practical fe techniques using msc.nastran

S1-1NAS105, Section 1, July 2003

SECTION 1

INTRODUCTION AND SETS IN MSC.NASTRAN


TABLE OF CONTENTS

Section Page

PURPOSE OF THE SEMINAR 1-5COMPANY OVERVIEW 1-6WHAT IS MSC.NASTRAN? 1-7MSC.NASTRAN MAINTENANCE AND UPDATES 1-8MSC.NASTRAN DOCUMENTATION 1-12REFERENCES ON THE FINITE ELEMENT METHOD 1-16SETS IN MSC.NASTRAN 1-17


PURPOSE OF THE SEMINAR Provide practical advice and guidance in using finite

elements Explain special MSC.NASTRAN modeling aids Review advanced diagnostic methods


COMPANY OVERVIEW

The MSC Software Corporation (formerly MacNeal-Schwendler Corporation) has been supplying sophisticated computer-aided engineering (CAE) tools since 1963.

MSC is the developer, distributor, and supporter of the most complete and widely used structural analysis program in the world, MSC.NASTRAN .

NASTRAN development was initiated in 1966 under the sponsorship of the National Aeronautics and Space Administration based on known requirements of the aerospace industry for structural analysis. MSC has been involved in NASTRAN since its inception, and has marketed its own enhanced, proprietary version, MSC.NASTRAN, since 1971.

NASTRAN is a registered Trademark of the National Aeronautics and Space Administration


WHAT IS MSC.NASTRAN? MSC.NASTRAN is a general-purpose, finite element analysis

program. The MSC.NASTRAN program is Based on the basic principles of engineering and the finite

element method Efficient due to its use of modern data base technology and use

of modern sparse matrix and numerical analysis techniques Capable of solving a large variety of engineering problems

Static and dynamic structural analysis Heat transfer Aeroelasticity Hydroelasticity Field-type problems (acoustics, electromagnetism) Optimization

Extensively documented and quality assurance tested Continually being enhanced by the addition of new capabilities


MSC.NASTRAN MAINTENANCE AND UPDATES

New delivery and re-releases on a regular basis Many errors corrected, plus new capabilities added

Error reports published periodically Latest (Released) Version: MSC.NASTRAN V2001

Some Interesting Features recently added to MSC.NASTRAN New Superelements (BEGIN SUPER)– V69 New External superelements (PARAM,EXTOUT)– V70 Pre–loaded structures in dynamic analysis (STATSUB)– V69 Static pre–load in transient analysis (PARAM,ADSTAT)– V70


MSC.Nastran MAINTENANCE AND UPDATES (Cont.)

Some Interesting Features recently added to MSC.NASTRAN (cont.) Beam Library – (PBEAML, PBARL) – V69 Shell to Solid connector (RSSCON) – V69 Beam and shell p–elements – V69 Frequency–dependent elements (CBUSH) –V69 Fluid damping – V69 Adaptive loading frequencies (FREQ3, FREQ4, FREQ5) – V69 Mode – tracking in optimization (MODTRAK) – V69 Customization – (ISHELL) – V70 Residual Vectors in dynamics (PARAM,RESVEC) – V70


MSC.NASTRAN MAINTENANCE AND UPDATES (Cont.)

Nonlinear radial gap element (NLRGAP) – V70.5 Major Aeroelastic enhancements – V70.5 Interface elements (p–elements) – V69 and V70.5 GAPs in SOL 101 (PARAM,CDITER) – V70.5 Geometric nonlinear spring–damper (CBUSH1D) – V70.5

Material and Connectivity Sensitivity Optimization – V70.7 Distributed Memory Parallel Computing (DMP) – v70.7

Simplified Loading Definition for Dynamics – V2001 Simplified Enforced Motion Dynamics – V2001 Modal Effective Mass – V2001


MSC.NASTRAN MAINTENANCE AND UPDATES (Cont.)

Some Interesting Features recently added to MSC.NASTRAN (cont)

P-Element Dynamics –V2001 Quadratic (10-Node) Tetrahedron Element for Nonlinear Analysis –

V2001 CWELD Element – V2001 Shell Output in Material Coordinate System - V2001 Closed Hat Section in Beam Section Library – V2001 Multiple DMIG Input – V2001 Multiple Transfer Functions – V2001 Fully Stressed Design – V2001 Discrete Variable Optimization – V2001 Enhancements in Dynamic Response for Optimization – V2001


MSC.NASTRAN DOCUMENTATION


MSC.NASTRAN DOCUMENTATION (Cont.) The MSC.NASTRAN Reference Manual contains primarily

reference-oriented, non-solution sequence dependent material and is highly subdivided for ease of use. The MSC.NASTRAN Reference Manual is typically not required for day-to-day use and can therefore be shared by a group.

The MSC.NASTRAN Quick Reference Guide contains descriptions of the NASTRAN Statement, File Management statements, Executive Control Statements, Case Control commands, Bulk Data entries, and parameters. The MSC.NASTRAN Quick Reference Guide is recommended for all users.

User’s Guides are designed to be single point sources of information on the full range of MSC.NASTRAN topics and are highly tutorial in nature. The following User’s Guides are available: Getting Started with MSC.NASTRAN User’s Guide MSC.NASTRAN Basic Dynamic Analysis User’s Guide MSC.NASTRAN Linear Static Analysis User’s Guide MSC.NASTRAN Design Sensitivity and Optimization User’s Guide MSC.NASTRAN Numerical Methods User’s Guide MSC.NASTRAN Thermal Analysis User’s Guide


MSC.NASTRAN DOCUMENTATION (Cont.) The following User’s Guides are available (Cont.):

MSC.NASTRAN Aeroelastic Analysis User’s Guide MSC.NASTRAN Advanced Dynamic Analysis User’s Guide

The following books are also available: MSC.NASTRAN Installations / Operation Guide – contains computer and

operating system-specific information. MSC.NASTRAN Release Guide – describes the new features available in each

version. Recommended for all users. MSC.NASTRAN Bibliography (Second Edition) – contains citations for

approximately 1900 MSC.NASTRAN-related papers, articles, and books, arranged by topic and author.

MSC.NASTRAN Common Questions and Answers – answers to commonly asked questions on a wide variety of analysis topics. Recommended for all MSC.NASTRAN users.

MSC.NASTRAN DMAP Module Dictionary (Version 68) – describes the DMAP modules formerly contained in Section 5 of the MSC.NASTRAN User’s Manual. This book is primarily intended for DMAP users.


MSC.NASTRAN DOCUMENTATION (Cont.)

The following User’s Guides are available from theMSC.Nastran 2001 and MSC.Nastran 2001 Documentation CD:

MSC.NASTRAN DMAP and Database User’s Guide MSC.NASTRAN Superelement Analysis User’s Guide


REFERENCES ON THE FINITE ELEMENT METHOD

K.-J. BatheFinite Element Procedure in Engineering Analysis Prentice-Hall, Inc, 1982

E.B. Becker, G.F. Carey, and J.T. OdenFinite Elements, Vol. 1-6Prentice Hall, 1981-1986

R. MacNealFinite Elements: their design and performanceDekker, 1994

R.D. Cook Concepts and Application of Finite Element AnalysisJohn Wiley and Sons, Inc., 1974

R.H. GallagherFinite Element Analysis FundamentalsPrentice-Hall, Inc., 1975

O.C. Zienkiewicz and R.L. TaylorThe Finite Element Method Volume 1-35th Edition, Butterworth-Heinemann, 2000


SETS IN MSC.NASTRAN Whenever you submit a job to MSC.NASTRAN, the program

defines “sets” of degrees of freedom (DOF) based on your input. The program first assembles vectors and matrices for all

structural degrees of freedom (excluding extra points). This is called the G-set (global) and contains 6 DOFs for each grid point and

1 DOF for each scalar point.

The following table provides examples of Bulk Data entries, which may define sets in the program.

Name Bulk Data Entry Name m MPC, MPCADD, MPCAX, POINTAX, RBAR, RBE1, RBE2, RBE3, RROD,

RSPLINE, RTRPLT, GMBC, GMSPC* sb SPC, SPC1, SPCADD, SPCAX, FLYSM, GMSPC*, BNDGRID, (PARAM,

AUTOSPC, YES) sg GRID, GRIDB, GRDSET (PS field) o OMIT, OMIT1, OMITAX, GRID (SEID field), SESET q QSET, QSET1 r SUPORT, SUPORT1, SUPAX c CSET, CSET1 b BSET, BSET1 e EPOINT sa CAEROi k CAEROi a ASET, ASET1, Superelements exterior degrees of freedom, CPSUPEXT

Degrees of Freedom Set Bulk Data Entries

*Placed in set only if constraints are not specified in the basic coordinate system


SETS IN MSC.NASTRAN (Cont.) Based on the input, DOF are placed into the appropriate sets These sets are used during the processing and have certain

requirements Mutually Exclusive sets – no DOF may belong to more than one

of these sets Combined sets – sets made up by combining other sets – a

DOF may belong to several combined sets

msbsg s

oq

cb

r

esak

papsp

g

nen

fef

datl

Mutually Exclusive Combined


SETS IN MSC.NASTRAN (Cont.) The mutually exclusive set names have the following

definitions:Mutually Exclusive Sets

Set Name Meaning m Points eliminated by multipoint constraints

sb Points eliminated by single-point constraints that are included in boundary conditions changes and by the automatic SPC feature

sg Points eliminated by single-point constraints that are listed on grid point Bulk Data entries

o Points omitted by structural matrix partitioning

q Generalized coordinates for dynamic reduction or component mode synthesis

r Reference points used to determine free body motion c Free body set for component mode synthesis or dynamic

reduction b Coordinates fixed for component mode analysis or dynamic

reduction

e Extra degrees of freedom introduced in dynamic analysis

sa Permanently constrained aerodynamic points

k Aerodynamic points


SETS IN MSC.NASTRAN (Cont.) The sets formed by the union of two mutually exclusive sets have

the following definitions:

Set Name Meaning s = sb + sg All Points eliminated by single- point constraints

L = b + c Structural coordinates remaining after the reference coordinates are removed (points left over)

t = L + r Total set of physical boundary points for superelements

a = t + q Set assembled in superelements analysis

d = a + e Set used in dynamic analysis by the direct method

f = a + o Unconstrained (free) structural points fe = f + e Free structural points plus extra points

n = f + s All structural points not constrained by multipoint constraints

ne = n + e All structural points not constrained by multipoint constraints plus extra points

g = n + m All structural (grid) points including scalar points

p = g +e All (physical) points

ps = p + sa Physical and constrained aerodynamic points

pa = ps + k Physical set for aerodynamics

fr = o + L Statically Independent set minus the statically determinate supports (o +L = f – q – r)

v = o + c + r Set free to vibrate in dynamic reduction and component mode synthesis

Combined Sets(+ indicates union of two sets)


SETS IN MSC.NASTRAN (Cont.) To help understand this, let us look at the most commonly used

sets and a simple example. Most commonly used sets – in terms of the problem solution: Create ”system” matrices

G – ”global” set – this set contains all DOF in the model. When the program works with matrices using the global set, it works using the ”displacement coordinate system” from the GRID entries.

Process multi–point constraint equations M – dependent DOF – defined by MPCs and R–type elements – these DOF

are the ”slave” DOF in multi–point constraint equations. Their motion is a combination of the motion of 1 or more other points. If they were kept in the matrices, the matrices would be ”rank deficient” and the problem could not be solved

N – independent DOF – any DOF in the model which are not in the M–set Apply constraints

S – constrained – any DOF constrained by you or by AUTOSPC belong in this set

F – not constrained – the set of DOF which remain after the constraints are applied


SECTION 2

ELEMENT TECHNIQUES


TABLE OF CONTENTSSection PageTHE MOST POPULAR MSC.NASTRAN ELASTIC ELEMENTS 2-7EVOLUTION OF PLATE FINITE ELEMENTS 2-8HIGHER ORDER VERSUS LOWER ORDER ELEMENTS 2-12MODERN FINITE ELEMENTS 2-13FEATURES USED BY THE CURRENT ELEMENTS 2-14FINITE ELEMENT VOLUME INTEGRATION 2-15ADDED STRAIN FUNCTIONS 2-16HEXA 2-17ELEMENT PERFORMANCE 2-18PATCH TEST – CLASSICAL 2-19PATCH TEST FOR SOLIDS 2-20STRAIGHT CANTILEVER BEAM 2-21TWISTED BEAM 2-22TYPES OF GEOMETRIC DISTORTION FROM A SQUARE PLATE 2-23RECTANGULAR PLATE 2-24RESULTS OF RECTANGULAR PLATE

CLAMPED SUPPORTS: CONCENTRATED LOAD 2-25SCORDELIS-LO ROOF 2-26RESULTS FOR SCORDELIS-LO ROOF 2-27SUMMARY OF SOME SAMPLE TEST PROBLEMS 2-28


TABLE OF CONTENTSSection PageRECOMMENDATIONS 2-29MPCS AND “R”-TYPE ELEMENTS 2-31MPC – BULK DATA ENTRY 2-36MULTIPOINT CONSTRAINT EXAMPLES 2-38“R”-TYPE ELEMENTS 2-46THE “R” ELEMENTS 2-47SAMPLE USES OF “R” ELEMENTS 2-48COMMONLY USED “R” ELEMENTS 2-49RBE2 AND RBAR 2-50RBAR: THE CONSTRAINT EQUATIONS 2-51RBAR – BULK DATA ENTRY 2-52RBE2 – BULK DATA ENTRY 2-54THE RBE2 2-56RBE2 EXAMPLE 2-57COMMON RBE2/RBAR USES 2-59RBE3 – THE “WIFFLETREE” 2-60RBE3 – BULK DATA ENTRY 2-61RBE3 DESCRIPTION 2-65RBE3 IS NOT RIGID 2-66RBE3: HOW IT WORKS 2-67


Section PageEXAMPLE 1 2-72EXAMPLE 1: FORCE THROUGH CG 2-73EXAMPLE 2 2-76EXAMPLE 2: LOAD NOT THROUGH CG 2-77RSPLINE – THE LINEAR SPLINE 2-78RSPLINE – BULK DATA ENTRY 2-80RSSCON – BULK DATA ENTRY 2-83

TABLE OF CONTENTS


THE MOST POPULAR MSC.NASTRAN ELASTIC ELEMENTS

Number of DOFs in element matrix

One Dimensional ROD Pin-ended rod 2 – 4 BAR Prismatic beam (can have 0.0 properties) 12

BEAM Straight beam with warping. (No zeros – based on force method. It inverts the flexibility.

12 – 14

BEND Curved beam or pipe 12

Two-Dimensional - No In-Plane Rotational Stiffness

TRIA3 Triangular plate 15 or 18* QUAD4 Quadrilateral plate 20 or 24* SHEAR 4-sided shear panel (unstable on its own) 4 or 8 TRIA6 Triangular shell 15 – 30

QUAD8 Quadrilateral shell 20 - 40

Three – Dimensional

- No Rotational Stiffness Terms

HEXA Solid with six quadrilateral faces 24 – 60 PENTA Solid with two triangular faces and three

quadrilateral faces 18 - 45

TETRA SOLID element with 4 triangular faces 12 – 3 0 Zero-Dimensional

ELAS Simple spring connecting any two degrees of freedom **

2

* With K6ROT

** The CBUSH element is recommended over the CELASI


EVOLUTION OF PLATE ELEMENTS It was years before “good” plate elements were available in finite element

programs. One of the first approximations was to use a beam simulation.

Triangles – the first plate elements were 3–noded triangles using polynomials to approximate the deformations.

The constants were evaluated in terms of the displacements of the corners. Unfortunately, there were not enough DOF available to allow the evaluation of a complete second–order polynomial and people had to improvise to try to correct for this.


EVOLUTION OF PLATE ELEMENTS (Cont.) Many of the early plate elements were made up by using a

combination of triangles.

Finally Irons came up with the idea of “Isoparametric” elements. These elements work using polynomials also, but use “Gauss

Integration” so that the integrals may be evaluated “exactly”, even for elements with curved edges, and faces.

Each isoparametric element is based on a “perfect” element. These “perfect” elements have dimensions of 2 units (non–

dimensional) and are “mapped” onto the physical geometry of the element.

Assumed polynomials


EVOLUTION OF PLATE ELEMENTS (Cont.) The following are examples of “perfect” elements

For each of these elements, the dimensions ζ and η vary from -1 to +1.

The reason for this is that it is easier to come up with shape functions to interpolate the deformations in terms of the “perfect” element. Gaussian integration is used to evaluate the stiffness and other element matrices.

Assuming Isoparametric Shapes(Mapping to Square)


EVOLUTION OF PLATE ELEMENTS (Cont.)

The “order” of an element is determined by the order of the polynomial used to represent the deformed shape.


HIGHER ORDER VERSUS LOWER ORDER ELEMENTS

According to Zienkiewicz 1 , “...A dramatic improvement of accuracy arises with the same number of degrees of freedom when complex elements are used. --- A considerable cost saving occurs.”

(Not necessarily true today; good lower order plate, shell, and solid elements exist.)

“Further, the data preparation is considerably reduced with complex elements.” (Not true, particularly if a mesh generator is used.)

“On the other side of the picture, it will be sometimes seen that the very much reduced number of complex elements may not be adequate to represent all of the local geometries of the real problem with the minimum number of elements.” (Therefore, the lower order elements are needed for special regions.)

“With the use of higher order elements, progressively, the departure from an easily conceived physical idealization occurs.” (True.)

“Probably the most serious economic problem of complex curvilinear elements is the computer time necessary for performing the numerical integration.” (True, but not important for large problems.)

1. O.C. Zienkiewicz, The Finite Element Method in Engineering Science, (Third Edition), McGraw-Hill, London, 1977.


MODERN FINITE MODELING Finite elements approximate engineering theory by

using polynomial displacement functions. Based on experience, people have found that using

“standard isoparametric” elements can give incorrect answers in many situations. Nastran elements are “customized” to provide the best performance possible.

The following pages show some of the methods used in “customizing” the elements.

REMEMBER – FINITE ELEMENTS ARE AN APPROXIMATION TO ENGINEERING THEORY AND ARE NOT EXACT!!!!!


FEATURES USED BY THE CURRENT ELEMENTS

1. Assumed strains at integration points 2. HEXA now has internal strain functions3. Reduced integration


FINITE ELEMENT VOLUME INTEGRATION Full integral

Reduced Integration


ADDED STRAIN FUNCTIONS

(Some people call them “Bubble Functions.”)

[ ] ∑+= bjiji UC εε

)X(F*)Y(G)Y(F*)X(G

y

x

==

εε


HEXA

Assumed strain functions

Lumped mass for higher order HEXA

{ } [ ][ ] [ ]{ }oogg CuC εε += INTOPT = 0

The theory gives negative mass at corners


ELEMENT PERFORMANCE

In 1985 Dr. R. MacNeal and R. Harder published (“Finite Elements in Analysis and Design”, North Holland Publishing Co., pp. 3-20) a series of proposed models to test the quality of finite elements.

These problems were designed to “punish” the elements based on their understanding of the elements and how they performed.

The following shows some of the models and results using the current elements in MSC.NASTRAN (V2001).


PATCH TEST CLASSICAL

Use regular shape, irregular pattern Non-redundant constraints Apply boundary forces or displacements for constant stress.

The patch tests check Convergence Compatibility


PATCH TEST FOR SOLIDS

X Y Z 1 .249 .342 .192 2 .826 .288 .288 3 .850 .649 .263 4 .273 .750 .230 5 .320 .186 .643 6 .677 .305 .683 7 .788 .693 .644 8 .165 .745 .702

Location of Inner Nodes

Outer Dimensions: unit cubeE = 1.0 x 106; v = 0.25


STRAIGHT CANTILEVER BEAM

Length = 6.0 E = 1.0 x 107

Width = 0.2 v = 0.30Depth = 0.1 Mesh = 6 x 1Loading: Unit forces at free endNote: All elements have equal volume.


TWISTED BEAM

Undeformed Geometry

Length = 12.0 E = 29.0 x 106

Width = 1.1 v = 0.22Depth = 0.32 Mesh = 12 x 2Twist = 90o (root to tip)Loading: unit forces at tip


TYPES OF GEOMETRIC DISTORTION FROM A SQUARE PLATE

Aspect ratio

Skew

Taper (2 directions)

Warp

Reasonable Limits

ba

Up to 10:1Normally < 4:1

3020 −≤δ

ah Up to ~5% is acceptable normally

No real limit, but element doesnot include warpage

Ta/Qa = 0.5 – 0.75Ta = Largest of the Areas of Triangles formed at each corner grids.

Qa = Area of the Quadrilateral.


RECTANGULAR PLATE

a = 2.0

b = 2.0 or 10.0 Thickness = 0.0001 (plates) Thickness = 0.01 (solids)

E = 1.7472 x 107 ν = 0.3

Boundaries = Simply supported or clamped Mesh = N x N (on ¼ of plate)

Loading: Uniform pressure q = 10-4 Or Central load P = 4.0 x 10-4


RESULTS OF RECTANGULAR PLATE CLAMPED SUPPORTS: CONCENTRATED

LOADSAspect Ratio = 1.0

Normalized Lateral Deflection at Center Number of Node Spaces per Edge of

Model

QUAD2 QUAD4 QUAD8 HEXA(8) HEX20 HEX20(R)

2 .979 .994 1.076 .855 .002 .983 4 1.008 1.010 .969 .972 .072 .433* 6 1.006 1.012 .992 .988 .552 .813 8 1.005 1.010 .997 .994 .821 .942

* This is the correct value

Normalized Lateral Deflection at Center Number of Node Spaces per Edge of

Model

QUAD2 QUAD4 QUAD8 HEXA(8) HEX20 HEX20(R)

2 .333 .519 .542 .321 .001 .363 4 .512 .863 .754 .850 .041 .447 6 .638 .940 .932 .927 .220 .721 8 .723 .972 .975 .957 .374 .867

Aspect Ratio = 5.0

Standard Isoparametric Reduced Integration


SCORDELIS-LO ROOF

Radius = 25.0 Length = 50.0 Thickness = 0.25 Loading = 90.0 per unit area in Z-direction on curved edges Mesh = N x N on shaded area


RESULTS FOR SCORDELIS-LO ROOFVertical Deflection at Midpoint of Free Edge (Results/Theory) Number of Node

Spaces per Edge of Model

QUAD2 QUAD4 QUAD6 HEXA(8) HEX20 HEXA20(R)

2 .784 1.376 1.021 1.320 .092 1.046 4 .665 1.050 .984 1.028 .258 .967 6 .781 1.018 1.002 1.012 .589 1.003 8 .854 1.008 .997 1.005 .812 .999

10 .897 1.004 .996 - - -


SUMMARY OF SOME SAMPLE TEST PROBLEMS

Problem Suitability of Problem for Element TypeBeam (1D) Plate (2D solid) Plate (Shell) Solid (3D)

Patch Test NO YES YES YES

Straight Cantilever Beam YES YES YES YES

Twisted Beam YES NO YES YES

Rectangular Plate NO NO YES YES*

Scordelis-Lo Roof NO NO YES YES

Spherical Shell NO NO YES YES

Thick walled Cylinder NO YES** NO YES

* only for Thicknes = 0.01, (NO for Thickness = 0.0001

** with Plane Strain Option


RECOMMENDATIONS Plates

QUAD4 Is not very sensitive to aspect ratios QUAD4 Membrane behavior can be sensitive to

irregular irregular shapes TRIA3 Is nearly as good but has some aspect ratio

sensitivity TRIA3 Tends to be stiff for membrane loads TRIA3 May have some local stress perturbations when

mixed with QUAD4 QUAD8 Has best results for curved shapes QUAD8 Should not span more than 20 – 30 degrees QUAD8 Should have midsize nodes placed accurately.


RECOMMENDATIONS (Cont.)

Solids HEXA Recommendation is eight nodes and strain

function integration (0) HEXA May be used as plates with strain function

integration HEXA Gives poor results with irregular shapes HEX20 Recommended for irregular shapes and curved

surfaces (use reduced integration) PENTA Recommended for irregular shapes PENTA May exhibit local irregular when mixed with

HEXA


MPCs AND “R”- TYPE ELEMENTS


MULTIPOINT CONSTRAINTS (MPC)

Each MPC entry is used to specify one displacement (Um ) as a linear combination of one or more other displacements(Un ).

Nastran divides the G-set into 2 sets.M = dependent DOFs,N = independent DOFs

Then performs the reduction from the G to N set.


MULTIPOINT CONSTRAINTS (MPC) (Cont.) General form for MPC equations:

where

The equations for all MPCs and R-type elements are assembled to form the constraint equations:

0UAUAj

NNMM jjii=+ ∑

=

=

=

=

j

i

j

i

N

M

N

M

UU

AA Scaling coefficient for the dependent DOFs

Scaling coefficient for the independent DOFs

Displacement of dependent DOFs

Displacement of independent DOFs

0URUR NNMM =+


MULTIPOINT CONSTRAINTS (MPC) (Cont.) This can be written as

The G-set matrices are rewritten as follows:

As explained in The MSC.NASTRAN REFERENCE GUIDE, Section 9.4.3, the equation

NMNN1

MM UGURRU =−=↑

−

Therefore, RM must not be singular

{ } NNN

MN

N

MG U

IG

UU

U

=

=

[ ][ ] [ ]gggg PUK =

=

=

M

N

M

N

MMTNM

NMNN

PP

UU

kkkk


MULTIPOINT CONSTRAINTS (MPC) (Cont.)

Becomes

where

[ ][ ] [ ]NNNN PUK =

[ ] [ ]{ } { }M

TMNN

MMMTM

TNM

TMMNMNNNN

PGPPGkGkGGkkk

+=

+++=


MPC – BULK DATA ENTRYDefines a multipoint constraint equation of the form

where uj represents degree of freedom Cj at grid scalar point Gj

0u Ajj

j =∑

Format: 1 2 3 4 5 6 7 8 9 10

MPC SID G1 C1 A1 G2 C2 A2 G3 C3 A3 -etc.-

Example:

MPC 3 28 3 6.2 2 4.29 1 4 -2.91

Field Contents SID Set identification number: (Integer > 0) Gj Identification number of grid or scalar point: (Integer > 0) Cj Component number. (Any one of the integer 1 through 6 for grid

points; blank or zero for scalar point). Aj Coefficient (Real; Default = 0.0 except A1 must be nonzero).


MPC – BULK DATA ENTRY (Cont.)Remarks:

1. Multipoint constraint sets must be selected with the Case Control command MPC = SID.

2. The first degree of freedom (G1, C1) in the sequence is defined to be the dependent degree of freedom assigned by one MPC entry cannot be assigned dependent by another MPC entry or by a rigid element.

3. Forces of multipoint constraint may be recovered in the linear structured solution sequences (101 – 200) with the MPCFORCE Case Control command and in SOL 24 with RF Alter RF23D24 (see the MSC.NASTRAN Reference Manual, Chapter 15).

4. The m-set degrees of freedom specified on this entry may not be specified on other entries that define mutually exclusive sets. See the MSC.NASTRAN Quick Reference Guide, Appendix B for a list of these entries.

5. By default, the grid point connectivity created by the MPC, MPCADD, and MPCAX entries is not considered during resequencing, (see the PARAM, NEWSEQ description in the MSC.NASTRAN quick Reference Guide, Section 6). In order to consider the connectivity during resequencing, SID must be specified on the PARAM, MPCX entry. Using the example above, specify PARAM, MPCX, 3.


MULTIPOINT CONSTRAINT EXAMPLE1. Thick plate with bars attached.

Using plate theory assumption (“plane sections remain plane”), we can write the equations for the in-plane motion of Grid Points 2 and 3 as function of the motion of Grid point 1.

113

112

611

611

u*5.uuu*5.uu

+=

−=

]2 1 . . u :[ 21 gridatcompdispNote =

1313

1212

6622

6622

;

;

uuuu

uuuu

==

==


MULTIPOINT CONSTRAINT EXAMPLE (Cont.)

Notes: Select MPC = 1 in the Case Control to use these entries.

The MPC equations are written using the displacementcoordinate system of the GRID points. If GRID points involvedin MPC have different coordinate system, be careful, to avoid“grounding” the structure.

MPC, 1, 2, 1, 1., 1, 1, -1., ,+MPC1A+MPC1A, , 1, 6, .5MPC, 1, 3, 1, 1., 1, 1, -1., ,+MPC1B+MPC1B, , 1, 6, -.5MPC, 1, 2, 2, 1., 1, 2, -1.MPC, 1, 3, 2, 1., 1, 2, -1.MPC, 1, 2, 6, 1., 1, 6, -1.MPC, 1, 3, 6, 1., 1, 6, -1.

MPC entries are:



1A. If GRID 1 has the following displacement coordinate system (CID =1), and GRID 2 and 3 have the basic displacement coordinate system, the MPC equations would look like:

XY CORD2R, 1, , 0., 0., 0., 0., 0., 1.,

, 0., 1., 0.GRID, 1, , 4., .5, 0., 1 $ CD = 1MPC, 1, 2, 1, 1., 1, 2, 1., ,&MPC1A+MPC1A, , 1, 6, 0.5MPC, 1, 3, 1, 1., 1, 2, 1., ,&MPC1B+MPC1B, , 1, 6, -0.5$MPC, 1, 2, 2, 1., 1, 1, -1.MPC, 1, 3, 2, 1., 1, 1, -1.$MPC, 1, 2, 6, 1., 1, 6, -1.MPC, 1, 3, 6, 1., 1, 6, -1.



2. Calculate the distance between two points:Example:There is a tolerance requirement on a structure. We want to know the clear distance between two specified points.

δo = 10.0 = Initial clearance



Total distance δ = δo + 1112uu −

Use a scalar point to represent the distance δ.

* Call out MPC = 10 in the Case Control Section

** Call out SPC = 1 in the Case Control Section

$ Component 1 of node 1001 is SPC'ed to 10.0 (the$ original distance between node 1 and node 2)GRID, 1001SPC, 1, 1001, 1, 10.0$ Create a scalar variable to hold (final) distance$ between node 1 and node 2SPOINT, 1000$ Set MPC: Ux1000 = Ux1001 + Ux2 - Ux1MPC, 10, 1000, 0, 1., 1001, 1, -1., ,+MPC10A+MPC10A, , 2, 1, -1., 1, 1, 1.



3. Average displacement

To get the average displacement of Grid Points 1 and 2 of the previous example, add the following:

221 11 uu +

=∆

$ Create a scalar variable (1002) to hold the average$ distance between node 1 and node 2SPOINT, 1002$MPC, 10, 1002, 0, 1., 1, 1, -0.5, ,+MPC10B+MPC10B, , 2, 1, -0.5



4 Enforce a relative gap. In order to do this, we wish to constrain the SPOINT to have the

desired gap. Therefore, the SPOINT must be independent on the MPC and we need to re-write the MPC entries.

* Call out MPC = 10 in the Case Control Section** Call out SPC = 1 in the Case Control Section

$ Create a scalar variable (1000) to hold the final$ distance (clearance) between node 1 and node 2SPOINT, 1000$ Relative gap (final) = (Ux2 - Ux1) + gap (initial)$ MPC Eq.: Ux1000 = Ux2 - Ux1 + Ux1001 input as:$ -Ux1 +Ux2 +Ux1001 -Ux1000 = 0.MPC, 10, 1, 1, -1., 2, 1, 1., ,+MPC10A+MPC10A, , 1001, 1, 1., 1000, 0, -1.$SPC, 1, 1001, 1, 0.02 $Initial gapSPC, 1, 1000, 0, 0.001 $Final gap to be allowed


MULTIPOINT CONSTRAINTSEXAMPLE (Cont.)

MPC at Selective Mesh Refinement

U8 = 0.5*(U7 + U9) ; U17 = 0.5*(U9 + U18)


“R” – TYPE ELEMENTS


THE “R” ELEMENTS Generate internal MPC equations that eliminate dependent

degrees of freedom. (The user selects the dependent points.) Are automatically included in the solution (the “MPC =”

command does not effect R–elements) “R” elements do not account for nonlinear, mass, thermal

expansion, or heat transfer effects. Internal forces and grid point forces are calculated for

output.(MPCFORCE case control request – output is by GRID point, not by R–element. The output at each GRID point is the summation from all R–elements and MPC’s connected to it. The MPCForces should sum to zero, if no structural element is attached at the grid.)


SAMPLE USES OF “R” ELEMENTS

When very stiff structure sections are inconvenient to model or numerically troublesome

When different pieces of the model are mismatched and the grid points cannot be connected with conventional elements

If connecting joints are free to slide and/or rotate in specific directions

When elements are offset from grid points To distribute input loading or enforced motions To connect incompatible elements


COMMONLY USED “R” ELEMENTS RBAR “Rigid” bar connecting two grid points with 6

independent and 1-6 dependent DOFs RBE2 “Rigid element with six independent DOF’s at one

grid point and any number of dependent DOF’s RBE3 “Interpolation” element with 1 – 6 dependent DOF’s

and any number of independent DOF’s. Used for distributing loads or obtaining average displacement.No stiffness is introduced by RBE3.

RSPLINE “Interpolation” element with any number of independent and dependent DOF’s. Uses the displacement pattern of a beam element based on the independent DOF’s to obtain displacements of the dependents DOF’s.

RSSCON “Interpolation” element used to connect shell elements to solid elements (added in v69).


RBE2 and RBAR These are “rigid elements“ Simpler versions of RBE1 RBAR connects to only two grid points with a total of

six independent degrees of freedom. The independent DOF are selected by you, but must be able to define any motion is space.

RBE2 has one independent grid point (all six degrees of freedom) and any number of dependent grid points.


RBAR: The CONSTRAINT EQUATIONS

Equations for RBAR A-SAVector Notation:

Component Notation:

][][][][ ASAASA ruu ×+= φ][][ ASA φφ =

zAzSA

yAySA

xAxSA

ASyAASxAzAzSA

ASxAASzAyAySA

ASzAASyAxAxSA

xxyyuuzzxxuuyyzzuu

φφφφφφ

φφφφφφ

=

==

−⋅−−⋅+=

−⋅−−⋅+=

−⋅−−⋅+=

)()(

)()(

)()(


RBAR – BULK DATA ENTRYDefines a rigid bar with six degrees of freedom at each end Format:

1 2 3 4 5 6 7 8 9 10 RBAR EID GA GB CNA CNB CMA CMB

Example:

RBAR 5 1 2 234 123

Field Contents

EID Element identification number.

GA, GB Grid point identification number of connection points. (Integer > 0).

CNA, CNB Component numbers of independent degrees of freedom in the global coordinate system for the element at grid points GA and GB. See Remark 1. See Remarks 2 and 3. (Integer 1 through 6 with no embedded blanks, or zero or blank).

CMA, CMB Component numbers of dependent degrees of freedom in the

global coordinate system assigned by the element at grid points GA and GB. See Remarks 2 and 3. (Integers 1 through 6 with no embedded blanks, or zero or blank).


RBAR – BULK DATA ENTRY (Cont.)Remarks:

1. The total number of components in CNA and CNB must equal six; for example, CNA = 1236, CNB = 34. Furthermore, they must jointly be capable of representing any general rigid body motion of the element.

2. If both CMA and CMB are zero or blank, all of the degrees of freedom not in CNA and CNB will be made dependent; i.e, they will be made members of the m-set.

3. The m-set coordinates specified on this entry may not be specified on other entries that define mutually exclusive sets. Se the MSC.NASTRAN Quick Reference Guide, Appendix B for a list of these entries.

4. Element identification numbers must be unique.5. Rigid elements, unlike MPC’s, are not selected through the Case Control Section.6. Forces of multipoint constraint may be recovered in the linear structured solution

sequences (101 – 200) with the MPCFORCE Case Control command and in SOL 24 with RF Alter RF24D24 (see the MSC.NASTRAN Reference Manual, Chapter 15).

7. Rigid elements are ignored in heat transfer problems.8. See the MSC.NASTRAN Reference Manual, Section 5.5, for a discussion of rigid

elements.


RBE2 – BULK DATA ENTRYDefines a rigid body whose independent degrees of freedom are specified at a single grid point and whose dependent degrees of freedom are specified at an arbitrary number of grid points. Format:

1 2 3 4 5 6 7 8 9 10 RBE2 EID GN CM GM1 GM2 GM3 GM4 GM5

GM6 GM7 GM8 -etc.-

Example: RBE2 9 8 12 10 12 14 15 16

20

Field Contents EID Element identification number. GN Identification number of grid point to which all six independent

degrees of freedom for the element are assigned. (Integer > 0) CM Component numbers of the dependent degrees of freedom in

the global coordinates system at grid points GMi. (Integers 1 through 6 with no embedded blanks).

GMi Grid point identification numbers at which dependent degrees of freedom are assigned. (Integer > 0).


RBE2 – BULK DATA ENTRY (Cont.)Remarks:

1. The components indicated by CM are made dependent (members of the m-set) at all grid points GMI.

2. Dependent degrees of freedom assigned by one rigid element may not also be assigned dependent by another rigid element or by a multipoint constraint.

3. Element identification numbers must be unique.4. Rigid elements, unlike MPCs are not selected through the Case Control

Section.5. Forces of multipoint constraints may be recovered in the linear structured

solution sequences (101 – 200) with MPCFORCE Case Control command and in SOL 24 with Alter RF24D24 (see the MSC.NASTRAN Reference Manual, Chapter 15).

6. Rigid elements are ignored in heat transfer problems.7. See the MSC.NASTRAN Reference Manual,Section 5.5, for a discussion of

rigid elements.8. The m-set coordinates specified on this entry may not be specified on other

entries that define mutually exclusive sets. See the MSC.NASTRAN Quick Reference Guide, Appendix B for a list of these entries.


The RBE2

One independent GRID (all 6 DOF) Multiple dependent GRID/DOFs

Note: Small Rotations assumed


RBE2 Example

Rigidly “weld” multiple GRIDs to one other GRID:

32RBE2 4110199 123456GM5GM3GM2RBE2 GM4GM1GNEID CM

13

2

101

4

No relative motion between GRIDs 1-4 !No deformation of element(s) between these GRIDs


RBE2 Example

Note: No relative motion between GRIDs 1-4 ! No deformation of element(s) between these GRIDs

32RBE2 4110199 123456GM5GM3GM2RBE2 GM4GM1GNEID CM

13

2

101

4


Common RBE2/RBAR Uses

RBE2 or RBAR between 2 GRIDs “Weld” 2 different parts together

6 DOF Connection “Ball Joint” 2 different parts together

3DOF connection

RBE2 “Spider” or “wagon wheel” connections Large mass/base-drive connection


RBE3 – THE “WIFFLETREE” The RBE3 is an “interpolation” element. Example:

Basic equations

Select one to six dependents from any DOF. Number of dependents = number of Uref .

ii

iiiref

iiref

WT

UrWT

UWTU

θ

θ

∑∑∑

+

×=

=

)( > >


RBE3 – BULK DATA ENTRYDefines the motion at a reference grid point as the weighted average of the motions at a set of other grid points.

Format:

1 2 3 4 5 6 7 8 9 10 RBE3 EID REFGRID REFC WT1 C1 G1,1 G1,2

G1,3 WT2 C2 G2,1 G2,2 -etc.- WT3 C3 G3,1 G3,2 -etc.- WT3 C4 G4,1 G4,2 -etc.- “UM” GM1 CM1 GM2 CM2 GM3 CM3 GM4 CM4 GM5 CM5 -etc.-

Example:

RBE3 14 100 1234 1.0 123 1 3 5 4.7 1 2 4 6 5.2 2 7 8 9 5.1 1 15 16 UM 100 14 5 3 7 2

Note: RBE3 degenerates into RBAR if there is only 1 Gij entry..


RBE3 – BULK DATA ENTRY (Cont.)Field Contents EID Element identification number. Unique with respect to other rigid

elements. (Integer > 0). REFGRID Reference grid point identification number. (Integer > 0).

REFC Component numbers at the reference grid point. (Any of the Integers 1 through 6 with no embedded blanks).

WTi Weighted factor for components of motion on the following entry at grid points Gi, j. (Real).

Ci Component numbers with weighted factor WTi at grid point Gi, j. (Any of the Integers 1 through 6 with no embedded blanks).

Gi, j Grid points whose components Ci have weighted factor WTi in the averaging equations. (Integer > 0).

“UM” Indicates the start of the degrees of freedom belonging to the m-set. The default action is to assign only the components in REFC to the m-set. (Character).

GMi Identification numbers of grid points with degrees of freedom in the m-set. (Integer > 0).

CMi Component numbers of GMi to be assigned to the m-set. (Any of the integers 1 through 6 with no embedded blanks).



1. It is recommended that for most applications only the translation components 123 be used for Ci. An exception is the case where the Gi,j are collinear. A rotation component may then be added to one grid point to stabilize its associated rigid body mode for the element.

2. Blank spaces may be left at the end of a Gi, j sequences.3. The default for “UM” should be used except in cases where the user wishes

to include some or all REFC components in displacement sets exclusive from the m-set. If the default is not used for “UM”.

a. The total number of components in the m-set (I.e.., the total number of dependent degrees of freedom defined by the element) must be equal to the number of components in REFC (four components in the example).

b. The components specified after “UM” must be a subset of the component specified under REFC and (Gi, j, Ci).

c. The coefficient matrix [Rm] described in the MSC.NASTRAN Reference Manual, Section 9.4.3 must be nonsingular. SOL 60 or PARAM, CHECKOUT in SOLs 101 – 200 may be used to check for this condition.

4. Dependent degrees of freedom assigned by one rigid element may not also be assigned dependent by another rigid element or by a multipoint constraint.



5. Rigid elements, unlike MPCs, are not selected through the Case Control section.

6. Forces of multipoint constraint may be recovered in the linear structured solution sequences (101 – 200) with the MPCFORCE Case Control command and SOL 24 with RF Alter RF24D24 (see the MSC.NASTRAN Reference Manual, Chapter 15).

7. Rigid elements are ignored in heat transfer problems.8. The m-set coordinates specified on this entry may not be specified

on other entries that define mutually exclusive sets. See the MSC.NASTRAN Quick Reference Guide, Appendix B for a list of these entries.

9. The formulation for RBE3 was change in V70.7. It now gives unit independent consistent answers. Only models that connected rotation DOF for Ci (ignoring the recommendation in Remark 1) are affected. The formulation prior to V70.7 may be obtained by setting SYSTEM(310)=1.


RBE3 Description

By default, the reference grid DOF will be the dependent DOF.

Number of dependent DOF is equal to the number of DOF on the REFC field.

Dependent DOF cannot be SPC’d, OMITted, SUPORTed or be dependent on other RBE/MPC elements.


RBE3 Is Not Rigid!

RBE3 vs. RBE2 RBE3 allows warping and 3D effects In this example, RBE2 enforces beam

theory (plane sections remain planar)

RBE3 RBE2


RBE3: How it Works?

Forces/Moments applied at reference grid are distributed to the master grids in same manner as classical bolt pattern analysis. Step 1: Applied loads are transferred to the CG of the weighted grid

group using an equivalent Force/Moment Step 2: Applied loads at CG transferred to master grids according

to each grid’s weighting factor

Note: If independent DOFs contain rotations, RBE3 does not work like classical bolt pattern analysis.


RBE3: How it Works?

If independent DOFs include rotations, moments at CG are mapped as equivalent force couples, and concentrated moments.

Step 1: Transform force/moment at reference grid to equivalent force/moment at weighted CG of master grids.

MCG=MA+FA*e

FCG=FA

CG

FCG

MCG

FA

MA

Reference Grid

e

CG


RBE3: How it Works?

Step 2: Move loads at CG to master grids according to their weighting values. Force at CG divided amongst master grids according to weighting

factors Wi

Moment at CG mapped as equivalent force couples on master grids according to weighting factors Wi


RBE3: How it Works?

Step 2: Continued…

CG

FCG

MCG

Total force at each master node is sum of...Forces derived from force at CG: Fif = FCG{Wi/ΣWi}

F1m

F3mF2m

Plus Forces derived from moment at CG: Fim = {McgWiri/(W1r1

2+W2r22+W3r3

2)}


RBE3: How it Works?

Masses on reference grid are smeared to the master grids similar to how forces are distributed Mass is distributed to the master grids according to their weighting

factors Motion of reference mass results in inertial force that gets

transferred to master grids Reference node inertial force is distributed in same manner as

when static force is applied to the reference grid.


Example 1

RBE3 distribution of loads when force at reference grid at CG passes through CG of master grids


Example 1: Force Through CG

Simply supported beam 10 elements, 11 nodes numbered 1 through 11

100 LB. Force in negative Y on reference grid 99



Load through CG with uniform weighting factors results in uniform load distribution



Comments… Since master grids are co-linear, the x rotation DOF is added so that

master grids can determine all 6 rigid body motions, otherwise RBE3 would be singular

RBE3, 11, ,99, 123456, 1., 123, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

*** USER FATAL MESSAGE 2038 (RBE3D)USER ACTION: ADD MORE DOFS TO THE CONNECTED POINTS TO

INSURE THAT THEY CAN CONSTRAIN ALL 6 RIGID BODY MODES OF THE ELEMENT.

Corrected RBE3 data (add DOF4 to one or more Master Grids):

RBE3, 11, ,99, 123456, 1., 1234, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11


Example 2

How does the RBE3 distribute loads when force on reference grid does not pass through CG of master grids?


Example 2: Load not through CG

The resulting force distribution is not intuitively obvious Note forces in the opposite direction on the left side of the beam.

Upward loads on left side of beam result from moment caused by movement of applied load to the CG of master grids.


RSPLINE – THE LINEAR SPLINE The RSPLINE is an “interpolation” element using the

equations of a beam – Example


RSPLINE – THE LINEAR SPLINE (Cont.)

Basic equations

In matrix notation

xU,

xU,U

xCxCxCC)x(U

2i

2i

i

3i

i3

2i

i2i

i1

i0

i

∂∂

∂∂

+++=

}]{[}{}{][}{ 1

CHUUHC

dd

ii

== −

match at grid points


RSPLINE – BULK DATA ENTRYDefines multipoint constraints for the interpolation of displacements at grid points.

Format:

1 2 3 4 5 6 7 8 9 10 SPLINE EID D/L G1 G2 C2 G3 C3 G4

C4 G5 C5 G6 -etc.-

Example: RSPLINE 73 .05 27 28 123456 29 30

123 75 123 71

Field Contents EID Element identification number. (Integer > 0).

D/L Ratio of the diameter of the elastic tube to the sum of the lengths of all segments. (Real > 0.0; Default = 0.1).

Gi Grid point identification number. (Integer >0). Ci Components to be constrained. See remark 2. (Blank or any

combination of the integers 1 through 6).


RSPLINE – BULK DATA ENTRY (Cont.)Remarks:

1. Displacements are interpolated from the equations of an elastic beam passing through grid points.

2. A blank field for Ci indicates that all six degrees of freedom of Gi are independent. Since G1 must be independent, no field is provided for C1. Since the last grid point must also be independent, the last field must be a Gi, not a Ci. For the example shown G1, G3, and G6 are independent. G2 has six constrained degrees of freedom while G4 and G5 each have three.

3. Dependent (I.e., constrained) degrees of freedom assigned by one rigid element may not be assigned dependent by another rigid element or by a multipoint constraint.

4. Degrees of freedom declared to be independent by one rigid body element can be made dependent by another rigid body element or by a multipoint constraint.

5. EIDs must be unique.6. Rigid elements (including RSPLINE), unlike MPCs, are not selected through

the Case Control Section.


RSPLINE – BULK DATA ENTRY (Cont.)Remarks: (Cont.)

7. Forces of multipoint constraint may be recovered in the linear structured solution sequences (101 –200) with the MPCFORCE Case Control command and in SOL 24 with RF Alter RF24D24 (see the MSC.NASTRAN Reference Manual, Chapter 15).

8. Rigid elements are ignored in heat transfer problems.9. See the MSC.NASTRAN Reference Manual, Section 5.5, for a discussion

of rigid elements. 10. The m-set coordinates specified on this entry may not be specified on other

entries that define mutually exclusive sets. See the MSC.NASTRAN Quick Reference Guide, Appendix B for a list of these entries.

11. The constraint coefficient matrix is affected by the order of the Gi Ci pairs on the RSPLINE entry. The order of the pairs should be specified in the same order that they appear along the line that joins the two regions. If this order is not followed then the RSPLINE will have folds in it that may yield some unexpected interpolation results.

12. The independent degrees of freedom which are the rotation components most nearly parallel to the line joining the regions should not normally be constrained.


RSSCON – BULK DATA ENTRY

Defines multipoint constraints to model clamped connections of shell-to-solid elements.

Format: 1 2 3 4 5 6 7 8 9 10

RSSCON RBID TYPE ES1 EA1 EB1 ES2 EA2 EB2

Example: RSSCON 110 GRID 11 12 13 14 15 16

RSSCON 111 GRID 31 74 75

RSSCON 115 ELEM 311 741


RSSCON – BULK DATA ENTRY (Cont.)Field Contents RBID Elements identification number. (Integer = 0 ) TYPE Type of connectivity

TYPE = “ELAM” connection is described with element identification numbers. TYPE = “GRID” connection is described with grid point identification numbers. (Character: “GRID” or “ELAM”; Default = “ELAM”)

ES1 Shell element identification number if TYPE = “ELAM”. Shell grid point identification number if TYPE = “GRID”. See Figure 1. (Integer = 0)

EA1 Solid element identification number if TYPE = “ELAM”. Solid grid point identification number if TYPE = “GRID”. (Integer = 0)

EB1 Solid grid point identification number for TYPE = “GRID”. (Integer = 0 or blank)

ES2 Shell grid point identification number if TYPE = “GRID”. (Integer = 0)

EA2 Solid grid point identification number if TYPE = “GRID”. (Integer = 0)

EB2 Solid grid point identification number if TYPE = “GRID”. (Integer = 0)


RSSCON – BULK DATA ENTRY (Cont.)Remarks:

1. RSSCON generates a multipoint constraint that models a clamped connection between a shell and a solid element. The shell degrees of freedom are put in the dependent set (m-set). The translational degrees of freedom of the shell edge are connected to the translational degrees of freedom of the upper and lower solid edge. The rotational degrees of freedom of the shell are connected to the translational degrees of freedom of the lower and upper edges of the solid element face. Poisson’s ratio effect are considered in the translational degrees of freedom.

2. The shell grid point must lie on the line connecting the two solid grid points. It can have an offset from this line, which can be not be more than 5% of the distance between the two solid grid points. The shell grid points that are out of the tolerance will not be constrained, and a fatal message will be issued. This tolerance is adjustable. Please see PARAM, TOLRSC, and PARAM, SEPIXOVR.

3. When using the TYPE = “ELAM” option.a. The elements may be p-elements. The solid elements are CHEXA, CPENTA, and

CTETRA with and without midside nodes. The shell elements are CQUAD4, CTRIA3, CQUADR, CTRIAR, CQUAD8, or CTRIA6.

b. In case of p-elements, the p-value of the shell element edge is adjusted to the higher of the p-value of the upper of lower solid p-element edge. If one of the elements is an h-elements, then the p-value of the adjacent edge is lowered to 1.


RSSCON – BULK DATA ENTRY (Cont.)Remarks: (Cont.)

c. Both the shell and solid elements have to belong to the same superelement. This restriction can be bypassed using SEELT entry to reassign the downstream boundary element to an upstream superelement.

d. When a straight shell p-element edge and a solid p-element are connected, the geometry of the shell edge is not changed to fit the solid face. When a curved shell p-element edge and a solid p-element are connected, the two solid edges and solid face are not changed to match the shell edge.

e. It is not recommended to connect more than one shell element to the same solid using the ELEM option. If attempted, conflicts in the multipoint constraint relations may leads to UFM 6692.

4. When using TYPE_”GRID” optiona. The GRID option does not verify that the grids used are valid shell and/or solid grids.b. The hierarchical degrees of freedom of p-element edges are not constrained. The

GRID option is therefore not recommended for p-elements.c. The grids in the GRID option can be different superelements. The shell grid must be

in the upstream superelement.5. It is recommended that the height of the solid element’s face is approximately

equal to the shell element’s thickness of the shell. The shell edge should then be placed in the middle of the solid face.


RSSCON – BULK DATA ENTRY (Cont.)Remarks: (Cont.)

6. The shell edge may coincide with the upper or lower edge of the solid face. 7. The RSSCON entry, unlike MPCs, cannot be selected through the Case

Control Section.8. Forces of multipoint constraints may be recovered in the linear structured

sequences (SOLs 101 through 200) with MPCFORCE Case Control command.

9. The RSSCON is ignored in heat-transfer problems.10. The shell edge may coincide with the upper or lower edge of the solid face.11. The RSSCON entry, unlike MPCs, cannot be selected through the Case

Control Section.12. Forces of multiple constraints may be recovered in the linear structure

solution sequences (SOLs 101 through 200) with the MPCFORCE Case Control command.

13. The RSSCON is ignored in heat-transfer problems.14. The m-set coordinates (shell degrees of freedom) may not be specified on

other entries that define mutually exclusive sets. See the MSC.NASTRAN Quick Reference Guide, Appendix B for a list of these entries.


RSSCON – BULK DATA ENTRY (Cont.)

