S4-1NAS105, Section 4, May 2005 SECTION 4 CONSTRAINTS AND BOUNDARY CONDITIONS

S4-2NAS105, Section 4, May 2005

S4-3NAS105, Section 4, May 2005 TABLE OF CONTENTS SectionPage FREE-FREE ANALYSIS-R CONSTRAINTS… … … … … … … … … … … … … … … 4-5 PROCESSING SUPPORT ENTRIES … … … … … … … … … … … … … … … … … …4-7 SUPPORT SAMPLE … … … … … … … … … … … … … … … … … … … … … … … …4-10 SAMPLE OF INERTIA RELIEF … … … … … … … … … … … … … … … … … … … … …4-11 SINGULARITIES … … … … … … … … … … … … … … … … … … … … … … … … … …4-13 METHODS OF DETECTING SINGULARITIES … … … … … … … … … … … … … … … …4-14 EVIDENCE OF ILL-CONDITIONED MATRICES … … … … … … … … … … … … … … … 4-15 AUTOSPC THEORY- (PARAM, AUTOSPC, YES) … … … … … … … … … … … … ……4-16 GPSP1 PARAMETERS … … … … … … … … … … … … … … … … … … … … … … … 4-18 HIDDEN SINGULARITIES THAT ARE NOT DETECTED BY AUTOSPC … … … … … … …4-19 EXTRANEOUS SINGULARITY MESSAGES FROM AUTOSPC … … … … … … … … … …4-20 IDENTIFICATION OF MECHANISM(MAXRATIO) … … … … … … … … … … … … … 4-21 SHALLOW SHELLS … … … … … … … … … … … … … … … … … … … … … … … … 4-23 PARAM, K6ROT, XX … … … … … … … … … … … … … … … … … … … … … … … … 4-24 EXERCISESMECHANISMS AND UNCONSTRAINED STRUCTURES … … … … … … 4-25 TYPES OF SYMMETRY … … … … … … … … … … … … … … … … … … … … … … 4-28

S4-4NAS105, Section 4, May 2005 TABLE OF CONTENTS (Cont.) SectionPage REFLECTIVE SYMMETRY … … … … … … … … … … … … … … … … … … … … … …4-31 DETERMINE BOUNDARY FROM THE LOADS … … … … … … … … … … … … … … …4-32 STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY … … … …

S4-5NAS105, Section 4, May 2005 FREEFREE ANALYSYSR CONSTRAINTS n In statics, if a model is not properly constrained, it is unstable, and a solution is not normally possible. n For this special case, a solution method called inertia relief is available n When using this approach, it is up to the user to select a statically determinate set of DOF which (if constrained) could prevent any possible rigid–body motion or mechanisms. These DOF must be listed on either SUPORT or SUPORT1 (selectable from case control) entries. In addition, for static analysis, PARAM,INREL,–1 should be specified. n A mass matrix must exist and there must be stiffness connecting the selected DOF to the rest of the model. n The structure is assumed to be in a state of equilibrium. That is a set of inertia loads are found which satisfy F = 0. and M = 0. These inertia loads are applied. n These inertia loading are applied and the DOF listed on the SUPORTi entry are constrained (the reaction forces will be zero if you have selected the DOF properly. n The solution will be the displacements of the model relative to the motion of the SUPORTed (R–set DOF)

S4-6NAS105, Section 4, May 2005 FREEFREE ANALYSYSR CONSTRAINTS n R constraints (Reference DOF) – Free-free structures u Dynamics – mathematically solves for rigid body modes in GIV, HOU, and INV (LANCZOS uses the calculated modes – can be over–ridden using checka_new.v2001) l In GIV, HOU, and INV, one eigenvalue is set to 0.0 for each SUPORT DOF (this is done whether the value calculated is 0.0 or not) l Lanczos will make a judgement as to whether the calculated frequencies are near to 0.0 and will set them to 0.0 if they are close. WARNING: Use of the SUPORT entry assumes that motion about the reference DOF is not constrained. Always check the EPSILON and STRAIN ENERGY table.

