Nastran

Copyright2007 MSC.Software Corporation

MSC.Software Corporation

MD Nastran - Dynamic Analysis

April 2007

NAS102 Seminar

NA*V2004*Z*Z*Z*SM-NAS102-NT1

Part Number: MDNA*R2*Z*Z*Z*SM-NAS102-NT1

EuropeMSC.Software GmbH

Am Moosfeld 1381829 Munich, Germany

Telephone: (49) (89) 43 19 87 0Fax: (49) (89) 43 61 71 6

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Telephone: (81) (3)-6911-1200Fax: (81) (3)-6911-1201


DISCLAIMER

MSC.Software Corporation reserves the right to make changes in specifications and other information contained in thisdocument without prior notice.

The concepts, methods, and examples presented in this text are for illustrative and educational purposes only, and are notintended to be exhaustive or to apply to any particular engineering problem or design. MSC.Software Corporation assumesno liability or responsibility to any person or company for direct or indirect damages resulting from the use of anyinformation contained herein.

User Documentation: Copyright 2007 MSC.Software Corporation. Printed in U.S.A. All Rights Reserved.

This notice shall be marked on any reproduction of this documentation, in whole or in part. Any reproduction or distributionof this document, in whole or in part, without the prior written consent of MSC.Software Corporation is prohibited.

MSC, MSC., MSC.Actran, MSC.ADAMS, MSC.Dytran, MSC.EASY5, MSC.Fatigue, MSC.FlightLoads, MSC.LaminateModeler, MSC.Marc, MSC.Marc Mentat, MSC.Mvision, MSC.Nastran, MSC.Patran, MSC.Robust Design, SimDesigner,SimManager, MSC.SOFY, the MSC.Software corporate logo, and Simulating Reality are trademarks or registeredtrademarks of the MSC.Software Corporation in the United States and/or other countries.

All other brand names, product names or trademarks belong to their respective owners.


1. Review of Fundamentals

2. Dynamic Modeling Input3. Normal Mode Analysis

4. Reduction in Dynamic Analysis

5. Rigid Body Modes

6. Damping7. Transient Response Analysis

8. Frequency Response Analysis

9. Direct Matrix Input

10. Dynamic Equations of Motion11. Residual Vector Methods

12. Enforced Motion

13. Shock and Response Spectrum Analysis

14. Random Response Analysis15. Complex Eigenvalue Analysis

16. Normal Mode Analysis Using Parts Superelement

17. Extra Points, Transfer Functions, and NOLINs

18. Normal Modes of Preloaded Structures19. Dynamic Design Optimization

20. Test-Analysis Correlation

TABLE OF CONTENTS


BLANK

NAS102, Section 1, March 2007Copyright2007 MSC.Software Corporation S1-1

SECTION 1

REVIEW OF FUNDAMENTALS


OVERVIEW OF DYNAMIC ANALYSISPROCESS

DynamicEnvironment

SystemDesign

Finite ElementModel

InterfacingMedia

ModalAnalysis

DynamicResponseAnalysis


m = mass (inertia)b = damping (energy dissipation)k = stiffness (restoring force)n = nonlinear restoring forcep = applied forceu = displacement of mass

= velocity of mass= acceleration of mass

u, u, u, and p are time varying (in general)m,b, and k are constantsn is a nonlinear function of u, u

m

kb

n

p(t)

u(t)

SINGLE DOF SYSTEM• Dynamic equation of motion:

)un(u,p(t)ku(t)(t)ub(t)um

uu


●Fundamental units●Length L (inch, meter)●Mass M (slug, kilogram)●Time T (second)●Length L (meter, millimeter)●Force F (Newton)●Time T (second)

●Fundamental and derived units●m M MassMass m FT2/L = F/(LT –2)

●b MT-1 DampingDamping b FT/L = F/(T/L)●k MT -2 StiffnessStiffness k F/L

●p MLT -2 ForceForce p F

●u L DisplacementDisplacement u L● LT -1 VelocityVelocity LT -1● LT -2 AccelerationAcceleration LT -2

tkgmmNs 110001

2

uu

uu

UNITS


• Engineering Units for Common Variables.Variable Dimensions* Common

English UnitsCommon

Metric UnitsLength L in mMass M lb-sec2/in kgTime T sec secArea L2 in2 m2

Volume L3 in3 m3

Velocity LT -1 in/sec m/secAcceleration LT -2 in/sec2 m/sec2

Rotation - rad radRotational Velocity T -1 rad/sec rad/secRotational Acceleration T -2 rad/sec2 rad/sec2

Circular Frequency T -1 rad/sec rad/secFrequency T -1 cps; Hz cps; HzEigenvalue T -2 rad2 /sec2 rad2/sec2

Phase Angle - deg degForce MLT -2 lb NWeight MLT -2 lb NMoment ML2T -2 in-lb N-mMass Density ML -3 lb-sec2/in4 kg/m3

Young’s Modulus ML-1T

-2lb/in

2Pa; N/m

2

Poisson’s Ratio - - -Shear Modulus ML -1T -2 lb/in2 Pa; N/m2

Area Moment of Inertia L4 in4 m4

Torsional Constant L4 in4 m4

Mass Moment of Inertia ML 2 in-lb-sec2 kg-m2

Stiffness MT -2 lb/in N/mViscous Damping Coefficient MT-1 lb-sec/in N-sec/mStress ML -1T -2 lb/in2 Pa; N/m2

Strain - - -

*L denotes lengthM denotes massT denotes time- denotes dimensionless

UNITS (Cont.)


● Use consistent units!

● Error in units is the number one cause of modelingerrors in dynamic analysis!

● Most common errors are in mass and damping units.

● MD Nastran contains no built-in set of units. Theanalyst must verify consistency.

● Consistent Units: e.g. N, t, mm, s or N, kg, m, s

UNITS (Cont.)


• Dynamic equation

• Solution

• Initial conditions

• Finally

tcosBtsinA)t(u nn

sec)/rad(frequencynaturalmk

n

sec)/cycles(frequencynatural2

f nn

)0t(uB

mu(t) + ku(t) = 0..

u(t=0)n

A =.

u(t) = sin nt + u(0) cos ntu(0)n

.

u(0) and u(0) are known..

n

nnnn

nn

Aω(0)u0)(tuBu(0)0)u(t

tsinωBωtcosωAω(t)u

tBcosωtAsinωu(t)

SINGLE DOF SYSTEM - UNDAMPED FREEVIBRATIONS


0. 0 .2 0.4 0 .6 0 .8

0. 0 .2 0.4 0 .6 0 .8 1 .00 .2E1

0 .1

0 .

-0.1

-0.2

TI ME

0 .2E1

0 .1

0 .

-0 .1

-0 .2

S DO F Osci lla to r- --- Non ze ro In i ti al Co n di tio ns

E 1

Disp lace m ent

SINGLE DOF SYSTEM - UNDAMPED FREEVIBRATIONS


SINGLE DOF SYSTEM - DAMPED FREEVIBRATIONS

●Dynamic Equation

●Critical Damping

●Fraction of critical damping

●The amount of damping determines the form of the solution.●Underdamped

●Damped natural frequency

nc m2km2b

cbb

cbb

t)ωBtω(Atζωet)ωBtω(Ambt/eu(t) ddn

dd cossincossin2

21 nd

0)t(Ku)t(ub)t(um .. .

b =b = **bbcc == *2m*2mnn


SINGLE DOF SYSTEM - DAMPED FREEVIBRATIONS (Cont.)

● Critically damped• b = bc

● No oscillation occurs.• u(t) = (A + Bt) e-bt/2m

● Overdamped• b > bc

• No oscillation occurs. The system gradually returns to equilibrium (atrest, undisplaced) position.

● The usual analysis case is underdamped.

● Structures typically have damping in the 0-10% range.


DAMPED FREE VIBRATIONS—UNDERDAMPED CASE

E0

T im e (Se co n d s )

0 .2E 1

0 .1

0 .

-0 .1

-0 .2

C D AM P E le m en t

- 0 .0 5

3 .02 .52 .01 .51 .00 .50.

D isp l ace m en t


n2n

2n )/1(

k/P)0t(uA

SINGLE DOF SYSTEM - UNDAMPEDSINUSOIDAL VIBRATIONS

Dynamic equation

where = forcing frequency

Solution

where

tsinP)t(uk)t(um ..

tsin/1k/PtcosBtsinA)t(u

2n

2nn

Initial Condition Steady-State

)0t(uB

....


SINGLE DOF SYSTEM - UNDAMPEDSINUSOIDAL VIBRATIONS

• Steady-state solution

• P/k is the static response.

• is the dynamic magnification factor.2n

2 /11


MAGNIFICATION FACTOR

E1

2.0E 1

1.5

1.0

0.5

0.

SD OF Syste mN o B

1.00.80. 60 .40. 20 .

Frequency (Hz)

Disp laceme ntMa gnit ude

P/knn


SINGLE DOF SYSTEM - DAMPEDSINUSOIDAL VIBRATIONS

● Dynamic equation•

● The transient solution decays rapidly and is of no interest.● Steady-state solution

•

•

● is phase lead => 180o 360o – for Nastran output

2n

22n

2 )/2()/1(

)tsin(k/P)t(u

2n

2n1

/1/2

tan

mu(t) + bu(t) + ku(t) = P sin t.. .


● For• Magnification factor 1 (static solution)• Phase angle 360° (response is in phase with the force)

● For

• Magnification factor 0 (no response)• Phase angle 180° (response has opposite sign of force)

● For• Magnification factor• Phase angle » 270°

1n

1n

1n

21

SINGLE DOF SYSTEM - DAMPEDSINUSOIDAL VIBRATIONS


E 1

F r e qu enc y ( Hz )

1. 2E 1

0. 6

0. 4

0. 2

0.1. 00. 80 .60.40. 20 .

0. 8

1. 0

D is p lac e m entM ag ni tud e

MAGNIFICATION FACTOR


● Now the equation of motion becomes

● where● {u} = displacement vector● {M}= mass matrix● {B}= damping matrix● {K}= stiffness matrix● {P}= forcing function● {N}= nonlinear vector of forces

[M] {u} + [B] {u} + [K] {u} = {P} + {N}.. .

MULTI-DEGREE OF FREEDOM SYSTEM


Deterministic

Periodic Transient

SimpleHarmonic

ShockSpectra

Random

StationaryNonstationary

Ergodic

MD NastranMD Nastran

CLASSIFICATION OF DYNAMICENVIRONMENTS


DYNAMIC EXCITATIONS


● Frequency range● Grid points/constraints/elements● Linear versus nonlinear behavior● Whole system versus superelement models● Interaction with adjacent media● Test/measured data integration● Damping

FINITE ELEMENT DYNAMIC MODELINGCONSIDERATIONS


● MD Nastran Documentation

• Quick Reference Guide• Release Guide• Installation and Operation Release Guide

● User’s Guides• Linear Static Analysis• Basic Dynamic Analysis• Advanced Dynamic Analysis• Design Sensitivity and Optimization• DMAP Programmer’s Guide• Numerical Methods• Aeroelastic Analysis• Thermal Analysis• Superelement• Implicit Nonlinear (SOL 600)• Explicit Nonlinear (SOL 700)• SOL 400 (MD Nastran)

MD NASTRAN DOCUMENTATION


● Other Documentation

• Reference Manual• Common Questions and Answers• Bibliography

http://www.mscsoftware.com/support/prod_support/nastran/biblio/index.cfm

● Documentation also available in online form

MD NASTRAN DOCUMENTATION (Cont.)


TEXT REFERENCES ON DYNAMIC ANALYSIS

• W. C. Hurty and M. F. Rubinstein, Dynamics of Structures, Prentice-Hall, 1964.

• R. W. Clough and J. Penzien, Dynamics of Structures, McGraw-Hill,1975.

• S. Timoshenko, D. H. Young, and W. Weaver, Jr., Vibration Problemsin Engineering, 4th Ed., John Wiley & Sons, 1974.

• K. J. Bathe and E. L. Wilson, Numerical Methods in Finite ElementAnalysis, Prentice-Hall, 1976.

• J. S. Przemieniecki, Theory of Matrix Structural Analysis, McGraw-Hill,1968.

• C. M. Harris and C. E. Crede, Shock and Vibration Handbook, 2ndEd., McGraw-Hill, 1976.

• L. Meirovitch, Analytical Methods in Vibrations, MacMillan, 1967.• L. Meirovitch, Elements of Vibration Analysis, McGraw-Hill, 1975.• M. Paz, Structural Dynamics Theory and Computation, Prentice-Hall,

1981.


TEXT REFERENCES ON DYNAMIC ANALYSIS(Cont.)

• W. T. Thomson, Theory of Vibrations with Applications, Prentice-Hall, 1981.

• R. R. Craig, Structural Dynamics: An Introduction to ComputerMethods, John Wiley & Sons, 1981.

• S. H. Crandall and W. D. Mark, Random Vibration in MechanicalSystems, Academic Press, 1963.

• J. S. Bendat and A. G. Piersol, Random Data Analysis andMeasurement Techniques, 2nd Ed., John Wiley & Sons, 1986.

• R. Gasch, K. Knothe: Strukturdynamik, Band 1+2. Springer, 1987


BLANK

NAS102,Section 2, March 2007Copyright2007 MSC.Software Corporation S2-1

SECTION 2

DYNAMIC MODELING INPUT


MD NASTRAN INPUT FILE SETUP

● FMS and NASTRAN Statements - File allocations and system cell● Executive Control Section - Solution type, time allowed, system diagnostics

● CEND - Required Delimiter

● Case Control Section - Output requests, selects certain Bulk Data items● BEGIN BULK - Required Delimiter

● Bulk Data Section - Structural model definition, solution conditions● ENDDATA - Required Delimiter


MD NASTRAN BULK DATA ENTRY FORMAT

● Fixed field format● GRID^^^^2^^^^^^^3^^^^^^^1.0^^^^^-2.0^^^^3.0^^^^^^^^^^^^^316

● Free format, same entry● GRID,2,3,1.0,-2.0,3.0,,316

● Replicators for repetitive input● This: GRID,1,,0.,0.,0.,,126

=, *(5),=,=,*(1.),===(3)

● Produces:

● Examples in this course will use free format and replicators.

GRID 1 0 0 0 126GRID 6 0 1 0 126GRID 11 0 2 0 126GRID 16 0 3 0 126GRID 21 0 4 0 126


FINITE ELEMENT ANALYSIS● The real world is not comprised exclusively of SDOF systems.● Finite elements are used to model the mass, damping, and stiffness of

complex systems and structures.● Degrees of freedom (DOF) are independent coordinates that describe

the motion of the structure at any instant in time.● GRIDs are used to model the continuous structure as a discrete entity.● Each GRID may have six DOFs: translation in the X, Y, and Z

directions and rotations about the X, Y, and Z axes.

● Bookkeeping is done via the matrices that define the relationshipsbetween the DOFs.

y

x

z

Time 1Time 2

12


COMMONLY USED ELASTIC ELEMENTSOne-Dimensional Geometry Number of DOFs

ROD Pin-ended rod 4 2 Grids x 2 DOFsBAR Prismatic beam 12 2 Grids x 6 DOFs

BEAM Straight beam with warping 14 2 Grids x(6+1)DOFsBEND Curved beam, pipe, or elbow 12 2 Grids x 6 DOFs

Two-Dimensional Geometry

TRIA3 / TRIAR Triangular plate 18 3 Grids x "6" DOFsQUAD4/ QUADR Quadrilateral plate 24 4 Grids x "6" DOFs

SHEAR 4-sided shear panel 8 4 Grids x 2 DOFsTRIA6 Triangular plate with midside nodes 30 6 Grids x 5 DOFs

QUAD8 Quadrilateral plate with midside nodes 40 8 Grids x 5 DOFsThree-Dimensional

Geometry

HEXA Solid with six quadrilateral faces 24-60 8-20 Grids x 3 DOFsTETRA Solid with four triangular faces 12-30 4-10 Grids x 3 DOFsPENTA Solid with two triangular faces and three quadrilateral faces 18-45 6-15 Grids x 3 DOFs

Zero-DimensionalGeometry

ELAS Simple spring connecting two degrees of freedom 2 1-2 Grids x 1DOFBUSH Frequency-dependent spring/damper connecting up to six

degrees of freedom 6 1-2 Grid x 6 DOFs


COUPLED VERSUS LUMPED MASS

● Coupled mass is generally more accurate than lumped mass.● Lumped mass is preferred for computational speed in dynamic analysis.● User-selectable coupled mass matrix for elements

● PARAM, COUPMASS, 1 to select coupled mass● The default is lumped mass.

● Elements which have either lumped or coupled mass:● BAR, BEAM, CONROD, HEXA, PENTA, QUAD4/QUADR, QUAD8, ROD, TETRA,

TRIA3/TRIAR, TRIA6, TRIAX6, TUBE

● Elements which have lumped mass only:● CONEAX, SHEAR – CONEAX obsolete (no longer documented)

● Elements which have coupled mass only:● BEND


COUPLED VERSUS LUMPED MASS (Cont.)Lumped mass contains only diagonal, translational components (no rotational

ones).● Coupled mass contains off-diagonal translational components as well as

rotations for BAR (though no torsion for BAR, by default), BEAM, and BENDelements.● Setting system cell 398 to 1 generates torsional mass for BAR● Setting system cell 398 to 1 and param,coupmass,1 generates consistent axial

mass for BAR

NASTRAN SYSTEM(398) = 1 or NASTRAN BARMASS = 1● Neglected inertia may result in massless mechanisms.


ROD FINITE ELEMENT

● Stiffness matrix:

● Classical Consistent mass:

L

12

34 Length L

Area ATorsional constant JYoung’s modulus EShear modulus G

Mass density Polar moment of inertia I

LGJ

LGJ

LAE

LAE

LGJ

LGJ

LAE

LAE

00

00

00

00

k

dArI

AI

AI

AI

AI

AL 2where

30

60

031

061

60

30

0610

31

ρm


ROD FINITE ELEMENT (Cont.)

● Classical and MD Nastran lumped mass:

● MD Nastran coupled mass:

● The axial translational terms represent the average of lumped massand classical consistent mass. This average is found to be best forROD and BAR elements.

000002100000000021

ALm

000001250121000001210125

ALm

5/12 =5/12 = ½½(1/2 + 1/3)(1/2 + 1/3)

1/12 =1/12 = ½½( 0 + 1/6)( 0 + 1/6)


JUSTIFICATION FOR MD NASTRANCOUPLED MASS CONVENTION

● Consider a fixed-free rod

● Exact quarter-wave natural frequency

u(t)

Single Element Model

2

1

L

L5708.1

L241

EE


JUSTIFICATION FOR MD NASTRANCOUPLED MASS CONVENTION (Cont.)

● Different approximations● Lumped mass

● Classical consistent mass

● MD Nastran● Coupled mass

LL

E

414.1E

2LM

%)10(

LL

E

732.1E

3C

%)10(

LL

E

549.1E

512N

%)4.1(


MASS UNITS● MD Nastran assumes you are providing a consistent set of units.● Weight units may be input instead of mass units if this is more

convenient. You must then convert them to mass units usingPARAM,WTMASS.

● Weight-to-mass conversion:Mass = (1/G) Weight (G = Gravity Acceleration)Mass Density = (1/G) Weight Density

● PARAM,WTMASS, factor performs conversion with factor = 1/G. Thedefault value for factor is 1.0.

● Example:● Input RHO = 0.3 for steel weight density.● Use PARAM, WTMASS, 0.00259 to multiply 0.3 for G = 386.4 in/sec2.

● PARAM,WTMASS is used once per run and multiplies all weight/massinput (including MASSi, CONMi, and nonstructural mass input). Do notmix input types. Use all mass or all weight inputs.


MASS INPUT● Material density

● MATi entries

● Scalar mass● CMASSi, PMASS

● Grid point mass● CONM1 (6x6 mass matrix) - The user defines half of the

terms, symmetry is assumed.● CONM2 (concentrated mass)

1 2 3 4 5 6 7 8 9 10MAT1 MID E G NU RHO A TREF GEMAT1 2 30.0E6 0.3 7.76E-4

33I32I31I

22I21I11I

M.SYMM

M


MASS INPUT (Cont.)

● Structural mass (e.g., CONM2, etc.) are always included inthe model

● Nonstructural mass● Mass input on element property entry which is not associated with

geometric properties of element. Input as mass/length for lineelements and mass/area for elements with 2-D geometry.

● Non structural mass can be specified on many property entries(e.g., NSM on the PSHELL entry)● This type of nonstructural mass is always included in the model

● Some nonstructural mass are Case Control selectable● NSM Case Control selects NSM, NSM1, NSML, NSML1 entries

Bulk Data EntriesBulk Data Entries


NONSTRUCTURAL MASS ENHANCEMENTS

● Five nonstructural mass Bulk Data entries—NSM,NSM1, NSML, NSML1, and NSMADD—are available● Distribute nonstructural mass by element lists or specific

property lists associated with property entries.● Case Control selectable with NSM Case Control command

● NSM callout can be different betweensuperelements—but only one per superelement


NONSTRUCTURAL MASS ENHANCEMENTS (Cont.)

● NSM and its alternate form NSM1 allows the user toallocate an NSM_value to selected sets of elements.

Example:

● Either one of the above examples assigns a 0.022non-structural mass per unit area to PSHELL ID 15.Additive if both are present

.02215PSHELL3NSM

15.022PSHELL3NSM1

SID TYPE ID VALUE ID VALUE ID VALUESID TYPE ID VALUE ID VALUE ID VALUE

SID TYPE VALUE ID ID ID ID IDSID TYPE VALUE ID ID ID ID ID

TYPE =TYPE = PSHELL, PCOMP, PSHEARPSHELL, PCOMP, PSHEARPBAR, PBARL, PBEAM, PBEAML, PBENDPBAR, PBARL, PBEAM, PBEAML, PBENDPROD, CONROD, PTUPEPROD, CONROD, PTUPEPCONEAX, PRAC2DPCONEAX, PRAC2D


NONSTRUCTURAL MASS ENHANCEMENTS (Cont.)● The NSML and NSML1 entries compute a

nonstructural mass coefficient value for “area”elements (e.g., CQUAD4)

● The NSML and NSML1 entries compute anonstructural mass coefficient value for“line” elements(e.g., CBAR)

● NSML and NSML1 cannot mix “area” and “line”element on the same entry

● NSML and NSML1 are then converted to NSM andNSM1


NONSTRUCTURAL MASS ENHANCEMENTS (Cont.)

Example:

The above example assigns a 0.95 lump mass distributed evenlyto the area elements referenced by PSHELL 100. It can also berepresented by the following entry:

● NSMADD--combines NSM, NSM1, NSML, and NSML1 sets andsum their results to selected sets of elements.

.95100PSHELL3NSML

100.95PSHELL3NSML1

SID TYPE ID VALUE ID VALUE ID VALUESID TYPE ID VALUE ID VALUE ID VALUE

SID TYPE VALUE ID ID ID ID IDSID TYPE VALUE ID ID ID ID ID

MSMADD SID S1 S2 S3 S4 S5 S6 S7MSMADD SID S1 S2 S3 S4 S5 S6 S7S8 S9 S10 etc.S8 S9 S10 etc.


Defines a concentrated mass at a grid point.

Format:

1 2 3 4 5 6 7 8 9 10CONM2 EID G CID M X1 X2 X3

I11 I21 I22 I31 I32 I33

Example:CONM2 2 15 6 49.7

16.2 16.2 7.8

Field ContentsEID Element identification number. (Integer > 0)

G Grid point identification number. (Integer > 0)

CID

M Mass value. (Real)

X1, X2, X3

Iij

Remarks:1.2.3.4.

Concentrated Mass Element Connection, Rigid Body Form

(Continued)Bulk Data Entry

CONM2

Element identification numbers should be unique with respect to all other element identification numbersFor a more general means of defining concentrated mass at grid points, see the CONM1 entry description.The continuation is optional.If CID = -1, offsets are internally computed as the difference between the grid point location and X1, X2, X3. Thegrid point locations may be defined in a nonbasic coordinate system. In this case, the values of Iij must be in acoordinate system that parallels the basic coordinate system.

Coordinate system identification number. For CID of -1; see X1, X2, X3 below. (Integer > -1; Default = 0)

Offset distances from the grid point to the center of gravity of the mass in the coordinate system defined in field4, unless CID = -1, in which case X1, X2, X3 are the coordinates, not offsets, of the center of gravity of the massin the basic coordinate

Mass moments of inertia measured at the mass center of gravity in the coordinate system defined by field 4. IfCID=-1, the basic coordinate system is implied. (For I11, I22, and I33; Real > 0.0; for I21, I31, and I32; Real)


5.

where

6.

7.

The form of the inertia matrix about its center of gravity is taken as:

Concentrated Mass Element Connection, Rigid Body Form

Bulk Data Entry

CONM2

and x1, x2, x3 are components of distance from the center of gravity in the coordinate system defined in field 4.

The negative signs for the off-diagonal terms are supplied automatically. A warning message is issued if theinertia matrix is nonpositive definite, since this may cause fatal errors in dynamic analysis modules.

If CID > 0, then X1, X2, and X3 are defined by a local Cartesian system, even if CID references a spherical orcylindrical coordinate system. This is similar to the manner in which displacement coordinate systems aredefined.

See the MSC.Nastran Reference Manual, Section 4.2 for a definition of coordinate system terminology.

úúúúúúú

û

ù

êêêêêêê

ë

é

---

33I32I31I

22I21I11I

M

symmetricM

M

dVxx32I

dVxx31I

dVxx21I

dV)xx(33I

dV)xx(22I

dV)xx(11I

dVM

32

31

21

2

2

2

1

2

3

2

1

2

3

2

2

òòòòòò

ò

=

=

=

+=

+=

+=

=

r

r

r

r

r

r

r


BASIC MD NASTRAN SET OPERATIONS

● See the MD Nastran Quick Reference Guide.Grid Set (G) = N + M

Multipoint Constraints

Independent DOF (N) = F + S

Constraints

Unconstrained DOF (F) = A + O

Static Condensation

Analysis Set A = L + R

Free-Body Partitioning

Solve A-Set Modes

Reverse Process for Data Recovery to G-Set

LL –– Left overLeft over

RR –– Rigid BodyRigid Body

OO -- omittedomitted


BASIC MD NASTRAN SET DEFINITIONSet

Namemsb*

sg*

oqrc

b degree of freedom fixed during component mode analysis or dynamic reductione

sak

* Strictly speaking, sb and sg are not exclusive w ith respect to one another. Degrees of freedom may exist in bothsets simultaneously. Since these sets are not used explicitly in the solution sequences, this need not concern theuser. How ever, those w ho use these sets in their ow n DMAPs should avoid redundant specif ications w hen usingthese sets for partitioning or merging operations. That is, a degree of f reedom should not be specif ied on both a PSfield and a GRID entry (sg set) and on a selected SPC entry (sb set). Redundant specif ications w ill cause UFM 2120in the VEC module and behavior listed in MSC.Nastran DMAP Module Dictionary for the UPARTN module. These setsare exclusive, how ever, f rom the other mutually exclusive sets.

Each degree of freedom is also a Each degree of freedom is also a member of one or morecombined sets called "supersets." Supersets have the following definitions:

degree of freedom eliminated by single-point constraints that are included in boundarycondition changes and by the AUTOSPC feature

degrees of freedom eliminated by single-point constraints that are specified on the PSfield on GRID Bulk Data entries

Each degree of freedom is a member of one mutually exclusive "set". Set names have thefollowing definitions:

Definition

aerodynamic degrees of freedompermanently constrained aerodynamic degrees of freedomextra degrees of freedom introduced in dynamic analysis

degrees of freedom which are free during component mode synthesis or dynamicreduction

reference degrees of freedom used to determine free body motiongeneralized degrees of freedom for dynamic reduction or component mode synthesisdegrees of freedom omitted by structural matrix partitioning

degree of freedom eliminated by multipoint constraints

setmsbsgoqrcbesak

papspgnen

s

fefd

atl

supersets


BASIC MD NASTRAN SET DEFINITION (Cont.)Set Names = sb + sg all degrees of freedom eliminated by single point constraints

= b + c

t = + r the total set of physical boundary degrees of freedom for superelementsa = t + q the set assembled in superelement analysisd = a + e the set used in dynamic analysis by the direct methodf = a + o unconstrained (free) structural degrees of freedomfe = f +e free structural degrees of freedom plus extra degrees of freedomn = f + s all structural degrees of freedom not constrained by multipoint constraintsne = n + e

g = n + m all structural (grid) degrees of freedom including scalar degrees of freedomp = g + e all physical degrees of freedomps = p + sa physical and constrained aerodynamic degrees of freedompa = ps + k physical set for aerodynamicsfr = o + statically independent set minus the statically determinate supports (fr = f - q - r)v = o + c + r the set free to vibrate in dynamic reduction and component mode synthesis

The a-set and o-set created in the following ways:

1.

2.

3.

4.

= rigid body (zero frequency) modal degrees of freedom= finite frequency modal degrees of freedom= + , the set of all modal degrees of freedom

One Vector set is defined that combines physical and modal degrees of freedom:

u h = + u e, the set of all modal degrees of freedom

Meaning (+ indicates union of two sets)

the structural degrees of freedom remaining after the reference degrees of freedom areremoved (degree of freedom left over)

The membership of each degree of freedom can be printed by use of the Bulk Data entries PARAM,USETPRT and PARAM, USETSEL.

If only OMITi entries are present then the o-set consists of degrees of freedom listedexplicitly on OMITi entries. The remaining f-set degrees of freedom are placed in the b-set which is a subset of the a-set.If ASETi or QSETi are present, then the a-set consists of all degrees of freedom listedon ASETi entries and any entries listing its subsets, such as QSETi, SUPORTi, CSETi,and BSETi entries. Any OMITi entries are redundant. The remaining f-set degrees

If there are no ASETi, QSETi, or OMITi entries present but there are SUPORTi, BSETi,or CSETi entries present then the entire f-set is placed in the a-set and the o-set is notcreated.There must be at least one explicitly ASETi, QSETi, or OMITi entry for the o-set toexist, even if the ASETi, QSETi, or OMITi entry is redundant.

all structural degrees of freedom not constrained by multipoint constraints plus extradegrees of freedom

In dynamic analysis, additional vector sets are obtained by a modal transformation derived from realeigenvalue analysis of the a-set. These sets are as follows:

O

f

i O f

i


BASIC MD NASTRAN SET DEFINITION (Cont.)Degree of Freedom Set Bulk Data Entries

Degrees of freedom are placed in sets as specified by the user on the following Bulk Data entries:

Namem

sb SPC, SPC1, SPCADD, SPCAX, FLSYM, GMSPC*, BNDGRID, (PARAM,AUTOSPC,YES)sg GRID, GRIDB, GRDSET (PS field)o OMIT, OMIT1, OMITAX, GRID (SEID field), SESETq QSET, QSET1r SUPORT, SUPORT1, SUPAXc CSET, CSET1b BSET, BSET1e EPOINTsa CAEROik CAEROia ASET, ASET1, Superelement exterior degrees of freedom, CSUPEXT

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

MPC, MPCADD, MPCAX, POINTAX, RBAR, RBE1, RBE2, RBE3, RROD, RSPLINE,RTRPLT, GMBC, GMSPC*

In superelement analysis, the appropriate entry names are preceded by the letter SE, and have a fieldreserved for the superelement identification number. This identification is used because a boundary(exterior) grid point may be in one mutually exclusive set in one superelement and in a different set in theadjoining superelement. The SE-type entries are internally translated to the following types of entry forthe referenced superelement:

Bulk Data Entry Name

Entry Type Equivalent TypeSEQSETi QSETi

SUPORTCSETiBSETi

SESUPSECSETiSEBSETi


TIPS ON MODEL VERIFICATION

● PARAM, GRDPNT, V1 (V1 > 0)● Grid point weight generator

● PARAM, USETPRT, V1 (V1 = 0, 1, 2, 10, 11, 12)● MD Nastran set tables

● Various rigid body and equilibrium checks (e.g.,GROUNDCHECK)

● Engineering judgment

V1 : ID of reference grid pointV1 : ID of reference grid point

V1=0 : The location [0 0 0] in theV1=0 : The location [0 0 0] in thebasic coordinate system willbasic coordinate system willbe used as reference pointbe used as reference point

See example on next (additional) pageSee example on next (additional) page

The membership of each degreeThe membership of each degree--ofof--freedom can befreedom can beprinted.printed.


WEIGHT GENERATOR TABLE IN .F06 FILE

MO - Rigid Body Mass Properties

S - Identity Matrix when the mass is the same in each coordinate direction

I(S) - Inertia Matrix of structure for C.G. with respect to the basic coordinate system

I(Q) - Corresponding Principal Moments of Inertia Matrix

Q - Transformation from the principal direction to the basic coordinate system

PARAM,GRDPNT,1


SECTION 3

NORMAL MODE ANALYSIS


REASONS TO COMPUTE NATURALFREQUENCIES AND NORMAL MODES

● Assess the dynamic characteristics of the structure. For example, if rotatingmachinery is going to be installed on a certain structure, to avoid excessivevibrations, it might be necessary to see if the frequency of the rotating mass isclose to one of the natural frequencies of the structure.

● Assess the possible dynamic amplification of the loads.● Use the natural frequencies and normal modes to guide subsequent dynamic

analysis (transient response, response spectrum analysis), i.e., what shouldthe appropriate time step be for integrating the equations of motion in transientanalysis?

● Use the natural frequencies and mode shapes for subsequent dynamicanalysis, i.e., transient analysis of the structure using modal expansion.

● Guide the experimental analysis of structures, i.e., location of accelerometers,etc.

● Evaluate the design changes.


THEORETICAL RESULTS

M x·· Kx+ 0=

x eit

=

x·· 2

eit–=

2

M eit

– Keit

+ 0=

K 2

M – 0=

● Consider(1)

● Assume a harmonic solution of the form(2)

● (Physically, this means that all the coordinates perform synchronousmotions. The system configuration does not change its shape duringmotion, only its amplitude.)

● From Equation 2 (3)

● Substituting Equations 2 and 3 into Equation 1, we obtain

● which (after dividing by eit) simplifies to

● This is then an eigenvalue problem.


THEORETICAL RESULTS (Cont.)

● Therefore, there are two cases:1. If det , the only possibility (from Equation 4) is

which is the so-called trivial solution and is not interesting from aphysical point of view.

2. Otherwise, we need in order to have anontrivial solution for .

● The eigenvalue problem reduces to solve the following:

or

where

0])M[]K([ 2 0}{

}{

0])M[]K([det 2

0])M[]K([det

2

0])M[]K([det 2



● If the structure has N degrees of freedom with attachedmass, there will be N that are solutions of theeigenvalue problem. These s (1,2,…,n) are naturalfrequencies, characteristic frequencies, fundamentalfrequencies, or resonance frequencies.

● The eigenvector associated with the natural frequencyis called the normal mode or mode shape. The

normal mode corresponds to the deflected shape patternsof the structure.

● When a structure is vibrating, its shape at any given timeis a linear combination of its normal modes.

j}{

s'

j}{



ExampleSimply Supported Beam

Mode 1

Mode 2

Mode 3

etc.

1

3

2


IMPORTANT FACTS AND RESULTS REGARDINGNORMAL MODES AND NATURAL FREQUENCIES

jiIf0}]{M[}{j

T

i

jiIf0}]{K[}{j

T

i

}]{M[}{}]{K[}{

j

T

j

j

T

j2

j

2)ondsec/radian(

)hertz(f j

j

Rayleigh Quotient

● When [K] and [M] are symmetric and real (this is true for all thestandard structural finite elements), the following orthogonalityproperty holds:

● and

● also

● The natural frequencies (1,2,…) are expressed inradians/seconds. They can also be expressed in hertz(cycles/seconds), using


IMPORTANT FACTS AND RESULTS REGARDING NORMALMODES AND NATURAL FREQUENCIES (Cont.)

1 0= 1 11

=

m

k

m

x1 x2

● Example: The following unconstrained structure has arigid body mode.

● If a structure is not totally constrained, i.e., if it admits arigid body mode (stress-free mode) or a mechanism, thenat least one natural frequency will be zero.


● The scaling of normal modes is arbitrary. For example,

represent the same “mode of vibration.”

m

m x1

x2

1 1

0.5

1 ,300150

= = 1 0.660.33

=,,



● For practical considerations, modes should be normalizedby a chosen convention. In MSC.Nastran there are threenormalization choices (except when using Lanczos):● The unit value of generalized mass (default)

● The unit value of the largest A-set component in each mode● The unit value of a specific component (not recommended)

● In the Lanczos method, normalization is to a unit value ofgeneralized mass and to a unit value of the largestcomponent.

i T

Mi 1.0=



ADDITIONAL MODAL PROPERTIES

● Since strains, internal loads and stresses develop when astructure deforms, we may recover additional useful modalinformation utilizing● Strain-displacement relationships

● Stress-strain relationships

● Static force - displacement relationships

● Element strain energy relationships

Ku u=

K =

Pst Ku=

Ve 1 2 ue T Kee ue =


ADDITIONAL MODAL PROPERTIES (Cont.)

u i i=

i

Ku i i=

i

K Ku i i=

P i

Ki i=

Vei

12---ei

T

Kee ei

i2

=

● Thus, for a given modal displacement● we have

● Modal strain

● Modal stress

● Modal force

● Modal strain energy


METHODS OF COMPUTATIONMD Nastran provides 3 types of methods for eigenvalue extraction:● Tracking methods (see Appendix A)

Eigenvalues (or natural frequencies) are determined one at a time using aniterative technique. Two variations of the inverse power method are provided,INV and SINV. This approach is more convenient when few naturalfrequencies are to be determined. In general, SINV is more reliable than INV.

● Transformation methods (see Appendix A)The original eigenvalue problem is transformed to the form

where

K M – 0=

A =

A M 1– K=


METHODS OF COMPUTATION (Cont.)Then the matrix A is transformed into a tridiagonal matrix usingeither the Givens technique or the Householder technique.Finally, all the eigenvalues are extracted at once using the QRalgorithm. Two variations of the Givens technique and twovariations of the Householder technique are provided: GIV, MGIV,HOU, and MHOU. These methods are more efficient when alarge proportion of eigenvalues are needed.

●Lanczos Method (recommended method)

This method is a combined tracking-transformation method.


STURM SEQUENCE THEORY● Choose .● Factor .● The number of negative terms on the factor diagonal is the

number of eigenvalues below .

K iM– LDL

T into

0.0

No. Neg No. Neg

(must be in the range)8

=i2

Terms=7 Terms=8


LANCZOS METHOD

● Block, shifted, inverted Lanczos● Random starting vectors● Automatic shift logic● Partial and selective orthogonalization● Sturm sequence diagnosis● Givens plus QL eigensolution● Can be used for both buckling and normal modes analysis● Mass and largest component normalize only


USER INTERFACE FOR LANCZOS METHOD

Default = 7, can beincreased up to 15


USER INTERFACE FOR ALL METHODS


USER INTERFACE (Cont.)


SOLUTION CONTROL FOR NORMAL MODES

● Executive Control Section● SOL 103

● Case Control Section● METHOD (required - selects EIGRL / EIGR entry)

● Bulk Data Section● EIGRL / EIGR (Lanczos method)


CASE CONTROL OUTPUT

● Grid output● DISPLACEMENT (or VECTOR)● GPFORCE● GPSTRESS● SPCFORCE● GPKE

● Element output● ELSTRESS (or STRESS)● ESE● EKE● ELFORCE (or FORCE)● STRAIN

● Special entry● OMODES – selects output for selective modes


PROBLEM #1

MODAL ANALYSIS OF A FLAT PLATE


PROBLEM #1 - MODAL ANALYSIS OF A FLATPLATE

For this problem, use Lanczos method to find the first ten naturalfrequencies and mode shapes of a flat rectangular plate. Below is a finiteelement representation of the rectangular plate. It also contains thegeometric dimensions and the loads and boundary constraints. Table 3Acontains the necessary parameters to construct the input file.

Grid Coordinates and Element Connectivities

5

2


WORKSHOP #1 - MODAL ANALYSIS OF A FLATPLATE

Loads and Boundary Conditions

Length (a) 5 inHeight (b) 2 inThickness 0.100 inWeight Density 0.282 lbs/in3Mass/Weight Factor 2.59E-3 sec2/inElastic Modulus 30.0E6 lbs/in2Poisson’s Ratio 0.3

Table 3A.