Figure 1. Shell Elements Connected to the Faces of Solid Elements


SECTION 3

TRICKS


Section PageTRICKS 3-5ELEMENT FORCE DISTRIBUTIONS FOR UNIFORM STRESS 3-6DISPLACEMENT COORDINATES SYSTEMS 3-9GRIDS POINTS 3-15OFFSETS ON BARS, BEAMS AND PLATES 3-16HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS 3-18SHEAR CENTER – BEAMS AND BARS 3-32BEAM VS. BAR 3-34HOW ARE OFFSETS IMPLEMENTED ON PLATES 3-35HANDY HINTS 3-36MORE HANDY HINTS 3-40COMPATIBILITIES 3-41MODELING CORNERS WITH PLATES AND BEAMS 3-43EXAMPLE OF POOR PRACTICE 3-45EQUIVALENT ROD 3-46EQUIVALENT ROD EXAMPLE RESULTS 3-49

TABLE OF CONTENTSTABLE OF CONTENTS


TABLE OF CONTENTSSection PagePLATES AND SHELLS IN-PLANE ROTATIONAL STIFFNESS 3-50

MESH TRANSITIONS 3-54

BEAM-TO-PLATE ELEMENTS 3-56

MORE MESH TRANSITIONS 3-62

SOME POSSIBLE PLATE-TO-SOLID TRANSITION 3-64

STRESS SORTING 3-66

DTI, INDTA – BULK DATA ENTRY 3-68

STRESS SORTING – SAMPLE 3-72


TRICKS

For the sake of discussion, let us define a trick as something you would not expect unless someone told you or you had already experienced it.


ELEMENT FORCE DISTRIBUTIONS FOR UNIFORM STRESS

Pressure loadings on elements are replaced by forces at the corners. For a simple element, such as the QUAD4 or TRIA3, the force is simply

the total force divided by the number of connecting GRID points. For higher order elements, the forces are found based on the element

formulation, and may not agree with user’s physical intuition or common sense.

Example Grid point loads equivalent to a uniform pressure

= =


ELEMENT FORCE DISTRIBUTIONS FOR UNIFORM STRESS

Different element types have different order polynomials used for their displacement fields.

As a result, it is possible that local discontinuities will occur when you mix different element types.


ELEMENT FORCE DISTRIBUTIONS FOR UNIFORM STRESS (Cont.)

Method of Testing Put SPCs on boundary, to induce uniform stress. Measure SPCF output.


DISPLACEMENT COORDINATE SYSTEMS When you define a GRID entry in MSC.NASTRAN, there are two fields used to select

coordinate systems.

CP is the “position” coordinate system, or the system in which X1, X2, and X3 are measured.

CD is the “displacement” coordinate system. This defines the coordinate system which is used to measure the displacements at the GRID point. All constraints, MPCs, SPCs, and the BAR/BEAM offsets and orientation vector use this coordinate system.

Coordinate systems may be rectangular (X,Y,Z) cylindrical (R, θ, Z) or spherical (R, θ, Φ)

Format: 1 2 3 4 5 6 7 8 9 10

GRID ID CP X1 X2 X3 CD PS SEID

Example: GRID 2 3 1.0 -2.0 3.0 316


DISPLACEMENT COORDINATE SYSTEMS (Cont.)

In MSC.NASTRAN the following coordinate systems may be used. Basic Coordinate System – Implicitly defined reference rectangular coordinate

system (Coordinate System 0). Orientation of this system is defined by the user through specifying the components of grid point locations.

Alternate (local) Coordinate Systems – Alternate systems can be defined to facilitate geometric input. Each local system must be related directly or indirectly to the basic coordinate system. The six possible alternate coordinate systems are:

All angular coordinates are input in degrees. Output associated for these coordinates is in radians.

Rectangular

Cylindrical

Spherical

CORD1RCORD2R

CORD1CCORD2C

CORD1SCORD2S



U1 = x directionU2 = y directionU3 = z direction

Note: A, B, and C are points used to define the local coordinate system.P is a grid point defined in the local system.

Rectangular Coordinate System (x, y,z)



Point A = local system originPoint P = grid point defined in local cylindrical systemPoint C = reference point in the r-z plane at θ = 0Point B = reference point for z axis direction(Ur ,Uθ ,Uz) = displacement components of P in local system

Cylindrical Local Coordinate System (R, θ,Z)



Point A = local system originPoint P = grid point defined in local spherical systemPoint C = reference point in the x-z plane at (φ = 0)Point B = reference point for z axis direction (θ = 0)(Ur ,Uθ ,Uφ) = displacement components of P in local systemNote: θcylindrical ≠θspherical

Spherical Local Coordinate System (R, θ, φ)



Sample – define a cylindrical systemCORD2C,1, ,0., 0., 0., 0., 0., 1.

, 1.,0.,0.Define Grid Points 10 and 20 on a circle.

GRID, 10, 1, 10., 45., 0., 0GRID, 20, 1, 10., 135., 0., 0

where field 3 (CP) references cylindrical coordinate system 1 (defined elsewhere), and field 7 (CD) references basic coordinate system 0.With this definition, all grid point output associated with Grids 10 and 20 will be oriented as shown.


GRIDS POINTS (Cont.)Now, in a separate model, define Grid Points 10 and 20 as :

GRID, 10, 1, 10., 45., 0., 1GRID, 20, 1, 10., 135., 0., 1

With both CP and CD referencing cylindrical coordinate system 1, all output grid information at Grids 10 and 20 will also be in terms of radial and tangential directions.

Remember – your output is in the displacement coordinate system


OFFSETS ON BARS, BEAMS AND PLATES


OFFSETS ON BARS, BEAMS AND PLATES

GRID points connecting BAR/BEAM elements may NOT lie on the beam neutral axis.

GRID points connecting plate elements mayNOT lie on the plate mid-surface.

Shear center of Beam sections mayNOT lie on the beam neutral axis.


HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS?

Bar and Beam elements may be offset from the connecting GRID points.

The offsets are entered on the continuation of the CBAR or CBEAM entry and are in the displacement coordinate system of the GRID points

Offset Bar Sample – Axial Load

Let us look at this problem two different ways Case 1: GRID points using the basic coordinate system as the

displacement coordinate system. Case 2: Define, and Use a coordinate system 100 as the displacement

coordinate system on the GRID points


HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS? (Cont.)

First problem – using the basic coordinate systemID OFFSET, BAR MODELSOL 101CENDTITLE = TEST OF OFFSET BAR MODEL – FILE offbar.datSUBT = SHOW WHERE ELEMENT FORCE OUTPUT ISLOAD = 1DISP = ALLELFORCE = ALLGPFORCE = ALLset 999 – 7oload = 999BEGIN BULK$$ DEFINE MODEL$GRID,1,,0.,0.,0.,,123456GRID,2,,2.,0.,0.=,*(1),=,*(2.),===(4)CBAR,1,1,1,2,0.,1.,0.,,+CB1=,*(1),=,*(1),*(1),=,=,=,=,*(1)=(4)$$ OFFSET 1 UNIT IN THE Y DIRECTION – IN GRID OUTPUT COORD$+CB1,,,0.,1.,0.,0.,1.,0.*(1),===(4)PBAR,1,1,1.,12.,12.,24.,MAT1,1,30.+6,,.3pload1,1,6,FX,FR,1.,–1.ENDDATA


SAMPLE OF OFFSET BAR AXIAL LOAD


SAMPLE OF OFFSET BAR AXIAL LOAD (Cont.)



The following summarizes the preceding output OLOAD RESULTANT – this is the summation of all applied

loads about PARAM, GRDPNT (about the BASIC origin if GRDPNT is not specified) in the basic coordinate system. In this case, it verifies that the applied load was –1.0 units in the X–direction

and is offset by 1.0 units (Mz = 1.0 = the load multiplied by the offset) ALWAYS VERIFY THAT THE OLOAD RESULTANT IS

CORRECT!!! SPCFORCE RESULTANT – Similar output for the reaction

forces. This should be equal and opposite to the OLOAD RESULTANT. If not, Use GROUNDCHECK Case Control Command.

DISPLACEMENT VECTOR – displacement at the GRID point in the displacement coordinates. These displacements are in the BASIC coordinate system and verify that

the loading is an axial load (there is only axial displacement). Notice that the OLOAD RESULTANT correctly shows a resultant moment about the origin, but the element is offset by 1.0 units.



LOAD VECTOR – is the loading at the GRID points in the displacement coordinate system. In this case, we can now see both the axial component (–1.0 in the X–

direction) and the moment due to the offset.

FORCE DISTRIBUTION IN BAR ELEMENTS – forces in the elements (in the element coordinate system) Once again, this verifies that the loading is an axial loading only, with the

element output showing only axial force. NOTE – the PLOAD1 is a loading applied on the element. In

MSC.NASTRAN, the element is considered to begin and end at the offset locations. Therefore, the applied load is along the axis of the element.

GRID POINT FORCE BALANCE – Where is the moment?

Remember that this output is in the displacement coordinate system of the grids. Each element is offset from the GRID points by 1.0 units (internally a rigid offset), and the elements transfer the axial force and the resulting moment to the GRID points.



Case 2: GRID points using the coordinate system 100 as the displacement coordinate system.

ID OFFSET, BAR MODELSOL 101CENDTITLE = TEST OF OFFSET BAR MODEL – FILE offbar100.datSUBT = USE DISPLACEMENT COORDINATE SYSTEMLOAD = 1DISP = ALLELFORCE = ALLGPFORCE = ALLset 999 – 7oload = 999BEGIN BULK$CORD2R,100,,0.,0.,0.,0.,0.,1.,,0.,1.,0.GRID,1,,0.,0.,0.,100,123456GRID,2,,2.,0.,0.,100=,*(1),=,*(2.),===(4)CBAR,1,1,1,2,1.,0.,0.,,+CB1=,*(1),=,*(1),*(1),=,=,=,=,*(1)=(4)$ OFFSET 1 UNIT IN THE BASIC Y DIRECTION –$ THIS IS IN THE +X IN GRID OUTPUT COORD+CB1,,,1.,0.,0.,1.,0.,0.*(1),===(4)PBAR,1,1,1.,12.,12.,24.,MAT1,1,30.+6,,.3pload1,1,6,FX,FR,1.,–1.ENDDATA



Since coordinate system 100 is the displacement coordinate system, the element offsets will be in the x-direction.

NOTE – For this example, orientation vectors of< 1, 0, 0> are used for the BAR elements. Although this appears to be parallel to the elements, it is not, because the orientation vector of BAR and BEAM elements is defined using the displacement coordinate system of the first GRID it connects to.


OFFSET BAR USING DISPLACEMENT COORDINATES


OFFSET BAR USING DISPLACEMENT COORDINATES (Cont.)


HOW ARE OFFSETS IMPLEMENTED ON BARS AND BEAMS?

The preceding output is similar to that obtained in the original run.

The OLOAD RESULTANT, SPCFORCE RESULTANT, and ELEMENT FORCES are identical. The RESULTANTs are in BASIC, and the ELEMENT FORCES are in the

element coordinate system.

The DISPLACEMENTs are identical, however, they are transformed into the displacement coordinate system (100), so they now show up as Y–translations.

The LOAD VECTOR is also in the displacement coordinate system We still see the 1.0 unit load and the moment due to the offset, but now

they are in system 100.

GRID POINT FORCE BALANCE – now we see the identical results as before, but they are in the displacement coordinate system (system 100).


SHEAR CENTER – BEAMS AND BARS One of the most common modeling errors is to ignore

the offset between the shear center and the neutral axis on a BAR or BEAM element when the cross-section is not doubly-symmetric.

Most users use the BAR element. The BAR element assumes that the shear center and the neutral axis are coincident.


SHEAR CENTER – BEAMS AND BARS (Cont.)

The BEAM element allows the shear center and the neutral axis to be offset from each other.

The offset on the CBEAM entry is from the GRID points to the shear center. The PBEAM entries (N1A, N2A, N1B,N2B) are the offset from the shear center to the neutral axis.

NOTE – the CI, DI, EI, and FI (stress recovery location) are relative to the shear center –they are NOT relative to the neutral axis.

Element loads located at the shear center; not at the neutral axis.

Once again, if the section is doubly-symmetric, or if there is no load effecting the offset, BAR will work fine in linear analysis. Otherwise use BEAM.


BEAM Vs. BAR

FEATURES CBEAM CBARVariable X-Section YES NOWarping X-Section YES NOShear Relief YES NOShear Center Offset YES NOMass Moment of Inertia YES NOGeometric Nonlinear YES NOPlastic Hinges YES NO


HOW ARE OFFSETS IMPLEMENTED ON PLATES

GRID points connecting plate elements may NOT lie on the plate mid-surface.

The plate offsets (ZOFFS) are entered on the CQUADi, and CTRIAi cards. A positive value of ZOFFS implies that the element reference plane is offset a distance of ZOFFS along the positive z-axis of the the element coordinate system. Material matrices and stress fiber locations are relative to the reference plane.


HANDY HINTS Element strain energy (ESE) is a good tool for

determining where to make changes to obtain maximum benefits.

Example – two springs in seriesWhich is the best one to stiffen to reduce the tip deflection?

where K1 = 10K2 = 1P = 1

2

21

∆= KESE

201

21

21,K

1

221111 ==∆=

KPKESEfor

11

KPwhere =∆


HANDY HINTS (Cont.)

Most of the ESE is in the smaller spring. Therefore, stiffening it is the most efficient way to reduce the tip deflection.

Normal modes analysis is similar in that changing the stiffness of the elements with the most ESE in a mode is often the most efficient way to shift the frequency of that mode.

( )2

1212

212222 K

PKP

KPK

21K

21ESE,K for

−+=−= ∆∆

21

KPK

21

2

22 =

=

11 K

P where =∆

212 K

PKP

+=∆


HANDY HINTS (Cont.) Determine mesh size based on behavior – in areas of

high stress variation, place extra elements from stiffener on a stiffened cylinder – normally 3+ elemrt61.

Plot of Moment versus Distancefor a Pressurized Stiffened Cylindrical Shell


HANDY HINTS (Cont.) Curved Shells

Normally QUAD4 can cover a 5–10° angle on a cylindrical surface.

QUAD8 can cover a 10–25° angle.

QUAD8 without midside points uses linear interpolation. This is much less accurate than QUAD4.

A simple rule – for buckling and normal modes, there should be at least 5 GRID points per half sine wave of the deformed shape. .


MORE HANDY HINTS Plates should have same orientation for stress output.

Plate output is usually in the element coordinate system. Pressure loads on plate are applied as point loads 1/4 at each

corner on a QUAD4. The direction is based on plate orientation. A positive pressure acts in the

positive element Z direction. Plates at a corner are much softer than the actual structure

Each one does not have in-plane rotational stiffness. The model has stability, but does not properly transfer loads.


COMPATABILITIES

Plate Beam Membrane – Beam

(To improve the compatibility use MPC on θz.)


COMPATABILITIES (Cont.)

Corners


MODELING CORNERS WITH PLATES AND BEAMS

Problem – A corner is much stiffer than is represented by two plate elements coming together

Poor Solution


MODELING CORNERS WITH PLATES AND BEAMS (Cont.)

Good Solutions

(recommended properties: calculated based on ½ of each attached plate elements – use only I1 ad I2 on the PBAR or PBEAM)


EXAMPLE OF POOR PRACTICE

Preferred Practice


EQUIVALENT ROD

Assume

( ) aAE

WAEK eq 22

==

a2u

ax1

ax1

o

o

o

=

−=

−=

ε

εε

σσ

22

Length of Rod (L) = wArea of Rod (A) = ??

2

2


EQUIVALENT ROD (Cont.)Energy

Stiffness

dxax1a2dA

dA E2tdU 2

−=

= ε

20

23a

0

20 4

Etadx ax1EtU εε =

−= ∫

82

2 EtuUK =

∂∂

=

2To match this stiffness by a rod of length w , and area A:

EA/(w ) = Et/8A = wt /82

2


EQUIVALENT ROD (Cont.)

Compare with constant strain results

Factor of 2 difference!

4Et

uUK

a2u

2Eta

a E2tU

a Area a2

u

2

2

22

220

20

=∂∂

=

=

=

===

ε

εε


EQUIVALENT ROD EXAMPLE RESULTS

E = 6.9 X 104

KEQ = 2156.25 N / mnA = 2.21

Case U τmax θ 1 TRIA 4.29 77.85 .0214

16 TRIAs 5.25 319.3 .0263 1 ROD 5.03 294.5 .0251

Results


PLATES AND SHELLS IN-PLATE ROTATIONAL STIFFNESS

Q: Why are QUAD plates represented with no in-plane rotational stiffness when, in fact, the real structure does have this stiffness?

A: A real plate does have in-plane rotational stiffness, but, this is implicitly represented by the in-plane translational degrees of freedom (T1, T2) in a plate finite element. Consequently, MSC.Nastran, like most general purpose finite element programs, uses either a zero (0.0), or a relatively small value to represent the stiffness with respect to the redundant in-plane rotational DOF (R3) at a plate node. It is left up to the user to either SPC this DOF, at nodes where the connected elements have parallel, or nearly parallel normals.


PLATES AND SHELLS IN-PLATE ROTATIONAL STIFFNESS (Cont.)

PARAMS: AUTOSPC, K6ROT, SNORM:

Beside the PARAM, AUTOSPC option, Nastran also has an option (PARAM, K6ROT) for giving a relatively small stiffness to this DOF so the problem can run, without encountering singularity, or round-off problems. However, the user has to be careful against using too large a value for K6ROT, and against errors hidden by the auto-stiffness feature (see example on the following page).

PARAMETERS DEFAULT VALUES

AUTOSPC YES (all, except SOL 4,106,129,153,159)NO (in SOL 4,106,129,153,159)

K6ROT 0.0 (all except SOL 106,129,153, 159)100. (in SOL 106,129,153, 159)



The model is to predict first two natural frequencies that are compared to test results. The first mode (in and out of paper) is satisfactory; the second mode (rocking back and forth) does not match the test (too low a frequency).

Problem The plates had small in-plane rotational stiffness added using K6ROT. This unknowingly resulted in a

pin-pin end condition for the beams supporting the vibrating mass and, therefore, the low frequency.

Possible Solutions Extend the beam one grid into the plates at both ends of the beam. This takes the beam moment out

as a shear couple. Use an MPC to connect the rotational DOF to adjacent translations

Example:


MESH TRANSITIONS In general, mesh transitions are handled by modern

pre–processors (e.g. MSC.Patran), and are not as much of a concern as they were in the past.

General rules for mesh transitions Keep transitions away from areas of interest. Try to use compatible elements. If compatible elements cannot be used, use “R”-type elements to

approximate the dominant behavior.

Coarse and Fine Mesh Elements of Different Types


MESH TRANSITIONS (Cont.)

Non-Conforming Element Types

Mismatched Shapes


BEAM TO PLATE ELEMENTS Situation: Your pre–processors may not handle beam–to–plate,

beam–to–solid, or plate–to–solid connections automatically. If you have any of these connections in your model, they require

special modeling efforts. Example:

Have existing Grid Points 1 through 12 and QUADs Q1 through Q6. It is desired to attach beam 10 to QUAD mesh using Grid Point 20. Several solutions are discussed here.(Note: DOF 6 of Grid Points 1 through 12 might be constrained since QUAD plates d not have stiffness in this direction.)


BEAM TO PLATE ELEMENTS (Cont.)Option 1 Grid 20 is not added. Use offsets for BAR.

Beam 10 goes from Grid 30 to Grid 4 with offset from Grid 4 to Beam 10 center line.

Problems Unrealistic moment in plates is due to the beam offset. The in-plane rotation must be handled. Otherwise, it is a “pinned connection”

for that DOF.


BEAM TO PLATE ELEMENTS (Cont.)Option 2

Add a grid and two beams

Beam properties approximated by section of a QUAD half width and its thickness

Problems May have added extra stiffness at edge due to beams

May lose some local effects where the beam attaches to the plates

Note: DOFs 1 through 6 refer to XYZ coordinate system as defined here. In applying these solutions to another problem, note which DOFs are the out-of-plane and in-plane stiffness.


BEAM TO PLATE ELEMENTS (Cont.)Option 3

Add a grid and three triangles

Problems

Need to add RBE3 DOF 6 from Grid Point 20 to Grid Points 4, 5, 7, and 8 in DOFs 1 and 2

Must add two more elastic elements and one rigid element

Be careful not to constrain DOF 6 at GRID point 20


BEAM TO PLATE ELEMENTS (Cont.)Option 4 Add a grid and an RBE3.

RBE3 DOF 1 through 6 at Grid Point 20 to DOFs 1, 2, 3, and 5 of Grid Points 4, 5, 7, 8

Problems Lose some local effects near the beam connection


BEAM TO PLATE ELEMENTS (Cont.) Situation where the beam attaches to an existing grid

Beam 10 extends from Grid Point 30 to Grid Point 7 RBE3 Grid Point 7 DOF 6 to DOF 1, 2, and 3 of Points 4, 8, and

10 Do not SPC DOF 6 at Grid Point 7. Problems

Handling the in-plane rotation


MORE MESH TRANSITIONSSolid Plate*

RBE2 = enforce plate theory at transitionRSSCON = easy way to make the connection, especially if your

preprocessor supports it.

Use MPCs or RBEs


MORE MESH TRANSITIONS (Cont.)

Higher Order – Lower Order

If you do this, always do it away from areas of interest.


SOME POSSIBLE PLATE-TO-SOLID TRANSITION

Split Plate

Extra Element


SOME POSSIBLE PLATE-TO-SOLID TRANSITION (Cont.)

Split Solid

Note: The plate may be the same thickness as the solid.


STRESS SORTING A number of solutions (including 109 and 112) in MSC.NASTRAN offer an

option to sort and filter the resultant stresses based on user-defined criteria. This option requires PARAM, S1,0 along with the following optional entries.

DTI, INDTA – Optional, shown later

The Following parameters may be defined on PARAM entries: NUMOUT An integer parameter with a default value of –1. A positive value,

N, specifies that the output will be limited to the N elements of each type that sustain the largest magnitude of stress. (See Remark 1.) If NUMOUT = -1, the parameter BIGER controls the number of elements of each type that will appear in the stress output. In this case the stresses will be sorted. If NUMOUT = -2, the parameter BIGER still controls those elements of each type that will be output, but the stresses will be unsorted.

BIGER A real parameters with a default value of 0.0. Only elements of each type that sustain stresses whose magnitude is greater than the value of BIGER will be output. This parameter will be ignored if NUMOUT is set to a positive integer value.

NOOLD An integer parameter with a default value of –1. If this parameter is set to a positive value, the standard or unsorted form of stress output will be provided by this RF Alter. The default value of –1 suppresses the standard form of stress output.

SRTOPT Controls the scanning option to be performed


STRESS SORTING (Cont.)

Remarks1. The user may override the existing default quantity on which the sort is to be

preformed or he may specify the quantity on which the sort is to be performed for elements that have no default. This option is exercised through the following procedure:

a. Include DTI,INDTA Bulk Data entry in the Bulk Data Section (see the MSC.NASTRAN Quick Reference Guide).

2. Sorting large amounts of data can be expensive.3. The parameters SRTELTYP refers to element type numbers that are provided in the

table given.

SRTOPT Description 0 Filter/sort on maximum magnitude 1 Filter/sort on minimum magnitude 2 Filter/sort on maximum algebraic value

3 Filter/sort on minimum algebraic value

SRTELTYP Controls the element type to be processed SRTELTYP Description

0 All elements types will be processed

>0 Only element type SRTELTYP will be processed


DTI, INDTA – BULK DATA ENTRYSpecifies or overrides default item codes for the sorting and filtering of element stresses, strains, and forces.Format:

1 2 3 4 5 6 7 8 9 10 DTI INDTA “0”

To specify/override items for a sort of stress quantities.

DTI INDTA “1” B1 C1 B2 C2 “ENDREC”

To specify/override items for a sort of force quantities.

DTI INDTA “2” B1 C1 B2 C2 “ENDREC”

Example: DTI INDTA 0

To specify/override item for a sort of stress quantities.

DTI INDTA 1 64 18 75 18 ENDREC

To specify/override items for a sort of force quantities.

DTI INDTA 2 34 2 2 4 ENREC


DTI, INDTA – BULK DATA ENTRY (Cont.)

Remarks:1. This table is recognized only in SOLs 1, 3, 5, 14, 15, 16, 101, 103, 105, 106, 108,

109, 111, 112, 114, 115, 144, 153, and for stress quantities only. One or more of the user parameters S1, S1G, or S1M must be specified with a value greater then or equal to zero in order to request sorting and/or filtering. See also parameters S1AG, and S1AM in the MSC.NASTRAN Quick Reference Guide.

2. If the Ci value is 1, the element type will be suppressed on the output file. An example of this feature could be as follows: If an element type is to be sorted on two different values and output twice, this can be accomplished by two calls to the STRSORT module with two unique DTI tables. However, other element types will be printed twice. This additional print can be suppressed by setting their sort codes to 1.

Field Contents Bi Element type identification number. See the table in the

MSC.NASTRAN Quick Reference Guide, Appendix A for allowable values. (Integer = 0)

Ci Item code identification number for the stress, strain, or force quantity on which the sort or filter is to be performed. See the table in the MSC.NASTRAN Quick Reference Guide, Appendix A for allowable values. (Integer)


DTI, INDTA – BULK DATA ENTRY (Cont.)Remarks: (Cont.)

3. Table 1 lists the elements currently that are sort able. In addition, the element type identification number, the default stress output quantity, and the associated stress code identification numbers are provided. If this entry is not specified, then the stresses are sorted based on the default quantity given in Table 1.

The Following Should Be NotedA. The element type identification number is used internally by the program to

differentiate element types.B. The stress code identification number is merely the word number in the

standard printed output for the stress quantity of interest. For example, the thirteenth word of stress output for the CHEXA element is the octahedral shear stress. For this element type, the element identification number and the grid point ID each count as a separate word. Stress codes for the elements are tabulated in the MSC.NASTRAN Quick Reference Guide, Appendix A.

C. By default, stress sorting for the membrane and plate elements will be performed on the Hencky-von Mises stress. For maximum shear stress, the STRESS (MAXS) Case Control command should be specified.


DTI, INDTA – BULK DATA ENTRY (Cont.) Default Stress Output

Quantity and Identification Number Element Element

Type ID

Number

Quantity Stress Codes

ID Number

CBAR 34 Maximum stress at end B 14 CBEAM 2 Maximum stress at end B 105 CBEND 69 Maximum stress at end B 20 CONROD 10 Axial stress 2 CELAS1 11 Stress 2 CELAS2 12 Stress 2 CELAS3 13 Stress 2 CHEXA 67 Henky-von-Mises or Octahedral stress 13 CQUAD4 33 Maximum shear or Henky-von-Mises stress at Z2 17 CQUAD4* 144 Maximum shear or Henky-von-Mises stress at Z2 19 CQUAD8 64 Maximum shear or Henky-von-Mises stress at Z2 19 CQUADR 82 Maximum shear or Henky-von-Mises stress at Z2 19 CPENTA 68 Octahedral stress 13 CROD 1 Axial stress 2 CSHEAR 4 No default -- CTETRA 39 No default -- CTRIA3 74 Maximum shear or Henky-von-Mises stress at Z2 17 CTRIA6 75 Maximum shear or Henky-von-Mises stress at Z2 19 CTRIAR 70 Maximum shear or Henky-von-Mises stress at Z2 19 CTIAX6 53 No default -- CTUBE 3 Axial Stress 2 *CORNER output


STRESS SORTING – SAMPLE

C A S E C O N T R O L D E C K E C H OCARDCOUNT

1 TITLE = S.E. SAMPLE PROBLEM 12 SUBTITLE = S.E. STATICS - RUN 1 - MULTIPLE LOADS3 DISP = ALL4 SEALL = ALL5 PARAM,GRDPNT,16 SUBCASE 1017 SUPER = 1,1 $ SUPERELEMENT 1 - RESIDUAL LOAD 18 LABEL = S.E. 1 PRESSURE LOAD9 LOAD = 10110 STRESS = ALL11 PARAM,S1,112 PARAM,NUMOUT,1013 $20 SUBCASE 20121 SUPER = 1,2 $ SUPERELEMENT 1 - RESIDUAL LOAD 222 LABEL = S.E. 1 - 2= NORMAL LOAD23 LOAD = 20124 $30 SUBCASE 30131 LABEL = S.E. 1 - OPPOSING LOADS32 SUPER = 1,3 $ SUPERELEMENT 1 - RESIDUAL LOAD 333 LOAD = 30134 $40 UBCASE 100141 LABEL = RES STR LOAD 1 - PRESSURE42 SPCFORCES = ALL


STRESS SORTING – SAMPLE (Cont.)


SECTION 4

CONSTRAINTS AND BOUNDARY CONDITIONS


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


TABLE OF CONTENTS (Cont.)Section Page

REFLECTIVE SYMMETRY 4-31

DETERMINE BOUNDARY FROM THE LOADS 4-32

STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY 4-33


FREE—FREE ANALYSYS—”R CONSTRAINTS”

In statics, if a model is not properly constrained, it is unstable, and a solution is not normally possible.

For this special case, a solution method called “inertia relief” is available 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.

A mass matrix must exist and there must be stiffness connecting the selected DOF to the rest of the model.

The structure is assumed to be in a state of “equilibrium”. That is a set of inertia loads are found which satisfy ΣF = 0. and ΣM = 0. These inertia loads are applied.

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.

The solution will be the displacements of the model relative to the motion of the SUPORTed (R–set DOF)


FREE—FREE ANALYSYS—”R CONSTRAINTS”

“R” constraints (Reference DOF) – Free-free structures 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) 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) 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.


PROCESSING SUPPORT ENTRIESWhen a SUPORT entry is used, rigid body vectors are calculated in MSC.NASTRAN by the following method: Step 1: “a-set” partitioning

Step 2: Stiffness matrix operations

Note: Pr is not actually applied!

where

This may be used to construct a set of rigid body vectors.

=

rrrrr

r

Puu

KKKK 0~

=ru

uu }{ a

}]{[}{ ruDu m=

rm

KKD -1−=][

=

r

m

ID

RIG][ aφ


PROCESSING SUPPORT ENTRIES (Cont.) Step 3: Mass matrix operations

where [Mr] is not diagonal in general 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

=

r

maa

T

r

mrr I

DM

ID

M ][][

}{][}{}{ lT

mrr PDPq −=

}{]][[]][[}{}{ r-1

rrlrmllli

l qMMDMPP +−=

}{}u]{[ illll PK =


PROCESSING SUPPORT ENTRIES (Cont.) 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 [Mr ] is orthogonalized by the transformation , i.e.,

Step 5: Rigid body mode construction

with the property

*If the structure truly has rigid body modes

][ Troφ

]][][[][ rorTroo MM φφ=

=

ro

romro

DR

φφ

φ IG][

][]][[][ *0IG]][[][

IG oRaaaTRIGa

rraaaTRIGa

MM

KRK

=

==

φφ

φφ


SUPORT SAMPLE

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 membershipPARAM,USETSEL,8 $ print only R–setPARAM,INREL,–1SUPORT,1,12345SUPORT,2,2$GRDSET,,,,,,,6$GRID,1,,-.4,0.,0.,,123456GRID,1,,-.4,0.,0.$GRID,2,,.4,0.,0.,,123456GRID,2,,.4,0.,0.

Sample Data – SOL101


SAMPLE OF INERTIA RELIEFS. E. SAMPLE PROBLEM 1 OCTOBER 22, 1998 MSC.NASTRAN 10/ 22/ 98 PAGE 12SUPORT ENTRY USED FOR INERTIA RELIEFSUBCASE 1U 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 KLRSUPPORT PT. NO. EPSILON STRAIN ENERGY EPSILONS LARGER THAN .001 ARE FLAGGED WITH ASTERISKS1 2.7329080E– 13 –4.6566129E– 102 2.7329080E– 13 –6.8452209E– 083 2.7329080E– 13 4.3655746E– 114 2.7329080E– 13 6.6747816E– 095 2.7329080E– 13 5.6525096E– 106 2.7329080E– 13 –1.2054807E– 07S. E. SAMPLE PROBLEM 1 OCTOBER 22, 1998 MSC.NASTRAN 10/ 22/ 98 PAGE 14

SUPORT ENTRY USED FOR INERTIA RELIEF SUBCASE 1INTERMEDIATE MATRIX ... QRR

COLUMN 11 1.583215E–03 1.363031E–02 0.000000E+00 0.000000E+00 0.000000E+00 –1.363031E–02 6

COLUMN 21 –4.273799E–16 7.916075E–04 0.000000E+00 0.000000E+00 0.000000E+00 7.916075E–04 6

COLUMN 31 0.000000E+00 0.000000E+00 1.583215E–03 1.090425E–02 –6.332861E–04 0.000000E+00 6

COLUMN 41 0.000000E+00 0.000000E+00 1.090425E–02 8.319332E–02 –4.361699E–03 0.000000E+00 6

COLUMN 51 0.000000E+00 0.000000E+00 –4.051360E–14 –4.119826E–13 1.562481E–02 0.000000E+00 6

COLUMN 61 –1.090425E–02 –1.235227E–01 0.000000E+00 0.000000E+00 0.000000E+00 1.235227E–01 61


SAMPLE OF INERTIA RELIEF (CONT)S. E. SAMPLE PROBLEM 1 OCTOBER 22, 1998 MSC.NASTRAN 10/ 22/ 98 PAGE 15SUPORT ENTRY USED FOR INERTIA RELIEFSUBCASE 1INTERMEDIATE MATRIX ... QRLCOLUMN 11 0.000000E+ 00 0.000000E+ 00 3.200000E+ 01 2.560000E+ 02 –1.280000E+ 01 0.000000E+ 00 6COLUMN 21 0.000000E+ 00 0.000000E+ 00 –4.000000E+ 00 –4.000000E+ 01 1.600000E+ 00 0.000000E+ 00 6COLUMN 31 0.000000E+ 00 0.000000E+ 00 6.221512E– 11 6.189182E– 10 –2.080000E+ 01 0.000000E+ 00 61INTERMEDIATE MATRIX ... URACOLUMN 11 0.000000E+ 00 0.000000E+ 00 1.009351E+ 04 –4.400139E+ 03 3.919970E– 10 0.000000E+ 00 6COLUMN 21 0.000000E+ 00 0.000000E+ 00 –8.071358E+ 03 1.538730E+ 03 1.058153E– 10 0.000000E+ 00 6COLUMN 31 0.000000E+ 00 0.000000E+ 00 6.203110E– 09 –1.660236E– 09 1.331216E+ 03 0.000000E+ 00 6 ***USER INFORMATION MESSAGE 5293 (SSG3A)FOR DATA BLOCK KLLLOAD SEQ. NO. EPSILON EXTERNAL WORK EPSILONS LARGER THAN .001 ARE FLAGGED WITH ASTERISKS1 –1.6002654E– 11 1.6844358E– 022 –1.2612771E– 11 2.8055164E– 013 –2.2015322E– 11 1.9882633E– 011 S. E. SAMPLE PROBLEM 1 OCTOBER 22, 1998 MSC.NASTRAN 10/ 22/ 98 PAGE 17SUPORT ENTRY USED FOR INERTIA RELIEF

0 PRESSURE LOAD SUBCASE 1D 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 R31 G .0 .0 .0 .0 .0 .02 G .0 .0 5.906156E– 06 –5.431463E– 16 –1.476539E– 05 .03 G .0 .0 –7.107476E– 05 –2.240753E– 04 6.748205E– 05 .04 G .0 .0 –6.516860E– 05 –2.240753E– 04 –8.224744E– 05 .05 G .0 .0 –6.123342E– 04 –1.125225E– 03 1.358658E– 04 .06 G .0 .0 –6.064281E– 04 –1.125225E– 03 –1.506312E– 04 .07 G .0 .0 –2.447977E– 03 –3.128961E– 03 5.067131E– 04 .08 G .0 .0 –2.442071E– 03 –3.128961E– 03 –5.214784E– 04 .09 G .0 .0 –1.956982E– 02 –1.011539E– 02 –6.370683E– 03 .0


SINGULARITIES

How are they Detected

GPSP (AUTOSPC)

Hidden singularities


METHODS OF DETECTING SINGULARITIES

Shown in order of increasing “cost”

Grid Point Singularity Table (GPSP1 Module)

Warning messages from the DECOMP module

Unreasonable displacements with reasonable forces

“Epsilon/Strain Energy” message

Zero frequency eigenvalues

Nonlinear divergence


EVIDENCE OF ILL-CONDITIONED MATRICES

1. After matrix assembly,Grid Point Singularity Table—Warning Message, AUTOSPC if requested

2. Before decomposition,Null columns Identified—Fatal Message [Needs PARAM ASING=-1]

3. After decomposition,

Information Message – Row NumberLog 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 uKu – P = δP (can be printed using PARAM, IRES)

flags possible loss of accuracy due to “numerical ill conditioning”.

MAXDiagonal FactorDiagonal MatrixR

=

PuPu

T

T

⋅⋅

=∈δ


AUTOSPC THEORY(PARAM, AUTOSPC, YES)

Operations performed by GPSP1 (AUTOSPC) Pick up [K_33] as the 3x3 diagonal partition [K].

If [K_33] is a null matrix, add the associated 3 DOFs to the singularity list.

Compute the eigenvalues (λ i) of [K_33]

If any (λ i /λ max) < ε0, add global direction nearest to the eigenvector corresponding to that λ i, to the singularity list.


AUTOSPC THEORY(PARAM, AUTOSPC, YES)

Find principal stiffness directions for a 3x3 matrix at a GRID point

If , then the global direction nearest i is called

singular. If PARAM, AUTOSPC, YES, then singular DOFs are constrained with SPCs if possible.

Test can be changed with PARAM, EPZERO or PARAM, EPPRT.

)10(EPSK

K 8

MAX

i −<


GPSP1 PARAMETERS

Parameter Default Use

AUTOSPC (Depends on SOL) "YES" or "NO"

Provides SPCs

EPPRT 10-8 List Problem DOFEPZERO 10-8 Constrain Problem DOFPRGPST YES List Problem DOF

Not Constrained SPCGEN 0 If = 1 Punch SPC1 Data


HIDDEN SINGULARITIES THAT ARE NOT DETECTED BY AUTOSPC


EXTRANEOUS SINGULARITY MESSAGES FROM AUTOSPC

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


IDENTIFICATION OF MECHANISMS (MAXRATIO)

Each grid point is stable itself (will not be constrained by AUTPSPC), but the assembly is capable of

This is found during matrix decomposition (module DECOMP or DCMP)

][ 0010

111

02

0][L [D] ][][ T

TLKK

KKKKK

KKLK

≅

−−=

−−−

−

=

Strain-FreeRigid-BodyDecoupled

etc.

Motion


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” = Kii/di.If any |Kii/di| > MAXRATIO (Default =107), print all terms over MAXRATIO.If MSC.Nastran finds a“matrix diagonal/factor diagonal ratio” > MAXRATIO,the action of the program depends on the parameterBAILOUT.

BAILOUT = 0 (Default) stops execution of MSC.NastranBAILOUT = -1 lets MSC.Nastran continue with the analysis

BAILOUT = -1 is not recommended, except for model checking.


SHALLOW SHELLS

If θ ~ 0°, θ N is nearly singular (use K6ROT or SNORM). If θ > 5° , θ N is weak but acceptable (use small K6ROT or use

SNORM). If θ ~ 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 “cure”corners or connections to other element types)


PARAM, K6ROT, XX

is based on displacements.K is a stiffness X 10-6.(In general, K6 ≤ 104 .)

There is no “recommended” value for K6ROT (some people recommend 1.0, the default in linear solution is 0.0, the default in nonlinear solutions is 100.0) A good idea if you are using this is to try a small value, then re-run with a large value (such as 10000.0) 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.

2)Energy NN6 G)(T,K(A,K θθ −⋅=

θ


EXERCISES—MECHANISMS AND UNCONSTRAINED STRUCTURES

How should the following “solid” model be constrained for stability?

How should the following connection be modeled?

Resting on a flat surface Hint – Grids 1-4 DOFs 3+(?)


EXERCISES—MECHANISMS AND UNCONSTRAINED STRUCTURES (Cont.)

TITLE = Mechanism check of a cubic block$ Note: R. Rotation about the Z-Axis ?ECHO = SORTSUBCASE 1

SUBTITLE=Pressure loaded blockLOAD = 1DISPLACEMENT(SORT1,REAL)=ALLSPCFORCES(SORT1,REAL)=ALLSTRESS(SORT1,REAL,VONMISES,BILIN)=ALL

BEGIN BULKCHEXA, 1, 1, 1, 2, 3, 4, 5, 6,+CH01+CH01, 7, 8GRID, 1, , 0., 0., 0., , 123456GRID, 2, , 10., 0., 0., , 3456GRID, 3, , 10., 10., 0., , 3456GRID, 4, , 0., 10., 0., , 3456GRID, 5, , 0., 0., 10., , 456GRID, 6, , 10., 0., 10., , 456GRID, 7, , 10., 10., 10., , 456GRID, 8, , 0., 10., 10., , 456MAT1 1 1.+7 .3 PSOLID 1 1 0 PARAM AUTOSPC NO PLOAD4, 1, 1, 10., , , , 6, 8 ENDDATA


EXERCISES—MECHANISMS 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 1.00000E+07

OR HAVE NEGATIVE TERMS ON THE FACTOR DIAGONAL.

USER INFORMATION:

GRID ID DEGREE OF FREEDOM MATRIX/FACTOR DIAG. RATIO MATRIX DIAGONAL

7 T2 9.59062E+14 2.00969E+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.

POSSIBLE FIX: Fix (SPC) DOF T2 at GRID 2


TYPES OF SYMMETRYReflective

Axisymmetric

Cyclic


TYPES OF SYMMETRY (Cont.)Repetitive Symmetry

Symmetric Model


TYPES OF SYMMETRY (Cont.)

for 0_th harmonic mode only.


REFLECTIVE SYMMETRY


DETERMINE BOUNDARY FROM THE LOADS

Example: Symmetric Loads

0=−= 'XBXB UU

Connect


STATIC ANALYSIS OF A SIMPLY SUPPORTED PLATE USING SYMMETRY

A half-plate model will be used to show several approaches to the same solution. This problem will be solved using both the “SUBCASE – SUBCOM” approach in Case Control. In a typical analysis, only one of these approaches would be used.

The plate will be 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.

Load Cases 1 and 3 being symmetric loadings can be analyzed by symmetric boundary conditions.

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.


STATIC ANALYSIS OF A SIMPLY-SUPPORTED PLATE USING SYMMETRY (Cont.)

Loading 2 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.

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.



The results obtained using this half model are identical to those obtained using a model of the complete structure.



SOL 101CEND$TITLE = HALF PLATE (16X8) MODEL, USING SYM., AND ANTI-SYM. DISPLACEMENT(SORT1,REAL)=ALLSPCFORCES(SORT1,REAL)=ALLSTRESS(SORT1,REAL,VONMISES,BILIN)=ALL$ Load Case 1: Pressure on Full PlateSUBCASE = 1SUBTITLE =LOAD CASE 1 - Full Pressure on whole plateSPC = 1 $ SS AND SYMMETRIC BCLOAD = 1

$ Load Case 2: Pressure on half Plate$ Use SUBCASE, and SYMCOM for Pressure on half plateSUBCASE = 11SUBTITLE =Case 2a: Sym. BC with .5 Psi on half plateSPC = 1LOAD = 2

SUBCASE = 12SUBTITLE =Case 2b: Anti-Sym. BC with .5 Psi on half plateSPC = 2LOAD = 2


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


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


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


SECTION 5

MODEL CHECKOUT


TABLE OF CONTENTSSection PageCOMMON TYPES OF ERRORS 5-5COMMON MODELING ERRORS 5-7DIAGNOSIS OF A NEW MODEL – PARAMs 5-8DIAGNOSIS OF A NEW MODEL – DIAGs 5-9F04 OUTPUT 5-10DIAG 8 F04 OUTPUT – MATRIX TRAILERS 5-11SPECIAL OUTPUTS 5-12GRID POINT STRESS AND STRESS DISCONTINUITIES 5-13SURFACE DATA 5-14MINIMUM RECOMMENDED MODEL CHECKS 5-15STIFFNESS MATRIX CHECKS 5-22OUTPUT FROM ground_check_1a 5-25DESCRIPTION OF ground_check_1a OUTPUT 5-27ground_check_1b-MODEL WITH A BAD ELEMENT 5-30OUTPUT FROM ground_check_1b 5-32RESULTS OF ground_check_1b 5-33


TABLE OF CONTENTS (CONT)Section Pageground_check_1c – MODEL WITH A BAD MPC 5-34

ground_check_1c – IMPROPER MPC 5-35

OUTPUT FROM ground_check_1c 5-36

DISCUSSION OF ground_check_1c RESULTS 5-38

ground_check_2a – MODEL ERROR 5-39

HOW TO AVOID SERIOUS MODELING MISTAKES 5-40

CHECK FOR BAD MODES 5-41

SOME ADDITIONAL DEBUGS FOR DYNAMICS 5-42

SAMPLE OF SHRINK PLOTS 5-43

SOME RECOMMENDATIONS 5-44


COMMON TYPES OF ERRORS Mistakes in engineering judgment Approximations to physical behavior

Engineering theory Finite element theory Finite element implementation Modeling

Bolted connection Welded connection Corners Transitions

Modeling errors Connections

Beam to plate Beam to solid Plate to solid

Beam orientation Beam releases Loading (how well do you know the loading yourself?)


COMMON TYPES OF ERRORS (CONT) Finite element error Round-off error (can be disastrous when it occurs)

Computers use binary arithmetic (If you enter .1, internally it may be .099999998)

Program bugs (nobody’s perfect) A list of known errors is maintained and distributed

Eternal Vigilance is the Price of a Good Analysis


COMMON MODELING ERRORS Plates not lining up = zipper

Any connections depending on in-plane rotational stiffness of plates, or any rotational stiffness on solids

Instabilities – for example, releasing both ends of a beam in torsion Offsets of elements in wrong coordinate system (should be in the

output coordinate systems of the grid points for Bars and Beams) Member properties wrong (beam orientation) – also plates – membrane

only – left out bending Beam end releases – are they local or global Element force output is normally in element coordinate system


DIAGNOSIS OF A NEW MODEL - PARAMs

PARAM OperationAUTOSPC, EPPRT, MAXRATIO Check relative magnitudes of

matrix termsFIXEDB Solve superelements individually

Statics = fixed-boundary solutionDynamics = calculated component modes

IRES Load ErrorGRDPNT Check mass, CG, M. Moment of InertiaUSETPRT Print set tablesSESEF Strain energy fractions

(superelements – SOL 103)TINY Minimum percentage value of element

strain energy forprintout (Values not printed are not available for post-processing)


DIAGNOSIS OF A NEW MODEL - DIAGsDIAG Operation

8 Print matrix trailers14 Print DMAP listing15 Print table trailers56 List Qualifier changes as the solution progresses – also,

list all DMAP statements executed on the .f04 file (normally only modules are listed)

MSC.NASTRAN DATA BLOCK NAME CONVENTION FOR MATRICES

where Y = type: A, D, 4 I,J = col, row setsK = stiffness M = massB = viscous damping G = transformationD = rigid body transformation U = displacementP = load Q = force of constraint

[ ]YIJKKYIJ =


Prints matrix trailers as the matrices are createdDAY TIME ELAPSED I/O SEC DEL_I/O CPU SEC DEL_CPU SUB_DMAP/DMAP_MODULE MESSAGES16:56:39 0:37 2.9 .0 8.9 .0 SEPREP2 17 GP1 BEGN16:56:40 0:38 2.9 .0 9.5 .6 SEPREP2 17 GP1 END

subDMAPElapsed Time for Job (used for “time” limit)

Wall Clock – Elapsed Seconds

F04 OUTPUT:Time Log and DMAP Trace Format

Time of Day

File Operations

DMAP Sequence ID

Module Name


DIAG 8 F04 OUTPUT – Matrix Trailers Sample printout using DIAG8

DefinitionsF(orm) 1=square, 2=rectangle, 3=diagonal, 6=symmetric, etc.T(ype) 1=single precision real, 2=double precision real,

3=single precision complex, 4=double precision complexNzwds Largest Number of nonzero words among all columnsDen Density, (number of nonzero terms) ÷ (Rows x Columns))BlockT Number of GINO blocks (1 block = “buffsize -1” words)

DAY TIME ELAPSED I/O MB DEL_MB CPU SEC DEL_CPU SUB_DMAP/DMAP_MODULE MESSAGES14:16:23 0:16 18.6 .0 3.5 .0 SEMG 28 EMG BEGN*8** Module DMAP Matrix Cols Rows F T NzWds Density BlockT StrL NbrStr EndAvg BndMax NulCol

EMG 28 KELM 1 300 2 2 600 1.00000D+00 3 300 1 300 300 0 *8**14:16:24 0:17 18.6 .0 3.5 .0 SEMG 111 EMG BEGN*8** Module DMAP Matrix Cols Rows F T NzWds Density BlockT StrL NbrStr EndAvg BndMax NulCol

EMG 111 KJJM 48 48 6 2 48 2.50000D+00 3 3 192 22 45 24 *8**


SPECIAL OUTPUTSStrain energy density “ESE” – use to determine where to

make changes most efficiently.Grid point forces “GPFORCE” – use to “trace” loads

thru the structure and verify load paths.Stress sorting PARAMS: S1, NUMOUT, BIGER,

SRTOPTGrid point stresses “GPSTRESS”Stress Discontinuities “GPSDCON”, “ELSDCON”


GRID POINT STRESS AND STRESS DISCONTINUITIES

(ALL CASE CONTROL) Use GPSTRESS, ELSDCON, or GPSDCON to select surfaces and

volumes. Use “OUTPUT (POST)” or “SETS DEFINITION” with

“SURFACE” data to define surfaces “VOLUME” data to define volumes

SURFACE Entry Definition

[ ][ ] { } [ ]

MESSAGEBREAK

,NOMESSAGE

MESSAGEBRANCH,

maxtheta0.0TOLERENCE

GEOMETRICLTOPOLOGICA

3X2X1X

R

MNORMAL,3X2X1X

AXIS,cidCORD

BASICELEMENT SYSTEM

MID2Z1Z

,

MIDZ2Z1

ALL

FIBRE , {sid} SET {id} SURFACE


SURFACE DATA Recommendations for discontinuous structures

Use tolerance and branch tests to handle discontinuous stresses Use local coordinate systems for orientation Use “GEOMETRIC” method when element sizes differ Try dividing into smaller “surfaces” Use several options in one run and compare them

Remember that the elements are isoparametric, that is, “ideal” elements are mapped onto the real elements in the model. If the grid point stresses differ when different options are used (or if the discontinuities are too large), it may indicate any of the following conditions: Mesh too coarse Elements badly shaped Modeling errors


MINIMUM RECOMMENDED MODEL CHECKS

Pre-Analysis Understand the structure and the elements

Make small models – understand the problem Use pilot models in areas of uncertainty If you are not familiar with using the element type or SOLution you

expect to use, make simple models and compare the answers to theoretical results (with a simple model, you should be able to obtain excellent correlation with theoretical results).

Model checks before the analysis Geometry

Pre-processor (or Undeformed plots) Look at connections between different element types

Based on knowledge of elements Based on loads Look at corners (QUAD plates)

Shrink plots


MINIMUM RECOMMENDED MODEL CHECKS (CONT)

Elements Beam and bar

Check that I1 and I2 have proper orientation and values Check all end releases (in member coordinates) Verify all offsets (in output coordinate system of GRIDs) Material – need E, ν (or G), and ρ

Plates and Shells Check aspect ratios, taper, and warpage Check orientation – Z, surfaces consistent Check attachments – especially any depending on in-plane rotational

stiffness, any corners, and “shells” Verify any offsets (in element coordinate system) Material – need E, ν (or G), and ρ Property entry – be sure to get the correct properties. (One of the most

commonly made errors is not specifying MID2 for “bending” plates



Solids Check aspect ratios Check taper Check attachments. If any attachments depend on rotational

stiffness, special modeling effort is required Material – need E, ν (or G), and ρ

Mass properties Check ρ on MATi entries Check NSM on property entries

Bars, beams = mass/unit length Plates = mass/unit area

Submit with PARAM, GRDPNT, xxxxwhere xxxx = ID of GRID point to calculate mass properties about Always check center of gravity and total weight (mass) versus known

values



Loadings: Verify they are correct (OLOAD RESULTANT)

Constraints: Verify that they are defined (often they are forgotten) Verify they are correct (location and orientation – in output coordinate

system of the GRID points) Verify that they are applied (SPC CASE CONTROL command)

Static Checks – ALWAYS RUN STATICS FIRST!!! Apply 1–g in X, Y, and Z directions independently

Check load paths (GPFORCE) Check reactions (SPC FORCE)

Does total = applied load? Are the reactions at the correct locations and do they have the correct orientation?

In Dynamics, approximate frequency:

where d = center of gravity displacement in direction of applied g-load

g = acceleration due to gravity

dgf

π21

=



Equilibrium check – verify model is not over-constrained Run free-free. Remove known constraints and check for

unconstrained motion under applied loads or imposed displacements.or

Use the Case Control Command GROUNDCHECK, to check for over-constrained systems.