S4-7NAS105, Section 4, May 2005 PROCESSING SUPPORT ENTRIES When a SUPORT entry is used, rigid body vectors are calculated in MSC.NASTRAN by the following method: u Step 1: a-set partitioning u Step 2: Stiffness matrix operations Note: P r is not actually applied! where This may be used to construct a set of rigid body vectors.

S4-8NAS105, Section 4, May 2005 PROCESSING SUPPORT ENTRIES (Cont.) n Step 3: Mass matrix operations where [M r ] is not diagonal in general n Step 4: Static solution of unconstrained structure in equilibrium First the determinate forces of reaction are calculated (equivalent reactions if the R-set is held constrained under the action of the applied loads) Then the combined applied loads and inertia loads for static equilibrium are calculated The solution for the static equation of the system in equilibrium with the SUPORT constrained (used as a reference point) is then found

S4-9NAS105, Section 4, May 2005 PROCESSING SUPPORT ENTRIES (Cont.) n Step 4 for Dynamic Solutions – system modes and/or dynamic response of free-free structures (optional) In GIV, HOU, and INV, Gram-Schmidt orthogonalization is used (in the READ module), the matrix [M r ] is orthogonalized by the transformation, i.e., n Step 5: Rigid body mode construction with the property * If the structure truly has rigid body modes

S4-10NAS105, Section 4, May 2005 SUPORT SAMPLE n In statics, the SUPORT DOFs are used as reference DOFs and have displacements of 0.0. $ $ REMOVE CONSTRAINTS AND REPLACE BY A SUPORT $ USE DOF 1-5 AT GRID POINT 1 AND DOF 2 AT GRID POINT 2 $ (SINCE THE PLATES HAVE NO STIFFNESS FOR DOF 6 IN THIS MODEL) PARAM,USETPRT,0 $ print set membership PARAM,USETSEL,8 $ print only R–set PARAM,INREL,–1 SUPORT,1,12345 SUPORT,2,2 $GRDSET,,,,,,,6 $GRID,1,,-.4,0.,0.,, GRID,1,,-.4,0.,0. $GRID,2,,.4,0.,0.,, GRID,2,,.4,0.,0. Sample Data – SOL101

S4-11NAS105, Section 4, May 2005 SAMPLE OF INERTIA RELIEF S. E. SAMPLE PROBLEM 1 OCTOBER 22, 1998 MSC.NASTRAN 10/ 22/ 98 PAGE 12 SUPORT ENTRY USED FOR INERTIA RELIEF SUBCASE 1 U S E T D E F I N I T I O N T A B L E ( I N T E R N A L S E Q U E N C E, R O W S O R T ) R DISPLACEMENT SET –1– –2– –3– –4– –5– –6– –7– –8– –9– –10– 1= 1– 1 1– 2 1– 3 1– 4 1– 5 2– 2 *** USER INFORMATION MESSAGE 7310 (VECPRN) ORIGIN OF ASSEMBLY BASIC COORDINATE SYSTEM WILL BE USED AS REFERENCE LOCATION. SUBCASE 1 *** SYSTEM INFORMATION MESSAGE 6916 (DFMSYN) DECOMP ORDERING METHOD CHOSEN: BEND, ORDERING METHOD USED: BEND *** USER INFORMATION MESSAGE 3035 (SOLVER) FOR DATA BLOCK KLR SUPPORT PT. NO. EPSILON STRAIN ENERGY EPSILONS LARGER THAN.001 ARE FLAGGED WITH ASTERISKS E– 13 – E– E– 13 – E– E– E– E– E– E– E– E– 13 – E– 07 S. E. SAMPLE PROBLEM 1 OCTOBER 22, 1998 MSC.NASTRAN 10/ 22/ 98 PAGE 14 SUPORT ENTRY USED FOR INERTIA RELIEF SUBCASE 1 INTERMEDIATE MATRIX... QRR COLUMN E– E– E E E+00 – E–02 6 COLUMN 2 1 – E– E– E E E E–04 6 COLUMN E E E– E–02 – E– E+00 6 COLUMN E E E– E–02 – E– E+00 6 COLUMN E E+00 – E–14 – E– E– E+00 6 COLUMN 6 1 – E–02 – E– E E E E–01 61