GEOMETRIC DESCRIPTION OF PLATE$$ plate.bdf$$ geometric input file for plate model$PSHELL 1 1 .1 1 1CQUAD4 1 1 1 2 13 12CQUAD4 2 1 2 3 14 13CQUAD4 3 1 3 4 15 14CQUAD4 4 1 4 5 16 15CQUAD4 5 1 5 6 17 16CQUAD4 6 1 6 7 18 17CQUAD4 7 1 7 8 19 18CQUAD4 8 1 8 9 20 19CQUAD4 9 1 9 10 21 20CQUAD4 10 1 10 11 22 21CQUAD4 11 1 12 13 24 23CQUAD4 12 1 13 14 25 24CQUAD4 13 1 14 15 26 25CQUAD4 14 1 15 16 27 26CQUAD4 15 1 16 17 28 27CQUAD4 16 1 17 18 29 28CQUAD4 17 1 18 19 30 29CQUAD4 18 1 19 20 31 30CQUAD4 19 1 20 21 32 31CQUAD4 20 1 21 22 33 32

CQUAD4 21 1 23 24 35 34CQUAD4 22 1 24 25 36 35CQUAD4 23 1 25 26 37 36CQUAD4 24 1 26 27 38 37CQUAD4 25 1 27 28 39 38CQUAD4 26 1 28 29 40 39CQUAD4 27 1 29 30 41 40CQUAD4 28 1 30 31 42 41CQUAD4 29 1 31 32 43 42CQUAD4 30 1 32 33 44 43CQUAD4 31 1 34 35 46 45CQUAD4 32 1 35 36 47 46CQUAD4 33 1 36 37 48 47CQUAD4 34 1 37 38 49 48CQUAD4 35 1 38 39 50 49CQUAD4 36 1 39 40 51 50CQUAD4 37 1 40 41 52 51CQUAD4 38 1 41 42 53 52CQUAD4 39 1 42 43 54 53CQUAD4 40 1 43 44 55 54$MAT1 1 3.+7 .3 .282$

Weight Densitysee PARAM,WTMASS


GEOMETRIC DESCRIPTION OF PLATE (Cont.)$GRID 1 0. 0. 0.GRID 2 .5 0. 0.GRID 3 1. 0. 0.GRID 4 1.5 0. 0.GRID 5 2. 0. 0.GRID 6 2.5 0. 0.GRID 7 3. 0. 0.GRID 8 3.5 0. 0.GRID 9 4. 0. 0.GRID 10 4.5 0. 0.GRID 11 5. 0. 0.GRID 12 0. .5 0.GRID 13 .5 .5 0.GRID 14 1. .5 0.GRID 15 1.5 .5 0.GRID 16 2. .5 0.GRID 17 2.5 .5 0.GRID 18 3. .5 0.GRID 19 3.5 .5 0.GRID 20 4. .5 0.GRID 21 4.5 .5 0.GRID 22 5. .5 0.GRID 23 0. 1. 0.GRID 24 .5 1. 0.GRID 25 1. 1. 0.GRID 26 1.5 1. 0.GRID 27 2. 1. 0.GRID 28 2.5 1. 0.GRID 29 3. 1. 0.GRID 30 3.5 1. 0.

GRID 31 4. 1. 0.GRID 32 4.5 1. 0.GRID 33 5. 1. 0.GRID 34 0. 1.5 0.GRID 35 .5 1.5 0.GRID 36 1. 1.5 0.GRID 37 1.5 1.5 0.GRID 38 2. 1.5 0.GRID 39 2.5 1.5 0.GRID 40 3. 1.5 0.GRID 41 3.5 1.5 0.GRID 42 4. 1.5 0.GRID 43 4.5 1.5 0.GRID 44 5. 1.5 0.GRID 45 0. 2. 0.GRID 46 .5 2. 0.GRID 47 1. 2. 0.GRID 48 1.5 2. 0.GRID 49 2. 2. 0.GRID 50 2.5 2. 0.GRID 51 3. 2. 0.GRID 52 3.5 2. 0.GRID 53 4. 2. 0.GRID 54 4.5 2. 0.GRID 55 5. 2. 0.$SPC1 1 12345 1 12 23 34 45


WORKSHOP # 1

● Start with the partial input file on the following page andmodify

● Executive Control Section● Add the solution sequence

● Case Control Section● Add eigenvalue method callout● Add eigenvector printout

● Bulk Data Section● Add weight/mass conversion factor● Add eigenvalue method


WORKSHOP # 1$$ wkshp1.dat$CENDTITLE = NORMAL MODES EXAMPLEECHO = UNSORTEDSUBCASE 1

SUBTITLE= USING LANCZOSSPC = 1

BEGIN BULKparam,post,-1PARAM COUPMASS 1$include 'plate.bdf'$ENDDATA


SOLUTION FILE FOR PROBLEM #1$$ soln1.dat$ID SEMINAR, PROB1SOL 103CENDTITLE = NORMAL MODES EXAMPLEECHO = UNSORTEDSUBCASE 1

SUBTITLE= USING LANCZOSMETHOD = 1SPC = 1VECTOR=ALL

BEGIN BULKparam,post,0PARAM COUPMASS 1PARAM WTMASS .00259EIGRL 1 10$include 'plate.bdf'$ENDDATA


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

*** USER INFORMATION MESSAGE 5010 (LNCILD)STURM SEQUENCE DATA FOR EIGENVALUE EXTRACTION.TRIAL EIGENVALUE = 8.507511D+07, CYCLES = 1.467984D+03 NUMBER OF EIGENVALUES BELOW THIS VALUE = 3

*** USER INFORMATION MESSAGE 5010 (LNCILD)STURM SEQUENCE DATA FOR EIGENVALUE EXTRACTION.TRIAL EIGENVALUE = 1.908154D+09, CYCLES = 6.952274D+03 NUMBER OF EIGENVALUES BELOW THIS VALUE = 10

0E I G E N V A L U E A N A L Y S I S S U M M A R Y (READ MODULE)

BLOCK SIZE USED ...................... 7NUMBER OF DECOMPOSITIONS ............. 2NUMBER OF ROOTS FOUND ................ 10NUMBER OF SOLVES REQUIRED ............ 7

1 NORMAL MODES EXAMPLE MARCH 9, 2007 MD NASTRAN 3/ 1/07 PAGE 90 SUBCASE 1

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

1 1 7.055894E+05 8.399937E+02 1.336891E+02 1.000000E+00 7.055894E+052 2 1.877186E+07 4.332651E+03 6.895628E+02 1.000000E+00 1.877186E+073 3 2.811177E+07 5.302053E+03 8.438480E+02 1.000000E+00 2.811177E+074 4 1.929422E+08 1.389036E+04 2.210720E+03 1.000000E+00 1.929422E+085 5 2.221657E+08 1.490523E+04 2.372240E+03 1.000000E+00 2.221657E+086 6 2.328451E+08 1.525926E+04 2.428587E+03 1.000000E+00 2.328451E+087 7 6.832396E+08 2.613885E+04 4.160127E+03 1.000000E+00 6.832396E+088 8 9.600053E+08 3.098395E+04 4.931249E+03 1.000000E+00 9.600053E+089 9 1.365293E+09 3.694987E+04 5.880754E+03 1.000000E+00 1.365293E+0910 10 1.850317E+09 4.301531E+04 6.846099E+03 1.000000E+00 1.850317E+09

PARTIAL OUTPUT FOR PROBLEM #1

i = i2 i fi = iTMi = i

TKi

[rad/sec] [Hz]


PARTIAL OUTPUT FOR PROBLEM #1 (Cont.)

0 SUBCASE 11 NORMAL MODES EXAMPLE MARCH 9, 2007 MD NASTRAN 3/ 1/07 PAGE 11

USING LANCZOS0 SUBCASE 1

EIGENVALUE = 7.055894E+05CYCLES = 1.336891E+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 -1.255791E-152 G -1.556529E-15 -1.386242E-15 -9.732725E-01 -1.099511E+00 4.003941E+00 -2.834058E-153 G -2.915758E-15 -3.505252E-15 -4.168619E+00 -1.599071E+00 8.686669E+00 -4.892793E-154 G -3.230860E-15 -6.176625E-15 -9.445514E+00 -1.530855E+00 1.230146E+01 -5.718867E-155 G -3.941097E-15 -9.087511E-15 -1.636256E+01 -1.365152E+00 1.522518E+01 -4.092859E-156 G -2.401380E-15 -1.049688E-14 -2.455516E+01 -1.080930E+00 1.740494E+01 -6.150749E-157 G -2.307428E-15 -1.346573E-14 -3.367780E+01 -8.040399E-01 1.895419E+01 -3.032976E-158 G -9.768109E-16 -1.680592E-14 -4.342896E+01 -5.477144E-01 1.993718E+01 -1.908953E-159 G -2.291250E-15 -1.610515E-14 -5.355173E+01 -3.472317E-01 2.046533E+01 -3.178451E-1510 G -3.957610E-15 -2.128953E-14 -6.384777E+01 -2.192121E-01 2.066150E+01 -8.190604E-1511 G -3.857835E-15 -2.386429E-14 -7.419273E+01 -1.687060E-01 2.069914E+01 -8.426097E-1512 G 0.0 0.0 0.0 0.0 0.0 -1.208914E-1513 G -3.253508E-16 -1.184481E-15 -1.263801E+00 -1.865870E-01 4.860682E+00 -2.281848E-15


Blank


SECTION 4

REDUCTION IN DYNAMIC ANALYSIS


INTRODUCTION TO DYNAMIC REDUCTION

Definition

● Dynamic reduction means reducing a given dynamic math model to one with fewerdegrees of freedom.

Why Reduction for Dynamics?

● The math model may be too big to be solved without reduction.● The math model has more detail than required.● Dynamic reduction allows the deletion of selective local modes.● Dynamic reduction is more accurate (and probably cheaper) than constructing a

separate, smaller dynamic model.


REDUCTION METHODS FOR DYNAMICSAVAILABLE WITH MD NASTRAN

● Guyan reduction (static condensation)● Generally not recommended except for correlation with test data

● Modal reduction

● Component mode synthesis (superelement option) - see Section 16.


STATIC CONDENSATION (INTERNALCALCULATION)

● Let {uf} be the set of the unconstrained (free) structural coordinates.● Partition

where ua = analysis set uo = omitted set

uf uauo-------

=

Degrees of freedom removed during Guyan reduction

User-selected dynamic degrees of freedomOO--SetSet

AA--SetSet


STATIC CONDENSATION (Cont.) (INTERNALCALCULATION)

● Form a static equation for uf and partition the stiffness matrix into the O-set andthe A-set.

● Assume Po is zero and solve for uo in terms of ua

● Transformation from the A-set to F-set is

● O-set is dependent upon the A-set via Equation 2. The motion of the O-set is alinear combination of the A-set motions. The columns of Goa are the static shapevectors.

uo

ua

Goa

I=={uf} {ua}



● The equations of motion for the F-set are written in terms of the A-set

● Dynamics problems are solved in terms of the reduced coordinates (A-set). O-set components arerecovered using Equation 2.

● O-set mass, damping, and stiffness is spread to the A-set.

● The largest cost is associated with the formulation of Maa and Baa, particularly for nondiagonal (coupledmass) Mff.

● The resulting Kaa, Baa, and Maa are small and dense (i.e., matrix bandedness is destroyed).

or

TMf{ua} + TBf{ua} + TKf{ua} = TPf

.. .

Maaua + Baaua + Kaaua = Pa.. .



SUMMARY●Separate free degrees of freedom (Uf) into the omitted set (U0) and

the analysis set (UA) by means of OMIT entries or ASET entries.●Retain only a small fraction of the DOFs (typically 10% or less) in

the analysis set because the computer costs for staticcondensation increase rapidly with the size of the analysis set.Otherwise, retain all of the DOFs.

●Retain DOFs with large concentrated masses in the analysis set.●Retain DOFs that are loaded (in transient and frequency response

analysis).●Retain DOFs to adequately describe deflected shape or modes of

interest.


USER INTERFACE

Either

and/or

or

OMIT, OMIT1

Specify either the A-set (with ASET entry) or the O-set (with OMIT entry). The remaining DOFsautomatically are placed in the complementary set.

1 2 3 4 5 6 7 8 9 10ASET ID C ID C ID C ID CASET 1 123 2 12 4 1 5 1

ASET1 C G G G G G G GASET1 123 1 2 3 4 5


SOLUTION CONTROL FOR GUYANREDUCTION

● Executive Control Section● Any SOL

● Case Control Section● No special commands required

● Bulk Data Section● ASET (optional* - specifies A-set)● OMIT (optional* - specifies O-set)

*Components not specified are placed in the complementary set. If bothASET and OMIT are present, components not specified are placed in theO-set.


DIFFICULTIES WITH GUYAN REDUCTION

● User effort in selecting A-set points● Accuracy depends on the user’s skill in selecting A-set points● Regardless of user’s skill, high accuracy requires a large number

of A-set points (cost consideration) - 2 to 5 times number ofaccurate modes wanted

● Stiffness reduction is exact; mass and damping reductions areonly approximate

● No loss in accuracy of modes occurs when omitting masslessdegrees of freedom

● Errors are most pronounced in higher modes● Local modes may be missed altogether● Not generally recommended, except when performing test-

analysis correlation (see Section 20)


Difficulties With Guyan Reduction (Cont.)

● The static condensation approximation may miss thelocal dynamic effects.

0

Local Dynamic Effects

Physical Variables

Static Transformation

Loads on O-set Components

uo Goa Ua= + Uoo

Koo Po=oou -1


WORKSHOP #2

NORMAL MODES ANALYSIS USINGGUYAN REDUCTION


For this example, use Guyan Reduction to reduce the model used inWorkshop #1. Then find the first five natural frequencies and modeshapes using the Lanczos method. Use the points indicated in Figure4B for the A-set.

Figure 4A. Grid Coordinates and Element Connectivities.

WORKSHOP #2 - MODAL ANALYSIS OF AFLAT PLATE USING STATIC REDUCTION

22

55


WORKSHOP #2 - MODAL ANALYSIS OF A FLAT PLATEUSING GUYAN REDUCTION (Cont.)

Figure 4B. Loads and Boundary Conditions.


WORKSHOP # 2

$$ wkshp2.dat$SOL 103CENDTITLE = REDUCTION PROCEDURES, NORMAL MODES EXAMPLESUBTITLE = USING STATIC REDUCTIONECHO = UNSORTEDSUBCASE 1

SUBTITLE=USING LANCZOSMETHOD = 1SPC = 1VECTOR=ALL

BEGIN BULKparam,post,-1EIGR,1,AHOU,,,,5PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259INCLUDE 'plate.bdf'$$ SELECT A-SET, STATIC REDUCTION IS DONE AUTOMATICALLY$ENDDATA

● Start with the following partial input file and add request forGuyan reduction. Compare the results with workshop # 1


SOLUTION FILE FOR WORKSHOP #2$$ soln2.dat$ID SEMINAR, PROB2SOL 103CENDTITLE = REDUCTION PROCEDURES, NORMAL MODES EXAMPLESUBTITLE = USING STATIC REDUCTIONECHO = UNSORTEDSUBCASE 1

SUBTITLE=USING LANCZOSMETHOD = 1SPC = 1VECTOR=ALL

BEGIN BULKparam,post,0EIGR,1,LAN,,,,5PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259INCLUDE 'plate.bdf'$$ SELECT A-SET, STATIC REDUCTION IS DONE AUTOMATICALLY$ASET1,345,3,5,7,9,11ASET1,345,25,27,29,31,33ASET1,345,47,49,51,53,55ENDDATA


PARTIAL OUTPUT FILE FOR WORKSHOP #2

1 REDUCTION PROCEDURES, NORMAL MODES EXAMPLE SEPTEMBER 28, 2006 MD NASTRAN 8/22/03 PAGE 9

USING STATIC REDUCTION0 SUBCASE 1

R E A L E I G E N V A L U E S

MODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZEDNO. ORDER MASS STIFFNESS

1 1 7.056352E+05 8.400210E+02 1.336935E+02 1.000000E+00 7.056352E+052 2 1.879631E+07 4.335472E+03 6.900117E+02 1.000000E+00 1.879631E+07

3 3 2.817725E+07 5.308225E+03 8.448301E+02 1.000000E+00 2.817725E+074 4 1.953805E+08 1.397786E+04 2.224645E+03 1.000000E+00 1.953805E+08

5 5 2.367517E+08 1.538674E+04 2.448875E+03 1.000000E+00 2.367517E+08.

.

.

0 SUBCASE 1EIGENVALUE = 7.056352E+05

CYCLES = 1.336935E+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 0.0

2 G 0.0 0.0 -9.732642E-01 -1.099629E+00 4.004109E+00 0.03 G 0.0 0.0 -4.168889E+00 -1.599175E+00 8.687232E+00 0.0

4 G 0.0 0.0 -9.445539E+00 -1.529646E+00 1.230197E+01 0.05 G 0.0 0.0 -1.636362E+01 -1.365241E+00 1.522617E+01 0.0

6 G 0.0 0.0 -2.455527E+01 -1.077562E+00 1.740565E+01 0.07 G 0.0 0.0 -3.367999E+01 -8.040918E-01 1.895542E+01 0.0


MODAL REDUCTION● All MD Nastran linear dynamic solutions have two versions.

● Direct - The solution is solved in terms of A-set coordinates.● Modal - The solution is solved in terms of modal coordinates (H-set).

● In the modal solution sequences the A-set coordinates are written in terms ofmodal coordinates.

● Modal vectors (mode shapes) are solutions to the undamped eigenvalue problem(A-set coordinates)

Modal Coordinates

Matrix of Mode Shapes

a=ua

[Maa] {ua} + [Kaa] {ua} = 0..


MODAL REDUCTION (Cont.)● Equations of motion for the A-set are written in terms of modal

coordinates (H-set notation, modal coordinates are handled internally.)(Note: E-set DOFs are not shown here for clarity.)

If [f] is mass normalized and there are no K2PP, M2PP, B2PP, or TF,then:

Note: A-set matrices may be reduced matrices from Guyanreduction or GDR. Transformation from modal coordinates to the F-setwould require two transformations.

[aT] [Maa] {a} {} + [T] [Baa] {a} {} + [a

T] [Kaa] {a} {}a

.. .

= [aT] {Pa}

[] {} + [aT] Baa {a} {} + [ W2] {} = [a

T] {Pa}.. .

{uf} = [] {ua}

{ua} = [a] {}

{uf} = [] {a} {}...

GeneralizedGeneralizedMass matrixMass matrix

GeneralizedGeneralizedStiffnessStiffnessmatrixmatrix


SOLUTION CONTROL FOR MODALREDUCTION

● Executive Control Section●Any modal dynamic analysis SOL

● Case Control Section●METHOD (required - selects Bulk Data EIGR or EIGRL entry)

● Bulk Data Section●EIGR or EIGRL (required - selects parameters for eigenanalysis)


SECTION 5

RIGID BODY MODES


RIGID BODY MODES AND RIGID BODY VECTORSTHEORETICAL CONSIDERATIONS

• A structure has the ability to displace without developing internal loads ofstresses if it is not sufficiently grounded. Examples of this are:

• In cases (a) and (b), the structure can displace as a rigid body.

No Constraints

P

(a)

P

Partial Constraints(b)

P

Mechanism(s)(c)


RIGID BODY MODES AND RIGID BODY VECTORSTHEORETICAL CONSIDERATIONS (Cont.)

• The presence of rigid body and/or mechanism modes is evidenced byzero frequency values in the solution of the eigenvalue problem.

• On the assumption that the mass matrix [M] is positive definite, zeroeigenvalues result from a positive semi-definite stiffness, i.e.,

• SUPORT does not constrain the structure. It simply defines the R-setcomponents. In normal modes analysis, rigid body modes arecalculated using the R-set as reference degrees of freedom.

K M =

RIGT MRIG 0>

RIGT KRIG 0=


• If R-set is present, rigid body modes are calculated in MD Nastranby the following method for methods other than the Lanczosmethod:

• The Lanczos method calculates rigid body modes directly

Step 1: “a”-set partitioning ul ua = urStep 2: Solve for u l in terms of ur .

Note: Pr is not actually applied!

CALCULATION OF RIGID BODY MODES

K ll Klr

Krl K̃rr

ulur

0Pr

=

ll-- set (left over set)set (left over set)

rr-- set (set (suportsuport DOF)DOF)


ul = Dm ur

where [ Dm ] = - [ Kll ]-1 [ Klr ]

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

RIG DmIr

=

CALCULATION OF RIGID BODY MODES(Cont.)


CALCULATION OF RIGID BODY MODES(Cont.)

Step 3: Mass matrix operations

where [Mr] is not diagonal in general• Using Gram-Schmidt orthogonalization (in the READ module), the matrix

[Mr] is orthogonalized by the transformation [ro], that is,

Step 4: Rigid body mode construction

with the property:

___________________________________________________________________________________________

*Only if R-set DOFs truly support rigid body modes

Mr DmIr

TMaa

DmIr

=

Mo roT Mr ro=

a RIG

Dmroro

=

a RIGT

Kaa a RIGKrr 0= = *

a RIGT

Maa a RIGMo =


SELECTION OF “SUPORT” DEGREES OFFREEDOM

• Care must be taken when selecting SUPORT DOFs.• SUPORT DOFs must be able to displace independently without

developing internal stresses (statically determinate).

3

Bad Selection

(The independent displacement of 1and 4 may produce internal stress.)

Good Selection

2 5

6

14

2 5

3 4

1

6


CHECKING OF “SUPORT” DOFsMD Nastran calculates internal strain-energy (work) for each rigid body vector.

Kll Klr D[ X ] = [ DT Ir ]

Krl Krr Ir

[ X ] = [ Krr ] + [ Klr ]T D

• If actual rigid body modes exist, the strain-energy is 0 .• Note that [X] is also the transformation of the stiffness matrix [Kaa] to R-set

coordinates, which by definition of rigid body (zero frequency) vector properties,should be null.

• MD Nastran also calculates the rigid body error ratio

• where means Euclidian norm of the matrix

• Note: Only one value of is calculated using [X], and [Krr] based on all SUPORTDOFs.

XKrr

--------------=

xij2

j

i=


CHECKING OF “SUPORT” DOFS (Cont.)

• Except for round-off errors, the rigid body error ratio and the strain energyshould be zero if a compatible set of statically determinate supports arechosen by the user. These quantities may be nonzero for any of the followingreasons:• Round-off error accumulation• The ur set is over determined leading to redundant supports (high strain energy).• The ur set is underspecified leading to a singular reduced stiffness matrix (high rigid

body error ratio).• The multipoint constraints are incompatible (high strain energy and high rigid body

error ratio).• There are too many single-point constraints (high strain energy and high rigid body

error ratio).• Krr is null (unit value for rigid body error but low strain energy). This is an

acceptable condition and may occur when generalized dynamic reduction is used.


Rigid Body Modes and Rigid Body Vectors

• In MD Nastran, flexible body modes associated with the A-set massand stiffness matrices are calculated. The first N modes calculated bythe eigen analysis (where N is the number of DOFs in the R-set) arediscarded. The N rigid body modes are substituted in their place.

• Note: MD Nastran does not check that discarded modes are rigidbody modes (i.e., = 0).

• When this transformation is applied to the dynamic system and themodes are unit mass normalized, we obtain

IRIG 0

0 IFLEX

··RIG

··FLEX

TB· RIG

· FLEX

0 0

0 FLEX2

RIGFLEX

+ +

RIGT P

FLEXT P

RIG

T

FLEXT

N Q+ +=


RIGID BODY MODES AND RIGID BODYVECTORS (Cont.)

• As a result of the transformation, the following consequencesoccur:

• Constraint forces are not externally active, i.e.,

• If damping elements are not connected to ground, then

Thus,

RIGT

FLEXT

Q 0=

RIGT B 0=

RIGT

FLEXT

BRIG FLEX0 0

0 FLEXT BFLEX

=


RIGID BODY MODES AND RIGID BODYVECTORS (Cont.)

• If damping is “proportional,” then

• The modal dynamic equations are fully uncoupled.

RIGT

FLEXT

BRIG FLEX0 00 2ii

=


SECTION 6

DAMPING


DAMPING

● Damping represents energy dissipation observed in structures.● Damping is difficult to accurately model since damping results from many

mechanisms:● Viscous effects (dashpot, shock absorber)● External friction (slippage in structural joints)● Internal friction (characteristic of material type)● Structural nonlinearities (plasticity)

● Analytical conveniences used to model damping● Viscous damping force

● Structural damping force

fv bu·=

mu·· bu· ku p=

fs igku= i 1–=whereg = structural damping coefficient

mu·· 1 ig+ ku+ p=

..

..

..

++


STRUCTURAL DAMPING VERSUS VISCOUSDAMPING

Assume sinusoidial response:

Then:● Viscous damping:

● Structural damping:

u ueit=

u· iueit= u·· 2ueit–=

mu·· bu· ku+ =

m 2ueit– b iueit kueit+ + p t=

2mueit– ibueit kueit+ + p t=

mu·· 1 ig+ ku+ p t=

m 2ueit– 1 ig+ kueit+ p t=

2mueit– igkueit kueit+ + p t=

p(t)+


STRUCTURAL DAMPING VERSUS VISCOUSDAMPING (Cont.)

● Both equations are identical if:

Therefore, if structural damping g is to be modeled using viscousdamping b, then the equality holds at only one frequency w3 (or w4).

if

but

gk b b gk-------= =

b gk

-------=

nkm-----= =

b gkn-------- gnm= =

bc 2mn=


STRUCTURAL DAMPING VERSUS VISCOUSDAMPING (Cont.)

then

= critical damping ratio (percent critical damping)

g = = structural damping factor

Q = quality factor or magnification factor

bbc------- g

2---= =

1Q----


STRUCTURAL DAMPING VERSUS VISCOUS DAMPING(CONSTANT DISPLACEMENT)

● Viscous and structural damping are equivalent at frequency3 (or 4).

Structural Damping, fs = igkuDamping

EquivalentViscous

b = gk/3 (or4)

3 (or 4)

fv bu·=

ForceForce


DAMPING SUMMARY

● Viscous damping force proportional to velocity● Structural damping force proportional to displacement● Critical damping ratio● Quality factor Q inversely proportional to energy

dissipated per cycle of vibration● At resonance

●= g/2● Q = 1/(2)● Q = 1/g

n

crb/b


STRUCTURAL DAMPING

● Structural damping● MATi entries

● PARAM,G, factor (Default = 0.0)● Overall structural damping coefficient to multiply entire system

stiffness matrix● PARAM,W3, factor (Default = 0.0)

● Converts overall structural damping to equivalent viscous damping● PARAM,W4, factor (Default = 0.0)

● Converts element structural damping to equivalent viscous damping● Units for W3,W4 are radians/unit time● If PARAM,G is used, PARAM,W3 must be given a setting greater

than zero; otherwise, PARAM,G is ignored in transient responseanalysis (see Section 7 for more information).

1 2 3 4 5 6 7 8 9 10MAT1 MID E G NU RHO A TREF GEMAT1 2 30.0E6 0.3 0.10


VISCOUS DAMPING

● Scalar viscous damping

CDAMP1

CDAMP2

CDAMP3

CDAMP4

CVISC

CBUSH

Element damper between two grid points; references a property entry(PVISC).

Generalized spring and damper element that may also be frequencydependent.

Scalar damper between two DOFs with reference to a property entry.

Scalar damper between two DOFs without reference to a property entry(PDAMP).

Scalar damper between two scalar points (SPOINT) with reference to aproperty entry (PDAMP).

Scalar damper between two scalar points (SPOINT) without reference toa property entry.


VISCOUS DAMPING (Cont.)Scalar Damper Connection

Defines a scalar damper element.

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

CDAMP1 EID PID G1 C1 G2 C2

Example:CDAMP1 19 6 0 23 2

Field ContentsEID Unique element identification number. (Integer > 0)PID Property identification number of a PDAMP property entry. (Integer > 0; Default = EID)G1, G2 Geometric grid point identification number. (Integer > 0)C1, C2

1.

2.3.4.5.6.7.

A scalar point specified on this entry need not be defined on an SPOINT entry.If Gi refers to a grid point then Ci refers to degrees of freedom(s) in the displacement coordinatesystem specified by CD on the GRID entry.

Component number. (0 < Integer < 6; 0 or up to six unique integers, 1 through 6 maybe specified in the field with no embedded blanks. 0 applies to scalar points and 1through 6 applies to grid points.)

Scalar points may be used for G1 and/or G2, in which case the corresponding C1 and/or C2 mustbe zero or blank. Zero or blank may be used to indicate a grounded terminal G1 or G2 with acorresponding blank or zero C1 or C2. A grounded terminal is a poin

Element identification numbers should be unique with respect to all other element identificationThe two connection points (G1, C1) and (G2, C2), must be distinct.

CDAMP1

Remarks:

For a discussion of the scalar elements, see the MSC.Nastran Reference Manual, Section 5.6.When CDAMP1 is used in heat transfer analysis, it generates a lumped heat capacity.


VISCOUS DAMPING (Cont.)PDAMP Scalar Damper PropertySpecifies the damping value of a scalar damper element using defined CDAMP1 or CDAMP3 entries.

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

PDAMP PID1 B1 PID2 B2 PID3 B3 PID4 B4

Example:PDAMP 14 2.3 2 6.1

Field ContentsPIDi Property identification number. (Integer > 0)Bi Force per unit velocity. (Real)

Remarks:1.

2.3. Up to four damping properties may be defined on a single entry.

· For a discussion of scalar elements, see the MSC.Nastran Reference Manual, Section 5.6..

Damping values are defined directly on the CDAMP2 and CDAMP4 entries, and therefore do notrequire a PDAMP entry.

A structural viscous damper, CVISC, may also be used for geometric grid points.


VISCOUS DAMPING (Cont.)CDAMP2 Scalar Damper Property and Connection

Defines a scalar damper element without reference to a material or property entry.

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

CDAMP2 EID B G1 C1 G2 C2

Example:CDAMP2 16 2.98 32 1

Field ContentsEID Unique element identification number. (Integer > 0)B Value of the scalar damper. (Real)G1, G2 Geometric grid point identification number. (Integer > 0)C1, C2

Remarks:1.

2.3.4.5.6.7.

When CDAMP2 is used in heat transfer analysis, it generates a lumped heat capacity.A scalar point specified on this entry need not be defined on an SPOINT entry.If Gi refers to a grid point then Ci refers to degrees of freedom(s) in the displacement coordinatesystem specified by CD on the GRID entry.

Component number. (0 < Integer < 6; 0 or up to six unique integers, 1 through 6 maybe specified in the field with no embedded blanks. 0 applies to scalar points and 1through 6 applies to grid points.)

Scalar points may be used for G1 and/or G2, in which case the corresponding C1 and/or C2 mustbe zero or blank. Zero or blank may be used to indicate a grounded terminal G1 or G2 with acorresponding blank or zero C1 or C2. A grounded terminal is a poin

Element identification numbers should be unique with respect to all other element identification numbers.The two connection points (G1, C1) and (G2, C2), must be distinct.For a discussion of the scalar elements, see the MSC.Nastran Reference Manual, Section 5.6.


VISCOUS DAMPING (Cont.)

CDAMP3 Scalar Damper Connection to Scalar Points Only

Defines a scalar damper element that is connected only to scalar points.

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

CDAMP3 EID PID S1 S2

Example:CDAMP3 16 978 24 36

Field ContentsEIDPIDS1, S2

Remarks:1.2.

3.4.5.6.

Unique element identification number. (Integer > 0)Property identification number of a PDAMP entry. (Integer > 0; Default = EID)Scalar point identification numbers. (Integer > 0; )

S1 or S2 may be blank or zero, indicating a constrained coordinate.

A scalar point specified on this entry need not be defined on an SPOINT entry.

Element identification numbers should be unique with respect to all other element identificationnumbers.Only one scalar damper element may be defined on a single entry.For a discussion of the scalar elements, see the MSC.Nastran Reference Manual, Section 5.6.When CDAMP3 is used in heat transfer analysis, it generates a lumped heat capacity.

S1 S2



CDAMP4 Scalar Damper Property and Connection to Scalar Points Only

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

CDAMP4 EID B S1 S2

Example:CDAMP4 16 -2.6 4 9

Field ContentsEID Unique element identification number. (Integer > 0)B Scalar damper value. (Real)S1, S2 Scalar point identification numbers. (Integer > 0; )

Remarks:1. S1 or S2 may be blank or zero, indicating a constrained coordinate.2. Element identification numbers should be unique with respect to all other element identification numbers.3. Only one scalar damper element may be defined on a single entry.4. For a discussion of the scalar elements, see the MSC.Nastran Reference Manual, Section 5.6.5. If this entry is used in heat transfer analysis, it generates a lumped heat capacity.

Defines a scalar damper element that connected only to scalar points and without reference to a material orproperty entry.

S1 S2



PVISC Viscous Damping Element Property

Defines properties of a one-dimensional viscous damping element (CVISC entry).

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

PVISC PID1 CE1 CR1 PID2 CE2 CR2

Example:PVISC 3 6.2 3.94

Field ContentsPIDi Property identification number. (Integer > 0)CE1, CE2 Viscous damping values for extension in units of force per unit velocity. (Real)CR1, CR2 Viscous damping values for rotation in units of moment per unit velocity. (Real)

Remarks:1. Viscous properties are material independent; in particular, they are temperature independent.2. One or two viscous element properties may be defined on a single entry.


MODAL DAMPING

CASE CONTROL

SDAMP = n $ selects the modal damping table to be used.

BULK DATA

TABDMP1,n,CRIT ,x1,y1,x2,y2,..endt

$ Lists damping values (in "G", "CRIT", or "Q") versus $ frequencies.


Rayleigh Damping

● Proportional to either the mass or stiffness matrix● Also known as proportional damping● Proportional to mass matrix (param,alpha1,x)● Proportional to stiffness matrix (param,alpha2,y)● Available in transient and frequency response analysis● Scale factors applied to d-set (direct) and h-set (modal)● Added to the viscous damping matrix as follows:

[B’] = [B] + alpha1 * [M] + alpha2 * [K]


Rayleigh Damping (cont.)

● ALPHA1 and ALPHA2 are complex parameters, e.g.

PARAM, ALPHA2, 1.25E-4, 0.

● Modal Damping Matrix reads

221

21

ΩIKφφMφφBφφ TTT

Real , ImaginaryReal , Imaginary


Blank


SECTION 7

TRANSIENT RESPONSE ANALYSIS


INTRODUCTION TO TRANSIENT RESPONSEANALYSIS

● Compute response to time-varying excitation.

● Excitation is explicitly defined in the time domain. All ofthe applied forces are known at each instant in time.

● Computed response usually includes nodal displacementsand accelerations, and element forces and stresses.

● Two categories of analysis - direct and modal.


DIRECT TRANSIENT RESPONSE

● Dynamic equation of motion

● Response solved at discrete times with a fixed● Using central finite difference representation for and

at discrete times

● Note: These equations are also used by MSC.Nastran tocompute velocity and acceleration output.

[M] {u(t)} + [B] {u(t)} +[K] {u(t)} = {P(t)}.. .

t

{un} = {un+1 - un-1}.

{un} = {un+1 - 2un + un-1}..

12 t

1t2

{u(t)}.

{u(t)}. .


DIRECT TRANSIENT RESPONSE (Cont.)

● Numerical integration (Central Difference type method)(except “smear” force over 3 adjacent time points)

m

t2

--------- un 1+ 2un– uu 1–+ b2t--------- un 1+ un 1–– +

k3--- un 1+ un un 1–+ + 1

3--- Pn 1+ Pn Pn 1–+ + =+

Time Average “Filters”

t

un 1+ u n un 1–+ +

3------------------------------------------------------

pn 1+ p n pn 1–+ +

3------------------------------------------------------

p t

t

in MDin MDNastranNastran



● Solution

● Solve by decomposing A1 and applying it to the right-hand side ofthe above equation.

● Similar to classical Newmark-Beta method

A1 un 1+ A2 A3 un A4 un 1– + +=

Initial Conditions,from Previous

Dynamic Matrix

Applied Force

where A1 M t2

B 2t K 3+ + =

A2 1 3 Pn 1+ Pn Pn 1–+ + =

A3 2M t2

K 3– =

A4 M t2

B 2t K 3–+– =



● M, B, and K do not change with time.

● A1 needs to be decomposed only once if t is unchangedthroughout the entire solution. If t is changed, A1 must beredecomposed (which may be a costly operation).

● The output time interval may be greater than the solutiontime interval (i.e., use solution t of 0.001 second andoutput results every fifth time step or with output t of0.005 second).


DAMPING IN DIRECT TRANSIENT RESPONSE

● Damping matrix B comprised of several matrices:

where B1 = damping elements (VISC,DAMP) + B2GGB2 = B2PP direct input matrix + transfer functionsG = overall structural damping coefficient (PARAM,G)W3 = frequency of interest - rad/sec (PARAM,W3)K1 = global stiffness matrixGe = element structural damping coefficient (GE on MATi entry)W4 = frequency of interest - rad/sec (PARAM,W4)KE = element stiffness matrix

● Transient analysis does not permit complex coefficients. Therefore,structural damping is included by means of equivalent viscous damping.

● The default values for W3, W4 are 0.0. In this case, they cause associateddamping terms to be ignored.

B B1 B2 GW3--------K1 1

W4-------- GEKE++ +=


MODAL TRANSIENT RESPONSE

● Transform from physical to modal coordinates.(1)

● Temporarily remove damping. The equation of motion becomes

(2)● Substitute Equation 1 into Equation 2 to obtain

(3)● Pre-multiply by [T] to obtain

(4)

where TM = modal mass matrix (diagonal) TK = modal stiffness matrix (diagonal) TP = modal force vector

u =

M u·· Ku+ P t =

M ·· K+ P t =

T M ·· T K+ T P t =


MODAL TRANSIENT RESPONSE (Cont.)

● Equation 4 can be written as decoupled SDOF systems:

(5)

where mi = ith modal mass ki = ith modal stiffness pi = ith modal force

mi·· ki+ pi t=ii ii


DAMPING IN MODAL TRANSIENT RESPONSE

● If damping matrix B exists, then the assumption is made that it is notdiagonalized by :

● The coupled problem is solved using modal coordinates utilizing the directtransient response Newmark-Beta-type numerical integration.

● where

TB diagonal

A1 n 1+ A2 A3 n A4 n 1– + +=

Initial Conditions,from Previous

Dynamic Matrix

Applied Force

A1 T M

t2--------- B

2t--------- K

3----+ + =

A2 13--- T Pn 1+ Pn Pn 1–+ + =

A3 T 2M

t2--------- K

3----– =

A4 T M

t2---------– B

2t--------- K

3----–+ =


DAMPING IN MODAL TRANSIENT RESPONSE(Cont.)

● If modal damping is used, then each mode hasdamping bi.

● The equations of motion become uncoupled



● Use Duhamel’s integral to solve for modal response asdecoupled SDOF systems.

● Duhamel’s integral:



● Most efficient to use modal damping ratios since equations aredecoupled

● TABDMP1 Bulk Data entry defines the modal damping ratios.

● Type = G (default), CRIT, or Q

● Example: for 10% critical damping

1 2 3 4 5 6 7 8 9 10TABDMP1 ID TYPE +ABC

+ABC f1 g1 f2 g2 f3 g3 f4 g4 +DEF

+DEF f5 g5 ... ... ENDT

b bcr G 2= =

Q 1 2 =

= 1/G

CRIT = 0.10

Q = 5.0

G = 0.2



● The TABDMP1 Bulk Data entry is selected with the SDAMPING CaseControl command.

● fi (units: Hz) and gi define pairs of frequencies and dampings. Straight-line interpolation is used for modal frequencies between consecutive fis.Linear extrapolation is used at the ends of the table. ENDT ends thetable input.

● Example: Assume modes at 1.0, 2.5, 3.6, and 5.5 Hz.