Thermal equilibrium check – if thermal loads are to be considered. Check α on MATi entries Check for unconstrained thermal expansion – on a copy of your

model Apply a determinate set of constraints Use the same α for all materials Apply a uniform ∆T to the structure. It should expand “freely,” that is, with

no reactions, element forces, or stresses



After the Analysis Statics

Check EPSILON and MAXRATIO Epsilon > 10-9 may indicate trouble MAXRATIO > 106 may indicate trouble

Check reactions. Do they equal the applied loads (Σ applied loads are printed as “OLOAD RESULTANT” in superelement solutions)?

Check load paths – use grid point force balance to “trace” loads Check stress contours for “consistency”

“Sharp” corners indicate bad modeling Use different options (i.e., topological and geometric) and compare results Check stress discontinuities Compare values to “hand calc” or small model results



Dynamics – normal modes Check frequencies. Are they in the expected range? (Did you forget

WTMASS???) If free-free, are there six “rigid-body” (f=0.0) modes? Are there any mechanisms (f=0.0)?

More than six “rigid-body” modes in free-free? Any “rigid-body” modes in constrained modes?

Check mode shapes, and Identify modes Plots and/or animation Effective weight and kinetic energy (Case Control Commands

MEFFMASS and EKE) help to identify “significant” modes


STIFFNESS MATRIX CHECKS The model (stiffness and mass matrices) should be

checked to verify that the elements are not (obviously) bad and that the model is not over-constrained Sample – CELASi between non coincident points or CELASi to

ground

This check can be performed at various stages during the analysis – at each stage, a potential problem is checked G-set – at this stage of the solution, the elements (including

GENELs and K2GG) are checked for grounding N-set – at this stage, the MPC equations are checked A-set – (free-free only) check that the SPC’s do not over-constrain

the structure

Use the Case Control Control Command GROUNDCHECK


STIFFNESS MATRIX CHECKS Sample Model 1 – Cantilever Beam

Properties:I1 = 10I2 = 10J = 5A = 1E = 10,000,000.ν = .3ρ = .1WTMASS = .002588


STIFFNESS (AND MASS) CHECKS (CONT)

SOL 103CENDTITLE = Cantilever Beam Modeled with 8 CBAR elementsGROUNDCHECK=YESSUBCASE 1

SUBTITLE=DefaultLABEL = Perform Model ChecksMETHOD = 1SPC = 1VECTOR(SORT1,REAL)=ALL

BEGIN BULKMAT1 1 1.+7 .3 .1 PBAR, 1, 1, 1., 10., 10., 5.CBAR, 1, 1, 1, 2, 0., 1., 0.CBAR, 2, 1, 2, 3, 0., 1., 0.CBAR, 3, 1, 3, 4, 0., 1., 0.CBAR, 4, 1, 4, 5, 0., 1., 0.CBAR, 5, 1, 5, 6, 0., 1., 0.CBAR, 6, 1, 6, 7, 0., 1., 0.CBAR, 7, 1, 7, 8, 0., 1., 0.CBAR, 8, 1, 8, 9, 0., 1., 0.GRID 1 0.00 0. 0. GRID 2 1.25 0. 0. GRID 3 2.50 0. 0. GRID 4 3.75 0. 0. GRID 5 5.00 0. 0. GRID 6 6.25 0. 0. GRID 7 7.50 0. 0. GRID 8 8.75 0. 0. GRID 9 10.00 0. 0. PARAM GRDPNT 0 PARAM WTMASS .002588PARAM AUTOSPC YES SPC1 1 123456 1EIGRL 1 5ENDDATA

Input File: ground_check_1a.bdf


OUTPUT FROM ground_check_1aCANTILEVER BEAM WITH 8 CBAR

M O D E L S U M M A R Y

NUMBER OF GRID POINTS = 9

NUMBER OF CBAR ELEMENTS = 8

O U T P U T F R O M G R I D P O I N T W E I G H T G E N E R A T O RREFERENCE POINT = 0

M O* 1.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 1.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 5.000000E+00 ** 0.000000E+00 0.000000E+00 1.000000E+00 0.000000E+00 -5.000000E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 -5.000000E+00 0.000000E+00 3.359375E+01 0.000000E+00 ** 0.000000E+00 5.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 3.359375E+01 *

S* 1.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 1.000000E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 1.000000E+00 *

DIRECTIONMASS AXIS SYSTEM (S) MASS X-C.G. Y-C.G. Z-C.G.

X 1.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00Y 1.000000E+00 5.000000E+00 0.000000E+00 0.000000E+00Z 1.000000E+00 5.000000E+00 0.000000E+00 0.000000E+00

I(S)* 0.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 8.593750E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 8.593750E+00 *

I(Q)* 0.000000E+00 ** 8.593750E+00 ** 8.593750E+00 *

Q* 1.000000E+00 0.000000E+00 0.000000E+00 ** 0.000000E+00 1.000000E+00 0.000000E+00 ** 0.000000E+00 0.000000E+00 1.000000E+00 *


OUTPUT FROM ground_check_1a (CONT)*** USER INFORMATION MESSAGE 7570 (GPWG1D)

RESULTS OF RIGID BODY CHECKS OF MATRIX KGG (G-SET) FOLLOW:PRINT RESULTS IN ALL SIX DIRECTIONS AGAINST THE LIMIT OF 1.228800E-01

DIRECTION STRAIN ENERGY PASS/FAIL--------- ------------- ---------1 1.862645E-09 PASS2 5.960464E-08 PASS3 5.960464E-08 PASS4 9.313226E-10 PASS5 5.714595E-06 PASS6 5.714595E-06 PASS

SOME POSSIBLE REASONS MAY LEAD TO THE FAILURE:1. CELASI ELEMENTS CONNECTING TO ONLY ONE GRID POINT;2. CELASI ELEMENTS CONNECTING TO NON-COINCIDENT POINTS;3. CELASI ELEMENTS CONNECTING TO NON-COLINEAR DOF;4. IMPROPERLY DEFINED DMIG MATRICES;

*** SYSTEM INFORMATION MESSAGE 6916 (DFMSYN)DECOMP ORDERING METHOD CHOSEN: BEND, ORDERING METHOD USED: BEND

*** USER INFORMATION MESSAGE 5010 (LNCILD)STURM SEQUENCE DATA FOR EIGENVALUE EXTRACTION.TRIAL EIGENVALUE = 8.697012D+08, CYCLES = 4.693590D+03NUMBER OF EIGENVALUES BELOW THIS VALUE = 2

*** USER INFORMATION MESSAGE 5010 (LNCILD)STURM SEQUENCE DATA FOR EIGENVALUE EXTRACTION.TRIAL EIGENVALUE = 1.787403D+10, CYCLES = 2.127803D+04NUMBER OF EIGENVALUES BELOW THIS VALUE = 6

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GNERALIZEDNO. ORDER MASS STIFFNESS

1 1 4.709041E+08 2.170032E+04 3.453714E+03 1.000000E+00 4.709041E+082 2 4.709041E+08 2.170032E+04 3.453714E+03 1.000000E+00 4.709041E+083 3 9.503416E+08 3.082761E+04 4.906367E+03 1.000000E+00 9.503416E+084 4 8.335352E+09 9.129815E+04 1.453055E+04 1.000000E+00 8.335352E+095 5 1.786391E+10 1.336559E+05 2.127200E+04 1.000000E+00 1.786391E+10


DESCRIPTION OF ground_check_1a OUTPUT Grid Point Weight Output (GPWG module)

The scale factor entered with parameter WTMASS is applied to the assembled element mass before GPWG. The GPWG module, however, converts mass back to the original input units that existed prior to the scaling effect of the parameter WTMASS

GPWG is performed on the g-size mass matrix, which is the mass matrix prior to the processing of the rigid elements, MPCs, and SPCs

Any masses at scalar points and fluid-related masses are not included in the GPWG calculation

GPWG for a superelement does not include the mass form upstream superelements. Therefore, GPWG for the residual structure includes only the mass of the residual (not any upstream superelements). The center of gravity location is also based on the mass of the current superelement only

The output from the GPWG is for information purposes only and is not used in the analysis

The rigid-body mass matrix [MO] is computed with respect to the reference grid point in the basic coordinate system. The Grid point to be used is specified using PARAM, GNDPNT

For further information see the MSC.NASTRAN Linear Static Analysis User’s Guide (V2001), Appendix B


DESCRIPTION OF ground_check_1a OUTPUT (CONT)

Stiffness Check Output These checks are performed by multiplying the stiffness matrix by a

set of rigid-body vectors(Rb) which are based on the geometry (calculated about PARAM, GRDPNT)

The rigid-body strain energy checks are calculated as (note that the factor of ½ is not included in the calculation)

This check is performed on the G-, N-, A-set matrices (I in CHKii is the set being checked)

If any term in the resulting “CHK” matrix exceeds the value of PARAM, CHECKTOL (default value is calculated based in the stiffness of your model), the results of the check are printed

“Reaction forces” are calculated, normalized to a minimum of 1.0, filtered, and printed (if CHECKTOL is exceeded)

CHKKiiKRR bTb =

REACiKRb =


DESCRIPTION OF ground_check_1a OUTPUT (CONT)

Stiffness Check Output (Cont.) Note that “full” data recovery is not performed, and if a DOF which

does not belong to the remaining set is constrained, the nearest point (by connection) in the remaining set is indicated. See results for CHKKAA—point 1 is constrained, but does not belong to the A-set, therefore, the constraint shows up at point 2

Mass Check Output These checks are performed by multiplying the mass matrix by a

set of rigid-body vectors(Rb) which are based on the geometry (calculated about PARAM, GRDPNT)

The calculation is similar to that performed on the stiffness matrix The results at the G-set should match Grid Point Weight Generator The checks at the N- and A-set check if MPCs and constraints

remove (or re-distribute) mass


Ground_check_1b – MODEL WITH A BAD ELEMENT

Same model as before, only now connect a CELAS2 element between DOF 2 at Grid Points 2 and 3 (this will cause “grounding”), as the direction of the stiffness terms is not along the line connecting the GRID points)

Samples of CELASi elements which cause “grounding”

Connected to “Ground” Geometric mismatch Kθ to ground


STIFFNESS CHECKS (CONT)

SOL 103CENDTITLE = Cantilever Beam with 8 CBAR + 1 CELAS2GROUNDCHECK=YESSUBCASE 1

SUBTITLE=DefaultLABEL = Perform Model ChecksMETHOD = 1SPC = 1VECTOR(SORT1,REAL)=ALL

BEGIN BULKMAT1 1 1.+7 .3 .1 PBAR, 1, 1, 1., 10., 10., 5.CBAR, 1, 1, 1, 2, 0., 1., 0.CBAR, 2, 1, 2, 3, 0., 1., 0...CBAR, 8, 1, 8, 9, 0., 1., 0.$ Add a CELAS2 incorrectly specifiedCELAS2, 99, 1000., 2, 2, 3, 2GRID 1 0.00 0. 0. GRID 2 1.25 0. 0. .. GRID 9 10.00 0. 0. PARAM GRDPNT 0 PARAM WTMASS .002588PARAM AUTOSPC YES SPC1 1 123456 1EIGRL 1 5ENDDATA

Input File – ground_check_1b.dat


OUTPUT FROM ground_check_1bCANTILEVER BEAM WITH 8 CBAR + 1 CELAS2

*** USER INFORMATION MESSAGE 7570 (GPWG1D)RESULTS OF RIGID BODY CHECKS OF MATRIX KGG (G-SET) FOLLOW:PRINT RESULTS IN ALL SIX DIRECTIONS AGAINST THE LIMIT OF 1.228801E-01

DIRECTION STRAIN ENERGY PASS/FAIL--------- ------------- ---------

1 1.862645E-09 PASS2 5.960464E-08 PASS3 5.960464E-08 PASS4 9.313226E-10 PASS5 5.714595E-06 PASS6 7.812500E+02 FAIL

SOME POSSIBLE REASONS MAY LEAD TO THE FAILURE:1. CELASI ELEMENTS CONNECTING TO ONLY ONE GRID POINT;2. CELASI ELEMENTS CONNECTING TO NON-COINCIDENT POINTS;3. CELASI ELEMENTS CONNECTING TO NON-COLINEAR DOF;4. IMPROPERLY DEFINED DMIG MATRICES;

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GNERALIZEDNO. ORDER MASS STIFFNESS1 1 4.709041E+08 2.170032E+04 3.453714E+03 1.000000E+00 4.709041E+082 2 4.709118E+08 2.170050E+04 3.453742E+03 1.000000E+00 4.709118E+083 3 9.503416E+08 3.082761E+04 4.906367E+03 1.000000E+00 9.503416E+084 4 8.335352E+09 9.129815E+04 1.453055E+04 1.000000E+00 8.335352E+095 5 1.786391E+10 1.336559E+05 2.127200E+04 1.000000E+00 1.786391E+10


RESULTS OF ground_check_1b At the G-set, the structural matrices are grounded when

the alter attempts to rotate the model about the z-axis This is indicated by the large term in the CHKKGG

matrix for DOF 6 By looking at the REACGNRM matrix – this matrix

represents the forces (normalized to a maximum of 1.0) preventing the model from moving as a rigid body. The column associated with DOF 6 (z-rotation) contains terms for DOF 2 of grid points 2 and 3, indicating that a modeling error exists in that area

This is the location of the CELAS2


Ground_check_1c – MODEL WITH A BAD MPC

Same model as before, only now connect an MPC between DOF 2 at Grid Points 2 and 3 (since the points are not coincident, this will cause “grounding”)

The MPC states that the y-direction translation of the Grid Point 2 must equal the y-direction translation of Grid Point 3


Ground_check_1c – IMPROPER MPC

SOL 103CENDTITLE = Cantilever Beam with 8 CBAR, and 1 MPCGROUNDCHECK(SET=(G,N))=YESSUBCASE 1

SUBTITLE=DefaultLABEL = Perform Model ChecksMETHOD = 1MPC = 1SPC = 1VECTOR(SORT1,REAL)=ALL

BEGIN BULKMAT1 1 1.+7 .3 .1 PBAR, 1, 1, 1., 10., 10., 5.CBAR, 1, 1, 1, 2, 0., 1., 0.CBAR, 2, 1, 2, 3, 0., 1., 0...CBAR, 8, 1, 8, 9, 0., 1., 0.$ Add an MPC Equation (incorrectly specified)MPC, 1, 2,2,1., 3, 2, -1.GRID 1 0.00 0. 0. GRID 2 1.25 0. 0. .. GRID 9 10.00 0. 0. PARAM GRDPNT 0 PARAM WTMASS .002588PARAM AUTOSPC YES SPC1 1 123456 1EIGRL 1 5ENDDATA

Input File: ground_check_1c


OUTPUT FORM ground_check_1cCANTILEVER BEAM WITH 8 CBAR, AND 1 MPC

*** USER INFORMATION MESSAGE 7570 (GPWG1D)RESULTS OF RIGID BODY CHECKS OF MATRIX KGG (G-SET) FOLLOW:PRINT RESULTS IN ALL SIX DIRECTIONS AGAINST THE LIMIT OF 1.228800E-01


1 1.862645E-09 PASS2 5.960464E-08 PASS3 5.960464E-08 PASS4 9.313226E-10 PASS5 5.714595E-06 PASS6 5.714595E-06 PASS

*** USER INFORMATION MESSAGE 7570 (GPWG1D)RESULTS OF RIGID BODY CHECKS OF MATRIX KNN (N-SET) FOLLOW:PRINT RESULTS IN ALL SIX DIRECTIONS AGAINST THE LIMIT OF 1.228800E-01


1 1.862645E-09 PASS2 5.960464E-08 PASS3 5.960464E-08 PASS4 9.313226E-10 PASS5 5.714595E-06 PASS6 9.600000E+08 FAIL

SOME POSSIBLE REASONS MAY LEAD TO THE FAILURE:1. MULTIPOINT CONSTRAINT EQUATIONS WHICH DO NOT SATISFY RIGID-BODY MOTION;2. RBE3 ELEMENTS FOR WHICH THE INDEPENDENT DEGREE-OF-FREEDOM CANNOT DESCRIBE

ALL POSSIBLE RIGID-BODY MOTIONS.


OUTPUT FORM ground_check_1c (Contd.)

CANTILEVER BEAM WITH 8 CBAR, AND 1 MPC

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GNERALIZEDNO. ORDER MASS STIFFNESS

1 1 4.709041E+08 2.170032E+04 3.453714E+03 1.000000E+00 4.709041E+082 2 9.503416E+08 3.082761E+04 4.906367E+03 1.000000E+00 9.503416E+083 3 1.233946E+09 3.512757E+04 5.590726E+03 1.000000E+00 1.233946E+094 4 8.335352E+09 9.129815E+04 1.453055E+04 1.000000E+00 8.335352E+095 5 1.786391E+10 1.336559E+05 2.127200E+04 1.000000E+00 1.786391E+10


Discussion of ground_check_1c Results At the G-set, the structural matrices pas the rigid-body tests,

since the CELAS2 which caused the problem in ground_check_1b has been removed.

Matrix KNN fails the rigid-body test due to the incorrectly-specified MPC equation. This is indicated by the large term in the CHKKNN matrix at DOF 6.

By looking at the REACNCOL matrix – this matrix represents the forces (normalized to a maximum of 1.0) preventing the model from moving as a rigid body. The 6th column contains terms for GRID points 1 and 3, indicating that a modeling error exists in that area.

This is the location of MPC (NOTE – since the test is performed on the N-set, GRID 2 DOF 2 no longer exists, since it is in the M-set and has been removed).


Ground_check_2a – MODEL ERROR

SOL 101CENDTITLE = Groundcheck for an Inclined RodECHO = SORTGROUNDCHECK(GRID=1, SET=(G,N+AUTOSPC))=YESSUBCASE 1

SUBTITLE=DefaultSPC = 1LOAD = 1DISPLACEMENT(SORT1,REAL)=ALLSPCFORCES(SORT1,REAL)=ALLSTRESS(SORT1,REAL,VONMISES,BILIN)=ALL

BEGIN BULKMAT1 1 1.+7 PROD 1 1 1. CROD 1 1 1 2 CROD 2 1 2 3 FORCE, 1, 3, 0, 1000., 0.866025, 0.5, 0. GRID 1 0. 0. 0. GRID 2 .866025 0.5 0. GRID 3 1.732051 1. 0. PARAM AUTOSPC YES PARAM GRDPNT 0 SPC1 1 123456 1 SPC1 1 3456 2 3 ENDDATA

Input File: ground_check_2a

Question: What is wrong with this rod model?


HOW TO AVOID SERIOUS MODELING MISTAKES

Take the time to understand the structure and how it behaves under load. Perform hand analysis or use a simple model first

Take the time to understand MSC.NASTRAN (particularly the elements). Run small samples each time you try something new

Use independent checks (if available) Estimate the cost (labor and computer costs) before

you start


CHECK FOR BAD MODES Identify your modes using one or more of the

following: Plot your eigenvectors (either using the MSC.NASTRAN plotter or

MSC.PATRAN) and identify them Try setting NORM=MAX on EIGRL entry and look at modal masses.

Small values may indicate singularities or local modes (not recommended).

Use Case Control Commands EKE, and MEFFMASS to print kinetic energy and modal effective mass .

Watch for warnings on orthogonality checks Look for extraneous low frequency modes – these

often indicate incorrect modeling (for example plate elements without MID2 on the PSHELL entry)


SOME ADDITIONAL DEBUGS FOR DYNAMICS

In dynamic analysis, use normal modes as a diagnostic tool Simulate statics in modal dynamic solutions and compare the results to

a static solution (this is a way to determine if your nodes are capable of representing the solution) In Transient analysis, apply a constant loading, and damping In frequency response, apply the load at 0.0 Hz, and remove structural damping

Use sssalter “modevala.vxx” to see if your modes can represent the solution if the loads are applied “statically” (although you are looking at a dynamic solution, it is hard for the modes to represent the dynamic solution under loading if they cannot represent the static solution)

Selecting time or frequency set selection can have a major impact on the solution accuracy In Transient response, the accuracy is directly related to the integration time step (A

central difference is used to calculate the velocity and acceleration). If you are using a direct solution, run using different integration time steps to see if the answers change

In Frequency Response, the peak responses normally at or near occur at resonance. Use a modal solution with FREQ3, FREQ4, and/or FREQ5 entries to guarantee that the solution is obtained with reasonable accuracy near the resonance frequencies.


SAMPLE OF SHRINK PLOTSStiffened Plate with Error in Modeling


SOME RECOMMENDATIONS Understand the important things BEFORE you get into

trouble!!! Understand your structure and how you expect it to perform Understand your loading Understand your model Understand how to use the program Understand the limitations of the method Use simple sample problems (preferably with known solutions) to

understand the MSC.Nastran solution.

ALWAYS perform a static solution first, then progress to the more complicated solutions.


SECTION 6

DYNAMIC ANALYSIS


TABLE OF CONTENTSSection Page

RECOMMENDATIONS FOR DYNAMIC SOLUTIONS 6-5

POSSIBLE USES FOR GUYAN REDUCTION (ASET, OMIT) 6-6

MODIFYING DYNAMIC BEHAVIOR 6-7

APPROXIMATING FREQUENCIES USING STATIC ANALYSIS 6-10

SIMULATE STATICS IN MODAL SOLUTIONS 6-11

SELECTING TRANSIENT PARAMETERS 6-12

USING RESIDUAL VECTOR TO IMPROVE ACCURACY INMODAL SOLUTIONS 6-13

SAMPLE 7-STATIC RESIDUAL VECTORS FOR A FLAT PLATE 6-18


RECOMMENDATIONS FOR DYANMIC SOLUTIONS

For Modal Analysis Lanczos method is recommended. Determine the correct frequency range in advance

Hopefully someone else will set it Based on the frequency content of the applied loading Wild Guess??????

For Response Calculations Use direct solution for small problems and/or when many modes lie in

the frequency range. Use modal method for most problems for efficiency. Use a combination (superelements direct with mode synthesis upstream)

when detailed results are needed at few selected points.

Don’t forget damping !!!!!!!!!!!!!!!!!!!!!


POSSIBLE USES FOR GUYAN REDUCTION (ASET, OMIT)

Remove local dynamic effects.Example: Remove panel modes

Evaluate dynamic test setups.Example: ASET=Accelerometer and excitation degrees

of freedom Isolate numerical problems.

Examples: 1. Stiff joints and couplers in a truss2. Normal rotations in a shallow shell


MODIFYING DYNAMIC BEHAVIOR Understanding the loading

In transient, create a shock spectrum on the input to determine frequency content (see the MSC.NASTRAN Advanced Dynamics User’s Guide).

In modal solutions, determine modes are being excited by applied loads.

use alters to print out PHDH = modal forces

Let x = {Φ}ζΦ = mode shapesζ = modal coordinates

Premultiply by φΤ

φΤF = modal forces = PHDHCompile SEMTRAN alter ‘call.*gma.*phdh’Matprn phdh//$

FKBM =++ φζζφζφ

FKxxBxM =++

FKBM TTTT φφζφζφφζφφ =++


MODIFYING DYNAMIC BEHAVIOR (Cont.) Determine which modes are contributing to the response

Frequency Response –

Use modconta.v2001 to determine modal contributions at selected DOF

Use mfreqea.v2001 to determine modal contributions to the “total” solution

Transient – use mtranea.v2001 to determine modal contributions to the “total” solution

Either solution – request SDISP in the Case Control – this provides the solution in modal coordinates

Modify modes with high response by shifting frequencies away from Input load peaks.

Print ESE for modes of interest.

Look at this output for modes with high response.

Change element properties in areas with high ESE in those modes.

Note: Lowering the frequency might also reduce the response. Look at the spectrum (frequency content) for the applied load.


MODIFYING DYNAMIC BEHAVIOR (Cont.)Shock Spectrum of Applied Loading

If peak response occurs in a mode at 3 Hz, raising the frequency (up to5 Hz) increases the response; but lowering the frequency reduces theresponse of that mode.


APPROXIMATING FREQUENCIES USING STATIC ANALYSIS

In statics, solve

For normal modes, solve

Perform the following steps to approximate the primary frequencies of a model:

Perform static analysis with a. Three separate loadings (1-g; x,y,z)b. Param, Grdpnt to locate the center of gravity

Obtain the displacement at the (∆) for each loadCalculate approximate frequencies using the following substitutions:

for an equivalent SDOF oscillator

This provides an estimate of the primary frequencies of the model.Note: Since 1-g checkout runs are recommended, this is a “no-cost”

estimate.

KxF =

0x)KM( 2 =−ω

∆== kmgG ∆= // gmk∴

∆== // gmk

ω


SIMULATE STATICS IN MODAL SOLUTION

Use modevala.v2001 to determine how well the modes can represent the static solution or:

In frequency response analysis run with F = 0.0 Use RLOAD=F (static load). Watch out for “G” damping. Do not try this on a “free-free” structure

In transient analysis run with big delta-time Use TLOAD=F (static load). Use PARAM, RESVEC, YES if modal transient (SOL 112) High frequencies are suppressed. Only two steps are necessary. Does not work (without damping) with NOLINi data.


SELECTING TRANSIENT PARAMETERS

Time step: Use at least 12 steps/period for Important modes

Add damping: No such thing as an “undamped” system! (add structural damping by using parameters G and W3, and/or GE on MATi, and W4; viscous damping by CDAMPi, CVISC, CBUSH, CBUSH1D, TABDMP1)


USING RESIDUAL VECTORS TO IMPROVE ACCURACY IN MODAL SOLUTIONS

Whenever a modal approach is used, an approximation is made that the modes are capable of representing the complete dynamic solution to the applied loadings.

This assumption may be dangerous - if the modes are not capable of representing the dynamic response, incorrect answers will result with no warning.

There is no method to verify that sufficient modes have been obtained. The following approaches are commonly used to try to account for this: Compare to a direct solution (expensive). Solve again with more modes of appropriate kind and compare. If there is a

change, repeat the process until the results do not change (expensive). Use Mode-Acceleration data recovery to attempt to correct for the high-

frequency truncation (may be expensive and is not available for superelements).

Append static residual vectors (RESVEC) to the modes and solve.


USING RESIDUAL VECTORS TO IMPROVE ACCURACY IN MODAL SOLUTIONS (Cont.)

The idea of residual vectors is that any “residuals” of the static solutions which the modes cannot represent are appended onto the modes as “pseudo-modes” and orthogonalized to the modes

These static solution are generated by the program based on any loadings defined using the LOADSET-LSEQ approach and any DOF defined in the “U6” set (a separate unit load applied to each of these DOFs).

The resulting set of modal coordinates is capable of exactly representing the static solution to the applied loading and therefore is much more likely to provide correct answers in the modal solution.


USING RESIDUAL VECTORS TO IMPROVE ACCURACY IN MODAL SOLUTIONS (Cont.)User Interface Bulk Data

PARAM, RESVEC, YES (default = NO) = this parameter enables residual vectors = MANDATORY if you want residual vectors.


USING RESIDUAL VECTORS TO IMPROVE ACCURACY IN MODAL SOLUTIONS (Cont.)

Output Standard output for the solution plus the following: After the eigenvalue (before augmentation by the residual vector) summary table, the program prints a second eigenvalue summary table that includes one or more additional mode (normally high-frequency) .


USING RESIDUAL VECTORS TO IMPROVE ACCURACY IN MODAL SOLUTIONS (Cont.)LimitationsThis approach does not work on structures with singular matrices. If the singularity occurs in the residual structure, then a SUPORT entry may be used in a manner similar to inertia relief when calculating the residual vectors.This approach does not support automatic restart. (If a loading change is made, the new vectors are not found unless the superelement eigenvalue solution is reprocessed for another reason).You can manually force the re-calculation of residual vectors by doing the following:

It is recommended to include modal damping for the high-frequency modes that represent the static residuals. If none is present, high-frequency oscillation about the correct answer may occur.

param, serst, manual

set 999 = ….. $ superelements to re-process residual vectors for

selg = 999

selr = 999

semr = 999


SAMPLE 7 - STATIC RESIDUAL VECTORS FOR A FLAT PLATE

The problem is a flat rectangular cantilevered plate with a load applied at the center of the free end

The applied loading can be described by the x, y, and z components: Fx = 100.

Fy = 100Fz = 1.

The x and y components of the loading are in-plane, while the z-component is out-of-plane

The load is a ½ sine pulse at 10hz.


SAMPLE 7 – STATIC RESIDUAL VECTORS FOR A FLAT PLATE (Cont.)



First step – see if modes can represent the static solution

SOL, 101include 'pchdispa.v2001'CENDTITLE = sample7a - static solution of plate with point loadsSUBTITLE=Static Case ControlSPC = 1DISP(PLOT)=ALLSUBCASE 1

LOAD = 1SUBCASE 2

LOAD = 2SUBCASE 3

LOAD = 3SUBCASE 4

LOAD = 4BEGIN BULKinclude 'plate.dat'PARAM POST 0PARAM AUTOSPC YESFORCE, 1, 105, 0, 1., 100., 100., 1.FORCE, 2, 105, 0, 1., 100., , 0.FORCE, 3, 105, 0, 1., , 100., 0.FORCE, 4, 105, 0, 1., , , 1.ENDDATA

Input for static run



See if modes can represent the static solution to the applied loading

Input for “modeval” run (sssalter – modevala.vxx)

SOL, 103include 'modevala.v2001'CENDTITLE = sample7b - Evaluate ability to represent static solutionECHO = SORTDISP(PLOT)=ALLSUBCASE 1

SPC = 1METHOD = 1

BEGIN BULKinclude 'sample7a.pch'include 'plate.dat'PARAM AUTOSPC YESPARAM, EVAL, -2EIGRL, 1, -1., , 20ENDDATA


OUTPUT FROM MODEVAL RUN –SAMPLE 7B

SAMPLE7B - EVALUATE ABILITY TO REPRESENT STATIC SOLUTION

R E A L E I G E N V A L U E SMODE EXTRACT EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 4.411126E+00 2.100268E+00 3.342680E-01 1.000000E+00 4.411126E+002 2 1.168239E+02 1.080851E+01 1.720228E+00 1.000000E+00 1.168239E+023 3 1.709162E+02 1.307349E+01 2.080711E+00 1.000000E+00 1.709162E+024 4 1.180099E+03 3.435257E+01 5.467381E+00 1.000000E+00 1.180099E+035 5 1.342210E+03 3.663619E+01 5.830831E+00 1.000000E+00 1.342210E+036 6 4.011240E+03 6.333435E+01 1.007997E+01 1.000000E+00 4.011240E+037 7 5.108886E+03 7.147647E+01 1.137583E+01 1.000000E+00 5.108886E+038 8 7.089261E+03 8.419775E+01 1.340049E+01 1.000000E+00 7.089261E+039 9 1.009674E+04 1.004825E+02 1.599229E+01 1.000000E+00 1.009674E+0410 10 1.047183E+04 1.023320E+02 1.628664E+01 1.000000E+00 1.047183E+0411 11 1.433233E+04 1.197177E+02 1.905366E+01 1.000000E+00 1.433233E+0412 12 1.814985E+04 1.347214E+02 2.144157E+01 1.000000E+00 1.814985E+0413 13 2.176740E+04 1.475378E+02 2.348137E+01 1.000000E+00 2.176740E+0414 14 3.063447E+04 1.750271E+02 2.785642E+01 1.000000E+00 3.063447E+0415 15 3.230971E+04 1.797490E+02 2.860794E+01 1.000000E+00 3.230971E+0416 16 4.180963E+04 2.044740E+02 3.254305E+01 1.000000E+00 4.180963E+0417 17 4.869267E+04 2.206642E+02 3.511979E+01 1.000000E+00 4.869267E+0418 18 5.114825E+04 2.261598E+02 3.599445E+01 1.000000E+00 5.114825E+0419 19 5.515309E+04 2.348470E+02 3.737705E+01 1.000000E+00 5.515309E+0420 20 6.210560E+04 2.492100E+02 3.966299E+01 1.000000E+00 6.210560E+04



^^^ ^^^ RESULTS FOR LOADING NUMBER 1 ^^^ ^^^ STRAIN ENERGY FRACTIONS FOR EACH MODE ^^^ MODE NO 1 = 8.986632E-01 ^^^ MODE NO 2 = 1.024669E-22 ^^^ MODE NO 3 = 2.337658E-02 ^^^ MODE NO 4 = 4.355374E-26 ^^^ MODE NO 5 = 3.175627E-03 ^^^ MODE NO 6 = 9.185600E-25 ^^^ MODE NO 7 = 1.137585E-03 ^^^ MODE NO 8 = 3.113251E-04 ^^^ MODE NO 9 = 3.177580E-26 ^^^ MODE NO 10 = 3.889483E-04 ^^^ MODE NO 11 = 2.074733E-04 ^^^ MODE NO 12 = 2.021724E-04 ^^^ MODE NO 13 = 1.668908E-25 ^^^ MODE NO 14 = 2.550916E-04 ^^^ MODE NO 15 = 1.996148E-06 ^^^ MODE NO 16 = 5.934722E-29 ^^^ MODE NO 17 = 1.514302E-22 ^^^ MODE NO 18 = 1.397200E-04 ^^^ MODE NO 19 = 1.069760E-22 ^^^ MODE NO 20 = 2.596889E-05 ^^^ ^^^ TOTAL STRAIN ENERGY IN INPUT VECTOR = 1.738667E+00 ^^^ TOTAL STRAIN ENERGY REPRESENTED BY MODES = 1.613285E+00 ^^^ ^^^ TOTAL FRACTION FOR ALL MODES = 9.278858E-01


OUTPUT FROM MODEVAL RUN –SAMPLE 7B (Cont.)

^^^^^^ RESULTS FOR LOADING NUMBER 2 ^^^ ^^^ STRAIN ENERGY FRACTIONS FOR EACH MODE ^^^ MODE NO 1 = 4.384733E-40 ^^^ MODE NO 2 = 7.393978E-39 ^^^ MODE NO 3 = 3.596012E-41 ^^^ MODE NO 4 = 1.600690E-37 ^^^ MODE NO 5 = 7.561645E-38 ^^^ MODE NO 6 = 1.042006E-41 ^^^ MODE NO 7 = 3.473819E-42 ^^^ MODE NO 8 = 2.815545E-40 ^^^ MODE NO 9 = 4.678964E-40 ^^^ MODE NO 10 = 3.770136E-39 ^^^ MODE NO 11 = 6.328972E-37 ^^^ MODE NO 12 = 9.143374E-40 ^^^ MODE NO 13 = 1.037771E-33 ^^^ MODE NO 14 = 2.508102E-30 ^^^ MODE NO 15 = 7.002179E-32 ^^^ MODE NO 16 = 7.722814E-29 ^^^ MODE NO 17 = 6.833245E-25 ^^^ MODE NO 18 = 1.700329E-23 ^^^ MODE NO 19 = 1.277280E-24 ^^^ MODE NO 20 = 4.071197E-22 ^^^ ^^^ TOTAL STRAIN ENERGY IN INPUT VECTOR = 6.858309E-03 ^^^ TOTAL STRAIN ENERGY REPRESENTED BY MODES = 2.922214E-24 ^^^ ^^^ TOTAL FRACTION FOR ALL MODES = 4.260837E-22



^^^ ^^^ RESULTS FOR LOADING NUMBER 3 ^^^ ^^^ STRAIN ENERGY FRACTIONS FOR EACH MODE ^^^ MODE NO 1 = 5.871140E-38 ^^^ MODE NO 2 = 2.983206E-37 ^^^ MODE NO 3 = 1.262346E-36 ^^^ MODE NO 4 = 3.109134E-36 ^^^ MODE NO 5 = 1.596317E-36 ^^^ MODE NO 6 = 4.041079E-39 ^^^ MODE NO 7 = 9.878081E-37 ^^^ MODE NO 8 = 1.202587E-35 ^^^ MODE NO 9 = 2.032995E-34 ^^^ MODE NO 10 = 1.464245E-34 ^^^ MODE NO 11 = 1.091518E-32 ^^^ MODE NO 12 = 4.412275E-33 ^^^ MODE NO 13 = 3.150745E-30 ^^^ MODE NO 14 = 1.592329E-26 ^^^ MODE NO 15 = 1.561463E-28 ^^^ MODE NO 16 = 4.725095E-24 ^^^ MODE NO 17 = 3.840482E-21 ^^^ MODE NO 18 = 6.340798E-20 ^^^ MODE NO 19 = 3.064197E-21 ^^^ MODE NO 20 = 1.999937E-18 ^^^ ^^^ TOTAL STRAIN ENERGY IN INPUT VECTOR = 1.175639E-01 ^^^ TOTAL STRAIN ENERGY REPRESENTED BY MODES = 2.433872E-19 ^^^

^^^ TOTAL FRACTION FOR ALL MODES = 2.070254E-18



^^^ RESULTS FOR LOADING NUMBER 4 ^^^ ^^^ STRAIN ENERGY FRACTIONS FOR EACH MODE ^^^ MODE NO 1 = 9.679300E-01 ^^^ MODE NO 2 = 1.103648E-22 ^^^ MODE NO 3 = 2.517839E-02 ^^^ MODE NO 4 = 4.691076E-26 ^^^ MODE NO 5 = 3.420397E-03 ^^^ MODE NO 6 = 9.893605E-25 ^^^ MODE NO 7 = 1.225268E-03 ^^^ MODE NO 8 = 3.353213E-04 ^^^ MODE NO 9 = 3.422500E-26 ^^^ MODE NO 10 = 4.189275E-04 ^^^ MODE NO 11 = 2.234648E-04 ^^^ MODE NO 12 = 2.177554E-04 ^^^ MODE NO 13 = 1.802378E-25 ^^^ MODE NO 14 = 2.747534E-04 ^^^ MODE NO 15 = 2.150006E-06 ^^^ MODE NO 16 = 3.464232E-25 ^^^ MODE NO 17 = 1.609257E-23 ^^^ MODE NO 18 = 1.504893E-04 ^^^ MODE NO 19 = 1.834015E-23 ^^^ MODE NO 20 = 2.797052E-05 ^^^ ^^^ TOTAL STRAIN ENERGY IN INPUT VECTOR = 1.614245E+00 ^^^ TOTAL STRAIN ENERGY REPRESENTED BY MODES = 1.613285E+00 ^^^ ^^^ TOTAL FRACTION FOR ALL MODES = 9.994051E-01



As you can see from the previous output, the modes can represent better than 99% of the solution in the z-direction, but less than 1% of the solution in the x- or y-direction. The modes are capable of representing 93% of the static solution to the total applied loading. This indicates that an answer found using these modes may be incorrect for any x- or y-direction response.

Input file (sample7c) for modal transient without residual vectors

SOL, 112CENDTITLE =sample7c -Modal Transient with NO RESVECDISP(PLOT)=ALLSUBCASE 1

SPC = 1METHOD = 1DLOAD = 20SDAMP = 30TSTEP = 40LOADSET = 50

BEGIN BULKinclude 'plate.dat'PARAM POST 0PARAM AUTOSPC YES

$Define Half Sine WaveTLOAD2, 20, 100, , ,0., 0.05, 10., 90.$Use LSEQ, to convert static load to dynamicLSEQ, 50, 100, 1TABDMP1, 30, CRIT, .1, 0.01, 200., 0.01, ENDTTSTEP,40, 400, 0.001, 2EIGRL, 1, -1., , 20FORCE, 1, 105, 0, 1., 100., 100., 1.FORCE, 2, 105, 0, 1., 100., , 0.FORCE, 3, 105, 0, 1., , 100., 0.FORCE, 4, 105, 0, 1., , , 1.ENDDATA



Results of Modal Transient Analysis with 20 modes (NO RESVEC)


SAMPLE 7 – STATIC RESIDUAL VECTORS FOR A FLAT PLATE (Cont.)Input file (sample7d) to include static

residual vectorsSOL, 112CENDTITLE = sample7d -Modal Transient with RESVECECHO = SORTDISP(PLOT)=ALLSUBCASE 1

SPC = 1METHOD = 1DLOAD = 20SDAMP = 30TSTEP = 40LOADSET = 50

BEGIN BULKinclude 'plate.dat'PARAM POST 0PARAM AUTOSPC YESPARAM RESVEC YES

$ Default in V2001 for RESVEC is NO$Define Half Sine WaveTLOAD2, 20, 100, , ,0., 0.05, 10., 90.$Use LSEQ, to convert static load to dynamicLSEQ, 50, 95, 2LSEQ, 50, 96, 3LSEQ, 50, 97, 4LSEQ, 50, 100, 1TABDMP1, 30, CRIT, .1, 0.01, 200., 0.01, ENDTTSTEP,40, 400, 0.001, 2EIGRL, 1, -1., , 20FORCE, 1, 105, 0, 1., 100., 100., 1.FORCE, 2, 105, 0, 1., 100., , 0.FORCE, 3, 105, 0, 1., , 100., 0.FORCE, 4, 105, 0, 1., , , 1.ENDDATA



In this case, the static deformed shape due to the dynamic loading is appended into the modes as “pseudo-modes”.

In the output on the following pages you will see: The original eigenvalue summary table. A final eigenvalue summary table including 1 new eigenvector

which represents the static residual vector.


OUTPUT FROM RUN WITH RESIDUAL VECTOR

(BEFORE AUGMENTATION OF RESIDUAL VECTORS)

MODE EXTRACT EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 4.411126E+00 2.100268E+00 3.342680E-01 1.000000E+00 4.411126E+002 2 1.168239E+02 1.080851E+01 1.720228E+00 1.000000E+00 1.168239E+023 3 1.709162E+02 1.307349E+01 2.080711E+00 1.000000E+00 1.709162E+024 4 1.180099E+03 3.435257E+01 5.467381E+00 1.000000E+00 1.180099E+035 5 1.342210E+03 3.663619E+01 5.830831E+00 1.000000E+00 1.342210E+036 6 4.011240E+03 6.333435E+01 1.007997E+01 1.000000E+00 4.011240E+037 7 5.108886E+03 7.147647E+01 1.137583E+01 1.000000E+00 5.108886E+038 8 7.089261E+03 8.419775E+01 1.340049E+01 1.000000E+00 7.089261E+039 9 1.009674E+04 1.004825E+02 1.599229E+01 1.000000E+00 1.009674E+0410 10 1.047183E+04 1.023320E+02 1.628664E+01 1.000000E+00 1.047183E+0411 11 1.433233E+04 1.197177E+02 1.905366E+01 1.000000E+00 1.433233E+0412 12 1.814985E+04 1.347214E+02 2.144157E+01 1.000000E+00 1.814985E+0413 13 2.176740E+04 1.475378E+02 2.348137E+01 1.000000E+00 2.176740E+0414 14 3.063447E+04 1.750271E+02 2.785642E+01 1.000000E+00 3.063447E+0415 15 3.230971E+04 1.797490E+02 2.860794E+01 1.000000E+00 3.230971E+0416 16 4.180963E+04 2.044740E+02 3.254305E+01 1.000000E+00 4.180963E+0417 17 4.869267E+04 2.206642E+02 3.511979E+01 1.000000E+00 4.869267E+0418 18 5.114825E+04 2.261598E+02 3.599445E+01 1.000000E+00 5.114825E+0419 19 5.515309E+04 2.348470E+02 3.737705E+01 1.000000E+00 5.515309E+0420 20 6.210560E+04 2.492100E+02 3.966299E+01 1.000000E+00 6.210560E+04


OUTPUT FROM RESIDUAL VECTOR DMAP ALTER (Cont.)

(AFTER AUGMENTATION OF RESIDUAL VECTORS)

MODE EXTRACT EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 4.411126E+00 2.100268E+00 3.342680E-01 1.000000E+00 4.411126E+002 2 1.168239E+02 1.080851E+01 1.720228E+00 1.000000E+00 1.168239E+023 3 1.709162E+02 1.307349E+01 2.080711E+00 1.000000E+00 1.709162E+024 4 1.180099E+03 3.435257E+01 5.467381E+00 1.000000E+00 1.180099E+035 5 1.342211E+03 3.663619E+01 5.830831E+00 1.000000E+00 1.342211E+036 6 4.011240E+03 6.333435E+01 1.007997E+01 1.000000E+00 4.011240E+037 7 5.108886E+03 7.147647E+01 1.137583E+01 1.000000E+00 5.108886E+038 8 7.089260E+03 8.419775E+01 1.340049E+01 1.000000E+00 7.089260E+039 9 1.009674E+04 1.004825E+02 1.599229E+01 1.000000E+00 1.009674E+0410 10 1.047183E+04 1.023320E+02 1.628664E+01 1.000000E+00 1.047183E+0411 11 1.433233E+04 1.197177E+02 1.905366E+01 1.000000E+00 1.433233E+0412 12 1.814985E+04 1.347214E+02 2.144157E+01 1.000000E+00 1.814985E+0413 13 2.176740E+04 1.475378E+02 2.348137E+01 1.000000E+00 2.176740E+0414 14 3.063447E+04 1.750271E+02 2.785642E+01 1.000000E+00 3.063447E+0415 15 3.230971E+04 1.797490E+02 2.860794E+01 1.000000E+00 3.230971E+0416 16 4.180963E+04 2.044740E+02 3.254305E+01 1.000000E+00 4.180963E+0417 17 4.869267E+04 2.206642E+02 3.511979E+01 1.000000E+00 4.869267E+0418 18 5.114824E+04 2.261598E+02 3.599445E+01 1.000000E+00 5.114824E+0419 19 5.515309E+04 2.348470E+02 3.737705E+01 1.000000E+00 5.515309E+0420 20 6.210560E+04 2.492099E+02 3.966299E+01 1.000000E+00 6.210560E+0421 21 6.037606E+05 7.770203E+02 1.236666E+02 1.000000E+00 6.037606E+05



Results with 20 modes plus residual vector



Results of Direct Transient Analysis



Comparison of Methods: sample7c (No RESVEC),

sample7d (RESVEC), and Sampl7e ( Direct Transient)


SECTION 7

SUPERELEMENT ANALYSIS


TABLE OF CONTENTSSection PageWHAT IS A SUPERELEMENT? 7-9

ADVANTAGES OF SUPERELEMENT ANALYSIS 7-10

DISADVANTAGES OF SUPERELEMENT ANALYSIS 7-12

HOW ARE SUPERELEMENTS DEFINED IN MSC.NASTRAN? 7-13

MAIN BULK DATA SUPERELEMENT DEFINITION 7-15

MAIN BULK DATA GRID POINT PARTITIONING 7-16

BULK DATA USED TO DEFINE PARTS 7-17

BULK DATA USED TO DEFINE SUPERELEMENTS 7-18

BULK DATA USED TO CONNECT PARTS 7-19

SEBNDRY ENTRYelement Boundary-Point Definition 7-20

SECONCT ENTRY 7-21

SEEXCLD ENTRY 7-23

SEBULK ENTRY 7-24

SAMPLE PROBLEM- STEEL STAMPING 7-26SAMPLE PROBLEM—STEEL STAMPING SAMPLE SUPERELEMENT 1 7-28


TABLE OF CONTENTSSection PageSAMPLE PROBLEM—STEEL STAMPING SESET ENTRIES FOR MAIN

BULK DATA SUPERELEMENTS 7-33

PARTITIONED SOLUTIONS 7-34

THEORY OF STATIC CONDENSATION 7-36

CONVENTIONAL ANALYSIS 7-38

SUPERELEMENT ANALYSIS 7-41

BULK DATA FOR STATIC LOADS ON SUPERELEMENTS 7-48

SINGLE-POINT CONSTRAINSTS ON SUPERELEMENTS 7-50

MPCs AND RIGID ELEMENTS IN SUPERELEMENTS 7-52

RIGID CONNECTION OF TWO SUPERELEMENTS 7-53

SUPERELEMENT CASE CONTROL COMMANDS 7-54

SUPER COMMAND 7-55

EXPANDED VERSUS CONDENSED 7-56

SUPER COMMAND EXAMPLE—ONE LOADING CONDITION 7-58

MULTIPLE LOADING CONDITIONS IN SUPERELEMENT CASE CONTROL – OPTION 1 7-59


TABLE OF CONTENTSSection PageMULTIPLE LOADING CONDITIONS IN SUPERELEMENT

CASE CONTROL – OPTION 2 7-60

REASONS TO USE OPTION 2 FOR MULTIPLE LOADINGS 7-62

MULTIPLE LOADINGS – SAMPLE OF OPTION 1 7-63

MULTIPLE LOADINGS – SAMPLE OF OPTION 2 7-64

PARAMETERS IN CASE CONTROL 7-65

SAMPLE SUPERELEMENT STATIC RUN INPUT 7-66

SAMPLE SUPERELEMENT STATIC RUN INPUT —USING PARTS 7-68

SUPERELEMENT REDUCTION METHODS AVAILABLE IN DYNAMIC ANALYSIS 7-71

DEGREES OF REDUCTION 7-72

COMPARISON OF REDUCTION METHODS 7-73

ADVANTAGES OF EACH REDUCTION METHOD 7-74

CALCULATION OF NORMAL MODES USING STATIC REDUCTION ONLY 7-75

CALCULATION OF NORMAL MODES USING DYNAMIC REDUCTION FOR SUPERELEMENT 7-76


TABLE OF CONTENTSSection Page“FIXED BOUNDARY” SOLUTIONS (PARAM, FIXEDB, -1) 7-78

PROCEDURES FOR SUPERELEMENT DYNAMIC REDUCTION 7-79

SPECIFICATION OF FIXED AND FREE BOUNDARY DEGREES OF FREEDOM 7-81

REFERENCES FOR CMS 7-82

SUPERELEMENT DYNAMICS EXAMPLE 7-83

APPENDIX – THE CRAIG-BAMPTON MEDTHOD –HAND-SOLVED EXAMPLE 7-100

DEFAULT CMS METHOD “FIXED BOUNDARY” CMS 7-101

SOLUTION BY HAND 7-104

SOLUTION USING MSC.NASTRAN SOL 103 7-114

SELECTED OUTPUT FROM MSC.NASTRAN 7-115



EXTERNAL SUPERELEMENTS 7-117

CREATING AN EXTERNAL SUPERELEMENT 7-118

ATTACHING AN EXTERNAL SUPERELEMENT 7-120

DATA RECOVERY FOR AN EXTERNAL SUPERELEMENT 7-122

ATTACHING AN EXTERNAL SUPERELEMENT 7-123

SAMPLE PROBLEM 7-125

SAMPLE PROBLEM USING MATRIXDB 7-126

SAMPLE PROBLEM USING DMGIOP2 PROGRAM 7-132


WHAT IS A SUPERELEMENT? Physical and mathematical representation

Physical – a superelement is a substructure: a finite element model of a portion of a structure

Mathematical – the physical model is replaced with “boundary matrices”: loads, mass, damping, and stiffness reduced from the interior points to the exterior or boundary points

When a model is divided into superelements, it is best to think of each superelement as having it’s own unique bulk data set (internally).

Each superelement is processed and the finite element model is replaced by reduced matrices which represent the mass, damping, stiffness and loadings on the superelement as seen by any adjacent parts of the model

Once a superelement is processed, the bulk data used to define it, along with all matrices used in processing it are not needed until performing data recovery on the superelement, or performing modifications on it. This data may be “archived” to reduce disk usage.

The reduced matrices are used to replace the physical model of the superelement


ADVANTAGES OF SUPERELEMENT ANALYSIS

Large problems (i.e., allows solving problems that exceed your hardware capabilities)

Less CPU or wall clock time per run (reduced risk since each superelement may be processed individually)

Partial redesign requires only partial solution (cost). Allows more control of resource usage Partitioned input desirable

Organization Repeated components

Partitioned output desirable Organization Comprehension

Components may be modeled by subcontractors.


ADVANTAGES OF SUPERELEMENT ANALYSIS (Cont.)

Multi-step reduction for dynamic analysis

Zooming (or global-local analysis)

Allows for efficient configuration studies (“What if...”)


DISADVANTAGES OF SUPERELEMENT ANALYSIS

Increased overhead due to DMAP compilation and database manipulation and storage

Mandatory static condensation may cancel other cost savings for small models.

All superelements must be linear. Approximations must be made in dynamics for mass

and damping through static, component mode, or generalized dynamic reduction.


HOW ARE SUPERELEMENTS DEFINED IN MSC.NASTRAN?

Superelements are identified using numbers (SEID). Each superelement (SEID > 0) is defined with its own set of grids,

elements, constraints, loads, etc. There are two ways to define superelements in MSC.NASTRAN, Main

Bulk Data Superelements and PARTS (not currently supported for nonlinear analysis), which allow partitioned input files.

Main Bulk Data superelements are easiest thought of as a “cookie–cutter” approach. All data provided in the “Main Bulk Data” section (Between the “BEGIN BULK” and either

the first “BEGIN SUPER = i” or ENDDATA entry) is partitioned (divided) into a separate set for each superelement based on GRID point assignments made by the user

Partitioned bulk data superelements (PARTs) are defined in separate (self–contained) sections of the input file. The separate PARTs are “assembled” together based on coincident points. Each PART is defined in a “self–contained” section which begins with a “BEGIN SUPER=i”

entry and ends with either the next “BEGIN SUPER=j” entry of the ENDDATA


HOW ARE SUPERELEMENTS DEFINED IN MSC.NASTRAN?

The residual structure is a superelement that contains grid points, elements, etc. (in the Main Bulk Data), which are not assigned to any other superelement. Last superelement (SEID = 0) to be processed Superelement on which the assembly analysis (nonlinear, transient

response, frequency response, buckling, system modes, etc.) is performed

A superelement may also be defined as an image of a superelement or obtained from outside MSC.NASTRAN.