S4-12NAS105, Section 4, May 2005 SAMPLE OF INERTIA RELIEF (CONT) S. E. SAMPLE PROBLEM 1 OCTOBER 22, 1998 MSC.NASTRAN 10/ 22/ 98 PAGE 15 SUPORT ENTRY USED FOR INERTIA RELIEF SUBCASE 1 INTERMEDIATE MATRIX... QRL COLUMN E E E E+ 02 – E E COLUMN E E+ 00 – E+ 00 – E E E COLUMN E E E– E– 10 – E E INTERMEDIATE MATRIX... URA COLUMN E E E+ 04 – E E– E COLUMN E E+ 00 – E E E– E COLUMN E E E– 09 – E– E E *** USER INFORMATION MESSAGE 5293 (SSG3A) FOR DATA BLOCK KLL LOAD SEQ. NO. EPSILON EXTERNAL WORK EPSILONS LARGER THAN.001 ARE FLAGGED WITH ASTERISKS 1 – E– E– 02 2 – E– E– 01 3 – E– E– 01 1 S. E. SAMPLE PROBLEM 1 OCTOBER 22, 1998 MSC.NASTRAN 10/ 22/ 98 PAGE 17 SUPORT ENTRY USED FOR INERTIA RELIEF 0 PRESSURE LOAD SUBCASE 1 D I S P L A C E M E N T V E C T O R POINT ID. TYPE T1 T2 T3 R1 R2 R3 1 G G E– 06 – E– 16 – E– G.0.0 – E– 05 – E– E– G.0.0 – E– 05 – E– 04 – E– G.0.0 – E– 04 – E– E– G.0.0 – E– 04 – E– 03 – E– G.0.0 – E– 03 – E– E– G.0.0 – E– 03 – E– 03 – E– G.0.0 – E– 02 – E– 02 – E– 03.0

S4-13NAS105, Section 4, May 2005 SINGULARITIES n How are they Detected n GPSP (AUTOSPC) n Hidden singularities

S4-14NAS105, Section 4, May 2005 METHODS OF DETECTING SINGULARITIES n Shown in order of increasing cost u Grid Point Singularity Table (GPSP1 Module) u Warning messages from the DECOMP module u Unreasonable displacements with reasonable forces u Epsilon/Strain Energy message u Zero frequency eigenvalues u Nonlinear divergence

S4-15NAS105, Section 4, May 2005 EVIDENCE OF ILL-CONDITIONED MATRICES 1. After matrix assembly, Grid Point Singularity TableWarning Message, AUTOSPC if requested 2. Before decomposition, Null columns IdentifiedFatal Message [Needs PARAM ASING=-1] 3. After decomposition, Information Message – Row Number Log R is thought to indicate how many significant digits may have been lost during the decomposition. 4. After equation solution, in statics, we solve Ku = P for u Ku – P = P (can be printed using PARAM, IRES) flags possible loss of accuracy due to numerical ill conditioning.

S4-16NAS105, Section 4, May 2005 AUTOSPC THEORY (PARAM, AUTOSPC, YES) n Operations performed by GPSP1 (AUTOSPC) u Pick up [K_33] as the 3x3 diagonal partition [K]. u If [K_33] is a null matrix, add the associated 3 DOFs to the singularity list. Compute the eigenvalues ( i ) of [K_33] If any ( i / max ) < EPS, add global direction nearest to the eigenvector corresponding to that i, to the singularity list.