● May add nonmodal damping (PARAM, G; VISC; DAMP; GE on MATi entry)● Computational cost due to coupled B causing direct integration to

be used● Recommended practice: Use only modal damping (TABDMP1) in modal

transient response analysis. If discrete damping is desired, use directtransient response analysis.

f g f g2.0 0.10 1.0 0.023.0 0.18 2.5 0.144.0 0.13 3.6 0.156.0 0.13 5.5 0.13

Entered Computed Calculated byCalculated byExtrapolationExtrapolation

Calculated byCalculated byInterpolationInterpolation


DATA RECOVERY IN MODAL TRANSIENTRESPONSE

● Recover physical response as the summation of the modalresponses.

● Not as large a computational penalty for changing t inmodal transient response as in direct. However, theconstant is still recommended.

● The output time interval may be greater than the solutiontime interval.

u =


MODE TRUNCATION

● May not need all of computed modes. Often only thelowest few will suffice for dynamic response calculation.

● PARAM,LFREQ specifies the lower limit on the frequencyrange of retained modes.

● PARAM,HFREQ specifies the upper limit on the frequencyrange of retained modes.

● PARAM,LMODES specifies the number of the lowestmodes to be retained.

● Truncating high-frequency modes truncates high-frequency response.


SELECTIVE MODES DELETION

● By default all modes calculated are included in theresponse analysis

● Parameters available for excluding modes at low or highend● Param,lfreq,value--modes below frequency “value” are not included

● Param,hfreq,value--modes above frequency “value” are not included

● Param,lmodes,number—only the lowest “number” modes are included

● Similar set of parameters are available for fluid modes

Default value = 0.0Default value = 0.0

Default value = 1.E+30Default value = 1.E+30

Default number = 0Default number = 0 --> the retained modes are> the retained modes aredetermined by parameter LFREQ and HFREQdetermined by parameter LFREQ and HFREQ


DELETING HIGH/LOW MODES

● Command to delete modes below and/or above certainfrequencies

FLSFSEL LFREQ = , HFREQ = ,

LMODES =

0.00.0

fsfs11

1.+301.+30

fsfs22

0.00.0

msms

FLSFEL is a CASE CONTROL commandFLSFEL is a CASE CONTROL command

fsfs11 = lower freq range for= lower freq range forstructure (real number)structure (real number)

fs2 = upper freq range forfs2 = upper freq range forstructure (real number)structure (real number)

ms = number of lowestms = number of lowestmodes to use for structuremodes to use for structureportion of modelportion of modelExample:Example:

FLSFSEL HFREQ = 4.FLSFSEL HFREQ = 4.


SELECTIVE MODES DELETION (Cont.)

● Previous commands remove high/low modes● MODESELECT allows you to remove selective mode

● Use to either:● Include a selective set of modes ( n > 0 )

or● Excludes a selective set of modes ( n < 0 )

MODESELECT is a CASE CONTROL commandMODESELECT is a CASE CONTROL command


SELECTIVE MODES DELETION (Cont.)

● Note that the modes that are deleted will not participate inthe response● This may lead in incorrect results if the wrong modes are deleted

Examples:1. Selects all 10 modes excluding modes 6 and 7.

SET 100 = 1 THRU 10 EXCEPT 6,7MODESELECT = 100orSET 200 = 6,7MODESELECT = -200


TRANSIENT EXCITATION

● Define force as a function of time.

● Several methods in MD Nastran:

● TLOAD1 “Brute force”; ordered time, force pairs table input

● TLOAD2 Efficient definition for analytical-type loadings

● LSEQ Generates dynamic loads from static loads (notrecommended)


TLOAD1 ENTRY● Defines excitation in the form:

Where A = spatial load distribution and scale factor(DAREA, static load, thermal load, or LSEQ)

= DELAY entry (Integer) or time delay (Real),Default=0.

F(t-) = TABLEDi entry

● DELAY defines DOFs and time delay● TID – specifies TABLEDi for defining time and force pairs.● Selected by DLOAD Case Control command.

P t AF t – =

Delay SID P1 C1 T1 P2 C2 T2Delay SID P1 C1 T1 P2 C2 T2PiPi –– GRID numberGRID numberCiCi –– CompenentCompenent numbernumberTiTi –– Time delay for designated point Pi and componentTime delay for designated point Pi and component CiCi (real)(real)


TLOAD1 ENTRY (Cont.)

● Excitation is defined by TYPE.

● Only loads (first row) will be discussed in this section. Forenforced motion, see section 12.


TLOAD2 ENTRY

● Defines excitation in the form of

where

A Defined as a spatial load distribution and scale factor(DAREA, static load, thermal load, or LSEQ)

Defined on a DELAY entry TYPE Defined as in the TLOAD1 T1,T2 Time constants (T2>T1) F Frequency (Hz) P Phase angle (degrees) C Exponential coefficient B Growth coefficient

● Selected by the DLOAD Case Control command

P(t) =

t̃ 0< t̃ T2 T1–>or

At̃BeCt̃ 2Ft̃ P+ cos

0

t̃ t T1 ––=

t<(T1+t<(T1+) or) ort>(T2+t>(T2+))(T1+(T1+)) tt (T2+(T2+))


LOAD SET COMBINATION - DLOAD

● The applied load PC is constructed from a combination ofcomponent load sets PK

where SC = overall scale factorSK = scale factor for k-th load setPK = SID of TLOAD

● TLOAD1s and TLOAD2s must have unique SIDs.● Use the DLOAD entry to combine TLOADs.● The DLOAD Bulk Data entry is selected by DLOAD Case

Control command.

PC SC SKPKK=

1 2 3 4 5 6 7 8 9 10

DLOAD SID SC SI P1 S2 P2 -etc-


DAREA ENTRY

● Defines the degree of freedom where the dynamic load isto be applied to the scale factor.

● Can point to static load directly, e.g., FORCE, PLOAD4,etc. (recommended) instead of DAREA

● Relationship to other input:DLOAD

TLOAD

DelayTime Lag

DAREAScaleFactor

Dynamic

Temporal Spatial

Case ControlBulk Data

,,SIDSID,P1,C1,A1,P2,C1,A2,P1,C1,A1,P2,C1,A2

i,SID,i,SID,EXCITEIDEXCITEID,,DELAYDELAY

LoadLoad


EXAMPLE

4000.22935TLOAD1

TIDTYPEDELAYEXCITEIDSIDTLOAD1

1.5.23029FORCE

N3N2N1FCIDGSIDFORCE

1.53.02.00.21.50.10.0.

ENDT1.54.0

40TABLED1

Y4X4Y3X3Y2X2Y1X1

YAXISXAXISIDTABLED1

Result is the load specified by the TLOAD1, scaled by 5.2, delayedby 0.2 seconds, and applied to grid point 30, component T1

Type = 0 => StructuralType = 0 => StructuralLoadLoad


STATIC LOAD – INDIRECT METHOD● Defines static loads that are being applied dynamically.● The LSEQ Bulk Data entry is selected by the LOADSET Case Control

command.● Contains a EXCITEID entry to identify the loadset for use with the

TLOAD entries.● Relationship to other input

DLOAD LOADSETCase Control

Bulk Data TLOAD LSEQ

Dynamic EXCITEID Static Load

Temporal Cross Reference Spatial


STATIC LOAD – INDIRECT METHOD (Cont.)

DLOAD = 25 LOADSET = 27

TLOAD1 25 28 5

LSEQ 27 28 100

PLOAD4 100 ….

TABLED1 5 ….

LOADSET must appear above allLOADSET must appear above all subcasessubcases..Only one LOADSET may be specified perOnly one LOADSET may be specified persuperelementsuperelement


● Defines static loads that are being applied dynamically.

● The EXCITEID references the static load (e.g., PLOAD4)directly

DLOADCase Control

Bulk Data TLOAD

Dynamic EXCITEID

Temporal Static Load

STATIC LOAD – DIRECT METHOD


STATIC LOAD – DIRECT METHOD (Cont.)

DLOAD = 25

TLOAD1 25 100 5

PLOAD4 100 …..

TABLED1 5 ….


● Remember the 1/3 “smearing” of applied loads. This will smooth the force anddecreases apparent frequency content.

● Avoid discontinuous forces. These may cause different results on different computers.

● If NDt causes a solution at ABC, then MD Nastran should select the average force B.● However, due to numerical roundoff, NDt on one computer may be at time A- and will

give force A. On another computer, NDt may be at time C+ and will give force C.● The integration results will differ depending on whether the force at NDt is A, B, or C.

TRANSIENT EXCITATION CONSIDERATIONS

Force

Time

C+

A-A

B

C


TRANSIENT EXCITATION CONSIDERATIONS(Cont.)

● Smooth a discontinuous force over one t.

Force

Time

= original force= smoothed force


INITIAL CONDITIONS● May impose initial displacement and/or velocity in transient response

via the TIC Bulk Data entry.● The IC Case Control command selects the TIC entry.● Be careful - initial conditions for unspecified DOFs are set to zero.● Initial conditions may be specified only for A-set DOFs.● Initial conditions are used to determine the values of {u0}, {u-1}, {P0},

and {P-1} used in calculating {u1}. The acceleration for all points isassumed to be zero for t < 0 (constant velocity).

● The load specified by the user at t = 0 is replaced by:

u 1– u0 u·0 t–=

P 1– Ku 1– Bu·0 +=

P0 Ku0 Bu· 0 +=


INITIAL CONDITIONS (Cont.)

● The recommended practice for any type of dynamic excitation is to use atleast one time step of zero excitation prior to applying the dynamic force.

Force

Time


Transient Initial Condition TIC

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

TIC SID G C U0 V0

Example:TIC 1 3 2 5.0 -6.0

Field ContentsSID Set identification number. (Integer > 0)G Grid, scalar, or extra point identification number. (integer > 0)C

U0 Initial displacement. (Real)V0 Initial velocity. (Real)

Remarks:1.

2. If no TIC set is selected in the Case Control Section, all initial conditions are assumed to be zero.3. Initial conditions for coordinates not specified on TIC entries will be assumed zero.

Bulk Data Entry

Defines values for the initial conditions of variables used in structural transient analysis. Both displacementand velocity values may be specified at independent degrees of freedom. This entry may not be used for heattransfer analysis.

Component numbers. (Integer zero or blank for scalar or extra points, any one of the Integers 1through 6 for a grid point.)

Transient initial condition sets must be selected with the Case Control command IC = SID. Notethe use of IC in the Case Control command versus TIC on the Bulk Data entry. For heat transfer,the IC Case Control command selects TEMP or TEMPD entries for

IC = SID in Case ControlIC = SID in Case Control

.. ..


TSTEP ENTRY● Select integration time step for direct and modal transient response.

● Integration errors increase with increasing natural frequency.● Recommended t is to use at least eight solution time steps per period (cycle) of

response.

● The TSTEP Bulk Data entry controls solution and output t, and is selected bythe TSTEP Case Control command.

● The cost of integration is directly proportional to the number of time stepswhen t is constant.

● Use an adequate length of time to properly capture long-period (lowfrequency) response.

● User may change t during a run. is assumed constant for t < Nt1 . calculates new initial conditions for the integration based on {un} and the

calculated velocity and acceleration at the transition. The assumption ofuniform acceleration assures a smooth transition when t is changed.

u·0 1t1--------- uN uN 1–– =

u··0 1

t12

--------- uN 2uN 1–– uN 2–+ =

u··

u··

Recommended:TT 1./(10*1./(10*fmaxfmax))


TSTEP ENTRY (Cont.)

● The initial conditions for the new integration are:

● Note: New matrices A1 - A4 (page S7-7) areassembled, and the new A1 must be decomposed.

UniformAcceleration

u0 uN =

u 1– uN t2 u0·

– 12---t2

2 u··0 –=

P0 PN =

P 1– Ku 1– Bu 1–·

M u··

1– + +=

K= u 1– Bu0·

t2u··0–

M u··0 + +


Defines time step intervals at which a solution will be generated and output in transient analysis.

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

TSTEP SID N1 DT1 NO1N2 DT2 NO2-etc.-

Example:TSTEP 2 10 0.001 5

9 0.01 1

Field ContentsSID Set identification number. (Integer > 0)Ni Number of time steps of value DTi. (integer 1)DTi Time increment. (Real 0.0)NOi Skip factor for output. Every NOi-th step will be saved for output. (Integer > 0; Default = 1)

Remarks:1.2.

3.

4.

5.

6.

Bulk Data Entry

Transient Time Step

See the MSC.Nastran Basic Dynamic Analysis User’s Guide, Chapter 10 for a discussion ofconsiderations leading to the selection of time steps.In modal frequency response analysis (SOLs 111 and 146), this entry is required only whenTLOADi is requested; i.e., when Fourier methods are selected.The maximum and minimum displacement at each time step and the SIL numbers of thesevariables can be printed by altering DIAGON(30) before the transient module TRD1 and by alteringDIAGOFF(30) after the module. This is useful for runs that terminate dueFor hear transfer analysis in SOL 159, use the TSTEPNL entry.

TSTEP

TSTEP entries must be selected with the Case Control command TSTEP = SID.Note that the entry permits changes in the size of the time step during the course of the solution.Thus, in the example shown, there are 10 time steps of value .001 followed by 9 time steps of value.01. Also, the user has requested that output be reco

TSTEP = SID in CASE ControlTSTEP = SID in CASE Control


DYNAMIC DATA RECOVERY

● Three options for recovering displacements and stresses in modal solutions:mode displacement method, matrix method, and mode acceleration method

● The mode displacement method computes physical displacements directly frommodal displacements and then computes element stresses from the physicaldisplacements. The number of operations is proportional to number of time steps(T).

● The matrix method computes displacements per mode and element stresses permode and then computes physical element stresses as the summation of modalelement stresses. Costly operations are proportional to the number of modes (H).

● Since H is usually << T, the matrix method is cheaper.

● The matrix method is the default and is the recommended method for most cases.The mode displacement method can be selected via PARAM, DDRMM, -1.

● The mode acceleration method automatically accounts for the quasi-staticresponse of all high frequency modes (See Appendix C ). (Requires modedisplacement method)

TH

methodntdisplacememodeofCostmethodmatrixofCost

Number of ModesNumber of Modes

Number of Time StepsNumber of Time Steps


MODAL TRANSIENT VERSUS DIRECT TRANSIENT

Modal DirectSmall Model XLarge Model XFew Time Steps XMany Time Steps XHigh Frequency Excitation XNonlinearities XInitial Conditions X XModal Damping X


SOLUTION CONTROL FOR TRANSIENT ANALYSIS

● Executive Control Section● SOL (for required input see below)

● Case Control Section● DLOAD (both - required)● LOADSET (both - optional)● METHOD (modal - required)● SDAMPING (modal - optional)● IC (Physical) (both - optional)● IC (Modal) (modal – optional)● TSTEP (both - required)

Method StructuredSolution

SequencesDirect 109Modal 112


SOLUTION CONTROL FOR TRANSIENTANALYSIS (Cont.)

● Bulk Data Section ASET,OMIT (both - optional) EIGRL or EIGR (modal - required) TSTEP (both - required) TIC (both - optional) TLOADi (both - required) LSEQ (both - optional) DAREA (both – optional) TABLEDi (both - optional) DELAY (both - optional) DLOAD (both - optional) TABDMP1 (modal - optional)


CASE CONTROL OUTPUT

● Grid output● ACCELERATION● DISPLACEMENT (or VECTOR)● GPSTRESS● NLLOAD (nonlinear load output)● OLOAD (output applied load)● SACCELERATION (solution set output - A-set in direct● SDISPLACEMENT solutions, modal variables in● SVELOCITY modal solutions)● SVECTOR (A-set eigenvector)● SPCFORCES● VELOCITY● MPCFORCE

● Element output● ELSTRESS (or STRESS)● ELFORCE (or FORCE)● STRAIN

● Special● OTIME (controls solution output times)


WORKSHOP #3

DIRECT TRANSIENT RESPONSE


WORKSHOP #3 - DIRECT TRANSIENT RESPONSE

Using the direct method, determine the transient response of the flat rectangular plate,created in Workshop 1, subject to time-varying excitation. This example structure isexcited by 1 psi pressure load over the total surface of the plate varying at 250 Hz. Inaddition, a 50 lb force is applied at a corner of the tip also varying at 250 Hz, but 180o

out-of-phase with the pressure load. Both time dependent dynamic loads are applied forthe duration of 0.008 seconds only. Use structural damping of g=0.06 and convert thisdamping to equivalent viscous damping at 250 Hz. Carry the analysis for 0.04 seconds.Below is a finite element representation of the flat plate. It also contains the loads andboundary constraints.

Figure 7-1. Loads and Boundary Conditions.

1 ps i ov e r the tota l s u rfa ce

5 0.0 0


WORKSHOP # 3$ wkshp3.dat$$SOL 109CENDTITLE= TRANSIENT RESPONSE WITH TIME DEPENDENT PRESSURE AND POINT LOADSSUBTITLE= USE THE DIRECT METHODECHO= PUNCHSPC= 1SET 1= 11, 33, 55DISPLACEMENT= 1SUBCASE 1DLOAD= 700 $ SELECT TEMPORAL COMPONENT OF TRANSIENT LOADINGTSTEP= 100 $ SELECT INTEGRATION TIME STEPS$OUTPUT (XYPLOT)XGRID=YESYGRID=YESXTITLE= TIME (SEC)YTITLE= DISPLACEMENT RESPONSE AT LOADED CORNERXYPLOT DISP RESPONSE / 11 (T3)YTITLE= DISPLACEMENT RESPONSE AT CENTER TIPXYPLOT DISP RESPONSE / 33 (T3)YTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNERXYPLOT DISP RESPONSE / 55 (T3)$BEGIN BULKparam,post,0PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$INCLUDE 'plate.bdf'$ENDDATA

• Use the following file as the starting point


SOLUTION FILE FOR WORKSHOP #3$$ soln3.dat$ID SEMINAR, PROB3SOL 109CENDTITLE= TRANSIENT RESPONSE WITH TIME DEPENDENTPRESSURE AND POINT LOADSSUBTITLE= USE THE DIRECT METHODECHO= PUNCHSPC= 1SET 1= 11, 33, 55DISPLACEMENT= 1SUBCASE 1DLOAD= 700 $ SELECT TEMPORAL COMPONENT OFTRANSIENT LOADINGTSTEP= 100 $ SELECT INTEGRATION TIME STEPS$OUTPUT (XYPLOT)XGRID=YESYGRID=YESXTITLE= TIME (SEC)YTITLE= DISPLACEMENT RESPONSE AT LOADED CORNERXYPLOT DISP RESPONSE / 11 (T3)YTITLE= DISPLACEMENT RESPONSE AT CENTER TIPXYPLOT DISP RESPONSE / 33 (T3)YTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNERXYPLOT DISP RESPONSE / 55 (T3)$BEGIN BULKparam,post,0PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$INCLUDE 'plate.bdf'$

$$ SPECIFY STRUCTURAL DAMPING$ 3 PERCENT AT 250 HZ. = 1571 RAD/SEC.$PARAM, G, 0.06PARAM, W3, 1571.$$ TIME VARYING PRESSURE LOAD (250 HZ)$TLOAD2, 200, 400, , 0, 0., 8.E-3, 250., -90.PLOAD2, 400, 1., 1, THRU, 40$$ APPLY POINT LOAD OUT OF PHASE WITH PRESSURE LOAD$TLOAD2, 500, 600, , 0, 0., 8.E-3, 250., 90.$DAREA, 600, 11, 3, 1.$$ COMBINE LOADS$DLOAD, 700, 1., 1., 200, 50., 500$$ SPECIFY INTERGRATION TIME STEPS$TSTEP, 100, 100, 4.0E-4, 1$ENDDATA


PARTIAL OUTPUT FILE FOR WORKSHOP #30 SUBCASE 1ML

0SUBCASE 1

POINT-ID = 11D I S P L A C E M E N T V E C T O R

TIME TYPE T1 T2 T3 R1 R2 R30.0 G 0.0 0.0 0.0 0.0 0.0 0.04.000000E-04 G 0.0 0.0 -2.173954E-02 1.105127E-02 1.051570E-02 0.08.000000E-04 G 0.0 0.0 -7.205779E-02 2.849259E-02 2.853552E-02 0.01.200000E-03 G 0.0 0.0 -1.433632E-01 4.084704E-02 4.916111E-02 0.01.600000E-03 G 0.0 0.0 -2.060064E-01 3.055025E-02 6.223300E-02 0.02.000000E-03 G 0.0 0.0 -2.459063E-01 1.403890E-03 6.812792E-02 0.0

0 SUBCASE1


TIME TYPE T1 T2 T3 R1 R2 R30.0 G 0.0 0.0 0.0 0.0 0.0 0.04.000000E-04 G 0.0 0.0 -1.122321E-02 9.220317E-03 6.136194E-03 0.08.000000E-04 G 0.0 0.0 -4.424645E-02 2.577100E-02 2.014592E-02 0.01.200000E-03 G 0.0 0.0 -1.030766E-01 3.820221E-02 3.921979E-02 0.01.600000E-03 G 0.0 0.0 -1.756299E-01 2.929389E-02 5.577315E-02 0.02.000000E-03 G 0.0 0.0 -2.443358E-01 1.775214E-03 6.761315E-02 0.02.400000E-03 G 0.0 0.0 -2.882814E-01 -2.449635E-02 7.435711E-02 0.02.800000E-03 G 0.0 0.0 -2.843486E-01 -3.428704E-02 7.112917E-02 0.0

0 SUBCASE1


TIME TYPE T1 T2 T3 R1 R2 R30.0 G 0.0 0.0 0.0 0.0 0.0 0.04.000000E-04 G 0.0 0.0 -2.850083E-03 7.786633E-03 4.615276E-03 0.08.000000E-04 G 0.0 0.0 -1.992840E-02 2.321643E-02 1.681915E-02 0.01.200000E-03 G 0.0 0.0 -6.642522E-02 3.539676E-02 3.502940E-02 0.0


PARTIAL OUTPUT FILE FOR WORKSHOP #300 SUBCASE 10 X Y - O U T P U T S U M M A R Y ( R E S P O N S E )0 SUBCASE CURVE FRAME CURVE ID./ XMIN-FRAME/ XMAX-FRAME/ YMIN-FRAME/ X FOR YMAX-FRAME/ X FOR

ID TYPE NO. PANEL : GRID ID ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX0 1 DISP 1 11( 5) 0.000000E+00 4.000000E-02 -2.623262E-01 2.400000E-03 2.823412E-01 5.200000E-03

0.000000E+00 4.000000E-02 -2.623262E-01 2.400000E-03 2.823412E-01 5.200000E-030 1 DISP 2 33( 5) 0.000000E+00 4.000000E-02 -2.882814E-01 2.400000E-03 3.220591E-01 5.200000E-03

0.000000E+00 4.000000E-02 -2.882814E-01 2.400000E-03 3.220591E-01 5.200000E-030 1 DISP 3 55( 5) 0.000000E+00 4.000000E-02 -3.166115E-01 2.800000E-03 3.570641E-01 5.200000E-03

0.000000E+00 4.000000E-02 -3.166115E-01 2.800000E-03 3.570641E-01 5.200000E-03


PARTIAL OUTPUT FILE FOR WORKSHOP #3(Cont.)


WORKSHOP #4

MODAL TRANSIENT RESPONSE


WORKSHOP #4 - MODAL TRANSIENTRESPONSE

Using the Modal Method, determine the transient response of the flat rectangular plate, created inWorkshop 1, subject to time-varying excitation. This example structure is excited by a 1 psi pressureload over the total surface of the plate varying at 250 Hz. In addition, a 25 lb force is applied at acorner of the tip also varying at 250 Hz, but starting 0.004 seconds after the pressure load begins.Both time-dependent dynamic loads are applied for a duration of 0.008 seconds. Use a modaldamping of = 0.03 for all modes. Carry out the analysis for 0.04 seconds.

Below is a finite element representation of the flat plate. It also contains the loads and boundaryconstraints.

XY

Z

12345

25.0

1234512345

1234512345 1.0 psi over the total surface

XY

Z


WORKSHOP # 4● Use the following file as a starting point

BEGIN BULKPARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE PROBLEM$INCLUDE 'plate.bdf'$ENDDATA

$$ wkshp4.dat$SOL 112diag 8CENDTITLE = TRANSIENT RESPONSE WITH TIME DEPENDENT PRESSURE

AND POINT LOADSSUBTITLE = USE THE MODAL METHODECHO = UNSORTEDSPC = 1SET 111 = 11, 33, 55DISPLACEMENT(SORT2) = 111SDAMPING = 100SUBCASE 1METHOD = 100DLOAD = 700TSTEP = 100$OUTPUT (XYPLOT)$


SOLUTION FILE FOR WORKSHOP #4ID SEMINAR, PROB4$$ soln4.dat$ID SEMINAR, PROB4SOL 112diag 8CENDTITLE = TRANSIENT RESPONSE WITH TIME DEPENDENTPRESSURE AND POINT LOADSSUBTITLE = USE THE MODAL METHODECHO = UNSORTEDSPC = 1SET 111 = 11, 33, 55DISPLACEMENT(SORT2) = 111SDAMPING = 100SUBCASE 1METHOD = 100DLOAD = 700TSTEP = 100$OUTPUT (XYPLOT)XGRID=YESYGRID=YESXTITLE= TIME (SEC)YTITLE= DISPLACEMENT RESPONSE AT LOADED CORNERXYPLOT DISP RESPONSE / 11 (T3)YTITLE= DISPLACEMENT RESPONSE AT TIP CENTERXYPLOT DISP RESPONSE / 33 (T3)YTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNERXYPLOT DISP RESPONSE / 55 (T3)$

BEGIN BULKPARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLEPROBLEM$INCLUDE 'plate.bdf'$$ EIGENVALUE EXTRACTION PARAMETERS$EIGRL, 100, , ,5$$ SPECIFY MODAL DAMPING$TABDMP1, 100, CRIT,+, 0., .03, 10., .03, ENDT$$ TIME VARYING PRESSURE LOAD (250 HZ)$TLOAD2, 200, 400, , 0, 0., 8.E-3, 250., -90.PLOAD2, 400, 1., 1, THRU, 40$$ APPLY POINT LOAD (250 HZ)$TLOAD2, 500, 600,610, 0, 0.0, 8.E-3, 250., -90.$DAREA, 600, 11, 3, 1.DELAY, 610, 11, 3, 0.004$$ COMBINE LOADS$DLOAD, 700, 1., 1., 200, 25., 500$$ SPECIFY INTERGRATION TIME STEPS$TSTEP, 100, 100, 4.0E-4, 1$ENDDATA



R E A L E I G E N V A L U E S(BEFORE AUGMENTATION OF RESIDUAL VECTORS)


1 1 7.055894E+05 8.399937E+02 1.336891E+02 1.000000E+00 7.055894E+052 2 1.877186E+07 4.332651E+03 6.895628E+02 1.000000E+00 1.877186E+073 3 2.811177E+07 5.302053E+03 8.438480E+02 1.000000E+00 2.811177E+074 4 1.929422E+08 1.389036E+04 2.210720E+03 1.000000E+00 1.929422E+085 5 2.221657E+08 1.490523E+04 2.372240E+03 1.000000E+00 2.221657E+08

R E A L E I G E N V A L U E S(AFTER AUGMENTATION OF RESIDUAL VECTORS)


1 1 7.055894E+05 8.399937E+02 1.336891E+02 1.000000E+00 7.055894E+052 2 1.877186E+07 4.332651E+03 6.895628E+02 1.000000E+00 1.877186E+073 3 2.811176E+07 5.302053E+03 8.438479E+02 1.000000E+00 2.811176E+074 4 1.929422E+08 1.389036E+04 2.210720E+03 1.000000E+00 1.929422E+085 5 2.221657E+08 1.490523E+04 2.372240E+03 1.000000E+00 2.221657E+086 6 2.351324E+08 1.533403E+04 2.440486E+03 1.000000E+00 2.351324E+087 7 7.974902E+08 2.823987E+04 4.494515E+03 1.000000E+00 7.974902E+088 8 1.453224E+09 3.812117E+04 6.067173E+03 1.000000E+00 1.453224E+099 9 2.625274E+09 5.123743E+04 8.154690E+03 1.000000E+00 2.625274E+0910 10 4.154733E+09 6.445722E+04 1.025868E+04 1.000000E+00 4.154733E+0911 11 4.205890E+09 6.485284E+04 1.032165E+04 1.000000E+00 4.205890E+0912 12 3.216783E+10 1.793539E+05 2.854506E+04 1.000000E+00 3.216783E+10


PARTIAL OUTPUT FILE FOR WORKSHOP #4 (Cont.)POINT-ID = 11

D I S P L A C E M E N T V E C T O R

TIME TYPE T1 T2 T3 R1 R2 R30.0 G 0.0 0.0 0.0 0.0 0.0 0.04.000000E-04 G -5.468071E-16 8.320540E-17 2.038250E-04 -3.199117E-06 2.623944E-05 2.230269E-168.000000E-04 G -2.152597E-15 6.019959E-15 1.980818E-03 -1.510274E-05 -2.074038E-04 3.010193E-151.200000E-03 G -4.731214E-15 1.976858E-14 6.911292E-03 4.029346E-07 -1.653521E-03 9.216824E-151.600000E-03 G -8.185502E-15 3.999204E-14 1.407448E-02 2.646134E-05 -3.815677E-03 1.824066E-142.000000E-03 G -1.174987E-14 6.113124E-14 2.121053E-02 4.361140E-05 -5.847335E-03 2.759835E-14


TIME TYPE T1 T2 T3 R1 R2 R30.0 G 0.0 0.0 0.0 0.0 0.0 0.04.000000E-04 G -8.130047E-16 5.783684E-16 2.019039E-04 1.682611E-08 2.577562E-05 8.783539E-168.000000E-04 G -5.437500E-15 9.431799E-15 1.971792E-03 -1.131783E-09 -2.092973E-04 7.575778E-151.200000E-03 G -1.469855E-14 2.928089E-14 6.911632E-03 -3.483805E-08 -1.653002E-03 2.206479E-141.600000E-03 G -2.785101E-14 5.819658E-14 1.409060E-02 -1.660608E-08 -3.811871E-03 4.290151E-142.000000E-03 G -4.146778E-14 8.828006E-14 2.123710E-02 1.814818E-10 -5.841007E-03 6.439503E-142.400000E-03 G -4.989242E-14 1.075860E-13 2.615920E-02 1.753988E-08 -7.446125E-03 7.806971E-142.800000E-03 G -4.913322E-14 1.067912E-13 2.613206E-02 2.261718E-08 -7.588134E-03 7.724409E-14


TIME TYPE T1 T2 T3 R1 R2 R30.0 G 0.0 0.0 0.0 0.0 0.0 0.04.000000E-04 G -2.313598E-15 1.279507E-15 2.038511E-04 3.216275E-06 2.622230E-05 1.886289E-158.000000E-04 G -1.738001E-14 1.431151E-14 1.980840E-03 1.514255E-05 -2.074704E-04 1.465446E-141.200000E-03 G -4.910303E-14 4.295687E-14 6.911171E-03 -5.569029E-07 -1.653289E-03 4.203939E-141.600000E-03 G -9.447002E-14 8.441534E-14 1.407446E-02 -2.646987E-05 -3.815674E-03 8.127868E-142.000000E-03 G -1.412618E-13 1.273955E-13 2.121054E-02 -4.360946E-05 -5.847338E-03 1.216724E-132.400000E-03 G -1.707585E-13 1.548346E-13 2.611754E-02 -6.863064E-05 -7.455783E-03 1.473061E-132.800000E-03 G -1.686479E-13 1.534346E-13 2.608514E-02 -7.741245E-05 -7.598883E-03 1.456237E-133.200000E-03 G -1.287358E-13 1.173798E-13 1.986141E-02 -5.680266E-05 -5.762664E-03 1.111591E-133.600000E-03 G -5.511781E-14 5.048424E-14 8.672326E-03 -2.997875E-05 -2.603049E-03 4.766799E-14



X Y - O U T P U T S U M M A R Y ( R E S P O N S E )

0 SUBCASE CURVE FRAME CURVE ID./ XMIN-FRAME/ XMAX-FRAME/ YMIN-FRAME/ X FOR YMAX-FRAME/ X FORID TYPE NO. PANEL : GRID ID ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX

0 1 DISP 1 11( 5) 0.000000E+00 4.000000E-02 -1.666260E-01 9.200000E-03 1.482653E-01 6.400000E-030.000000E+00 4.000000E-02 -1.666260E-01 9.200000E-03 1.482653E-01 6.400000E-03



NAS102,Seminar 8, March 2007Copyright2007 MSC.Software Corporation Sem 8-1

SECTION 8

FREQUENCY RESPONSE ANALYSIS


INTRODUCTION TO FREQUENCY RESPONSEANALYSIS

• Compute the response to oscillatory excitation.• Excitation explicitly defined in the frequency domain -

all of the applied forces are known at each forcingfrequency.

• Computed response usually includes nodaldisplacements and element forces and stresses.

• The computed responses are the complex numbersdefined as magnitude and phase (with respect to theforcing) or as real and imaginary components.

• Two categories of analysis - direct and modal.


DIRECT FREQUENCY RESPONSE

• Dynamic equation of motion:(1)

• PARAM,G and GE on MATi entry do not form a damping matrix. They form acomplex stiffness matrix

where K1 = global stiffness matrixG = overall structural damping coefficient (PARAM,G)KE = element stiffness matrixGE = element structural damping coefficient (GE on MATi entry)

• Contrast this with transient response analysis

• Solve the equation by inserting ω to form a complex left-hand side, and thensolve it similar to a static problem (using complex arithmetic).

2

M– iB K+ + u P =

K 1 iG+ K1 i GEkE+=

BTRANS B1

B2 G

W3---------K

1 1W4--------- GEkE+ + +=


MODAL FREQUENCY RESPONSE

• Convert to modal coordinates and solve as decoupledSDOF systems

• Much quicker to solve this equation than in directmethod

• Decoupled procedure can be used only if either nodamping is present or if modal damping alone (viaTABDMP1) is used. Otherwise, use the less efficientdirect approach (on smaller modal coordinatematrices) if non-modal damping (VISC,DAMP) ispresent.

i

Pi

mi2– ib i ki++

--------------------------------------------------=


EXCITATION DEFINITION

• Define force as a function of frequency.• Several methods in MD Nastran:

• RLOAD1 (defines frequency-dependent load in real and imaginaryforms)

• RLOAD2 (defines frequency-dependent load in magnitude andphase forms)

• LSEQ (generates dynamic loads from static loads)

• DLOAD Bulk Data entries are used to combinefrequency-dependent forces.

• RLOADi entries are selected by DLOAD Case Controlcommands.


FREQUENCY RESPONSE CONSIDERATIONS

• Exciting an undamped (or modal damped) system at 0.0 Hz gives the same resultsas a static analysis. Therefore, if the maximum excitation frequency is much lessthan the lowest resonant frequency of the system, a static analysis is sufficient.

• Very lightly-damped structures exhibit large dynamic responses for excitationfrequencies near resonant frequencies. A small change in the model (or runningit on another computer) may give large changes in such response.

• Use a fine-enough frequency step size (f) to adequately predict peak response.Use at least 5 points per half-power bandwidth.

• For maximum efficiency, use an uneven frequency step size. Use smaller Df inregions around resonant frequencies and larger Df in regions far away fromresonant frequencies.

Peak Response

Half-Power

Response

Frequency

Peak / = Half-Power Point

f1 f2


FREQi ENTRIESApplicable for Direct and Modal Method

• Select frequency step size.• The FREQ entry defines discrete excitation frequencies.• The FREQ1 entry defines fSTART, frequency increment, and the number of

increments.• The FREQ2 entry defines fSTART, fEND, and the number of logarithmic

intervals.

Applicable for Modal Method

• The FREQ3 entry defines F1, F2, and the number of frequencies inbetween using either a linear or log interpolation. Biased towards endpoints or center.

• The FREQ4 entry specifies a frequency at each resonant frequency andthe number of equally spaced excitation frequencies within the spread.

• The FREQ5 entry specifies a frequency range and fractions of the naturalfrequencies within that range.

• The FREQ3, FREQ4, and FREQ5 entries are available only for the modalmethod.


FREQi ENTRIES (CONT.)

• FREQi Bulk Data entries are selected by theFREQUENCY Case Control commands.

• All FREQi Bulk Data entries with the same set ID areused. Therefore, FREQ, FREQ1, FREQ2, FREQ3,FREQ4, and FREQ5 entries may all be used in ananalysis.


Defines a set of frequencies to be used in the solution of frequency response problems.

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

FREQ SID F1 F2 F3 F4 F5 F6 F7F8 F9 F10 -etc.-

Example:FREQ 3 2.98 3.05 17.9 21.3 25.6 28.8 31.2

29.2 22.4 19.3

Field ContentsSID Set identification number. (Integer > 0)Fi Frequency value in units of cycles per unit time. (Real 0.0)

Remarks:1. Frequency sets must be selected with the Case Control command FREQUENCY = SID.2.

3.

Bulk Data Entry

FREQFrequency List

All FREQi entries with the same frequency set identification numbers will be used. Duplicatefrequencies will be ignored. fN and fN-1 are considered duplicated if |fN - fN-1|<DFREQ*|fMAX - fMIN|,where DFREQ is a user parameter, with default of 10-5. f

In modal analysis, solutions for modal DOFs from rigid body modes at zero excitation frequenciesmay be discarded. Solutions for nonzero modes are retained.


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

FREQ1 SID F1 DF NDF

Example:FREQ1 6 2.9 0.5 13

Field ContentsSID Set identification number. (Integer > 0)F1 First frequency set. (Real 0.0)DF Frequency increment. (Real > 0.0)NDF Number of frequency increments. (Integer > 0; Default = 1)

Remarks:1. FREQ1 entries must be selected with the Case Control command FREQUENCY = SID.2. The units for F1 and DF are cycles per unit time.3. The frequencies defined by this entry are given by

where i = 1 to (NDF + 1)4.

5.

FREQ1Frequency List, Alternate Form 1


fi = F1 + DF * (i - 1)

All FREQi entries with the same frequency set identification numbers will be used. Duplicatefrequencies will be ignored. fN and fN-1 are considered duplicated if |fN - fN-1|<DFREQ*|fMAX - fMIN|,where DFREQ is a user parameter, with default of 10-5. fMAX and fMIN are the maximum andminimum excitation frequencies of the combined FREQi entries.In modal analysis, solutions for modal DOFs from rigid body modes at zero excitation frequenciesmay be discarded. Solutions for nonzero modes are retained.

Defines a set of frequencies to be used in the solution of frequency response problems by specification of astarting frequency, frequency increment, and the number of increments desired..


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

FREQ2 SID F1 F2 NF

Example:FREQ2 6 1 8 6

Field ContentsSID Set identification number. (Integer > 0)F1 First frequency. (Real > 0.0)F2 Last frequency. (Real > 0.0, F2 > F1)NF Number of logarithmic intervals. (Integer > 0; Default = 1)

Remarks:1. FREQ2 entries must be selected with the Case Control command FREQUENCY = SID.2. The units for F1 and F2 are cycles per unit time.3. The frequencies defined by this entry are given by

where d = (1/NF) * ln (F2/F1) and i = 1, 2, . . . , (NF + 1)

4.

5.

Bulk Data Entry

Defines a set of frequencies to be used in the solution of frequency response problems by specification of astarting frequency, final frequency, and the number of logarithmic increments desired..


fi = F1 * e (i - 1)d

In the example above, the list of frequencies will be 1.0, 1.4142, 2.0, 2.8284, 4.0, 5.6569, and 8.0cycles per unit timeAll FREQi entries with the same frequency set identification numbers will be used. Duplicatefrequencies will be ignored. fN and fN-1 are considered duplicated if |fN - fN-1|<DFREQ*|f MAX - fMIN|,where DFREQ is a user parameter, with default of 10-5. f


(Continued)

FNd

i

i

FFe

ff

1

1

21


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

FREQ3 SID F1 F2 TYPE NEF CLUSTER

Example:FREQ3 6 20 2000 LINEAR 10 2

Field ContentsSID Set identification number. (Integer > 0)F1 Lower bound of modal frequency range in cycles per unit time. (Real > 0.0)F2 Upper bound of modal frequency range in cycles per unit time. (Real>0.0, F2F1, Default = F1)TYPE

NEF

CLUSTER

Remarks:1.

