BLOCK CARTesian 11 1 0 0 2 1 0 0 h 47 CHAPTER 8 CONTACT PROBLEM 48 2 b h EBOUnd 1 00 1 1 1 a 0 1 1 1 EFORce 2 h O pr MATE 1 FRAMe ELAStic ISOTropic 20000 CROSs section 0 1 1 FINIte MATE 2 FRAMe ELAStic ISOTropic 20000 CROSs section 0 1 1 FINIte END A plot of the mesh is shown in Fig 8 1 VA VA AV VA Y a 4 Time 5 0 Figure 8 1 Mesh of two beams The solution of a contact problem requires the specification of each individual surface to be considered and each pair of surfaces for which possible contacts are to be checked Each surface is defined using a right hand rule to specify the outward pointing normal In the present example the contact between the two parallel beams is to be defined The lower beam is designated as contact surface number 1 and surface facets are defined from the right end to the left end to satisfy the right hand rule giving a outward normal pointing up on the
44. solved Chapter 6 Thermal Problem In this example we consider the solution of a linear thermal problem The domain for the solution is a square with side lengths of 5 units A steady state and a transient thermal analysis are to be performed For the steady state analysis a temperature of T 1 is applied on the entire left boundary and the right boundary is restrained to have temperature zero For the transient problem the left side has a specified unit temperature T 1 suddenly applied at time zero and held constant and the right boundary is insulated q 0 The thermal material parameters are set as follows k 10 c 1 p 0 1 The problem is solved using a 10 x 10 uniform mesh of 9 node quadrilateral elements For the above properties the material data is specified as MATErial 1 THERmal FOURIER ISOTROPIC 10 0 1 0 DENSITY MASS 0 10 The mesh is generated using the block command with the data FEAP Oo 0 O 2 1 9 BLOCK CARTESIAN 20 2000109 1 0 0 0 0 2 5 0 0 0 3 5 0 5 0 4 0 0 5 0 EBOUnd 37 CHAPTER 6 THERMAL PROBLEM 38 1 0 1 1 5 1 Use for steady state problem only EDISp1 1 0O 1 MATE 1 material properties as above END A plot of the mesh is shown in Fig 6 1 ir m Figure 6 1 Mesh of 9 node elements The solution to the steady state heat problem is given by xv and is exactly captured by the finite element solution as indicated in Fig 6
45. vertical displacements along the top and bottom The possibility of having different materials for the top and bottom halves is anticipated and thus the entire mesh needs to be constructed Figure 5 1 Tension Strip with Hole and Slit To construct the mesh a set of 8 blocks of nodes and elements will be constructed and merged to form the final analysis Since the mesh has a considerable amount of symmetry it is proposed to generate the two blocks for one quadrant using the BLOCk mesh command and then use the TRANsformation command to form the other quadrants The steps to form the mesh may be summarized as follows 1 Assign REGIon 1 as the first quadrant 2 Use the BLOCk command to form the two blocks for the first quadrant Save the mesh for the first quadrant in a file called IHQUAD 30 CHAPTER 5 STRIP WITH HOLE AND SLIT 31 3 Use an INCLude IHQUAD in a problem data file e g file ISTRIP to import the data for quadrant 1 see Chapter 4 of the User Manual for more information on use of the include option 4 Assign the second and third quadrants to REGIon 2 5 Set the TRANsformation to reflect the x axis and use an INCLude tt IHQUAD to generate the 2 blocks for the second quadrant Note that since the coordi nate transformation is not a rotation the generated elements may have negative volume 6 Set the TRANsformation to reflect the x axis and y axes Use an INCLude IHQUAD to generate the 2 blocks for the t
46. whose coordinates are specified as shown above The first record states that the master element coordinates will be specified in cartesian form other options are POLAr and SPHErical the block will be divided into m subdivisions in the 1 direction and n subdivisions in the 2 direction an m x n mesh of quadrilaterals with the first node element and material set numbers set to 1 The values for m and n are assigned prior to the BLOCK specification using the PARAmeter command followed by the description of parameters Each parameter may only be one character between a and z FEAP converts all data to lower case internally hence only 26 parameters are available for use at any one time Parameters may however be redefined at later stages of the data Parameters for the sine and cosine of the angles at the middle of the circular part of blocks are defined by PARAmeters p atan 1 21f user mesh functions are employed this feature may not work correctly CHAPTER 4 CIRCULAR DISK 25 gt n 1 5 2 a 6 Figure 4 4 Node Sequence to Define Master Nodes on a Block sin 0 5 p cos 0 5 p Blank termination record Alternatively the inputs may be specified directly in degrees using the expressions PARAmeters s sind 22 5 c cosd 22 5 Blank termination record Consult Chapter 4 of the User Manual for a list of all functions which may be used in expressions The additional blocks needed to define the circula
47. y be solved using the solution command language Two modes of execution are available A batch mode where FEAP enters a solution mode and processes all data without user intervention and an interactive CHAPTER 4 CIRCULAR DISK 28 mode where the user issues each command or command group and receives prompts for additional data An analysis may use multiple batch and or interactive solutions in the same analysis To enter a batch execution the user inserts a command BATCh after the mesh END record and any other commands which manipulate the mesh e g TIE whereas to enter an interactive mode the command INTEractive is inserted The BATCh execution command must be immediately followed by the other solution commands and terminates with an END command To perform a batch mode static steady state solution for the example problem the commands BATCh TANGent Form tangent K FORM Form residual R SOLVe Solve K du R END End of batch execution may be included in the input data file after the TIE command Alternatively the command INTEractive may be included and the other commands given sequentially after receiving the Macro 1 gt prompt With either mode FEAP process the sequence of commands to produce a solution however no report of the results is provided unless specifically requested To output all the nodal displacement values for the current solution state the command DISPlacement ALL may be given Similarly the co
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