MAIN BULK DATA SUPERELEMENT DEFINITION

Each superelement(SEID > 0) defined in the Main Bulk Data section is defined with its own set of grids, elements, constraints, loads, etc. Interior grid points are assigned (partitioned) to a superelement by

the user. Exterior grid points, elements, loads, and constraints are

automatically partitioned by the program based on interior grid point assignments.


MAIN BULK DATA GRID POINT PARTITIONING

Bulk Data Entries

Only interior points need to be defined. SESET takes precedence over GRID.

For the example shown above, Grid Point 47 will belong to the residual structure (SEID=0).

Elements, constraints, loads, etc., are automatically partitioned. SESET “THRU” option allows “open sets.” Points not assigned to any superelement belong to the residual

structure by default. A model with no grid point assignments is defined as a residual structure-only model.

1 2 3 4 5 6 7 8 9 10GRID GID ETC. SEIDGRID 47 2

1 2 3 4 5 6 7 8 9 10SESET SEID G1 "THRU" G2SESET 0 47 THRU 57

Superelements are identified by an integer.


BULK DATA USED TO DEFINE PARTS Each PART is defined in a separate section of the input file The section containing the data for a PART will begin with:

BEGIN (BULK) SUPER = i where i is the superelement id to be defined by the following input

The section containing the data for a PART will end with either:BEGIN (BULK) SUPER = j

where j is the superelement defined in the next section of the input fileor

ENDDATA which indicates the end of the input file

The Bulk Data for each PART must be ’self–contained’ It must contain all data defining elements, properties, materials, and

loadings for that PART Different PARTs may use the same id numbers for elements and GRID

points, since each is in a “self–contained” input section.


BULK DATA USED TO DEFINE SUPERELEMENTS

ID test, problemSOL 101CENDTITLE = SAMPLE INPUT FILE DEMONSTRATING PART INPUTSUBCASE 1LOAD = 1DISP = ALLBEGIN BULK$$ MAIN BULK DATA – may be omitted if desired$ contains data defining residual structure and also any Main Bulk Data$ superelements$$ any superelements defined in this section will be defined by$ using SESET entries or field 9 on the GRID entries$BEGIN SUPER = 1$$ model data for PART 1$BEGIN SUPER = 2$$ model data for PART 2$ENDDATA

Sample input stream


BULK DATA USED TO CONNECT PARTS Since PARTs are “self–contained”, it is necessary to “connect” them to

each other and the Main Bulk Data superelements The Program will automatically determine coincident grid points

between each PART and any other PARTs or Main Bulk Data superelements

If desired, the automatic connection logic may be modified or overridden by using the following entries in the Main Bulk Data section

SEBNDRY – defines a set of points for a PART which may be used in the automatic search for attachments

SECONCT – Allows definition of a tolerance for connection and (if desired) manual listing of the grid points being connected

SEEXCLD – Allows you to provide a list of points to be excluded from the boundary search

SEBULK – the “METHOD” field on this entry controls whether the AUTO or MANUAL connection logic is used.


SEBNDRY ENTRYDefines a list of grid points in a partitioned superelement for the automatic boundary search between a specified superelement or between all other superelements in the model.Format:

Example 1:

Example 2:

Field ContentsSEIDA Superelement Identification number. See Remark 2. (Integer ≥ 0)SEIDB Superelement Identification. See Remark 3. (Integer ≥ 0 or Character

“All” ; Default = “ALL” )GIDAI Identification number of a boundary grid point in superelement SEIDA.

Remarks:1. SEBNDRY may only be specified in the main Bulk Data Section and is not

recognized after the BEGIN SUPER = n.2. SEIDA AND SEIDB may reference partitioned superelements or superelements in

the main Bulk Data Section

1 2 3 4 5 6 7 8 9 10SEBNDRY SEIDA SEIDB GIDA1 GIDA2 GIDA3 GIDA4 GIDA5 GIDA6

GIDA7 GIDA8 etc.

SEBNDRY 400 4 10 20 30 40

SEBNDRY 400 4 10 20 30 40


SECONCT ENTRYExplicitly defines grid and scalar point connection procedures for a partitioned superelement. Format:

Example:

Field ContentsSEIDA Partitioned superelement Identification number. See Remark 2.

(Integer > 0)SEIDB Identification number of superelement for connection to SEIDA.

(Integer ≥ 0)TOL Location tolerance to be used when searching for or checking

boundary grid points. (Real; Default = 10E –5 )LOC Coincident location check option for manual connection.

(Character; “YES” or “NO”; Default = “YES”)GIDAI Identification number of a grid or scalar point in superelement

SEIDA, which will be connected to GIDBI.GIDBI Identification number of a grid or scalar point in superelement

SEIDB, which will be connected to GIDAI.

1 2 3 4 5 6 7 8 9 10SECONCT SEIDA SEIDB TOL LOC

GIDA1 GIDB1 GIDA2 GIDB2 GIDA3 GIDB3 etc.

SECONCT 10 20 1.00E-04 YES1001 4001 1002 4002 2222 4444


SECONCT ENTRY (Cont.)Remarks:

1. SECONCT can only be specified in the main Bulk Data Section and is ignored after the BEGIN SUPER = n command.

2. TOL and LOC can be used to override the default values specified on the SEBULK entries.

3. The continuation entry is optional.4. The (GIAI, GIBI) pair must both be grids or scalar points.5. All six degrees of freedom of grid points will be defined as

boundary degrees of freedom.


SEEXCLD ENTRYDefines grids that will be excluded during the attachment of a partitioned superelement. Format:

Example:

Field ContentsSEIDA Partitioned superelement Identification number. See Remark 2.

(Integer > 0)SEIDB Superelement Identification. (Integer > 0 or Character = “ALL” )GIDAI Identification number of a grid in superelement SEIDA to be

executed from connection to superelement SEIDB.Remarks:

1. SEEXCLD can only be specified in the main Bulk Data Section and is ignored after the BEGIN SUPER = n command.

2. SEIDA and SEIDB may reference partitioned superelements or superelements defined in the main Bulk Data Section.

1 2 3 4 5 6 7 8 9 10SEEXCLD SEIDA SEIDB GIDA1 GIDA2 GIDA3 GIDA4 GIDA5 GIDA6

GIDA7 GIDA8 etc.

SEEXCLD 110 10 45 678 396


SEBULK ENTRYDefines superelement boundary search options and a repeated, mirrored, or collector superelement. Format:

Example:

Field ContentsSEID Superelement identification number. (Integer 0)TYPE Superelement type. (Character; No Default)

PRIMARY PrimaryREPEAT IdenticalMIRROR MirrorCOLLCTR CollectorEXTERNAL External

RSEID Identification number of the reference superelement, used if TYPE “REPEAT” and “MIRROR”. (Integer 0; Default 0)

METHOD Method to be used when searching for boundary grid points. (Character: “AUTO” or “MANUAL”; Default = “AUTO”)

TOL Location tolerance to be used when searching for boundary grid points. (Real; Default 10E–5)

LOC Coincident location check option for manual connection option. (Character: “YES” or “NO”; Default = “YES”)

1 2 3 4 5 6 7 8 9 10SEBULK SEID TYPE RSEID METHOD TOL LOC

SEBULK 14 REPEAT 4 AUTO 1.00E-03


SEBULK ENTRY (Cont.)Remarks:

1. The TYPE = “REPEAT” or “MIRROR” does not include superelements upstream of the reference superelement. A repeated or mirrored superelement can have boundaries, loads, constraints, and reduction procedures that are different than the reference superelement.2. METHOD = “MANUAL” requires SECONCT entries. SEBNDRY and SEEXCLD, which reference SEID, will produce a fatal message.3. SECONCT, SEBNDRY, and SEEXCLD entries can be used to augment the search procedure and/or override the global tolerance.4. For combined automatic and manual boundary search, the METHOD = “AUTO” should be specified and connections should be specified on a SECONCT entry.5. TOL and LOC are the default values that can be modified between two superelements by providing the required tolerance on the SECONCT entry.6. TYPE = “MIRROR” also requires specification of a SEMPLN entry.7. TYPE = “COLLCTR” indicates a collector superelement, which does not contain any grids or scalar points.8. For TYPE = “EXTERNAL”, see also PARAM, EXTOUT, etc. description in Section 6 of the MSC.NASTRAN Quick Reference Guide.


SAMPLE PROBLEM- STEEL STAMPING


SAMPLE PROBLEM- STEEL STAMPING (Cont.)

Grid Points 1 and 2 fixed Material properties:

Steel t = 0.05”E = 29 x 106 psiν = 0.3ρ = 0.283 lb/in3 (weight density)

Applied loads 1 psi pressure on square portions Normal force of 2 lb on Grids 93 and 104 Opposing normal force of 2 lb on Grids 93 and 104


SAMPLE PROBLEM—STEEL STAMPINGSAMPLE SUPERELEMENT 1


SAMPLE PROBLEM – STEEL STAMPING (Cont.)

Grids 1 and 2 are fixedSteel D = .06”

E = 20 x 106 psiD = . 3ρ = .283 lb./In3 (weight density)

Applied Loads1. Pressure on square portions of 1 psi2. Normal force of 2 lb on Grids points 93 and 1043. Opposing normal forces of 2lb on Grid points 93 and 104

SE# Elements 1 18 – 42 2 43 – 87 3 14 – 15 4 16 – 17 5 6 – 9 6 10 – 13 7 1 – 4 0 5


MODEL DEFINITION FOR SAMPLE PROBLEM

BEGIN BULK$$

*******************************************************************

$ BASIC MODEL DEFINITION - SAME FOR ALL RUNS

$ *******************************************************************

$GRDSET,,,,,,,6GRID,1,,-.4,0.,0.,,123456GRID,3,,-.4,0.9,0.=,*2,=,=,*.9,===1GRID,2,,.4,0.,0.,,123456GRID,4,,.4,0.9,0.=,*2,=,=,*.9,===1GRID,9,,-3.6,3.6,0.=,*1,=,*.8,===8GRID,19,,-3.6,4.4,0.=,*1,=,*.8,===8GRID,29,,-3.6,5.2,0.GRID,30,,-2.8,5.2,0.GRID,31,,2.8,5.2,0.GRID,32,,3.6,5.2,0.GRID,33,,-5.2,6.,0.=,*1,=,*.8,===4

GRID,39,,1.2,6.,0.=,*1,=,*.8,===4GRID,45,,-5.2,6.8,0.=,*1,=,*.8,===4GRID,51,,1.2,6.8,0.=,*1,=,*.8,===4GRID,57,,-5.2,7.6,0.=,*1,=,*.8,===4GRID,63,,1.2,7.6,0.=,*1,=,*.8,===4GRID,69,,-5.2,8.4,0.=,*1,=,*.8,===4GRID,75,,1.2,8.4,0.=,*1,=,*.8,===4GRID,81,,-5.2,9.2,0.=,*1,=,*.8,===4GRID,87,,1.2,9.2,0.=,*1,=,*.8,===4GRID,93,,-5.2,10.,0.=,*1,=,*.8,===4GRID,99,,1.2,10.,0.=,*1,=,*.8,===4


MODEL DEFINITION FOR SAMPLE PROBLEM (Cont.)

$$ ELEMENTS$CQUAD4,1,1,1,2,4,3=,*1,=,*2,*2,*2,*2=1CQUAD4,4,1,7,8,14,13CQUAD4,6,1,9,10,20,19=,*1,=,*1,*1,*1,*1=2CQUAD4,5,1,13,14,24,23CQUAD4,10,1,14,15,25,24= *1,=,*1,*1,*1,*1=2CQUAD4,14,1,19,20,30,29CQUAD4,15,1,29,30,36,35CQUAD4,16,1,27,28,32,31CQUAD4,17,1,31,32,42,41CQUAD4,18,1,33,34,46,45=,*1,=,*1,*1,*1,*1=3CQUAD4,23,1,45,46,58,57=,*1,=,*1,*1,*1,*1=3CQUAD4,28,1,57,58,70,69=,*1,=,*1,*1,*1,*1=3CQUAD4,33,1,69,70,82,81=,*1,=,*1,*1,*1,*1=3

CQUAD4,38,1,81,82,94,93=,*1,=,*1,*1,*1,*1=3CQUAD4,43,1,39,40,52,51=,*1,=,*1,*1,*1,*1=3CQUAD4,48,1,51,52,64,63=,*1,=,*1,*1,*1,*1=3CQUAD4,53,1,63,64,76,75=,*1,=,*1,*1,*1,*1=3CQUAD4,58,1,75,76,88,87=,*1,=,*1,*1,*1,*1=3CQUAD4,63,1,87,88,100,99=,*1,=,*1,*1,*1,*1=3MAT1,1,30.+6,,.3,.283PARAM,WTMASS,.00259PSHELL,1,1,.05,1,,1$$ LOADINGS$$ LOAD CASE 1 - PRESSURE LOAD$PLOAD2,101,-1.,18,THRU,42PLOAD2,101,-1.,43,THRU,67$


MODEL DEFINITION FOR SAMPLE PROBLEM (Cont.)

$ LOAD CASE 2 - 2 POINT LOADS AT CORNERS

$

FORCE,201,93,,2.,0.,0.,1.

FORCE,201,104,,2.,0.,0.,1.

$

$ LOAD CASE 3 - OPPOSING POINT LOADS AT CORNERS

$

FORCE,301,93,,2.,0.,0.,1.

FORCE,301,104,,2.,0.,0.,-1.

$ *******************************************************************

$ END OF BASIC MODEL DEFINITION

$ *******************************************************************

ENDDATA


SAMPLE PROBLEM—STEEL STAMPINGSESET ENTRIES FOR MAIN BULK DATA

SUPERELEMENTS$ FILE SESET.DAT$$ DEFINE S.E. MEMBERSHIP OF GRID POINTS FOR SINGLE–LEVEL SUPERELEMEMT$ SAMPLE PROBLEM$SESET,1,33,34,37,38SESET,1,45,THRU,50SESET,1,57,THRU,62SESET,1,69,THRU,74SESET,1,81,THRU,86SESET,1,93,THRU,98$SESET,2,39,40,43,44SESET,2,51,THRU,56SESET,2,63,THRU,68SESET,2,75,THRU,80SESET,2,87,THRU,92SESET,2,99,THRU,104$SESET,3,29,30$SESET,4,31,32$SESET,5,21,22SESET,5,9,THRU,12$SESET,6,25,26SESET,6,15,THRU,18$SESET,7,1,THRU,8$


PARTITIONED SOLUTIONS For each superelement, its degrees-of-freedom (DOFs) are

divided into two subsets: Exterior DOFs (called the A-set): Designates the analysis DOFs, which are

retained for subsequent processing (for Superelement 1, Grid Points 35 and 36)

Interior DOFs: Designates the DOFs that are reduced out during superelement processing and are omitted in subsequent processing (for Superelement 1 of the sample problem, Grid Points 33, 34, 37,38, 45–50, 57–62, 69–74, 81–86, 93–9 8).

The Main Bulk Data is partitioned by superelement (although the following operations are performed using tables, it is easier to think of them in terms of the Bulk Data). All Bulk Data unique to the superelement is removed from the original input

and placed into a unique set for the superelement. Bulk Data that is ‘shared’ or used by more than one superelement (ex:

PSHELL, MAT1, etc.) is copied for each applicable superelement.

PARTs are already separated.


PARTITIONED SOLUTIONS (Cont.) For each superelement, the program produces a description in

matrix terms of its behavior as seen at the boundary or exterior degrees of freedom. A set of ‘G’-sized matrices is produced for each superelement based on the

input data. These matrices are reduced down to matrices representing the properties of the

superelement as seen by the adjacent (attached) structure.

At the residual structure, the program combines and assembles the boundary matrices. The BULK DATA for the RESIDUAL consists of all ‘residual’ Main Bulk Data

not assigned to any superelement plus any common data.

Solve for the residual structure displacements. For each superelement, expand boundary (exterior)

displacements to obtain its interior displacements.


THEORY OF STATIC CONDENSATIONAfter generating matrices and applying MPCs and SPCs,

O-Set = Interior points (to be condensed out by the reduction)A-Set = exterior (or boundary) points (which are retained for

further analysis)Partition

Extract upper equation and pre-multiply by

Let (Boundary Transformation)

and (Fixed Boundary Displacements)

then (Total Interior Displacements)

ffff PUK =

=

a

o

a

o

aaToa

oaoo

PP

UU

KKKK

1ooK −

o1

ooaoaooo1

oo PK]UKU[KK −− =+

o1

oooo

oa1

oooa

PKU

KKG−

−

=

−=

aoaooo UGUU +=


THEORY OF STATIC CONDENSATION(Cont.)

Substitute expression for U o in the lower equation

then (Boundary Stiffness)

and (Boundary Loads)

Solve for residual structure

(Boundary displacements)

aaaaooaoa

Toa PUK]UU[GK =++

aaoaToaaa KGKK +=

aoToaa PPGP +=

a-1aaa PKU =


CONVENTIONAL ANALYSIS

Flowchart

Generation

Solution


CONVENTIONAL ANALYSIS(Cont.)

Generation

−−+−

−+−−+−

−

=

4545

45453434

34342323

23231212

1212

GG

KK000KKKK000KKKK000KKKK000KK

][K

−−−

−−−−

−

=

100000

00000001

KGG

1121

121121

1


CONVENTIONAL ANALYSIS(Cont.)Apply Constraints and Solve

+−−+−

−+=

−

4

3

21

453434

34342323

232312

4

3

2

PPP

KKKKKKK

KKK

UUU

−−−

−=

−

321

UUU 1

4

3

2

210121

012

=

3.54.02.5

UUU

4

3

2


SUPERELEMENT ANALYSISFlowchart

DO LABELAI = 1, NSE

Phase IGenerationAssemblyReduction

LABELA

Phase IISolution

DO LABELBI = 1, NSE

Phase IIIData Recovery

LABELB


SUPERELEMENT ANALYSIS (Cont.)

Generation –SEID = 1

Residual Structure

−−+−

−=

2323

23231212

12121

gg

KK0KKKK0KK

][K


SUPERELEMENT ANALYSIS (Cont.)

Reduction – SEID = 1Eliminate constraints:

Compute boundary transformation:

=

=

010

PPP

}{P13

2

11

g

=

−

−+=

aaao

oaoo

2323

2323121ff

KKKK

KKKKK

][K

0.5KK

K

][K][K][G

2312

23

oa1

oo1

oa

=+

=

−= −


SUPERELEMENT ANALYSIS (Cont.) Compute boundary stiffness:

Compute boundary loading:

0.5KK

KKK

]GKK[][K

2312

23121aa

oaToaaa

1aa

=+

=

+=

=

=

=

a

0

13

21f

PP

01

PP

}{P

0.5PKK

KPP

}PGP{}{P

22312

2313

13

oToaa

1a

=+

+=

+=0


SUPERELEMENT ANALYSIS (Cont.) Similarly – SEID = 2

=

=

−−+−

−=

030

PPP

}{P

KK0KKKK0KK

][K

5

4

23

2g

4545

45453434

34342

gg

.5PKK

K

0.5KK

KK

0.5KK

K

44534

34

4534

4534

4534

34

1PP

]K[

]G[

23

23

2aa

2oa

=+

+=

=+

=

=+

=

0


SUPERELEMENT ANALYSIS (Cont.)Residual Structure

Assembly

Solution4PPPP

}PP{P}{P

1KKK

]KK[K][K

03

23

13

0g

2a

1aa

21

0gg

2aa

1aaaa

=++=

++=

=+=

++=

4KPU

}{P][K}{U

03

a1

aaa

==

= −

0


SUPERELEMENT ANALYSIS (Cont.)Data Recovery – SEID = 1Enforce (transform) boundary motion.

Compute fixed-boundary motion.

Compute total motion.

2.0UKK

KU

}]{U[G}{U

32312

2332

aoaao

=⋅+

=

=

0.5PKK

1U

}{P][K}{U

22312

o2

o-1

oooo

=+

=

=

2.5KK

PUKU

}{U][U}{U

2312

23232

ao

ooo

=+

+⋅=

+=


BULK DATA FOR STATIC LOADS ON SUPERELEMENTS

Main Bulk Data Superelements: Loads applied to interior grid points are assigned to the

superelement. Loads applied to exterior grid points are assigned to the most

downstream superelement, that is, the superelement for which the grid point is interior.

Loads applied to elements (PLOADi) are assigned in the same manner as elements.Note: A PLOAD entry may not reference the interior points of more

than one superelement.

Partitioned Superelements: Any Loading–related entries must be defined in the partitioned data

(in the area of the input file beginning with ’BEGIN SUPER =’)


STATIC LOADS ON MAIN BULK DATA SUPERELEMENTS

Example

SESET, 1, 4, 5, 6 Grids 4, 5, and 6 are interior points to Superelement 1. Point 3 is exterior to Superelement 1. P2 is assigned to Superelement 0. W and P1 is assigned to Superelement 1.

Superelement 0Superelement 1


MAIN BULK DATA SUPERELEMENTS Constraint entries applied to the interior points of a superelement

are assigned to that superelement. Constraint entries applied to the exterior points of a superelement

are “sent” downstream. Multiple boundary conditions are allowed for the residual structure

only For multiple boundary conditions, place grid points that will be

constrained interior to the residual structure. Each superelement may have only one SPC set per run.

PARTITTIONED SUPERELEMENTS All constraint–related bulk data entries for the interior points of a

PART must be defined in the partitioned bulk data(BEGIN SUPER=).

SINGLE-POINT CONSTRAINSTS ON SUPERELEMENTS


SINGLE-POINT CONSTRAINSTS ON SUPERELEMENTS (Cont.)

SESET, 1, 4,5, 6 Grid Points 4, 5, and 6 are interior to Superelement 1. Point 3 is exterior to Superelement 1. SPC at 3 is assigned to Superelement 0. SPC at 6 is assigned to Superelement 1.

Superelement 0 Superelement 1


MPCs AND RIGID ELEMENTS IN SUPERELEMENTS

Rigid elements and MPCs that connect only interior points are modeled conventionally.

Dependent degrees of freedom may not be exterior. For MPCs and rigid elements that connect two

superelements, Place the upstream degrees of freedom in the dependent set. Place the downstream degrees of freedom in the independent set.

Multiple multipoint constraint conditions are allowed for the residual structure only For multiple multipoint constraints, place grid points that will be

specified on these interior to the residual structure. Each superelement may have only one MPC set per run. (Note:

MPCADD may be used.)


RIGID CONNECTION OF TWO SUPERELEMENTS

METHOD 1 – RBAR

METHOD 2 – MPC

$ SEID GP1 GP2 GP3 ETC.SESET 1 11 12 13$ EID GA GB CNA CNB CMA CMBRBAR 100 3 11 123456 123456

$ SEID GP1 GP2 GP3 ETC.SESET 1 11 12 13$ SID G C A G C AMPC 100 11 1 1. 3 1 –1.= = = *(1) = = *(1) ==(4)

CSUPEXT 1 3

$ EID G1 G2PLOTEL 100 3 11

or

Residual Structure ? SEID = 1

CBAR Rigid Connection


SUPERELEMENT CASE CONTROL COMMANDS

SE-type (manual processing) – SEMG, SELG, SEKR, SELR, SEMR, SEDR, and SEALL – appear above the first SUBCASE if used Control solution sequence execution Make no requests for loads, constraints, or output SEALL combines SEMG, SELG, SEKR, SELR, and SEMR Not necessary in SOL 101 and higher (default is SEALL=ALL, which implies

that all necessary processing will be performed)

Superelement processing order control – appear above the first SUBCASE if used SEFINAL – Last superelements to be processed before residual structure –

not recommended SEEXCLUDE – Superelements not to be assembled downstream

Case Control partitioning – SUPER Assigns a subcase(s) to a specific superelement(s) Appears above or below subcase level


SUPER COMMAND Partitions (assigns) a subcase to a superelement(s) Associates a superelement(s) with requests for parameters, loads,

constraints, and output Pre–V69 – If the Case Control Section does not contain a SUPER

command, then loads, constraints, and output requests are applied to the residual structure only (the old default was SUPER = 0).

V69 The new default is SUPER=ALL. if no SUPER command is present, the subcases are assumed to apply to ALL superelements (if any SUPER commands occur in the Case Control, the default reverts to SUPER=0 for upward compatibility).

The SUPER command may reference a superelement or a SET of superelements.Note: The SET ID must be unique with respect to any superelement IDs.

Form of SUPER commandSUPER = i, j

where i = superelement ID or set of superelementsj = load sequence number (a counter on loading conditions)


EXPANDED VERSUS CONDENSED

Conventional Case Control Expanded

Condensed

SUBCASE 10SET 1 = 101 THRU 110DISP = 1LOAD = 100



SET 1 = 101 THRU 110SET 3 = 201 THRU 210DISP = 1LOAD = 200SUBCASE 10LOAD = 100SUBCASE 20SUBCASE 30DISP = 3


EXPANDED VERSUS CONDENSED (Cont.)Superelement Case Control Expanded – one loading condition

Condensed

$ model with superelements 10, 20, 0DISP = ALLSUBCASE 1 $ SE 10

SUPER = 10LOAD = 100

SUBCASE 2 $ SE 20SUPER = 20LOAD = 100

SUBCASE 101 $ RESIDUAL STRUCTURESET 999 = 0SUPER = 999LOAD = 100

BEGIN BULK

SUBCASE 1

DISP = ALL

LOAD = 100

BEGIN BULK


SUPER COMMAND EXAMPLE—ONE LOADING CONDITION

$ SEIDS 1, 2, 3, 4, 5, 0DISP = ALLSUBCASE 10

SET 101 = 1, 4SUPER = 101SPC = 12

SUBCASE 20SET 103 = 2, 5SUPER = 103SET 15 = 7, 9ELFOR = 15LOAD = 9

SUBCASE 30SUPER = 0ELSTRE = ALL

SPC LOAD DISP ELSTRE ELFOR1 X X2 X X X34 X X5 X X X0 X X

SEIDConditions Output Requests


MULTIPLE LOADING CONDITIONS IN SUPERELEMENT

CASE CONTROL – OPTION 1

Appears identical to conventional Case Control For each loading, create one subcase (use the

default SUPER=ALL) Option 1 requires

All superelements must use the same loading, SPC, and MPC sets.


MULTIPLE LOADING CONDITIONS IN SUPERELEMENT

CASE CONTROL – OPTION 2 For the residual structure

Define a subcase for each loading condition.

For each superelement (or set of superelements) Define a subcase for each loading condition using a SUPER command

identifying the superelement (or a set of superelements) and the loading sequence number.

SUBCOMs are treated as a new load sequence and, therefore, must have a SUPER command and the residual structure must have a corresponding subcase or subcom.

REPCASEs must immediately follow the subcase they reference and contain the same SUPER=i,j command.


MULTIPLE LOADING CONDITIONS EXAMPLE -- OPTION 2

SEALL = ALLDISP = ALLSPC = 10SUBCASE 1 $ SEID 10 LOAD SEQ 1

SUPER = 10, 1LOAD = 100

SUBCASE 2 $ SEID 10 LOAD SEQ 2SUPER = 10, 2ELFORCE = ALL

SUBCASE 12 $ SEID 20 LOAD SEQ 2SUPER = 20, 2LOAD = 200

SUBCASE 101 $ R.S. LOAD SEQUENCE 1SUPER = 0,1GPFOR = ALL

SUBCASE 102 $ R.S. LOAD SEQUENCE 2SUPER = 0,2LOAD = 1000

DISP ELSTRE ELFOR10 100 X200 X X

10 X X20 200 X0 1000 X

1

2

Load SEQ

Output RequestsSEID

Load Set ID


REASONS TO USE OPTION 2 FOR MULTIPLE LOADINGS

It allows different LOAD, SPC, MPC IDs, etc., for each superelement.

Each superelement may have unique output requests. It may be the only way to perform an analysis if

groups have not coordinated their efforts.


MULTIPLE LOADINGS – SAMPLE OF OPTION 1

Coordinated input allows for simple Case Control P1 and W1 are applied for loading 1 P2 is applied for loading 2

SOL 101TIME 5CENDTITLE = SAMPLE OF OPTION 1 FOR MULTIPLE LOADINGSDISP = ALL $ DEFAULT CASE CONTROL BEFORE FIRST$ SUPER = ALL is now the defaultSUBCASE 1LOAD = 1SUBCASE 2LOAD = 2BEGIN BULK..ENDDATA



MULTIPLE LOADINGS – SAMPLE OF OPTION 2

Uncoordinated input forces complicated Case Control P1 and W 1 are applied for loading 1 for Superelement 1.

P2 is applied for loading 1 on the residual structure. P1 is applied on the residual structure for loading 2.

SOL 101TIME 5CENDTITLE = UNCOORDINATED INPUT FORCES COMPLEX CASE CONTROLDISP = ALLSET 99 = 0SUBCASE 1SUPER = 1,1 $ S.E. 1, LOAD CONDITION 1LOAD = 1SUBCASE 2SUPER = 99,1 $ R.S., LOADING 1LOAD = 2SUBCASE 11SUPER = 1,2 $ S.E. 1, LOAD CONDITION 2$ NO LOADS APPLIED DIRECTLY ON S.E. 1 – SUBCASE ONLY FOR$ DATA RECOVERYSUBCASE 12SUPER = 99,2 $ R.S., LOAD CONDITION 2LOAD = 1BEGIN BULK.ENDDATA



PARAMETERS IN CASE CONTROL Allows changes between superelements on same run Most, but not all, can be used in Case Control. There is a hierarchical rule for what value used will

be. Subcase value first Above subcase level value if not in a subcase Bulk Data value if not in either of the above Default value if not in any of the above

The default is taken from the main subDMAP if one exists. If not in main subDMAP from the called subDMAP If NDDL, the default is from the NDDL default table.

Recommendations Specify the parameter value for each subcase (safe).

or Specify the default value above the subcase level and exceptions

within subcases.


SAMPLE SUPERELEMENT STATIC RUN INPUT

ID SE, SAMPLE PROBLEM SOL 101$$ SUPERELEMENT STATICS – SAMPLE PROBLEM – STATIC SOLUTION$ USING SIMPLE CASE CONTROL$SOL 101 $ SUPERELEMENT STATICS – SINGLE LEVEL TREETIME 15CENDTITLE = S.E. SAMPLE PROBLEM 1SUBTITLE = S.E. STATICS – RUN 1 – MULTIPLE LOADSDISP = ALLPARAM,GRDPNT,0SUBCASE 101LABEL = PRESSURE LOADLOAD = 101$SUBCASE 201LABEL = 2# NORMAL LOADSLOAD = 201$SUBCASE 301LABEL = OPPOSING LOADSLOAD = 301$$BEGIN BULKPARAM,POST,0$INCLUDE ’seset.dat’INCLUDE ’model.dat’INCLUDE ’load1.dat’$ENDDATA

File – se1s101.dat


SAMPLE SUPERELEMENT STATIC RUN INPUT (Cont.)

$ FILE LOAD1.DAT$$ LOADINGS – FOR RUN SHOWING CONVENTIONAL CASE CONTROL$$ LOAD CASE 1 – PRESSURE LOAD$$ NOTE: THRU RANGE SHOULD INCLUDE ELEMENTS OF ONLY ONE SUPERELEMENT$PLOAD2,101,–1.,18,THRU,42PLOAD2,101,–1.,43,THRU,67$$ LOAD CASE 2 – 2 POINT LOADS AT CORNERS$FORCE,201,93,,2.,0.,0.,1.FORCE,201,104,,2.,0.,0.,1.$$ LOAD CASE 3 – OPPOSING POINT LOADS AT CORNERS$FORCE,301,93,,2.,0.,0.,1.FORCE,301,104,,2.,0.,0.,–1.$

File – load1.dat


SAMPLE SUPERELEMENT STATIC RUN INPUT —USING PARTS

$ file – se1s101p.datSOL 101CENDTITLE = S.E. SAMPLE PROBLEM 1 USING PARTsSUBTITLE = S.E. STATICS – RUN 1 – MULTIPLE LOADSDISP = ALLstress = allPARAM,GRDPNT,0PARAM,WTMASS,.00259SUBCASE 101LABEL = PRESSURE LOADLOAD = 101$SUBCASE 201LABEL = 2# NORMAL LOADSLOAD = 201$SUBCASE 301LABEL = OPPOSING LOADSLOAD = 301BEGIN BULKinclude ’part0.dat’ $ main bulk data sectionbegin super=1$include ’loadprt1.dat’include ’part1.dat’begin super=2

File – se1s101p.DAT$include ’loadprt2.dat’include ’part2.dat’begin super=3$include ’part3.dat’begin super=4$include ’part4.dat’begin super=5$include ’part5.dat’begin super=6$include ’part6.dat’begin super=7$include ’part7.dat’enddata



$$ file – loadprt1.dat$ loads on s.e. 1$$ LOAD CASE 1 – PRESSURE LOAD$PLOAD2,101,–1.,18,THRU,42$$ LOAD CASE 2 – 2 POINT LOADS AT CORNERS$FORCE,201,93,,2.,0.,0.,1.$$ LOAD CASE 3 – OPPOSING POINT LOADS AT CORNERS$FORCE,301,93,,2.,0.,0.,1.$

File – loadprt1.dat



$$ file – loadprt2.dat$ loads on s.e. 2$$ LOAD CASE 1 – PRESSURE LOAD$PLOAD2,101,–1.,43,THRU,67$$ LOAD CASE 2 – 2 POINT LOADS AT CORNERS$FORCE,201,104,,2.,0.,0.,1.$$ LOAD CASE 3 – OPPOSING POINT LOADS AT CORNERS$FORCE,301,104,,2.,0.,0.,–1.$

File – loadprt2.dat


SUPERELEMENT REDUCTION METHODS AVAILABLE IN DYNAMIC ANALYSIS

Static reduction Static condensation of stiffness and Guyan reduction of mass

Static reduction is the default

Dynamic reduction Generalized dynamic reduction (GDR) (not recommended) Component modal synthesis (CMS)

Analytical (All SE dynamic SOLs)


DEGREES OF REDUCTION Static reduction (default)

Interior masses relumped to boundary (Guyan) Rigid body properties preserved Important masses must be made exterior (boundary)

Generalized dynamic reduction – in addition to static reduction Interior masses represented by approximate eigenvectors Approximate natural frequencies and mode shapes may be output

Component mode reduction – in addition to static reduction Interior masses represented by calculated eigenvectors of the component Eigensolutions for each superelement may be output

All reductions are performed using a set of transformation vectors –these vectors are best thought of as Ritz vectors


COMPARISON OF REDUCTION METHODS

Static reduction

Generalized dynamic reduction

Approximate eigenvectors are used to represent the interior motion.

Component mode reduction

Exact eigenvectors are used to represent the interior motion.

}{U}]{u[G}{U oototo +=

0 Local dynamic effects are ignored.

=q

toqoto U

U]G [G}{U

=q

toqoto U

U]G [G}{U


ADVANTAGES OF EACH REDUCTION METHOD

Advantages of Component Mode Reduction over Static Reduction Can use experimental results More accurate for the same number of dynamic DOFs Ideal for highly coupled and uncoupled structures

Advantages of Static Reduction over Component Mode Reduction Cheaper Less sophisticated


CALCULATION OF NORMAL MODES USING

STATIC REDUCTION ONLY This is the default method used to reduce

superelements is always be performed Superelement mass, damping, and stiffness are

reduced statically to exterior DOFs. Case Control is similar to static analysis with the

addition of a METHOD command under the residual structure subcase.


CALCULATION OF NORMAL MODES USING DYNAMIC REDUCTION FOR

SUPERELEMENT Dynamic reduction of superelements is optional and is performed in

addition to static (Guyan) reduction if requested The behavior of a superelement is represented by its real modes in

addition to the static shapes. The superelement stiffness, mass, and damping are transformed using

both physical and modal variables. The superelement modes are computed if a METHOD command appears

under the superelement subcase and SEQSETi entries are specified for the superelement (QSETi or SENQSET for PARTs).

The number of superelement modes computed (modal truncation) is controlled by the EIGRL entry.

The number of superelement modes sent downstream is controlled by the number of Q–set DOFs provided.

SEQSETi entries can reference GRID points or SPOINTs By default, superelement modes are computed with all exterior degrees of

freedom fixed (in the B-set). This is better known as the Craig-Bampton method.


CALCULATION OF NORMAL MODES USING DYNAMIC REDUCTION FOR

SUPERELEMENT (Cont.) For free-free superelement component modes, all exterior DOFs should

be specified on SECSETi entries (use of the SESUP is not recommended). The rigid-body modes (f=0.0 Hz) are a linear combination of the static vectors and

should not be included in the reduction.Either:

Do not calculate them (F1>0.0 on the EIGR or EIGRL entry). Calculate them and hope that the program will remove them (see PARAM,ERSRC in the

MSC.NASTRAN User’s Manual). Calculate them and remove them by using the SESUP entry. (For every exterior DOF listed on

the SESUP entry, one eigenvector is “thrown away.”)

Mixed-boundary modes may be calculated by using the SECSETi and SEBSETi entries to describe the exterior DOFs to be unconstrained and constrained during CMS. If 0.0 Hz mixed boundary modes exist, they must be handled in a similar manner to

those in the free-free case. For most problems, the default (Craig–Bampton) method will be

adequate. The accuracy of the transformation is dependent on the number of component modes used, no matter which dynamic reduction method is used.


“FIXED BOUNDARY” SOLUTIONS (PARAM, FIXEDB, -1)

Statics

Allows output of the superelement component modes in dynamics

where z implies superelement component modesv indicates the v-set ( 0 + R + C)

Allows checkout of one superelement at a time – displacements, stresses, deformed plots, etc. – any standard data recovery option.

In SOL 63 after checkout, PARAM,RESDUAL,–1 may be used to restart for system (residual structure) modes.

}]{[G }{U }{ ooo ao UU +=

0.0 if FIXEDB = -1

Motion Due to Boundary

Displacements

Motion Due to Interior LoadsTotal Motion of

Interior Points

0]][MK[ vzww =− φλSuperelement Modes (Are Printed if FIXEDB = -1)


PROCEDURES FOR SUPERELEMENT DYNAMIC REDUCTION

Component boundary conditions Fixed-fixed

Default – All exterior DOFs are automatically placed in the B-set. Free-free

Specify all exterior DOFs in C-set. Specify PARAM,INRLM,–1 in Case Control for more accuracy.

Mixed Define exterior DOFs in C- and B-sets as desired.


PROCEDURES FOR SUPERELEMENT DYNAMIC REDUCTION (Cont.)

Residual structure If static reduction is desired, specify selected physical DOFs in the

A-set.Note: If CMS has been performed for upstream superelements, the

generalized coordinates from the superelements should be in the A-set in order to be included in the final solution.

If GDR or residual structure CMS is used, no physical DOFs in A-set are required.


SPECIFICATION OF FIXED AND FREE BOUNDARY DEGREES OF FREEDOM

Set DefinitionB Fixed during GDR or CMRC Free during GDR or CMR

Entry Type

SECSETi No No Yes YesSEBSETi No Yes No YesUndefined Exterior DOFs Placed In

B C B B

Present?


REFERENCES FOR CMS W. C. Hurty, “Dynamic Analysis of Structural Systems Using

Component Modes,” AIAA Journal, Vol. 3, No. 4, April 1965 (Based upon JPL Tech. Memo 32-530, January 1964).

R. H. MacNeal, “A Hybrid Method of Component Mode Synthesis,” Computers & Structures, Vol. 1, 1971.

R. R. Craig and M. C. C. Bampton, “Coupling of Substructures for Dynamic Analysis,” AIAA Journal, Vol. 6, No. 7, July 1968.

W. A. Benfield and R. F. Hruda, “Vibration Analysis of Structures by Component Mode Substitution,” presented at AIAA/ASME 11th Structures, Structural Dynamics, and Materials Conference, Denver, CO, April 1970.

S. Rubin, “An Improved Component-Mode Representation,” presented at AIAA/ASME 15th Structures, Structural Dynamics, and Materials Conference, Las Vegas, NV, April 1974.

R. R. Craig, Structural Dynamics: An Introduction to Computer Methods, John Wiley and Sons, New York, 1981.

E. D. Bellinger, “Component Mode Synthesis for External Superelements,” MSR-71, Los Angeles, May 1981, (SOLs 41, 42, 43).


SUPERELEMENT DYNAMICS EXAMPLE

Cantilever beam modeled with two superelements

Beam propertiesA = 5 in2

I = 50.66059 in4

Material propertiesE = 10,000,000 psip = 0.01 lb-sec2 / in4


SUPERELEMENT DYNAMICS EXAMPLE (Cont.)

Compute first five system modes using the following techniques: Static reduction Assume fixed exterior points.

Generalized dynamic reduction (GDR) Component mode reduction (CMR) GDR and CMR

Assume all free exterior points with CMR.



$ FILE SEDYNBLK.DAT$DYNRED,1,100.EIGR,37,MGIV,,,,5SPC1,10,26,1001SPC1,10,1345,1001,THRU,1011SPC1,10,1345,2001,THRU,2016RBE2,1001,1011,26,2001GRID,1001,,0.=,(1),=,(2.),===(9)GRID,2001,,20.=,(1),=,(2.),===(14)CBAR,111,10,1001,1002,,1.=,(1),=,(1),(1),===(8)CBAR,211,10,2001,2002,,1.=,(1),=,(1),(1),===(13)PBAR,10,10,5.,50.66059,12.66516MAT1,10,1.+4,,.3,.01PARAM,COUPMASS,1

Bulk Data Input



Static Reduction Only For accuracy, assign six evenly-spaced points along the beam to the residual

structure. Without an ASETi entry, ALL DOFs in the residual structure belong to the A-set.

The SUPER=ALL and METHOD commands tell MSC.NASTRAN to perform CMS on all superelements, but the lack of SEQSET prevents it and a static reduction is performed. (System modes are found at the residual.)

$ FILE = SEDYN1.DAT$SOL 103TIME 5CENDTITLE = SUPERELEMENT CMS SAMPLE – RUN 1SPC = 10SEALL = ALL $ ONLY REQUIRED IF SOL<101SUPER=ALLMETHOD = 37BEGIN BULKSESET,0,1001,1006,1011SESET,0,2001,2006,2011,2016SESET,100,1002,THRU,1005SESET,100,1007,THRU,1010SESET,200,2002,THRU,2005SESET,200,2007,THRU,2010SESET,200,2012,THRU,2015$INCLUDE SEDYNBLK.DATENDDATA



Static Reduction Only (Cont.)SUPERELEMENT CMS SAMPLE – RUN 1 MAY 2, 1990 MSC.NASTRAN 1/ 4/ 89 PAGE 20

SUPERELEMENT 0

R E A L E I G E N V A L U E S

MODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZED

NO. ORDER MASS STIFFNESS

1 1 2.004165E+01 4.476790E+00 7.125033E–01 1.000000E+00 2.004165E+01

2 2 7.878806E+02 2.806921E+01 4.467353E+00 1.000000E+00 7.878806E+02

3 3 6.215354E+03 7.883752E+01 1.254738E+01 1.000000E+00 6.215354E+03

4 4 2.425566E+04 1.557423E+02 2.478715E+01 1.000000E+00 2.425566E+04

5 5 6.681884E+04 2.584934E+02 4.114050E+01 1.000000E+00 6.681884E+04



Generalized Dynamic Reduction Assign to the residual structure only the

superelement endpoints that are assumed to be fixed for GDR.

Specify Q-set (SEQSET1) along with the corresponding variables (SPOINT).

Request GDR (DYNRED) for both superelements and eigensolution (METHOD) for the residual structure.



$ FILE = SEDYN2.DAT$SOL 103TIME 5CENDTITLE = SUPERELEMENT CMS SAMPLE – RUN 2$SEALL = ALL $ ONLY REQUIRED IF SOL<101SPC = 10SUBCASE 1 $ SUPERELEMENTSSET 47 = 100,200SUPER = 47DYNRED = 1SUBCASE 2METHOD = 37BEGIN BULK$ definition of SE 0 not requiredSESET,0,1001,1011,2001,2016SESET,100,1002,THRU,1010SESET,200,2002,THRU,2015$$ MODAL VARIABLES$SPOINT,101,THRU,125SPOINT,201,THRU,230SEQSET1,100,0,101,THRU,125SEQSET1,200,0,201,THRU,230INCLUDE SEDYNBLK.DATENDDATA

For accuracy, it is recommended to use the Lanczos method to perform CMS, rather than GDR.

Generalized Dynamic Reduction (Cont.)



SUPERELEMENT CMS SAMPLE – RUN 2 JUNE 26, 1990 MSC. NASTRAN 10/ 20/ 89 PAGE 15 SUPERELEMENT 100

*** USER INFORMATION MESSAGE––– PROCESSING OF SUPERELEMENT 100 IS NOW INITIATED.^^^ PHASE 1 – SUPERELEMENT GENERATION, ASSEMBLY AND REDUCTION.*** USER INFORMATION MESSAGE 4158––– STATISTICS FOR SYMMETRIC DECOMPOSITION OF

DATA BLOCK SCRATCH FOLLOW NUMBER OF NEGATIVE TERMS ON FACTOR DIAGONAL = 2*** USER INFORMATION MESSAGE 4181––– NUMBER OF ROOTS BELOW 0.1000E+ 03 CYCLES IS 2

NUMBER OF GENERALIZED COORDINATES SET TO 6SUPERELEMENT CMS SAMPLE – RUN 2 JUNE 26,1990 MSC.NASTRAN 10/20/ 89 PAGE 16

SUPERELEMENT 200*** USER INFORMATION MESSAGE––– PROCESSING OF SUPERELEMENT 200 IS NOW INITIATED.^^^ PHASE 1 – SUPERELEMENT GENERATION, ASSEMBLY AND REDUCTION.*** USER INFORMATION MESSAGE 4158––– STATISTICS FOR SYMMETRIC DECOMPOSITION OF

DATA BLOCK SCRATCH FOLLOWNUMBER OF NEGATIVE TERMS ON FACTOR DIAGONAL = 3

*** USER INFORMATION MESSAGE 4181––– NUMBER OF ROOTS BELOW 0.1000E+ 03 CYCLES IS 3

Generalized Dynamic Reduction (Cont.)



NUMBER OF GENERALIZED COORDINATES SET TO 6SUPERELEMENT CMS SAMPLE – RUN 2 JUNE 26, 1990 MSC.NASTRAN 10/ 20/ 89 PAGE

19SUPERELEMENT 0

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 2.004111E+01 4.476730E+00 7.124937E–01 1.000000E+00 2.004111E+012 2 7.870997E+02 2.805530E+01 4.465139E+00 1.000000E+00 7.870997E+023 3 6.171406E+03 7.855830E+01 1.250294E+01 1.000000E+00 6.171406E+034 4 2.370312E+04 1.539582E+02 2.450320E+01 1.000000E+00 2.370312E+045 5 6.478747E+04 2.545338E+02 4.051031E+01 1.000000E+00 6.478747E+046 6 1.446095E+05 3.802755E+02 6.052273E+01 0.0 0.07 7 2.829334E+05 5.319149E+02 8.465688E+01 0.0 0.08 8 5.007846E+05 7.076614E+02 1.126278E+02 0.0 0.09 9 8.291153E+05 9.105577E+02 1.449198E+02 0.0 0.010 10 1.302097E+06 1.141094E+03 1.816108E+02 0.0 0.011 11 1.942606E+06 1.393774E+03 2.218260E+02 0.0 0.012 12 2.978247E+06 1.725760E+03 2.746632E+02 0.0 0.013 13 5.330453E+06 2.308777E+03 3.674533E+02 0.0 0.014 14 1.028805E+07 3.207499E+03 5.104893E+02 0.0 0.015 16 2.174905E+07 4.663588E+03 7.422330E+02 0.0 0.016 15 3.803833E+07 6.167522E+03 9.815917E+02 0.0 0.0



Component Modal Synthesis Assign to the residual structure only the superelement endpoints that are assumed to

be fixed for calculation of component modes. Specify Q-set (SEQSET1) for each superelement along with the corresponding modal

variables (SPOINT). Request eigensolution (METHOD) for both superelements and the residual structure.

$ FILE = SEDYN3.DAT$SOL 103TIME 5CENDTITLE = SUPERELEMENT CMS SAMPLE – RUN 3SPC = 10$SEALL = ALL $ ONLY REQUIRED IF SOL<101SET 99 = 0,100,200SUPER = 99METHOD = 37BEGIN BULKSESET,0,1001,1011SESET,0,2001,2016SESET,100,1002,THRU,1010SESET,200,2002,THRU,2015$$ MODAL VARIABLES$SPOINT,101,THRU,120SPOINT,201,THRU,220SEQSET1,100,0,101,THRU,120SEQSET1,200,0,201,THRU,220INCLUDE SEDYNBLK.DATENDDATA



SUPERELEMENT CMS SAMPLE – RUN 3 MARCH 17, 1992 MSC.NASTRAN 11/ 20/ 91 PAGE 13SUPERELEMENT 100

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 3.170078E+04 1.780471E+02 2.833708E+01 1.000000E+00 3.170078E+042 2 2.409874E+05 4.909047E+02 7.812991E+01 1.000000E+00 2.409874E+053 3 9.275004E+05 9.630682E+02 1.532771E+02 1.000000E+00 9.275004E+054 4 2.542844E+06 1.594630E+03 2.537932E+02 1.000000E+00 2.542844E+065 5 5.709340E+06 2.389423E+03 3.802884E+02 1.000000E+00 5.709340E+06SUPERELEMENT CMS SAMPLE – RUN 3 MARCH 17, 1992 MSC.NASTRAN 11/ 20/ 91 PAGE 16

SUPERELEMENT 200R E A L E I G E N V A L U E S

MODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 6.261534E+03 7.912985E+01 1.259391E+01 1.000000E+00 6.261534E+032 2 4.758257E+04 2.181343E+02 3.471715E+01 1.000000E+00 4.758257E+043 3 1.829213E+05 4.276930E+02 6.806945E+01 1.000000E+00 1.829213E+054 4 5.001845E+05 7.072372E+02 1.125603E+02 1.000000E+00 5.001845E+055 5 1.117615E+06 1.057173E+03 1.682543E+02 1.000000E+00 1.117615E+06

SUPERELEMENT CMS SAMPLE – RUN 3 MARCH 17, 1992 MSC.NASTRAN 11/ 20/ 91 PAGE 20SUPERELEMENT 0

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 2.004112E+01 4.476730E+00 7.124937E–01 1.000000E+00 2.004112E+ 012 2 7.871165E+02 2.805560E+01 4.465187E+00 1.000000E+00 7.871165E+ 023 3 6.171481E+03 7.855878E+01 1.250302E+01 1.000000E+00 6.171481E+ 034 4 2.370327E+04 1.539587E+02 2.450328E+01 1.000000E+00 2.370327E+ 045 5 6.486245E+04 2.546811E+02 4.053375E+01 1.000000E+00 6.486245E+ 04

Component Modes Reduction (Cont.)



GDR and CMR Modified Case Control from GDR-only file set-up. In addition,

eigensolution is requested for both superelements and the residual structure.

$ FILE = SEDYN4.DAT$SOL 103TIME 5CENDTITLE = SUPERELEMENT CMS SAMPLE – RUN 4SEALL = ALL $ ONLY REQUIRED IF SOL<101SPC = 10METHOD = 37SUBCASE 1 $ SUPERELEMENTSSET 47 = 100,200SUPER = 47DYNRED = 1SUBCASE 2BEGIN BULKSESET,0,1001,1011,2001,2016SESET,100,1002,THRU,1010SESET,200,2002,THRU,2015$$ MODAL VARIABLES$SPOINT,101,THRU,125SPOINT,201,THRU,230SEQSET1,100,0,101,THRU,125SEQSET1,200,0,201,THRU,230INCLUDE SEDYNBLK.DATENDDATA



SUPERELEMENT CMS SAMPLE – RUN 4 JUNE 26, 1990 MSC.NASTRAN 10/ 20/ 89 PAGE 15SUPERELEMENT 100

USER INFORMATION MESSAGE––– PROCESSING OF SUPERELEMENT 100 IS NOW INITIATED.^^^ PHASE 1 – SUPERELEMENT GENERATION, ASSEMBLY AND REDUCTION.USER INFORMATION MESSAGE 4158––– STATISTICS FOR SYMMETRIC DECOMPOSITION OF DATA BLOCK SCRATCH FOLLOW NUMBER OF

NEGATIVE TERMS ON FACTOR DIAGONAL = 2USER INFORMATION MESSAGE 4181––– NUMBER OF ROOTS BELOW 0.1000E+ 03 CYCLES IS 2

NUMBER OF GENERALIZED COORDINATES SET TO 6USER INFORMATION MESSAGE 5458, MODIFIED GIVENS METHOD IS FORCED BY USER .SUPERELEMENT CMS SAMPLE – RUN 4 JUNE 26, 1990 MSC.NASTRAN 10/ 20/ 89 PAGE 16


MODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 3.170078E+04 1.780471E+02 2.833708E+01 1.000000E+00 3.170078E+042 2 2.409874E+05 4.909047E+02 7.812991E+01 1.000000E+00 2.409874E+053 3 9.275004E+05 9.630682E+02 1.532771E+02 1.000000E+00 9.275004E+054 4 2.542844E+06 1.594630E+03 2.537932E+02 1.000000E+00 2.542844E+065 6 5.709515E+06 2.389459E+03 3.802943E+02 1.000000E+00 5.709515E+066 5 1.367973E+07 3.698612E+03 5.886524E+02 0.0 0.0SUPERELEMENT CMS SAMPLE – RUN 4 JUNE 26, 1990 MSC.NASTRAN 10/ 20/ 89 PAGE 18

SUPERELEMENT 200USER INFORMATION MESSAGE––– PROCESSING OF SUPERELEMENT 200 IS NOW INITIATED.^^^ PHASE 1 – SUPERELEMENT GENERATION, ASSEMBLY AND REDUCTION.USER INFORMATION MESSAGE 4158––– STATISTICS FOR SYMMETRIC DECOMPOSITION OF DATA BLOCK SCRATCH FOLLOW

NUMBER OF NEGATIVE TERMS ON FACTOR DIAGONAL = 3USER INFORMATION MESSAGE 4181––– NUMBER OF ROOTS BELOW 0.1000E+ 03 CYCLES IS 3

NUMBER OF GENERALIZED COORDINATES SET TO 6USER INFORMATION MESSAGE 5458, MODIFIED GIVENS METHOD IS FORCED BY USER .SUPERELEMENT CMS SAMPLE – RUN 4 JUNE 26, 1990 MSC.NASTRAN 10/ 20/ 89 PAGE 19

SUPERELEMENT 200

GDR and CMR (Cont.)