2. FREQ3 entries must be selected with the Case Control command FREQUENCY = SID.3.

4.

Frequency List, Alternate Form 3 FREQ3

Defines a set of excitation frequencies for modal frequency response solutions by speciying a number ofexcitation frequencies between two modal frequencies.

LINEAR or LOG. Specifies linear or logarithmic interpolation between frequencies.(Character; Default = "LINEAR")

Number of excitation frequencies within each subrange including the end points. Thefirst subrange is between F1 and the first modal frequency within the bounds. Thesecond subrange is between first and second modal frequencies between the bounds.The

Specifies clustering of the excitation frequency near the end points of the range.SeeRemark3. (Real > 0.0; Default=1.0)

FREQ3 applies only to modal frequency-response solutions (SOLs 11, 111, 146, and 200) and isignored in direct frequency response solutions.

In the example above, there will be 10 frequencies in the interval between each set of modeswithin the bounds 20 and 2000, plus 10 frequencies between 20 and the lowest mode in the range,plus 10 frequencies between the highest mode in the range and 2000

Since the forcing frequencies are near structural resonances, it is important that some amount ofdamping be specified.



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

FREQ4 SID F1 F2 FSPD NFM

Example:FREQ4 6 20 200 0.3 21

Field ContentsSID Set identification number. (Integer > 0)F1 Lower bound of frequency range in cycles per unit time. (Real 0.0, Default=0.0)F2 Upper bound of frequency range in cycles per unit time. (Real>0.0, F2 F1, Default = 1.0E20)FSPD

NFM

Remarks:1.

2. FREQ4 entries must be selected with the Case Control command FREQUENCY = SID.3.

4.


Defines a set of frequencies used in the solution of modal frequency-response problems by specifying theamount of "spread" around each natural frequency and the number of equally spaced excitation frequencieswithin the spread.

Frequency spread, +/- the fractional amount specified for each mode which occurs unthe frequency range F1 to F2. (1.0 > Real > 0.0; Default = 0.10)


Number of evenly spaced frequencies per "spread"mode.. (Integer > 0; Default = 3; IfNFM is even, NFM + 1 will be used.)


There will be NFM excitation frequencies between (1-FSPD) * fN and (1+FSPD) * fN, for eachnatural frequency in the range F1 to F2.In the example above there will be 21 equally spaced frequencies across a frequency band of 0.7 *fN to 1.3 * fN for each natural frequency that occurs between 20 and 2000. See Figure 1 fordefintion of frequency spread.


5.



Excitation frequencies may be based on natural frequencies that are not within the range (F1 andF2) as long as the calculated excitation frequencies are within the calculated range. Similarly, anexcitation frequency calculated based on the natural frequencies within the range (F1through F2)may be excluded if it falls outside the range.

The frequency spread can also be used to define the half-power bandwidth. The half-poerbandwidth is given by 2 * x * fN, where x is the damping ratio. Therefore, if FSPD is specifiedequal to the damping ratio for the mode, NFM specifies the number of excitation frequency withinthe half-power bandwidth. See Figure 2 for the definition of half-power bandwidth.

(1-FSPD) * fN fN (1+FSPD )* fN

Figure 1. Frequency Spread Definition


Frequency List, Alternate Form 4

6.

7.

8.

9.

FREQ4

Since the forcing frequencies are near structural resonances, it is important that some amount ofdamping be specified.All FREQi entries with the same frequency set identification numbers will be used. Duplicatefrequencies will be ignored. fN and fN-1 are considered duplicated if

|f N - fN-1|<DFREQ*|fMAX - fMIN|,

Bulk Data Entry

where DFREQ is a user parameter, with default of 10-5. f MAX and fMIN are the maximum andminimum excitation frequencies of the combined FREQi entries.In design optimization, (SOL 200), the excitation frequencies generated from this entry are derivedfrom the natural frequencies computed in the first design cycle and the ecitation frequenciesremain fixed through subsequent design cycles. In other word


(Continued)

Peak Response

Half-PowerBandwidth

Am

plit

ude

FrequencyfN

Half-Power Point (.707 Peak)

2 * * fN

Figure 2. Half-Power Bandwidth Definition



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

FREQ5 SID F1 F2 FR1 FR2 FR3 FR4 FR5FR6 FR7 -etc.-

Example:FREQ5 6 20 200 1. 0.6 0.8 0.9 0.95

1.05 1.1 1.2

Field ContentsSID Set identification number. (Integer > 0)F1 Lower bound of frequency range in cycles per unit time. (Real 0.0, Default=0.0)F2 Upper bound of frequency range in cycles per unit time. (Real>0.0, F2 F1, Default = 1.0E20)FRi Fractions of the natural frequencies in the range F1 to F2 (Real > 0.0)

Remarks:1.

2. FREQ5 entries must be selected with the Case Control command FREQUENCY = SID.3. The frequencies defined by this entry are given by

where fNi are the natural frequencies in the range F1 through F2.

4.


Defines a set of frequencies used in the solution of modal frequency-response problems by speciification of afrequency range and fractions of the natural frequencies within that range.

FREQ5


In the example above, the list of frequencies will be 0.6, 0.8, 0.9, 0.95, 1.0, 1.05, 1.1, 1.2 timeseach natural frequency band between 20 and 2000. If this computation results in excitationfrequencies less than F1 and greater than F2, those computed frequencies are ignored.

fi = FRi * fNi



5.

6.

7.

8.

FREQ5Since the forcing frequencies are near structural resonances, it is important that some amount ofdamping be specified.

All FREQi entries with the same frequency set identification numbers will be used. Duplicatefrequencies will be ignored. fN and fN-1 are considered duplicated if

where DFREQ is a user parameter, with default of 10-5 . fMAX and fMIN are the maximum andminimum excitation frequencies of the combined FREQi entries.

Bulk Data Entry

In design optimization, (SOL 200), the excitation frequencies generated from this entry are derivedfrom the natural frequencies computed in the first design cycle and the ecitation frequenciesremain fixed through subsequent design cycles. In other words, the excitation frequencies will notbe readjusted even if the natural frequencies change.


|fN - fN-1|<DFREQ*|fMAX - fMIN|,

(Continued)


DYNAMIC DATA RECOVERY

• The matrix method and the mode displacement method are usedto recover data from modal frequency analysis.

where H = number of modesF = number of excitation frequencies

• The matrix method is the default and is cheaper for H < F and isthe recommended method for most cases.• PARAM,DDRMM,0 (default)

• The mode displacement method may be selected byPARAM,DDRMM,-1.

Cost of matrix methodCost of mode displacement method

HF----=


MODAL VERSUS DIRECT FREQUENCYRESPONSE

Modal Direct*Small Model XLarge Model XFew Excitation Frequencies XMany Excitation Frequencies XModal Damping X


SORT1 VERSUS SORT2 OUTPUT

• SORT1 is the listing of the output for each excitation frequency.• SORT2 is the listing of the output for each requested grid point or

element.• Primarily useful for frequency response analysis

• If SORT1 and SORT2 are mixed in a run, all output will default to SORT1for frequency response and SORT2 for transient response.

Direct Modal Direct ModalDefault 2 2 1 1Deformed Plot Requests 1 1 1 1XY Plot Requests 2 2 2 2

Transient Response Frequency Response


SOLUTION CONTROL FOR FREQUENCYRESPONSE

• Executive Control SectionSOL (for required input see below)

• Case Control SectionDLOAD (both - required)LOADSET (both - optional)METHOD (modal - required)SDAMPING (modal - optional)FREQUENCY (both - required)

Method StructuredSoultion

SequencesDirect 108Modal 111


SOLUTION CONTROL FOR FREQUENCYRESPONSE (Cont.)

• Bulk Data Section

• ASET,OMIT (both - optional)• EIGRL or EIGR (modal - required)• FREQ (both - required)• RLOADi (both – required)• LSEQ (both - optional)• DAREA (both - optional)• DELAY (both - optional)• DPHASE (both - optional)• TABDMP1 (modal - optional)• DLOAD (both - optional)• TLOADi (both - optional)


CASE CONTROL OUTPUT• Grid Point

• ACCELERATION• DISPLACEMENT (or VECTOR)• OLOAD• SACCELERATION• SDISPLACEMENT• SVELOCITY• SVECTOR• SPCFORCES• VELOCITY• MPCFORCE

• Element Output• ELSTRESS (or STRESS)• ELFORCE (or FORCE)• STRAIN• ESE• EKE• EDE

• Special• OFREQUENCY (control solution output frequencies)


WORKSHOP #5

DIRECT FREQUENCY RESPONSE


WORKSHOP #5 - DIRECT FREQUENCYRESPONSE

Using the direct method, determine the frequency response of the flatrectangular plate, created in Workshop 1, subject to frequency-varyingexcitation. The structure is excited by a unit load at a corner of the tip.Use a frequency step of 20 Hz between a range of 20 and 1000 Hz. Usestructural damping of g=0.06. Below is a finite element representation ofthe flat plate. It also contains the loads and boundary constraints.

1.01.0


WORKSHOP # 5

• Start with following input file$$ wkshp5.dat$SOL 108CENDTITLE = FREQUENCY RESPONSE DUE TO UNIT FORCE AT TIPECHO = UNSORTEDSPC = 1SET 111 = 11, 33, 55DISPLACEMENT(SORT2, PHASE) = 111SUBCASE 1DLOAD = 500FREQUENCY = 100$

OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NOXTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER,

MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASEXYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER,

MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, PHASEXYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$BEGIN BULKparam,post,0PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE$INCLUDE 'plate.bdf'$ENDDATA


SOLUTION FILE FOR WORKSHOP #5ID SEMINAR, PROB5SOL 108TIME30CENDTITLE = FREQUENCY RESPONSE DUE TO UNIT FORCE AT TIPECHO = UNSORTEDSPC = 1SET 111 = 11, 33, 55DISPLACEMENT(SORT2, PHASE) = 111SUBCASE 1DLOAD = 500FREQUENCY = 100$OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NOXTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASEXYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER,MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, PHASEXYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$

BEGIN BULKparam,post,0PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ PLATE MODEL DESCRIBED IN NORMAL MODESEXAMPLE$INCLUDE ’plate.bdf’$$ SPECIFY STRUCTURAL DAMPING$PARAM, G, 0.06$$ APPLY UNIT FORCE AT TIP POINT$RLOAD2, 500, 600, , ,310$DAREA, 600, 11, 3, 1.0$TABLED1, 310,, 0., 1., 1000., 1., ENDT$$ SPECIFY FREQUENCY STEPS$FREQ1, 100, 20., 20., 49$ENDDATA


PARTIAL OUTPUT FILE FOR WORKSHOP #5POINT-ID = 11

C O M P L E X D I S P L A C E M E N T V E C T O R(MAGNITUDE/PHASE)

FREQUENCY TYPE T1 T2 T3 R1 R2 R30 2.000000E+01 G .0 .0 8.817999E-03 6.435859E-04 2.632016E-03 .0

.0 .0 356.4954 176.5664 176.5000 .00 4.000000E+01 G .0 .0 9.404316E-03 6.434992E-04 2.795561E-03 .0

.0 .0 356.2596 176.5677 176.2785 .0...

0 9.799999E+02 G .0 .0 9.965085E-04 2.691742E-04 4.097779E-04 .0.0 .0 187.6832 7.8008 15.1581 .0

0 1.000000E+03 G .0 .0 8.803169E-04 2.354655E-04 3.317750E-04 .0.0 .0 186.9298 8.2146 14.6645 .0

.

.

.POINT-ID = 33

C O M P L E X D I S P L A C E M E N T V E C T O R(MAGNITUDE/PHASE)

FREQUENCY TYPE T1 T2 T3 R1 R2 R30 2.000000E+01 G .0 .0 8.183126E-03 5.993296E-04 2.443290E-03 .0

.0 .0 356.4899 176.5639 176.4950 .00 4.000000E+01 G .0 .0 8.768992E-03 6.006201E-04 2.606561E-03 .0

.0 .0 356.2376 176.5565 176.2581 .0...

0 9.799999E+02 G .0 .0 6.867234E-04 3.836353E-04 5.393046E-04 .0.0 .0 188.0180 5.5597 10.0794 .0

0 1.000000E+03 G .0 .0 6.062436E-04 3.454143E-04 4.648783E-04 .0.0 .0 186.8358 5.4959 8.8514 .0

.

.

.



1 FREQUENCY RESPONSE DUE TO UNIT FORCE AT TIP MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 24

0 SUBCASE 10 X Y - O U T P U T S U M M A R Y ( R E S P O N S E )0 SUBCASE CURVE FRAME CURVE ID./ XMIN-FRAME/ XMAX-FRAME/ YMIN-FRAME/ X FOR YMAX-FRAME/ X FOR

ID TYPE NO. PANEL : GRID ID ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX0 1 DISP 1 11( 5,--) 2.000000E+01 1.000000E+03 1.002672E-04 3.800000E+02 6.786681E-02 1.400000E+02

2.000000E+01 1.000000E+03 1.002672E-04 3.800000E+02 6.786681E-02 1.400000E+020 1 DISP 1 11(--, 11) 2.000000E+01 1.000000E+03 1.834657E+02 2.200000E+02 3.564954E+02 2.000000E+01

2.000000E+01 1.000000E+03 1.834657E+02 2.200000E+02 3.564954E+02 2.000000E+010 1 DISP 2 33( 5,--) 2.000000E+01 1.000000E+03 4.982490E-05 6.000000E+02 6.850390E-02 1.400000E+02




2.000000E+01 1.000000E+03 6.490854E+00 7.599999E+02 3.579576E+02 7.800000E+02



The display of PhaseThe display of Phasewill be different inwill be different inMDMD PatranPatran!!

--180180oo < Phase < 180< Phase < 180oo

Any Phase Lead largerAny Phase Lead larger180180oo will be displayedwill be displayedas ( Phaseas ( Phase -- 360360oo ))

Phase LeadPhase Leadcalculated bycalculated by

MD Nastran:MD Nastran:

00oo < Phase << Phase < 360360oo


WORKSHOP #6

MODAL FREQUENCY RESPONSE


WORKSHOP #6 - MODAL FREQUENCYRESPONSE

Using the modal method, determine the frequency response of the flatrectangular plate, created in Workshop 1, excited by a 0.1 psi pressure load overthe total surface of the plate and a 1.0 lb. force at a corner of the tip lagging 45o.Use a modal damping of = 0.03. Use a frequency step of 20 hz between a rangeof 20 and 1000 hz; in addition, specify five evenly spaced excitation frequenciesbetween the half power points of each resonant frequency between the range of20-1000 hz. Below is a finite element representation of the flat plate. It alsocontains the loads and boundary constraints.

1.0

0.1 psi over the total surface


WORKSHOP # 6

• Start with the following input file$$ wkshp6.dat$$ executive control add : modal frequency response solution

sequence$$ case control, add : eigenvalue callout$ frequency selection$ modal damping selection$ xyplot commands$$ bulk data, add : eigenvalue method$ modal damping$ loading$ frequencies of load application$CENDTITLE = FREQUENCY RESPONSE WITH PRESSURE AND POINT LOADSSUBTITLE = USING THE MODAL METHOD WITH LANCZOSECHO = UNSORTEDSEALL = ALLSPC = 1SET 111 = 11, 33, 55DISPLACEMENT(PHASE, PLOT) = 111SUBCASE 1DLOAD = 100$OUTPUT (XYPLOT)$

$BEGIN BULKPARAM,COUPMASS,1PARAM,WTMASS,0.00259$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE$INCLUDE 'plate.bdf'$ENDDATA


SOLUTION FOR WORKSHOP #6ID SEMINAR, PROB6$$ soln6.dat$ID SEMINAR, PROB6SOL 111CENDTITLE = FREQUENCY RESPONSE WITH PRESSURE AND POINT LOADSSUBTITLE = USING THE MODAL METHOD WITH LANCZOSECHO = UNSORTEDSEALL = ALLSPC = 1SET 111 = 11, 33, 55DISPLACEMENT(PHASE ) = 111METHOD = 100FREQUENCY = 100SDAMPING = 100SUBCASE 1DLOAD = 100$OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NOXTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASEXYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, PHASEXYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$

BEGIN BULKPARAM,COUPMASS,1PARAM,WTMASS,0.00259$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE$INCLUDE 'plate.bdf'$$ EIGENVALUE EXTRACTION PARAMETERS$EIGRL, 100, 10., 2000.$$ SPECIFY MODAL DAMPING$TABDMP1, 100, CRIT,+, 0., .03, 10., .03, ENDT$$ APPLY PRESSURE LOAD$RLOAD2, 400, 300, , ,310PLOAD2, 300, 1., 1, THRU, 40$TABLED1, 310,, 10., 1., 1000., 1., ENDT$$ POINT LOAD$RLOAD2, 500, 600, , -45., 310$FORCE,600,11,,1.0,,,1.0$$ COMBINE LOADS$DLOAD, 100, 1., .1, 400, 1.0, 500$$ SPECIFY FREQUENCY STEPS$FREQ1, 100, 20., 20., 49FREQ4, 100, 20., 1000., .03, 5$ENDDATA


POINT-ID = 11C O M P L E X D I S P L A C E M E N T V E C T O R

(MAGNITUDE/PHASE)

FREQUENCY TYPE T1 T2 T3 R1 R2 R30 2.000000E+01 G 1.647094E-12 2.530238E-12 1.122426E-02 6.413405E-04 3.278660E-03 2.135828E-12

146.8949 146.8692 325.6783 134.5891 144.8534 146.86980 4.000000E+01 G 1.770526E-12 2.719501E-12 1.199385E-02 6.408464E-04 3.493634E-03 2.295604E-12

146.2678 146.2447 325.1725 134.4988 144.4191 146.24530 6.000000E+01 G 2.022193E-12 3.105379E-12 1.356089E-02 6.394051E-04 3.931127E-03 2.621361E-12

145.4318 145.4127 324.5159 134.3886 143.8702 145.41310 8.000000E+01 G 2.520848E-12 3.869944E-12 1.666288E-02 6.353444E-04 4.797074E-03 3.266807E-12

144.1163 144.1025 323.4427 134.2561 142.9458 144.10280 1.000000E+02 G 3.674813E-12 5.639226E-12 2.383472E-02 6.228435E-04 6.799188E-03 4.760436E-12

141.4430 141.4358 321.0814 134.1330 140.7818 141.43590 1.200000E+02 G 8.109207E-12 1.243793E-11 5.136045E-02 5.649115E-04 1.448202E-02 1.049992E-11

131.6994 131.7002 311.7284 135.1967 131.6851 131.7000

POINT-ID = 55C O M P L E X D I S P L A C E M E N T V E C T O R

(MAGNITUDE/PHASE)

FREQUENCY TYPE T1 T2 T3 R1 R2 R30 2.000000E+01 G 2.299811E-12 2.628319E-12 1.003391E-02 5.635861E-04 3.021529E-03 2.074211E-12

326.8539 146.8715 326.9735 135.2839 145.7045 146.87060 4.000000E+01 G 2.471663E-12 2.824950E-12 1.079900E-02 5.666142E-04 3.235993E-03 2.229377E-12

326.2311 146.2468 326.3320 135.1997 145.1825 146.24610 6.000000E+01 G 2.822033E-12 3.225849E-12 1.235921E-02 5.723091E-04 3.672785E-03 2.545738E-12

325.4014 145.4144 325.4773 135.1333 144.5074 145.41380 8.000000E+01 G 3.516238E-12 4.020179E-12 1.545119E-02 5.823869E-04 4.537677E-03 3.172566E-12

324.0942 144.1037 324.1384 135.0829 143.4117 144.1033

SOLUTION FOR WORKSHOP #6


0 SUBCASE 1

0 X Y - O U T P U T S U M M A R Y ( R E S P O N S E )

0 SUBCASE CURVE FRAME CURVE ID./ XMIN-FRAME/ XMAX-FRAME/ YMIN-FRAME/ X FOR YMAX-FRAME/ X FOR

ID TYPE NO. PANEL : GRID ID ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX

0 1 DISP 1 11( 5,--) 2.000000E+01 1.000000E+03 3.651020E-04 4.200000E+02 1.699705E-01 1.336891E+02

2.000000E+01 1.000000E+03 3.651020E-04 4.200000E+02 1.699705E-01 1.336891E+02

0 1 DISP 1 11(--, 11) 2.000000E+01 1.000000E+03 1.400466E+02 1.000000E+03 3.256783E+02 2.000000E+01

2.000000E+01 1.000000E+03 1.400466E+02 1.000000E+03 3.256783E+02 2.000000E+01

0 1 DISP 2 33( 5,--) 2.000000E+01 1.000000E+03 2.236997E-04 6.400000E+02 1.700564E-01 1.336891E+02

2.000000E+01 1.000000E+03 2.236997E-04 6.400000E+02 1.700564E-01 1.336891E+02

0 1 DISP 2 33(--, 11) 2.000000E+01 1.000000E+03 1.384945E+02 1.000000E+03 3.263299E+02 2.000000E+01

2.000000E+01 1.000000E+03 1.384945E+02 1.000000E+03 3.263299E+02 2.000000E+01

0 1 DISP 3 55( 5,--) 2.000000E+01 1.000000E+03 1.824597E-04 1.000000E+03 1.696912E-01 1.336891E+02

2.000000E+01 1.000000E+03 1.824597E-04 1.000000E+03 1.696912E-01 1.336891E+02

0 1 DISP 3 55(--, 11) 2.000000E+01 1.000000E+03 1.702332E+01 6.999999E+02 3.577236E+02 7.102496E+02

2.000000E+01 1.000000E+03 1.702332E+01 6.999999E+02 3.577236E+02 7.102496E+02

OUTPUT FOR WORKSHOP #6 (Cont.)


• Specifies stiffness as a function of forcing frequency

• Specifies damping as a function of forcing frequency

• Different impedance in different direction

• CBUSH• Defines generalized spring, damper connectivity

• PBUSH• Defines nominal spring and damper values

• PBUSHT• Defines frequency-dependent spring and damper values

FREQUENCY-DEPENDENT SPRINGS ANDDAMPERS


• Specifies stiffness as a function of forcing frequency• Specifies damping as a function of forcing frequency• Different impedance in different direction• CBUSH

• Defines generalized spring, damper connectivity

• PBUSH• Defines nominal spring and damper values

• PBUSHT• Defines frequency-dependent spring and damper values

FREQUENCY-DEPENDENT SPRINGS ANDDAMPERS


Generalized Spring-and-Damper Connection

Bulk Data Entry

Figure 2. Definition of Offset S1, S2, S3

Figure 1. CBUSH Element.

CBUSH

zelem

yelem

xelem

GA

GB

v

S(1 - )

*S *

GA

GB

(S1, S2, S3)OCID

zelem

yelem

Note : 1. The material stiffness and damping properties of the elastomer are located at (S1, S2, S3).

2. The elastomer itself has zero length; i.e., GA and GB are coincident. It is shown here in an exploded view


12

11

P() = 2.0 P(f) sin t

mass 12 = 1.0

damper(f)

CBUSH 1000

spring(f)

ForcingFrequency

(Hz)K(f) B(f) P(f)

0.9 0.81 0.286479 0.811 1 0.31831 1

1.1 1.21 0.350141 1.21

Table of Impedance

* (1./2* (1./2))22

PARAM.WTMASSPARAM.WTMASS

ffii ffii22 ffii// ffii22

FREQUENCY DEPENDENT IMPEDANCESAMPLE


$$ cbush1.dat$TIME 10SOL 108CENDTITLE = VERIFICATION PROBLEM, FREQ. DEP. IMPEDANCE BUSHVERSUBTITLE = SINGLE DOF, CRITICAL DAMPING, 3 EXCITATION FREQUENCIESECHO = BOTHSPC = 1002DLOAD = 1DISP = ALLFREQ = 10ELFO = ALLBEGIN BULK$ CONVENTIONAL INPUT FOR MOUNTGRDSET, , , , , , , 23456 $ PS$ TIE DOWN EVERYTHING BUT THE 1 DOFGRID, 11, , 0., 0., 0.0 $ GROUND=, 12, =, =, =, , $ ISOLATED DOFSPC1, 1002 123456 11 $ GROUNDCONM2, 12, 12, , 1.0 $ THE ISOLATED MASS$$ EID PID GA GB GO/X1 X2 X3 CID$CBUSH 1000 2000 11 12 0$PBUSH 2000 K 1.0

B 0.0$PBUSHT 2000 K 2001

B 2002$TABLED1, 2001 $ STIFFNESS TABLE, 0.9 0.81, 1.0, 1.0, 1.1, 1.21 ENDTTABLED1 2002 $ DAMPING TABLE, 0.9 .2864789, 1.0, .318309, 1.1 , .3501409 ENDT$CONVENTIONAL INPUT FOR FREQUENCY RESPONSEPARAM, WTMASS, .0253303 $ 1/(2*PI)**2. GIVES FN=1.0DAREA, 1, 12, 1, 2. $CAUSES UNIT DEFLECTIONFREQ, 10, 0.9, 1.0, 1.1 $ BRACKET THE NATURAL FREQUENCYRLOAD1, 1, 1, , , 3TABLED1,3 $ TABLE FOR FORCE VS. FREQUENCY, 0.9, 0.81, 1., 1., 1.1, 1.21,ENDT $ P = KENDDATA

SAMPLE USING CBUSH ELEMENT


FREQUENCY = 9.000000E-01C O M P L E X D I S P L A C E M E N T V E C T O R

(REAL/IMAGINARY)

POINT ID. TYPE T1 T2 T3 R1 R2 R30 11 G 0.0 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.00 12 G -6.682744E-08 0.0 0.0 0.0 0.0 0.0

-1.000000E+00 0.0 0.0 0.0 0.0 0.0

1 VERIFICATION PROBLEM, FREQ. DEP. IMPEDANCE BUSHVER MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 10SINGLE DOF, CRITICAL DAMPING, 3 EXCITATION FREQUENCIES

0FREQUENCY = 1.000000E+00

C O M P L E X D I S P L A C E M E N T V E C T O R(REAL/IMAGINARY)


0.0 0.0 0.0 0.0 0.0 0.00 12 G -1.046835E-07 0.0 0.0 0.0 0.0 0.0

-9.999999E-01 0.0 0.0 0.0 0.0 0.01 VERIFICATION PROBLEM, FREQ. DEP. IMPEDANCE BUSHVER MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 11

SINGLE DOF, CRITICAL DAMPING, 3 EXCITATION FREQUENCIES0

FREQUENCY = 1.100000E+00C O M P L E X D I S P L A C E M E N T V E C T O R

(REAL/IMAGINARY)


0.0 0.0 0.0 0.0 0.0 0.00 12 G -6.855670E-08 0.0 0.0 0.0 0.0 0.0


SINGLE DOF, CRITICAL DAMPING, 3 EXCITATION FREQUENCIES

DISPLACEMENT OUTPUT FOR CBUSHELEMENT


FREQUENCY = 9.000000E-01C O M P L E X F O R C E S I N B U S H E L E M E N T S ( C B U S H )

(REAL/IMAGINARY)

ELEMENT-ID FORCE-X FORCE-Y FORCE-Z MOMENT-X MOMENT-Y MOMENT-Z0 1000 1.620000E+00 0.0 0.0 0.0 0.0 0.0



FREQUENCY = 1.000000E+00C O M P L E X F O R C E S I N B U S H E L E M E N T S ( C B U S H )

(REAL/IMAGINARY)


-1.000000E+00 0.0 0.0 0.0 0.0 0.01 VERIFICATION PROBLEM, FREQ. DEP. IMPEDANCE BUSHVER MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 15


FREQUENCY = 1.100000E+00C O M P L E X F O R C E S I N B U S H E L E M E N T S ( C B U S H )

(REAL/IMAGINARY)


-1.210000E+00 0.0 0.0 0.0 0.0 0.01 VERIFICATION PROBLEM, FREQ. DEP. IMPEDANCE BUSHVER MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 16


FORCE OUTPUT FOR CBUSH ELEMENT


BLANK PAGE


SECTION 9

DIRECT MATRIX INPUT


DIRECT MATRIX INPUT

● In addition to generating the mass, stiffness, and damping matrix automaticallyfrom the FE model, you can add mass, stiffness, and damping to specificDOFs.

● The DMIG Bulk Data entry is used to input a mass, damping, or stiffnessmatrix connecting specified DOFs.

● Two types of DMIGs

● Type 1 G-Type Matrices● G-type matrices are the size of the G-set.● G-type matrices are applied at the system level prior to any constraints operations.● G-type matrices are real, symmetric. They are selected in the Case Control Section via:

● M2GG = name of mass matrix addition● B2GG = name of damping matrix addition● K2GG = name of stiffness matrix addition

● G-type matrices can be added to the superelement and the residual structure.


DIRECT MATRIX INPUT (Cont.)

● Type 2 P-Type Matrices

● P-type matrices are the same size as the P-set (G-set plus E-setfor extra points).

● P-type matrices are not applied at the system level prior toconstraint operations.

● The P-type direct input matrices are processed through theconstraints and reduction procedures in parallel with the G-typematrices, then added to the reduced model (A- or H-set) beforethe analysis operation.

● Note that modal reduction (modal solutions) operations do notinclude the effects for P-type matrices.

● The load reduction operations to the analysis sets does notinclude the effects of P-type matrices.


DIRECT MATRIX INPUT (Cont.)

● P-type matrices need not be real or symmetric. They areselected in the Case Control Section via:● M2PP = name of mass matrix addition● B2PP = name of damping matrix addition● K2PP = name of stiffness matrix addition

● P-type matrices can only be added to the residualstructure. They cannot be added to superelements.● PARAM,WTMASS does not affect direct input matrices M2GG or

M2PP. PARAM,CM2 can be used to scale M2GG.


DMIG Direct Matrix Input at Points

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

DMIG NAME “0" IFO TIN TOUT POLAR NCOL

Column Entry Format:DMIG NAME GJ CJ G1 C1 A1 B1

G2 C2 A2 B2 -etc.-

Example:DMIG STIF 0 1 3 4DMIG STIF 27 1 2 3 3.+5 3.+3

2 4 2.5+10 0 50 1 0

Field ContentsNAME

IFO

1 = Square9 or 2 = Rectangular6 = Symmetric

TIN Type of matrix being input: (Integer)1 = Real, single precision (One field is used per element.)2 = Real, double precision (One field is used per element.)3 = Complex, single precision (Two fields are used per element.)4 = Complex, double precision (Two fields are used per element.)

TOUT Type of matrix that will be created: (Integer)0 = Set by precision system cell (Default)1 = Real, single precision2 = Real, double precision3 = Complex, single precision4 = Complex, double precision

Defines direct input matrices related to grid, extra, and/or scalar points. The matrix is defined by a single headerentry and one or more column entries. A column entry is required for each column with nonzero elements.

Form of matrix input. IFO = 6 must be specified for matrices selected by the K2GG, M2GG, andB2GG Case Control commands. (Integer)

Name of the matrix. See Remark 1. (One to eight alphanumeric characters, the first of which isalphabetic.)


Field ContentsPOLAR

NCOL Number of columns in a rectangular matrix. Used only for IFO = 9. (Integer > 0)

GJ Grid, scalar or extra point identification number for column index. (Integer > 0)

CJ Component number for grid point GJ. (0 < Integer £ 6; blank or zero if GJ is a scalar or extra point.)

Gi Grid, scalar, or extra point identification number for row index. (Integer > 0)

Ci Component number for Gi for a grid point. (0 < CJ £ 6; blank or zero if Gi is a scalar or extra point.)

Ai, Bi

Remarks:1.

2.

3. Field 3 of the header entry must contain an integer 0.

Real and imaginary (or amplitude and phase) parts of a matrix element. If the matrix is real (TIN =1 or 2), then Bi must be blank. (Real)

Matrices defined on this entry may be used in dynamics by selection in the Case Control withK2PP = NAME, B2PP = NAME, M2PP = NAME for [K pp], [B pp ], or [M pp ], respectively.Matrices may also be selected for all solution sequences by K2GG = NAME, B2GG = NAME, andM2GG = NAME. The g-set matrices are added to the structural matrices before constraints areapplied, while p-set matrices are added in dynamics after constraints are applied. Load matricesmay be selected by P2G = NAME for dynamic and superelement analyses.

The header entry containing IFO, TIN and TOUT is required. Each nonnull column is started witha GJ, CJ pair. The entries for each row of that column follows. Only nonzero terms need beentered. The terms may be input in arbitrary order. A GJ, CJ pai

Input format of Ai, Bi. (Integer=blank or 0 indicates real, imaginary format; Integer > 0 indicatesamplitude, phase format.)


4.

5.

6. If NCOL is not used for rectangular matrices, two different conventions are available:·

·

7. The matrix names must be unique among all DMIGs.8.

9.

10.

For symmetric matrices (IFO = 6), a given off-diagonal element may be input either below orabove the diagonal. While upper and lower triangle terms may be mixed, a fatal message will beissued if an element is input both below and above the diagonal.

On long-word machines, almost all matrix calculations are performed in single precision and onshort-word machines, in double precision. It is recommended that DMIG matrices also followthese conventions for a balance of efficiency and reliability. The recommended value for TOUT is0, which instructs the program to inspect the system cell that measures the machine precision atrun time and sets the precision of the matrix to the same value. TOUT = 0 allows the same DMIGinput to be used on any machine. If TOUT is contrary to the machine type specified (for example,a TOUT of 1 on a short-word machine), unreliable results may occur.If any DMIG entry is changed or added on restart then a complete re-analysis is performed.Therefore, DMIG entry changes or additions are not recommended on restart.

The recommended format for rectangular matrices requires the use of NCOL and IFO = 9. Thenumber of columns in the matrix is NCOL. (The number of rows in all DMIG matrices is alwayseither p-set or g-set size, depending on the context.) The GJ term i

If IFO = 9, GJ and CJ will determine the sorted sequence, but will otherwise beignored; a rectangular matrix will be generated with the columns submitted being in the1 to N positions, where N is the number of logical entries submitted (not counting th

If IFO = 2, the number of columns of the rectangular matrix will be equal to the indexof the highest numbered non-null column (in internal sort). Trailing null columns of theg- or p-size matrix will be truncated.

TIN should be set consistent with the number of decimal digits required to read the input dataadequately. For a single-precision specification on a short-word machine, the input will betruncated after about eight decimal digits, even when more digits are needed, a double precisionspecification should be used instead. However, note that a double precision specification requiresa “D” type exponent even for terms that do not need an exponent. For example, unity may beinput as 1.0 in single precision, but the longer form 1.0D0 is required for double precision.


SCALE FACTOR FOR X2GG AND X2PPMATRICES

● The K2GG, M2GG, B2GG, and P2G commands acceptthe specifications of multiple matrices and real scalefactor(s)

● K2PP, M2PP, and B2PP commands accept thespecifications of multiple matrices and complex or realscale factor(s)

Example:

K2GG = 1.25*KDMIGM2gg = 2.8*MDMIG1,3.9*MDMIG2

Example:

K2PP = KDMIGK2PP = KDMIG1, KDMIG2,KDMIG3K2PP = 2.04*KDMIG1,0.82*KDMIG2K2PP = (2.04,.5)*KDMIG1,(0.82,0.)*KDMIG2


DIRECT INPUT OF STRUCTURAL ELEMENTDAMPING MATRICES

● Can also add structural element damping by using theK42GG command and DMIG entries.

● Can also specify multiple matrices

Example:

K42GG = KDMIGK42GG = KDMIG1, KDMIG2,KDMIG3K42GG = 2.04*KDMIG1,0.82*KDMIG2


1 2 3 4 5 61 2 3 4 5 6

CBAR,1 CBAR,2 CBAR,3 CBAR,4CBAR,1 CBAR,2 CBAR,3 CBAR,4 DMIG = EL5DMIG = EL5

The stiffness and mass matrices from a contractor which connectThe stiffness and mass matrices from a contractor which connect thethegrid points 5 and 6. They contain twogrid points 5 and 6. They contain two DOFsDOFs (R2 and T3) per grid point.(R2 and T3) per grid point.

5,35,3 5,55,5 6,36,3 6,56,5

5,35,3 500039.500039. SYMMETRICSYMMETRIC

5,55,5 --250019250019 166680.166680.

6,36,3 --500039.500039. 250019.250019. 500039.500039.

6,56,5 --250019.250019. 83340.83340. 250019.250019. 166680.166680.

5,35,3 5,55,5 6,36,3 6,56,5

5,35,3 3.58293.5829 0.0. 0.0. 0.0.

5,55,5 0.0. 0.0. 0.0. 0.0.

6,36,3 0.0. 0.0. 3.58293.5829 0.0.

6,56,5 0.0. 0.0. 0.0. 0.0.

KK55 ==

MM55 ==

DMIG EXAMPLE


$$ DMIG EXAMPLE$ NORMAL MODES$SOL 103 $ NORMAL MODES ANALYSISTIME 10CENDTITLE = DMIG TO READ STIFFNESS AND MASS FOR ELEM 5SUBTITLE = PLANAR PROBLEM$SPC = 10$$ SPECIFY K2GG AND M2GGK2GG = EXSTIFM2GG = EXMASS$METHOD = 10$BEGIN BULK$$$EIGRL SID V1 V2 NDEIGRL 10 2$CBAR,1,1,1,2,10CBAR,2,1,2,3,10CBAR,3,1,3,4,10CBAR,4,1,4,5,10$$ HEADER ENTRY FOR STIFFNESSDMIG,EXSTIF,0,6,1$

DMIG,EXSTIF,5,3,,5,3,500039.+,5,5,-250019.,,6,3,-500039.+,6,5,-250019.$DMIG,EXSTIF,5,5,,5,5,166680.+,6,3,250019.,,6,5,83340.$DMIG,EXSTIF,6,3,,6,3,500039.+,6,5,250019.$DMIG,EXSTIF,6,5,,6,5,166680.$$ HEADER ENTRY FOR MASSDMIG,EXMASS,0,6,1$$ DATA ENTRIES FOR MASS$DMIG,EXMASS,5,3,,5,3,3.5829DMIG,EXMASS,6,3,,6,3,3.5829$GRID,1,,0.,0.,0.,,1246GRID,2,,1.,0.,0.,,1246GRID,3,,2.,0.,0.,,1246GRID,4,,3.,0.,0.,,1246GRID,5,,4.,0.,0.,,1246GRID,6,,5.,0.,0.,,1246GRID,10,,0.,0.,10.,,123456MAT1,1,7.1+10,,.33 2700.PBAR,1,1,2.654-3,5.869-7SPC1,10,123456,1$ENDDATA

INPUT FILE FOR THE DMIG EXAMPLE


BLANK


SECTION 10

DYNAMIC EQUATIONS OF MOTION


DYNAMIC MATRIX ASSEMBLY

● MD Nastran provides direct and modal methods forperforming transient and frequency response and complexmode analysis.

● The dynamic matrices are assembled differentlydepending on the analysis and method.


DIRECT METHODS

● The general dynamic equation used in the direct methods is:

where p = a derivative operatorud = the union of the analysis set ua and extra points ue

● For frequency response and complex eigenvalue analysis, the dynamicmatrices are:

● For transient response, the dynamic matrices are:

Mddp2 Bddp Kdd+ + ud Pd =

Kdd 1 ig+ Kdd1 Kdd

2 i Kdd4 + +=

Bdd Bdd1 Bdd

2 +=

Mdd Mdd1 Mdd

2 +=

Kdd Kdd1 Kdd

2 +=

Bdd Bdd1 Bdd

2 g3------ Kdd

1 14------ Kdd

4 + + +=

Mdd Mdd1 Mdd

2 +=


DYNAMIC MATRIX DEFINITIONS

[K1dd] is the reduced structural stiffness matrix plus the reduced direct input K2GG

(symmetric).

[K2dd] is the reduced direct input matrix K2PP plus the reduced transfer function

input (symmetric or unsymmetric).

[K4dd] is the reduced structural damping matrix obtained by multiplying the

stiffness matrix [Ke] of an individual structural element by an element dampingfactor ge and combining the results for all structural elements (symmetric).

[B1dd] is the reduced viscous damping matrix plus the reduced direct input B2GG

(symmetric).


DYNAMIC MATRIX DEFINITIONS (Cont.)

● [B2dd] is the reduced direct input matrix B2PP plus the reduced

transfer function input (symmetric or unsymmetric).