S4-17NAS105, Section 4, May 2005 AUTOSPC THEORY n Find principal stiffness directions for a 3x3 matrix at a GRID point n If, then the global direction nearest i is called singular. If AUTOSPC = YES, then singular DOFs are constrained with SPCs (or MPCs). n Default for EPS can be changed with AUTOSPC Case Control.

S4-18NAS105, Section 4, May 2005 HIDDEN SINGULARITIES THAT ARE NOT DETECTED BY AUTOSPC

S4-19NAS105, Section 4, May 2005 EXTRANEOUS SINGULARITY MESSAGES FROM AUTOSPC n Autospc may incorrectly identify DOF which are nearly singular, or Q–set DOF, which have not been used yet when AUTOSPC is performed Modal Coordinates (Q-set) in Component Modal Synthesis

S4-20NAS105, Section 4, May 2005 IDENTIFICATION OF MECHANISMS (MAXRATIO) Each grid point is stable itself (will not be constrained by AUTPSPC), but the assembly is capable of n This is found during matrix decomposition (module DECOMP or DCMP) Strain-Free Rigid-Body Decoupled etc. Motion

S4-21NAS105, Section 4, May 2005 IDENTIFICATION OF MECHANISMS (MAXRATIO) (CONT) A singularity is indicated by a value of 0.0 on the diagonal D. When is 0 really = 0? Ratio of factor diagonal = K ii /d i. If any |K ii /d i | > MAXRATIO (Default =10 7 ), print all terms over MAXRATIO. If MSC.Nastran finds a matrix diagonal/factor diagonal ratio > MAXRATIO, the action of the program depends on the parameter BAILOUT. BAILOUT = 0 (Default) stops execution of MSC.Nastran BAILOUT = -1 lets MSC.Nastran continue with the analysis BAILOUT = -1 is not recommended, except for model checking.

S4-22NAS105, Section 4, May 2005 SHALLOW SHELLS n If ~ 0, N is nearly singular (use K6ROT or SNORM). n If > 5, N is weak but acceptable (use small K6ROT or use SNORM). n If ~ 30, determine if it is a corner or a shell? (Use special modeling for corners, use SNORM or K6ROT for shells.) Use SNORM (PARAM, default=20 in V2001) to assist with curved shell problems (does not curecorners or connections to other element types)

S4-23NAS105, Section 4, May 2005 PARAM, K6ROT, XX is based on displacements. K is a stiffness X (In general, K ) u There is no recommended value for K6ROT (some people recommend 1.0, the default is 100.0) u A good idea if you are using this is to try a small value, then re-run with a large value (such as ) and see if the answers are noticeably different. If they are, then you probably have a load path depending on the in-plane stiffness and need to update the model to correct this.

S4-24NAS105, Section 4, May 2005 EXERCISESMECHANISMS AND UNCONSTRAINED STRUCTURES n How should the following solid model be constrained for stability? n How should the following connection be modeled? Resting on a flat surface Hint – Grids 1-4 DOFs 3+(?)

S4-25NAS105, Section 4, May 2005 EXERCISESMECHANISMS AND UNCONSTRAINED STRUCTURES (Cont.) TITLE = Mechanism check of a cubic block $ Note: R. Rotation about the Z-Axis ? ECHO = SORT SUBCASE 1 SUBTITLE=Pressure loaded block LOAD = 1 DISPLACEMENT(SORT1,REAL)=ALL SPCFORCES(SORT1,REAL)=ALL STRESS(SORT1,REAL,VONMISES,BILIN)=ALL BEGIN BULK CHEXA, 1, 1, 1, 2, 3, 4, 5, 6,+CH01 +CH01, 7, 8 GRID, 1,, 0., 0., 0.,, GRID, 2,, 10., 0., 0.,, 3456 GRID, 3,, 10., 10., 0.,, 3456 GRID, 4,, 0., 10., 0.,, 3456 GRID, 5,, 0., 0., 10.,, 456 GRID, 6,, 10., 0., 10.,, 456 GRID, 7,, 10., 10., 10.,, 456 GRID, 8,, 0., 10., 10.,, 456 MAT PSOLID PARAM AUTOSPC NO PLOAD4, 1, 1, 10.,,,, 6, 8 ENDDATA