R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 6.261534E+03 7.912985E+01 1.259391E+01 1.000000E+00 6.261534E+032 2 4.758257E+04 2.181343E+02 3.471715E+01 1.000000E+00 4.758257E+043 3 1.829213E+05 4.276930E+02 6.806945E+01 1.000000E+00 1.829213E+054 4 5.001845E+05 7.072372E+02 1.125603E+02 1.000000E+00 5.001845E+055 5 1.117615E+06 1.057173E+03 1.682543E+02 1.000000E+00 1.117615E+066 6 2.185548E+06 1.478360E+03 2.352883E+02 0.0 0.0SUPERELEMENT CMS SAMPLE – RUN 4 JUNE 26, 1990 MSC.NASTRAN 10/ 20/ 89 PAGE 23


MODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS 1 1 2.004112E+01 4.476730E+00 7.124937E–01 1.000000E+00 2.004112E+012 2 7.871165E+02 2.805560E+01 4.465187E+00 1.000000E+00 7.871165E+023 3 6.171481E+03 7.855878E+01 1.250302E+01 1.000000E+00 6.171481E+034 4 2.370327E+04 1.539587E+02 2.450328E+01 1.000000E+00 2.370327E+045 5 6.486243E+04 2.546810E+02 4.053374E+01 1.000000E+00 6.486243E+046 6 1.447435E+05 3.804517E+02 6.055077E+01 0.0 0.07 7 2.831655E+05 5.321330E+02 8.469160E+01 0.0 0.08 8 5.018220E+05 7.083940E+02 1.127444E+02 0.0 0.09 9 8.432106E+05 9.182650E+02 1.461464E+02 0.0 0.010 10 1.317067E+06 1.147635E+03 1.826518E+02 0.0 0.011 11 2.444493E+06 1.563487E+03 2.488367E+02 0.0 0.012 12 4.798879E+06 2.190634E+03 3.486503E+02 0.0 0.013 14 1.099342E+07 3.315632E+03 5.276993E+02 0.0 0.014 13 1.587894E+07 3.984839E+03 6.342068E+02 0.0 0.0

GDR and CMR (Cont.)



CMS with Free-Free Components Specify exterior points, which are unconstrained during CMS, with SECSET1

entries. Recommend not using the SESUP entry or calculating 0.0 Hz component

modes $ FILE = SEDYN5.DAT$SOL 103TIME 5CENDTITLE = SUPERELEMENT CMS SAMPLE – RUN 5SEALL = ALL $ ONLY REQUIRED IF SOL<101SPC = 10SUBCASE 1 $ SUPERELEMENTSSET 98 = 100,200SUPER = 98METHOD = 38SUBCASE 2METHOD = 39BEGIN BULK$ DO NOT COMPUTE RIGID BODY MODES IN SUPERELEMENTSEIGR,38,MGIV,.001,1000.$ COMPUTE R.S. MODESEIGR,39,MGIV,0.,1000.$ FREE–FREE BOUNDARIES FOR CMSSECSET1,100,26,1001,1011SECSET1,200,26,2001,2016$SESET,0,1001,1011,2001,2016SESET,100,1002,THRU,1010SESET,200,2002,THRU,2015$$ MODAL VARIABLES$SPOINT,101,THRU,120SPOINT,201,THRU,220SEQSET1,100,0,101,THRU,120SEQSET1,200,0,201,THRU,220INCLUDE SEDYNBLK.DATENDDATA



SUPERELEMENT CMS SAMPLE – RUN 5 MAY 2, 1990 MSC.NASTRAN 1/ 4/ 89 PAGE 16SUPERELEMENT 100

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 3.170071E+04 1.780469E+02 2.833705E+01 1.000000E+00 3.170071E+ 042 2 2.409813E+05 4.908985E+02 7.812891E+01 1.000000E+00 2.409813E+ 053 3 9.273744E+05 9.630028E+02 1.532666E+02 1.000000E+00 9.273744E+ 054 6 2.541635E+06 1.594251E+03 2.537329E+02 1.000000E+00 2.541635E+ 065 7 5.702000E+06 2.387886E+03 3.800439E+02 1.000000E+00 5.702000E+ 066 8 1.121551E+07 3.348956E+03 5.330029E+02 1.000000E+00 1.121551E+ 077 9 2.010154E+07 4.483474E+03 7.135671E+02 1.000000E+00 2.010154E+ 078 10 3.353399E+07 5.790854E+03 9.216431E+02 1.000000E+00 3.353399E+ 07

SUPERELEMENT CMS SAMPLE – RUN 5 MAY 2, 1990 MSC.NASTRAN 1/ 4/ 89 PAGE 19SUPERELEMENT 200

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 6.261533E+03 7.912984E+01 1.259391E+01 1.000000E+00 6.261533E+ 032 2 4.758247E+04 2.181341E+02 3.471711E+01 1.000000E+00 4.758247E+ 043 3 1.829191E+05 4.276904E+02 6.806904E+01 1.000000E+00 1.829191E+ 054 6 5.001635E+05 7.072224E+02 1.125579E+02 1.000000E+00 5.001635E+ 055 7 1.117487E+06 1.057113E+03 1.682447E+02 1.000000E+00 1.117487E+ 066 8 2.184376E+06 1.477963E+03 2.352252E+02 1.000000E+00 2.184376E+ 067 9 3.883847E+06 1.970748E+03 3.136542E+02 1.000000E+00 3.883847E+ 068 10 6.435782E+06 2.536884E+03 4.037577E+02 1.000000E+00 6.435782E+ 069 11 1.010135E+07 3.178262E+03 5.058361E+02 1.000000E+00 1.010135E+ 0710 12 1.518767E+07 3.897136E+03 6.202484E+02 1.000000E+00 1.518767E+ 0711 13 2.204906E+07 4.695643E+03 7.473347E+02 1.000000E+00 2.204906E+ 0712 14 3.106928E+07 5.573983E+03 8.871269E+02 1.000000E+00 3.106928E+ 07

CMR with Free-Free Components (Cont.)



CMR with Free-Free Components (Cont.)SUPERELEMENT CMS SAMPLE – RUN 5 MAY 2, 1990 MSC.NASTRAN 1/ 4/ 89 PAGE 23SUPERELEMENT 0

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 2.004113E+01 4.476732E+00 7.124940E–01 1.000000E+00 2.004113E+ 012 2 7.871033E+02 2.805536E+01 4.465149E+00 1.000000E+00 7.871033E+ 023 3 6.171444E+03 7.855854E+01 1.250298E+01 1.000000E+00 6.171444E+ 034 4 2.370205E+04 1.539547E+02 2.450265E+01 1.000000E+00 2.370205E+ 045 5 6.477100E+04 2.545015E+02 4.050517E+01 1.000000E+00 6.477100E+ 046 6 1.445857E+05 3.802443E+02 6.051776E+01 1.000000E+00 1.445857E+ 057 7 2.821152E+05 5.311451E+02 8.453437E+01 1.000000E+00 2.821152E+ 058 8 5.004866E+05 7.074507E+02 1.125943E+02 1.000000E+00 5.004866E+ 059 9 8.265909E+05 9.091704E+02 1.446990E+02 1.000000E+00 8.265909E+ 0510 10 1.290264E+06 1.135898E+03 1.807838E+02 1.000000E+00 1.290264E+ 0611 11 1.929138E+06 1.388934E+03 2.210557E+02 1.000000E+00 1.929138E+ 0612 12 2.779604E+06 1.667214E+03 2.653454E+02 1.000000E+00 2.779604E+ 0613 13 3.891916E+06 1.972794E+03 3.139799E+02 1.000000E+00 3.891916E+ 0614 14 5.313539E+06 2.305112E+03 3.668699E+02 1.000000E+00 5.313539E+ 0615 15 7.082243E+06 2.661248E+03 4.235508E+02 1.000000E+00 7.082243E+ 0616 16 9.299721E+06 3.049544E+03 4.853501E+02 1.000000E+00 9.299721E+ 0617 19 1.199213E+07 3.462966E+03 5.511481E+02 1.000000E+00 1.199213E+ 0718 20 1.527929E+07 3.908873E+03 6.221165E+02 1.000000E+00 1.527929E+ 0719 21 1.925098E+07 4.387594E+03 6.983073E+02 1.000000E+00 1.925098E+ 0720 24 2.384324E+07 4.882954E+03 7.771462E+02 1.000000E+00 2.384324E+ 0721 23 2.949947E+07 5.431341E+03 8.644248E+02 1.000000E+00 2.949947E+ 0722 22 3.708353E+07 6.089625E+03 9.691939E+02 1.000000E+00 3.708353E+ 07

CMR with Free-Free Components (Cont.)


APPENDIX 6A–THE CRAIG-BAMPTON MEDTHOD –

HAND-SOLVED EXAMPLE


DEFAULT CMS METHOD “FIXED BOUNDARY” CMS

Description of Methodology (better known as Craig-Bampton CMS) The superelement matrices are partitioned into two sets of degrees

of freedom (DOFs). The first set (the B-set) represents the boundary points. The second set is the interior DOFs (the O-set).

A set of “constraint” modes is generated. Each “constraint” mode represents the motion of the model resulting from moving one boundary DOF 1.0 unit, while holding the other boundary DOF fixed. Therefore, there is one “constraint” mode for each boundary DOF (these vectors are known as GOAT in MSC.NASTRAN)

In matrix form,

(Pb is not actually applied.) The first line gives

=

bbb

ob

bbbo

oboo

P0

IKKKK φ

)(G }{ OAT}]{I[K][K bbob1

ooob−−=φ


DEFAULT CMS METHOD “FIXED BOUNDARY” CMS (Cont.)

giving the following “constraint” modes:

Now the O-set equations are solved for the “fixed-boundary” modes (known as GOAQ in MSC.NASTRAN).

As many fixed-boundary modes as are desired are found. Then they are concatenated with the “constraint” modes to form the generalized coordinates.

The mass and stiffness matrices are pre- and postmultiplied by these modes to obtain the “generalized” mass and stiffness

where the F-set is the union of the B- and O-sets.

=bb

obb I

φφ }{

0}]{[}]{[2k =+− oooooooo KM φφω

=0Ibb

ooobG

φφφ }{

}]{[}{][}]{[}{][

T

T

GffGG

GffGG

MM

KK

φφ

φφ

=

=


DEFAULT CMS METHOD “FIXED BOUNDARY” CMS (Cont.)

These “generalized” matrices contain physical DOFs representing the boundaries and “modal” coordinates representing the “fixed-boundary” component modes.

At this point, these matrices can be treated like any other structural matrices, and data recovery can be performed for the component in a manner similar to using modal coordinates. That is, the displacements of the generalized coordinates are multiplied by the associated vectors and added together to obtain the component displacements.

The calculated modes for each superelement are internally scaled to have a maximum displacement = 1.0 in MSC.NASTRAN (regardless of the scaling requested by the user).


SOLUTION BY HANDComponent Modal Synthesis Sample

Spring Stiffness = 1.Each Mass = 1.

SESET,1,4,5SESET,2,2SPOINT, 1001,THRU,1010SEQSET1,1,1001,1002SEQSET1,2,1005

Theoretical solution for frequencies

i 1 2 3 4fi 0.0553 0.1592 0.2438 0.2991

0.1206 1.00 2.3473 3.53212ωλ =i


SOLUTION BY HAND (Cont.) Superelement 1

Mass at Grid Point 3 “belongs” to the residual structure and is therefore exterior.

Grid Point 3 is the “boundary” point; solve for “constraint” modes.

where

=

−−−

−=

5

4

3

gggg

UUU

100010000

M 110121

011K

=

−−−

−

00P

UU1

110121

011 b

5

4

=

−=

−

−=

−

2111

K

01

K

1112

K

1oo

ob

oo


SOLUTION BY HAND (Cont.)where

Solve for “fixed-boundary” modes.Note: Internally MSC.NASTRAN uses component modes scaled to a maximum deformation of 1.0. Output for the component modes is based on the normalization performed by the eigenvalue solution.

=

=

−

−=

111

11

01

2111

b

ob

φ

φ

}{1112

00.0}]{KM[ oo2

2

oooooo2 φ

ωω

φω

−

−+

−−

==+−


SOLUTION BY HAND (Cont.)

where 1001 and 1002 are scalar points used to represent Superelement 1’s modes.

011

12det 2

2=

−−−−ω

ωHz .2575 Hz,.098 f

2.618 ,3819.2

==ω

−

−=

=

−=

−

=

−

=

−

=

=

=

3820.103820.03820.16180.1

3820.6180.10.2}]{M[}{

uuu

6180.30005279.0000

}]{K[}{

6180.0.110.1618.1

0016180.0.1

0.16180.5257.8506.

6180.

0.18506.5257.

0.1

6180.

GggT

G

1002

1001

3

GggT

G

G

oo

22

11

φφ

φφ

φ

φ

φφ

φφ Normalized to unit mass


SOLUTION BY HAND (Cont.)Superelement 2

where 1005 is a scalar point used to represent Superelement 2’s mode

}1{}{1,1

000010000

110121

011

oo =φ==

=

−−−

−=

MooKoo

MK gggg

2251.f0.22

==ω

=

01012/12/1001

}{ Gφ

uuu

0.150.50.

50.25.25.50.25.25.

}]{M[}{

uuu

0.200

05.5.05.5.

}]{K[}{

1005

3

1

GggT

G

1005

3

1

GggT

G

=

−

−=

φφ

φφ

=

102/12/1

01

bφ


SOLUTION BY HAND (Cont.)Residual Structure Before adding superelement:

1005

1002

1001

3

1

gg

gg

UUUUU

0000000000000000001000001

M

0000000000000000000000000

K

=

=


SOLUTION BY HAND (Cont.)

Add Superelement 1

0000003820.103820.0003820.16180.1003820.6180.13000001

M

000000618.3000005279.000000000000

K

gg

gg

−

−=

=


SOLUTION BY HAND (Cont.) Add Superelement 2

0.1005.5.03820.103820.0003820.16180.105.3820.6180.125.325.5.0025.25.1

M

0.200000618.3000005279.000005.5.0005.5.

K

gg

gg

−

−=

−

−

=


SOLUTION BY HAND (Cont.) Apply constraints at DOF 1.

Solve which gives

Data recovery (grid point displacement for mode 1)

Residual Structure

1005

1002

1001

3

ffff

UUUU

0.1005.03820.103820.003820.16180.15.3820.6180.125.3

M

20000618.300005279.00005.

K

−

−

=

=

0}}{MK{ fff2

ff =− φω.5321.3 ,3473.2 ,00.1 ,1206.2 =ω

7568.7705.2887.0137.7012.05464.0986.00572.8619.3188.0937.12315.

6565.2280.5773.4285.

f

−−−

−−−

=φ

UU

4285.

0

3

1

=φ


SOLUTION BY HAND (Cont.) Superelement 2

for exterior points

Superelement 1for exterior points

1005

3

1

G2

UUU

0137.4285.

0

=φ

1002

1001

3

G1

UUU

00572.2315.4285.

−=φ

3

2

1

G22G21

uuu

4285.2280.

0

0137.4285.

0

01012/12/1001

}}{{

=

== φφφ

5

4

3

G11G11

uuu

6565.5773.4285.

00572.2315.4285.

6180.1116180.1

001}}{{

=

−

−== φφφ


SOLUTION USING MSC.NASTRAN SOL 103ID CMS1, SAMPLE PROBLEM FOR CMS

SOL 103TIME 10CENDTITLE = SAMPLE PROBLEM FOR CMSSPC = 1SUBCASE 1DISP = ALLLABEL = CMS OF SUPERELEMENTSSET 1000 = 1,2SUPER =1000METHOD=2 $ GET 2 MODESSUBCASE 2LABEL=SOLVE FOR RESIDUAL STRUCTURE MODES IF DESIREDMETHOD = 1DISP = ALLBEGIN BULKPARAM,FIXEDB,–1PARAM,GRDPNT,0EIGRL,1,,,10EIGRL,2,,,2$ ADD MODAL COORDINATES FOR S.E. 1SPOINT,1001,THRU,1010SEQSET1,1,0,1001,THRU,1004SEQSET1,2,0,1005,THRU,1010GRID,1,,0.,0.,0.=,(1),=,(10.),===(3)CELAS2,1,1.,1,1,2,1CELAS2,2,1.,2,1,3,1CELAS2,3,1.,3,1,4,1CELAS2,4,1.,4,1,5,1$ DEFINE SUPERELEMENTSSESET,1,4,5SESET,2,2PARAM,AUTOSPC,YESSPC1,1,123456,1CONM2,11,1,,1.CONM2,12,2,,1.CONM2,13,3,,1.CONM2,14,4,,1.CONM2,15,5,,1.ENDDATA

The input data was run in MSC.NASTRAN:


SELECTED OUTPUT FROM MSC.NASTRAN

SAMPLE PROBLEM FOR CMS OCTOBER 3, 1989 MSC.NASTRAN 1/ 4/ 89 PAGE 18SUPERELEMENT 1

R E A L E I G E N V A L U E SMODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 3.819660E–01 6.180340E–01 9.836316E–02 1.000000E+00 3.819660E–012 2 2.618034E+00 1.618034E+00 2.575181E–01 1.000000E+00 2.618034E+00


MODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 2.000000E+00 1.414214E+00 2.250791E–01 1.000000E+00 2.000000E+00


MODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS1 1 1.206148E–01 3.472964E–01 5.527393E–02 1.000000E+00 1.206148E–012 2 1.000000E+00 1.000000E+00 1.591549E–01 1.000000E+00 1.000000E+003 4 2.347296E+00 1.532089E+00 2.438395E–01 1.000000E+00 2.347296E+004 3 3.532089E+00 1.879385E+00 2.991135E–01 1.000000E+00 3.532089E+00

SUPERELEMENT 0SOLVE FOR RESIDUAL STRUCTURE MODES IF DESIRED SUBCASE 2EIGENVALUE = 1.206148E– 01CYCLES = 5.527393E– 02 R E A L E I G E N V E C T O R N O . 1

POINT ID. TYPE T1 T2 T3 R1 R2 R31 G 0.0 0.0 0.0 0.0 0.0 0.03 G 4.285251E–01 0.0 0.0 0.0 0.0 0.01001 S 2.315487E–01 –5.720218E–03 0.0 0.0 1.375089E–02 0.01007 S 0.0 0.0 0.0 0.0


SELECTED OUTPUT FROM MSC.NASTRAN (Cont.)

SAMPLE PROBLEM FOR CMS OCTOBER 3, 1989 MSC.NASTRAN 1/ 4/ 89 PAGE 43SUPERELEMENT 1

CMS OF SUPERELEMENT 1 SUBCASE 1EIGENVALUE = 2.347296E+ 00

CYCLES = 2.438395E–01 R E A L E I G E N V E C T O R N O . 3POINT ID. TYPE T1 T2 T3 R1 R2 R3

3 G –2.280134E–01 0.0 0.0 0.0 0.0 0.04 G –5.773503E–01 0.0 0.0 0.0 0.0 0.05 G 4.285251E–01 0.0 0.0 0.0 0.0 0.0

1001 S 3.188475E–01 5.463955E–01 0.0 0.0SAMPLE PROBLEM FOR CMS OCTOBER 3, 1989 MSC.NASTRAN 1/ 4/ 89 PAGE 48

SUPERELEMENT 2CMS OF SUPERELEMENT 1 SUBCASE 1EIGENVALUE = 1.206148E–01

CYCLES = 5.527393E–02 R E A L E I G E N V E C T O R N O . 1POINT ID. TYPE T1 T2 T3 R1 R2 R3

1 G 0.0 0.0 0.0 0.0 0.0 0.02 G 2.280134E–01 0.0 0.0 0.0 0.0 0.03 G 4.285251E–01 0.0 0.0 0.0 0.0 0.0

1005 S 1.375089E–02 0.0 0.0 0.0 0.0 0.0


EXTERNAL SUPERELEMENTS

In V70, the ability to use external superelements, complete with data recovery was added for SOLs 101 and 103.

In V70.5, these new external superelements have been extended into SOL’s 101 thru 159 and data recovery for them exists in SOLs 101, 103, and 107 thru 112.

The procedure for this is as follows: Create reduced model.

Read in reduced model as an external superelement.

Perform solution and data recovery of assembly.

Perform data recovery on external superelement.


CREATING AN EXTERNAL SUPERELEMENT

A separate model file is used to create an external superelement.

The component must be modeled as the residual structure in this file. upstream superelements are allowed in this file, but the residual structure

(assembly) is the component with reduced matrices will be available for as an external superelement in subsequent runs.

Interface dof must be identified using ASETi, BSETi, and/or CSETi entries.

If you are using component modal synthesis, QSETi dof must be provided to represent the component modes.

Only one boundary condition may be used. Only one SUBCASE is required.

If you are performing a static solution, multiple residual structure SUBCASEs may be specified, but they must be in the correct order for use when the component is attached.


CREATING AN EXTERNAL SUPERLEMENT (Cont)

There are 4 ways the reduced data may be stored for use in future runs.

The format of the reduced data is controlled by PARAM,EXTOUT: MATRIXDB = the reduced matrices are stored on the database.

They do not contain connectivity data. DMIGDB = the reduced matrices are stored on the database using

DMIG format and can be “automatically” attached. DMIGOP2 = the reduced matrices are written using OUTPUT2

format to a file (specified by PARAM,EXTUNIT – default=30). The matrices are stored using DMIG format.

DMIGPCH = the reduced matrices are written to the “.pch” file using DMIG format.


ATTACHING AN EXTERNAL SUPERELEMENT

External superelements may be attached using partitioned bulk data or by using the CSUPER entry.

If you use the partitioned bulk data method to attach an external superelement: You need an EXTRN entry in the partitioned bulk data section. you need to provide the GRID and SPOINTs to attach the external

superelement to. (Be careful to align the displacement coordinate systems properly – there is no checking).

If EXTOUT was MATRIXDB or DMIGDB when the superelement was created: use FMS to attach the database and locate the matrices:

ASSIGN SExxx=’run1.MASTER’DBLOCATE DATABLK=(EXTDB), convert(SEID=xx),

LOGICAL=SExxx If EXTOUT was DMIGOP2, then

you must assign the OUTPUT2 file in the FMS: ASSIGN INPUTT2=’run1.OP2’, unit=i

specify PARAM,EXTUNIT,i to point to the file


ATTACHING AN EXTERNAL SUPERELEMENT (Cont)

If EXTOUT was DMIGPCH, include the “.pch” file from the previous run and use the following case control for the superelement:

K2GG= KAAXP2G = PAXM2GG = MAAXB2GG = BAAx

At this point, the run will proceed normally, attaching the external superelement and solving the problem.

Standard data recovery is available for all superelements (except the external ones) during the solution run.

Data recovery for the external superelement run requires saving the database from the assembly run and performing a “data recovery” restart on the external superelement. This is controlled by PARAM,EXTDROUT: EXTDROUT=MATRIXDB – solution for boundary displacements stored in database using the

sequencing of the assembly model EXTDROUT = DMIGDB – solution stored in database using DMIG (only applicable if EXTOUT was

set to DMIGDB or DMIGOP2) EXTDROUT = DMIGOP2 – writes DMIG to OUTPUT2 file selected by PARAM,EXTDRUNT (default =

unit 31) – available only for EXTOUT=DMIGOP2 or DMIGDB


DATA RECOVERY FOR AN EXTERNAL SUPERELEMENT

Performing data recovery on the external superelement requires using a restart from the run which created the reduced matrices.

The run requires the following FMS (or similar): ASSIGN SE10=’run1.MASTER’

RESTART, LOGICAL=SE10 $ read–only restart – not requiredASSIGN RESID=’run2.MASTER’DBLOCATE DATABLK=(EXTDB), LOGICAL=RESID

The run also requires PARAM,EXTDR,YES


ATTACHING AN EXTERNAL SUPERELEMENT

EXTRN bulk data entry:Defines a boundary connection for an external superelement.Format:

Example:

Field ContentsGIDi Grid identification number to which the exterior

superelement matrices will be connected.Ci Component numbers. (Integer 0, blank, or 1 for scalar points;

Integers 1 through 6 with no embedded blanks for grids.)

1 2 3 4 5 6 7 8 9 10EXTRN GID1 C1 GID2 C2 GID3 C3 GID4 C4

etc. GID6 "THRU" GID7 C6 etc.

EXTRN 1001 123 1120 123456 1201 123


ATTACHING AN EXTERNAL SUPERELEMENT (Cont)

Remarks:1. EXTRN can only be specified in partitioned Bulk Data Sections and

is ignored in the main Bulk Data Section.2. Connection grids must be specified in the partitioned Bulk Data

Section following BEGIN SUPER = SEID.3. “THRU” may be specified in fields 3, 5, or 7.4. Pairs of blank fields may be entered to allow easier modification of

the EXTRN entry.


SAMPLE PROBLEM

The solution will be a modal transient, SOL 112. The loading on this model is a pressure on the elements in Superelement 10

(the external superelement). The solution will consist of 5 runs and will use the MATRIXDB method (the

other approaches would also work fine). Run1 – process SE 10 Run 2 – Read in SE 10 as external Run3 – define and process internal SE 11 Run 4 – Define and solve residual structure Run 5 – data recovery on external SE 10


SAMPLE PROBLEM USING MATRIXDBSOL 112 $ superelement SSS modal transientCENDTITLE = Generate data to be attached as SE 10PARAM,EXTOUT,MATRIXDBSUBCASE 1loadset = 15 $ Define loadingMETHOD = 10 $ request cmsparam,resvec,yes $ request residual vectorsSPC = 1BEGIN BULK$ define loads$lseq,15,1001,101lseq,15,2001,201pload2,101,1.,97,thru,112force,201,1108,,1.,10.,0.,0.$$ define modal coordinates for CMS$SPOINT 91001 THRU 91006QSET1 0 91001 THRU 91006$$ define which dofs will be retained (i.e. which dofs will form the$ attachment to the system model when we bring it in as an external se)$ASET1 123456 1100 THRU 1104$$ print dof map for connecting the external superelement to the$ system model, in se10.dat, with EXTRN entry. The MATRIXDB option$ requires the dofs specified in the subsequent se10.dat run be in$ ASET ascending order. This is obtained with these parameters in$ the f06$ usetsel =128 will print only ASET dof$PARAM USETPRT 0PARAM USETSEL 128$EIGRL 10 4PSHELL 1 1 .01 1 1CQUAD4 97 1 1100 1101 1106 1105$$ model description occurs here...$GRID 1122 5.5 3. 0.GRID 1123 5.75 3. 0.GRID 1124 6. 3. 0.ENDDATA

Run1 – file run1_se10.dat


SAMPLE PROBLEM USING MATRIXDB (Cont)

ASSIGN SE10DB = ’run1_se10.MASTER’DBLOCATE DB=(EXTDB), CONVERT(SEID=10), LOGI=SE10DB$SOL 103TIME 600CENDTITLE = Add external data and call it SE 10SET 99 = 10SEALL = 99 $ process only SE 10SUBCASE 1SUPER = 10 $ process only SE 10param,resvec,yesloadset=15METHOD = 10BEGIN BULK$ declare SE 10 as external$SEBULK 10 EXTERNALBEGIN SUPER = 10$$ set flag for data recovery$PARAM EXTDROUTMATRIXDB$ dynamic loading definitionlseq,15,1001,101lseq,15,2001,201$pload2,101,1.,97,thru,112$force,201,1108,,1.,10.,0.,0.SPOINT 10001 THRU 10006QSET1 0 10001 THRU 10006ASET1 123456 1030 THRU 1034$$ Connect external superelement to the system model:$ Note that for the MATRIXDB option the order of the$ grids must be in ASET ASCENDING order$EXTRN 1030 123456 1031 123456 1032 123456 1033 1234561034 123456 10001 0 10002 0 10003 010004 0EIGRL 10 4GRID 1030 5. 2. 0.GRID 1031 5.25 2. 0.GRID 1032 5.5 2. 0.GRID 1033 5.75 2. 0.GRID 1034 6. 2. 0.ENDDATA

Run2 – file run2_se10ln.dat – read

SE 10 as external



ASSIGN MASTER=’run2_se10in.MASTER’RESTART, VERSION=1, KEEPSOL 112TIME 600CENDTITLE = Add in SE 11ECHO = NONEMAXLINES = 999999999SET 99 = 10,11SEALL = 99SUBCASE 1SUPER = 10METHOD = 10param,resvec,yesloadset = 15SUBCASE 2SUPER = 11 $ process only SE 11METHOD = 11param,resvec,yesloadset = 15BEGIN BULKBEGIN SUPER = 11$$ dynamic loading definitionlseq,15,1001,101lseq,15,2001,201$ define non–existant loads to allow upstream loads$ as place holdersforce,101,1007,,0.,1.,0.,0.force,201,1007,,0.,1.,0.,0.$ define modal coordinates for CMS$SPOINT 11001 THRU 11006QSET1 0 11001 THRU 11006$$ define attachment points to the next SE –$ optional if they already exist in the model$ASET1 123456 1000 THRU 1004$EIGRL 11 4PSHELL 1 1 .01 1 1CQUAD4 81 1 1000 1001 1006 1005$ model of SE 11......GRID 1023 5.75 2. 0.GRID 1024 6. 2. 0.ENDDATA

Run3 – file run2_se11.dat –define and process SE11



ASSIGN MASTER=’run2_se10in.MASTER’RESTART, VERSION=2, KEEPSOL 112TIME 600$$ insert dmap avoidance for error 32074 – see next page$CENDTITLE = Solve residual structureSUBCASE 1SUPER = 10METHOD = 10loadset = 15param,extdrout,matrixdbSUBCASE 2SUPER = 11METHOD = 11loadset = 15SUBCASE 3SUPER = 0 $ process only the residualMETHOD = 90tstep = 35SPC = 1loadset = 15dload = 25SPCFORCES(plot)=ALLBEGIN BULK$tstep,35,100,.01tload2,25,1001,,,0.,100.,10.,90.lseq,15,1001,101lseq,15,2001,201force,101,1,,0.,1.,0.,0.force,201,1,,0.,1.,0.,0.$EIGRL 90 4SPC1 1 123456 1 18 35 52 69PSHELL 1 1 .01 1 1CQUAD4 1 1 1 2 19 18CQUAD4 2 1 2 3 20 19GRID 104 5.75 1. 0.GRID 105 6. 1. 0.ENDDATA

Run4 – file run4_resid.dat – define residual structure and solve



ASSIGN EXT10=’run1_se10.MASTER’RESTART, LOGI=EXT10$ASSIGN SYSTEM=’run2_se10in.MASTER’DBLOCATE DB=(EXTDB), WHERE(SEID=10),LOGI=SYSTEM$SOL 112TIME 600diag 56CENDTITLE = Data Recovery for external dataECHO = NONEMAXLINES = 999999999$$ tell NASTRAN this is a data recovery run for the external data$PARAM,EXTDR,YES$DISP = ALLSUBCASE 1METHOD = 10param,resvec,yesloadset = 15tstep = 35dload = 25SPCFORCES(plot)=ALL$BEGIN BULKENDDATA

Run5 – file run5_dr10.dat – perform data recovery on se 10



Plot of displacement GRID 1020


SAMPLE PROBLEM USING DMGIOP2 PROGRAM

assign file for use by dmigop2$assign output2=’ext10.op2’, unit=30, delete$SOL 112 $ superelement SSS modal transient$ include alter for OTM – optionalinclude ’alteria.v705’CENDTITLE = Generate data to be attached as SE 10$param,extout,dmigop2$SUBCASE 1loadset = 15METHOD = 10param,resvec,yes $ request residual vectorsSPC = 1disp = allstress = allforce = allBEGIN BULK$$ parameter to create OTM using alter1ia.v705param,drmh,yes$$ define loadings – used for residual vectors (also stored in database)lseq,15,1001,101lseq,15,2001,201pload2,101,1.,97,thru,112force,201,1108,,1.,10.,0.,0.$$ define modal coordinates for CMS – allow for 6 modes$SPOINT 91001 THRU 91006QSET1 0 91001 THRU 91006$$ define which dofs will be retained (i.e. which dofs will form the$ attachment to the system model when we create SE10 in se10.dat)$ASET1 123456 1100 THRU 1104$EIGRL 10 4$ model goes here....$ENDDATA$

Run 1 + file run1`_se.dat


SAMPLE PROBLEM USING DMGIOP2$ run 2 – se10.dat – locate external data and attach as$ superelement 10$ ––––––$ attach file containing reduced matrices and OTMASSIGN inputt2=’ext10.op2’, unit=30$SOL 103diag 8,15,56include ’alteria.v705’CENDTITLE = Add external data and call it SE 10SET 99 = 10SEALL = 99SUBCASE 1SUPER = 10 $ process only SE 10param,resvec,yesloadset=15METHOD = 10BEGIN BULK$ declare SE 10 as externalSEBULK 10 EXTERNALBEGIN SUPER = 10$ point to file used for INPUTT2param,extunit,30$ set flag for data recoveryPARAM EXTDROUTDMIGOP2param,extdrunt,31$ dynamic loading definitionlseq,15,1001,101lseq,15,2001,201$pload2,101,1.,97,thru,112$force,201,1108,,1.,10.,0.,0.EXTRN 1100 123456 1101 123456 1102 123456 1103 1234561104 123456 91001 0 91002 0 91003 091004 0 91005 0 91006 0$ identify exterior points (not needed if coincident points elsewhereASET1 123456 1100 THRU 1104$ define modal coordinates for CMSSPOINT 91001 THRU 91006QSET1 0 91001 THRU 91006GRID 1100 5. 2. 0.GRID 1101 5.25 2. 0.GRID 1102 5.5 2. 0.GRID 1103 5.75 2. 0.GRID 1104 6. 2. 0.ENDDATA

Run2 – file run2_se10in.dat


SAMPLE PROBLEM USING DMGIOP2$ SE 10 is all ready in the run2_se10in.DBALL database$$RESTART LOGI=SE10MASSIGN MASTER=’run2_se10in.MASTER’RESTART, VERSION=1, KEEPSOL 112TIME 600include ’alteria.v705’CENDTITLE = Add in SE 11SET 99 = 10,11SEALL = 99SUBCASE 1SUPER = 10METHOD = 10param,resvec,yesloadset = 15$ add new subcase (let the auto–restart logic work it out)SUBCASE 2SUPER = 11 $ process only SE 11METHOD = 11param,resvec,yesloadset = 15BEGIN BULKBEGIN SUPER = 11lseq,15,1001,101lseq,15,2001,201$ define non–existant loads to allow upstream loads$ as place holdersforce,101,1007,,0.,1.,0.,0.force,201,1007,,0.,1.,0.,0.$ define modal coordinates for CMSSPOINT 11001 THRU 11006QSET1 0 11001 THRU 11006$$ attachment points to the next SE not needed if coincident points exist$ASET1 123456 1000 THRU 1004$EIGRL 11 4$ model of se 11ENDDATA

Run3 – file run3 se11.dat


SAMPLE PROBLEM USING DMGIOP2$ run4_resid.dat – add residual data and solve$ assign file for boundary solution$assign output2=’se10bndry.op2’, unit=31, delete$$ SE 10 and 11 are in the run2_se10 database.$ for a read–only restart (not required)assign oldrun=’run2_se10in.MASTER’restart, logical=oldrun$SOL 112TIME 600include ’alteria.v705’CENDTITLE = Solve residual structuredisp(plot)=allSUBCASE 1disp=allstress = allforce = allSUPER = 10METHOD = 10loadset = 15param,extdrout,dmigop2param,extdrunt,31SUBCASE 2SUPER = 11METHOD = 11loadset = 15SUBCASE 3SUPER = 0 $ process only the residualMETHOD = 90tstep = 35SPC = 1loadset = 15dload = 25BEGIN BULKtstep,35,100,.01tload2,25,1001,,,0.,100.,10.,90.lseq,15,1001,101lseq,15,2001,201force,101,1,,0.,1.,0.,0.force,201,1,,0.,1.,0.,0.EIGRL 90 4$ residual structure modelENDDATA

Run4 – file run4-including data recovery of SE10 (OTM)


SAMPLE PROBLEM USING DMGIOP2$ run 5_dr10.dat – data recovery for external data$ ==================================$$ Features demonstrated:$ ––––––––––––––––––––––$ Data recovery for the external data (that became SE 10)$$ Notes:$ ––––––$ This deck must be run in MSC.NASTRAN version 70.5 or above.$ASSIGN EXT10=’run1_se10.MASTER’RESTART, LOGI=EXT10$assign inputt2=’se10bndry.op2’, unit=31$SOL 112TIME 600diag 56CENDTITLE = Data Recovery for external data$$ tell NASTRAN this is a data recovery run for the external data$PARAM,EXTDR,YESparam,extdrunt,31$param,extdr,yesDISP = ALLSUBCASE 1METHOD = 10spc = 1param,resvec,yesloadset=15tstep = 35dload = 25SPCFORCES(plot)=ALLOUTPUT(XYPLOT)XTITLE = TIME IN SECSXGRID LINES = YESYGRID LINES = YESYTITLE = Z DISPLACEMENTTCURVE = DISPL of GRID 1120XYPLOT DISP RESP/1120(T3)BEGIN BULKENDDATA

Run5 – file run5_dr10.dat –optional data recovery of SE 10


SECTION 8

NONLINEAR ANALYSIS


TABLE OF CONTENTSSection PageNONLINEAR ANALYSIS 7-7IS THE PROBLEM NONLINEAR? 7-8LINEAR VERSUS NONLINEAR STRUCTURAL ANALYSIS 7-9TYPES OF NONLINEAR ANALYSIS 7-11USER INTERFACE FOR NONLINEAR ANALYSIS 7-20SUMMARY OF NONLINEAR ANALYSIS 7-22OVERVIEW OF TRANSIENT ANALYSIS 7-23CAN THE PROBLEM BE SOLVED IN A LINEAR SOLUTION? 7-25GAP CONSTRAINTS IN SOL 101 7-26NONLINEAR LOADS IN DYNAMICS 7-37EXAMPLE PROBLEM USING NOLIN 7-44HINTS WHEN USING NOLINs 7-52NONLINs IN A MODAL TRANSIENT ANALYSIS 7-53TRANSFER FUNCTION – ADDS TERMS DIRECTLY INTO THE MATRICES 7-56BASICS OF NONLINEAR ANALYSIS 7-57TYPES OF NONLINEAR ELEMENTS 7-58


TABLE OF CONTENTS (CONT)Section PageGAP ELEMENT 7-60NLPARM BULK DATA ENTRY 7-72ADVANCING SCHEMES IN MSC.NASTRAN 7-73GAPS IN SOL 106 7-74NONLINEAR TRANSIENT ANALYSIS USING SOL 129 7-78HINTS AND RECOMMENDATIONS WHEN USING SOL 129 7-80NONLINEAR TRANSIENT USER INTERFACE 7-82TYPICAL INPUT FILE SETUP FOR SOL 129 7-85NONLINEAR TRANSIENT SOLUTION STRATEGY 7-86EXAMPLE PROBLEM USING GAP ELEMENTS 7-91EXAMPLE PROBLEM USING GAP ELEMENTS 7-923-D SLIDELINE CONTACT 7-96BCONP BULK DATA ENTRY 7-104BLSEG BULK DATA ENTRY 7-106BFRIC BULK DATA ENTRY 7-108BWIDTH BULK DATA ENTRY 7-109BOUTPUT BULK DATA ENTRY 7-111BOUTPUT CASE CONTROL COMMAND 7-112


TABLE OF CONTENTS (CONT)Section PagePARAM ADPCON 7-113PRINTOUT FOR SOLUTION STRATEGY 7-114EXAMPLE PROBLEM 7-115RESTARTS FOR NONLINEAR TRANSIENT ANALYSIS 7-121NONLINEAR TRANSIENT ANALYSIS USING SUPERELEMENTS 7-122EXAMPLE PEOBLEM USING SUPERELEMENTS 7-125


NONLINEAR ANALYSISWith the advent of both hardware and software, nonlinear analysis has become an integral part of structural analysis If linear analysis is used to simulate a nonlinear structural behavior,

the results may be meaningless. However, there are many times when a linear approximation will suffice

Since the nonlinear analysis is an iterative process, the time it takes to perform a nonlinear analysis is in general substantially longer than that required for a linear analysis

Before starting a nonlinear analysis, you should ask yourself the following questions: Is my structure truly nonlinear? Can I idealize the nonlinear behavior in a linear solution? Can I take advantage of superelements to isolate the nonlinear regions?


IS THE PROBLEM NONLINEAR?The following “rules of thumb” help decide: Before the analysis look for

Gaps that might close Areas where the separation or sliding might occur Is the problem a “snap-through” buckling problem? Does the problem depend on large displacements? (Examples: cables

out-of-plane loads on membranes) Is the material nonlinear?

Always run a linear analysis first After the analysis check for

Stresses (strains) exceeding yield (can use PARAM, BIGER) Areas with large displacements (ex. ∆ > (t / 15) for plates Gaps that might have opened (closed)


LINEAR VERSUS NONLINEAR STRUCTURAL ANALYSIS

What Constitutes a Nonlinear Analysis? Geometric nonlinear analysis:

The kinematic relationship is nonlinear. The displacements and rotations are large. Equilibrium is satisfied in deformed configuration.

Follower forces:Loads are a function of displacements.

Large strain analysis:The element strains are a nonlinear function of the element deformations.

Material nonlinear analysis:Element constitutive relationship is nonlinear. Elements may yield.Element forces are no longer equal to stiffness times displacements (Kee· Ue)


LINEAR VERSUS NONLINEAR STRUCTURAL ANALYSIS (CONT)

What Constitutes a Nonlinear Analysis? (Cont.) Buckling analysis:

Force transformation matrix is not the transpose of displacement transformation matrix. The equilibrium is satisfied in the perturbed configuration.

Contact (interface) analysis:Gap closure and opening, and relative sliding of different components.

Boundary conditions may change during the solution. It follows that:

Displacements are not directly proportional to loads. Results for different loads cannot be superimposed. The solution is a function of the loading path.


TYPES OF NONLINEAR ANALYSISGeometric Nonlinearity Large displacements and large rotations

Element deformations are a nonlinear function of the grid point displacements (nonlinear displacement transformation matrix).

Large displacements Deflection of highly-loaded thin flat plates (geometric stiffening).


TYPES OF NONLINEAR ANALYSISGeometric Nonlinearity (Cont.)

Large rotations

Both compatibility and equilibrium are satisfied in a deformed configuration.

Effects of initial stress (geometric or differential stiffness) are included.

The follower force effect can be included. Examples: cable net, thin shells, tires, water hose, etc.


TYPES OF NONLINEAR ANALYSIS (CONT)

Material Nonlinearity Element stiffness matrix is not constant. Two reasons for variable stiffness matrix:

1. Stress-strain relationship is nonlinear (I.e., matrix D changes), but strains are small (I.e., matrix B is linear).Example: Yielding structure (nonlinear elastic or plastic), creepUser Interface: MATS1 and CREEP Bulk Data entries

Nonlinear elastic

Isotropic (MATS1, MAT1)


TYPES OF NONLINEAR ANALYSIS (CONT)Material Nonlinearity (cont) Elastic-plastic

Isotropic (MATS1, MAT1) Anisotropic (MATS1 with MAT2 or MAT9)

Viscoelastic

Isotropic (CREEP, MAT1)


TYPES OF NONLINEAR ANALSYSIS (CONT)

Material Nonlinearity (cont) Visco-elastic-plastic

Isotropic (CREEP, MATS1, MAT1)

2. Strains are large (I.e., strain deformation matrix B is nonlinear). In general, stress-strain relationships and displacement transformation relationships are also nonlinear.Example: Rubber materialsUser Interface: MATHP, PLPLANE, and PLSOLID Bulk Data entries

Large Strains Element strains are nonlinear functions of element deformations.


TYPES OF NONLINEAR ANALYSIS (CONT)Material Nonlinearity (cont) Hyperelastic (large strain)

Isotropic (MATHP)

Rubber Bearing (Hyperelastic Material)



Material Nonlinearity (cont)Temperature-Dependant Material Properties Linear elastic materials (MATT1, MATT2, and MATT9). Elastic

Isotropic (MAT1, MATT1) Orthotropic (MAT2, MATT2) Anisotropic (MAT2, MATT2, MAT9, MATT9)



Material Nonlinearity (cont) Nonlinear elastic materials (MATS1, TABLES1, TABLEST). Nonlinear Elastic

Isotropic (MAT1, MATS1, MATT1)

Note: Nonlinear elastic composite materials cannot be temperature dependant.



Contact (Interface) Analysis Treated by gap and 3-D slideline contact. Example: O-rings, rubber springs in the auto and

aerospace industry, auto or bicycle brakes, and rubber seals in disc brakes, etc.


USER INTERFACE FOR NONLINEAR ANALYSIS

Compatible with linear analysis Analysis types

Nonlinear static analysis: SOL 106 Quasi-static (creep) analysis: SOL 106 Linear buckling analysis: SOL 105 Nonlinear buckling analysis: SOL106 (PARAM, BUCKLE) Nonlinear transient response analysis: SOL 129

Subcase structure Allows changes in loads, boundary conditions (SOL 106), and

methods. Allows changes in output requests.


USER INTERFACE FOR NONLINEAR ANALYSIS (CONT)

Bulk Data classification Geometric data Element data Material data Boundary conditions Loads and enforced motion Selectable in Subcases Solution Strategy


SUMMARY OF NONLINEAR ANALYSIS In nonlinear analysis:

Any one or more of the following relationships may be nonlinear: Kinematics Element compatibility Constitutive relationship Equilibrium

Loads may be functions of displacements. Opening and closing of different components. Boundary conditions may change.

Nonlinear Solution Sequences: SOL 106: Nonlinear static analysis (geometric, material,

large strain, buckling, surface contact, and constraint changes).

SOL 129: Nonlinear transient analysis (geometric, material, large strain, and surface contact). No constraint changes are allowed.


OVERVIEW OF TRANSIENT ANALYSIS Static analysis:

Compute a solution U that satisfies the equilibrium equation:

F(U) = P Transient analysis:

Compute a solution u that satisfies the equilibrium equation:

For a linear system

For a general nonlinear system Mass of the system may change

{ } { } ( ) ( )U,tP t,UF t,UD t,UI +++

InertiaForces

DampingForces

ElementForces

ExternalLoad

( )tP KU UB UM =++


OVERVIEW OF TRANSIENT ANALYSIS (CONT)

Damping may change Stiffness may change Load may be function of system response In MSC.NASTRAN mass and damping (except for CBUSH and

CBUSH1D) cannot change. Therefore, the equilibrium equation is

Nonlinear Transient Analysis Nonlinear transient analysis proceeds by dividing the time into a

number of small time steps.

( )( ) ( )U,tP tUF UB UM =++


CAN THE PROBLEM BE SOLVED IN A LINEAR SOLUTION?

Certain class of nonlinear problems can be solved with a linear solution (e.g. SOL 101 or 109).

The following criteria must be satisfied. The structure must not yield. The displacements and the rotations of the structure are small. Boundary condition does not change. The nonlinearity is localized (recommended). Ideal for modeling nonlinear springs and dampers.

In Statics, SOL 101 has the ability to model gaps (added in V70.5). Using PARAM, CDITER and SUPORT entries.

In transient solutions, MSC.NASTRAN has a series of nonlinear loads which can be applied to any transient solution to simulate this type of nonlinear problem. These types of nonlinear loads can be used to simulate nonlinear springs and dampers. These nonlinear loads are applied on the right hand side of the equation as follows:

N(t) can also be a function of displacement and/or velocity.

( ) ( )tN tP KU UB UM +=++


GAP CONSTRAINTS ON SOL 101 The sssalter cda.v70 has been implemented as part of SOL 101 This feature allows you to model GAPs in SOL 101, rather than

having to use SOL 106 with GAP elements. The feature is actually a “GAP constraint”. It constrains the

displacement of selected dof to be >=0.0 (or the reaction force may not be negative).

This works well for points which may come into contact with the ground.

The initial opening is set to 0.0 and you determine whether the gap is assumed to open or closed.

The constraints are satisfied by an iterative procedure which is built into SOL 101. The process starts with a random vector, which assumes certain GRIDs to be in contact and others to be open.

A solution is obtained when all of the GAP constraints are satisfied (there are no negative reaction forces at the selected dof).

Multiple SUBCASEs are allowed, each is solved separately.


GAP CONSTRAINTS ON SOL 101 (CONT) User Interface:

Input You need to use the SUPORT Bulk Data entry, some new parameters and

a special DMIG entry named CDSHUT. SUPORT (required) Selects constrained degrees-of-freedom. These points

must be in the a-set of the residual structure. This means they must not be dependent in an MPC equation, constrained by SPC, partitioned by OMIT, or in an upstream superelement.

PARAM CDITER,I (required) Constraints will be applied if CDITER is greater than zero. The value is the maximum number of iterations allowed. (Default=0).

PARAM CDPRT Controls the printing of constraint violations during iterations. The sparse matrix printer prints UR (negative displacements) and QR (negative forces of constraint) for constrained degrees-of-freedom. (Default=‘YES’)

PARAM CDPCH Controls the PUNCH output of DMIG CDSHUT records for the final state of the constrained degrees-of-freedom. (Default=‘NO’)

DMIG CDSHUT Optional input of the vector defining the state of the constrained degrees-of-freedom. A one indicates closed and a zero means open. See PARAM CDPCH for a method to have MSC.NASTRAN create these records. (Default=all closed)


GAP CONSTRAINTS ON SOL 101 (CONT) Output

The output is standard for SOL 101, and all existing postprocessors will work. Forces for closed degrees-of-freedom are in the SPCFORCE output. In addition there is information in the “.f06” file which shows diagnostic information for the iterations. A final state vector may be output in the “.pch” file.

Guidelines The finite element model input looks just like input to SOL 101 with the

addition of input records shown above. It is recommended that the “.f06” file be examined to ensure that the iterations have converged, since the results of the last iteration will be output. The last iteration should have zero changes.

If the constraint is between a finite element model and a fixed boundary, then arrange to have one of the degrees-of-freedom at the boundary grid points represent motion perpendicular to the boundary. A positive displacement represents motion away from the boundary. If, on the other hand, the constraint represents relative motion between two bodies, MPC equations are needed to define a relative motion degree-of-freedom, which is then constrained to have a non-negative displacement.


GAP CONSTRAINTS ON SOL 101 (CONT) Limitations

The only nonlinearity allowed is the constrained displacements. There is no gap stiffness and no sliding friction. Free bodies can not be analyzed using SUPORT to define rigid

body modes and have constrained degrees-of-freedom in the same model. Parameters INREL and CDITER are mutually exclusive. A fatal message is issued if both parameters are not present.

No constraint changes are allowed between subcases. There is no guarantee that the solution will converge or that all

systems will follow the same path.


GAP CONSTRAINTS ON SOL 101 (CONT) Example: A cantilever beam with a GAP along the span

Since the GAP constraint only checks to see if a displacement is >=0., we define a dof (by GRID points, 60 and 61) to measure the opening.

Point 60 will “measure” the gap. Point 61 will have the initial opening as a constraint. The following equation will be used (as an MPC) to accomplish this:

If the displacement of GRID 60 becomes less than 0.0, the GAP will be closed and the displacement will be zero (0.0).

61660 UUU +=


GAP CONSTRAINTS ON SOL 101 (CONT)

Example: Input file gap101.dat –ID CBAR, TEST GAP IN SOL 101SOL 101CENDTITLE = LINEAR STATICS WITH A GAPLABEL = CANTILEVER BEAM WITH GAPSPC = 1LOAD = 12MPC = 100set 999 = 1spcforce = ALLMPCFORCE = ALLGPFORCE = ALL$SUBCASE 1LABEL = TIP LOADDISP = ALLBEGIN BULKPARAM,POST,0GRID,1,,0.,0.,0.=,*(1),=,*(10.),===(8)GRID,11,,0.,1.,0.,,123456CBAR,1,1,1,2,11=,*(1),=,*(1),*(1),===(7)$PBAR,1,1,1.,10.,10.,10.PARAM,AUTOSPC,YESMAT1,1,30.+6,,,.283SPC1,1,123456,1FORCE,12,10,,2.,0.,–1.,0.$

$ MODEL .05 GAP AT POINT 6$$ GRID 60 = GAP MEASUREMENT$ GRID 61 = INITIAL GAP$ U60 = U61+U6$GRID,60GRID,61MPC,100,6,2,–1.0,60,2,+1.0, ,61,2,–1.0SUPORT,60,2SPC,1,61,2,.05PARAM,CDITER,10ENDDATA


OUTPUT FROM GAP 101


GAP CONSTRAINTS ON SOL 101 (CONT)

Example: gap101.dat – results From the output, we see that the GAP has not closed

under this loading. For the first iteration, the gaps are assumed closed. Matrix QRI indicates that we have a negative force in

GRID 60 from the gap in the initial iteration, so the gap is “opened” for the second iteration, which converges.