● [M1dd] is the reduced mass matrix plus the reduced direct input M2GG

(symmetric).

● [M2dd] is the reduced direct input matrix M2PP plus the reduced

transfer function input (symmetric or unsymmetric).

● g, 3, 4 are the constants specified by the user.


DYNAMIC MATRIX DEFINITIONS (Cont.)

● The structural matrices , , , and are expanded by theaddition of zeroes in the rows and columns corresponding to extrapoints to form , , , and .

● Only the , , and matrices can reference extra points.● The direct input matrices , , and are processed through

the multipoint and single-point constraint elimination and any reductionprocedure.

● Note: The extra points are unaffected by any constraint or reductionprocedures. The constraint and reduction procedures can onlyeliminate grid or scalar point DOFs but not extra points.

● The matrices , , and are examined to identify rows andcolumns which are null in all three matrices. For transient andfrequency response, is augmented by placing unity in each nullrow and column. In complex eigenvalue analysis, null rows andcolumns are discarded from , , and .

Kaa Kaa4 Maa

Kdd1 Kdd

4 Mdd1 Bdd

1

Kdd2 Bdd

2 Mdd2

Kpp2 Bpp

2 Mpp2

Kdd Bd d Mdd

K dd

Kdd Bd d Mdd

[B[Baaaa]]


MODAL METHODS

● The general dynamic equation used in the modal methods is:

where p = a derivative operatoruh = the union of the modal coordinates i and extra points ue.

● The transformation between i and ua is:

where [ai] is the matrix of eigenvectors obtained in real eigenvalue analysis.

● The transformation from uh to ud is obtained by augmenting [ai] to include the extrapoints.

where

Mhhp 2 Bhhp Khh+ + uh ph =

ua ai i =

ud dh uh =

dh ai 0

0 Iee=

uh iue

=


MODAL METHODS (Cont.)

● For frequency response and complex eigenvalue analysis the dynamicmatrices are:

where [mi] = a diagonal matrix with terms[bi] = a diagonal matrix with terms bii = ig(i)m ii ,

where i = is the radian frequency of the i-th normal mode andg(i) is a damping factor obtained from interpolation ofa user-supplied table (TABDMP1)

[ki] = a diagonal matrix with terms kii = 2imii

● If parameter

Khh ki dh T ig Kdd1 Kdd

2 i Kdd4 + + dh +=

Bhh bi dh T Bdd1 Bdd

2 + dh +=

Mhh mi dh T Mdd2 dh +=

mii ai T Maa ai =

KDAMP = -1, then

mii = m iibii = 0kii 1 ig i + kii=

( Default: PARAM,KDAMP,1 )( Default: PARAM,KDAMP,1 )


MODAL METHODS (Cont.)

● g(wi) is a damping factor obtained from the interpolation of a user-supplied table (TABDMP1).

● [mi], [bi], and [ki] are expanded by the addition of zeros to the rows andcolumns corresponding to the extra points (ue).

● For transient response the dynamic matrices are:

● If only [mi], [bi], and [ki] are present in any modal dynamic analysis,then the modal dynamic equations are uncoupled.

Khh ki dh T Kdd2 dh +=

Bhh bi dh T Bdd1 Bdd

2 g3------- Kdd

1 14------- Kdd

1 + + + dh +=

Mhh mi dh T Mdd2 dh +=


BLANK


SECTION 12

ENFORCED MOTION


● Used to analyze constrained structures with base inputacceleration, displacement, and velocity.

● Common examples are earthquakes (for transientanalysis), swept-sine shaker test simulation (for frequencyresponse analysis) and automobile suspensions.

ENFORCED MOTION IN DYNAMIC ANALYSIS


ANALYSIS METHODS

● Four available methods:

● 1. Direct Specification of enforced displacement, velocity, oracceleration (recommended method).

● 2. Large Mass (See Appendix B)

● 3. Large Stiffness (Enforced displacement only, see appendix B)

● 4. Lagrange Multiplier (See Appendix B)


● Method 1 is the recommended method and it is themethod discussed in this seminar.

● This method of enforced motion can be specified by thedirect specification of displacements, velocities oraccelerations via SPC / SPC1 and SPCD Bulk Dataentries.

● Interface very similar to enforced motion in static analysis.

METHOD 1


GG--Set: 6 Degrees of Freedom per GridSet: 6 Degrees of Freedom per Grid

NN--SetSet MM--Set:Set:Eliminated byEliminated bymmultipointultipointconstraintsconstraints((MPCsMPCs, R, R--typetypeelements)elements)

SS--Set:Set:Prescribed byPrescribed by ssingleingle--pointpointconstraints (constraints (SPCsSPCs,,AUTOSPC, PS)AUTOSPC, PS)

FF--SetSet

● In the direct method, enforced motion is applied to thedegrees of freedom in the S-set.

DEGREE OF FREEDOM SETS


METHODS FOR DIRECT ENFORCED MOTION

● Two approaches are available for performing this type ofenforced motion

● Absolute motion (v2001)

● Relative motion (v2004+)


sqpp

uu

KKKK

uu

BBBB

uu

MMMM

s

f

s

f

sssf

fsff

s

f

sssf

fsff

s

f

sssf

fsff

sp

su

sq Constraint Forces (unknown)

Prescribed Displacements (known)

Applied Loads (known),fp

....

........ 11

BASIC EQUATIONS FOR USING ABSOLUTEMOTION

● For the N-set, the equation of motion is as follows:

where


sfssfssfsffffffffff uKuBuMpuKuBuM

s

fsssf

s

fsssf

s

fsssfss u

uKK

uu

BBuu

MMpq

....

....

....

.... ..

..

.. .. 22

33

● Equation 1 can be solved for the F-set displacements:

● Subsequently, the constraint forces are obtained from thesecond matrix equation:

EQUATIONS FOR TRANSIENT RESPONSE


EQUATIONS FOR FREQUENCY RESPONSE

sfsfsfsffffffff ii UKBMPUKBM 22

s

fsssfsssfsssfss i

UU

KKBBMMPQ 2

44

55

● In frequency response analysis, the F-set displacementsare obtained from

● The constraint forces read


● { Uf } = { Yf } + { X }

where { Uf } = absolute motion{ Yf } = relative motion{ X } = base motion

● { X } - motion of f-set DOFs due to specified enforcedmotion of the s-set DOFs (computed purely from staticconsiderations)

● { X } = -[ Kff ]–1 [ Kfs ] { Us} = [ Z ] { Us }

RELATIVE MOTION APPROACH

66

77

KKffffUUff ++ KKfsfsUUss = P= Pff

KKffffYYff ++ KKffffXX ++ KKfsfsUUss = P= PffRelRel. Motion:. Motion: KKffffYYff = P= Pff

KKffffXX ++ KKfsfsUUss = 0= 0

X =X = --KKffff--11 KKfsfsUUss


● Substituting equations 6 and 7 into equation 2, we get:

● Solve for Yf, Yf, and Yf

● Uf, Uf, and Uf are obtained using equation 6● This is the default output

● To get Yf, Yf, and Yf, set PARAM,ENFMOTN,REL

s sfsffffffffffu uMpYKYBYM

.... ...... ..MffZ

fsB Bff Z

....

....

....

..

..

..

PARAM,ENFMOTN,ABSPARAM,ENFMOTN,ABS -- absolute motion of the model (default)absolute motion of the model (default)PARAM,ENFMOTN,RELPARAM,ENFMOTN,REL -- relative motion to the enforced motionrelative motion to the enforced motion

of the baseof the base

RELATIVE MOTION APPROACH


● SPC / SPC1 Bulk Data entries are used to identify the degrees offreedom with enforced motion. These entries are activated by an SPCCase Control command.

● SPCD Bulk Data entries are used to define the enforced motion.These entries are referenced by the EXCITEID field of TLOADi orRLOADi Bulk Data entries.

● The TYPE field of the TLOADi or RLOADi Bulk Data entries indicatesthe type of enforced motion.

● For modal method, residual vectors should always be included whichis the default

USER INTERFACE


● The type of excitation is defined in field 5 of TLOADi BulkData entries or in field 8 of RLOADi Bulk Data entries:

● The character fields may be shortened to as little as asingle character

THE TYPE FIELD


SOL 11SOL 1111CENDCEND$$TITLE =Example for Direct Enforced MotionTITLE =Example for Direct Enforced MotionSUBTITLE=Modal Frequency Response AnalysisSUBTITLE=Modal Frequency Response Analysis$$SPC =1SPC =1METHOD =10METHOD =10FREQUENCY=20FREQUENCY=20$$SET 1 = 1000,1001SET 1 = 1000,1001ACCELERATION(SORT2,PRINT,PHAS)=1ACCELERATION(SORT2,PRINT,PHAS)=1$$SUBCASE 1SUBCASE 1

LABEL=Unit Acceleration in xLABEL=Unit Acceleration in x--DirectionDirectionDLOAD=100DLOAD=100

$$SUBCASE 2SUBCASE 2

LABEL=Unit Acceleration in yLABEL=Unit Acceleration in y--DirectionDirectionDLOAD=200DLOAD=200

$$

EXAMPLE: EXECUTIVE AND CASE CONTROL


EXAMPLE: BULK DATA DECKBEGIN BULKBEGIN BULK$$PARAM, G, 0.02 $ 2% Structural DampingPARAM, G, 0.02 $ 2% Structural DampingSPC1, 1, 3456, 1000 $SPC1, 1, 3456, 1000 $ zz--DisplDispl. and. and RotationsRotations areare fixedfixedSPC1, 1, 12, 1000SPC1, 1, 12, 1000 $ x$ x-- and yand y--Accelerations areAccelerations are pprescribedrescribed$$$ Modal Reduction$ Modal ReductionEIGRL, 10,, 150.EIGRL, 10,, 150. $ Modes up to 150Hz$ Modes up to 150Hz

$$$ Base Motion Excitation$ Base Motion Excitation$$RLOAD1,RLOAD1, 100100,, 10011001,,,,,, 1010,,,, AA $ Load of$ Load of SubcaseSubcase 1:1:SPCDSPCD,, 10011001,, 10001000, 1, 1. $ Unit x, 1, 1. $ Unit x--AccelerationAcceleration$$RLOAD1,RLOAD1, 200200,, 10021002,,,,,, 1010,,,, AA $ Load of$ Load of SubcaseSubcase 2:2:SPCDSPCD,, 10021002,, 10001000, 2, 1. $ Unit y, 2, 1. $ Unit y--AccelerationAcceleration$$TABLED1,TABLED1, 1010 $ Constant for all Frequencies$ Constant for all Frequencies

, 0., 1., 100., 1., ENDT, 0., 1., 100., 1., ENDTFREQ1, 20, 1., 1., 49 $ Frequency Range from 1Hz toFREQ1, 20, 1., 1., 49 $ Frequency Range from 1Hz to 50Hz50Hz$$INCLUDE 'INCLUDE 'tower.bdftower.bdf' $ Structural Model' $ Structural Model$$ENDDATAENDDATA


INITIAL CONDITION SPECIFICATION FORENFORCED MOTION (SPC/SPCD)

● Enforced acceleration or velocity in transient analysis(using SPC/SPCD) requires integration to computecorresponding enforced velocities and/or displacements

● Above integration involves the use of initial conditions

● The US0 (initial displacement) and VS0 (initial velocity)fields on the TLOADi entries accomplish this.


VS0US0TIDTYPEDELAYEXCITEIDSIDTLOAD1

UV0US0BC

PFT2T1TYPEDELAYEXCITEIDSIDTLOAD2

INITIAL CONDITION SPECIFICATION FORENFORCED MOTION (SPC/SPCD)

● Format of TLOAD1 is shown below

● Format of TLOAD2 is shown below

● US0—initial displacement● VS0—initial velocity● US0 and VS0 can be used in conjunction with other initial

condition (TIC) at independent DOFs


EXAMPLE SPECIFYING INITIAL DISPLACEMENT

● A simple 2 dof system is used to illustrate the use of initialdisplacement.

● The following enforced displacement is applied to grid 1, xdirection

● d(t) = cos (2* * 5 * t)● Note that d(0) = 1.0

● The following enforced velocity is applied to grid 1, ydirection

● v(t) = -10 * * sin (2 * *5 * t)● Note that v(0) = 0

● The following enforced acceleration is applied to grid 1, zdirection

● a(t) = -100 * 2 * cos ( 2 * * 5 * t )


d(t) = cos (2* * 5 * t)

EXAMPLE SPECIFYING INITIALDISPLACEMENT (Cont.)

● Applied enforced displacement at grid 1 in x direction



v(t) = -10 * * sin (2 * *5 * t)

● Applied enforced velocity at grid 1 in y direction



a(t) = -100 * 2 * cos ( 2 * * 5 * t )

● Applied enforced acceleration at grid 1 in z direction


● Without US0, the enforced displacement is off when usingenforced velocity and acceleration● This affects the subsequent displacement, velocity, and

acceleration output

● With the appropriate US0 added to the enforced velocityand enforced acceleration input, the correct enforceddisplacement is applied● The product of US0 and the value specified on the SPCD should

equal to the enforced value



begin bulkparam,post,0tstep,40,500,.001,1grid,1,,,,,,123grid,2dload,5000,1.0,1.0,30,1.0,300,1.0,3000$$ Enforced displacement at Grid 1 - Component 1$ Corresponding response is at Grid 2 - Component 1$celas2,100,986.9604,1,1,2,1cmass2,200,1.,2,1spcd,20,1,1,1.tload2,30,20,,d,0.,1000.,5.$$ Enforced velocity at Grid 1 - Component 2$ Corresponding response is at Grid 2 - Component 2$celas2,1000,986.9604,1,2,2,2cmass2,2000,1.,2,2spcd,200,1,2,-31.4159tload2,300,200,,v,0.,1000.,5.,-90.,,,-3.183-2$$ Enforced acceleration at Grid 1 - Component 3$ Corresponding response is at Grid 3 - Component 3$celas2,10000,986.9604,1,3,2,3cmass2,20000,1.,2,3spcd,2000,1,3,-986.960tload2,3000,2000,,a,0.,1000.,5.,,,-1.013-3enddata




● Displacement output at grid 1



● Velocity output at grid 1



● Acceleration output at grid 1



● Displacement output at grid 2



● Velocity output at grid 2



● Acceleration output at grid 2


WORKSHOP # 7A

DIRECT TRANSIENT RESPONSE WITHENFORCED ACCELERATION


XY

Z

12451245

12451245

1245

XY

Z

Large Mass1000 lbsDrive PointDrive Point

WORKSHOP # 7A - DIRECT TRANSIENTRESPONSE WITH ENFORCED ACCELERATION

Using the direct method, determine the transient response due to aunit acceleration sine pulse of 250 Hz applied at the base in the z-direction. Use a structural damping coefficient of g = 0.06 and convertthis damping to equivalent viscous damping at 250 Hz.Below is a finite element representation of the flat plate. It alsocontains the loads and boundary constraints.


$$ wkshp7a.dat$$ add : boundary conditions at left end$ rbe2 for left edge in the vertical direction$ loading and enforced motion$SOL 109CENDTITLE = TRANSIENT RESPONSE WITH ENFORCED MOTIONSUBTITLE = USING DIRECT TRANSIENT METHODECHO = UNSORTEDSPC = 200SET 111 = 23, 33set 1000 = 23spcf = 1000DISPLACEMENT (SORT2) = 111VELOCITY (SORT2) = 111ACCELERATION (SORT2) = 111SUBCASE 1DLOAD = 500TSTEP = 100$

$OUTPUT (XYPLOT)XGRID=YESYGRID=YESXTITLE= TIME (SEC)YTITLE= BASE ACCELERATIONXYPLOT ACCELERATION RESPONSE / 23 (T3)YTITLE= BASE DISPLACEMENTXYPLOT DISP RESPONSE / 23 (T3)YTITLE= TIP CENTER DISPLACEMENT RESPONSEXYPLOT DISP RESPONSE / 33 (T3)$BEGIN BULK$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE$param,post,0INCLUDE 'plate.bdf'PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ SPECIFY STRUCTURAL DAMPING$PARAM, G, 0.06PARAM, W3, 1571.$TSTEP, 100, 200, 2.0E-4, 1$ENDDATA

WORKSHOP 7A● Start with the following input file


ID SEMINAR, PROB7ASOL 109CENDTITLE = TRANSIENT RESPONSE WITH BASE EXCITATIONSUBTITLE = USING DIRECT TRANSIENT METHOD, NOREDUCTIONECHO = UNSORTEDSPC = 200SET 111 = 23, 33DISPLACEMENT (SORT2) = 111VELOCITY (SORT2) = 111ACCELERATION (SORT2) = 111SUBCASE 1DLOAD = 500TSTEP = 100$OUTPUT (XYPLOT)XGRID=YESYGRID=YESXTITLE= TIME (SEC)YTITLE= BASE ACCELERATIONXYPLOT ACCELERATION RESPONSE / 23 (T3)YTITLE= BASE DISPLACEMENTXYPLOT DISP RESPONSE / 23 (T3)

YTITLE= TIP CENTER DISPLACEMENT RESPONSEXYPLOT DISP RESPONSE / 33 (T3)$BEGIN BULK$$ PLATE MODEL DESCRIBED IN NORMAL MODESEXAMPLE$INCLUDE ’plate.bdf’PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ SPECIFY STRUCTURAL DAMPING$PARAM, G, 0.06PARAM, W3, 1571.$$ APPLY EDGE CONSTRAINTS$SPC1, 200, 12456, 1, 12, 23, 34, 45

SOLUTION FOR WORKSHOP # 7A


$$ APPLY ACCELERATION TO THE BASE$SPC1, 200, 3, 23SPCD, 600, 23, 3, 1.0TLOAD2, 500, 600, , ACCE, 0.0, 0.004, 250., -90.$$ RBE MASS TO REMAINING BASE POINTS$RBE2, 101, 23, 3, 1, 12, 34, 45$$ SPECIFY INTEGRATION TIME STEPS$TSTEP, 100, 200, 2.0E-4, 1$ENDDATA

SOLUTION FOR WORKSHOP # 7A (Cont.)



ID TYPE NO. PANEL : GRID ID ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX0 1 ACCE 1 23( 5) 0.000000E+00 4.000000E-02 -1.000000E+00 3.000000E-03 1.000000E+00 9.999999E-04

0.000000E+00 4.000000E-02 -1.000000E+00 3.000000E-03 1.000000E+00 9.999999E-040 1 DISP 2 23( 5) 0.000000E+00 4.000000E-02 6.123031E-25 0.000000E+00 2.525501E-06 4.000000E-03

0.000000E+00 4.000000E-02 6.123031E-25 0.000000E+00 2.525501E-06 4.000000E-030 1 DISP 3 33( 5) 0.000000E+00 4.000000E-02 -4.217794E-07 9.400000E-03 5.628117E-06 5.800000E-03

0.000000E+00 4.000000E-02 -4.217794E-07 9.400000E-03 5.628117E-06 5.800000E-03

SOLUTION FOR WORKSHOP # 7A (Cont.)


BaseBaseAcceleration:Acceleration:Grid Point 23Grid Point 23

PARTIAL OUTPUT FILE FOR WORKSHOP # 7A(Cont.)


BaseBaseDisplacement:Displacement:Grid Point 23Grid Point 23

PARTIAL OUTPUT FILE FOR WORKSHOP #7A (Cont.)


Tip CenterTip CenterDisplacement:Displacement:Grid Point 33Grid Point 33



WORKSHOP # 7B

MODAL TRANSIENT RESPONSE WITHENFORCED ACCELERATION


XY

Z

12451245

12451245

1245

XY

Z

Large Mass1000 lbsDrive PointDrive Point

Using the modal method, determine the transient response due to aunit acceleration sine pulse of 250 Hz applied at the base in the z-direction. Use a structural damping coefficient of g = 0.06 and convertthis damping to equivalent viscous damping at 250 Hz. Be sure toinclude residual vector.Below is a finite element representation of the flat plate. It alsocontains the loads and boundary constraints.

WORKSHOP # 7B - MODAL TRANSIENTRESPONSE WITH ENFORCED ACCELERATION


WORKSHOP 7B

● Use solution from workshop 7A as a starting point andmake appropriate changes.


ID SEMINAR, PROB7BSOL 112CENDTITLE = TRANSIENT RESPONSE WITH BASE EXCITATIONSUBTITLE = USING MODAL TRANSIENT METHOD, NO REDUCTIONECHO = UNSORTEDSPC = 200METHOD = 1000SET 111 = 23, 33DISPLACEMENT (SORT2) = 111VELOCITY (SORT2) = 111ACCELERATION (SORT2) = 111SUBCASE 1DLOAD = 500TSTEP = 100$OUTPUT (XYPLOT)XGRID=YESYGRID=YESXTITLE= TIME (SEC)YTITLE= BASE ACCELERATIONXYPLOT ACCELERATION RESPONSE / 23 (T3)YTITLE= BASE DISPLACEMENTXYPLOT DISP RESPONSE / 23 (T3)YTITLE= TIP CENTER DISPLACEMENT RESPONSEXYPLOT DISP RESPONSE / 33 (T3)$

BEGIN BULK$$ PLATE MODEL DESCRIBED IN NORMAL MODES $EXAMPLE$INCLUDE ’plate.bdf’PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ SPECIFY STRUCTURAL DAMPING$PARAM, G, 0.06PARAM, W3, 1571.$$ APPLY EDGE CONSTRAINTS$SPC1, 200, 12456, 1, 12, 23, 34, 45

SOLUTION FOR WORKSHOP # 7B (Cont.)


EIGRL, 1000, , , 10$$ APPLY ACCELERATION TO THE BASE$SPC1, 200, 3, 23SPCD, 600, 23, 3, 1.0TLOAD2, 500, 600, , ACCE, 0.0, 0.004, 250., -90.$$ RBE MASS TO REMAINING BASE POINTS$RBE2, 101, 23, 3, 1, 12, 34, 45$$ SPECIFY INTEGRATION TIME STEPS$TSTEP, 100, 200, 2.0E-4, 1$ENDDATA




ID TYPE NO. PANEL : GRID ID ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX0 1 ACCE 1 23( 5) 0.000000E+00 4.000000E-02 -1.000000E+00 3.000000E-03 1.000000E+00 9.999999E-04

0.000000E+00 4.000000E-02 -1.000000E+00 3.000000E-03 1.000000E+00 9.999999E-040 1 DISP 2 23( 5) 0.000000E+00 4.000000E-02 6.123031E-25 0.000000E+00 2.525501E-06 4.000000E-03

0.000000E+00 4.000000E-02 6.123031E-25 0.000000E+00 2.525501E-06 4.000000E-030 1 DISP 3 33( 5) 0.000000E+00 4.000000E-02 -4.332059E-07 9.400000E-03 5.631920E-06 5.800000E-03

0.000000E+00 4.000000E-02 -4.332059E-07 9.400000E-03 5.631920E-06 5.800000E-03

SOLUTION FOR WORKSHOP # 7B


BaseBaseAcceleration:Acceleration:Grid Point 23Grid Point 23

PARTIAL OUTPUT FILE FOR WORKSHOP #7B (Cont.)


BaseBaseDisplacement:Displacement:Grid Point 23Grid Point 23



Tip CenterTip CenterDisplacement:Displacement:Grid Point 33Grid Point 33



WORKSHOP # 8A

DIRECT FREQUENCY RESPONSE WITHENFORCED DISPLACEMENT


Using the direct method, determine the frequency response of the flatrectangular plate, created in Workshop 1, subject to a 0.1 enforceddisplacement at the corner of the tip. Use a frequency step of 20 Hz inthe range of 20 to 1000 Hz. Use a structural damping of g = 0.06.Below is a finite element representation of the flat plate. It alsocontains the loads and boundary constraints.

WORKSHOP # 8A – DIRECT FREQUENCYRESPONSE WITH ENFORCED DISPLACEMENT


$$ wkshp8a.dat$$ add : enforced motion input$ frequencies of loading$SOL 108CENDTITLE= FREQUENCY RESPONSE DUE TO .1 DISPLACEMENT

AT TIPSUBTITLE= DIRECT METHODECHO= UNSORTEDSPC= 1SET 111= 11, 33, 55DISPLACEMENT(PHASE, SORT2)= 111$SDISP(PHASE, SORT2)= ALLset 222 = 11OLOAD= 222SUBCASE 1DLOAD= 500FREQUENCY= 100$

$OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NOXTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER,

MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASEXYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER,

MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, PHASEXYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$BEGIN BULK$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE$param,post,0INCLUDE 'plate.bdf'PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ SPECIFY STRUCTURAL DAMPING$PARAM, G, 0.06$ENDDATA

● Use the following input file as a starting pointWORKSHOP 8A


ID SEMINAR, PROB8ASOL 108CENDTITLE= FREQUENCY RESPONSE DUE TO .1 DISPLACEMENT ATTIPSUBTITLE= DIRECT METHODECHO= UNSORTEDSPC= 1SET 111= 11, 33, 55DISPLACEMENT(PHASE, SORT2)= 111$SDISP(PHASE, SORT2)= ALLset 222 = 11OLOAD= 222SUBCASE 1DLOAD= 500FREQUENCY= 100$OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NO

XTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER,MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASEXYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER,MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER,PHASEXYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$BEGIN BULK$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE$INCLUDE ’plate.bdf’PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ SPECIFY STRUCTURAL DAMPING

SOLUTION FOR WORKSHOP # 8A


SOLUTION FOR WORKSHOP # 8A (cont.)

$PARAM, G, 0.06$$ APPLY UNIT DISPLACEMENT AT TIP POINT$SPC1, 1, 3, 11SPCD, 600, 11, 3, 0.1$RLOAD2, 500, 600, , ,310, , DISP$TABLED1, 310, 0., 1., 10000., 1., ENDT$$$ SPECIFY FREQUENCY STEPS$FREQ1, 100, 20., 20., 49$ENDDATA







2.000000E+01 1.000000E+03 2.310271E-03 6.000000E+02 8.439928E-01 3.800000E+020 1 DISP 2 33(--, 11) 2.000000E+01 1.000000E+03 3.306744E-01 9.799999E+02 3.599946E+02 2.000000E+01

2.000000E+01 1.000000E+03 3.306744E-01 9.799999E+02 3.599946E+02 2.000000E+010 1 DISP 3 55( 5,--) 2.000000E+01 1.000000E+03 2.443192E-02 1.000000E+03 1.623458E+00 3.800000E+02

2.000000E+01 1.000000E+03 2.443192E-02 1.000000E+03 1.623458E+00 3.800000E+020 1 DISP 3 55(--, 11) 2.000000E+01 1.000000E+03 3.672138E+00 1.000000E+03 3.599891E+02 2.000000E+01

2.000000E+01 1.000000E+03 3.672138E+00 1.000000E+03 3.599891E+02 2.000000E+01


WORKSHOP # 8B

MODAL FREQUENCY RESPONSE WITHENFORCED DISPLACEMENT


Using the modal method, determine the frequency response of the flatrectangular plate, created in Workshop 1, subject to a 0.1 enforceddisplacement at the corner of the tip. Use a frequency step of 20 Hz inthe range of 20 to 1000 Hz. Use a structural damping of g = 0.06. Besure to include residual vectors.Below is a finite element representation of the flat plate. It alsocontains the loads and boundary constraints.

WORKSHOP # 8B – MODAL FREQUENCYRESPONSE WITH ENFORCED DISPLACEMENT


WORKSHOP 8B

● Use solution from workshop 8A as a starting point andmake appropriate changes.


ID SEMINAR, PROB8BSOL 111CENDTITLE= FREQUENCY RESPONSE DUE TO .1 DISPLACEMENTAT TIPSUBTITLE= MODAL METHODECHO= UNSORTEDSPC= 1SET 111= 11, 33, 55DISPLACEMENT(PHASE, SORT2)= 111$SDISP(PHASE, SORT2)= ALLset 222 = 11OLOAD= 222SUBCASE 1METHOD= 1000DLOAD= 500FREQUENCY= 100$OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NO

XTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASEXYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER,MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, PHASEXYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$BEGIN BULK$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE$INCLUDE ’plate.bdf’PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$$ SPECIFY STRUCTURAL DAMPING



$PARAM, G, 0.06EIGRL, 1000, , , 10$$ APPLY UNIT DISPLACEMENT AT TIP POINT$SPC1, 1, 3, 11SPCD, 600, 11, 3, 0.1$RLOAD2, 500, 600, , ,310, , DISP$TABLED1, 310, 0., 1., 10000., 1., ENDT$$$ SPECIFY FREQUENCY STEPS$FREQ1, 100, 20., 20., 49$ENDDATA







2.000000E+01 1.000000E+03 2.307091E-03 6.000000E+02 8.439932E-01 3.800000E+020 1 DISP 2 33(--, 11) 2.000000E+01 1.000000E+03 3.325545E-01 9.799999E+02 3.599946E+02 2.000000E+01

2.000000E+01 1.000000E+03 3.325545E-01 9.799999E+02 3.599946E+02 2.000000E+010 1 DISP 3 55( 5,--) 2.000000E+01 1.000000E+03 2.447906E-02 1.000000E+03 1.623457E+00 3.800000E+02

2.000000E+01 1.000000E+03 2.447906E-02 1.000000E+03 1.623457E+00 3.800000E+020 1 DISP 3 55(--, 11) 2.000000E+01 1.000000E+03 3.658197E+00 1.000000E+03 3.599891E+02 2.000000E+01

2.000000E+01 1.000000E+03 3.658197E+00 1.000000E+03 3.599891E+02 2.000000E+01



BLANK


SECTION 13

SHOCK AND RESPONSE SPECTRUMANALYSIS


The following graph

is generated from

Response

Resonant Frequency (Hz)

x(t)FMAX

FN = f1 f2 f3

uB (t)

Point on larger, vibrating structure

RESPONSE SPECTRUM

● Response spectrum depicts the maximum response of an SDOFsystem as a function of its resonant frequency for base excitation.


● The peak response of each SDOF oscillator is calculated from its X(t).The oscillator base motion UB is derived from the force or baseexcitation applied to a larger structure.

● Example: An earthquake time history is applied to a power plant.Response spectra are calculated at the locations of the floors to beused in the design of components (i.e., machinery and pipingsystems).

● An implicit assumption is that the oscillator’s mass is very smallrelative to the larger, vibrating mass. Therefore, no dynamicinteraction occurs between the two. (Consequently, the responsespectrum analysis is decoupled from the transient analysis.)

RESPONSE SPECTRUM (Cont.)


● Analysis is repeated for several damping values to generate a family of curves.

Damping applies to each oscillator, not to the vibrating structure.● Maximum displacement response from X(t) is calculated for each oscillator. The maximum relative

displacement between each oscillator and its base (a point on the vibrating structure) is alsocomputed.

● Relative velocity and absolute acceleration are approximately related to the relative displacementsby

● For design, useful variables are Xr, , and . Design spectra are usually in terms of thesevariables.


rr XX

Response

Resonant Frequency (Hz)

1 damping = 0% critical2 damping = 3% critical

12

3

Xr = maximum relative displacement

X = maximum intertial (absolute) displacement

..

......

r2XX

rX X


ZETA=.05 SDOF RESPONSE TO 4 HZ SINE PULSE

SINE

0.0 1.490

978.2

0.0

-1.1450.0 1.490

2 HZ SYS 2.327

1.000

0.0

-1.000

-2.000-2.726

4 HZ SYS

0.0 1.49010 HZ SYS 20 HZ SYS 40 HZ SYS

1.490 0.0 1.490 0.0 1.490

1.039

0.0

-1.016

1.068

0.0

-999.7m



SHOCK SPECTRUM OF 4 HZ PULSE FOR ZETA = .05

100.0 1.00 10.0 20.0

2.009 4.025 9.928 20.41



0

X 0Xr UB

X UBXr

0.. ....


● For very low oscillator frequencies

● For very high oscillator frequencies

● The approximate relationships between Xr, Xr, and X are not valid atvery high or very low frequencies or for large damping values.

● Note that only the magnitudes of response are computed with nophase information.


● Response spectra may be generated in any transientsolution (e.g., SOLs 109, 112).

● The transient response for selected DOFs in the model isused as the input time history for the generation of theresponse spectra curves.

● Further information is in the MSC.Nastran AdvancedDynamics User’s Guide.



Required input

● Executive Control SectionSOL command selecting a transient solution (e.g., SOL 109)

● Case Control SectionXYPLOT SPECTRAL Compute spectraXYPUNCH SPECTRAL Punch spectra

Example:

XYPUNCH ACCELERATION SPECTRAL 1/1(T1RM)

This XYPLOT command creates a set of absolute (RM) accelerationspectra based on record number 1 on the DTI,SPSEL entry using themotion of Grid Point 1 in the x (T1) direction.

RESPONSE SPECTRUM GENERATION


● Bulk Data Section

PARAM,RSPECTRA,0 Requests calculation of spectraDTI,SPSEL Correlates frequency and damping requestsFREQ Used to specify frequencies and damping

values (one FREQ set each)

● Example Input:

RESPONSE SPECTRUM GENERATION (Cont.)


● Available in SOL 103

● “Poor man’s transient.” The input spectra are used to determinethe peak response of each mode.

● These peak modal responses are combined to obtain the systemresponse (the only problem is that the timing of each mode’s peakis not known).

● Three methods of combining the modal responses are available(ABS, SRSS, NRL).

APPLYING SPECTRA


Procedure

● A model of the structure to be analyzed is created with the inputpoints identified as ‘SUPORT’ DOFs.

● A “large mass” (usually 103 to 106 times the structural mass) isattached to the ‘SUPORT’ DOFs.

● System modes are obtained for the model (including the 0.0 Hzmodes) with the ‘SUPORT’ DOFs unconstrained.

● This approximates the “cantilevered” modes of the model attachedto the “exciting” structure.

APPLYING SPECTRA (Cont.)



● The 0.0 Hz modes (Dm) approximate the ‘static’ motion the modelexperiences when the supporting structure moves statically.

● “Participation Factors” (PF) are calculated using the followingexpression:

TMDm

● where = the set of “elastic” modes

● PF is used in conjunction with the spectra described as shown in theinput section to calculate the peak response for each mode.

● Data recovery quantities (displacements, stresses, forces, etc.) arethen calculated for each mode based on its peak motion.

● These quantities are then combined for the modes using the selectedmethod (ABS, SRSS, NRL), and the results are printed.


x··r gx·r 2xr+ + u·· r t=

uk t ikirxr i g i t

r

i=

uk i k irxr i i gi

r

i

xr i i gi max xr i i gi t =

● Xr, response of a single DOF oscillator due to the base motion iscalculated as follows:

● The actual transient response at a physical point is

● ABSOLUTE Option

● where● and i Represents a mode● and r Represents a direction



uk i ki 2

i

i

i ir xr i g i 2

r

uk jkj iki 2

i j+

jkj


● SRSS option

where the average peak modal magnitude, is

● NRL option

where is the peak modal magnitude



Required Input● Executive Control Section

SOL statement selecting SOL 103● Case Control Section

SDAMP To select modal damping ratiosDLOAD To select input spectraMETHOD To select eigenvalue solver

● ExampleMETHOD = 1 Selects eigenvalue solution method 1 from the Bulk

Data (be sure that the range includes 0.0)SDAMP = 1 Selects modal damping to be used for the calculated

modes. Refers to a TABDMP1 entry in the Bulk DataDLOAD = 1 Selects DLOAD Bulk Data entry that describes which

spectra are applied at which ‘SUPORT’ DOF


Bulk Data

PARAM,SCRSPEC,0 Requests application of responsespectra

DLOAD Selects spectra and ‘SUPORT’ DOF atwhich to apply them

DTI,SPECSEL Selects spectra, states associateddamping and type of spectrum

TABLED1 Provides input spectraSUPORT Selects spectrum input locationsTABDMP1 Describes modal damping for the

calculated modesPARAM,OPTION Selects modal combination method



SUPORT 1 3$$ Define input dof for the spectra - in this case, dof 3 for GRID 1$ is selected$CONM2 1001 3 0 1000000.$ apply large masses in the directions of the spectra inputTABDMP1 1 CRIT +DMP1+DMP1 0.0 .01 100. .01 100.01 .02 1000. .02 +DMP2+DMP2 ENDT$$ Select damping ratios for the calculated modes - in this case, a ratio of$ 1% of ctrtical $ is used for all modes from 0hz to 100hz and 2% of critical$ is used for all modes above $ 100.01hz$$PARAM,SCRSPEC,0$ Tells MSC.Nastran to perform shock spectrum analysis$ DEFINE WHERE AND HOW TO APPLY SPECTRA$$ NOTE THAT SPECTRA ARE APPLIED USING INTERNAL SORT...NOT ASCENDING ORDER$$ SID S S1 L1 S2 L2 ....DLOAD 1 1.0 1.0 1$$ Define where spectra are to be applied - this entry is called from$ Case Control by a $ ’DLOAD=1’ command - for this entry, an overall$ scale factor of 1.0 (S) is applied, $ a factor of 1.0 (S1) is used$ to apply spectrum 1 (L1) at ’SUPORT’ dof 1.$$ (It should be noted that the order of the ’SUPORT’ dof used on this$ entry is the MSC.Nastran internal sort. If only one GRID point is used,$ this is no problem, but if more than one GRID point is used,$ then PARAM,USETPRT,1 should be used to obtain the internal order)$

● Sample Input



DTI SPECSEL 0

DTI SPECSEL 1 A 2 0.0 3 .01 +SP1

+SP1 4 .02

$

$ This table defines the relationship between the input tables

$ (from the spectrum creation run) and the spectra sets. For example,

$ record 1 defines a spectra set representing acceleration spectra,

$ containing spectrum 2 for 0% of critical damping, spectrum 3 for 1% of

$ critical damping, and spectrum 4 for 2% of critical damping.

$ The program will interpolate between the spectra if a mode has a

$ damping value other than those defined in the table.

$

$ GRID Component (XYPLOT terminology)

$ACCE 0 1 3 1

$ 0.000000E+00

TABLED1 2

.5 3.156-4 1.0 .001263 1.5 .002842 2. .005056

2.5 .007905 3. .011393 3.5 .015524 4. .020303

4.5 .025738 5. .031839 5.5 .038615 6. .046073

6.5 .054219 7. .063052 7.5 .072569 8. .082766

.

.