S4-26NAS105, Section 4, May 2005 EXERCISESMECHANISMS AND UNCONSTRAINED STRUCTURES (Cont.) *** USER INFORMATION MESSAGE 4158 (DFMSA) ---- STATISTICS FOR SPARSE DECOMPOSITION OF DATA BLOCK KLL FOLLOW NUMBER OF NEGATIVE TERMS ON FACTOR DIAGONAL = 0 MAXIMUM RATIO OF MATRIX DIAGONAL TO FACTOR DIAGONAL = 9.6E+14 AT ROW NUMBER 14 *** USER WARNING MESSAGE 4698 (DCMPD) STATISTICS FOR DECOMPOSITION OF MATRIX KLL. THE FOLLOWING DEGREES OF FREEDOM HAVE FACTOR DIAGONAL RATIOS GREATER THAN E+07 OR HAVE NEGATIVE TERMS ON THE FACTOR DIAGONAL. USER INFORMATION: GRID ID DEGREE OF FREEDOM MATRIX/FACTOR DIAG. RATIO MATRIX DIAGONAL 7 T E E+07 ^^^ USER FATAL MESSAGE 9050 (SEKRRS) ^^^ RUN TERMINATED DUE TO EXCESSIVE PIVOT RATIOS IN MATRIX KLL. ^^^ USER ACTION: CONSTRAIN MECHANISMS WITH SPCI OR SUPORTI ENTRIES OR SPECIFY PARAM,BAILOUT,-1 TO CONTINUE THE RUN WITH MECHANISMS. n POSSIBLE FIX: Fix (SPC) DOF T2 at GRID 2

S4-27NAS105, Section 4, May 2005 TYPES OF SYMMETRY Reflective Axisymmetric Cyclic

S4-28NAS105, Section 4, May 2005 TYPES OF SYMMETRY (Cont.) Repetitive Symmetry Symmetric Model

S4-29NAS105, Section 4, May 2005 TYPES OF SYMMETRY (Cont.) for 0_th harmonic mode only.

S4-30NAS105, Section 4, May 2005 REFLECTIVE SYMMETRY

S4-31NAS105, Section 4, May 2005 DETERMINE BOUNDARY FROM THE LOADS Example: Symmetric Loads Connect

S4-32NAS105, Section 4, May 2005 STATIC ANALYSIS OF A SIMPLY SUPPORTED PLATE USING SYMMETRY n The plate is subjected to three different loading conditions. The three loadings are as follows: 1. Uniform pressure of 1.0 psi over the surface. 2. Uniform pressure of 1.0 psi acting on half the surface. 3. A point loading of 10.0 pounds acting at the center of the plate. n With a half (left) model, the problem will be solved using the SUBCASE – SUBCOM approach in Case Control. n Load Case 2, though not symmetric can, however, be divided into symmetric and anti-symmetric components, and the results combined (using linear superposition) to get the solution. n Note: Load Case 1, and 2 (being symmetric) could have been solved without using SUBCOM.

S4-33NAS105, Section 4, May 2005 STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY (Cont.) n Loading 2 n The pressure load on half of the plate may be described as a combination of two different loads, a symmetric and an anti-symmetric loading. The symmetric load is a uniform pressure of 0.5 psi over the entire surface, and the anti-symmetric load is a pressure of 0.5 psi acting downward on left half, and upward on the right half. n Each of the individual loads (symmetric and anti-symmetric) may be applied to the half model and solved using the appropriate symmetry conditions. The results of these analyses may be combined to obtain the solution for the entire plate.

S4-34NAS105, Section 4, May 2005 STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY (Cont.) n The results obtained using this half model are identical to those obtained using a model of the complete structure.