Example: gap101close.dat – change the loading to 2000.0 – more than enough to close the gap.


OUTPUT FROM GAP101CLOSE


NONLINEAR LOADS IN DYNAMICS These nonlinear loads are applied on the right hand side of the

equation as follows:

where: N(t) is function of displacement and/or velocity. Allows for the specification of load at a particular degree of freedom to

be the function of displacement and velocity at another degree of freedom.Example:

Load at grid point 1, displacement component 2 as a function of the displacement component 1 at grid point 3.

Useful for specifying nonlinear springs and nonlinear damping. Nonlinear loads are specified using NONLINi entries.

( ) ( )tN tP KU UB UM +=++


NONLINEAR LOADS IN DYNAMICS (CONT) Nonlinear loads are selected via the NONLINEAR Case Control

command. Nonlinear loads cannot be selected via the DLOAD Case Control

command. All degrees of freedom referenced on NOLINi entry must be members

of the solution set. Velocity for an independent degree of freedom (for the purpose of

loads) is calculated as

Note: This may be different from that calculated in the integration scheme. But it is acceptable.

In all NOLINi entries a degree of freedom is specified by the grid number and its component number.

All loads generated with NOLINi entries lag behind by one time step ∆t in the linear solutions, they are updated at each time step in SOL 129.

tUUU ttt

t ∆∆−−

=


NONLINEAR LOADS IN DYNAMICS (CONT)NOLIN1 Bulk Data EntryDescription: Defines nonlinear transient functions of the form.

Function of displacement: (1)Function of velocity: (2)

where and are the displacement and velocity at point GJ in the direction of CJ.Format:

Example:

Field ContentsSID Nonlinear load set identification number (Integer > 0)GI Grid, scalar, or extra point identification number at which nonlinear

load is to be applied. (Integer > 0)CI Component number for GI. (0 < Integer < 6; blank or zero if GI is a

scalar or extra point)S Scale factor (Real)

( ))t(uT*S)t(P ji =( ))t(uT*S)t(P ji =

)t(u j )t(u j

1 2 3 4 5 6 7 8 9 10NOLIN1 SID GI CI S GJ CJ TID

NOLIN1 21 3 4 2.1 3 10 6


NONLINEAR LOADS IN DYNAMICS (CONT)NOLIN1 Bulk Data Entry (CONT)

GJ Grid, Scalar, or extra point identification number (Integer > 0)

CJ Component number for GJ according to the following table:

TID Identification number of a TABLEDi entry. (Integer > 0)

Type of Point Displacement Velocity

Grid 1 < Integer < 6 11 < Integer < 16

Scalar Blank or zero Integer = 10

Extra Blank or zero Integer = 10


NONLINEAR LOADS IN DYNAMICS (CONT)

NOLIN2 Bulk Data EntryDescription: Defines nonlinear transient forcing functions of

the form:

where Xj(t) and Xk(t) can be either displacement or velocity at points GJ and GK in the directions of CJ and CK.

Format:

Example:

1 2 3 4 5 6 7 8 9 10NOLIN2 SID GI CI S GJ CJ GK CK

NOLIN2 14 2 1 2.9 2 1 2

(t)X * (t)X *S (t)P kji =



NONLIN3 Bulk Data EntryDescription: Defines nonlinear transient forcing functions of

the form:

where Xj(t) may be a displacement or a velocity at point GJ in the direction of CJ.

Format:

Example:

1 2 3 4 5 6 7 8 9 10NOLIN3 SID GI CI S GJ CJ A

NOLIN3 4 102 -6.1 2 15 -3.5

( ) ( )[ ] ( )( )

≤

>=

0tX, 0 0tX,tx*S

tPj

jA

ji



NONLIN4 Bulk Data EntryDescription: Defines nonlinear transient forcing functions of

the form:

where Xj(t) may be a displacement or a velocity at point GJ in the direction of CJ.

Format:

Example:

( ) ( )[ ] ( )( )

≥

<−−=

0tX, 0 0tX,tx*S

tPj

jA

ji

1 2 3 4 5 6 7 8 9 10NOLIN4 SID GI CI S GJ CJ A

NOLIN4 2 4 6 2 101 16.3


EXAMPLE PROBLEM USING NOLIN The following nonlinear problem can be solved using a linear solution (e.g.,

SOL 109)

Detailed behavior of the nonlinear spring stopper

A = 0.314 in2

I = 0.157 in4

ρ = 0.3 lb/in3

20 Beam Elements

Forcing Function


EXAMPLE PROBLEM USING NOLIN (CONT)



Input Loading:



Response at GRID 10005



Response at GRID 10010



Nonlinear load:


HINTS WHEN USING NOLINs

Use smaller time steps than normal linear transient analysis.

Start with initial stiffness or damping value and use NOLIN to add or subtract stiffness or damping value rather than defining the whole range on the NOLINs directly.

Verify the nonlinear loads (using NLLOAD). Easier to use in a direct solution.


NOLINs IN MODAL TRANSIENT ANALYSIS


NOLINs IN MODAL TRANSIENT ANALYSIS (CONT)


TRANSFER FUNCTION – ADDS TERMS DIRECTLY IN THE MATRICES

Transfer functions are used to add terms directly into the dynamic matrices. The form of these is:

where ud = dof determining the row in the matricesbi = term to be added to the diagonal terms in the mass

(b2), damping (b1), or stiffness (b0) matrices ai = term to be added in the mass, damping, or stiffness

column associated with the ui dof

Equivalent to P-type DMIG matrices (M2PP, B2PP, K2PP)

Defined by the TF Bulk Data entry and selected by the TFL Case Control command (residual structure only)

( ) ( )∑=

=+++++N

1iii

2210d

2210 0upapaaupbpbb


BASICS OF NONLINEAR ANALYSIS Basic User Interface:

Solution strategy: Solution strategy nonlinear static analysis NLPARM Arc length increments for nonlinear static analysis NLPCI Solution strategy for nonlinear transient analysis TSTEPNL Displacement-increment analysis SPCD, SPC

Nonlinear materials: Nonlinear elastic and plastic MATS1 Creep materials CREEP Hyperelastic (rubber-like) materials MATHP Temperature-dependant elastic materials MATT1, MATT2, MATT9 Temperature-dependant nonlinear elastic materials TABLEST, TABLES1

Geometric nonlinear PARAM, LGDISP

Follower forces FORCE1, FORCE2, MOMENT1, MOMENT2, PLOAD, PLOAD2, PLOADX1, and RFORCE

Nonlinear buckling analysis: PARAM, BUCKLE, in SOL106

Contact (interface): gap and 3-D slideline contact

Boundary changes: SPC, SPCD, and MPC (in static nonlinear)


TYPES OF NONLINEAR ELEMENTS Physical elements which support geometric and material nonlinear

(selected elements from table 1 from Section 5.1 in the Reference Manual)


TYPES OF NONLINEAR ELEMENTS (CONT)

Small strain – less than 10% (ROD, CONROD, BEAM, QUAD4, TRIA3, HEXA, PENTA, TETRA).

Large strain – hyperelastic elements (QUAD4, QUAD8, QUAD, QUADX, TRIA3, TRIA6, TRIA6, TRIAX, HEXA, PENTA, TETRA)

Contact (interface) elements GAP 3-D slideline


GAP ELEMENT Connects two grid points with the orientation (gap direction). Opening or closing (contact) is determined in the gap direction. Uses hard surface contact, I.e., no penetration of grid points is

allowed in the gap direction. Can specify friction between the two points Uses the penalty method for both contact and friction. Can have a large opening between the two points. No large relative slipping between the two points is permitted. No large rotation for the two points (relative or rigid).

NoPGAPCGAP

Geometric Nonlinearity

MaterialPropertyConnectivity


GAP ELEMENT (CONT)CGAP Bulk Data EntryDefines a gap or frictional element for nonlinear analysis.Format:

Example:

Alternate Format and Example:

If CID is present, then CID identifies the element coordinate system T1, T2, and T3 of the CID are the element x-, y-, and z- axis,

respectively. If CID field is blank and GA and GB are not coincident (distance from A

to B > 10-4), then the GAP element coordinate system is defined as follows: GA – GB defines the x-axis. Orientation vector is given by x1, x2, and x3, (like beam element) or GA – GO defines the

x-y plane.

1 2 3 4 5 6 7 8 9 10CGAP EID PID GA GB X1 X2 X3 CID

CGAP 17 2 110 112 5.2 0.3 -6.1

CGAP EID PID GA GB G0 CID

CGAP 17 2 110 112 13


GAP ELEMENT (CONT) For coincident grid points GA and GB,

If CID is blank, the job is terminated with a fatal message

CGAP Element Coordinate System

Note: KA and KB is this figure are from the PGAP entry.


GAP ELEMENT (CONT)PGAP Bulk Data EntryDefines the properties of the gap element (CGAP entry)

Format:

Example:

1 2 3 4 5 6 7 8 9 10PGAP PID U0 F0 KA KB KT MU1 MU2

TMAX MAR TRMIN

PGAP 2 0.025 2.5 1.00E+06 1.00E+06 0.25 0.25


GAP ELEMENT (CONT)

GAP Element Force-Deflection Curve for Nonlinear Analysis.

Shear Force for GAP Element.


GAP ELEMENT (CONT) There are 2 kinds of GAP element:

New and adaptive (TMAX > 0., preferred choice). New GAP can force bisection and stiffness updates.

Old and non-adaptive (TMAX = -1.0)

New adaptive GAP element is recommended. Initial GAP opening is defined by U0, not by the

distance between GA and GB. Preload is defined by F0 (not recommended). Closed stiffness Ka is used when Ua – Ub > U0

The default for open stiffness Kb = 10-14 Ka


GAP ELEMENT (CONT) The transverse shear stiffness KT becomes active upon contact.

(The default = µ1 * Ka)

The continuation line is applicable for adaptive features of the new GAP element only.

Adaptive features are specified by TMAX > 0. If the penetration is greater than TMAX, the penalty value is

increased by an order of magnitude. If the penetration is less that TRMIN * TMAX, the penalty value

is decreased by an order of magnitude. MAR defines the lower and upper bounds for the penalty value

adjustment ratio.

0.0µ1

Static FrictionKinetic Friction

µ1µ2

DefaultNew


GAP ELEMENT (CONT) Proper Estimation of Gap Stiffness

The stiffness of the beam at points 1 and 2

The recommended GAP stiffness:

The recommended stiffness acts rigid when closed, and free when open with an error of 0.1%.

Factors (103 or 10-3) may be reduced to facilitate convergence at the expense of accuracy.

Recommended stiffness is based on the decoupled stiffnesses.

1LEI3K 31 == 16

LEI48K 32 ==

( )( ) 001.0K,KMIN10K

10 x 16K,KMAX1000K

213-

B

321A

=∗≤

=∗≥


GAP ELEMENT (CONT) Friction Features Friction effect is turned off with Kt = 0 Static and kinetic frictions are allowed. Frictional gap problem is path dependent. Sticking with elastic stiffness Kt before slipping Slipping is similar to plasticity. Subincremental process similar to plasticity is used for the new

gap. No subincremental process for the old gap. Accuracy deteriorates if the increment produces large changes

in the displacements with friction. The slip locus is generalized by an ellipse:

( )2xs2

z2

y FFF µ≤+

( )2xk2

z2

y FFF µ>+

Closed and Sticking

Closed and Slipping


GAP ELEMENT (CONT) Caution for Using GAP Element Large displacement or rotation capability is not implemented. When used for linear analysis, GAP stays linear with the initial

stiffness. The penalty values (Ka and Kt) should be as small as possible

for solution efficiency, but large enough for acceptable accuracy. Penalty values are constants while the structural stiffness in the

adjacent structure changes continuously during loading. Avoid friction unless its effect is significant. Use smaller increments if friction is involved. Avoid complications by using isotropic friction (for old gap). Typical coefficients of friction:

Steel on steel (dry) 0.4 to 0.6 Steel on steel (greasy) 0.05 to 0.1 Brake lining on cast iron 0.3 to 0.4 Tire on pavement (dry) 0.8 to 0.9


GAP ELEMENT (CONT)


GAP ELEMENT (CONT) Output is obtained by a STRESS output request in the Case

Control Section. Output quantities are in the element coordinates. Output shows GAP status: open, slide, stick, slip. Positive Fx is a compression force. Total displacement is from the original position. Slip displacement for the sticking or slipping condition is the slip

from the current contact position or slip center.

Slip displacement for the open or sliding condition is the same as the total displacement.


NLPARM BULK DATA ENTRYNLPARM with all its field is shown belowParameters for Nonlinear Static Analysis ControlDefines parameters for nonlinear static analysis iteration strategy.

Format:

Example:

1 2 3 4 5 6 7 8 9 10NLPARM ID NINC DT KMETHOD KSTEP MAXITER CONV INTOUT

EPSU EPSP EPSW MAXDIV MAXQN MAXLS FSTRESS LSTOLMAXBIS MAXR RTOLB

NLPARM 15 5


ADVANCING SCHEMES IN MSC.NASTRAN Constant load increments Constant displacement increments Arc-length increments

Constant Load Increment

Field ContentsID Identification number (Integer > 0)NINC Number if increments (0 < Integer < 1000)Example:

1 2 3 4 5 6 7 8 9 10NLPARM ID NINC


GAPS IN SOL 106 Example: Cantilever beam with a GAP (the same

problem as was shown in earlier using SOL 101)

Stiffness of the beam at point 6

The recommended GAP stiffness000,200,7LEI3K 36 ==

200,7K10K10 x 2.7K1000K

3B

9A

=∗≤

=∗≥−


GAPS IN SOL 106 (CONT)ID CBAR, TEST GAP IN SOL 106SOL 106CENDTITLE = LINEAR STATICS WITH A GAPSUBCASE 1LABEL = TIP LOADDISP = ALLSPC = 1LOAD = 12nlparm = 9set 999 = 100stress = 999force = 999set 998 = 6,60spcforce = allGPFORCE = allBEGIN BULKnlparm,9,1GRID,1,,0.,0.,0.=,*(1),=,*(10.),===(8)GRID,11,,0.,1.,0.,,123456CBAR,1,1,1,2,11=,*(1),=,*(1),*(1),===(7)PBAR,1,1,1.,10.,10.,10.PARAM,AUTOSPC,YESMAT1,1,30.+6,,,.283SPC1,1,123456,1FORCE,12,10,,2000.,0.,–1.,0.$$ MODEL .05 GAP AT POINT 6grid,60,,50.,–.5,0.,,123456CGAP,100,100,6,60,1.,0.,0.PGAP,100,.05,,7.2+9,7.2ENDDATA


GAP ELEMENTS (CONT)


NONLINEAR TRANSIENT ANALYSIS USING SOL 129

GENERAL FEATURES Transient, material nonlinear, geometric nonlinear,

combined geometric and material nonlinear, and contact problems can be solved using this solution sequence.

Linear superelements can be used to simplify the nonlinear solution.

Modal reduction (SEQSET, EIGRL) and generalized dynamic reduction (DYNRED) are available for the linear superelements.

Parameter-controlled Restarts are available from the end of any SOL129 subcase, or from SOL 106.


NONLINEAR TRANSIENT ANALYSIS USING SOL 129 (CONT)

GENERAL LIMITATIONS No constraint changes after first subcase – including restart. No time–varying thermal loads (except using LOADSET–LSEQ)

or enforced displacements. Reduction (GDR, Guyan reduction, component modes) only for

superelements. PARAM “G” damping only applies to linear elements (requires

PARAM,W3 also). Nonlinear element damping provided by GE on MAT Bulk Data

entries (PARAM “W4” must also be used) only for initial K. Damping stays linear (except for damping via CBUSH1D). No element force output for nonlinear elements. Upstream loads are ignored in the superelement data recovery. No grid point stresses for nonlinear elements. Mass is assumed to remain constant.


HINTS AND RECOMMENDATIONS WHEN USING SOL 129

Identify the type of nonlinearity. Localize nonlinear region. Put linear regions in upstream superelement(s). Divide time history by subcases for convenience and efficiency. Each subcase should not have more than 200 time steps

(recommendation). Select default values to start – TSTEPNL Pick time step size for highest frequency of interest. Use 12 or

more steps per cycle based on frequency content of input as starting ∆t.

Use the adaptive time stepping algorithm. Adaptive time stepping is based on the response of the model,

not the dynamic loading. Consequently, if sudden change in loading occurs, the adaptive time step may miss it.


HINTS AND RECOMMENDATIONS WHEN USING SOL 129 (CONT)

Care should be taken as to ensure that loading history is properly traced with the adaptive time stepping.

Some damping is desirable for numerical stability There is no such thing as an undamped structure.

Avoid massless degrees of freedom. Choose GAP stiffness carefully. Increase MAXITER on the TSTEPNL entry if

convergence is poor.


NONLINEAR TRANSIENT USER INTERFACE

Solution sequence SOL 129.

Solution strategy TSTEPNL Bulk Data entry. TSTEPNL Case Control command (always required). PARAM,LGDISP,1 – enable large displacement nonlinearities

(default is –1 = no large displacement effects included)

Mass specification RHO field in MATi Bulk Data entries. CMASSi Bulk Data entries for scalar mass elements. CONMi Bulk Data entries for concentrated mass elements. PARAM,COUPMASS, to specify the generation of coupled rather

than lumped mass matrices for elements with coupled mass capability.

PARAM,WTMASS.


NONLINEAR USER INTERFACE (CONT) Damping specification

CVISC Bulk Data entry for the viscous damper element Field GE in MATi Bulk Data entries for nonlinear element damping PARAM, G for overall structural damping PARAM, W3 to convert structural damping to equivalent viscous

damping PARAM, W4 to convert element damping to equivalent viscous

damping PARAM, NDAMP to specify numerical damping

Initial conditions specification (same as linear transient) TIC Bulk Data entry IC Case Control command


NONLINEAR USER INTERFACE (CONT) Additional entries for nonlinear analysis

Similar to nonlinear static analysis Material nonlinear

MATS1 Geometric nonlinear

PARAM, LGDISP, +1 Contact (interface)

CGAP/PGAP BCONP, BLSEG, BWIDTH, BFRIC, BOUTPUT

Combined material and geometric nonlinear MATS1 PARAM, LGDISP, +1


TYPICAL INPUT FILE SETUP FOR SOL 129


NONLINEAR TRANSIENT SOLUTION STRATEGY

Specified by TSTEPNL Bulk Data entrySelected by TSTEPNL Case Control commandTSTEPNL Bulk Data EntryDescription: Defines parametric controls and data for nonlinear transient analysisFormat:

Examples:

Field ContentsID Identification number (Integer > 0)NDT Number of times steps of value DT (Integer > 4)DT Time increment (Real > 0.0)NO Time step interval for output. Every NO-th step will be saved for output

(Integer > 0; Default = 1)METHOD Method for dynamic matrix update and the direct time integration strategies

(Character: “AUTO”, “TSTEP”, or “ADAPT”; Default = “ADAPT”)

1 2 3 4 5 6 7 8 9 10TSTEPNL ID NDT DT NO METHOD KSTEP MAXITER CONV

EPSU EPSP EPSW MAXDIV MAXQN MAXLS FSTRESSMAXBIS ADJUST MSTEP RB MAXR UTOL RTOLB

TSTEPNL 250 1 ADAPT 2 -10 PW1.00E-02 1.00E-03 1.00E-06 2 10 2 0.02

5 5 0 0.75 16 0.1 20


NONLINEAR TRANSIENT SOLUTION STRATEGY (CONT)

KSTEP If METHOD = “TSTEP”, then KSTEP is the time step interval for stiffness Updates. If METHOD = “ADAPT”, then KSTEP is the number of converged bisection solutions between stiffness updates (Integer > 0; Default = 2)

MAXITER Limit on number of iterations for each time step. (Integer ≠ 0; Default = 10)CONV Flags to select convergence criteria (Character: “U”, “P”, “W”, or any

combination; Default = “PW”)EPSU Error tolerance for displacement (U) criterion (Real > 0.0; Default = 1.0E-2)EPSP Error tolerance for load (P) criterion (Real > 0.0; Default = 1.0E-3)EPSW Error tolerance for work (W) criterion (Real > 0.0; Default = 1.0E-6)MAXDIV Limit on the number of diverging conditions for a time step before the

solution is assumed to diverge (Integer > 0; Default =2)MAXQN Maximum number of quasi-Newton correction vectors to be saved on the

database (Integer > 0; Default = 10)MAXLS Maximum number of line searches allowed per iteration (Integer > 0;

Default = 2)FSTRESS Fraction of effectiveness stress (σ ) used to limit the subincrement size in

the material routines (0.0 < Real < 1.0; Default = 0.2)



MAXBIS Maximum number if bisections allowed for each time step ( -9 < Integer < 9; Default = 5)

ADJUST Time step skip factor for automatic time step adjustment (Integer > 0; Default = 5)

MSTEP Number of steps to obtain the dominant period response (10 < Integer < 200; Default = variable between 20 and 40)

RB Define bounds for maintaining the same time step for the stepping function if METHOD = “ADAPT”. (0.1 < Real < 1.0; Default = 0.75)

MAXR Maximum ratio for the adjusted incremental time relative to DT allowed for time step adjustment (1.0 < Real < 32.0; Default = 0.1)

UTOL Tolerance on displacement increment beneath which there is no time step adjustment (0.001 < Real < 1.0; Default = 0.1)

RTOLB Maximum value of incremental rotation (in degrees) allowed per iteration to activation bisection (Real > 2.0; Default = 20.0)



Various Methods ADAPT (NLTRD2 module)

Recommended method Two-point integration scheme PARAM, BETA is ignored KSTEP is the number of converged bisection solutions between stiffness updates Time step is automatically adjusted Use ADJUST = 0 to deactivate the time step adjustment Stiffness is automatically updated to improve convergence Time step can be bisected

AUTO (NLTRD module) Three-point integration scheme with BETA = 1/3 PARAM, BETA can be used to change the value of BETA. However, the

recommendation is not to change the value of BETA KSTEP has no meaning ??? Time step remains constant



Stiffness is automatically updated to improve convergence. No time step bisection (MAXBIS is not relevant)

TSTEP (NLTRD module) Three-point integration scheme with BETA = 1/3 PARAM, BETA can be used to change the value of BETA. However,

the recommendation is not to change the value of BETA KSTEP is the time step interval for the stiffness update (??? ask John) Time step remains constant No time step bisection (MAXBIS is not relevant)


EXAMPLE PROBLEM USING GAP ELEMENTS

Solve the problem described earlier in page 7-43, using nonlinear transient solution (SOL 129).

Use a GAP element to simulate the stopper, instead of NONLIN.

Adaptive numerical integration time steps are used.


EXAMPLE PROBLEM USING GAP ELEMENTS

ID NOLIN EXSOL 129CENDTITLE = DIRECT TRANSIENT RESPONSE, USING GAP ELEMENTSUBTITLE = SINUSOIDAL PULSE FOR ONE PERIODSPC=1002DLOAD=30TSTEPNL=20SET 1=10005SET 2=10010DISPL=1ACCEL=2OLOAD=1OUTPUT(XYPLOT)..BEGIN BULK$....... 2....... 3....... 4....... 5....... 6....... 7....... 8....... 9....... 10....$ MODELING INFORMATION FOR BEAM ONLYCBAR 101 100 10000 10001 0.0 0.0 1.CBAR 102 100 10001 10002 0.0 0.0 1.CBAR 103 100 10002 10003 0.0 0.0 1.CBAR 104 100 10003 10004 0.0 0.0 1.CBAR 105 100 10004 10005 0.0 0.0 1.CBAR 106 100 10005 10006 0.0 0.0 1.CBAR 107 100 10006 10007 0.0 0.0 1.CBAR 108 100 10007 10008 0.0 0.0 1.CBAR 109 100 10008 10009 0.0 0.0 1.CBAR 110 100 10009 10010 0.0 0.0 1.CBAR 111 100 10010 10011 0.0 0.0 1.CBAR 112 100 10011 10012 0.0 0.0 1.CBAR 113 100 10012 10013 0.0 0.0 1.CBAR 114 100 10013 10014 0.0 0.0 1.CBAR 115 100 10014 10015 0.0 0.0 1.CBAR 116 100 10015 10016 0.0 0.0 1.CBAR 117 100 10016 10017 0.0 0.0 1.CBAR 119 100 10018 10019 0.0 0.0 1.


EXAMPLE PROBLEM USING GAP ELEMENTS (CONT)

CBAR 120 100 10019 10020 0.0 0.0 1.CONM2 12 10010 .1GRID 10 50. –1.GRID 10000 0.0 0.0 0.0 1246GRID 10001 5. 0.0 0.0 1246GRID 10002 10. 0.0 0.0 1246GRID 10003 15. 0.0 0.0 1246GRID 10004 20. 0.0 0.0 1246GRID 10005 25. 0.0 0.0 1246GRID 10006 30. 0.0 0.0 1246GRID 10007 35. 0.0 0.0 1246GRID 10008 40. 0.0 0.0 1246GRID 10009 45. 0.0 0.0 1246GRID 10010 50. 0.0 0.0 1246GRID 10011 55. 0.0 0.0 1246GRID 10012 60. 0.0 0.0 1246GRID 10013 65. 0.0 0.0 1246GRID 10014 70. 0.0 0.0 1246GRID 10015 75. 0.0 0.0 1246GRID 10016 80. 0.0 0.0 1246GRID 10017 85. 0.0 0.0 1246GRID 10018 90. 0.0 0.0 1246GRID 10019 95. 0.0 0.0 1246GRID 10020 100. 0.0 0.0 1246MAT1 1000 3.+7 .3 .3PARAM WTMASS .002588PBAR 100 1000 0.31416 0.15708SPC 1002 10 123456SPC 1002 10020 3 10000 3TLOAD2 30 33 0.0 .11451–187.33 –90.DAREA 33 10005 3 47.2$TSTEP 20 299 .0002 1TSTEPNL 20 299 .0002 1CGAP 200 210 10 10010 0. 1. 0.PGAP 210 .02 1000. 10. 100. .1 . 1ENDDATA


3-D SLIDLINE CONTACTConcept Allows contact between two deformable bodies in a plane

One of the bodies is called the master and the other the slave. The master/slave line is the region where contact can occur.


3-D SLIDLINE CONTACT (CONT) A master/slave segment is the line joining two

consecutive nodes. Master/slave nodes are the grid points in the contact

region. The slideline plane is the plane in which the master

and the slave nodes must lie. The master and slave nodes can have large relative

motion within the slideline plane. Relative motions outside the slideline plane are

ignored. Therefore, they must be small. Contact is determined between the slave nodes and

the master line.


3-D SLIDLINE CONTACT (CONT)3-D Slideline Element Consists of 3 nodes: slave s, and master m1 and m2.

where s, m1, m2 = slave, master node 1, master node 2, respectively a, a0 = current and previous surface coordinategn = penetration of slave node into the master segmentgt = sliding of the slave node on the master segmentn = normal direction for the master segment


3-D SLIDLINE CONTACT (CONT) The element tangential (x) direction is the direction from master node 1

to master node 2. The element normal (y) direction is perpendicular to the tangential

direction in the slideline plane. The element z-direction is the slideline plane vector. Normal direction (y) is obtained by z x x. The normal direction must point toward the slave node. The penetration or gap is calculated by measuring how close the slave

node is to the master segment in the normal direction. The slave node slides on the master segment until a tensile force

develops. The surface coordinate is the parametric projection (0 to 1) of the slave

node onto the master segment. A 3-D slideline element is created for each slave node. The master nodes to which a slave node connects change continually. The only way an internal element can be identified is by the external grid

number of the slave node.


3-D SLIDLINE CONTACT (CONT)


3-D SLIDLINE CONTACT (CONT)General Features Can have as many slideline contact regions as desired. Contact is determined only for slave nodes and the master line. May specify symmetric penetration, i.e., contact is determined for

both slave and master nodes into master and slave line, respectively.

Initial penetration of slave nodes into master line is not allowed. User Warning Message 6315 is issued, if the initial penetration is

less than 10% of the master segment length. Coordinates of the slave node are changed internally to preclude

penetration. User Fatal Message 6314 is issued, if initial penetration for any

slave node is greater then 10% of the master segment length.


3-D SLIDLINE CONTACT (CONT)General Features(Cont.)

The master and slave nodes must be in the slideline plane in the initial geometry; otherwise Fatal message 6312 is issued.

During the analysis, no check is made to ensure that the master and slave nodes are in the slideline plane.

The slave or master nodes need not be attached to the physical element (model rigid surface).

Ensure that the contact is properly defined so that there are no erroneous overhangs.

Output can be requested in SORT1 or SORT2.


3-D SLIDLINE CONTACT (CONT)User Interface Bulk Data entries:

BCONP Defines the parameters for a contact region and its properties.

BLSEG Defines the grid points on the master/slave line.BFRIC Defines the frictional properties.BWIDTH Defines the width/thickness associated with each

slave node.BOUTPUT Defines the output requests for slave nodes in a

slideline contact region. Case Control Command:

BOUTPUT Selects contact region for output. DMAP parameter:

ADPCON Adjusts penalty values on restart.


BCONP BULK DATA ENTRYDescription: Defines the parameters for a contact region and its propertiesFormat:

Example:

Field ContentsID Contact region identification number (Integer > 0)SLAVE Slave region identification number (Integer >0)MASTER Master region identification number (Integer >0)SFAC Stiffness scaling factor. The factor is used to scale the penalty

values automatically calculated by the program (Real >0. or blank; Default = 1.0).

FRICID Contact friction identification number (Integer > 0. or blank)PTYPE Penetration type (Integer = 1 or 2; Default =1)

1: Unsymmetrical (slave penetration only by default)2: Symmetrical

CID Coordinate system ID to define the slide line plane vector and the slide line plane or contact (Integer > 0 or blank; Default = 0 which means the basic coordinate system)

1 2 3 4 5 6 7 8 9 10BCONP ID SLAVE MASTER SFAC FRICID PTYPE CID

BCONP 95 10 15 1 33 1


BCONP BULK DATA ENTRY (CONT)

Can have as many contact regions as desired. Penalty values are automatically selected based on the diagonal

terms of grid points. In symmetrical penetration, both the slave and master nodes are

checked for penetration into the master and the slave surface, respectively.

The t3 direction of CID is the z-direction of all the 3-D slideline elements (one corresponding to each slave node and also to each master node for symmetric penetration) of the contact region.


BLSEG BULK DATA ENTRYDescription Defines a curve which consists of a number of line

segments via grid numbers that may come in contact with another body. A line segment is defined between every two consecutive grid points. Thus, number of line segments defined is equal to the number of grid points specified minus 1. A corresponding BWIDTH Bulk Data entry may be required to define the width/thickness. Otherwise,the widh/thickness for the corresponding line segment will beassumed to be unity.

Format:

Examples:

Field ContentsID Line segments identification number (Integer > 0)Gi Grid number on a curve in continuous topological order so

that the normal to the segment points towards the other curve.

1 2 3 4 5 6 7 8 9 10BLSEG ID G1 G2 G3 G4 G5 G6 G7

G8 THRU G9 BY G10 G11 G12

BLSEG 15 5 THRU 21 BY 4 27 30 32 33 35 THRU 44


BLSEG BULK DATA ENTRY (CONT) Grid points must be specified in topological order. Normals to (z x t) of the master segments must face toward the

slave line for unsymmetric penetration. Normals of master and slave segments must face each other for

symmetric penetration. These conditions are accomplished by traversing

counterclockwise or clockwise from the master line to the slave line depending on whether the slideline vector forms the right-hand rule or the left hand rule.

The master line must have at least two grid points. The slave line may have only one grid point for unsymmetrical

penetration. Two grid points in a line cannot be the same or coincident

except for the first point and the last point, which signify a close region.


BFRIC BULK DATA ENTRYDescription: Defines frictional properties between two bodies in contact.Format:

Example(s):

Field ContentsFID Friction identification number (Integer > 0)STIF Frictional stiffness in stick (Real > 0.0; Default = auto

select by program)MU1 Coefficient of static friction (Real > 0)

(Note: no distinction made between static and kinetic friction.)

1 2 3 4 5 6 7 8 9 10BFRIC FID STIF MU1

BFRIC 33 0.3


BWIDTH BULK DATA ENTRYDescription: Defines width/thickness for line segments in 2-D / 3-D slideline

contact defined in the corresponding BLSEG BULK Data entry. This entry may be omitted if the width/thickness of each segment defined in the BLSEG entry is unity. Number of thicknesses to be specified is equal to the number of segments defined in the corresponding BLSEG entry. If there is no corresponding BLSEG entry, the width/thickness specified in the entry are not used by the program.

Format:

Examples:

Field ContentsID Width/thickness set identification number (Integer > 0)Wi Width/thickness values for the corresponding line segments

defined in the BLSEG entry (Real > 0.0)

1 2 3 4 5 6 7 8 9 10BWIDTH ID W1 W2 W3 W4 W5 W6 W7

W8 THRU W9 BY W10 W11 W12

BWIDTH 15 2.0 THRU 5.0 BY 1.02.0 2.0 2.0 2.03.0 THRU 12.0


BWIDTH CULK DATA ENTRY (CONT)

ID is the same as the slave line (BLSEG) ID. Widths/thicknesses are specified for slave nodes only.

Default = unity. Widths/thicknesses are used for calculating contact

stresses. Each slave node is assigned a contributory area. The number of widths to be specified is equal to the

number of slave nodes minus one. For only one slave node, specify the area in W1 field.


BOUTPUT BULK DATA ENTRYDescription: Defines the slave nodes at which the output is

requested.Format:

Example:

Field ContentsID Boundary identification number for which output is

desired (Integer > 0)Gi Slave node numbers for which output is desiredNote: The ID is the same as the corresponding BCONP ID. This

entry can selectively specify the slave grid points for whichOUTPUT is desired.

1 2 3 4 5 6 7 8 9 10BOUTPUT ID ALL

G1 G2 G3 G4 G5 G6 G7 G8G8 THRU G9 BY G10

BOUTPUT 15 ALL


BOUTPUT CASE CONTROL COMMANDDescription: Selects slave nodes specified in the Bulk Data entry BOUTPUT for history

output.Format:

Example:BOUTPUT = ALLBOUTPUT = 5

Field ContentsSORT1 Output is presented as a tabular listing of slave nodes for each load or time depending

on the solution.SORT2 Output is presented as a tabular listing of load or time for each slave node.PRINT The print file (Fortran I/O unit 6) is the output media.PUNCH The punch file is the output media.PLOT Generate salve node results history but do not print.ALL Histories of all the slave nodes listed in all the BOUTPUT bulk data entries are output. If

no BOUTPUT bulk data entries are specified, histories of all the slave nodes in all the contact regions are output.

n Set identification of previously appearing set command. Only contact regions whose identification numbers appear on the set command are selected for output. If there is a BOUTPUT bulk data entry for a contact region selected via the set command, histories for all the slave nodes in that contact region are output.

None Results histories for the slave nodes are not calculated or output.

=

Nonen

ALL

PLOT PUNCH SORT2,PRINT SORT1,

BOUTPUT

NOTE: This command selects the contact region for which output is desired


PARAM ADPCON User interface

PARAM, ADPCON, (real value) On restart, ADPCON can be used to increase or decrease the

penalty vales for all the line contact regions. A negative value of ADPCON implies that penalty values are

calculated at the beginning of a subcase only. This is useful for contact between elastic bodies.

Penalty values for a line contact region are given by

whereks = number calculated automatically for a slave node by NastranSFAC = scale factor specified in BCONP

ADPCONSFACks ∗∗


PRINTOUT FOR SOLUTION STRATEGY

This is printed only for the converged iteration

This is printed only for the converged iteration.


EXAMPLE PROBLEMPurpose To illustrate the use of slideline contact and nonlinear transient

analysis in bumper crash applications.Problem Description A rigid barrier moving at 5 mph impacts a bumper fixed at the

bumper brackets. Plot the deformed shape of the bumper after 20 msec of contact.


EXAMPLE PROBLEM (Cont.)Solution Five separate contact regions are defined with the

barrier as the master and the bumper as the slave. Each master region consists of two master nodes. Each slave region consists of 23 slave nodes.


3-D SLIDELINE CONTACT (Cont.) Bumper Input


3-D SLIDELINE CONTACT (Cont.) Bumper Input (Cont.)


3-D SLIDELINE CONTACT (Cont.)Concluding Remarks Penalty method is used for both contact and friction. Penalty values automatically calculated. Multiple slideline contacts can be specified. Relative motions outside the slideline plane are ignored during

the analysis The master and slave nodes must be in the slideline plane in the

initial geometry. Initial penetration is not allowed and is checked. Only Coulomb friction (slick and slip) with equal static and

kinematic coefficient is available for relative sliding. Only hard surface contact is available for opening and closing.


3-D SLIDELINE CONTACT (Cont.)Summary of Capabilities Applicable for 2-D plane strain/stress, axisymmetric,

and 3-D models. Can be used with all element types. Can be used with geometric and material

nonlinearities. With and without friction Nonlinear static solution (SOL 106) Nonlinear transient dynamic solution (SOL 129) More than 2-D surface contact and less than 3-D

surface contact


RESTARTS FOR NONLINEAR TRANSIENT ANALYSIS

Starting from a previous nonlinear transient analysis Restarts are allowed only from the end of subcases. Set parameters:

PARAM, LOOPID, I I = loop number on printoutPARAM, STIME, T0 T0 = starting value of time

T0 should be the last printed value for subcase I The database will be modified starting from LOOPID+1, T = T0

Starting from a previous nonlinear static analysis Set parameter

PARAM, SLOOPID, I I = loop number on SOL 106 run Initial transient load should be identical to static loads at restart state (SPC,

etc., may change).

Caution: The database will be completely overwritten. Transient analysis will destroy the static analysis database.


NONLINEAR TRANSIENT ANALYSIS USING SUPERELEMENTS

Why Use Superelement in Nonlinear Transient Analysis? All the benefits of using superelements in a linear

analysis can also be realized. See section on Superelement Analysis.

In many nonlinear problems, the nonlinearity may be localized. Substantial computational savings can be obtained by putting the

linear portion of the structure as superelements.

The portion of the structure that is nonlinear must be placed in the residual structure.

Restrict the iterations to the nonlinear regions of the structure.


NONLINEAR TRANSIENT ANALYSIS USING SUPERELEMENTS (Cont)

Linear assumptions – only the residual structure is allowed to be nonlinear (material or geometric).

Nonlinear superelement analysis can be restarted from linear analysis (databases from SOL 101, and SOL 109).

Restarts – No recalculations are required for upstream superelements if there is no change in superelements.

Load vectors for the upstream elements must be generated before the nonlinear solutions.

Case Control command SUPER is used to partition the proper subcase to a superelement.


NONLINEAR TRANSIENT ANALYSIS USING SUPERELEMENTS (Cont)

All the subcases should include the SUPER command (default, SUPER = 0) except when SUPER = ALL is specified above the subcases.

Component modes information is passed down to the residual structure using SPOINT and SEQSET1 - similar to conventional superelement analysis.

Since PARAM, AUTOSPC does not constrain singular DOF in the residual structure in nonlinear analysis, specify the exact number of generalized coordinates needed.

If feasible, assign load application points to the residual structure.


EXAMPLE PROBLEM USING SUPERELEMENT

Problem Description Analyze the same problem as before using

superelements. Put grid points 10, 10005, and 10010 in the residual

structure. Place the rest of the model in a superelement Note that for this small problem, you may not realize

much savings, if any. For large problems, the savings can be substantial.


EXAMPLE PROBLEM USING SUPERELEMENT (Cont)


SECTION 9CYCLIC SYMMETRY


TABLE OF CONTENTSSection PageCYCLIC SYMMETRY ANALYSIS 9-5TYPES OF SYMMETRY 9-6WHY CYCLIC SYMMETRY? 9-7ROTATIONAL SYMMETRY 9-8DIHEDRAL SYMMETRY 9-9CYCLIC SYMMETRY SOLUTION SEQUENCES 9-10INPUT SECTION REQUIREMENTS SPECIFIC TO CYCLIC SYMMETRY 9-11

SOLUTION SEQUENCEMSC.NASTRAN DOCUMENTATION FOR CYCLIC SYMMETRY 9-13CYCLIC SYMMETRY BULK DATA ENTRIES 9-14CYCLIC SYMMETRY RESTRICTIONS ON GRID POINTS 9-21CODED SUBCASE IDENTIFICATION NUMBERS 9-22SAMPLES SHOWING CYCLIC SYMMETRY FEATURES 9-23FULL MODEL 9-25CYCLIC MODEL 9-26EXECUTIVE AND CASE CONTROL FOR SAMP1 9-27EXECUTIVE AND CASE CONTROL FOR SAMP2 9-31EXECUTIVE AND CASE CONTROL FOR SAMP3 9-37


CYLCIC SYMMETRY ANALYSIS Types of Cyclic Symmetry

Reflective Rotational symmetry Dihedral symmetry (reflective plus rotational)

Solutions available Static's Normal modes Buckling Direct frequency response

I/O in either physical or cyclic components User models only primary segment Variable loads on each segment


TYPES OF SYMMETRYReflective Symmetry

Axisymmetric Symmetry

Cyclic Symmetry


WHY CYCLIC SYMMETRY? Traditional

MSC.NASTRAN Techniques

Cyclic Symmetry Option

Model size May be reduced by one-half for each plane of reflective symmetry

Much smaller size possible symmetry

Boundary conditions at planes of symmetry

Analyst must define explicitly

Automatic

Loads Analyst must decompose loads into symmetric and ant symmetric cases

Decomposition is automatic

Data recovery Analyst must combine solutions via SUBCOM or SYMCOM requests

Response for entire structure provided automatically

Rotational symmetry option

Not available Completely supported

Axisymmetry Requires special elements Uses ordinary elements


ROTATIONAL SYMMETRY

1. The user models one segment.2. Each segment has its own coordinate system that rotates with

the segment.3. Segment boundaries may be curved surfaces.


DIHEDRAL SYMMETRY

1. The user models one-half segment (an R segment). The L half segments are mirror images of the R half segments.

2. Each half segment has its own coordinates system that rotates with the segment. The L half segments use left-hand coordinate systems.

3. Segment boundaries must be planar.


CYCLIC SYMMETRY SOLUTION SEQUENCES

Available for static, modes, buckling, and frequency response analysis

Analysis Class SOL Name Static Analysis SOL 14

Modes SOL 15 Buckling SOL 16* Cyclic Static’s with Alternate Reduction Option

SOL 114*

Cyclic Normal Modes SOL 115* Cyclic Direct Frequency Response SOL 118 * Recommended


INPUT SECTION REQUIREMENTS SPECIFIC TO CYCLIC SYMMETRY

SOLUTION SEQUENCES Executive Control Section

No special requirements.

Case Control Section HARMONICS command required. NOUTPUT HOUTPUT DSYM Required if overall symmetry of the model is

to be used in DlH-type symmetry problems

Required if physical segment output or harmonic output is desired.


INPUT SECTION REQUIREMENTS SPECIFIC TO CYCLIC SYMMETRY SOLUTION SEQUENCES (Cont.)

Bulk Data Section CYJOlN Required. CYSYM Required. CYAX Required if model contains points on the axis of

rotation. CYSUP Required if free body supports are present in

the model. LOADCYH LOADCYN LOADCYT

Required if harmonic, physical or tabular form of loading is to be applied to the model


MSC.NASTRAN DOCUMENTATION FOR CYCLIC SYMMETRY

Reference Manual Section 15.6 Cyclic Symmetry

Quick Reference Guide Section 5

Bulk Data Entries

Quick Reference Guide Section 4

Case Control Commands

Reference Manual Section 13.2.2

i1,i2 Subcase ID for Plot

Linear Static Section 16 Cyclic Symmetry Users Guide

The following obsolete documents are still useful. Application Manual Section 3.2

Cyclic Symmetry

Application Manual Section 5 (6/82)

Secondary Superelements

Verification Manual Section 3

For SOLs 47 48 77 78


CYCLIC SYMMETRY BULK DATA ENTRIES

Input Data Entry: CYSYM Cyclic Symmetry ParametersDescription: Selects parameters for cyclic symmetry

problems.

Format and Example: 1 2 3 4 5 6 7 8 9 10

CYSYM NSEG STYPE CYSYM 6 ROT

Field Contents NSEG Number of segments (Integer > 0) STYPE Symmetry type (ROT or DIH or AXI).


CYCLIC SYMMETRY BULK DATA ENTRIES (Cont.)

Input Data Entry: CYJOIN Cyclic Symmetry Boundary PointsDescription: Defines the boundary points of a segment in cyclic

symmetry problems.Format and Example:

1 2 3 4 5 6 7 8 9 10 CYJOIN SIDE C G1 G2 G3 G4 G5 G6

G7 G8 G9 -etc.-

CYJOIN 1 7 9 16 THRU 33 64 72 THRU

Field Contents SIDE Side identification (Integer 1 or 2) C Type of coordinate system used on boundaries of dihedral or

axisymmetric problems. (Character: “T1”, “T2”, “T3”, “R”, “C”, “S”)

Gi Grid or scalar point identification numbers.



Input Data Entry: CYAX Grid Points on Axis for Cyclic Symmetry Problems

Description: Lists grid points that lie on the axis of symmetry in cyclic symmetry problems.


CYAX G1 G2 G3 G4 G5 G6 G7 G8 CYAX 27 152 THRU 160 192 11

Field Contents G1,G2,etc A list of grid points on the axis of symmetry.



Input Data Entry: CYSUP Fictitious Supports for Cyclic Symmetry Problems

Description: Defines fictitious supports for cyclic symmetry problems.


CYSUP GID C CYSUP 16 1245

Field Contents GID Grid point identification number C Component numbers



Input Data Entry: LOADCYN Physical Load Input for Cyclic Symmetry Problems

Description: Defines a physical static or dynamic load for use in cyclic symmetry analysis.


LOADCYN SID S SEGID SEGTYPE S1 L1 S2 L2 LOADCYN 10 1.0 1 R 0.5 17

Field Contents SID Load set identification number S Scale factor SEGID Segment ID SEGTYPE Segment type. Si Scale factor Li Load set ID numbers



Input Data Entry: LOADCYH Harmonic Load Input for Cyclic Symmetry Problems

Description: Defines the harmonic coefficients of a static or dynamic load for use in Cyclic Symmetry Analysis.


LOADCYH SID S HID HTYPE S1 L1 S2 L2 LOADCYH 10 1.0 7 C 0.5 15

Field Contents SID Load set identification number S Scale factor HID Harmonic identification number HTYPE Harmonic type (C, S, CSTAR, SSTAR, GRAV or RFORCE or

blank). Si Scale factor on Li Li Load set ID numbers



Input Data Entry: LOADCYT Table Load Input for AXl Type Cyclic Symmetry Problems

Description: Specifies loads as a function of azimuth angle by references to tables that define scale factors of loads versus azimuth angles.


LOADCYT SID TABLEID LOADSET METHOD TABLEID LOADSET METHOD

LOADCYH 10 19 27 21 26 1

Field Contents SID Load set identification number TABLEID Table ID for table load input for load set Li LOADSET Load set Li METHOD Method of interpolation to be used


CYCLIC SYMMETRY RESTRICTIONS ON GRID POINTS

ROT DIH AXI Number of Side Points

Same for Sides One and Two

No Restriction Same for Grid Sides One and Two

Side Grid Point Location

Advance by 360/NSEG

Lie Within Radial Plane

Advance by 360/NSEG

Side Displacement Coordinate System

Conforms with Next Segment

One Component Perpendicular

One Component Perpendicular

Axis Displacement Coordinate System

T3 Positive Axis Symmetry

T3 Positive Axis Symmetry T1 inPlane

T3 Positive Axis Symmetry

Interior Grid Point

No Restriction No Restriction Not Allowed


CODED SUBCASE IDENTIFICATION NUMBERS

Normally not required. Used for:

For physical (i.e., NOUTPUT) requests: SSSSNNNT

For harmonic (i.e., HOUTPUT) requests: SSSSHHHT

SORT2 output,Plotting Punched Temperature DataRandom Analysis

[ ]

+++

DIH(L) for 2DIH(R) for 1

AXI ROT, for 0]ID[SEGMENT*10SUBCASEID*1000

++−

*S for 4*C for3

S for 2C for 1

C]ID[HARMONI*10]ID[SUBCASE*1000


SAMPLES SHOWING CYCLIC SYMMETRY FEATURES

Sample Problem Using Feature FEATURE SAMP1 SAMP2 SAMP3 SAMP4 SAMP5 SAMP6

Solution Sequence Static Heat X - - - - - Static Analysis - X - - - X

Vibration Modes - - X - - - Buckling - - - X - - Freq. Response - - - - X - Superelement - - - - - X Cyclic Components Harmonic X X X X X X Dsym - - - X - -

NOUTPUT X X - - X X HOUTPUT - - X X - -


SAMPLES SHOWING CYCLIC SYMMETRY FEATURES (Cont.)

Sample Problem Using Feature FEATURE SAMP1 SAMP2 SAMP3 SAMP4 SAMP5 SAMP6

Symmetry Type ROT X X - - - - AXI - - X - X - DIH - - - X - X

Grid Type CYJOIN X X X X X X CYAX X X X X X X CYSUP - X - - - -

Load Selection LOADCYH - X - X - - LOADCYN X X - - - X LOADCYT - - - - X - NOTE: Only samples 1,2, and 3 shown in this seminar – the others are in the cyclic symmetry seminar.


FULL MODEL

MSC.NASTRAN Cyclic Symmetry Demonstration Full Undeformed Shape


CYCLIC MODEL

MSC.NASTRAN Cyclic Symmetry Demonstration Cyclic Undeformed Shape


EXECUTIVE AND CASE CONTROL FOR SAMP1

ID CYC SAMP1APP HEATTIME 5 $ C.P.U. MINUTESSOL 114 $ CYCLIC STATIC ANALYSISCENDTITLE = MSC.NASTRAN CYCLIC SYMMETRY DEMONSTRATION SAMPLE 1ECHO = UNSORTHARMONIC = ALLSPC = 210NOUTPUT = ALLTHERM(PRINT,PUNCH) = ALLSUBCASE 10LABEL = THERMAL LOAD ONELOAD = 311SUBCASE 20LABEL = THERMAL LOAD TWOLOAD = 321BEGIN BULK


BULK DATA FOR SAMP1


BULK DATA FOR SAMP1 (Cont.)


SELECTED OUTPUT FROM SAMP1



ID CYC SAMP2TIME 5 $ C.P.U. MINUTESSOL 114 $ CYCLIC STATIC ANALYSISCENDTITLE = MSC.NASTRAN CYCLIC SYMMETRY DEMONSTRATION SAMPLE 2ECHO = UNSORTHARMONIC = ALLSET 1 = 3NOUTPUT = 1DISPLACE = ALLELFORCE = ALLSUBCASE 30LABEL = POINT LOADLOAD = 331SUBCASE 40LABEL = GRAVITY LOADLOAD = 341SUBCASE 50LABEL = CENTRIFUGAL LOADLOAD = 351SUBCASE 60LABEL = THERMAL LOAD ONETEMP(LOAD) = 361SUBCASE 70LABEL = THERMAL LOAD TWOTEMP(LOAD) = 371BEGIN BULK


BULK DATA FOR SAMP2


SELECTED OUTPUT FROM SAMP2 (Cont.)



ID CYC SAMP3TIME 15 $ C.P.U. MINUTESSOL 115 $ CYCLIC VIBRATION ANALYSISCENDTITLE = MSC.NASTRAN CYCLIC SYMMETRY DEMONSTRATION SAMPLE 3ECHO = UNSORTSET 3 = 0 1 2HARMONIC = 3METHOD = 400HOUTPUT = ALLDISPLACE = ALLBEGIN BULK


BULK DATA FOR SAMP3


SECTION 10

COMPOSITES



OVERVIEW 10-5

CLASSICAL LAMINATION THEORY (CLT) 10-6

COMPOSITE MATERIALS 10-7

CLASSIFICATION OF COMPOSITE MATERIALS 10-8

UNIDIRECTIONAL FILAMENTARY LAMINA 10-9

LAMINATE CONSTRUCTION 10-10

LAMINA ARRANGEMENT IN A 0/90/0 LAMINATE 10-11

COMPOSITES 10-12

ROTATION TO MATERIAL COORDINATE SYSTEM 10-13

CALCULATION OF COMPOSITE ELEMENT PROPERTIES 10-14

SYMMETRY IN COMPOSITES 10-19

FINITE ELEMENT ANALYSIS OF COMPOSITE MATERIAL STRUCTURES 10-20

MSC.NASTRAN INPUT FOR COMPOSITE ANALYSIS 10-21

TYPICAL ELEMENT – PCOMP – MAT RELATIONSHIP 10-22

SAMPLE LAMINATE DEFINITION 10-23

SPECIFICATION OF REFERENCE DIRECTION 10-24

PROPERTY AND MATERIAL INPUT 10-25


TABLE OF CONTENTSSection PagePLATE ELEMENTS FOR COMPOSITE MATERIAL ANALYSIS 10-30ELEMENT INPUT 10-31OUTPUT 10-35PLY STRESS AND STRAIN OUTPUT 10-36ELEMENT FORCE AND STRAIN OUTPUT 10-37COMPOSITE FAILURE INDICES 10-38ALLOWABLE STRESSES 10-39FAILURE THEORIES FOR COMPOSITE MATERIALS 10-40HILL’S THEORY 10-41HOFFMAN’S THEORY 10-42HOFFMAN FAILURE THEORY 10-43TENSOR POLYNOMIAL THEORY (TSAI-WU THEORY) 10-44TSAI-WU THEORY 10-45INTERLAMINAR SHEAR FAILURE INDEX 10-46CONCLUSION 10-47REFERENCES 10-48


OVERVIEW Classical lamination theory is used. Family of plate elements, QUAD4, QUAD8, TRIA3,

and TRIA6 available for modeling composites. User input is simple. Stress output for user-requested plies is available. Can be used in optimization (SOL 200) Failure indices for elements can be requested.