100.5 3.87229ENDT

$ Table representing the input spectra



WORKSHOP #9 (PART I)

Generates The Shock Spectrum InputGenerate a shock spectrum of the plate due to a 2.0 in/sec2

sine pulse applied at the clamped edge for 0, 2, and 4 percentdamping.


$$ wkshp9a.dat$$ case control, add : generate response spectrum curve at grid point 3000$ using xyplot/xypunch commands$ bulk data, add : appropriate parameters$ create response spectrum curves for appropriate damping$ and frequencies at grid point 3000$ID SEMINAR, PROB9aSOL 109TIME 30CENDTITLE= TRANSIENT RESPONSESUBTITLE= USING DIRECT TRANSIENT METHODLABEL= SHOCK SPECTRUM CALCULATIONECHO= UNSORTEDSPC= 100SET 111= 3000DISPLACEMENT (SORT2)= 111 $ AT LEAST DISP AND VEL MUST APPEARVELOCITY (SORT2)= 111ACCELERATION ()= 111DLOAD= 500TSTEP= 100$

WORKSHOP # 9A● Start with the following partial input file


$OUTPUT (XYPLOT)$$ SHOCK RESPONSE IS ONLY AVAILABLE IN PLOT OR PUNCH OUTPUT. THEREFORE,$ THE `OUTPUT(XYPLOT)' SECTION OF THE CASE CONTROL MUST BE USED.$XGRID=YESYGRID=YESXYPLOT ACCE / 3000(T1)XLOG= YESYLOG= YES$$ RELATIVE SHOCK RESPONSES ARE CONTAINED IN THE IMAGINARY/PHASE$ COMPONENTS OF THE OUTPUT$ ABSOLUTE SHOCK RESPONSES ARE CONTAINED IN THE REAL/MAGNITUDE$ COMPONENTS OF THE OUTPUT$$BEGIN BULK$$ DEFINE GRID POINT$GRID, 3000, ,0.,0.,0., ,23456$$ DEFINE MASS$CMASS2, 100, 1.0, 3000, 1$$ APPLY LOADING TO MASS$TLOAD2, 500, 600, , 0, 0., 0.004, 250., -90.$DAREA, 600, 3000, 1, 1.$$ SPECIFY INTEGRATION TIME STEPS$TSTEP, 100, 100, 4.0E-4, 1$ENDDATA

WORKSHOP # 9A


ID SEMINAR, PROB9aSOL 109TIME 30CENDTITLE= TRANSIENT RESPONSESUBTITLE= USING DIRECT TRANSIENT METHODLABEL= SHOCK SPECTRUM CALCULATIONECHO= UNSORTEDSPC= 100SET 111= 3000DISPLACEMENT (SORT2)= 111 $ AT LEAST DISP AND VEL MUST APPEARVELOCITY (SORT2)= 111ACCELERATION ()= 111DLOAD= 500TSTEP= 100$OUTPUT (XYPLOT)$$ SHOCK RESPONSE IS ONLY AVAILABLE IN PLOT OR PUNCH OUTPUT. THEREFORE,$ THE ‘OUTPUT(XYPLOT)’ SECTION OF THE CASE CONTROL MUST BE USED.$XGRID=YESYGRID=YESXYPLOT ACCE / 3000(T1)XLOG= YESYLOG= YES$$ RELATIVE SHOCK RESPONSES ARE CONTAINED IN

$IMAGINARY/PHASE$ COMPONENTS OF THE OUTPUT$ ABSOLUTE SHOCK RESPONSES ARE CONTAINED IN THE REAL/MAGNITUDE$ COMPONENTS OF THE OUTPUT$

XTITLE= FREQUENCY (CYCLES/SEC)YTITLE= RELATIVE DISPLACEMENTXYPLOT DISP SPECTRAL 1 / 3000 (T1IP)YTITLE= RELATIVE VELOCITYXYPLOT VELOCITY SPECTRAL 1 / 3000 (T1IP)YTITLE= ABSOLUTE ACCELERATIONXYPLOT ACCELERATION SPECTRAL 1 / 3000 (T1RM)$$ PUNCH SHOCK SPECTRUM FOR LATER USE$XYPUNCH ACCELERATION SPECTRAL 1 / 3000(T1RM)$BEGIN BULK$$ DEFINE GRID POINT$GRID, 3000, ,0.,0.,0., ,23456

SOLUTION FILE FOR WORKSHOP #9 (PART I)


$$ DEFINE MASS$CMASS2, 100, 1.0, 3000, 1$$ APPLY LOADING TO MASS$TLOAD2, 500, 600, , 0, 0., 0.004, 250., -90.$DAREA, 600, 3000, 1, 1.$$ SPECIFY INTEGRATION TIME STEPS$TSTEP, 100, 100, 4.0E-4, 1$$ PARAMETER TO CALCULATE SHOCK SPECTRUM$PARAM, RSPECTRA, 0$$ SPECIFY FREQUENCY AND DAMPING VALUES FOR$ THE SDOF OSCILLATORS AT GRID 3000$DTI, SPSEL, 0DTI, SPSEL, 1, 111, 222, 3000$ 1= SUBCASE... 111= DAMPING... 222= FREQUENCIES... 3000= GRID NUMBER$$ DAMPING INFORMATION FOR OSCILLATORS$FREQ, 111, 0., 0.02, 0.04$$ NATURAL FREQUENCIES OF OSCILLATORS$FREQ1, 222, 20., 20., 49$ENDDATA

SOLUTION FILE FOR WORKSHOP #9 (PART I)(Cont.)


PARTIAL OUTPUT FILE FOR WORKSHOP #9(PART I)


PARTIAL OUTPUT FILE FOR WORKSHOP #9(PART I) (Cont.)


WORKSHOP #9 (PART II)

Applies The Shock SpectrumApply the shock spectrum generated in Part I and sum the response using theSRSS option. Include modes up to 1000 Hz using 3% critical damping.


$$ wkshp9b.dat$$ case control, add: appropriate solution sequence$$ bulk data, add: apply the response spectrum curves generated in run 9a.$ use the srss summing method.$ID SEMINAR, PROB9bTIME 30CENDTITLE= RESPONSE SPECTRUM ANALYSISSUBTITLE= USING CALCULATED SHOCK RESPONSELABEL= SHOCK WILL BE INPUT IN Z DIRECTIONECHO= UNSORTEDSET 111= ALLDISPLACEMENT= 111SPC= 200SUBCASE 1METHOD= 100SDAMP= 200DLOAD= 500$

● Use the following partial input file as a starting point

WORKSHOP 9 (PART 2)


BEGIN BULK$$ PLATE MODEL DESCRIBED IN NORMAL MODES EXAMPLE$INCLUDE 'plate.bdf'PARAM,COUPMASS,1PARAM,WTMASS,0.00259$$ BOUNDRY CONDITIONS FOR `CLAMPED' MODES$SPC1, 200, 1245, 1, 12, 23, 34, 45$$ PLACE BIG FOUNDATION MASS (BFM) AT BASE$ TO STIMULATE `CLAMPED' MODES$CMASS2, 110, 1000., 23, 3$$ RBE MASS TO REMAINING BASE POINTS$RBE2, 101, 23, 3, 1, 12, 34, 45$$ SUPORT CARD TO IDENTIFY EXCITATION DOFS$SUPORT, 23, 3$$ EIGENVALUE EXTRACTION$ MUST BE MASS NORMALIZED (DEFAULT)$eigrl,100,0.,1000.$$ TABLE TO SPECIFY DAMPING FOR USE IN THE ANALYSIS$TABDMP1, 200, CRIT,, 0., 0.03, 1000., 0.03, ENDT$$ SPECIFICATION OF SHOCK SPECTRUM TO BE USED$$ MODAL FREQUENCY RANGE CAN BE SELECTED USINGPARAM, LFREQ, 0.1PARAM, HFREQ, 1000.$ENDDATA

WORKSHOP 9 (PART 2)


ID SEMINAR, PROB9bSOL 103TIME 30CENDTITLE= RESPONSE SPECTRUM ANALYSISSUBTITLE= USING CALCULATED SHOCK RESPONSELABEL= SHOCK WILL BE INPUT IN Z DIRECTIONECHO= UNSORTEDSET 111= ALLDISPLACEMENT= 111SPC= 200SUBCASE 1METHOD= 100SDAMP= 200DLOAD= 500$BEGIN BULK$$ PLATE MODEL DESCRIBED IN NORMAL MODESEXAMPLE$INCLUDE ’plate.bdf’PARAM,COUPMASS,1PARAM,WTMASS,0.00259$$ BOUNDRY CONDITIONS FOR ‘CLAMPED’ MODES$SPC1, 200, 1245, 1, 12, 23, 34, 45$

$$ PLACE BIG FOUNDATION MASS (BFM) AT BASE$ TO STIMULATE ‘CLAMPED’ MODES$CMASS2, 110, 1000., 23, 3$$ RBE MASS TO REMAINING BASE POINTS$RBE2, 101, 23, 3, 1, 12, 34, 45$$ SUPORT CARD TO IDENTIFY EXCITATION DOFS$SUPORT, 23, 3$$ EIGENVALUE EXTRACTION$ MUST BE MASS NORMALIZED (DEFAULT)$eigrl,100,0.,1000.$$ TABLE TO SPECIFY DAMPING FOR USE IN THE ANALYSIS

$TABDMP1, 200, CRIT,, 0., 0.03, 1000., 0.03, ENDT$$ SPECIFICATION OF SHOCK SPECTRUM TO BE USED$DLOAD, 500, 1.0, 2.0, 1$$ DLOAD, ID, OVERALL SCALE, SCALE FOR R-SET DOF# 1, SHOCK TABLE FOR$DOF# 1,$ SCALE FOR R-SET DOF# 2, SHOCK TABLE FOR DOF# 2, ETC.$

SOLUTION FILE FOR WORKSHOP #9 (PART II)


$ SELECT SHOCK RESPONSE CALCULATION1 RESPONSE SPECTRUM ANALYSIS MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 7

USING CALCULATED SHOCK RESPONSE0 SHOCK WILL BE INPUT IN Z DIRECTION0 I N P U T B U L K D A T A E C H O

. 1 .. 2 .. 3 .. 4 .. 5 .. 6 .. 7 .. 8 .. 9 .. 10 .$PARAM, SCRSPEC, 0$$ SELECT SUMMATION OPTION$PARAM, OPTION, SRSS$$ MODAL FREQUENCY RANGE CAN BE SELECTED USINGPARAM, LFREQ, 0.1PARAM, HFREQ, 1000.$$ SPECIFICATION FOR SHOCK TABLES$DTI, SPECSEL, 0DTI, SPECSEL, 1, , A, 2, 0., 3, 0.02,, 4, 0.04, ENDREC$$ DTI, SPECSEL, SHOCK TABLE NUMBER, , [(A)CCELERATION, (V)ELOCITY, OR (D)ISP],$ TABLED1 POINTER, DAMPING FOR TABLE, ETC.$$ PUNCH OUTPUT FOR SHOCK SPECTRUM CALCULATION$$ ACCE 4 3000 3 1$ 0.000000E+00$TABLED1 2

20. .038683 40. .152539 60. .33511 80. .576059100. .862049 120. 1.17619 140. 1.50169 160. 1.82018180. 2.11404 200. 2.36801 220. 2.56617 240. 2.70027260. 2.76275 280. 2.75073 300. 2.74632 320. 2.61887340. 2.4218 360. 2.39068 380. 2.24931 400. 2.02296420. 1.78538 440. 1.70355 460. 1.57056 480. 1.40493500. 1.22608 520. 1.20483 540. 1.17631 560. 1.14097580. 1.10048 600. 1.05582 620. 1.00818 640. .958761660. .908725 680. .859158 700. .827667 720. .782127740. .728996 760. .694088 780. .668602 800. .635044820. .598496 840. .571831 860. .563072 880. .550499

SOLUTION FILE FOR WORKSHOP #9 (PART II)(Cont.)


900. .528854 920. .509281 940. .500534 960. .498016980. .488793 1000. .468321 ENDT

$ACCE 4 3000 3 52$ 2.0000000E-02TABLED1 3

20. .037708 40. .143365 60. .314936 80. .541342100. .80976 120. 1.10506 140. 1.40671 160. 1.69567180. 1.98167 200. 2.22217 220. 2.35249 240. 2.53055260. 2.56231 280. 2.55577 300. 2.58668 320. 2.45921340. 2.29411 360. 2.25956 380. 2.12901 400. 1.92605420. 1.68656 440. 1.61355 460. 1.4968 480. 1.35263500. 1.19796 520. 1.17707 540. 1.14947 560. 1.11613580. 1.07807 600. 1.03637 620. .992124 640. .946383

1 RESPONSE SPECTRUM ANALYSIS MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 8USING CALCULATED SHOCK RESPONSE

0 SHOCK WILL BE INPUT IN Z DIRECTION0 I N P U T B U L K D A T A E C H O

. 1 .. 2 .. 3 .. 4 .. 5 .. 6 .. 7 .. 8 .. 9 .. 10 .660. .900171 680. .854434 700. .810016 720. .767647740. .727923 760. .691288 780. .658039 800. .628311820. .602091 840. .579207 860. .559362 880. .542128900. .526973 920. .51329 940. .500403 960. .487602980. .474171 1000. .459408 ENDT

$ACCE 4 3000 3 103$ 4.0000000E-02TABLED1 4

20. .039336 40. .137673 60. .297382 80. .511244100. .764891 120. 1.04406 140. 1.31588 160. 1.58461180. 1.85678 200. 2.10175 220. 2.19165 240. 2.3921260. 2.39929 280. 2.42782 300. 2.44263 320. 2.317340. 2.17923 360. 2.14283 380. 2.0227 400. 1.8407420. 1.62279 440. 1.53417 460. 1.43168 480. 1.30597500. 1.17212 520. 1.15165 540. 1.12513 560. 1.09349580. 1.05768 600. 1.01868 620. .977462 640. .934986660. .892143 680. .849752 700. .808538 720. .769114740. .731968 760. .69746 780. .665814 800. .637115820. .611319 840. .588261 860. .567655 880. .549125900. .532205 920. .516369 940. .501047 960. .485644980. .469568 1000. .452243 ENDT

$ENDDATA

SOLUTION FILE FOR WORKSHOP #9 (PART II)(Cont.)



0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 1


1 1 0.0 0.0 0.0 1.000000E+00 0.02 2 7.057114E+05 8.400663E+02 1.337007E+02 1.000000E+00 7.057114E+053 3 1.877186E+07 4.332651E+03 6.895628E+02 1.000000E+00 1.877186E+074 4 2.811329E+07 5.302197E+03 8.438708E+02 1.000000E+00 2.811329E+07


0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 1^^^ USER INFORMATION MESSAGE 9047 (POSTREIG)^^^ SCALED RESPONSE SPECTRA FOR RESIDUAL STRUCTURE ONLY*** USER INFORMATION MESSAGE 7588 (GKAM)

BASED ON THE USER PARAMETERS LMODES, LFREQ OR HFREQ, ONLY 3 OF THE 4 COMPUTED STRUCTURE MODES(MODES 2 THROUGH 4) WILL BE USED IN THIS MODAL TRANSIENT RESPONSE ANALYSIS(DETAILS OF THE EIGENVALUE DATA FOR THE MODES USED ARE GIVEN BELOW)


0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 1

R E A L E I G E N V A L U E S(ACTUAL MODES USED IN THE DYNAMIC ANALYSIS)


2 2 7.057114E+05 8.400663E+02 1.337007E+02 1.000000E+00 7.057114E+053 3 1.877186E+07 4.332651E+03 6.895628E+02 1.000000E+00 1.877186E+074 4 2.811329E+07 5.302197E+03 8.438708E+02 1.000000E+00 2.811329E+07


0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 11 RESPONSE SPECTRUM ANALYSIS MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 16

USING CALCULATED SHOCK RESPONSE0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 10 MATRIX FN (GINO NAME 101 ) IS A DB PREC 1 COLUMN X 3 ROW RECTANG MATRIX.0COLUMN 1 ROWS 1 THRU 3 --------------------------------------------------

ROW1) 1.3370D+02 6.8956D+02 8.4387D+02

0THE NUMBER OF NON-ZERO TERMS IN THE DENSEST COLUMN = 30THE DENSITY OF THIS MATRIX IS 100.00 PERCENT.1 RESPONSE SPECTRUM ANALYSIS MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 17

USING CALCULATED SHOCK RESPONSE0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 10 PSIT

POINT VALUE POINT VALUE POINT VALUE POINT VALUE POINT VALUE

COLUMN 123 T3 -2.11561E-02

COLUMN 223 T3 9.14315E-13

COLUMN 323 T3 1.18597E-02


0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 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 SET0 -1- -2- -3- -4- -5- -6- -7- -8- -9- -10-

1= 23-30 SCALED SPECTRAL RESPONSE, SRSS OPTION, DLOAD = 500 CLOSE = 1.00

PARTIAL OUTPUT FILE FOR WORKSHOP #9(PART II)


PARTIAL OUTPUT FILE FOR PROBLEM #9 (PART II) (Cont.)1 RESPONSE SPECTRUM ANALYSIS MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 19

USING CALCULATED SHOCK RESPONSE0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 10 MATRIX UHVR (GINO NAME 101 ) IS A REAL 3 COLUMN X 3 ROW SQUARE MATRIX.0COLUMN 1 ROWS 1 THRU 3 --------------------------------------------------

ROW1) 7.6204E-08 8.1011E-20 4.8920E-10

0COLUMN 2 ROWS 1 THRU 3 --------------------------------------------------ROW

1) 6.4017E-05 3.5099E-16 2.5938E-060COLUMN 3 ROWS 1 THRU 3 --------------------------------------------------

ROW1) 5.3778E-02 1.5207E-12 1.3753E-02

0THE NUMBER OF NON-ZERO TERMS IN THE DENSEST COLUMN = 30THE DENSITY OF THIS MATRIX IS 100.00 PERCENT.1 RESPONSE SPECTRUM ANALYSIS MARCH 12, 2007 MD NASTRAN 3/ 1/07 PAGE 20

USING CALCULATED SHOCK RESPONSE0 SHOCK WILL BE INPUT IN Z DIRECTION SUBCASE 1

SCALED SPECTRAL RESPONSE, SRSS OPTIOND 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 6.222940E-10 0.0 0.0 7.935161E-192 G 1.656899E-18 7.935161E-19 7.362168E-08 8.380406E-08 3.053683E-07 1.051286E-183 G 3.038017E-18 1.358321E-18 3.172597E-07 1.218669E-07 6.622535E-07 1.255057E-184 G 4.072633E-18 2.361632E-18 7.194942E-07 1.166965E-07 9.376346E-07 1.666253E-185 G 4.328228E-18 3.551627E-18 1.246647E-06 1.041301E-07 1.160351E-06 1.080113E-186 G 3.627554E-18 4.017149E-18 1.870941E-06 8.255398E-08 1.326449E-06 9.153620E-19

49 G 4.083018E-18 2.280663E-18 1.246647E-06 1.041301E-07 1.160351E-06 1.558670E-1850 G 6.021022E-18 3.094015E-18 1.870941E-06 8.255398E-08 1.326449E-06 3.331682E-1951 G 8.089630E-18 3.015453E-18 2.566087E-06 6.151257E-08 1.444593E-06 1.727660E-1852 G 9.842749E-18 1.590179E-18 3.309156E-06 4.199305E-08 1.519643E-06 4.018735E-1853 G 1.094405E-17 9.800442E-19 4.080598E-06 2.667464E-08 1.560018E-06 5.598891E-1854 G 1.141588E-17 4.045981E-18 4.865306E-06 1.686023E-08 1.575035E-06 6.242042E-1855 G 1.153212E-17 7.227172E-18 5.653790E-06 1.297877E-08 1.577920E-06 6.353067E-18

PARTIAL OUTPUT FILE FOR WORKSHOP #9(PART II) (Cont.)


SECTION 14

RANDOM RESPONSE ANALYSIS


Deterministic

Periodic Transient

SimpleHarmonic

ShockSpectra

Random

StationaryNonstationary

Ergodic

MD NastranMD Nastran

CLASSIFICATION OF DYNAMICENVIRONMENTS



● Random vibration is vibration that can be described only in a statisticalsense. Its instantaneous magnitude at any time is not known; rather,the probability of its magnitude exceeding a certain value is given.

● Examples include earthquake ground motion, ocean wave heights andfrequencies, wind pressure fluctuations on aircraft and tall buildings,and acoustic excitation due to rocket and jet engine noise.

● MD Nastran performs random response analysis as post processing tofrequency response. Inputs include the output from frequencyresponse analysis as well as user-supplied loading conditions in theform of auto- and cross spectral densities. Outputs are responsepower spectral densities, autocorrelation functions, number of zerocrossings with positive slope per unit time, and the RMS values ofresponse.

● The theory is described in Random Vibration in Mechanical Systems,by S. H. Crandall and W. D. Mark, Academic Press, 1963.

● Further information is in the MSC.Nastran Advanced Dynamics User’sGuide.


RANDOM ANALYSIS THEORY

● There are several conventions used to define randomanalysis quantities. Care must be taken to use MDNastran random capability properly (see the MSC.NastranAdvanced Dynamics User’s Guide for details and Bendatand Piersol (Reference 13) to understand theconventions).

● MD Nastran random analysis assumes ergodic processes.● The concepts of autocorrelation, autospectrum (power

spectrum), cross-correlation, and cross-spectrum must bedefined.

● The mean square value and apparent frequency are thekey statistical quantities.


STATIONARY RANDOM

-3.242

3.363

1.751

-1.449

0.0 10.23

0.0 10.23

NON-STATIONARY RANDOM

EXAMPLES OF RANDOM DYNAMICENVIRONMENT


EXAMPLE OF ENSEMBLE OF ERGODICRANDOM DATA


● Autocorrelation function:

● Note: Rj(o) is the mean-square value of uj(t).● Autospectrum function:

● Fourier transform pairs:

● Mean square value:

Sj 2T---

T lim uj te

it–td

0

T

2

=

Rj 12------ Sj cos d

0=

uj t2

Rj o 12------ Sj d

0

= =

AUTOCORRELATION AND AUTOSPECTRUM


N 20

2 2

Sj d0

Sj d0

-----------------------------------------------------=

● Apparent frequency N0 (zero crossings):

AUTOCORRELATION AND AUTOSPECTRUM


Fa ()

Uj ()

uj Hja Fa=

uj Hja Fa Hjb F

b ++=

CALCULATION OF LINEAR SYSTEM RESPONSETO ERGODIC RANDOM EXCITATION

● From frequency response analysis

● where Hja(w) is the frequency response or transfer functionrelating output uj to input Fa.

● If we have several inputs, then


● In matrix form we haveFa ()

uj() = [ Hja() Hjb() … ] Fb ()..

● The output autospectrum is

Fa () H*ja

Sujuj = T[ Hja Hjb … ] Fb () [F*a()F*b()…] H*jb

. .

. .

DEFINITION OF MULTIPLE INPUT-OUTPUT SPECTRALRELATIONSHIP FOR A LINEAR SYSTEM


TFa Fa Saa =

TFa Fb Sab =

TFb Fb Sbb =

● The multiple input-output spectral relationship is therefore:

Sujuj Hj T

Saa Sab

Sba Sbb Hj = .

.

.

.

.

.

.

.

.

DEFINITION OF MULTIPLE INPUT-OUTPUT SPECTRALRELATIONSHIP FOR A LINEAR SYSTEM (Cont.)

● The individual input spectra are


The input cross-spectral matrix is

Saa() Sab() …[ S ]in = Sba() Sbb() …

. .

. .

and it has the special properties

Hj T

HjaHjb =whereHj

HjaHjb

=...

Sab Sba=

Sbb real 0=Saa () ,



Sujuj Hja 2Saa =

Sujuj Hja 2Saa Hjb 2Sbb + +=


● Commonly used special cases● Single input analysis (fully correlated inputs)

● Uncorrelated multiple inputs


Hja uj Fa =

● It is assumed that the output from the frequency responsecalculations is Hja(). It does not calculate

● If Hja() is desired, use F() = 1.0.

RANDOM ANALYSIS AS IMPLEMENTED INMD NASTRAN


StructuredSolution

Direct 108Modal 111

● Executive Control SectionSOL (required)

● Case Control SectionRANDOM (selects Bulk Data RANDPS, RANDT entries and entries

for frequency response, and must be above thesubcases)

● Bulk Data SectionRANDPS (PSD specification)RANDT1 (autocorrelation time lag entries for frequency response)

INPUT REQUIRED FOR RANDOM RESPONSEANALYSIS


RANDPS ENTRY


TABRND1 ENTRY


RANDOM RESPONSE OUTPUT

● Output option 1:● Auto-PSDF, auto correlation functions, N0, and CRMS (cumulative root

mean square) print and punch output are available using “normal” CaseControl command.

● Available for acce, disp, velocity, force, oload, spcf, mpcf, stress, andstrain output, using the RPRINT and RPUNCH optionsFormat for Disp:

● Output option 2:● Auto-psdf, auto correlation functions, and N0 (# of positive crossings) are

available using the xyplot/xypeak/xypunch options


RANDOM RESPONSE OUTPUT (CONT.)

● Format for Disp (cont.):where

PSDF—request output for auto power spectral density functionATOC—request output for auto correlation functionCRMS—request output for cumulative root mean squareRALL—request output for psdf, atoc, and crmsRPRINT—request printed output in the f06 fileRPUNCH—request punch outputNORPRINT—none of the above output

● Log-Log option available when computing RMS, N0, and CRMS● Param,rmsint,log-log


● Additional Output:● cross power spectral● cross-correlation functions● These output requires

● RCROSS (and RANDOM) Case Control commands

PSDF—request output for cross power spectral density functionCROF—request output for cross correlation functionRALL—request output for both psdf and crof

● RCROSS (and RANDPS) Bulk Data Entries

RANDOM RESPONSE OUTPUT (Cont.)



Example:● The following simplified car model is subjected to random loadings that are fully

correlated at the front and back wheels

● Output request:● The auto psdf disp, RMS, CRMS, N0 at grid points 1,2, and 5● The cross psdf displacement output between grid 1 (t2 ) and grid 2 (t2)


SOL 108 $CENDTITLE = SIMPLE CAR WITH RANDOM INPUTSPC = 100FREQUENCY = 1000set 50 = 1,2,5disp(phase,psdf,crms) = 50rcross(phase,psdf) = 100$random = 1000SUBCASE 1DLOAD = 111

SUBCASE 2DLOAD = 112

$output (xyplot)xtitle = frequency (hz)ytitle = disp psd at grid pt 5xypunch disp psdf /5(t2)$BEGIN BULK

$$ 2 3 4 5 6 7 8 9 10RCROSS 100 DISP 1 2 DISP 2 2$FREQ1 1000 0.1 .05 40$$ DEFINE THE INPUT PSD$ 2 3 4 5 6 7 8 9 10RANDPS 1000 1 1 1. 0. 145RANDPS 1000 2 2 1. 0. 145RANDPS 1000 1 2 1. 0. 146RANDPS 1000 1 2 0. 1. 147TABRND1 145

.1 .1 5. 1. 10. .05 ENDT$..

ENDDATA

Example (cont):

RANDOM RESPONSE ANALYSIS (Cont.)


Example (cont):



EXAMPLE(cont.):



RANDOM ANALYSIS RECOMMENDATIONS

● Most spectra are given as a log function. Use the log features on theTABRND1 entry if PSD is given in log scale.

● Always generate the output PSD at the input location if possible.

● Plot the output PSD. Do not use the summary output blindly.

● Use several frequencies in the vicinity of each mode. For the modalmethod, a combination of FREQ1 (or FREQ2) and FREQ4 usuallyworks best.

● For low frequencies (<20 Hz), use many frequencies since thedisplacement spectra is changing rapidly for a constant inputacceleration.


WORKSHOP #10

RANDOM RESPONSE WITH SINGLE INPUT


55

33

y

x

9999Frequency (Hz) G2/Hz

20 0.130 1100 1500 0.1

1000 0.1

Autospectra of theBase Excitation

WORKSHOP #10 - RANDOM RESPONSEWITH SINGLE INPUT

● For the plate model below, apply a base motion in the z-direction using the following power spectral density,(PSD).

● Connect the left edge with an RBE2 to grid point 9999 andapply the enforced motion at grid point 9999

● Use modal solution


WORKSHOP #10 - RANDOM RESPONSEWITH SINGLE INPUT (Cont.)

● Assume a constant critical damping ratio of 3% across thewhole frequency range.

● Use a log-log input for the PSD.● Determine the acceleration PSD response at the drive

point (grid point 9999) and at the corner and center of thefree edge (grid points 33 and 55)

● Request output in both print and xyplot format


$$ wkshp10.dat$$ case control, add : plot command for psd output$$ bulk data, add : enforced motion at grid point 9999$ add forcing frequencies$ add random input$ residual vector$ID SEMINAR, PROB10SOL 111CENDTITLE= RANDOM ANALYSIS - BASE EXCITATIONSUBTITLE= USING THE MODAL METHOD WITH LANCZOSECHO= UNSORTEDSPC= 101SET 111= 33, 55, 9999ACCELERATION(SORT2, PHASE)= 111METHOD= 100FREQUENCY= 100SDAMPING= 100RANDOM= 100DLOAD= 100$


WORKSHOP # 10


$OUTPUT(XYPLOT)XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESXTITLE= FREQUENCYYTTITLE= ACCEL RESPONSE BASE, MAGNITUDEYBTITLE= ACCEL RESPONSE AT BASE, PHASEXYPLOT ACCEL RESPONSE / 9999 (T3RM, T3IP)YTTITLE= ACCEL RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= ACCEL RESPONSE AT TIP CENTER, PHASEXYPLOT ACCEL RESPONSE / 33 (T3RM, T3IP)YTTITLE= ACCEL RESPONSE AT OPPOSITE CORNER, MAGNITUDEYBTITLE= ACCEL RESPONSE AT OPPOSETE CORNER, PHASEXYPLOT ACCEL RESPONSE / 55 (T3RM, T3IP)$$ PLOT OUTPUT IS ONLY MEANS OF VIEWING PSD DATA$BEGIN BULKparam,post,0PARAM,COUPMASS,1PARAM,WTMASS,0.00259$INCLUDE 'plate.bdf'$GRID, 9999, , 0., 1., 0.$RBE2, 101, 9999, 12345, 1, 12, 23, 34, 45$SPC1, 101, 12456, 9999$$ EIGENVALUE EXTRACTION PARAMETERS$EIGRL, 100 , , 2000.$$ SPECIFY MODAL DAMPING$TABDMP1, 100, CRIT,+, 0., .03, 10., .03, ENDT$ENDDATA

WORKSHOP # 10


ID SEMINAR, PROB10SOL 111CENDTITLE= RANDOM ANALYSIS - BASE EXCITATIONSUBTITLE= USING THE MODAL METHOD WITH LANCZOSECHO= UNSORTEDSPC= 101SET 111= 33, 55, 9999ACCELERATION (rall, PHASE)= 111METHOD= 100FREQUENCY= 100SDAMPING= 100RANDOM= 100DLOAD= 100$OUTPUT(XYPLOT)XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESXTITLE= FREQUENCYYTTITLE= ACCEL RESPONSE BASE, MAGNITUDEYBTITLE= ACCEL RESPONSE AT BASE, PHASEXYPLOT ACCEL RESPONSE / 9999 (T3RM, T3IP)YTTITLE= ACCEL RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= ACCEL RESPONSE AT TIP CENTER, PHASEXYPLOT ACCEL RESPONSE / 33 (T3RM, T3IP)YTTITLE= ACCEL RESPONSE AT OPPOSITE CORNER, MAGNITUDEYBTITLE= ACCEL RESPONSE AT OPPOSETE CORNER, PHASEXYPLOT ACCEL RESPONSE / 55 (T3RM, T3IP)$$ PLOT OUTPUT IS ONLY MEANS OF VIEWING PSD DATA$XGRID= YESYGRID= YESXLOG= YESYLOG= YESYTITLE= ACCEL P S D AT LOADED CORNERXYPLOT ACCEL PSDF / 9999(T3)YTITLE= ACCEL P S D AT TIP CENTERXYPLOT ACCEL PSDF / 33(T3)YTITLE= ACCEL P S D AT OPPOSITE CORNERXYPLOT ACCEL PSDF / 55(T3)$

BEGIN BULKPARAM,COUPMASS,1

PARAM,WTMASS,0.00259$INCLUDE 'plate.bdf'

$GRID, 9999, , 0., 1., 0.$RBE2, 101, 9999, 12345, 1, 12, 23, 34, 45

$SPC1, 101, 12456, 9999$$ EIGENVALUE EXTRACTION PARAMETERS$

EIGRL, 100 , , 2000.$$ SPECIFY MODAL DAMPING

$TABDMP1, 100, CRIT,+, 0., .03, 10., .03, ENDT$

$ POINT LOADING AT TIP CENTER$RLOAD2, 100, 600, , , 310,,Aspcd,600,9999,3,1.0spc1,101,3,9999

$TABLED1, 310,+, 10., 1., 1000., 1., ENDT

$$ SPECIFY FREQUENCY STEPS$FREQ,100,30.

FREQ1,100,20.,20.,50FREQ4,100,20.,1000.,.03,5$$ SPECIFY SPECTRAL DENSITY$

RANDPS, 100, 1, 1, 1., 0., 111$TABRND1, 111,LOG,LOG

+, 20., 0.1, 30., 1., 100., 1., 500., .1,+, 1000., .1, ENDT$ENDDATA

SOLUTION FILE FOR WORKSHOP # 10


POINT-ID = 33A C C E L E R A T I O N V E C T O R( POWER SPECTRAL DENSITY FUNCTION )

FREQUENCY TYPE T1 T2 T3 R1 R2 R32.000000E+01 G 8.869005E-23 3.857560E-21 1.072273E-01 4.070925E-20 9.165858E-06 2.384971E-213.000000E+01 G 4.759690E-21 2.069922E-19 1.171296E+00 2.184254E-18 4.929607E-04 1.279754E-194.000000E+01 G 1.637099E-20 7.118072E-19 1.328467E+00 7.510322E-18 1.700663E-03 4.400866E-196.000000E+01 G 1.078228E-19 4.685393E-18 1.937051E+00 4.941785E-17 1.129788E-02 2.896886E-18

A C C E L E R A T I O N V E C T O R( ROOT MEAN SQUARE )

POINT ID. TYPE T1 T2 T3 R1 R2 R333 G 7.193928E-08 4.649358E-07 9.228086E+01 1.510409E-06 4.589051E+01 3.657763E-0755 G 4.272782E-07 4.836801E-07 9.166651E+01 1.969691E+00 4.610466E+01 3.796657E-07

9999 G 0.0 0.0 1.561982E+01 0.0 0.0 0.0

A C C E L E R A T I O N V E C T O R( NUMBER OF ZERO CROSSINGS )

POINT ID. TYPE T1 T2 T3 R1 R2 R333 G 2.561972E+02 2.110519E+02 3.924331E+02 2.173109E+02 7.319445E+02 2.122896E+0255 G 2.176310E+02 2.103150E+02 3.846914E+02 8.502588E+02 7.328254E+02 2.098791E+02

9999 G 0.0 0.0 3.991722E+02 0.0 0.0 0.0

PARTIAL OUTPUT FILE FOR WORKSHOP #10


X Y - O U T P U T S U M M A R Y ( R E S P O N S E )0 SUBCASE CURVE FRAME CURVE ID./ XMIN-FRAME/ XMAX-FRAME/ YMIN-FRAME/ X FOR YMAX-FRAME/ X FOR

ID TYPE NO. PANEL : GRID ID ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX0 1 ACCE 1 9999( 5,--) 2.000000E+01 1.020000E+03 9.999999E-01 1.356945E+02 1.000000E+00 1.296785E+02

2.000000E+01 1.020000E+03 9.999999E-01 1.356945E+02 1.000000E+00 1.296785E+020 1 ACCE 1 9999(--, 11) 2.000000E+01 1.020000E+03 0.000000E+00 2.000000E+01 0.000000E+00 2.000000E+01

2.000000E+01 1.020000E+03 0.000000E+00 2.000000E+01 0.000000E+00 2.000000E+010 1 ACCE 2 33( 5,--) 2.000000E+01 1.020000E+03 1.008510E+00 3.800000E+02 2.621251E+01 1.336891E+02

2.000000E+01 1.020000E+03 1.008510E+00 3.800000E+02 2.621251E+01 1.336891E+020 1 ACCE 2 33(--, 11) 2.000000E+01 1.020000E+03 1.044886E+01 1.020000E+03 3.599818E+02 2.000000E+01

2.000000E+01 1.020000E+03 1.044886E+01 1.020000E+03 3.599818E+02 2.000000E+010 1 ACCE 3 55( 5,--) 2.000000E+01 1.020000E+03 1.000853E+00 3.800000E+02 2.617639E+01 1.336891E+02

2.000000E+01 1.020000E+03 1.000853E+00 3.800000E+02 2.617639E+01 1.336891E+020 1 ACCE 3 55(--, 11) 2.000000E+01 1.020000E+03 1.055883E+01 1.020000E+03 3.599818E+02 2.000000E+01

2.000000E+01 1.020000E+03 1.055883E+01 1.020000E+03 3.599818E+02 2.000000E+01

0 X Y - O U T P U T S U M M A R Y ( A U T O O R P S D F )0 PLOT CURVE FRAME CURVE ID./ RMS NO. POSITIVE XMIN FOR XMAX FOR YMIN FOR X FOR YMAX FOR X FOR*TYPE TYPE NO. PANEL : GRID ID VALUE CROSSINGS ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX

0PSDF ACCE 4 9999( 5) 1.561982E+01 3.991722E+02 2.000E+01 1.020E+03 1.000E-01 9.800E+02 1.000E+00 3.000E+01



XYPLOT OUTPUT FOR WORKSHOP #10


OUTPUT FILE FOR WORKSHOP #10 (Cont.)


WORKSHOP #11

RANDOM RESPONSE WITH MULTIPLEINPUTS


Frequency (Hz) psi/Hz Frequency (Hz) lb/Hz20 0.1 20 0.530 1 30 2.5100 1 500 2.5500 0.1 1000 01000 0.1

Autospectra of Pressure Load Auto Spectra of Corner Load

Frequency (Hz) Real Part Imaginary Part20 -0.099619 0.007816100 -0.498097 0.043579500 0.070711 -0.070711

1000 0 0

Cross-Spectrum of Pressure and Corner Loads Real/Imaginary

WORKSHOP #11 - RANDOM RESPONSEWITH MULTIPLE INPUTS

Using the modal method, determine the displacement responsespectrum of the tip center point due to the input spectrum of thepressure and point loads listed below. Use the complex matrixrepresentation (SAB) for the cross spectrum.


● Request Auto psdf and CRMS displacement output atgrid points 11, 33, and 55

● Request cross spectrum displacement output betweengrid point 11 direction 3 and grid point 55 direction 3

WORKSHOP #11 - RANDOM RESPONSEWITH MULTIPLE INPUTS


$$ wkshp11.dat$$ add: random input plus corresponding output$SOL 111CENDTITLE= FREQUENCY RESPONSE WITH PRESSURE AND POINT LOADSSUBTITLE= USING THE MODAL METHOD WITH LANCZOSECHO= UNSORTEDSPC= 1SET 111= 11, 33, 55DISPLACEMENT(PLOT, PHASE)= 111ACCELERATION(PLOT,PHASE) = 111METHOD= 100FREQUENCY= 100SDAMPING= 100DISP(PSDF,CRMS,PHASE)=111SUBCASE 1LABEL= PRESSURE LOADDLOAD= 100SUBCASE 2LABEL = CORNER LOADDLOAD= 200$


WORKSHOP # 11


OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NOXTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASEXYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, PHASEXYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$$ PLOT OUTPUT IS ONLY MEANS OF VIEWING PSD DATA$XGRID= YESYGRID= YESXLOG= YESYLOG= YESYTITLE= DISP P S D AT LOADED CORNERXYPLOT DISP PSDF / 11(T3)YTITLE= DISP P S D AT TIP CENTERXYPLOT DISP PSDF / 33(T3)YTITLE= DISP P S D AT OPPOSITE CORNERXYPLOT DISP PSDF / 55(T3)$BEGIN BULK

PARAM,COUPMASS,1PARAM,WTMASS,0.00259$$ MODEL DESCRIBED IN NORMAL MODES EXAMPLE$INCLUDE 'plate.bdf'$$ EIGENVALUE EXTRACTION PARAMETERS$EIGRL, 100, 10., 2000.$$ SPECIFY MODAL DAMPING$TABDMP1, 100, CRIT,+, 0., .03, 10., .03, ENDT$$ FIRST LOADING$RLOAD2, 100, 400, , , 310$TABLED1, 310,+, 10., 1., 1000., 1., ENDT$$ UNIT PRESSURE LOAD TO PLATE$PLOAD2, 400, 1., 1, THRU, 40$$ SECOND LOADING$RLOAD2, 200, 600, , , 310$$ POINT LOAD AT TIP CENTER$FORCE,600,11,,1.,0.,0.,1.$$ SPECIFY FREQUENCY STEPS$FREQ1, 100, 20., 20., 49$ENDDATA

WORKSHOP # 11


$$ soln11.dat$ID SEMINAR, PROB11SOL 111CENDTITLE= FREQUENCY RESPONSE WITH PRESSURE AND POINT LOADSSUBTITLE= USING THE MODAL METHOD WITH LANCZOSECHO= UNSORTEDSPC= 1SET 111= 11, 33, 55DISPLACEMENT(PLOT, PHASE)= 111ACCELERATION(PLOT,PHASE) = 111METHOD= 100FREQUENCY= 100SDAMPING= 100RANDOM= 100DISP(PSDF,CRMS,PHASE)=111RCROSS(PSDF,PHASE)=1000SUBCASE 1LABEL= PRESSURE LOADDLOAD= 100SUBCASE 2LABEL = CORNER LOADDLOAD= 200$

$OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NOXTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASEXYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, PHASEXYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$$ PLOT OUTPUT IS ONLY MEANS OF VIEWING PSD DATA$XGRID= YESYGRID= YESXLOG= YESYLOG= YESYTITLE= DISP P S D AT LOADED CORNERXYPLOT DISP PSDF / 11(T3)YTITLE= DISP P S D AT TIP CENTERXYPLOT DISP PSDF / 33(T3)YTITLE= DISP P S D AT OPPOSITE CORNERXYPLOT DISP PSDF / 55(T3)

SOLUTION FILE FOR WORKSHOP11


$BEGIN BULKPARAM,COUPMASS,1PARAM,WTMASS,0.00259$$ MODEL DESCRIBED IN NORMAL MODES EXAMPLE$INCLUDE 'plate.bdf'$$ EIGENVALUE EXTRACTION PARAMETERS$EIGRL, 100, 10., 2000.$$ SPECIFY MODAL DAMPING$TABDMP1, 100, CRIT,+, 0., .03, 10., .03, ENDT$$ FIRST LOADING$RLOAD2, 100, 400, , , 310$TABLED1, 310,+, 10., 1., 1000., 1., ENDT$$ UNIT PRESSURE LOAD TO PLATE$PLOAD2, 400, 1., 1, THRU, 40$$ SECOND LOADING$RLOAD2, 200, 600, , , 310$$ POINT LOAD AT TIP CENTER$FORCE,600,11,,1.,0.,0.,1.

$$ SPECIFY FREQUENCY STEPS$FREQ1, 100, 20., 20., 49$RCROSS,1000,DISP,11,3,DISP,55,3$$ SPECIFY SPECTRAL DENSITY$RANDPS, 100, 1, 1, 1., 0., 100RANDPS, 100, 2, 2, 1., 0., 200RANDPS, 100, 1, 2, 1., 0., 300RANDPS, 100, 1, 2, 0., 1., 400$TABRND1, 100,+, 20., 0.1, 30., 1., 100., 1., 500., .1,+, 1000., .1, ENDT$TABRND1, 200,+, 20., 0.5, 30., 2.5, 500., 2.5, 1000., 0.,+, ENDT$TABRND1, 300,+, 20., -.099619, 100., -.498097, 500., .070711, 1000., 0.,+, ENDT$TABRND1, 400,+, 20., .0078158, 100., .0435791, 500., -.70711, 1000., 0.,+, ENDT$ENDDATA

SOLUTION FOR WORKSHOP11 (Cont.)