S4-35NAS105, Section 4, May 2005 STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY (Cont.) SOL 101 CEND $ TITLE = HALF PLATE (16X8) MODEL, USING SYM., AND ANTI-SYM. DISPLACEMENT(SORT1,REAL)=ALL SPCFORCES(SORT1,REAL)=ALL STRESS(SORT1,REAL,VONMISES,BILIN)=ALL $ Load Case 1: Pressure on Full Plate SUBCASE = 1 SUBTITLE =LOAD CASE 1 - Full Pressure on whole plate SPC = 1 $ SS AND SYMMETRIC BC LOAD = 1 $ Load Case 2: Pressure on half Plate $ Use SUBCASE, and SYMCOM for Pressure on half plate SUBCASE = 11 SUBTITLE =Case 2a: Sym. BC with.5 Psi on half plate SPC = 1 LOAD = 2 SUBCASE = 12 SUBTITLE =Case 2b: Anti-Sym. BC with.5 Psi on half plate SPC = 2 LOAD = 2

S4-36NAS105, Section 4, May 2005 STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY (Cont.) SUBCOM = 21 SUBTITLE = Load Case 2: Solution for left side SUBSEQ = 0., 1., 1. $ 1.*(SUBCASE 11) +1.*(SUBCASE 12) SUBCOM = 22 SUBTITLE = Load Case 2: Solution for right side SUBSEQ = 0., 1., -1. $ 1.*(SUBCASE 11) -1.*(SUBCASE 12) $ $ Load Case 3: Point Load at center of Plate $ SUBCASE = 103 SUBTITLE =LOAD CASE 3 - Point Load at Center SPC = 1 $ SS AND SYMMETRIC BC LOAD = 3 BEGIN BULK PARAM POST 0 PARAM AUTOSPC YES PARAM,NOCOMPS,-1 PARAM PRTMAXIM YES $ MAT1, 1, 1.+7,,.3 $ PSHELL, 1, 1,.1, 1,, 1

S4-37NAS105, Section 4, May 2005 STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY (Cont.) CQUAD4, 1, 1, 1, 2, 11, 10 CQUAD4, 2, 1, 2, 3, 12, 11. $ GRID, 1,, 0., 0., 0. GRID, 2,, 1., 0., 0.. $ $ Combine SPC for ss, and symmetric bc. SPCADD, 1, 101, 102 $ $ Combine SPC for ss, and anti-symmetric bc. SPCADD, 2, 101, 103 $ $ Simply Supported BC at outer edges SPC1, 101, 123, 1, THRU, 10 SPC1, 101, 123, 19, 28, 37, 46, 55, 64 SPC1, 101, 123, 73, 82, 91, 100, 109, 118 SPC1, 101, 123, 127, 136 SPC1, 101, 123, 145, THRU, 153

S4-38NAS105, Section 4, May 2005 STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY (Cont.) $ Symmetric BC at X=8 centerline SPC1, 102, 156, 9, 18, 27, 36, 45, 54 SPC1, 102, 156, 63, 72, 81, 90, 99, 108 SPC1, 102, 156, 117, 126 SPC1, 102, 156, 135, 144, 153 $ Anti-Symmetric BC at X=8 centerline SPC1, 103, 346, 9, 18, 27, 36, 45, 54 SPC1, 103, 346, 63, 72, 81, 90, 99, 108 SPC1, 103, 346, 117, 126 SPC1, 103, 346, 135, 144, 153 $ Load for Pressure on full plate LOAD, 1, 1., 2., 101 $ Load for Pressure on half plate LOAD, 2, 1., 1., 101 $ Nodal Forces of Load Load Case 3 FORCE, 3, 81, 0, 5., 0., 0., -1. $ Pressure Load $ Note: half the pressure specified here PLOAD4, 101, 1, -.5,,,, THRU, 128 $ ENDDATA