CLASSICAL LAMINATION THEORY (CLT) By this theory, equations for laminate are derived from those of

laminas. Each individual lamina is in plane stress. The laminate is presumed to consist of perfectly bonded lamina.

allowing no relative slip between layers. A distinct feature of MSC.NASTRAN plate elements is the

provision for including transverse shear stiffness:

The effective transverse shear stiffness matrix (G3) for composite plate elements is evaluated on the assumption of the applicability of elementary beam theory type equations for plates. This introduces an approximation that the effects of twisting moments are negligible. In the vast majority of cases such an approximation is satisfactory.

{ } { }xz

yz

x

y

qq3

VV ]G[=


COMPOSITE MATERIALS Composite material is defined as one where two or

more materials are combined on a macroscopic scale. This is done to obtain the best qualities of the

constituent materials (and in some cases, additional qualities that the constituents do not have).

The following properties are improved and are of major interest:

StrengthStiffnessLower weightTailored properties


CLASSIFICATION OF COMPOSITE MATERIALS

Lamina is a group of unidirectional fibers (or sometimes woven fibers) arranged to form a flat or curved load resisting member by the use of a matrix.

The principal material axes are parallel and perpendicular to the fiber directions.

The principal directions are also referred to as: fiber direction, longitudinal direction or 1-direction matrix direction, transverse direction or 2-direction


UNIDIRECTIONAL FILAMENTARY LAMINA


LAMINATE CONSTRUCTION A laminate is a stack of lamina arranged with the

principal directions of each lamina at different orientations so as to obtain the desired strength and stiffness properties.

The various layers of a laminate are bonded together by the same matrix material that is used in the lamina.

Curing bonds the lamina together, usually in the presence of heat and pressure.


LAMINA ARRANGEMENT IN A 0/90/0 LAMINATE


COMPOSITES Composites generally specify lamina properties as a 2-D

orthotropic material The stress-strain relations in principal lamina material directions

are

There are four independent constants in the relationship,E1, E2 , ν12 (or ν21), G12

In many references, this is also written as

γεε

νν−νν−ν

νν−ν

νν−

=

τσσ

12

2

1

12

2112

2

2112

122

2112

211

2112

1

12

2

1

00

011

011

G

EE

EE

εεε

=

σσσ

6

2

1

66

2212

1211

6

2

1

0000

QQQQQ

}]{Q[}{.,e.i εσ =


ROTATION TO MATERIAL COORDINATE SYSTEM

To form laminate properties, first the lamina properties are rotated to the element material coordinate system [xm Ym]

Using the equation:

where = lamina properties rotated to material coordinate system.[u] = the stress-transformation matrix for transforming stresses from

the 1-2 system to the x-y system that is given by

]U][Q[]U[]Q[ T=

Q

−−−=

θθθθθθθθθθ

θθθθ

22

22

22

sincoscossin2cossin2cossincossin

cossinsincos]U[


CALCULATION OF COMPOSITEELEMENT PROPERTIES

The for the stacked lamina are then integrated through the thickness, to relate the curvatures and mid-surface strains with Forces and Moments:

where

]Q[

=

χε 0

DBBA

MF

=

662616

262212

161211

AAAAAAAAA

]A[

=

662616

262212

161211

BBBBBBBBB

]B[

=

662616

262212

161211

DDDDDDDDD

]D[


CALCULATION OF COMPOSITEELEMENT PROPERTIES (Cont.)

and where Aij – Extensional Stiffness

Bij – Coupling Stiffness

Dij – Bending Stiffness

)zz()Q(A 1kkkijN1ij −−= ∑

kkijN1ij t)Q(A ∑=

)zz()Q(21B 2

1k2kkij

N1ij −−= ∑

kkkijN1ij zt)Q(B ∑=

)zz()Q(31D 3

1k3

kijN1ij k −−= ∑

)12

()(3

21

ktztQD kkkij

Nij += ∑



MSC.NASTRAN uses the G1, G2, G3, and G4 matrices to define element properties. The relation between forces and strains used for MSC.NASTRAN plate elements is

where membrane forces per unit length

γχ

ε

=

M

sGTIGGT

GTTG

QMF

3

242

42

1

0000

{ } ,FFF

F

xy

y

x

=

{ } ,MMM

M

xy

y

x

=

{ } ,QQ

Qy

x

=

{ } ,

xy

y

x

M

=

εεε

ε

{ } ,

xy

y

x

=

χχχ

χ { } ,y

x

=γγ

γ

bending moments per unit length

transverse shear force per unit length

membrane strains in mean plane

curvatures; transverse shear strains



Note that

where = Laminate thickness

= I = Bending inertia

ijij A1TG =TA

1G ijij =

2ij

ij TB

4G =

−=

12TD

2G 3ij

ij

ij2 4GT

ijij

3

D2G12T

−=

12T 3

T



Note that MSC.NASTRAN’S Mx,My, and Mxy terms are reversed from classical lamination theory.

NASTRAN Forces In Plate Elements Classical Lamination Theory Forces In Plate Elements


SYMMETRY IN COMPOSITES [B] is zero for symmetric laminator symmetry that

occurs if for each lamina above the midplane, there is an identical ply (in properties and orientation) located at the same distance below the midplane.

In MSC.NASTRAN, [G4] is similar to [B]. Examples of symmetric laminates

+45/0/ +45 +45/ +45/ -45/ -45/ -45/ -45/ +45/ +45/ +45/ -45/ +45/ -45/ -45/ +45/ -45/ +45/


FINITE ELEMENT ANALYSIS OF COMPOSITE MATERIAL STRUCTURES

2-D analysis using lamination theory is found to give good results where the laminate is thin relative to its length.

Otherwise, a full 3-D anisotropic material analysis is desirable.

3-D analysis is also needed near free edges. 3-D analysis uses HEXA elements to represent either

single lamina or sets of lamina


MSC.NASTRAN INPUT FOR COMPOSITE ANALYSIS

Executive Control – No changes required. Case Control – No changes required. Bulk Data

Plate elements QUAD4, QUAD8, TRIA3, TRIA6 referring toPCOMP property entry.

PCOMP may refer to MAT1, MAT2, or MAT8 material propertyentries. However, currently QUAD8 and TRIA6 only works withisotropic materials (MAT1) only.


TYPICAL ELEMENT –PCOMP – MAT RELATIONSHIP

Output

If

ECHO = SORT

CQUAD

PCOMP

MAT1 MAT2 MAT8

EQUIV PSHELL

MID1

MAT2

G1

MID2

MAT2

G2

MID3

MAT2

G3

MID4

MAT2

G4


SAMPLE LAMINATE DEFINITION

2

ply material

CQUAD4, 101, 1, A, B, C, D, 5PCOMP, 1, , ,5000., STRN, , , ,++, 2,.003, 0.,YES, 1,.005, 45.,YES,++, 1,.005, 0.,YES, 1,.005,-45.,YES,++, 3,.007,90.,YESMAT8, 1, 1.+7, 1.+7, .05, 1.+6, 1.+6, 1.+6, ,++, , , , .007, .006, .007, .006, .001MAT8, 2, 2.+7, 2.+6, .35, 1.+6, 1.+6, 1.+6, ,++, , , , .007, .006, .007, .006, .001MAT8, 3, 8.+6, 8.+6, .05, 7.+5, 7.+5, 1.+6, ,++, , , , .006, .005, .006, .005, .001CORD2R, 5, ,0., 0., 0., 0., -1., 0.,+CORD+CORD, 0., 0., 1.


SPECIFICATION OF REFERENCE DIRECTION

Specify angle theta in element connection entries Provision to specify coordinate system ID in theta

field The X axis is projected onto the element to define the direction of

the X axis of the element material coordinate system. The Z axis of the material coordinate system is defined by the

element coordinate system Z axis (in other words, by the grid order in the element).


PROPERTY AND MATERIAL INPUTInput Data Entry PCOMP: Layered Composite Element PropertyDescription: Defines the properties of an n-ply composite material

laminate.$1------$2------$3------$4------$5------$6------$7------$8------$9------$10-----

PCOMP PID Z0 NSM SB FT TREF GE LAM +AM1

+AM1 MID1 T1 THETA1 SOUT1 MID2 T2 THETA2 SOUT2 +AM2

+AM2 MID3 T3 THETA3 SOUT3 ..ETC

1/2 line per ply definition

• PID Property ID

• Z0 Offset from grid points to first ply (default -1/2 element thickness)

• NSM Non-Structural Mass

• SB Allowable inter-laminar shear

• FT Failure Theory, Hill, Hoffman, Tsai-Wu, max strain

• TREF Reference temperature. (Overrules MATi TREF)

• GE Element damping coefficient

• LAM SYM indicates only 1/2 plies defined and symmetry is used

• MIDi, Ti, THETAi Material, thickness and angle of ply

• SOUTi YES gives stress output at this ply


PROPERTY AND MATERIAL INPUT(Cont.)

Example:

1 2 3 4 5 6 7 8 9 10

PCOMP PID Zo NSM Sb F.T. TREF GE LAM

PCOMP 181 -0.224 7.45 10000.0 HOFF

MID1 T1 THETA1 SOUT1 MID2 THETA2 ? 2 SOUT2

171 0.056 0. YES 45.

MID3 T3 THETA3 SOUT3 MID4 T4 THETA4 SOUT4

-45. 90.


PROPERTY AND MATERIAL INPUT (Cont.)Input Data Entry MAT8 Material Property Definition, Form 8

Description: Defines the material property for a 2-D orthotropic material

Format and Example:1 2 3 4 5 6 7 8 9 10

MAT8 MID E1 E2 V12 G12 G1,z G2,z RHO

MAT8 171 30.+6 1.+6 0.3 2.+6 3.+6 1.5+6 0.056

A1 A2 TREF Xt Xc Yt Yc S

28.-6 1.5-6 155 1.+4 1.5+4 2.+2 8.+2 1.+3

GE F12

1.-4


PROPERTY AND MATERIAL INPUT (Cont.)Input Data Entry PSHELL Shell Element PropertyDescription: Defines the membrane, bending,

transverse shear, and coupling properties of thin shell elements


PSHELL PID MID1 T MID2 12l/T3 MID3 TS/T NSM

PSHELL 203 204 1.90 205 1.2 206 0.8 6.32

Z1 Z2 MID4

+.95 -.95


PROPERTY AND MATERIAL INPUT (Cont.)Input Data Entry MAT2 Material Property Definition, Form 2

Description: Defines the material property for a 2-D orthotropic material


MAT2 MID G11 G12 G13 G22 G23 G33 RHO

MAT2 13 6.2.+3 6.2.+3 5.1+3 0.056 ABC

a A1 A2 A12 T0 GE ST SC SS

+BC 6.5-6 6.5-6 -500.0 0.002 20.+5 DEF

MCSID

+DE 1003


PLATE ELEMENTS FOR COMPOSITE MATERIAL ANALYSIS

QUAD4

TRIA3

QUAD8LINEAR ANALYSIS with isotropic materials only

TRIA6

LINEAR AND GEOMETRIC NONLINEAR ANALYSIS


ELEMENT INPUTInput Data Entry CQUAD4 Quadrilateral Element Connection

Description: Defines a quadrilateral plate element (QUAD4) of the structural model. This is an isoparametric membrane-bending element.


CQUAD4 EID PID G1 G2 G3 G4 θ ZOFFS

CQUAD4 111 203 31 74 75 32 2.6 0.3 ABC

T1 T2 T3 T4

+BC 1.77 2.04 2.09 1.80


ELEMENT INPUT (Cont.)Input Data Entry CTRIA3 Triangular Element Connection

Description: Defines a triangular plate element (TRIA3) of the structural model. This is an isoparametric membrane-bending element.

Format and Example:

T1 T2 T3

+BC 1.77 2.04 2.09

1 2 3 4 5 6 7 8 9 10

CTRIA3 EID PID G1 G2 G3 θ ZOFFS

+BC 1.77 2.04 2.09


ELEMENT INPUT (Cont.)Input Data Entry CQUAD8 Quadrilateral Element Connection

Description: Defines a curved quadrilateral shell element (QUAD8) with 8 grid points. Currently CQUAD8 works with

isotropic materials only.Format and Example:

G7 G8 T1 T2 T3 T4 θ ZOFFS

+BC 53 72 0.125 0.025 0.030 .025 30. .03

1 2 3 4 5 6 7 8 9 10

CQUAD8 EID PID G1 G2 G3 G4 G5 G6

CQUAD8 207 3 31 33 73 71 32 51 ABC


ELEMENT INPUT (Cont.)Input Data Entry CTRIA6 Triangular Element Connection

Description: Defines a curved triangular shell element (TRIA6) with 6 grid points. Currently CTRIA6 works with isotropic materials only.

Format and Example: θ ZOFFS T1 T2 T3

+BC 45. .03 .020 .025 .025

1 2 3 4 5 6 7 8 9 10

CTRIA6 EID PID G1 G2 G3 G4 G5 G6

CTRIA6 302 3 31 33 71 32 51 52 ABC


OUTPUT Stresses in individual lamina including approximate

interlaminar shear stresses Failure Index table Element strains Element forces


PLY STRESS AND STRAIN OUTPUT To obtain ply stresses and strains, use the following

Case Control Commands respectively: STRESS = STRAIN =

To obtain Failure Index Table, allowables (Xt, Xc , Yt ,Yc , S) must be supplied in MAT8 Bulk Data, and Sb can be supplied in PCOMP Bulk Data entry.

Interlaminar shear stresses are output between the lamina.

Individual lamina stresses can be sorted (use NUMOUT1 and BIGER1 parameters).

Failure Index table can be sorted (use NUMOUT2 and BIGER2 parameters).

Available in SOL101 and 103 without alters.


ELEMENT FORCE AND STRAIN OUTPUT

Element force output and strain output available. Calculation of the element property data (e.g A, B, D

matrices) from user supplied data for lamina is available as printed or punched output.


COMPOSITE FAILURE INDICES

Composite failures are checked at the lamina level. Failure index of a lamina checks whether the state of

stress can cause a failure. If the failure index of the element is less than or equal

to 1.0, stresses in all laminas are within or on the respective failure envelopes.

If the failure index is greater than 1.0 in at least one lamina, then the element is assumed to fail.


ALLOWABLE STRESSESXt = Allowable stress or strain in tension in longitudinal direction (or

1-direction or fibre direction)Xc = Allowable stress or strain in compression in longitudinal

direction (positive sign will be used for Xc)Yt = Allowable stress or strain in tension in transverse direction (or

2-direction or matrix direction)Yc = Allowable stress or strain in compression in transverse

direction (positive sign will be used for Yc) S = Allowable stress in shear (positive or negative shear has the

same allowable)Sb= Allowable shear stress of bonding material (allowable inter-

laminar shear stress)

Xt , Xc , Yt , Yc , and S are supplied in MAT8 Bulk Data entry. Sb can be supplied in PCOMP Bulk Data entry.


FAILURE THEORIES FOR COMPOSITE MATERIALS

HILL’S THEORY HOFFMAN’S THEORY TSAI-WU THEORY INTERLAMINAR SHEAR


HILL’S THEORY

where X = Xt if σ1 is tensileX = Xc if σ1 is compressiveY = Yt if σ2 is tensileY = Yc if σ2 is compressive

For the product term, X = Xt if σ1 and σ2 are of the same sign; X = Xc otherwise

Basically, the equation represents a failure envelope in the stress space.

If the state of stress in the orthotropic lamina (σ1 , σ2 , σ12 ) is such that the stress point is within or on the envelope, the lamina is said to be “safe.” If the point is outside, the lamina is said to have “failed.”

1SXYX 2

212

221

2

22

2

21 =+−+

σσσσσ


HOFFMAN’S THEORY

Hill’s theory does not take into account the differing tensile and compressive strengths in the fiber and matrix directions.

This equation can be thought of as having been derived from Hill’s theory by adding linear terms to account for differing strengths in tension and compression.

111112

21221

22

21

21 =σ

+σσ

−σ

+σ

+σ

−+σ

−

SXXYYXXYYXX ctctctctct


HOFFMAN’S FAILURE THEORY

Is an ellipsoid in σ1, σ2 , τ12 space:


TENSOR POLYNOMIAL THEORY(TSAI-WU THEORY)

The theory of strength for anisotropic materials, proposed by Tsai and Wu, specialized to the case of an orthotropic lamina in a general state of plane stress is

where

and F12 is to be determined experimentally.

1FFF2FFF 2666

22222112

21112211 =+++++ σσσσσσσ

ct1 X

1X1F −=

ct2 Y

1Y1F −=

ct11 XX

1F =

ct22 YY

1F =

266 S1F =


TSAI-WU THEORY The magnitude of F12 is constrained by the following inequality

that is called the “stability criterion” associated with the theory

The need to satisfy the stability criterion together with the requirement that F12 be determined experimentally from a combined stress-state poses difficulties.

In the absence of experimental value, Tsai recommends using:

Geometrically, this condition ensures that the strength envelope is closed. That is, the shape of the envelope must be ellipsoidal rather than parabolic or hyperbolic ensuring that the material has finite strength in all directions.

221112 FF21F −=

02122211 >− FFF


INTERLAMINAR SHEAR FAILURE INDEX

where = Shear stress between the i lamina and the i+1 lamina in

the X direction of the element material coordinate system.= Shear stress between the i lamina and the i+1 lamina in the

Y direction of the element material and coordinate system. = Allowable interlaminar shear stress that is input on the

PCOMP entry.

b

zi1

Sτ

b

zi2

Sτ

zi1τ

zi2τ

bS

,


CONCLUSION

MSC.NASTRAN layered composite analysis capability Is user friendly Is easy to use Has simple input Allows stresses in individual plies to be sorted and output Provides failure index for individual plies


REFERENCES MSC.NASTRAN Reference Manual, Sections 6.5 and

15.2.

“Mechanics of Composite Materials,” R.M. Jones; Scripta Book Co., Washington D.C., 1975.

“Mechanics of Composite Materials,” R.M. Christensen; John Wiley & Sons, New York, 1979.

“Primer on Composite Materials Analysis,” J.E. Ashton, J.C. Halph, P.H. Petit; Technomic Publishing Co., Inc., Stamford, Connecticut, 1969.


SECTION 11

OPTIMIZATION



WHAT IS “DESIGN OPTIMIZATION”? 11-7

BASIC FEATURE IMPLEMENTED IN MSC.NASTRAN 11-9

STRENGTHES OF MSC.NASTRAN STRUCTURE OPTIMIATION 11-10

EXTENT OF MSC.NASTRAN OPTIMIZATION CAPABLITIES 11-11

CONCEPT OF THE DESIGN MODEL 11-12

DESIGN MODEL DEFINITION PROCEDURE 11-14

CHALLENGES IN DESIGN MODELING 11-15

SOME STRUCTURAL RESPONSES 11-16

RESPONSES FOR COMPOSITE MATERIALS 11-17

RESPONSE QUANTITIES – CONSTRAINTS/OBJECTIVE FUNCTION 11-18

OBJECTIVE AND CONSTRAINTS 11-19

DESIGN MODELING INPUT DATA 11-20

SAMPLE PROBLEM 11-22

DESVAR 11-24

DVPREL1 11-25



DRESP1 11-28

DESOBJ 11-39

DCONSTR 11-40

DESGLB 11-44

ANALYSIS 11-45

DESSUB 11-46

SAMPLE OPT1 – CONSTANT CROSS -SECTION CANTILEVER BEAM 11-47

SAMPLE OPT2 – VARIABLE CROSS - SECTION CANTILEVER BEAM 11-62

WORKSHOPS 11-76

AVANCED FEATURES OF OPTIMIZATION IN MSC.NASTRAN 11-77

TYPE-2 STRUCTURAL RESPONSES 11-78

TYPE-2 RESPONSE FORMULATION 11-79

RESTRICTIONS IN FORMING SYNTHETIC RESPONSES 11-80

DEQATIN 11-81

DTABLE 11-86

DRESP2 11-87



DRESP2 BULK DATA ENTRY 11-94

DOPTPRM 11-95

DYNAMIC RESPONSE OPTIMIZATION 11-101

SAMPLE PROBLEM XXX – MINIMIZATION OF DRIVER RESPONSETO A ROTATING IMBALANCE FOR THE CAR MODEL 11-103

INPUT FOR OPTIMIZATION 11-104

SHAPE OPTIMIZATION 11-112

HOW DO I CREATE SHAPE DESIGN VARIABLES? 11-115

MANUAL GRID VARIATION 11-116

INPUT FILE STRUCTURE: MANUAL GRID VARIATION 11-119

INPUT FOR SHAPE OPTIMIZATION 11-122

OPTIMIZATION INPUT 11-127

SHAPE-OPTIMIZATION RESULTS 11-129

OPTMIZATION SUMMARY 11-130


WHAT IS “DESIGN OPTIMIZATION”?Automated modifications of the analysis model parameters to achieve a desired objective while satisfying specified requirements

WHAT ARE THE POSSIBLE APPLICATIONS?

Structural design improvements (optimization)

Generation of feasible designs from infeasible designs

Model matching to produce similar structural responses

System parameter identification

Configuration evaluations

Others—(depends on designer’s creativity)


BASIC FEATURE IMPLEMENTED IN MSC.NASTRAN

Easy access to design synthesis capabilities Concept of design model

Flexibility for design model representation User-supplied equation interpretation capability

Efficient solution for problems of “any” size Number of finite element analysis as the measure of efficiency


STRENGTHES OF MSC.NASTRAN STRUCTURE OPTIMIATION

Efficient performance for small- to large-scale problems

Reliable convergence characteristics

Flexible user interface and user-defined equations

Full implementation of approximation concepts

Continuous enhancements

Results dependent on the proven reliability of MSC.NASTRAN analysis

Commercial level support as a part of MSC.NASTRAN

Access to the familiar analysis tools in MSC.NASTRAN


EXTENT OF MSC.NASTRAN OPTIMIZATION CAPABLITIES

Design Sensitivity and Optimization

Each SUBCASE may reference a different solution type by using the ANALYSIS command For example, for one static SUBCASE and one normal modes SUBCASE in

SOL 200:SUBCASE 1ANALYSIS = STATICLOAD = 100DISP = allSUBCASE 2ANALYSIS = MODESMETHOD = 1ESE = ALL

SOL Responses Design ModelStatics Sizing PropertiesNormal Modes (Superelements)BucklingDirect Frequency* ShapeModal Frequency*Modal Transient*Static Aeroelastic

200(Superelements with Manual Grid Variation)

* Including Acoustics


CONCEPT OF THE DESIGN MODEL

Given an initial design

Find X, that will minimize a scalar function F(X) while satisfying thefollowing:

( )N21 x,...,x,xX =

N,...,2,1i xxxK1,2,...,k 0)X(HJ1,2,...,j 0)X(G

)u(ii

(I)i

k

j

=≤≤

==

=≤

Hardwareor

Drawings

Finite ElementAnalysis Model

PredictedStructuralResponses

Design Model

1. Design Variables2. Objective Function3. Constraints


CONCEPT OF THE DESIGN MODEL (Cont)

Design constraints: Deformations within limits Local web buckling criteria

Structural responses Displacement at grids Stress computed at grids

Design Model

Design Variables

Analysis Model Parameters

Analysis Model

CBAR 101 21 … …

… … … … …

PBAR 21 6 2.36 …

,...J,I,I,I,A 1221

{ }fw t,t,b,hFor example:

( )( ) ( ) 1

3f

w3

wff

It2h12

tb12

hbAtt2htb2

=∗−∗−

−∗

=∗−+∗∗


DESIGN MODEL DEFINITION PROCEDURE

1. Select design variables. – DESVAR entry

2. Describe relations between design variables and analysis model parameters. – DVPREL1 and DVPREL2 entries

3. Define response to be used as an objective function. – DRESP1 and DRESP2 entries

4. Define responses which are to be constrained. – DRESP1 and DRESP2 entries

5. Specify bounds on constrained responses. – DCONSTR entry

6. Select design process control parameters, if necessary (I.e., DSCREEN, DOPTPRM)


CHALLENGES IN DESIGN MODELING

How to describe the relations between the design variables and the analysis model parameters?

How to describe the relations between the structural responses computed by MSC.NASTRAN and the design objective and constraints?

How to handle large-scale problems effectively? Several hundred design variables Thousands of constraints

The complexities required for the design model description are far more demanding than for the analysis model description. Therefore, no fixed data structure will be sufficiently general.


SOME STRUCTURAL RESPONSES

Displacement Specific displacement degree of freedom, identified by the grid ID and

component ID

The stress/strain/internal force The stress/strain/force component (described in Appendix A of the

MSC.NASTRAN Quick Reference Guide) is identified by the component ID, and either the element ID or the property ID.

If identified by a property ID, all elements that make reference to that property are covered.

Buckling load factor Always associated with the first static subcase, designated by the buckling

mode ID (lowest mode = 1)

Natural vibration eigenvalue or frequency Eigenvalue or frequency, identified by the mode ID (lowest mode = 1)

Weight, Volume


RESPONSES FOR COMPOSITE MATERIALS

CSTRESS and CSTRAIN These responses are the same as STRESSES or STRAIN, but are

imposed on specific ply laminas. The ATTB field of the DRESP1 entry is used to identify the specific laminas.

CFAILURE One of the four failure theories selected by the PCOMP entry is

applied. The response is normalized so that if the response value is less than 1.0, the laminas do not fail.


RESPONSE QUANTITIES FORCONSTRAINTS/OBJECTIVE FUNCTION

Weight Volume Eigenvalues or natural frequencies Buckling load factors Displacements, Velocities, Accelerations Stresses Strains Forces Failure indices for composites Lamina stresses for composites Lamina strains for composites Trim Stability Derivatives Flutter Analytic equations supplied by the user


OBJECTIVE AND CONSTRAINTS

Constraints Select the responses. Provide the lower and upper bounds using DCONSTR entries.

(Note: If possible, do not use 0.0 for the lower or upper bounds.) Select constraint sets in Case Control using DESGLB for global

constraints and/or DESSUB for subcase-dependent constraints.

ObjectiveUsing the DESOBJ Case Control Command: Select a response. Choose whether this response is to be minimized or maximized

(Default: Minimize).


DESIGN MODELING INPUT DATA Design variables

DESVAR Design variable definition* DLINK Definition of dependent design variable

Relation between design variables and analysis model parameters DVPREL1 Linear relations* DVPREL2 Nonlinear relations

Definition of structural responses DRESP1 Responses computed directly by analysis* DRESP2 Synthesized responses for design problems

_____________________________________________________*Presented in this section.


DESIGN MODELING INPUT DATA (Cont.) Definition of objective and constraint functions

DESOBJ Case Control Command Objective function definition* DCONSTR Constraint functions* DCONADD Constraint Set Combinations DESSUB Case Control Command Selection of Subcase-Dependent

Constraints* DESGLB Case Control Command Selection of Global Constraints*

Optimization control parameters and constants DSCREEN Measures of constraint screening DOPTPRM Optimization process control DTABLE Constants

User equation input DEQATN User-defined equation

_____________________________________________________*Presented in this section.


SAMPLE PROBLEM For an example, let us look at a cantilever beam with a tip load

Properties:L=100. P=10.Initial dimensions:

2 x 2 square BARE = 10,000,000.ν = .3ρ = .1WTMASS = 0.002588

We wish to minimize the weight, while keeping the tip deflection less than .01 (disp for initial design = .0333) and stress less than 1000.Let us use a PBARL to define the cross sectionLet us try two different approaches: Constant cross-section we will allow the depth of the section to vary (In order to do this properly, we

will use a different PBARL for each element – not recommended for large models, but it makes for a nice sample problem).


SAMPLE PROBLEM (Cont.) Basic Model with one PBARL – file cantbeam.dat

Let us now look at the entries need for the optimizer

PARAM GRDPNT 0PARAM WTMASS .002588PARAM POST 0PARAM AUTOSPC YES$GRID 1 0 0.0 0.0 0.0 0GRID 2 0 12.5 0.0 0.0 0GRID 3 0 25. 0.0 0.0 0GRID 4 0 37.5 0.0 0.0 0GRID 5 0 50. 0.0 0.0 0GRID 6 0 62.5 0.0 0.0 0GRID 7 0 75. 0.0 0.0 0GRID 8 0 87.5 0.0 0.0 0GRID 9 0 100. 0.0 0.0 0$CBAR 1 1 1 2 0. 1.CBAR 2 1 2 3 0. 1.CBAR 3 1 3 4 0. 1.CBAR 4 1 4 5 0. 1.CBAR 5 1 5 6 0. 1.CBAR 6 1 6 7 0. 1.CBAR 7 1 7 8 0. 1.CBAR 8 1 8 9 0. 1.$SPC 1 1 123456 0.0$PBARL,1,1,,BAR,2.,2.$MAT1 1 1.+7 .3 .1$


DESVARDefines a design variable for design optimizationFormat:

Example:

Field ContentsID Unique design variable identification number. (Integer > 0).LABEL User-supplied name for printing purposes. (Character).XINIT Initial value. (Real, XLB < XINIT < XUB).XLB Lower bound. (Real, default = -1 .0E+20).XUB Upper bound. (Real, default = +1 .0E+20).DELXV Fractional change allowed for the design variable during

approximate optimization. (Real > 0.0, for default see Remark 2).Remarks:

1. DELXV can be used to control the change in the design variable during one optimization cycle.

2. If DELXV is blank, the default is taken from the specification of the DELX parameter on the DOPTPRM entry. If DELX is not specified then the default is 1.0.

1 2 3 4 5 6 7 8 9 10

DESVAR ID LABEL XINIT XLB XUB DELXV

DESVAR 2 BARA1 35.0 10. 100. 0.2


DVPREL1Defines the relation between an analysis model property and design variables.Format:

Example:

Field ContentsID Unique identification number. (Integer > 0).TYPE Name of a property entry, such as “PBAR”, “PBEAM”, etc. (Character).PID Property entry identification number. (Integer > 0).FID Field position of the property entry, or word position in the element property

table of the analysis model. (Integer ≠ 0).PMIN Minimum value allowed for this property. If FID references a stress

recovery location, then the default value for PMIN is –1.0+35. PMIN must be explicitly set to a negative number for properties that may be less than zero (for example, field Z0 on the PCOMP entry). (Real; default = 0.001).

PMAX Maximum value allowed for this property. (Real; Default = 1.0E20).CO Constant term of relation. (Real; Default = 0.0)DVIDi DESVAR entry identification number. (Integer > 0).COEFi Coefficient of linear relation. (Real).

1 2 3 4 5 6 7 8 9 10

DVPREL1 ID TYPE PID FID PMIN PMAX C0

DVID1 COEF1 DVID2 COEF2 DVID3 -etc.-

DVPREL1 12 PBAR 612 6 0.2 3.0

4 0.25 20 20.0 5 0.3


DVPREL1 (Cont.)

Remarks:1. The relationship between the analysis model property and design

variables is given by:

2. The continuation entry is required.3. PTYPE = “PBEND” is not supported, either directly through FIDs or

indirectly via word positions in the element property table. 4. FID may be either a positive or a negative number. If FID > 0, it

identifies the field position on a property entry. If FID < 0, it identifies the word position of an entry in the element property table. For example, to specify the area of a PBAR, either FID = +4 or FID = -3 can be used. However, if PTYPE = “PBEAM”, FID must be negative. See the following element property table for the word positions for PBEAM.

DVIDiCOEFiCOPi

i ×+= ∑


DVPREL1 (Cont.) An analysis model property Aj is related to design variables as:

Dependent and Independent design variables are treated equally

FID is the field position in the entry if positive. For example, the second field of the third continuation entry has a field ID equal to 22. FID is the entry position in the element property table (see MSC.NASTRAN Programmer’s Manual), if negative. For example, the cross-sectional area of a PBAR entry may be designated by FID=-3. PBEAM must use negative FID.

...X*2COEFX*1COEFCoA 2DVID1DVIDj +++=


DRESP1Defines a set of structural responses that is used in the design either as constraints or as anobjective.Format:

Example:

Field ContentsID Unique entry identifier. (Integer > 0).LABEL User-defined label. (Character).RTYPE Response type. See table below. (Character).PTYPE Element flag (PTYPE = “ELEM”) or property entry name. Used with

element type responses (stress, strain, force, etc.) to identify the property type, since property entry IDs are not unique across property types. (Character: “ELEM”, “PBAR”, “PSHELL”, etc.).

REGION Region identifier for constraint screening. See Remark 10 for defaults. (Integer > 0).

ATTA, ATTB, ATTi Response attributes. See Table 1. (Integer > 0 or blank).

(Continued)

1 2 3 4 5 6 7 8 9 10

DRESP1 ID LABEL RTYPE PTYPE REGION ATTA ATTB ATT1

ATT2 -etc.-

DRESP1 1 DX1 STRESS PROD 2 3 15 102

103


DRESP1 (Cont.)

Table 1. Design Sensitivity Response Attributes.

ATTA (Integer>0) ATTB (Integer>0 or Real>0.0) ATTI (Integer>0)

WEIGHT Row Number (1<=ROW<=6) See Remark 25.

Column Number (1<=COL<=6) SEIDi or All or blank

VOLUME Blank Blank SEIDi or ALLEIGN Normal Modes Mode Number Approximation Code. See

Remark 19.Blank

CEIG Complex Eigenvalue Mode Number (Integer>0)

ALPHA or OMEGA (Default=ALPHA)

Blank

FREQ Normal Modes Mode Number See Remark 18.

Approximation Code. See Remark 19.

Blank

LAMA Buckling Mode Number Approximation Code. See Remark 19.

Blank

DISP Displacement Component Blank Grid IDSTRAIN Strain Item Code Blank Property entry

(PDI)ESE Strain Energy Item Code See

Remark 21.Blank Property entry

(PDI)STRESS Stress Item Code Blank Property entry

(PDI)FORCE Force Item Code Blank Property entry

(PDI)SPCFORCE SPE Force Component Blank Grid IDCSTRAIN Strain Item Code LAMINA Number (Integer;

Default=1)Property entry (PDI)

Response Type

(RTYPE)

Response Attributes


DRESP1 (Cont.)

Table 1. Design Sensitivity Response Attributes. (Cont.)

CSTRESS Stress Item Code LAMINA Number (Integer; Default=1)

Property entry (PDI)

CFAILURE Failure Criterion Item Code LAMINA Number (Integer; Default=1)


FRACCL Acceleration Component Frequency Value. (Blank, Real>0.0 or Character) See Remarks 15 and 20.

Grid ID

FRDISP Displacement Component Frequency Value. (Blank, Real>0.0 or Character) See Remarks 15 and 20.

Grid ID

PRES Acoustic Pressure Component Frequency Value. (Real > 0.0) See Remarks 15.

Grid ID

FRFORC Force Item Code Frequency Value. (Blank,Real>0.0 or Character) See Remarks 15. and 20.


FRSPCF SPC Force Component Frequency Value. (Blank,Real>0.0 or Character) See Remarks 15. and 20.

Grid ID

FRSTRE Stress Item Code Frequency Value. (Blank,Real>0.0 or Character) See Remarks 15. and 20.


FRVELO Velocity Component Frequency Value. (Blank,Real>0.0 or Character) See Remarks 15. and 20.

Grid ID

RMSDISP Displacement Component RANDPS ID Grid IDRMSVELO Velocity Component RANDPS ID Grid IDRMSACCL Acceleration Component RANDPS ID Grid ID


DRESP1 (Cont.)

Remarks:1. Stress, strain, and force item codes can be found in “Item Codes” in Appendix A,

in the MSC.Nastran Quick Reference Guide. For stress or strain item codes that have dual meanings, such as von Mises or maximum shear, the option specified in the Case Control Section will be used; i.e., STRESS(VONM) or STRESS(MAXS).

(Continued)

TDISP Displacement Component Time Value. (Blank, Real, or Character) See Remarks 16. and 20.

Grid ID

TVELO Velocity Component Time Value. (Blank, Real, or Character) See Remarks 16. and 20.

Grid ID

TACCL Acceleration Component Time Value. (Blank, Real, or Character) See Remarks 16. and 20.

Grid ID

TSPCF SPC Force Component Time Value. (Blank, Real, or Character) See Remarks 16. and 20.

Grid ID

TSTRE Stress Item Code Time Value. (Blank, Real, or Character) See Remarks 16. and 20.


TFORC Force Item Code Time Value. (Blank, Real, or Character) See Remarks 16. and 20.


TRIM AESTAT or AESURF Entry ID Blank BlankSTABDER AESTAT or AESURF Entry ID Restraint Flag. (Integer 0 or 1)

See Remark 13.Component

FLUTTER Blank Blank See Remark 14.


DRESP1 (Cont.)2. RTYPE = “CSTRESS”, “CSTRAIN”, and “CFAILURE” are used only with the PCOMP

entry. “CSTRESS”, “CSTRAIN” and “CFAILURE” are described in Appendix A in the MSC.Nastran Quick Reference Guide. Only force item codes that refer to failure indices of direct stress and interlaminar shear stress are valid.The CFAILURE response type requires the following specifications on the applicable entries:a. Failure theory in the FT field on PCOMP entryb. Allowable bonding shear stress in the SB field on PCOMP entryc. Stress limits in the ST, SC, and SS fields on the MATi entries

3. ATTB is used only for responses of composite laminae, dynamics, complex eigenvalue, and stability derivatives. For other responses, this field must be blank.

4. All grids associated with a DRESP1 entry are considered to be in the same region for screening purposes. Only up to NSTR displacement constraints (see DSCREEN entry) per group per load case will be retained in the design optimization phase.

5. DRESP1 identification numbers must be unique with respect to DRESP2 identification numbers.

6. If PTYPE = “ELEM”, the ATTi correspond to element identification numbers.7. If RTYPE = “DISP”, “TDISP”, “TVELO”, “TACCL” or “TSPC”, multiple component

numbers (any unique combination of the digits 1 through 6 with no embedded blanks) may be specified on a single entry. Multiple response components may not be used on any other response types.

(Continued)


DRESP1 (Cont.)8. If RTYPE = “FRDISP”, “FRVELO”, “FRACCL”, or “FRSPC” only one

component number may be specified in the ATTA field. Numbers 1 through 6 correspond to real (or magnitude) components and 7 through 12 imaginary (or phase) components. If more than one component for the same grid is desired, then a separate entry is required.

9. Real/imaginary representation is the default for complex response types. Magnitude/phase representation must be requested by the corresponding Case Control command; e.g., DISP(PHASE) = ALL.

10. REGION is used for constraint screening. The NSTR field on DSCREEN entries gives the maximum number of constraints retained for each region per load case. If RTYPE = “WEIGHT”, “VOLUME”, “LAMA”, “EIGN” or “FREQ”, no REGION identification number should be specified. For all other responses, if the REGION field is left blank, the default specified in Table 2 is used. Usually, the default value is appropriate. If the REGION field is not blank, all the responses on this entry as well as all responses on other DRESP1 entries that have the same RTYPE and REGION identification number will be grouped into the same region.


11.REGION is valid only among the same type of responses. Responses of different types will never be grouped into the same region, even if they are assigned the same REGION identification number by the user.

(Continued)

DRESP1 (Cont.)

Table 2. Default Regions for Design Sensitivity Response Types.

Response Type Default Region

WEIGHT No regionVOLUME No regionLAMA No regionEIGN No regionFREQ No regionDISP One region per DRESP1 entryFRDISP One region per DRESP1 entryPRES One region per DRESP1 entrySPCFORCE One region per DRESP1 entryFRVELO One region per DRESP1 entryFRACCL One region per DRESP1 entryFRSPCF One region per DRESP1 entryTDISP One region per DRESP1 entryTVELO One region per DRESP1 entryTACCL One region per DRESP1 entryTSPCF One region per DRESP1 entryFLUTTER One region per DRESP1 entry

OTHER

One Region per PROPERTY entry. If PTYPE="ELEM", then one region per DRESP1 entry


DRESP1 (Cont.)12. If RTYPE = “WEIGHT” or “VOLUME”, field ATTi = “ALL” implies total weight/volume of all

superelements except external superelements.13.RTYPE = “STABDER” identifies a stability derivative response. ATTB is the restraint flag

for the stability derivative. ATTB = 0 means unrestrained and ATTB = 1 means restrained. For example, ATTA = 4000, ATTB = 0, and ATT1 = 3, references the unrestrained Cz derivative for the AESTAT (or AESURF) entry ID = 4000.

14.RTYPE = “FLUTTER” identifies a set of damping responses. The set is specified by ATTi:ATT1 = identification number of a SET1 entry that specifies a set of modes.ATT2 = identification number of an FLFACT entry that specifies a list of densities. ATT3 = identification number of an FLFACT entry that specifies a list of Mach numbers.ATT4 = Identification number of an FLFACT entry that specifies a list of velocities.If the flutter analysis is type PKNL, it is necessary to put PKNL in the PTYPE field of this entry.

15.For RTYPE = “FRDISP”, “FRVELO”, “FRACCL”, FRSPC”, “FRFORC”, and “FRSTRE”, a real value for ATTB specifies a frequency value in cycles per unit time. If ATTB is specified, then the responses are evaluated at the closest frequency selected by the OFREQ command. The default for ATTB is all frequencies selected by the OFREQ command. See Remark 20 for additional ATTB options.

16.For RTYPE = “TDISP”, TVELO”, “TACCL”, “TSPC”, TFORC”, and “TSTRE”, ATTB specifies a time value. If ATTB is specified, then the responses are evaluated at the closest time selected by the OTIME command. The default for ATTB is all time steps selected by the OTIME command.


DRESP1 (Cont.)17.Intermediate station responses on CBAR elements due to PLOAD1 and / or

CBARAO entries may not be defined on the DRESP1 entry.18.RTYPE = “EIGN” refers to normal modes response in terms of eigenvalue

(radian / time)**2 while RTYPE = “FREQ” refers to normal modes response in terms of natural frequency or units of cycles per unit time.

19.For RTYPE = LAMA, EIGN or FREQ, the response approximation used for optimization can be individually selected using the ATTB field.(Approximation Code = 1 = direct linearization, = 2 = Inverse Linearization).

20.Character input for ATTB is available for RTYPE of FRDISP, FRVELO, FRACCL, FRSPCF, FRSTRE, FRFORC, TDISP, TVELO, TACCL, TSPCF, TSTRE and TFORC. The character input represents a mathematical function and the options for character input are SUM, AVG, SSQ, RSS,MAX and MIN. The expression of mathematical function is shown as follows:


DRESP1 (Cont.)MAX (X1, X2, …, Xn) = Largest value among xi(i=1 to n)MIN (X1, X2, …, Xn) = Smallest value among xi(i=1 to n)where Xi is the response for a forcing frequency. For exampleDRESP1,10,DX1,FRSTRE,ELEM,,3,AVG,10yields a response which is equal to the average stress for element 10. The average is done by first adding up the component 3 stress of element 10 for all forcing frequencies (all time steps for transient response), and then dividing by the number of forcing frequencies. Note that the response computed is considered as type 2 response. Therefore, if referenced on a DRESP2, the ID of such DRESP1 (ATTB with character input) must be listed following DRESP2 keyword.

21.Element strain energy item codes can be found under Table A-5 in “Element Strain Energy Item Codes” on page 1571 in Appendix A. Only element strain energy and element strain energy density can be referenced on a DRESP1 entry.

22.For RTYPE=CEIG, the allowable character inputs are ALPHA and OMEGA with ALPHA being the default.

23.For RTYPE=RMSDISP, RMSVELO, or RMSACCL the ATTB specifies the appropriate RANDPS ID.


DRESP1 (Cont.)24. Input other than 1 or 7 of ATTA field, acoustic pressure component, for PRES

response type will be reset to 1 (if less than 7) or 7 (if greater than 7 and less than 13)

25.Design response weight is obtained from Grid Point Weight Generator for a reference point GRDPNT (see parameter GRDPNT). If GRDPNT is either not defined, equal to zero, or not a defined grid point, the reference point is taken as the origin of the basic coordinate system. Fields ATTA and ATTB refer to the row and column numbers of the rigid body weight matrix, which is partitioned as follows:

6x6

The default values of ATTA and ATTB are 3, which specifies weight in the Z direction. Field ATT1 = “ALL” implies total weight of all superelements except external superelements. SEIDi refers to a superelement identification number. SEIDi = “0” refers to the residual superelement. The default of ATT1 is blank

z

y

x

z

y

x

IWWWWWWIWWWWWWIWWWWWWWWWWWWWWWWWWWWW

6564636261

5654535251

4645434241

3635343231

2625242321

1615141312


DESOBJSelects the DRESP1 or DRESP2 entry to be used as the design objective.

Format:

Examples:

DESOBJ = 10DESO = 25

Describer MeaningMIN Specifies that the objective is to be minimized.MAX Specifies that the objective is to be maximized.n Set identification of a DRESP1 or DRESP2 Bulk Data entry (Integer > 0).

Remarks:1. A DESOBJ command is required for a design optimization task and is optional for a

sensitivity task.2. If the DESOBJ command is specified within a SUBCASE, the identified DRESP1 Bulk Data

entry uses responses only from that subcase. If DESOBJ appears above all SUBCASE commands and there are multiple subcases, it uses global responses.

3. The referenced DRESP1 entry must define a scalar response.

nMINMAX

DESOBJ =


DCONSTRDefines design constraintsFormat:

Example:

Field ContentsDCID Design constraint set identification number (Integer > 0)RID DRESP1 entry identification number (Integer > 0)LALLOW Lower bound on the response quantity (Real, Default = -1E20)UALLOW Upper bound on the response quantity (Real, Default = 1E20)LOWFQ Low end of frequency range in Hertz (Real >=0.0, Default=0.0). See Remark 8HIGHFQ High end of frequency range in Hertz (Real >=LOWFQ, Default=1.E20).

Remarks:1. The DCONSTR entry may be selected in Case Control by the DESSUB or SESGLB command2. DCID may be referenced by the DCONNAD Bulk Data entry

1 2 3 4 5 6 7 8 9 10

DCONSTR DCID RID LALLOW UALLOW

DCONSTR 10 4 1.25


DCONSTR (Cont)3. For a given DCID, the associated RID can be referenced only once4. The units of LALLOW and UALLOW must be consistent with the referenced response

defined on the DRESP1 entry. If a RID refers to an eigenvalue response, the the imposed bounds must be expressed in units of eigenvalue, (radian/time)2

5. LALLOW and UALLOW are unrelated to the stress limits specified on the MATI entry

6. Constraints are computed as follows:

g = (LALLOW – r) / GNORM for lower bound constraintsg = (r – UALLOW) / GNORM for upper bound constraints

where r is the response defined on the DRESPi entry and

GSCAL is specified on the DOPTPRM entry (Default = 0.001).

7. As Remark 6 indicates, small values of UALLOW and LALLOW require special processing and should be avoided. Bounds of exactly zero are particularly troublesome. This can be avoided by using a DRESP2 entry that offsets the constrained response from zero.

>

>

=

otherwise GSCAL GSCALUALLOW ifbounds upper for UALLOW

GSCALLALLOW ifbounds lower for LALLOW GNORM


DCONSTR (Cont)8. LOWFQ and HIGHFQ fields are functional only for RTYPE with ‘FR’ prefix,

e.g., FRDISP. The bounds provided in LALLOW and UALLOW are applicable to a response only when the value of forcing frequency of the response falls between the LOQFQ and HIGHFQ. If the ATTB field of the DRESP1 entry is not blank, LOQFQ and HIGHFQ are ignored.


DCONSTR (Cont) It is possible that many constraints are generated by one

DECONSTR entry

The user should be aware of this structure when describing the design model.

Equality constraints are not supported directly at this time. If necessary, provide equivalent lower and upper bounds.


DESGLBSelects the design constraints to be applied at the global level indesign optimization task.Format:DESGLB = n

Example:DESGLB = 10DESG = 25

Describer Meaningn Set identification of a DCONSTR or DCONADD Bulk Data

entry identification number (Integer > 0).Remarks:

1. A DESGLB command is optional and invokes constraints that are to be applied independent of a particular subcase. These constraints could be based on responses that are independent of subcases (e.g., WEIGHT or VOLUME).

2. The DESGLB command can be used to invoke constraints that are not a function of DRESP1 entries; e.g., DRESP2 responses that are not functions of DRESP1 responses are subcase independent.


ANALYSISSpecifies the type of analysis being performed for the current subcaseFormat:

ANALYSIS = typeExamples:

ANALYSIS = STATICSANAL = MODES

Describer Meaningtype Analysis type. Allowable values and application solution sequences (Character):

STATICS StaticsMODES Normal ModesBUCK BucklingDFREQ Direct FrequencyMFREQ Modal FrequencyMTRAN Modal Transient SAERO Static AeroelasticFLUTTER FlutterDCEIG Direct Complex EigenvalueMCEIG Modal Complex EigenvalueDIVERGE Static Aeroelastic DivergenceHEAT Heat Transfer Analysis(SOLs 153 and 159 only)STRUCURE Structural Analysis (SOLs 153 and 159 only)

Remarks: In SOL 200, all subcases must be assigned by an ANALYSIS command. Also, all subcases assignedby ANALYSIS=MODES must contain a DESSUB request.

(SOL 200 only)


DESSUBSelect the design constraints to be used in design optimization task for the

current subcase

Format:DESSUB = nExamples:DESSUB = 10DESS = 25Describer Meaningn Set Identification of DCONTR or DCONADD Bulk Data

entry Identification number (Integer > 0)

Remarks:1. A DESSUB command is required for every subcase for which constraints are to be

applied.2. The DESSUB command in a given subcase is the default for all subsequent subcases.

In order to override the default, specify a new DESSUB = n or DESSUB = 0 if no responses are desired.


SAMPLE OPT1 – CONSTANT CROSS-SECTION CANTILEVER BEAM

Let us now create the optimization input. First define the response quantities which will be

input for the optimizer. We need to measure the tip displacement, the stress at the root,

and the weight of the beam.

$ create design response entries$$ total weight = objective will be to minimize this$DRESP1,1,TOTW,WEIGHT$$ tip displacement – we need to keep this less than .01$DRESP1,2000,tipdisp,DISP,,,2,,9$$ Stress at the root – select the element, rather than the property$ select the maximum stress at the start of the BAR$DRESP1,3000,rootstr,STRESS,ELEM,,8,,1$



Now let us define the design variables for the optimizer.

We want to vary the depth of the element, so we use DVPREL1 entries to “tie” the DESVAR entry to the fields on the PBARL.

$$ Create design variable$$ allow height to vary from .5 to 8.0$DESVAR,2,Height,2.,.5,8.0$$ Relate Design Variables to the Fields on the PBARL$DVPREL1,110,PBARL,1,13,.5,8.0,2,1.0$



Now we need to define the design constraints (stress and displacement constraints).

$ Define Design Constraints

$

$ limit tip displacement

$

DCONSTR,200,2000,–.01,.01

$

$ limit stress at root

$

DCONSTR,200,3000,–1000.,1000.

$



Now we need to supply the executive and case control

SOL 200TIME 5

CEND

TITLE = Practical Dynamics Seminar Sample Problem 1

SUBTITLE = Cantilever Beam

LABEL = Perform Model Checks

DISP = ALL

SUBCASE 1

SPC = 1

load = 100

ANALYSIS = STATICSDESOBJ = 1000 $ minimize weightDESSUB = 200 $ apply constraints in the optimizer$



The complete input fileSOL 200CENDTITLE = Practical Dynamics Seminar Sample Problem 1SUBTITLE = Cantilever BeamLABEL = Perform Model ChecksDISP = ALLSUBCASE 1SPC = 1load = 100ANALYSIS = STATICSDESOBJ = 1000 $ minimize weightDESSUB = 200 $ apply constraints in the optimizerBEGIN BULKFORCE,100,9,,10.,0.,–1.,0.include ’cantbeam.dat’DRESP1,1000,TOTW,WEIGHTDRESP1,2000,tipdisp,DISP,,,2,,9DRESP1,3000,rootstr,STRESS,ELEM,,8,,1DESVAR,2,Height,2.,.5,4.0DVPREL1,110,PBARL,1,13,.5,8.0,2,1.0DCONSTR,200,2000,–.01,.01DCONSTR,200,3000,–1000.,1000.ENDDATA


OUTPUT FROM OPT1


SAMPLE OPT2 – VARIABLE CROSS-SECTION CANTILEVER BEAM

Now let us allow the thickness to vary along the length. For this, we will need to define a different property for

each BAR element and allow the properties of each to vary.