X Y - O U T P U T S U M M A R Y ( R E S P O N S E )0 SUBCASE CURVE FRAME CURVE ID./ XMIN-FRAME/ XMAX-FRAME/ YMIN-FRAME/ X FOR YMAX-FRAME/ X FOR













2.000000E+01 1.000000E+03 7.973937E+00 7.599999E+02 3.594539E+02 2.000000E+01

PARTIAL OUTPUT FOR WORKSHOP11


POINT-ID = 11D I S P L A C E M E N T V E C T O R( POWER SPECTRAL DENSITY FUNCTION )

FREQUENCY TYPE T1 T2 T3 R1 R2 R32.000000E+01 G 2.047303E-24 4.813135E-24 7.947328E-05 2.155795E-07 6.016012E-06 3.429885E-244.000000E+01 G 3.063225E-23 7.202310E-23 1.189875E-03 1.059814E-06 8.969508E-05 5.132451E-236.000000E+01 G 3.794013E-23 8.919452E-23 1.465137E-03 1.070977E-06 1.104442E-04 6.356191E-238.000000E+01 G 5.574504E-23 1.310384E-22 2.142037E-03 1.086198E-06 1.618793E-04 9.338235E-23

D I S P L A C E M E N T V E C T O R( ROOT MEAN SQUARE )

POINT ID. TYPE T1 T2 T3 R1 R2 R311 G 2.217665E-10 3.399741E-10 1.376854E+00 8.629130E-02 3.851211E-01 2.870088E-1033 G 4.747619E-11 3.394890E-10 1.377021E+00 8.664112E-02 3.844410E-01 2.681251E-1055 G 3.086634E-10 3.532137E-10 1.378706E+00 8.410328E-02 3.849473E-01 2.787290E-10

1 MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 470 RANDOM 100

D I S P L A C E M E N T V E C T O R( NUMBER OF ZERO CROSSINGS )

POINT ID. TYPE T1 T2 T3 R1 R2 R311 G 1.364916E+02 1.362757E+02 1.401700E+02 6.491689E+02 1.503498E+02 1.362929E+0233 G 1.355889E+02 1.362910E+02 1.360215E+02 6.552164E+02 1.492609E+02 1.363054E+0255 G 1.361721E+02 1.362927E+02 1.432664E+02 6.581000E+02 1.498830E+02 1.362823E+02

1 MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 480 RANDOM 100

POINT-ID = 11D I S P L A C E M E N T V E C T O R( CUMULATIVE ROOT MEAN SQUARE )

FREQUENCY TYPE T1 T2 T3 R1 R2 R32.000000E+01 G 0.0 0.0 0.0 0.0 0.0 0.04.000000E+01 G 1.807749E-11 2.771935E-11 1.126653E-01 3.571265E-03 3.093721E-02 2.339966E-116.000000E+01 G 3.182011E-11 4.879076E-11 1.981000E-01 5.836253E-03 5.439213E-02 4.118748E-118.000000E+01 G 4.415168E-11 6.769688E-11 2.744364E-01 7.458793E-03 7.537731E-02 5.714762E-111.000000E+02 G 6.018009E-11 9.226939E-11 3.734311E-01 8.809525E-03 1.027079E-01 7.789153E-111.200000E+02 G 9.969239E-11 1.528460E-10 6.180981E-01 1.001494E-02 1.708699E-01 1.290305E-10



1 FREQUENCY RESPONSE WITH PRESSURE AND POINT LOADS MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 52USING THE MODAL METHOD WITH LANCZOS

0 PRESSURE LOAD RANDOM 100SEQUENTIAL CURVE-ID = 1

C O M P L E X C R O S S - P O W E R S P E C T R A L D E N S I T Y F U N C T I O N(MAGNITUDE/PHASE)

0 RCROSS RTYPE1 ID1 COMP1 RTYPE2 ID2 COMP2 CURID0 1000 DISP 11 3 DISP 55 3 0

FREQUENCY CPSDF FREQUENCY CPSDF2.000000E+01 7.782711E-05 / 359.7912 4.000000E+01 1.169207E-03 / 359.98166.000000E+01 1.446440E-03 / 359.9715 8.000000E+01 2.125043E-03 / 359.96511.000000E+02 4.256192E-03 / 359.9659 1.200000E+02 1.995366E-02 / 0.02061.400000E+02 5.931299E-02 / 0.0270 1.600000E+02 3.981690E-03 / 359.96411.800000E+02 1.110766E-03 / 359.8067 2.000000E+02 4.651507E-04 / 359.52292.200000E+02 2.361836E-04 / 359.0721 2.400000E+02 1.342108E-04 / 358.40122.600000E+02 8.205519E-05 / 357.4377 2.800000E+02 5.276163E-05 / 356.08123.000000E+02 3.515652E-05 / 354.1868 3.200000E+02 2.402327E-05 / 351.53853.400000E+02 1.670592E-05 / 347.8049 3.600000E+02 1.176429E-05 / 342.46553.800000E+02 8.387207E-06 / 334.7123 4.000000E+02 6.119115E-06 / 323.42144.200000E+02 4.722277E-06 / 307.6403 4.400000E+02 4.072001E-06 / 288.29074.600000E+02 4.034620E-06 / 269.2190 4.800000E+02 4.427422E-06 / 253.87995.000000E+02 5.106796E-06 / 242.7013 5.200000E+02 5.390143E-06 / 236.15715.400000E+02 5.879032E-06 / 229.9111 5.600000E+02 6.667265E-06 / 223.85395.800000E+02 7.941911E-06 / 217.8872 6.000000E+02 1.009844E-05 / 211.92926.200000E+02 1.407105E-05 / 205.9172 6.400000E+02 2.244283E-05 / 199.81126.600000E+02 4.367885E-05 / 193.5947 6.799999E+02 9.592211E-05 / 187.27146.999999E+02 7.964308E-05 / 180.8491 7.200000E+02 2.743963E-05 / 174.29287.400000E+02 1.057217E-05 / 167.3723 7.599999E+02 4.433168E-06 / 158.94037.800000E+02 1.351509E-06 / 138.3460 8.000000E+02 1.558493E-06 / 1.16148.200000E+02 5.536476E-06 / 343.4609 8.399999E+02 1.063914E-05 / 337.73128.600000E+02 6.950112E-06 / 334.6017 8.800000E+02 2.748527E-06 / 332.94028.999999E+02 1.156800E-06 / 332.4832 9.199999E+02 5.375278E-07 / 333.20769.400000E+02 2.648299E-07 / 335.2509 9.600000E+02 1.325349E-07 / 339.02119.799999E+02 6.395399E-08 / 345.7172 1.000000E+03 2.718711E-08 / 359.9906



X Y - O U T P U T S U M M A R Y ( A U T O O R P S D F )0 PLOT CURVE FRAME CURVE ID./ RMS NO. POSITIVE XMIN FOR XMAX FOR YMIN FOR X FOR YMAX FOR X FOR*

TYPE TYPE NO. PANEL : GRID ID VALUE CROSSINGS ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX0

PSDF DISP 7 11( 5) 1.376854E+00 1.401700E+02 2.000E+01 1.000E+03 2.720E-08 1.000E+03 5.926E-02 1.400E+020





PARTIAL OUTPUT FOR WORKSHOP #11(Cont.)


Partial Output For Workshop #11 (Cont.)


SECTION 15

COMPLEX EIGENVALUE ANALYSIS


COMPLEX EIGENVALUE ANALYSIS

● Used to assess the stability of systems modeled withtransfer functions (including servomechanisms androtating systems)

● Also used to compute damped modes of systems

● Mass and stiffness matrices may be unsymmetric and maycontain complex numbers.

● See the MSC.Nastran Advanced Dynamics User’s Guide.


COMPLEX EIGENSOLUTIONS - THEORY

Mp2

Bp K+ + u 0 =

g /

Imaginary

Real

● Equation of motion

● where p = + i● and = real part of solution● = imaginary part of solution● For stable systems, a < 0

● Damping coefficient


COMPLEX EIGENSOLUTIONS IN MD Nastran

● B matrix is the same as that used in frequency response analysis.● The direct method solves the equation with M,B,K matrices of D-size

(physical variables plus extra points).● Modal solves the equation with M,B,K matrices of H-size (modal

coordinates plus extra points). Undamped modes are first computedto transform matrices from D-size to H-size.

● Four methods of complex eigenanalysis: HESS, INV, DET, and CLAN● HESS (upper Hessenberg) is related to GIV. This method requires

nonsingular M and may be quite expensive for large problems.Therefore, the modal method is usually recommended except for smallproblems.

● HESS: solves canonical equation forms. There are two categories:● Systems with [B] = 0● Systems with [B] 0


● [B] = 0 case solves

● where[B] 0 case studies:

● CLAN - Similar to real Lanczos - hybrid of a transformation and a tracking method.● INV is related to INV for undamped eigenvalue analysis. The user must select the search

region in the complex plane. This method may be used for larger problems and a singular M isallowed. However, the method is expensive and less reliable than HESS.

● DET is not recommended because it is awkward and inefficient.● These methods with search regions are specified on the EIGC Bulk Data entry, which is

selected by the CMETHOD Case Control command.● Mode acceleration is available and is selected by PARAM,MODACC,0 and PARAM,DDRMM,-

1. Mode acceleration has no effect on calculated eigenvalues, it only affects the data recovery.

COMPLEX EIGENSOLUTIONS IN MD Nastran(Cont.)

A I– 0=

[A] = -[M-1K], = p2 A p I– 0=

uv

=where

A0 I

M 1– K– M 1– B–=


INPUT REQUIRED FOR COMPLEXEIGENANALYSIS

StructuredSolution

Direct 107Modal 110

● Executive Control SectionSOL (for required input see below)

● Case Control SectionCMETHOD (required for both)METHOD (required for modal)

● Bulk Data SectionEIGC (required for both)EIGR or EIGRL (required for modal)


WORKSHOP #12

COMPLEX EIGENANALYSISCompute the complex modes of the following pile driver

Ground

Pile

Exciter


WORKSHOP #12 - COMPLEX EIGENVALUE

1

2

3

C2K2

K1

m2

m1

m1 3.0 lb-sec2/in

m2 1.5 lb-sec2/in

K1 50,000 lb/inK2 12,500 lb/inC2 30 lb-sec/in


$$ wkshp12.dat$$ case control, add : direct complex eigenvalue callout$$ bulk data, add : Hess method$ ask for eigenvalue and eigenvector output$CENDTITLE= TWO-DOF MODEL (IMAC 8, PG 891)SUBTITLE= COMPLEX MODESSPC= 100$BEGIN BULK$$ COMPLEX EIGENVALUE EXTRACTION PARAMETERS$$$ DEFINE GRIDS, MASSES, AND STIFFNESSES$ GRID 1 = EXCITER (X=2, MASS=3) 50K STIFFNESS BETWEEN GRIDS 1 AND 2$ GRID 2 = PILE (X=1, MASS=3) 12.5K STIFFNESS BETWEEN GRIDS 2 AND 3$ GRID 3 = BASE (X=0, FIX BASE)$GRID, 1, , 2., 0., 0.GRID, 2, , 1., 0., 0.GRID, 3, , 0., 0., 0.GRDSET, , , , , , , 23456CELAS2, 1, 50000., 1, 1, 2, 1CELAS2, 2, 12500., 2, 1, 3, 1CONM2, 201, 1, , 3.0CONM2, 202, 2, , 1.5SPC, 100, 3, 1$$ DEFINE DAMPER OF 30 BETWEEN GRIDS 2 AND 3$CVISC, 101, 1, 2, 3PVISC, 1, 30.$ENDDATA

• Use the following partial input file as a starting pointWORKSHOP # 12


SOLUTION FOR WORKSHOP #12SOL 107CENDTITLE= TWO-DOF MODEL (IMAC 8, PG 891)SUBTITLE= COMPLEX MODESDISPLACEMENT= ALL $ DEFAULT= REAL, IMAGINARYSPC= 100CMETHOD= 99$BEGIN BULK$$ COMPLEX EIGENVALUE EXTRACTION PARAMETERS$EIGC, 99, HESS, , , , , 4$$ DEFINE GRIDS, MASSES, AND STIFFNESSES$ GRID 1 = EXCITER (X=2, MASS=3) 50K STIFFNESS BETWEEN GRIDS 1 AND 2$ GRID 2 = PILE (X=1, MASS=3) 12.5K STIFFNESS BETWEEN GRIDS 2 AND 3$ GRID 3 = BASE (X=0, FIX BASE)$GRID, 1, , 2., 0., 0.GRID, 2, , 1., 0., 0.GRID, 3, , 0., 0., 0.GRDSET, , , , , , , 23456CELAS2, 1, 50000., 1, 1, 2, 1CELAS2, 2, 12500., 2, 1, 3, 1CONM2, 201, 1, , 3.0CONM2, 202, 2, , 1.5SPC, 100, 3, 1$$ DEFINE DAMPER OF 30 BETWEEN GRIDS 2 AND 3$CVISC, 101, 1, 2, 3PVISC, 1, 30.$ENDDATA


PARTIAL OUTPUT FILE FOR WORKSHOP #121 TWO-DOF MODEL (IMAC 8, PG 891) MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 9

COMPLEX MODES0

C O M P L E X E I G E N V A L U E S U M M A R Y0 ROOT EXTRACTION EIGENVALUE FREQUENCY DAMPING

NO. ORDER (REAL) (IMAG) (CYCLES) COEFFICIENT1 2 -2.660969E+00 -4.983521E+01 7.931520E+00 1.067907E-012 1 -2.660969E+00 4.983521E+01 7.931520E+00 1.067907E-013 4 -7.339031E+00 -2.360312E+02 3.756553E+01 6.218695E-024 3 -7.339031E+00 2.360312E+02 3.756553E+01 6.218695E-02

1 TWO-DOF MODEL (IMAC 8, PG 891) MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 10COMPLEX MODES

01 TWO-DOF MODEL (IMAC 8, PG 891) MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 11

COMPLEX MODES0

COMPLEX EIGENVALUE = -2.660969E+00, -4.983521E+01C O M P L E X E I G E N V E C T O R NO. 1

(REAL/IMAGINARY)

POINT ID. TYPE T1 T2 T3 R1 R2 R30 1 G 1.000000E+00 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.00 2 G 8.514119E-01 0.0 0.0 0.0 0.0 0.0

1.591320E-02 0.0 0.0 0.0 0.0 0.00 3 G 0.0 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.01 TWO-DOF MODEL (IMAC 8, PG 891) MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 12

COMPLEX MODES0

COMPLEX EIGENVALUE = -2.660969E+00, 4.983521E+01C O M P L E X E I G E N V E C T O R NO. 2

(REAL/IMAGINARY)

POINT ID. TYPE T1 T2 T3 R1 R2 R30 1 G 1.000000E+00 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.00 2 G 8.514119E-01 0.0 0.0 0.0 0.0 0.0

-1.591320E-02 0.0 0.0 0.0 0.0 0.00 3 G 0.0 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.0


1 TWO-DOF MODEL (IMAC 8, PG 891) MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE13

COMPLEX MODES0

COMPLEX EIGENVALUE = -7.339031E+00, -2.360312E+02C O M P L E X E I G E N V E C T O R NO. 3

(REAL/IMAGINARY)

POINT ID. TYPE T1 T2 T3 R1 R2 R30 1 G -4.241094E-01 0.0 0.0 0.0 0.0 0.0

-3.768431E-02 0.0 0.0 0.0 0.0 0.00 2 G 1.000000E+00 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.00 3 G 0.0 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.01 TWO-DOF MODEL (IMAC 8, PG 891) MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE14

COMPLEX MODES0

COMPLEX EIGENVALUE = -7.339031E+00, 2.360312E+02C O M P L E X E I G E N V E C T O R NO. 4

(REAL/IMAGINARY)

POINT ID. TYPE T1 T2 T3 R1 R2 R30 1 G -4.241094E-01 0.0 0.0 0.0 0.0 0.0

3.768431E-02 0.0 0.0 0.0 0.0 0.00 2 G 1.000000E+00 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.00 3 G 0.0 0.0 0.0 0.0 0.0 0.0

0.0 0.0 0.0 0.0 0.0 0.0



Section 16

Normal Mode Analysis Using PartsSuperelement


WHAT IS A SUPERELEMENT?

● Physical and mathematical representation● Physical - substructure: a finite element model of a portion of a

structure● Mathematical - boundary matrices: loads, mass, damping, and

stiffness reduced from the interior points to the exterior orboundary points

● Other types of substructuring analysis● Cyclic symmetry analysis● GENEL and DMIG input

● There are two ways to define superelements● PARTS superelement (discussed in this section)● Main Bulk Data superelement (See Superelement seminar)


HOW ARE PARTS SUPERELEMENT DEFINEDIN MD NASTRAN

● The description for each superelement is completely self-containedwithin its own Bulk Data Section.

● This type of superelement is often referred to as parts in this section.● Each superelement begins with a

BEGIN [BULK] SUPER=mdelimiter and terminates with another

BEGIN [BULK] SUPER=nor ENDATA delimiter

● The Main Bulk Data Section contains the full description of theresidual structure and other superelements not described by parts.These superelements can be defined by either the SESET or SEELTentries. See Superelement Seminar for further description of SESET.


HOW ARE PARTS SUPERELEMENT DEFINEDIN MD NASTRAN (Cont.)

● By default, if the location of a grid point in one part fallswithin a specified tolerance of a grid point in another part,then these parts will be connected together at thatlocation.

● You can manually overwrite the connection tolerance.● Duplicate IDs (e.g., grid points, elements, properties, etc.)

are allowed across different parts.● Each part can have its own loadings and component

modes calculation.● Plotting of complete system modes is supported.● A superelement may also be defined as a copy of a

superelement or obtained from outside MD Nastran.


SAMPLE PROBLEM - STEEL STAMPING

33 34 35 36 37

38 39 40 41 42

28 29 30 31 32

23 24 25 26 27

18 19 20 21 22

94 95 96 97

86

74

62

81

69

57

9893

33 38

58 59 60 61 62

63 64 65 66 67

53 54 55 56 57

48 49 50 51 52

43 44 45 46 47

100 101 102 103

92

80

68

87

75

63

10499

39 4434 35 36 37 40 41 42 4315

14

17

16

7 8 9 5 10 11 12

4

3

2

21 22 23 24 25 26

30

20 27

32

28

18

29

19

4” 4”

4”

1.6”

1.6” 1.6”.8”

.8”

8

6

4

7

5

3

.8”

9 10 11 12

3.2” 3.2”

3.6”

15 16 171413

31

11

66 1313


SAMPLE PROBLEM - STEEL STAMPING(Cont.)

● Grid Points 1 and 2 fixed● Material properties:

Steel t = .05”E = 29 x 106 psin = .3r = .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


MODEL DEFINITION FOR SAMPLE PROBLEM$$ file - fs1.dat$$ all 7 s.e. brought in using begin super$ duplicate boundary grids id$ each s.e. contains its own property description withthe same id$ condensed subcase setup$$ -------------------------------------------------------$id allsep1 datSOL 101TIME 15CENDTITLE = S.E. SAMPLE PROBLEM 1SUBTITLE = S.E. STATICS - RUN 1 - MULTIPLE LOADSDISP = ALLstress = alloload = allSET 999 = 0,1,2,3,4,5,6,7SUPER = 999 $ ALL CASE CONTROL IS FOR ALL SUPERELEMENTSPARAM,GRDPNT,1SUBCASE 101LABEL = PRESSURE LOADLOAD = 101$SUBCASE 201LABEL = 2# NORMAL LOADSLOAD = 201$SUBCASE 301LABEL = OPPOSING LOADSLOAD = 301$

$include ’plot.blk’$BEGIN BULK$CQUAD4 5 1 13 14 24 23$GRDSET 6GRID 13 -.4 3.6 0.GRID 14 .4 3.6 0.GRID 23 -.4 4.4 0.GRID 24 .4 4.4 0.$include ’prop1.blk’$begin super=1$include ’loadse1.blk’include ’prop1.blk’include ’se1.blk’


MODEL DEFINITION FOR SAMPLE PROBLEM(Cont.)$

begin super=2$include ’loadse2.blk’include ’prop1.blk’include ’se2.blk’$begin super=3$include ’prop1.blk’include ’se3.blk’$begin super=4$include ’prop1.blk’include ’se4.blk’$begin super=5$include ’prop1.blk’include ’se5.blk’$begin super=6$include ’prop1.blk’include ’se6.blk’$begin super=7$include ’prop1.blk’include ’se7.blk’$enddata


MODEL DEFINITION FOR SAMPLE PROBLEM(Cont.)

seplot 5ptitle = superelement 5find scale, origin 1, set 1plot static deformation set 1 origin 1 label both$seplot 6ptitle = superelement 6find scale, origin 1, set 1plot static deformation set 1 origin 1 label both$seplot 7ptitle = superelement 7find scale, origin 1, set 1plot static deformation set 1 origin 1 label both

$seplot 0ptitle = superelement 0find scale, origin 1, set 1plot static deformation set 1 origin 1 label both$seupplot 0ptitle = full structureaxes x,mz,yfind scale, origin 1, set 1plot static deformation set 1 origin 1$

$$ plot.blk$output(plot)$set 1 = allaxes z,x,yview 0.,0.,0.seupplot 0ptitle = full structurefind scale, origin 1, set 1plot set 1 origin 1 label both$$ deform plots$seplot 1ptitle = superelement 1find scale, origin 1, set 1plot static deformation set 1 origin 1 label both$seplot 2ptitle = superelement 2find scale, origin 1, set 1plot static deformation set 1 origin 1 label both$seplot 3ptitle = superelement 3find scale, origin 1, set 1plot static deformation set 1 origin 1 label both$seplot 4ptitle = superelement 4find scale, origin 1, set 1plot static deformation set 1 origin 1 label both$



GRID 33 -5.2 6. 0.GRID 34 -4.4 6. 0.GRID 37 -2. 6. 0.GRID 38 -1.2 6. 0.GRID 45 -5.2 6.8 0.GRID 46 -4.4 6.8 0.GRID 47 -3.6 6.8 0.GRID 48 -2.8 6.8 0.GRID 49 -2. 6.8 0.GRID 50 -1.2 6.8 0.GRID 57 -5.2 7.6 0.GRID 58 -4.4 7.6 0.GRID 59 -3.6 7.6 0.GRID 60 -2.8 7.6 0.GRID 61 -2. 7.6 0.GRID 62 -1.2 7.6 0.GRID 69 -5.2 8.4 0.

GRID 70 -4.4 8.4 0.GRID 71 -3.6 8.4 0.GRID 72 -2.8 8.4 0.GRID 73 -2. 8.4 0.GRID 74 -1.2 8.4 0.GRID 81 -5.2 9.2 0.GRID 82 -4.4 9.2 0.GRID 83 -3.6 9.2 0.GRID 84 -2.8 9.2 0.GRID 85 -2. 9.2 0.GRID 86 -1.2 9.2 0.GRID 93 -5.2 10. 0.GRID 94 -4.4 10. 0.GRID 95 -3.6 10. 0.GRID 96 -2.8 10. 0.GRID 97 -2. 10. 0.GRID 98 -1.2 10. 0.

$$ se1.blk$CQUAD4 18 1 33 34 46 45CQUAD4 19 1 34 35 47 46CQUAD4 20 1 35 36 48 47CQUAD4 21 1 36 37 49 48CQUAD4 22 1 37 38 50 49CQUAD4 23 1 45 46 58 57CQUAD4 24 1 46 47 59 58CQUAD4 25 1 47 48 60 59CQUAD4 26 1 48 49 61 60CQUAD4 27 1 49 50 62 61CQUAD4 28 1 57 58 70 69CQUAD4 29 1 58 59 71 70CQUAD4 30 1 59 60 72 71CQUAD4 31 1 60 61 73 72CQUAD4 32 1 61 62 74 73CQUAD4 33 1 69 70 82 81CQUAD4 34 1 70 71 83 82CQUAD4 35 1 71 72 84 83CQUAD4 36 1 72 73 85 84CQUAD4 37 1 73 74 86 85CQUAD4 38 1 81 82 94 93CQUAD4 39 1 82 83 95 94CQUAD4 40 1 83 84 96 95CQUAD4 41 1 84 85 97 96CQUAD4 42 1 85 86 98 97$GRDSET 6$6$ boundary grids$GRID 35 -3.6 6. 0.GRID 36 -2.8 6. 0.$



$$ se2.blk$CQUAD4 43 1 39 40 52 51CQUAD4 44 1 40 41 53 52CQUAD4 45 1 41 42 54 53CQUAD4 46 1 42 43 55 54CQUAD4 47 1 43 44 56 55CQUAD4 48 1 51 52 64 63CQUAD4 49 1 52 53 65 64CQUAD4 50 1 53 54 66 65CQUAD4 51 1 54 55 67 66CQUAD4 52 1 55 56 68 67CQUAD4 53 1 63 64 76 75CQUAD4 54 1 64 65 77 76CQUAD4 55 1 65 66 78 77CQUAD4 56 1 66 67 79 78CQUAD4 57 1 67 68 80 79CQUAD4 58 1 75 76 88 87CQUAD4 59 1 76 77 89 88CQUAD4 60 1 77 78 90 89CQUAD4 61 1 78 79 91 90CQUAD4 62 1 79 80 92 91CQUAD4 63 1 87 88 100 99CQUAD4 64 1 88 89 101 100CQUAD4 65 1 89 90 102 101CQUAD4 66 1 90 91 103 102CQUAD4 67 1 91 92 104 103$GRDSET 6$$ boundary grids$GRID 41 2.8 6. 0.GRID 42 3.6 6. 0.$

$GRID 39 1.2 6. 0.GRID 40 2. 6. 0.GRID 43 4.4 6. 0.GRID 44 5.2 6. 0.$GRID 51 1.2 6.8 0.GRID 52 2. 6.8 0.GRID 53 2.8 6.8 0.GRID 54 3.6 6.8 0.GRID 55 4.4 6.8 0.GRID 56 5.2 6.8 0.GRID 63 1.2 7.6 0.GRID 64 2. 7.6 0.GRID 65 2.8 7.6 0.GRID 66 3.6 7.6 0.GRID 67 4.4 7.6 0.GRID 68 5.2 7.6 0.GRID 75 1.2 8.4 0.

GRID 76 2. 8.4 0.GRID 77 2.8 8.4 0.GRID 78 3.6 8.4 0.GRID 79 4.4 8.4 0.GRID 80 5.2 8.4 0.GRID 87 1.2 9.2 0.GRID 88 2. 9.2 0.GRID 89 2.8 9.2 0.GRID 90 3.6 9.2 0.GRID 91 4.4 9.2 0.GRID 92 5.2 9.2 0.GRID 99 1.2 10. 0.GRID 100 2. 10. 0.GRID 101 2.8 10. 0.GRID 102 3.6 10. 0.GRID 103 4.4 10. 0.GRID 104 5.2 10. 0.$



$$ se3.blk$CQUAD4 14 1 19 20 30 29CQUAD4 15 1 29 30 36 35$GRDSET 6$$ boundary grids$GRID 19 -3.6 4.4 0.GRID 20 -2.8 4.4 0.GRID 35 -3.6 6. 0.GRID 36 -2.8 6. 0.$GRID 29 -3.6 5.2 0.GRID 30 -2.8 5.2 0.$



$$ se4.blk$CQUAD4 16 1 27 2832 31CQUAD4 17 1 31 3242 41$GRDSET6$$ boundary grids$GRID 27 2.8 4.40.GRID 28 3.6 4.40.GRID 41 2.8 6.0.GRID 42 3.6 6.0.$GRID 31 2.8 5.20.GRID 32 3.6 5.20.$

$$ se5.blk$CQUAD4 6 1 9 10 20 19CQUAD4 7 1 10 11 21 20CQUAD4 8 1 11 12 22 21CQUAD4 9 1 12 13 23 22$GRDSET 6$$ boundary grids$GRID 19 -3.6 4.4 0.GRID 20 -2.8 4.4 0.GRID 13 -.4 3.6 0.GRID 23 -.4 4.4 0.$GRID 9 -3.6 3.6 0.GRID 10 -2.8 3.6 0.GRID 11 -2. 3.6 0.GRID 12 -1.2 3.6 0.GRID 21 -2. 4.4 0.GRID 22 -1.2 4.4 0.$



$$ se6.blk$CQUAD4 10 1 14 15 25 24CQUAD4 11 1 15 16 26 25CQUAD4 12 1 16 17 27 26CQUAD4 13 1 17 18 28 27$GRDSET6$$ boundary grids$GRID 27 2.8 4.4 0.GRID 28 3.6 4.4 0.GRID 14 .4 3.6 0.GRID 24 .4 4.4 0.$GRID 15 1.2 3.6 0.GRID 16 2. 3.6 0.GRID 17 2.8 3.6 0.GRID 18 3.6 3.6 0.$GRID 25 1.2 4.4 0.GRID 26 2. 4.4 0.$

$$ se7.blk$CQUAD4 1 1 1 2 4 3CQUAD4 2 1 3 4 6 5CQUAD4 3 1 5 6 8 7CQUAD4 4 1 7 8 14 13$GRDSET6$GRID 1 -.4 0. 0.123456GRID 2 .4 0. 0.123456GRID 3 -.4 0.9 0.GRID 4 .4 0.9 0.GRID 5 -.4 1.8 0.GRID 6 .4 1.8 0.GRID 7 -.4 2.7 0.GRID 8 .4 2.7 0.$$ boundary grids$GRID 13 -.4 3.6 0.GRID 14 .4 3.6 0.$



$$ prop1.blk$MAT1,1,30.+6,,.3,.283PARAM,WTMASS,.00259PARAM,AUTOSPC,YESPSHELL,1,1,.05,1,,1$

$$ file - loadse1.blk$ loads on s.e. 1$$ 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 ATCORNERS$FORCE,301,93,,2.,0.,0.,1.$

$$ file - loadse2.blk$ loads on s.e. 2$$ LOAD CASE 1 - PRESSURE LOAD$PLOAD2,101,-1.,43,THRU,67$$ LOAD CASE 2 - 2 POINT LOADS ATCORNERS$FORCE,201,104,,2.,0.,0.,1.$$ LOAD CASE 3 - OPPOSING POINT LOADSAT CORNERS$FORCE,301,104,,2.,0.,0.,-1.$


SAMPLE PROBLEM - STEEL STAMPINGSAMPLE SUPERELEMENT 1

33 34 35 36 37

38 39 40 41 42

28 29 30 31 32

23 24 25 26 27

18 19 20 21 22

94 95 96 97

86

74

62

81

69

57

9893

33 38

58 59 60 61 62

63 64 65 66 67

53 54 55 56 57

48 49 50 51 52

43 44 45 46 47

100 101 102 103

92

80

68

87

75

63

10499

39 4434 35 36 37 40 41 42 4315

14

17

16

7 8 9 5 10 11 12

4

3

2

21 22 23 24 25 26

30

20 27

32

28

18

29

19

4” 4”

4”

1.6”

1.6” 1.6”.8”

.8”

8

6

4

7

5

3

.8”

9 10 11 12

3.2” 3.2”

3.6”

15 16 171413

Part 1

31


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 Superelement1, Grid Points 35 and 36)

● Interior DOFs: Designates the DOFs that are reduced out duringsuperelement processing and are omitted in subsequentprocessing (for Superelement 1 of the sample problem, GridPoints 33, 34, 37, 38, 45-50, 57-62, 69-74, 81-86, 93-98).


PARTITIONED SOLUTIONS (Cont.)

● For each superelement, produce a description in matrix terms of itsbehavior 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, combine and assemble the boundarymatrices.● The BULK DATA for the RESIDUAL structure consists of all ‘residual’

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 CONDENSATION

After generating matrices and applying MPCs and SPCs,

Kff Uf = Pf

O-Set = interior points (to be condensed out bythe reduction)

A-Set = exterior (or boundary) points (which areretained for further analysis)

PartitionKoo Koa

KoaT Kaa

Uo

Ua

Po

Pa

=


THEORY OF STATIC CONDENSATION

● Extract upper equation and premultiply by

Let (Boundary Transformation)

and (Fixed Boundary Displacements)

then (Total Interior Displacements)aoaooo UGUU

o1

oooo PkU

oa1

oooa kkG

o1

ooaoaooo1

oo PkUkUkk

1ook


● Substitute expression for Uo in the lower equation

then (Boundary Stiffness)and (Boundary Loads)

● Solve for residual structure

Ua = Kaa-1 Pa (Boundary Displacments)

KoaT GoaUa Uo

o+ KaaUa+ Pa=

Kaa KoaT Goa Kaa+=

Pa GoaT Po Pa+=

THEORY OF STATIC CONDENSATION (Cont.)


ADVANTAGES OF SUPERELEMENTANALYSIS

● Large problems (i.e., allows solving problems that exceedyour hardware capabilities)

● Less CPU or wall clock time per run (reduced risk sinceeach superelement may be processed individually)

● Partial redesign requires only partial solution (cost).● Allows more control of resource usage● Partitioned input desirable

● Organization● Repeated components


ADVANTAGES OF SUPERELEMENTANALYSIS (Cont.)

● Partitioned output desirable● Organization● Comprehension

● Components may be modeled by subcontractors.

● Multi-step reduction for dynamic analysis

● Zooming (or global-local analysis)

● Allows for efficient configuration studies (“What if...”)


DISADVANTAGES OF SUPERELEMENTANALYSIS

● Increased overhead due to DMAP compilation anddatabase manipulation and storage

● Mandatory static condensation may cancel other costsavings for small models.

● Residual structure is not resequenced and its stiffnessmatrix is usually dense.

● All superelements must be linear.● Approximations must be made in dynamics for mass and

damping through static, component mode, or generalizeddynamic reduction.

● Automatic restarts are available in SOLs 101 and above.


CONVENTIONAL ANALYSIS

Generation

Solution

Flowchart


CONVENTIONAL ANALYSIS (Cont.)

Generation

1 2 3 4 5

Kxx = unit stiffness

P2 = 1 P3 = 2 P4 = 3

K12 K23 K34 K45

KGG

K12 K12– 0 0 0

K12– K12 K23+ K23– 0 0

0 K23– K23 K34+ K34– 0

0 0 K34– K34 K45+ K45–

0 0 0 K45– K45

=

KGG

1.0 1.0– 0 0 01.0– 2.0 1.0– 0 00 1.0– 2.0 1.0– 00 0 1.0– 2.0 1.0–0 0 0 1.0– 1.0

=


CONVENTIONAL ANALYSIS (Cont.)

Apply Constraints and Solve

U2

U3

U4

K12 K23+ K23–

K23– K23 K34+ K34–

K34– K34 K45+

1–P2

P3

P4

=

U2U3U4

2.0 1.0– 01.0– 2.0 1.0–0 1.0– 2.0

1– 1.02.03.0

=

U2

U3

U4

2.54.03.5

=


SUPERELEMENT ANALYSIS

FlowchartDO LABELA

Phase I

GenerationAssembly

LABELA

Phase IISolution

DO LABELB

Phase III

LABELB


SUPERELEMENT ANALYSIS (Cont.)

Generation - SEID = 1

1 2 3 4 5

Residual Structure

SEID = 1 SEID = 2

1 2 3

P2 = 1

K23K12

u2 u3

Kgg1

K12 K12– 0

K12– K12 K23+ K23–

0 K23– K23

=



Reduction - SEID = 1 Eliminate constraints:

Compute boundary transformation:

Pg 1

P1

P2

P31

010

= =

Koo Koa

Kao Kaa

=

Kgg1 K

12K

23+ K

23–

K23– K23

=

Goa1 Koo

1–Koa–=

K23K12 K23+---------------------------= 0.5=



Compute boundary stiffness:

Compute boundary loading:

Kaa1

Kaa KoaT Goa+=

Kaa1

K12K23

K12 K23+--------------------------- 0.5= =

P f 1 P2

P31

10

= =

Po

Pa

=

0

Pa 1

Pa GoaT Po+

=

P31 P3

1K23

K12 K23+---------------------------P2+ 0.5= =



Similarly - SEID = 2

Kgg2

K34 K34– 0

K34– K34 K45+ K45–

0 K45– K45

=

Pg 2 P3

2

P4P5

030

= =

P32 P3

2K34

K34 K45+---------------------------P2+ 1.5= =

0

Goa2 K34

K34 K45+--------------------------- 0.5= =

Kaa2 K34K45

K34 K45+--------------------------- 0.5= =



Residual Structure

Assembly

Solution

3 3

K

PP3

2P3

1

P30

K1K2

0Kaa Kaa

1 Kaa2 Kgg

0+ +=

K K1 K2+ 1= =

Pa

Pa1 Pa

2 Pg0+ +

=

P P31 P3

2 P30+ + 4= =

Ua

Kaa1–

Pa

=

U30 P

K---- 4= =



1 2 3

U oa

G oa U a

=

U23

K23K12 K23+---------------------------U3 2.0= =

U23 U3 4== ?

1 2 3

Uoo

Koo1–

Po =

U2o 1

K12 K23+---------------------------P2 0.5= =

U 20 = ?

1 2 3

Uo

Uoo Uo

a +=

U2K 23 U3 P2+

K12 K23+------------------------------------ 2.5= =U2 2.5= U3 4=

● Data Recovery - SEID = 1● Enforce (transform) boundary motion.

● Compute fixed-boundary motion.

● Compute total motion.