This is done in file cantbeam2.dat (on the following page).

Measured response quantities and constraints will be unchanged.

Only the properties, design variables, and DVPREL1 entries will change.


SAMPLE OPT2 – VARIABLE CROSS-SECTION CANTILEVER BEAM (Cont)



Let us define the new design variables for the optimizer. We want to be able to vary the depth of the elements, so we use

DVPREL1 entries to “tie” the DESVAR entry to the fields on the BARL.

DESVAR,21,Height1,2.,.5,8.0DESVAR,22,Height2,2.,.5,8.0DESVAR,23,Height3,2.,.5,8.0DESVAR,24,Height4,2.,.5,8.0DESVAR,25,Height5,2.,.5,8.0DESVAR,26,Height6,2.,.5,8.0DESVAR,27,Height7,2.,.5,8.0DESVAR,28,Height8,2.,.5,8.0$$ Relate Design Variables to the Fields on the PBARLDVPREL1,111,PBARL,1,13,.5,8.0,21,1.0DVPREL1,112,PBARL,2,13,.5,8.0,22,1.0DVPREL1,113,PBARL,3,13,.5,8.0,23,1.0DVPREL1,114,PBARL,4,13,.5,8.0,24,1.0DVPREL1,115,PBARL,5,13,.5,8.0,25,1.0DVPREL1,116,PBARL,6,13,.5,8.0,26,1.0DVPREL1,117,PBARL,7,13,.5,8.0,27,1.0DVPREL1,118,PBARL,8,13,.5,8.0,28,1.0



$ file opt2.at – cantilever beam static optimization with variable$ cross–sectionSOL 200CENDTITLE = Practical Dynamics Seminar Sample Problem 1 opt2.datSUBTITLE = Cantilever Beam with variable cross–sectionLABEL = Perform Model ChecksDISP = ALLSTRESS = ALLSUBCASE 1SPC = 1load = 100ANALYSIS = STATICSDESOBJ = 1000 $ minimize weightDESSUB = 200 $ apply constraints in the optimizer$BEGIN BULKFORCE,100,9,,10.,0.,–1.,0.include ’cantbeam2.dat’$ create design response entries$ total weight = objective will be to minimize this$DRESP1,1000,TOTW,WEIGHT$$ tip displacement – we need to keep this less than .01$DRESP1,2000,tipdisp,DISP,,,2,,9$ Stress at the root – select the element, rather than the property$ select the maximum stress at the start of the BAR$DRESP1,3000,rootstr,STRESS,ELEM,,8,,1

The complete input file



$ Create design variablesDESVAR,21,Height1,2.,.5,8.0DESVAR,22,Height2,2.,.5,8.0DESVAR,23,Height3,2.,.5,8.0DESVAR,24,Height4,2.,.5,8.0DESVAR,25,Height5,2.,.5,8.0DESVAR,26,Height6,2.,.5,8.0DESVAR,27,Height7,2.,.5,8.0DESVAR,28,Height8,2.,.5,8.0$ Relate Design Variables to the Fields on the PBARLDVPREL1,111,PBARL,1,13,.5,8.0,21,1.0DVPREL1,112,PBARL,2,13,.5,8.0,22,1.0DVPREL1,113,PBARL,3,13,.5,8.0,23,1.0DVPREL1,114,PBARL,4,13,.5,8.0,24,1.0DVPREL1,115,PBARL,5,13,.5,8.0,25,1.0DVPREL1,116,PBARL,6,13,.5,8.0,26,1.0DVPREL1,117,PBARL,7,13,.5,8.0,27,1.0DVPREL1,118,PBARL,8,13,.5,8.0,28,1.0$ Define Design Constraints$ limit tip displacementDCONSTR,200,2000,–.01,.01$ limit stress at rootDCONSTR,200,3000,–1000.,1000.DOPTPRM,DESMAX,25ENDDATA

The complete input file (Cont.)


OUTPUT FROM OPT2


WORKSHOPS Part 1: for the optimization example (opt1.dat) add the bar width

as a design variable and re-run the problem Part 2: using a copy of the file from part1, add a requirement

that the buckling eigenvalue must be greater that 1.5 Part 3: using a copy of the input file from part 2, add a

concentrated load weight of 10.0 on the end of the beam and then add a requirement that the first natural frequency be greater that 4.0hz (initial value is 4.45hz, but since the program will attempt to reduce the width, the first frequency will drop)

Part 4: try the same thing using opt2.dat as a starting point


ADVANCED FEATURES OF OPTIMIZATION IN MSC.NASTRAN


TYPE-2 STRUCTURAL RESPONSES

=

2) TYPE(RESPONSESYNTHETIC

)1 TYPE(RESPONSESSIMPLE

RESPONSESTRUCTURAL

Responses that are obtained directlyfrom MSC.NASTRAN (displacementsstress, strain, internal forces, bucklingloads, natural frequencies, compositefailure criteria, weight, volume

Responses that are synthesizedfrom simple responses, designvariables, and constants throughanalytic equations given by user


TYPE-2 RESPONSE FORMULATION

Type 2 Response(s) = Function of:

Note: Recall that one designation of type 1 response from the DRESP1 data entry may generate a large number of type1 responses of theyare stress, strain or internal forces. If the type1 responses are of that type, one type2 response data entry can generate the same number of type 2 responses.

sCoordinate Grid)Response(s 1Type

sConstraintTable Variables Design


RESTRICTIONS IN FORMING SYNTHETIC RESPONSES

Different types of responses (for example, weight and displacement) can be mixed in one equation, however, use caution. Each associated DRESP1 entry must generate a single response only.

Responses cannot be mixed across the subcases in one equation.

Multiple equations must be separated by a semi-colon (;) and no recursive reference is allowed.


DEQATNDefines one or more equations for use in design sensitivity or p-element

analysisFormat:

Example:

Field ContentsEQID Unique equation identification number. (Integer > 0)EQUATION Equation(s). See Remarks. (Character)Remarks:

1. EQUATION is a single equation, or a set of nested equations, and is specified in fields 3 through 9 on the first entry and may be continued on fields 2 through 9 on the continuation entries. On the continuation entries, no commas can appear in columns 1 through 8. All data in fields 2 through 9 must be specified in columns 9 through 72. The large field format is not allowedA single equation has the following format:

Variable1 (x1, x2, …, xn) = expression1

1 2 3 4 5 6 7 8 9 10

DEQATN EQID EQUATION

EQUATION (Cont)

DEQATN 14 F1(A,B,C,D,R) = A + B * C – (D * * 3 +10.0)

Sin (PI(1) * R) + A * * 2/(B – C); F = A + B – F1 * D


DEQATN (Cont)A set of nested equations is separated by semi-colons and has the format:

variable1 (x1, x2, …, xn) = expression1variable2 = expression2variable3 = expression3

etc.variable_m = expression_m

Expression_i is a collection of constants, real variables, and real functions, separated by operators, and must produce a single real value. (x1, x2, …, xn) is the list of all the variable names except variable_i, which appears in all expressions. Variable_i may be used in subsequent expressions. The last equation, variable_m = expression_m, provides the value that is returned to the Bulk Data entry that references EQUID; e.g., DRESP2. The example above represents the following mathematical equations:

where SIN and PI are intrinsic functions. See Remark 42. EQUATION may contain embedded blanks. EQUATION must contain less than 12,500

nonblank characters. This is equivalent to approximately 195 continuation entries.

( )CB

AR)1(PIsin)10D(CBA1F2

3

−+∗++−∗+=

DFBAF ∗−+= 1


DEQATN (Cont)3. The syntax of the expression follows FORTRAN language standards. The

allowable arithmetic operations are shown in Table 3 in the order of execution precedence. Parenthesis are used to change the order of precedence. Operations within parenthesis are performed first with the usual order of precedence maintained within the parenthesis

4. The expression may contain intrinsic functions. Table 4 contains the format and descriptions of functions which may appear in the expressions. The use of functions that may be discontinuous must be used with caution because they can cause discontinuous derivatives. These are ABS, DIM, MAX, MIN, and MOD. For examples and further details see the MSC.NASTRAN DMAP Programmer’s Guide.

Table 3. DEQATN Entry Operators


DEQATN (Cont)

Table 4. DEQATN Entry Functions


DEQATN (Cont)5. If the DEQATN entry is referenced by the:

a. DVPREL2 entry, then Xi represents the DVIDj and LABLk fieldsb. DRESP2 entry then xi represents the DVIDj, LABLk, NRm, Gp, DPIPq, DCICr,

DMIMs, DPI2Pt, DCI2Cu, DMI2Mv, and NRRw fields in that orderc. GMLOAD, GMBC, or TEMPF entries then

X1 represents x in the basic coordinate system,X2 represents y in the basic coordinate system, andX3 represents z in the basic coordinate system

d. GMCURV entry thenX1 represents line parameter u

e. GMSURF entry thenX1 represents surface parameter u andX2 represents surface parameter v

6. If the DEQATN entry is referenced by the GMLOAD, GMBC, TEMPF, GMCURV, or GMSURF entries and your computer has a short word length (e.g., 32 bits/word), the EQUATION is processed with double precision and constraints may be specified in double precision; e.g., 1.2DO. If your machine has a long word length (e.g., 64 bits/word) then EQUATION is processed in single precision and constants must be specified in single precision; e.g., 1.2.


DTABLEDefines a table of real constraints that are used in equations (see DEQATN entry).Format:

Example:

Field ContentsLABLi Label for the constant. (Character).VALUi Value of the constant. (Real).Remarks:

1. Only one DTABLE entry may be specified in the Bulk Data Section.2. LABLI are referenced by the LABI on the DVPREL2 or DRESP2 entries.

1 2 3 4 5 6 7 8 9 10

DTABLE LABL1 VALU1 LABL2 VALU2 LABL3 VALU3 LABL4 VALU4

LABL5 VALU5 LABL6 VALU6 -etc-

DTABLE PI 3.142 H 10.1 E 1.0E6 T 0.1

G 5.5E5 B 100.


DRESP2Defines equation responses that are used in the design, eitheras constraints or as an objective.Format:

1 2 3 4 5 6 7 8 9 10DRESP2 ID LABEL EQID or FUNC REGION

"DESVAR" DVID1 DVID2 DVID3 DVID4 DVID5 DVID6 DVID7DVID8 -etc.-

"DATABLE" LABL1 LABL2 LABL3 LABL4 LABL5 LABL6 LABL7LABL8 -etc.-

"DRESP1" NR1 NR2 NR3 NR4 NR5 NR6 NR7NR8 -etc.-

"DNODE" G1 C1 G2 C2 G3 C3G4 C4 etc.

"DVPREL1" DPIP1 DPIP2 DPIP3 DPIP4 DPIP5 DPIP6 DPIP7DPIP8 DPIP9 -etc.-

"DVCREL1" DCIC1 DCIC2 DCIC3 DCIC4 DCIC5 DCIC6 DCIC7DCIC8 DCIC9 -etc.-

"DVMREL1" DMIM1 DMIM2 DMIM3 DMIM4 DMIM5 DMIM6 DMIM7DMIM8 DMIM9 -etc.-

"DVPREL2" DPI2P1 DPI2P2 DPI2P3 DPI2P4 DPI2P5 DPI2P6 DPI2P7DPI2P8 DPI2P9 -etc.-

"DVCREL2" DCI2C1 DCI2C2 DCI2C3 DCI2C4 DCI2C5 DCI2C6 DCI2C7DCI2C8 DCI2C9 -etc.-

"DVMREL2" DMI2M1 DMI2M2 DMI2M3 DMI2M4 DMIM5 DMI2M6 DMI2M7DMI2M8 DMI2M9 -etc.-

"DRESP2" NRR1 NRR2 NRR3 NRR4 NRR5 NRR6 NRR7NRR8 -etc.-


DRESP2 (Cont.)Example:

Field ContentsID Unique identification number. (Integer > 0).LABEL User-defined label. (Character).EQID DEQUATN entry identification number. (Integer > 0).FUNC Function to be applied to the arguments. See Remark 8.

(Character)REGION Region identifier for constraint screening. See Remark 5.

(Integer > 0).“DESVAR” Flag indicating DESVAR entry identification numbers. (Character).DVIDi DESVAR entry identification number. (Integer > 0).

DRESP2 1 LBUCK 5 3DESVAR 101 3 4 5 1 205 209

201DTABLE PI YM LDRESP1 14 1 4 22 6 33 2DNODE 14 1 4 1 22 3

2 1 43 1DVPREL1 101 102DVCREL1 201 202DVMREL1 301DVPREL2 401 402DVCREL2 501DVMREL2 601 602 603DRESP2 50 51


DRESP2 (Cont.)“DTABLE” Flag indicating that the labels for the constants in a DTABLE entry follow.

(Character)

LABLj Label for a constant in the DTABLE entry. (Character)

“DRESP1” Flag indicating DRESP1 entry identification numbers. (Character)

NRk DRESP1 entry identification number. (Integer > 0)

“DNODE” Flag indicating grid point and component identification numbers. (Character)

Gm Grid point identification number. (Integer > 0)

Cm Component number of grid point Gm. (1 <= Integer <=3)

DVPREL1 Flag indicating DVPREL1 entry identification number. (Character)

DPIPi DVPREL1 entry identification number. (Integer > 0)

DVCREL1 Flag indication DVCREL1 entry identification number. (Character)

DCICi DVCREL1 entry identification number. (Integer > 0)

DVMREL1 Flag indicating DVPREL2 entry identification number. (Character)

DMIMi DVMREL1 entry identification number. (Integer > 0)

DVPREL2 Flag indicating DVPREL2 entry identification number. (Character)

DPI2Pi DVPREL2 entry identification number. (Integer > 0)

DVCREL2 Flag indicating DVCREL2 entry identification number. (Character)

DCI2Ci DVCREL2 entry identification number. (Integer > 0)


Remarks:1. DRESP2 entries may only reference DESVAR, DTABLE, DRESP1,

DNODE, DVPREL1, DVCREL1, DVMREL1, DVPREL2, DVCREL2, and DVMREL2 entries. They may also reference other DRESP2 entries. However, a DRESP2 entry cannot reference itself directly or recursively.

2. Referenced DRESP1 entries cannot span analysis types or superelements.

3. DRESP2 entries must have unique identification number with respect to DRESP1 entries.

4. The “DESVAR”, “DTABLE”, “DNODE”, “DVPREL1”, “DVCREL1” and “DVMREL1”, “DVPREL2”, “DVCREL2”, “DVMREL2”, and “DRESP2” flags in field 2 must appear in the order given above. Any of these words along with the identification numbers associated with them may be omitted if they are not involved in this DRESP2 relationship. However, at least one of these four types of arguments must exist.

DRESP2 (Cont.)DVMREL2 Flag indicating DVMREL2 entry identification number. (Character)

DMI2Mi DVMREL2 entry identification number. (Integer > 0)

DRESP2 Flag indicating other DRESP2 entry identification number. (Character)

NRRk DRESP2 entry identification number. (Integer > 0)


DRESP2 (Cont.)5. The REGION field follows the same rules as for the DRESP1 entries. DRESP1 and

DRESP2 responses will never be contained in the same region, even if they are assigned the same REGION identification number. The default is to put all responses referenced by one DRESP2 entry in the same region.

6. The variables identified by DVIDi, LABLj, NRk, and the Gm, Cm pairs, DPIPi, DCICm, DMIMn, DPI2Po, DCI2Cp, DMI2Mq, and NRRu are assigned (in that order) to the variable names (x1, x2, x3, etc.) specified in the left-hand side of the first equation on the DEQUATN entry, referenced by EQID. In the example below,

DESVARs 101 and 3 are assigned to arguments A and B.DTABLEs PI and YM are assigned to arguments C and D.Grid 14, Component 1 is assigned to argument R.

DRESP2 1 LBUCK 5 3

DESVAR 101 3

DTABLE P1 YM

DNODE 14 1

DEQATN 5 F1(A, B, C, D, R) = A + B * C – (D * * 3 + 10.0) + sin(C * R)


DRESP2 (Cont.)7. (Gm, Cm) refer to a any grid component and is no longer limited to a

designed grid component. 8. The FUNC attributes can be used in place of the EQID and supports the

functions shown in the following table:

When EQID has character input, the DEQATN entry is no longer needed. The functions are applied to all arguments on the DRESP2 regardless of the type.

9. The number of arguments of a DEQATN can be more than the number of values defined on the DRESP2 if the DRESP1s referenced have RTYPE with ‘FR’ prefix. Arguments are still positional. The extra arguments in the DEQATN must appear at the end of the argument list. The discrepancy is resolved internally with the forcing frequency(ies) associated with DRESP1s. An follows:

Function DescriptionAVG Average of the argumentsSSQ Sum of the squares of the argumentsRSS Square root of the sum of the squares of the argumentsMAX The larges value of the argument listMIN The smallest value of the argument list


DRESP1 10 FDISP1 FRDISP 1 10 1001DRESP1 20 FDISP2 FRDISP 1 20 1001DRESP2 30 AVGFD 100

DRESP1 10 20DEQATN 100 AVG(D1,D2,F1,F2) = (D1/F1+D2/F2)*0.5

DRESP2 (Cont.)

In the above example, the DEQATN has two more additional terms than have been defined on the DRESP2. The first additional term is the forcing frequency (in hertz) of the first DRESP1 ID on the DRESP2. The second additional term is the forcing frequency of second DRESP1 ID in the list. When all DRESP1s involved have the same frequency, the user is not required to name all the additional terms in the argument list of DEQATN.


DRESP2 BULK DATA ENTRY DRESP2 is used to define a synthetic response that

is not directly available from the MSC.NASTRAN analysis capabilities

Examples: Generation of a new stress or strain failure criterion that is not

available in MSC.NASTRAN Imposing local buckling criteria based on element sizes as well as

stress components Programming proprietary design-sizing equations Generation of nonlinear displacement responses such as

displacement magnitudes:

More complex relations among displacements could be formed by combining with MPC and RBE3 capabilities

( )2z

2y

2x uuuU ++=


DOPTPRMOverrides default values of parameters used in designoptimization.

Format:

Example:

Field ContentsPARAMi Name of the design optimization parameter. Allowable

names are given in Table 5. (Character.)VALi Value of the parameter. (Real or Integer, see Table 5).

Remarks:1. Only one DOPTPRM entry is allowed in the Bulk Data Section.

1 2 3 4 5 6 7 8 9 10

DOPTPRM PARAM1 VAL1 PARAM2 VAL2 PARAM3 VAL3 PARAM4 VAL4

PARAM5 VAL5 -etc.-

DOPTPRM PRINT 5 DESMAX 10

I


DOPTPRM (Cont.)Name Description, Type, and Default Value

APRCOD Approximation method to be used. 1 = Direct Linearization; 2 = Mixed Method based on response type; 3 = Convex Linearization. APRCOD = 1 is recommended for shape optimization problems. (Integer 1, 2, or 3; Default = 2)

CONV1 Relative criterion to detect convergence. If the relative change in objective between two optimization cycles is less than CONV1, then optimization is terminated. (Real > 0.0; Default = 0.001)

CONV2 Absolute criterion to detect convergence. If the absolute change in objective between two optimization cycles is less than CONV2, then optimization is terminated. (Real > 0.0; Default = 1.0E—20)

CONVDV Relative convergence criterion on design variables. (Real > 0.0; Default = 0.001)

CONVPR Relative convergence criterion on properties. (Real > 0.0; Default = 0.001)

CT Constraint tolerance. Constraint is considered active if current value is greater than CT. (Real < 0.0; Default = –0.03)

CTMIN Constraint is considered violated if current value is greater than CTMIN.(Real > 0.0; Default = 0.003)

DABOBJ Maximum absolute change in objective between ITRMOP consecutive iterations (see ITRMOP) to indicate convergence at optimizer level. F0 is the initial objective function value. (Real > 0.0; Default = MAX[0.001 * ABS(F0), 0.0001])

DELB Relative finite difference move parameter. (Real > 0.0; Default = 0.0001)

Table 5. PARAMi Names and Description.


DOPTPRM (Cont.)

Table 5. PARAMi Names and Description. (Cont.)

Name Description, Type, and Default Value

DELOBJ Maximum relative change in objective between ITRMOP consecutive iterations to indicate convergence at optimizer level. (Real > 0.0; Default = 0.001)

DELP Fractional change allowed in each property during any optimization design cycle. This provides constraints on property moves. (Real > 0.0; Default = 0.2)

DELX Fractional change allowed in each design variable during any optimization cycle. (Real > 0.0; Default = 1.0)

DESMAX Maximum number of design cycles (not including FSD cycle) to be performed. (Integer > 0; Default = 5)

DISCOD Discrete Processing Method: (Integer 1, 2, 3 or 4; Default = 1)1: Design of Experiments2: Conservative Discrete Design3: Rounding up to the nearest design variable4: Rounded off to the nearest design variable

DISBEG Design cycle ID for discrete variable processing initiation. Discrete variable processing analysis is carried out for every design cycle after DISBEG. (Integer >=0, default = 0 = the last design cycle)

DOBJ1 Relative change in objective attempted on the first optimization iteration. Used to estimate initial move in the one-dimensional search. Updated as the optimization progresses. (Real > 0.0; Default = 0.1)

DOBJ2 Absolute change in objective attempted on the first optimization iteration.(Real > 0.0; Default = 0.2 * (F0))

DPMIN Minimum move limit imposed. (Real > 0.0; Default = 0.01)

DX1 Maximum relative change in a design variable attempted on the first optimization iteration. Used to estimate the initial move in the one dimensional search. Updated as the optimization progresses. (Real > 0.0; Default = 0.01)


DOPTPRM (Cont.)


Specifies the number of Fully Stressed Design Cycles that are to be performed (Integer, Default = 0)

DX2 Absolute change in a design variable attempted on the first optimization iteration. (Real > 0.0; Default = 0.2 * MAX[X(I)])

DXMIN Minimum design variable move limit (Real > 0.0; Default = 0.05)

Name Description, Type, and Default Value

FSDALP Relaxation parameter applied in Fully Stressed Design (Real, 0.0 < FSDMAX <= 1.0, Default = 0.9)

FSDMAX

GMAX Maximum constraint violation allowed at the converged optimum. (Real > 0.0; Default = 0.005)

GSCAL Constraint normalization factor. See Remarks under the DSCREEN and DCONSTR entries. (Real > 0.0; Default = 0.001)

IGMAX If IGMAX = 0, only gradients of active and violated constraints are calculated. If IGMAX > 0, up toNCOLA gradients are calculated including active, violated, and near active constraints.(Integer > 0; Default = 0)

IPRINT Print control during approximate optimization phase. Increasing values represent increasing levels of optimizer information. (0 <= Integer <= 7; Default = 0)

0 no output (Default)1 internal optimization parameters, initial information, and results2 same, plus objective function and design variables at each iterations3 same, plus constraint values and identification of critical constraints4 same, plus gradients 5 same, plus search direction 6 same, plus scaling factors and miscellaneous search information7 same, plus one dimensional search information


DOPTPRM (Cont.)


ISCAL Design variables are rescaled every ISCAL iterations. Set ISCAL= –1 to turn off scaling. (Integer; Default = NDV (number of design variables))

IPRINT1 If IPRNT1 = 1, print scaling factors for design variable vector. (Integer 0 or 1; Default = 0)

IPRINT2 If IPRNT2 = 1, print miscellaneous search information. If IPRNT2 = 2, turn on print during one-dimensional search process. (Warning: This may lead to excessive output.) (Integer 0, 1, or 2; Default = 0)

ITMAX Maximum number of iterations allowed at optimizer level during each design cycle. (Integer; Default = 40)

ITRMOP Number of consecutive iterations for which convergence criteria must be satisfied to indicate convergence at the optimizer level. (Integer; Default = 2)

ITRMST Number of consecutive iterations for which convergence criteria must be met at the optimizer level to indicate convergence in the Sequential Linear Programming Method. (Integer > 0; Default = 2)

IWRITE FORTRAN unit for print during approximate optimization phase. Default value for IWRITE is set to the FORTRAN unit for standard output. (Integer > 0, Default = 6 or value of SYSTME (2).)

JTMAX Maximum number of iterations allowed at the optimizer level for the Sequential Linear Programming Method. This is the number of linearized sub-problems solved. (Integer ≥ 0; Default = 20)

JPRINT Sequential Linear Programming sub-problem print. If JPRINT > 0, IPRINT is turned on during the approximate linear sub-problem. (Default = 0)

JWRITE If JWRITE > 0, file number on which iteration history will be written. (Integer > 0; Default = 0)


DOPTPRM (Cont.)


METHOD Optimization Method: (Integer 1, 2, or 3; Default = 1)1: Modified Method of Feasible Directions. (Default)2: Sequential Linear Programming3: Sequential Quadratic Programming

P1 Print control items specified for P2. (Integer >= 0; Default = 0) Initial results are always printed prior to the first approximate optimization. If an optimization task is performed, final results are always printed for the final analysis unless PARAM,SOFTEXIT,YES is specified. These two sets of print are not controllable.n: Print at every n-th design cycle.

P2

PLVIOL Flag for handling of property limit violation. By default, the job will terminate with a user fatal message if the property derived from design model (DVPRELi, DVMRELi, DVCRELi) exceeds the property limits. Setting PLVIOL to a non-zero number will cause the program to issue a user warning message by ignoring the property limits violation and proceed with the analysis. (Integer; Default = 0)

Items to be printed according to P1: (Integer; Default = 1)0: No print.1: Print objective and design variables. (Default)2: Print properties.4: Print constraints.8: Print responses.10: Print weight as a function of a material ID (note that there is not a design quantity so that only inputs to the approximate design are available)n: Sum of desired items. For example, P2 = 10 means print properties and responses.

PTOL Maximum tolerance on differences allowed between the property values on property entries and the property values calculated from the design variable values on the DESVAR entry (through DVPRELi relations). PTOL is provided to trap ill-posed design models. (The minimum tolerance may be specified on user parameter DPEPS. (Real > 0.0; Default = 1.0E+35)

STPSCL Scaling factor for shape finite difference step sizes, to be applied to all shape design variables. (Real > 0.0; Default = 1.0)


DYNAMIC RESPONSE OPTIMIZATION Available dynamic analysis disciplines in Solution 200

Direct Frequency Modal Frequency Modal Transient Acoustic (Fluid-Structure Interaction)

Available response types: (see also the DRESP1 entry) Displacement Velocity Acceleration SPC Force Stress Element force Equations (DRESP2 + DEQATN) Weight Volume


DYNAMIC RESPONSE OPTIMIZATION(Cont.)

Limitations in Dynamic Response Sensitivity ∆F is assumed zero when calculating “sensitivities” in direct and

modal frequency and modal transientThis assumption is usually good, except for those situations in which the following may be significant: Gravity (or other mass-related) loads Follower forces (shape sensitivity) Thermal loading


SAMPLE PROBLEM XXX – MINIMIZATION OF DRIVER RESPONSE TO A ROTATING

IMBALANCE FOR THE CAR MODEL This uses the model from Sample 17a. The automobile has the

front left wheel out of balance. The amount of mass is .3 units and the radius to the mass is 10 units. Apply this rotating loading to the car and determine the response. The frequency range of interest is from 0.5-50hz.

We want to minimize the driver response to the input excitation.


INPUT FOR OPTIMIZATION Sample xxx – dynamic optimization of the car model Executive and Case Control

$ samplexxx.dat – linear bushings in the car model

$

SOL 200

TIME 200

CEND

TITLE = Samplexxx – dynamic analysis model

SUBTITLE = Rotating force due to tire out of balance

LABEL = perform optimization to minimize driver response

set 999 = 358,471

DISP(phase) = 999

SUBCASE 1

ANALYSIS = MFREQ

DESSUB = 100 $ constraints

DESOBJ(min) = 300 $ design objective – minimize driver response

DLOAD = 1

METHOD = 10

FREQ = 14

BEGIN BULK


INPUT FOR OPTIMIZATION (Cont.)

Model and loading$

$ following param will exit after calculating sensitivities

$

$param,optexit,4

$ get 40 modes

eigrl,10,,,40,0

include ’car.dat’

include ’linspring.dat’

$

DLOAD 1 1. 1. 11 1. 12

RLOAD1 11 20 111

RLOAD1 12 30 40 111

DPHASE 40 358 2 90.

DAREA 20 358 1 .03

DAREA 30 358 2 .03

TABLED4 111 0. 1. 0. 100.

0. 0. 39.478 ENDT

FREQ1 14 .5 .5 100

$


INPUT FOR OPTIMIZATION (Cont.) Design variables—spring stiffness and shock absorber damping

$$ data for design sensitivity$$ define design variables$$ allow more than 5 cycles in the optimization$doptprm,desmax,25$desvar,1,frntdamp,10.,1.,100.desvar,2,reardamp,5.,1.,100.desvar,3,frntstif,10.,4.,20.desvar,4,rearstif,8.,4.,20.desvar,5,frntstfx,10.,4.,20.desvar,6,rearstfx,10.,4.,20.$$ relation between properties and variables$dvprel1,101,pvisc,2001,3,1.,,,,+dv101+dv101,1,1.dvprel1,102,pvisc,2002,3,1.,,,,+dv102+dv102,2,1.dvprel1,103,prod,1001,4,400.,,,,+dv103+dv103,3,100.dvprel1,104,prod,1002,4,400.,,,,+dv104+dv104,4,100.dvprel1,105,pelas,2001,3,4000.,,,,+dv105+dv105,5,1000.dvprel1,106,pelas,2002,3,4000.,,,,+dv106+dv106,6,1000.$$


INPUT FOR OPTIMIZATION (Cont.) Responses – driver’s seat (GRID 471) and wheel (GRID 358)

Note that only one response is defined for the wheel displacement, while separate responses are defined for each loading frequency for the driver’s seat.

The wheel response will internally generate a response quantity for each excitation frequency, since no frequency is provided on the DRESP1 entry.

Since a DRESP2 is going to be used to define the SRSS value of the driver’s seat displacements, it is necessary to have a separate response specified for each excitation frequency. Note that only the first 51 are specified here– it may be necessary to also define responses for the higher frequencies.

$ select displacement Y at driver seat and mount point as$ response quantities$$ mount point – only one response – this will be constrained$dresp1,200,disp,frdisp,,,2,,358$$ define driver’s seat disp as a response –$ select the first 51 excitation frequencies$ this will be used as both a constraint and the objective$dresp1,201,driver,frdisp,,,2,.5,471=,*(1),=,=,=,=,=,*(.5),===(49)$


INPUT FOR OPTIMIZATION (Cont.) Constraints—

Wheel—displacement must be less than 0.5 Driver—measured displacements must be less than 0.25

$ add constraints$$ require that maximum tire displacement be .5 inches$dconstr,101,200,–.5,.5$$ require that maximum driver displacement be .25 inches$dconstr,102,201,–.25,.25=,*(1),*(1),===(49)$$ combine constraints into set 100$dconadd,100,101,102,103,104,105,106,107,+dc100a+dc100a,108,109,110,111,112,113,114,115,+dc100b*(1),*(8),*(8),*(8),*(8),*(8),*(8),*(8),*(8),*(8)=(3)+dc100f,148,149,150,151,152$


INPUT FOR OPTIMIZATION (Cont.)

Objective—minimize driver response – over the range 0–25hz

$$ define objective = minimize srss of response$dresp2,300,srss, 301,,,,,,+dr300a+dr300a,dresp1,201,202,203,204,205,206,207,+dr300b+dr300b,,208,209,210,211,212,213,214,+dr300c*(1),=,*(7),*(7),*(7),*(7),*(7),*(7),*(7),*(7)=(4)+dr300h,,250,251$deqatn 301 resp(a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,aa,bb,cc,dd,ee,ff,gg,hh,ii,jj,kk,ll,mm,nn,oo,pp,qq,rr,ss,tt,uu,vv,ww,xx,yy)=sqrt(a**2+b**2+c**2+d**2+e**2+f**2+g**2+h**2+j**2+i**2+l**2+m**2+n**2+o**2+p**2+q**2+r**2+s**2+t**2+u**2+v**2+w**2+x**2+y**2+z**2+aa**2+bb**2+cc**2+dd**2+ee**2+ff**2+gg**2+hh**2+ii**2+jj**2+kk**2+ll**2+mm**2+nn**2+oo**2+pp**2+qq**2+rr**2+ss**2+tt**2+uu**2+vv**2+ww**2+xx**2+yy**2)$$ end of optimization input


OPTIMIZATION RESULTS (Cont.) Results—graphical


SHAPE OPTIMIZATION MSC.NASTRAN allows shape changes in your model to be used as

design variables. For instance, consider the finite element model shown below with the

loads shown. Find a shape of the hole that will minimize weight but keep stresses below the yield point of the material.

In order to minimize weight, increase the size of the hole Tell the optimizer how to change the shape of the hole by defining shape

design variables.


SHAPE OPTIMIZATION (Cont.) The shape design variables shown below tell the optimizer that

elliptical shape changes to the hole in both the x and the y directions are allowable.


SHAPE OPTIMIZATION (Cont.)

The before and after optimization shape of the model are shown below:

The weight was decreased while the stresses were kept below the yield point.

Before After


HOW DO I CREATE SHAPE DESIGN VARIABLES?

Four different methods for generating shape basis vectors have been implemented: Manual grid variation Direct input of shapes Geometric boundary shapes Analytic boundary shapes

In these notes, only the manual grid variation is shown.


MANUAL GRID VARIATION Since the model is a coarse model, the manual input of shapes is used A DVGRID Bulk Data entry defines the direction and magnitude of a grid

variation for a given change in a design variable. Its format is:DVGRID, DVID, GID, CID, COEFF, N1, N2, N3

COEFF times the vector {N1,N2,N3} determines the direction and magnitude of a grid variation ∆G as shown in the figure below:

The figure below shows that with manual grid variation, if you want the horizontal movement of the left boundary of your model to be a shape variable, then only the specified grids will be allowed to be variable

Effect of manual grid variation shape variables

{G}I {∆G}i


MANUAL GRID VARIATION (Cont.) All possible shape design variable movements must be explicitly defined. A single

DVGRID Bulk Data entry is required for every grid movement for each shape design variable.

An example DVGRID Bulk Data entry is:dvgrid, 41, 116, 2, 1., 1., 0.0, 0.0

This data entry says that grid 116 is controlled by design variable 41. It can move in the direction of the {1.0, 0.0, 0.0} vector in coordinate system 2.

The DESVAR Bulk Data entry defines the controlling design variable. An example is:

This data entry says that design variable 41, labeled OHUBRAD, has an initial value of 8.68, a lower limit of 8.0, an upper limit of 9.6 and it can move in increments of .08, where .08 is the fraction of the current value. In this case the maximum movement at the next design cycle is .08 times 8.68 or .07. At .07 increments it will take three design cycles to exceed the upper limit of 9.6. If you want more cycles, you can decrease this value. If you put 0.0 as an initial value, then a small default movement will be used.


MANUAL GRID VARIATION (Cont.)

DVGRID entries can be used in combination with the direct input of shapes approach but not in combination with geometric boundary shapes and analytic boundary shapes.

Advantages: Since this can be considered the lowest-level approach, its strength lies in

its generality. Using DVGRIDs alone, the designer has direct control over every designed grid point in the model (that is, every grid point with a location that is to change during shape optimization).

Disadvantages: In all but the simplest of problems, the data input can be formidable without

a preprocessor. The resultant basis vectors are treated as constant and not updated with each design cycle. Therefore, the risks of mesh distortion are higher than with the geometric boundary shape and analytic methods.


INPUT FILE STRUCTURE: MANUAL GRID VARIATION

sol 200 (Executive)---------------------------------------------------------------------------Analysis = staticdessub = 20 (Case Control)desobj = 10

---------------------------------------------------------------------------dresp1, 10, wtmin, weight (Bulk Data)dconstr, 20, 31, -10000., 10000. dresp1, 31, Vmhexa, stress, psolid, ,13, ,2desvar, 41, ohubrad, 8.68, 8.0, 9.6, .08dvgrid, 41, 116, 2, 1., 1., 0.0, 0.0dvgrid, 41, 1141, 2, 1., 1., 0.0, 0.0---------------------------------------------------------------------------

Executive SectionOnly the SOL 200 statement is required


INPUT FILE STRUCTURE: MANUAL GRID VARIATION (Cont.)

Case Control Section The ANALYSIS request specifies which type of analysis you are going to

perform (in this case, static analysis). The DESOBJ request specifies that your objective is to minimize weight.

You can specify minimum or maximum, with minimum the default. The DESOBJ references the DRESP1 Bulk Data entry with rtype=weight.

The DESSUB references the DCONSTR Bulk Data entry, which in turn references a DRESP1 Bulk Data entry. These three Bulk Data entries define a constraint.

Bulk Data Section The DRESP1 Bulk Data entry, which invokes a weight response. The DCONSTR Bulk Data entry references another DRESP1 Bulk Data

entry, which together define a constraint such that the von Mises stress is less than 10000 psi for all the elements.

The shape variables are defined with the combination of a DESVAR Bulk Data entry and a set of DVGRID Bulk Data entries.


INPUT FILE STRUCTURE: MANUAL GRID VARIATION (Cont.)

Design variable 41 defines the change in thickness of the outer hub. As shown in the figure below, all the grids in the inner part of the hub are included in the set of DVGRIDS. In addition, the grids that represent the fillet from the stiffener to the hub are also included, and thus move with the hub. This keeps fillet radius constant as the hub and stiffener design variables move relative to one another.

Only the first and last DVGRID are shown for the design variable. In reality, all the grids (approximately 150) defining the part of the model that are moving for a particular design variable are present in the model file for this design variable alone.


INPUT FOR SHAPE OPTIMIZATION Use the size of the front beams as design variables. Two new design variables are to be created, one for the upper

beams and the other for the lower beams. Each variable will represent a change in the depth of the

member.


UPPER BEAMS


UPPER BEAMS (Cont.)


LOWER BEAMS


LOWER BEAMS (Cont.)


OPTIMIZATION INPUT

$$ define shape variables$desvar,7,topchnl,1.,–10.,10.desvar,8,botbox,1.,–10.,10.$$ define shapes$$ top beams$dvgrid,7,542,,.1,0.,1.,0.dvgrid,7,551,,.1,0.,1.,0.dvgrid,7,894,,.1,0.,1.,0.dvgrid,7,896,,.1,0.,1.,0.dvgrid,7,893,,.1,0.,1.,0.dvgrid,7,895,,.1,0.,1.,0.dvgrid,7,795,,.1,0.,1.,0.dvgrid,7,794,,.1,0.,1.,0.dvgrid,7,541,,.1,0.,1.,0.dvgrid,7,573,,.1,0.,1.,0.$dvgrid,7,48,,.1,0.,1.,0.dvgrid,7,39,,.1,0.,1.,0.dvgrid,7,429,,.1,0.,1.,0.dvgrid,7,427,,.1,0.,1.,0.dvgrid,7,428,,.1,0.,1.,0.dvgrid,7,426,,.1,0.,1.,0.dvgrid,7,323,,.1,0.,1.,0.dvgrid,7,324,,.1,0.,1.,0.dvgrid,7,70,,.1,0.,1.,0.dvgrid,7,38,,.1,0.,1.,0.


OPTIMIZATION INPUT (Cont.)

$$ bottom beams$dvgrid,8,361,,.1,0.,1.,0.dvgrid,8,360,,.1,0.,1.,0.dvgrid,8,366,,.1,0.,1.,0.dvgrid,8,367,,.1,0.,1.,0.dvgrid,8,362,,.1,0.,1.,0.dvgrid,8,363,,.1,0.,1.,0.dvgrid,8,370,,.1,0.,1.,0.dvgrid,8,371,,.1,0.,1.,0.dvgrid,8,343,,.1,0.,1.,0.dvgrid,8,363,,.1,0.,1.,0.dvgrid,8,342,,.1,0.,1.,0.dvgrid,8,365,,.1,0.,1.,0.dvgrid,8,827,,.1,0.,1.,0.dvgrid,8,828,,.1,0.,1.,0.dvgrid,8,834,,.1,0.,1.,0.dvgrid,8,833,,.1,0.,1.,0.dvgrid,8,830,,.1,0.,1.,0.dvgrid,8,829,,.1,0.,1.,0.dvgrid,8,838,,.1,0.,1.,0.dvgrid,8,837,,.1,0.,1.,0.dvgrid,8,744,,.1,0.,1.,0.dvgrid,8,810,,.1,0.,1.,0.dvgrid,8,746,,.1,0.,1.,0.dvgrid,8,809,,.1,0.,1.,0.$


SHAPE-OPTIMIZATION RESULTS


OPTMIZATION SUMMARY Selection of design variables should be based on an

understanding of the problem. Proper selection of design variables can result in noticeable

improvements in the design and reductions in weight. For the example shown, the shape variables did not have a

significant effect on the final response, but the weight of the model was reduced by changing the shapes of the beams.

MSC.NASTRAN optimization is a powerful tool which can result in significant improvement in the design without a noticeable amount of effort by the user.


SECTION 12

RESOURCE ESTIMATION



THE NECESSITY FOR MAKING ESTIMATES 8-5

PREDICTION TECHNIQUES 8-6

MSC/ESTIMATE 8-7

MSC/ESTIMATE KEYWORDS 8-9

MSC/ESTIMATE RULES 8-16

MSC/ESTIMATE SAMPLES 8-19


THE NECESSITY FOR MAKING ESTIMATES

Assure that enough disk space is available. Evaluate partitioning schemes against available

resources. Avoid having to run jobs overnight or in the slow

queue. Avoid fragmentation, disorganization, and chaos

associated with executing jobs without planning. Reduce turnaround time.


PREDICTION TECHNIQUES Type A – Before the event

Use MSC/Estimate Use experience from previous analyses – maintain a log. Be conservative.

Type B – During the event The database dictionary provides space facts. The performance summary table provides time facts. Some modules provide estimates, such as DCMP, FBS, and

MPYAD.

Type C – After the event Useful for predicting rerun requirements. Provides experience for future projects. Recommended follow up activity for every project.


MSC/ESTIMATE ESTIMATE may be used to estimate the memory and disk

requirements for MSC.NASTRAN jobs and make suggestions on improving the performance of these jobs.

ESTIMATE will read the input data file and estimate the job’s memory and disk requirements.

The ESTIMATE program is most accurate in predicting the requirements of static analyses that do not have excessive output requests.

The memory requirements for normal modes analysis using the Lanczos Method are reasonably accurate; however, the disk requirements are dependent upon the number of modes, this is a value that ESTIMATE cannot determine.

Memory and disk requirements for other solutions are less accurate. The basic format of the “estimate” command is:

msc2001 estimate input_file [keywords] where input_file is the name of the data file. If the file suffix of the input

data file is “.dat”, it may be omitted from the command line.Note: On restart runs, MSC/ESTIMATE may incorrectly calculate the estimates, as it only

looks at the data in the input file. In this case, use the estimate information from the initial run.


MSC/ESTIMATE ESTIMATE processes keywords using the following

precedence to resolve conflicts when keywords are duplicated: The bulk data file.

The command line.

The NASTRAN INI and RC files (if”nastrc=yes” is specified)

data-file-directory/.estimaterc (where data-file-directory is the directory containing the input data file).

$HOME/.estimaterc

Estimate.ini (in the directory containing the ESTIMATE executable).


MSC/ESTIMATE KEYWORDSapplication application=NASTRAN Default: NASTRAN

Defines the application. The default is based on the SOL or LINK name specified in the executive section.

bpool bpool=value Default: 27 (Cray and NEC); 37 (all others)Same as MSC.NASTRAN keyword.This keyword cannot appear in an ESTIMATE RC file if “nastrc=yes” is specified.

buffsize buffsize=value Default: 4097 (Cray and NEC); 2049 (all others)Same as MSC.NASTRAN keyword.This keyword cannot appear in an ESTIMATE RC file if “nastrc=yes” is specified.

dskco dskco=value Default: 1.0Allows you to specify a constant factor that is either more or less conservative than the default.Example: msc2001 estimate example dskco=2This will double the disk space estimate.

enable The “enable” keyword can be used to explicitly enable rules. This may be useful to enable a rule that was automatically suppressed when a value was assigned. For example, the following command will now calculate the estimated memory requirements for a job even though a value for memory was specified on the command line:Example: msc2001 estimate example memory = 5mb, enable = 10

Note: This keyword should only be set to “NASTRAN”.


MSC/ESTIMATE KEYWORDSestimatedof estimatedof={yes I no} Default: No

Indicates if the number of degrees of freedom are to be estimated. By default, ESTIMATE will count the DOF. This process takes time, but it is generally more accurate. Specifying “estimatedof=no” will result in a less accurate, but faster, estimate of the DOF. The presence of any MESH entries in the Bulk Data will force “estimatedof=yes”.

memco memco=value Default: 1.0Allows you to specify a constant factor that is either more or less conservative than the default.Example: msc2001 estimate example memco = 2This setting will double the memory estimate.

memory memory=memory_size Default: 4MWSame as MSC.NASTRAN keyword.This keyword cannot appear in an ESTIMATE RC file if “nastrc=yes” is specified.

method method=sid Default: NoneSelects a METHOD for dynamics jobs if a METHOD Case Control command is not present or multiple METHOD Case Control commands are present in the data file. By default, ESTIMATE will choose the first method found.


MSC/ESTIMATE KEYWORDSmode mode={estimate/suggest/modify} Default: suggest

Selects the program operating mode. Specifying “mode=estimate” will result in memory and disk estimates only. Specifying “mode=suggest”, the default, will estimate memory and disk requirements for the current job configuration, suggest modifications to improve the performance, and provide estimates for the memory and disk requirements of the suggested configuration. Specifying “mode=modify” does all that “mode=suggest” does plus actually make the suggested changes to your data file. See “out” to specify the new data file’s name and information on organizing your input file.

Example: msc2001 estimate example mode=estimateThe memory and disk requirements for the current job are displayed.Example: msc2001 estimate exampleThe memory and disk requirements for the current job, suggestions for improving performance, and memory and disk requirements for the suggested configuration are displayed.Example: msc2001 estimate example mode = modifyThe memory and disk requirements for the current job, suggestions for improving performance, and estimates of memory and disk requirements for the suggested configuration are displayed. If, and only if, modifications to “example.dat” aresuggested, the original input file is versioned (given indices) and the revised data file is written to “example.dat”.

mpc mpc=sid Default: NoneSelects an MPC if an MPC Case Control command is not present or multiple MPC Case Control commands are present in the data file. By default, ESTIMATE will choose the first MPC found.

Note: If “mode=modify” is specified, and ESTIMATE detects errors in the input file or encounters valid Bulk Data that is not understood by ESTIMATE, the program will revert to “mode=suggest”.


MSC/ESTIMATE KEYWORDSnastrc The “nastrc” keyword allows you to select the type of RC file processing invoked by the ESTIMATE utility. Setting

nastrc=yes”, the default, will process the standard MSC.NASTRAN RC files before the standard ESTIMATE RC files, i.e., $HOME/.esimaterc and “data-file- directory/.estimaterc”, are processed. Setting “nastrc=no” will only process the standard ESTIMATE RC files.

out out=pathname Default: input filenameSpecifies the name of the output file if “mode=modify” is specified and modifications of the data file are actually required. By default, the original file is versioned (given indices) and the revised data file is written to the original input file’s name. Example: msc2001 estimate example mode = modifyIf modifications to “example.dat” are suggested, the original input file is versioned (given indices) and the revised data file is written to “example.dat”Example: msc2001 estimate example mode = modify, out = modifiedThe revised data file is written to “modified”.

Note: In order to minimize the amount of data duplicated between the original input file and the modified file, MSC recommends that the Bulk Data that is not subject to modification by ESTIMATE (i.e., all Bulk Data except PARAM and EIGRL entries) be placed in an INCLUDE file.An example of the recommended input file organization is:

NASTRAN statementsFMS statementsExecutiveCENDCase ControlBEGIN BULKPARAM,…$EIGRL,…$INCLUDE file.bulk$ENDDATA


MSC/ESTIMATE KEYWORDSpause

real

report

smemory

spc

pause=keyword Default: noPause ESTIMATE before exiting to wait for the “Enter” or “Return” key to be pressed. This can be useful when ESTIMATE is embedded within another program. The values are “fatal”, “information”, “warning”, “yes”, and “no”. Setting “pause=yes” will unconditionally wait; “pause=fatal” will only wait if a fatal message has been issued by ESTIMATE; “pause=information” and “pause=warning” will similarly wait only if an information or warning message has been issued. The default is “pause=no”, I.e., do not wait when ESTIMATE ends.Same as MSC.NASTRAN keyword . This keyword cannot appear in an ESTIMATE RC file if “nastrc=yes” is specified.

report={normal/keyword} Default: NormalSpecifies the program’s report format. The “report=normal” format is intended to be read by you. The “report=keyword” format is intended to be read by a program.

smemory=value Default: 0 (Cray and NEC); 100 (all others)Same as MSC.NASTRAN keyword. This keyword cannot appear in an ESTIMATE RC file if “nastrc=yes” is specified.

spc=sid Default: NoneSelects an SPC if an SPC Case Control command is not present or multiple SPC Case Control commands are present in the data file. By default, ESTIMATE will choose the first SPC found.


MSC/ESTIMATE KEYWORDSsuppress suppress=list Default: None

verbose verbose={yes/no} Default: No

version version=string Default: 2001Specifies the version of MSC.NASTRAN for which the estimates are to be targeted. The version will affect the estimated memory requirements and the actions of various rules. This keyword cannot appear in an ESTIMATE RC file if “nastrc=yes” is specified.

Specifies rules that are to be suppressed when “mode=suggest” or “mode=modify” is specified. See page 15 for the list of Rules. If no value is specified, I.e., “suppress=“, then any rules previously suppressed are enabled. Multiple rules can be suppressed by using the keyword multiple times or by specifying a comma-separated list.Example: msc2001 estimate example suppress=1Suppress rule 1, the rule controlling BUFFSIZE.Example: msc2001 estimate example suppress=1,6

or msc2001 estimate example suppress=1 suppress=6or msc2001 estimate example suppress=2 suppress=6

Suppress rules 1 and 6.

Specifies the amount of information to be displayed. Specifying “verbose=yes” will generate a much larger amount of output. The additional information includes a more detailed summary of the input file, the parameters used in estimating the memory and disk requirements, and the estimates for the original file, even when “mode=suggest” or “mode=modify” is specified.

Note: Supported versions: All Versions >= 70.5


MSC/ESTIMATE KEYWORDSwordsize wordsize={32/64},[{32/64}] Default: 64(Cray and NEC); 32 (all

others) Specifies the word size of the estimate’s target computer. By default, ESTIMATE’s calculations will be appropriate to the current computer. This keyword may be used to specify estimates for a computer with a different word size. A comma-separated list of values may be specified when estimates and suggestions for multiple machines are desired. If “mod=modify” was specified, the modification are based on the last word size specified.


MSC/ESTIMATE RULES ESTIMATE has a fixed rule base that it uses to make

suggestions for improvement. Any of the rules may be suppressed with the “suppress” keyword. The current rules are:

1. Set recommended BUFFSIZE.BUFFSIZE=8193 DOF < 100000

BUFFSIZE 16385 10,000 < DOF ≤ 4000

BUFFSIZE 32769 DOF > 40,000

2. Use default BPOOL.BPOOL=37 if wordsize = 32BPOOL=20 if wordsize = 64, and version < 70.5BPOOL=27 if wordsize = 64, and version ≥ 70.5


MSC/ESTIMATE RULES3. Suppress symmetric decomposition if not enough memory for

sparseSYSTEM(166)=0

4. Make all open core available to modules.Delete HICORE.

5. Select the sparse solver.Delete SPARSE if density ≤ 12.0Delete USPARE if density ≤ 12.0SPARSE=1 if density > 12. USPARSE=0 if density > 12.0

6. Force default rank size.Delete SYSTEM(198)Delete SYTEM(205)

7. Do not sequence.PARAM, NEWSEQ, -1 version < 69


MSC/ESTIMATE RULES (Cont.)8. Use default Lanczos parameters.

EIGRL,…, V1=“”EIGRL,…, MAXSET=15

9. Use default SMEMORY.INIT SCRATCH (MEM=100) If wordsize = 32INIT SCRATCH (MEM=0) If wordsize = 64

10. Use estimated memory size.memory=estimated-memory

11. Use default RAM.INIT MASTER (RAM=30000)

12. Real.Delete REAL.

13. Do not use Supermodule.Delete PARAM, SM, YES.

14. Do not use Parallel Lanczos.Delete NUMSEG.


MSC/ESTIMATE SAMPLES The ESTIMATE program can be used in several ways. The

default mode will make suggestions on improving the performance of MSC.NASTRAN and estimate the resource requirements of the job assuming the suggested parameters.

msc2001 estimate example

To get an estimate of the job using the current parameters, use the command:

msc2001 estimate example mode=estimate other_estimate_keywords

To have a new input file generated with the suggested changes, use the command:

msc201 estimate example mode=modify other_estimate_keywords

To run MSC.NASTRAN with the memory estimated by ESTIMATE, use:

msc2001 nastran example memory=estimateother_nastran_keywords

practical fe techniques using msc.nastran

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