SUPERPOSITION OF PARTIAL SOLUTIONSApplied Loads

A. Assembled Structure

Superelement 1 Superelement 2 Superelement 3

Residual Structure

C. Partial Solution Due to External Loads

B. Partitioned Structure

D. Partial Solution Due to Boundary Motion

E. Assembled Solution

+

=


SUPERELEMENT REDUCTION METHODSAVAILABLE IN DYNAMIC ANALYSIS

● Static reduction● Static condensation of stiffness and Guyan reduction of mass

● Automatically done

● Dynamic reduction● Component modal synthesis (CMS)

● Analytical (SOL 103)


DEGREES OF REDUCTION

● Static reduction (default)● Interior masses relumped to boundary (Guyan)● Rigid body properties preserved● Important masses must be made exterior (boundary)

● Component mode reduction - in addition to static reduction● Interior masses represented by exact eigenvectors of the

component● Eigensolutions for each superelement may be output


GUYAN OR STATIC REDUCTION

Koo Koa

KoaT Kaa

Uo

Ua Po

Pa

=

Kaa KaaKoa

TGoa+=

Maa Maa MoaT

Goa GoaT

Moa GoaT

Moo Goa+ + +=

Uoo

● Based on stiffness● Perform static condensation on stiffness

If Po=0, then {uo} = [Goa] {ua}, where Goa = -[Koo-1] [Koa]

Use this transformation to go from the F-set to the A-set

However, (internal dynamic effects) is ignored● No approximation if no masses, damping, or applied loads are specified on O-

set (interior DOFs)● Good approximation if component frequencies are much higher than residual

structure frequencies and the excitation frequencies


● 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.

COMPARISON OF REDUCTION METHODS

uo Points

ut Points

XX

X

XXXXXXX

XXXXXXX

X

X

X

X

X

X

X

X

0 Local dynamiceffects are ignored.Uo

Got Ut

Uoo

+=

Uo

Got Goq

Ut

Uq

=

Uo

Got Goq

Ut

Uq

=


ADVANTAGES OF EACH REDUCTION METHOD

● Advantages of Component Mode Reduction over StaticReduction● 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 ModeReduction● Cheaper● Less sophisticated● Fewer problems


CALCULATION OF NORMAL MODES USINGSTATIC REDUCTION ONLY

● No generalized coordinates are needed for the superelements.(exception: residual structure, if component reduction at the residualstructure Is desired)

● Superelement mass, damping, and stiffness are reduced statically toexterior DOFs.

● ASETi and QSETi entries may be specified for residual structureDOFs only.

● If no ASETi entries are present, then all DOFs interior to the residualstructure are retained for the eigensolution.

● If ASETi entries are present, then only those DOFs specified on theASETi entries are retained for the eigensolution.

● Case Control is similar to static analysis with the addition of aMETHOD command under the residual structure subcase.


CALCULATION OF NORMAL MODES USINGDYNAMIC REDUCTION FOR SUPERELEMENTS● The behavior of a superelement is represented by its component 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.● The number of superelement modes computed (modal truncation) is controlled by the

EIGR or EIGRL entry.● The number of superelement modes sent downstream directly to the residual is done

one of three ways:● Param,autoqset,yes● SENQSET DOFs● QSETi and SPOINT DOFs.

● If the superelement modes are to be sent downstream to another superelement, QSETiand SPOINTs must be used. These SPOINTs must also be “connected” to theSPOINTs in the downstream superelement using the SECONCT entry.


CALCULATION OF NORMAL MODES USINGDYNAMIC REDUCTION FOR SUPERELEMENTS

(Cont.)

● By default, superelement modes are computed with allexterior degrees of freedom fixed (in the B-set). This isbetter known as the Craig-Bampton method.

● Superelement modes are computed in Phase I under theSEMR operation.

● Copied superelements need to have the same number ofexterior DOFs as their primary. If the primary has anSENQSET, then the image must have equivalent DOFs torepresent the modes.


Superelement Internal Generalized Degree of Freedom

Defines number of internally generated scalar points for superelement dynamic reduction.

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

SENQSET SEID N

Example:SENQSET 110 45

Field ContentsSEID Superelement identification number. See Remark 3.. (Integeru0 or Character=“ALL”)

N

Remarks:1.

2.

3.

4.

SEID=“ALL” will automatically generate N q-set degrees of freedom for all superelements, exceptthe residual structure (SEID=0). Specifying additional SENQSET entries for specificsuperelements will override the value of N specified on this entry.

If the user manually specifies q-set degrees of freedom using a SEQSETi or QSETi entries, thenthe internally generated scalar points will not be generated.LQ

Bulk Data Entry

SENQSET

Number of internally generated scalar points for dynamic reduction generalizedcoordinates. (Integer>0; Default=0)

SENQSET can only be specified in the main Bulk Data Section and is ignored after the BEGINSUPER=n command.

SENQSET is only required if the user wants to internally generated scalar points used for dynamicreduction.


Scalar Point Definition

Defines scalar points.

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

SPOINT ID1 ID2 ID3 ID4 ID5 ID6 ID7 ID8

Example:SPOINT 3 18 1 4 16 2

Alternate Format and Example:SPOINT ID1 “THRU” ID2SPOINT 5 THRU 649

Field ContentsIDi Scalar point identification number. (0<Integer<1000000; For “THRU” option, ID1<ID2)

Remarks:1.

2.

3.

4. If the alternate format is used, all scalar points ID1 through ID2 are defined.5. For a discussion of scalar points, see the MSC.Nastran Reference Manual , Section 5.6.

Bulk Data Entry

SPOINT

A scalar point defined by its appearance on the connection entry for a scalar element (see theCELASi, CMASSi, and CDAMPi entries) need not appear on an SPOINT entry.

All scalar point identification numbers must be unique with respect to all other structural, scalar,and fluid points. However, duplicate scalar point identification numbers are allowed in the input.

This entry is used primarily to define scalar points appearing in single-point or multipoint constraintequations to which no scalar elements are connected.


Generalized Degree of Freedom

Defines generalized degrees of freedom (q-set) to be used for dynamic reduction or component mode synthesis.

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

QSET ID1 C1 ID2 C2 ID3 C3 ID4 C4

Example:QSET 15 123456 1 7 9 2 105 6

Field ContentsIDi Grid or scalar point identification number. (Integer>0)

Ci

Remarks:1.

2. Degrees of freedom specified on QSET and QSET1 entries are automatically placed in the a-set.3.

Bulk Data Entry

QSET

Component number. (Integer zero or blank for scalar points or any unique combinationof the Integers 1 through 6 for grid points with no embedded blanks.)

Degrees of freedom specified on this entry form members of the mutually exclusive q-set. Theymay not be specified on other entries that define mutually exclusive sets. See the MSC.NastranQuick Reference Guide, Appendix B for a list of these entries.

When ASET, ASET1, QSET, and/or QSET1 entries are present, all degrees of freedom nototherwise constrained (e.g., SPCi or MPC entries) will be placed in the omitted set (o-set).


Generalized Degree of Freedom (Alternate Form of QSET Entry

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

QSET1 C ID1 ID2 ID3 ID4 ID5 ID6 ID7ID8 ID9 -etc.-

Example:QSET1 123456 1 7 9 22 105 6 22

52 53

Alternate Format and Example:QSET1 C ID1 “THRU” ID2QSET1 0 101 THRU 110

Field ContentsC

IDi Grid or scalar point identification number. (Integer>0; For THRU option, ID1<ID2.)

Remarks:1.

2. Degrees of freedom specified on QSET and QSET1 entries are automatically placed in the a-set.3.

QSET1

Bulk Data Entry

Defines generalized degrees of freedom (q-set) to be used for generalized dynamic reduction or componentmode synthesis.

Component number. (Integer zero or blank for scalar points or any unique combinationof the Integers 1 through 6 for grid points with no embedded blanks.)

Degrees of freedom specified on this entry form members of the mutually exclusive q-set. Theymay not be specified on other entries that define mutually exclusive sets. See the MSC.NastranQuick Reference Guide, Appendix B for a list of these entries.

When ASET, ASET1, QSET, and/or QSET1 entries are present, all degrees of freedom nototherwise constrained (e.g., SPCi or MPC entries) will be placed in the omitted set (o-set).


CALCULATION OF NORMAL MODES USINGDYNAMIC REDUCTION FOR SUPERELEMENTS

(Cont.)● For free-free superelement component modes, all exterior DOFs should be

specified on CSETi entries.● 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,EPSRC in the

MD Nastran Reference Manual).

● Mixed-boundary modes may be calculated by using the CSETi and BSETientries to describe the exterior DOFs to be unconstrained and constrainedduring CMS.● If 0.0 Hz mixed boundary modes exist, they must be handled in a similar manner to

those in the free-free case.


Fixed Analysis Degrees of Freedom

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

BSET ID1 C1 ID2 C2 ID3 C3 ID4 C4

Example:BSET 2 135 14 6

Field ContentsIDi Grid or scalar point identification number. (Integer>0)Ci

Remarks:1.

2.

3.

a.

b.

Bulk Data Entry

Defines analysis set (a-set) degrees of freedom to be fixed (b-set) during generalized dynamic reduction orcomponent mode synthesis calculations.

If there are no CSETi or BSETi entries present, all a-set points are considered fixed duringcomponent mode analysis. If there are only BSETi entries present, any a-set degrees of freedomnot listed are placed in the free boundary set (c-set). If there

Degrees of freedom specified on this entry form members of the mutually exclusive b-set. Theymay not be specified on other entries that define mutually exclusive sets. See the MSC.NastranQuick Reference Guide, Appendix B for a list of these entries.

If PARAM,AUTOSPC is YES, then singular b-set and c-set degrees of freedom will be reassignedas follows:

BSET

If there are no o-set (omitted) degrees of freedom, then singular b-set and c-setdegrees of freedom are reassigned to the s-set.

If there are o-set (omitted) degrees of freedom, then singular c-set degrees of freedomare reassigned to the b-set. Singular b-set degrees of freedom are not reassigned.

Component number. (Integer zero or blank for scalar points, or any uniquecombinations of the Integers 1 through 6 for grid points. No embedded blanks.)


Fixed Analysis Degrees of Freedom, Alternate Form of BEST Entry

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

BSET1 C ID1 ID2 ID3 ID4 ID5 ID6 ID7ID8 ID9 ID10 -etc.-

Example:BSET1 2 135 14 6 23 24 25 26

122 127

Alternate Format and Example:BSET1 C ID1 “THRU” ID2BSET1 3 6 THRU 32

Field ContentsC

IDi

Remarks:1.

2.

3.

a.

b.

Defines analysis set (a-set) degrees of freedom to be fixed (b-set) during generalized dynamic reduction orcomponent mode synthesis calculations.

BSET1

Component numbers. (Integer zero or blank for scalar points, or any uniquecombinations of the Integers 1 through 6 for grid points with no embedded blanks.)

Grid or scalar point identification numbers. (Integer>0; For “THRU” option, ID1<ID2)

If there are no o-set (omitted) degrees of freedom, then singular b-set and c-setdegrees of freedom are reassigned to the s-set.If there are o-set (omitted) degrees of freedom, then singular c-set degrees of freedomare reassigned to the b-set. Singular b-set degrees of freedom are not reassigned.

Bulk Data Entry

If there are no CSETi or BSETi entries present, all a-set points are considered fixed duringcomponent mode analysis. If there are only BSETi entries present, any a-set degrees of freedomnot listed are placed in the free boundary set (c-set). If there

Degrees of freedom specified on this entry form members of the mutually exclusive b-set. Theymay not be specified on other entries that define mutually exclusive sets. See the MSC.NastranQuick Reference Guide, Appendix B for a list of these entries.If PARAM,AUTOSPC is YES, then singular b-set and c-set degrees of freedom will be reassignedas follows:


Free Boundry Degree of Freedom

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

CSET ID1 C1 ID2 C2 ID3 C3 ID4 C4

Example:CSET 124 1 5 23 6 16

Field ContentsIDi Grid or scalar point identification number. (Integer>0)

Ci

Remarks:1.

2.

3.

a.

b.

Bulk Data Entry

Defines analysis set (a-set) degrees of freedom to be free (c-set) during generalized dynamic reduction orcomponent modes calculations.

Component numbers. (Integer zero or blank for scalar points, or any uniquecombination of the Integers 1 through 6 for grid points with no embedded blanks.)

If there are no CSETi or BSETi entries present, all a-set degrees of freedom are considered fixedduring component modes analysis. If there are only BSETi entries present, any a-set degrees offreedom not listed are placed in the free boundary set (c-set

Degrees of freedom specified on this entry form members of the mutually exclusive c-set. Theymay not be specified on other entries that define mutually exclusive sets. See the MSC.NastranQuick Reference Guide, Appendix B for a list of these entries.

CSET

If there are no o-set (omitted) degrees of freedom, then singular b-set and c-setdegrees of freedom are reassigned to the s-set.If there are o-set (omitted) degrees of freedom, then singular c-set degrees of freedomare reassigned to the b-set. Singular b-set degrees of freedom are not reassigned.

If PARAM,AUTOSPC is YES then singular b-set and c-set degrees of freedom will be reassignedas follows:


Free Boundry Degree of Freedom, Alternate Form of CSET Entry

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

CSET1 C ID1 ID2 ID3 ID4 ID5 ID6 ID7ID8 ID9 -etc.-

Example:CSET1 124 1 5 7 6 9 12 122

127

Alternate Formats and Examples:CSET1 C ID1 “THRU” ID2CSET1 3 6 THRU 32

CSET1 “ALL”CSET1 ALL

Field ContentsC

IDi Grid or scalar point identification number. (Integer>0; For THRU option, ID1<ID2)

Remarks:1.

Bulk Data Entry(Continued)

Defines analysis set (a-set) degrees of freedom to be free (c-set) during generalized dynamic reduction orcomponent modes calculations.

CSET1

Component number. (Integer zero or blank for scalar points, or any unique combinationof the Integers 1 through 6 for grid points with no embedded blanks.)

If there are no CSETi or BSETi entries present, all a-set degrees of freedom are considered fixedduring component modes analysis. If there are only BSETi entries present, any a-set degrees offreedom not listed are placed in the free boundary set (c-set


Free Boundry Degree of Freedom, Alternate Form of CSET Entry

2.

3.

a.

b. If there are o-set (omitted) degrees of freedom, then singular c-set degrees of freedomare reassigned to the b-set. Singular b-set degrees of freedom are not reassigned.

CSET1

Bulk Data Entry

Degrees of freedom specified on this entry form members of the mutually exclusive c-set. Theymay not be specified on other entries that define mutually exclusive sets. See the MSC.NastranQuick Reference Guide, Appendix B for a list of these entries.If PARAM,AUTOSPC is YES then singular b-set and c-set degrees of freedom will be reassignedas follows:

If there are no o-set (omitted) degrees of freedom, then singular b-set and c-setdegrees of freedom are reassigned to the s-set.


DEFAULT CMS METHOD “FIXEDBOUNDARY” 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 GOATin MD Nastran)

● In matrix form,

(Pb is not actually applied.)

● The first line gives

Koo Kob

Kbo Kbb

ob

Ibb

0

Pb

=

ob

Koo1–

Kob Ibb

–= (GOAT)


● giving the following “constraint” modes:ob

b = -------Ibb

● Now the O-set equations are solved for the “fixed-boundary” modes {oo} (known as GOAQ in MDNastran).

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.

DEFAULT CMS METHOD “FIXEDBOUNDARY” CMS (Cont.)

k2 Moo– oo

Koo oo

+ 0=

G ob oo

Ibb 0

=

KGG T Kff

G =

MG G T Mff G =


DEFAULT CMS METHOD “FIXEDBOUNDARY” CMS (Cont.)

● These “generalized” matrices contain physical DOFsrepresenting the boundaries and “modal” coordinatesrepresenting the “fixed-boundary” component modes.

● At this point, these matrices can be treated like any otherstructural matrices, and data recovery can be performedfor the component in a manner similar to using modalcoordinates. That is, the displacements of the generalizedcoordinates are multiplied by the associated vectors andadded together to obtain the component displacements.

● The calculated modes for each superelement are internallyscaled to have a maximum displacement = 1.0 in MDNastran (regardless of the scaling requested by the user).


SOLUTION BY HAND

Component Modal Synthesis Sample

k1 = k2 = k3 = k4 = 1.0 m1 = m2 = m4 =m5 = 1.0 ; m3 = 1.0

Theoretical solution for frequencies

51 2 3 4

Part 2 Part 1

k1 k2k3 k4

2 3 4

Part 2 Part 1

3 3

Residual

51

i 1 2 3 4fi 0.0553 0.1592 0.2438 0.2991

li = w2 0.1206 1 2.3473 3.5321


● Superelement 1

● Grid Point 3 is the “boundary” point; solve for “constraint”modes.

Where Koo =

Kob =

=

SOLUTION BY HAND (Cont.)

Kgg

1.0 1.0– 01.0– 2.0 1.0–0 1.0– 1.0

= Mgg

0 0 00 1.0 00 0 1.0

U3

U4

U5

=

1.0 1.0– 01.0– 2.0 1.0–0 1.0– 1.0

1.0U4

U5

Pb

00

=

2.0 1.0–1.0– 1.0

1.0–0

1.0 1.01.0 2.0

Koo-1


● Where

● Solve for “fixed-boundary” modes.● Note: Internally MD Nastran uses component modes scaled to a maximum deformation

of 1.0. Output for the component modes is based on the normalization performed bythe eigenvalue solution.


0.10.1

0.00.1

0.20.10.10.1

ob

2Moo– Koo+ oo 0. 2– 0

0 2–

2.0 1.0–1.0– 1.0

+

oo= =

0.1

0.10.1

b



where 1001 and 1002 are scalar points used to represent Superelement 1’smodes.

20.8506–0.5257

=

Normalized tounit mass

det 2 2– 1–

1– 1 2–0=

f = 0.098 Hz, 0.2575 Hz2 = 0.3819, 2.618

10.61801.0000

=

21.0000–0.6180

=

10.52570.8506

=

oo0.6180 1.0000–1.0000 0.6180

=

G

1.000 0 01.000 0.618 1.000–1.000 1.000 0.618

=

G

TKgg G

0 0 00 0.5279 00 0 3.6180

=u3u1001u1002

G

TMgg G

2.0000 1.6180 0.3820–1.6180 1.3820 00.3820– 0 1.3820

=u3u1001u1002



Superelement 2

where 1005 is a scalar point used to represent Superelement 2’smode.

f = 0.22512 = 2.0

b0.51.0

=

G0.5 1.01.0 0

=

GT

Kgg G0.5 00 2.0

=

G

TMgg G

0.25 0.500.50 1.00

=

Kgg

1.0 1.0– 01.0– 2.0 1.0–0 1.0– 1.0

= M gg

1.0 0 00 1.0 00 0 0

=

oo 1=

u3u1005

u3u1005

• Apply constraint to grid point 1.



Kgg

0 0 0 00 0 0 00 0 0 00 0 0 0

=

Mgg

1 0 0 00 0 0 00 0 0 00 0 0 0

=

u3u1001u1002u1005

Residual Structure● Before adding superelements:



Mgg

3.0000 1.6180 0.3820– 01.6180 1.3820 0 00.3820– 0 1.3820 0

0 0 0 0

=

u3u1001

u1002

u1005

Kgg

0 0 0 00 0.5279 0 00 0 3.618 00 0 0 0

=

● Add Superelement 1



Mgg

3.2500 1.6180 0.3820– 0.51.6180 1.3820 0 00.3820– 0 1.3820 0

0 0 0 1.0

=

Kgg

0.5 0 0 00 0.5279 0 00 0 3.618 00 0 0 2.0

=

u3u1001u1002u1005

● Add Superelement 2



● Solve {KFF - 2Mff}{f} which gives 2 = 0.1206, 1.0000, 2.3473, 3.5321.

● Data recovery (grid point displacement for mode 1)

Residual Structure

Mff

3.2500 1.6180 0.3820– 01.6180 1.3820 0 00.3820– 0 1.3820 0

0 0 0 1.0

=

u3u1001u1002

u1005

Kff

0.5 0 0 00 0.5279 0 00 0 3.618 00 0 0 2.0

=

f

0.42850 0.5773– 0.2280– 0.65650.23150 1.0937 0.3188 0.8619–0.00572– 0.0986 0.5464 0.70120.01370 0.2887– 0.7705 0.7568–

=

00.4285

u1

u3

=



● Superelement 2

● Superelement 1

for exterior points 2G

00.42850.0137

=u1u3u1005

21 G2 2G 1.0 0 00.5 0.5 1.00 1.0 0

00.42850.0137

0

0.22800.4285

= = =u1u2u3

for exterior points 1G

0.42850.23150.00572–

=u1u1001u1002

11 G1 1G 1.0 0 01.0 0.6180 1.0–1.0 1.0 0.6180

0.42850.23150.00572–

0.4285

0.57730.6565

= = =u3u4u5


SOLUTION USING MD NASTRAN SOL 103

$$ sesp2.dat$SOL 103CENDTITLE = SAMPLE PROBLEM FOR CMS USINGPARTSSPC = 1DISP = ALLPARAM,GRDPNT,0PARAM,USETPRT,0$SUBCASE 1LABEL = CMS FOR PART 1SUPER = 1METHOD=1 $$SUBCASE 2LABEL = CMS FOR PART 2SUPER = 2METHOD = 2$SUBCASE 100LABEL = SYSTEM MODESSUPER = 0METHOD = 100$BEGIN BULK$grid,3,,20.conm2,13,3,,1.0$EIGRL,100,,,4param,autoqset,yes

BEGIN SUPER = 1$EIGRL,1,,,2grid,3,,20.grid,4,,30.grid,5,,40.$CELAS2,3,1.,3,1,4,1CELAS2,4,1.,4,1,5,1CONM2,14,4,,1.

CONM2,15,5,,1.$BEGIN SUPER = 2$EIGRL,2,,,1grid,1,,0.grid,2,,10.grid,3,,20.$CELAS2,1,1.,1,1,2,1CELAS2,2,1.,2,1,3,1CONM2,11,1,,1.CONM2,12,2,,1.SPC1,1,123456,1$ENDDATA

The following input data was created and run in MD Nastran:


OUTPUT FROM SPRING MODEL CMS RUN1 SAMPLE PROBLEM FOR CMS USING PARTS MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 28

SUPERELEMENT 10 CMS FOR PART 1 SUBCASE 1



1 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


OUTPUT FROM SPRING MODEL CMS RUN(Cont.)

1 SAMPLE PROBLEM FOR CMS USING PARTS MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE43




1 1 2.000000E+00 1.414214E+00 2.250791E-01 1.000000E+00 2.000000E+00



1 SAMPLE PROBLEM FOR CMS USING PARTS MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 66SUPERELEMENT 0

0 SYSTEM MODES SUBCASE 100


1 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 3 2.347296E+00 1.532089E+00 2.438395E-01 1.000000E+00 2.347296E+004 4 3.532089E+00 1.879385E+00 2.991135E-01 1.000000E+00 3.532089E+00


0 SYSTEM MODES SUBCASE 1001 SAMPLE PROBLEM FOR CMS USING PARTS MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 68

SUPERELEMENT 00 SYSTEM MODES SUBCASE 100*** USER INFORMATION MESSAGE 7321 (SEDRDR)

DATA RECOVERY FOR SUPERELEMENT 0 IS NOW INITIATED.1 SAMPLE PROBLEM FOR CMS USING PARTS MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 69

SUPERELEMENT 00 SYSTEM MODES SUBCASE 100

EIGENVALUE = 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 R33 G 4.285251E-01 0.0 0.0 0.0 0.0 0.0



EIGENVALUE = 1.000000E+00CYCLES = 1.591549E-01 R E A L E I G E N V E C T O R N O . 2

POINT ID. TYPE T1 T2 T3 R1 R2 R33 G -5.773503E-01 0.0 0.0 0.0 0.0 0.0


0 SYSTEM MODES SUBCASE 100EIGENVALUE = 2.347296E+00

CYCLES = 2.438395E-01 R E A L E I G E N V E C T O R N O . 3

POINT ID. TYPE T1 T2 T3 R1 R2 R33 G 2.280134E-01 0.0 0.0 0.0 0.0 0.0


0 SYSTEM MODES SUBCASE 100EIGENVALUE = 3.532089E+00


POINT ID. TYPE T1 T2 T3 R1 R2 R33 G -6.565385E-01 0.0 0.0 0.0 0.0 0.0




DATA RECOVERY FOR SUPERELEMENT 1 IS NOW INITIATED.1 SAMPLE PROBLEM FOR CMS USING PARTS MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 75


EIGENVALUE = 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 R33 G 4.285251E-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 6.565385E-01 0.0 0.0 0.0 0.0 0.0




0 CMS FOR PART 1 SUBCASE 1EIGENVALUE = 1.000000E+00


POINT ID. TYPE T1 T2 T3 R1 R2 R33 G -5.773503E-01 0.0 0.0 0.0 0.0 0.04 G -1.110223E-16 0.0 0.0 0.0 0.0 0.05 G 5.773503E-01 0.0 0.0 0.0 0.0 0.0




POINT ID. TYPE T1 T2 T3 R1 R2 R33 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




POINT ID. TYPE T1 T2 T3 R1 R2 R33 G -6.565385E-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 -2.280134E-01 0.0 0.0 0.0 0.0 0.0




DATA RECOVERY FOR SUPERELEMENT 2 IS NOW INITIATED.




0 CMS FOR PART 2 SUBCASE 2EIGENVALUE = 1.206148E-01


POINT ID. TYPE T1 T2 T3 R1 R2 R31 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




POINT ID. TYPE T1 T2 T3 R1 R2 R31 G 0.0 0.0 0.0 0.0 0.0 0.02 G -5.773503E-01 0.0 0.0 0.0 0.0 0.03 G -5.773503E-01 0.0 0.0 0.0 0.0 0.0




POINT ID. TYPE T1 T2 T3 R1 R2 R31 G 0.0 0.0 0.0 0.0 0.0 0.02 G -6.565385E-01 0.0 0.0 0.0 0.0 0.03 G 2.280134E-01 0.0 0.0 0.0 0.0 0.0




POINT ID. TYPE T1 T2 T3 R1 R2 R31 G 0.0 0.0 0.0 0.0 0.0 0.02 G 4.285251E-01 0.0 0.0 0.0 0.0 0.03 G -6.565385E-01 0.0 0.0 0.0 0.0 0.0

1 SAMPLE PROBLEM FOR CMS USING PARTS MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 85


Blank Page


SECTION 17

EXTRA POINTS, TRANSFER FUNCTIONS,AND NOLINS


EXTRA POINTS

● Nonstructural DOFs are utilized to represent nonstructural variables.● Defined by EPOINT Bulk Data entry● Exist only in the residual structure for the dynamics solutions (E-set,

subset of D-set)● Unaffected by any reduction procedure including modal reduction.● Cannot be used as structural DOFs*● Cannot be constrained by MPCs or SPCs*● Can only be used in P-type direct input matrices or transfer functions● Can only undergo dynamic loading with the use of the DAREA Bulk

Data entry


TRANSFER FUNCTIONS

b0 b1p b2p2

+ + ud a0 a1p a2p

2+ +

iui

i 1=

N

+ 0=

● Transfer functions are used to define the dynamic functions of theform:

where ud = dependent coordinateui = independent degree of freedomp = differential operator (p = d/dt)

● Equivalent to P-type DMIG matrices (M2PP, B2PP, K2PP)● Function is added to other P-type matrix input. ud defines the row with

b0, b1, and b2 specifying the diagonal terms. ui defines the column forthe a0, a1, and a2 terms.

● Defined by the TF Bulk Data entry and selected by the TFL CaseControl command (residual structure only)


● Grid point-oriented nonlinearity

● N(t) is the nonlinear force and is a function of thedisplacement and the velocity.

● Defined on NOLINi Bulk Data entry that is selected by theNONLINEAR Case Control command.

● Available only in transient response solution sequences(must be in E-set for modal and D-set for directformulation).

NONLINEAR TRANSIENT FORCE

Mu··

t Bu·

t Ku t+ + P t N t+=


● Nonlinear force represents a deviation from linearity.

● Easiest to implement in direct solutions. Apply nonlinearforce directly to grid point. Modal solutions require transferfunctions and extra points since only E-set points mayhave nonlinear loads in a modal formulation.

NONLINEAR TRANSIENT FORCE

= +

= +

Force

Displacement

NonlinearElasticSwaybrace

ForceForce

DisplacementDisplacement

Mu··

t Bu·

t Ku t+ + P t N t+=


CONSIDERATIONS WHEN USINGNONLINEAR FORCES

Mu·· t Bu· t Ku t+ + P t N t t– +=

● Nonlinear forces are based on the solution at the previous time step.

● A smaller t will give a more accurate solution.● Easier to use in direct solution sequences than in modal solution

sequences● Nonlinear forces may be applied only to members of D-set (A-set + E-

set) for direct and H-set (modal set + E-set). DOFs cannot beOMITted nor can they be dependent upon the motion of other omittedDOFs.

● May need a local coordinate system for the nonlinear force applied ina non-basic axis direction. Verify this with a pilot model.


● NOLIN1 - Nonlinear transient load as a tabular function● Function of displacement

● Function of velocity

● NOLIN2 - Nonlinear transient load as the product of twovariables

where Xj and Xk can be either two displacements ortwo velocities or one of each

TYPES OF NOLINS

Pi tn ST uj tn =

Pi t ST u· j tn =

u· tn u tn u tn 1– –

t------------------------------------------=where

Pi tn SXj tn Xk tn =


TYPES OF NOLINS (Cont.)

Pi tn S Xj tn A

0

= , Xj (tn) > 0

, Xj (tn) < 0

Pi tn S– X– j tn A

0

= , Xj (tn) < 0

, Xj (tn) > 0

● NOLIN3 - Nonlinear transient load as a positive variable raised to apower

● where Xj can be a displacement or a velocity● NOLIN4 - Nonlinear transient load as a negative variable raised to a

power

● where Xj can be a displacement or a velocity


WORKSHOP #13

LINEAR TRANSIENT ANALYSIS USINGNOLINS


WORKSHOP #13 - CAR TRAVELING OVER ABUMP

Simulate the following car model travelling over a bump using theNOLIN element.

3864 lbs

120”50”

4” 3 4

1 5 2

u

Velocity = 100 in/sec

u 2.0in–u 2.0in–

k: 197.4 lb/in

394.8 lb/in

u· 0

u·

0

c: 1.88 lb/(in/sec)1.88 lb/(in/sec) + 0.3 lb/(in/sec)2


WORKSHOP #13 - CAR TRAVELING OVER ABUMP

F

Urel.

F

Urel..

-2.0”


WORKSHOP # 13

$$ wkshp13.dat$$ case control, add : nolin callout$ plot nonlin forces$$ bulk data, add : transfer function to monitor relative displacements$ nolin for springs and dampers$ apply appropriate enforced displacement$ID NAS102, WORKSHOP13SOL 109TIME 100CENDTITLE= SIMPLE CAR MODEL WITH NOLINEARSUBTITLE= SPRINGS AND DAMPERS RUNNING OVER A BUMPLABEL= SOL 109, CONSTANT DELTA TIMESPC= 100TFL= 100NONLINEAR = 100DLOAD = 100TSTEP = 100DISPLACEMENT(PLOT)= ALLNLLOAD(PLOT)= ALL$



WORKSHOP # 13 (Cont.)

$OUTPUT(XYPLOT)CSCALE=1.3XAXIS= YESYAXIS= YESXGRID LINES= YESYGRID LINES= YESXTITLE= TIME (SEC)YTITLE= VERTICAL DISPLACEMENT OF POINT 1XYPLOT DISP/1(T2)YTITLE= VERTICAL DISPLACEMENT OF POINT 2XYPLOT DISP/2(T2)YTITLE= VERTICAL DISPLACEMENT OF POINT 3XYPLOT DISP/3(T2)YTITLE= VERTICAL DISPLACEMENT OF POINT 4XYPLOT DISP/4(T2)YTITLE= VERTICAL DISPLACEMENT OF POINT 5XYPLOT DISP/5(T2)YTITLE= NONLINEAR FORCES AT POINT 1XYPLOT NONLINEAR/1(T2)YTITLE= NONLINEAR FORCES AT POINT 2XYPLOT NONLINEAR/2(T2)$

$BEGIN BULKPARAM,POST,-1$$ CARRIAGE POINTS$GRID, 1, , 0., 0., 0.GRID, 2, , 120., 0., 0.GRID, 5, , 60., 0., 0.$$ WHEEL POINTS$GRID, 3, , 0., -10., 0.GRID, 4, , 120., -10., 0.$$ CAR CARRIAGE$CBAR, 5, 11, 1, 5, 0., 1., 0.CBAR, 6, 11, 5, 2, 0., 1., 0.PBAR, 11, 12, 10., 10., 10.MAT1, 12, 3.0E+7, , .33$$ CONSTRAINTS TO ELIMINATE RIGID-BODY MODES$SPC1, 100, 1345, 1, 2, 5SPC1, 100, 13456, 3, 4$$$ INTEGRATION INFORMATIONTSTEP, 100, 200, .05, 1$ENDDATA



SOL 109CENDTITLE= SIMPLE CAR MODEL WITH NOLINEARSUBTITLE= SPRINGS AND DAMPERS RUNNING OVER ABUMPLABEL= SOL 109, CONSTANT DELTA TIMESEALL= ALLSPC= 100TFL= 100NONLINEAR = 100DLOAD = 100TSTEP = 100DISPLACEMENT(PLOT)= ALLNLLOAD(PLOT)= ALL$OUTPUT(XYPLOT)CSCALE=1.3XAXIS= YESYAXIS= YESXGRID LINES= YESYGRID LINES= YESXTITLE= TIME (SEC)YTITLE= VERTICAL DISPLACEMENT OF POINT 1XYPLOT DISP/1(T2)YTITLE= VERTICAL DISPLACEMENT OF POINT 2XYPLOT DISP/2(T2)YTITLE= VERTICAL DISPLACEMENT OF POINT 3XYPLOT DISP/3(T2)

YTITLE= VERTICAL DISPLACEMENT OF POINT 4XYPLOT DISP/4(T2)YTITLE= VERTICAL DISPLACEMENT OF POINT 5XYPLOT DISP/5(T2)YTITLE= NONLINEAR FORCES AT POINT 1XYPLOT NONLINEAR/1(T2)YTITLE= NONLINEAR FORCES AT POINT 2XYPLOT NONLINEAR/2(T2)$BEGIN BULKPARAM,POST,-1$$ CARRIAGE POINTS$GRID, 1, , 0., 0., 0.GRID, 2, , 120., 0., 0.GRID, 5, , 60., 0., 0.$

$WHEEL POINTS$GRID, 3, , 0., -10., 0.GRID, 4, , 120., -10., 0.$$ CAR CARRIAGE$CBAR, 5, 11, 1, 5, 0., 1., 0.CBAR, 6, 11, 5, 2, 0., 1., 0.PBAR, 11, 12, 10., 10., 10.MAT1, 12, 3.0E+7, , .33$


SOLUTION FOR WORKSHOP #13 (Cont.)$$ CONSTRAINTS TO ELIMINATE RIGID-BODY MODES$SPC1, 100, 1345, 1, 2, 5SPC1, 100, 13456, 3, 4$$ SYSTEM WILL HAVE A NATURAL FREQUENCY OF 1 HZ$ WITH CRITICAL DAMPING OF 1 PERCENT$CONM2, 10, 1, ,2.5CONM2, 15, 2, ,2.5CONM2, 20, 5, ,5.$CELAS2, 30, 197.4, 1, 2, 3, 2CELAS2, 40, 197.4, 2, 2, 4, 2$CDAMP2, 50, 1.88, 1, 2, 3, 2CDAMP2, 60, 1.88, 2, 2, 4, 2$$ DEFINE EXTRA POINTS TO HOLD DIFFERENCES$ BETWEEN WHEELS AND CARRIAGE$EPOINT, 101, 102$$ USE TRANSFER FUNCTIONS TO TRACK DIFFERENCES$ 101= V1 - V3$ 102= V2 - V4$TF, 100, 101, 0, 1., 0., 0.,, 1, 2, -1., 0., 0.,, 3, 2, 1., 0., 0.$TF, 100, 102, 0, 1., 0., 0.,, 2, 2, -1., 0., 0.,, 4, 2, 1., 0., 0.$$ ADD NONLINEAR PORTION OF SPRINGS$

NOLIN1, 100, 1, 2, 197.4, 101, 0, 111NOLIN1, 100, 2, 2, 197.4, 102, 0, 111TABLED2, 111, -2.0,, -1., 1., 0., 0., 1., 0.,ENDT$$ ADD NONLINEAR PORTION OF DAMPERS$NOLIN4, 100, 1, 2, -0.3, 101, 10, 2.NOLIN4, 100, 2, 2, -0.3, 102, 10, 2.$$ MOVE WHEELS OVER BUMP$TLOAD2, 100, 222, 333, D, 0., 0.5, 1., -90.SPCD, 222, 3, 2, 4.SPCD, 222, 4, 2, 4.SPC1,100,2,3,4

DELAY, 333, 4, 2, 1.2$$ INTEGRATION INFORMATIONTSTEP, 100, 200, .05, 1$ENDDATA


PARTIAL OUTPUT FILE FOR WORKSHOP #130 X Y - O U T P U T S U M M A R Y ( R E S P O N S E )0 SUBCASE CURVE FRAME XMIN-FRAME/ XMAX-FRAME/ YMIN-FRAME/ X FOR YMAX-FRAME/ X FOR

ID TYPE NO. CURVE ID. ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX0 1 NONLIN 1 1( 4) 0.000000E+00 1.000002E+01 0.000000E+00 0.000000E+00 2.975151E+02 6.000001E-01

0.000000E+00 1.000002E+01 0.000000E+00 0.000000E+00 2.975151E+02 6.000001E-010 1 NONLIN 2 2( 4) 0.000000E+00 1.000002E+01 0.000000E+00 0.000000E+00 4.661460E+02 1.749999E+00

0.000000E+00 1.000002E+01 0.000000E+00 0.000000E+00 4.661460E+02 1.749999E+00.....

0 X Y - O U T P U T S U M M A R Y ( R E S P O N S E )0 SUBCASE CURVE FRAME XMIN-FRAME/ XMAX-FRAME/ YMIN-FRAME/ X FOR YMAX-FRAME/ X FOR

ID TYPE NO. CURVE ID. ALL DATA ALL DATA ALL DATA YMIN ALL DATA YMAX0 1 DISP 3 1( 4) 0.000000E+00 1.000002E+01 -2.836541E+00 2.199999E+00 5.942877E+00 4.500000E-01

0.000000E+00 1.000002E+01 -2.836541E+00 2.199999E+00 5.942877E+00 4.500000E-010 1 DISP 4 2( 4) 0.000000E+00 1.000002E+01 -2.671688E+00 1.999999E+00 7.464219E+00 1.600000E+00

0.000000E+00 1.000002E+01 -2.671688E+00 1.999999E+00 7.464219E+00 1.600000E+000 1 DISP 5 3( 4) 0.000000E+00 1.000002E+01 0.000000E+00 0.000000E+00 4.000000E+00 2.500000E-01

0.000000E+00 1.000002E+01 0.000000E+00 0.000000E+00 4.000000E+00 2.500000E-010 1 DISP 6 4( 4) 0.000000E+00 1.000002E+01 0.000000E+00 0.000000E+00 4.000000E+00 1.450000E+00

0.000000E+00 1.000002E+01 0.000000E+00 0.000000E+00 4.000000E+00 1.450000E+000 1 DISP 7 5( 4) 0.000000E+00 1.000002E+01 -2.150819E+00 2.149999E+00 3.963917E+00 1.600000E+00

0.000000E+00 1.000002E+01 -2.150819E+00 2.149999E+00 3.963917E+00 1.600000E+00


Blank Page


SECTION 18

NORMAL MODES OF PRELOADEDSTRUCTURES


NORMAL MODES WITH DIFFERENTIAL STIFFNESS

Preloaded Structure

Fx

yx

Reaction

Differential Stiffness

yx

● Calculate the modes of structures with preloads, large changes ingeometry, and/or nonlinear materials.


NORMAL MODES WITH DIFFERENTIALSTIFFNESS (Cont.)

Large Geometry Changes

u1

Nonlinear Material

k1

F

k0

KK00 = k1= k1


NORMAL MODES WITH DIFFERENTIALSTIFFNESS (Cont.)

Procedures for obtaining frequencies of a preloadedstructure.

● Use SOL 103.● Material must be linear.● Two subcases are required.

● The first subcase is a static subcase calling out the preload.● The second subcase calculated the modes with a method = x

callout.● The second subcase must also contain a statsub = y command,

where y = subcase ID of the first subcase.


WORKSHOP #14

NORMAL MODES WITH PRELOAD


WORKSHOP #14 - MODES OF PRELOADEDSTRUCTURE

P

● Consider the simply supported beam as shown below.Calculate the first bending frequency:● Without preload● With preload using SOL 103


WORKSHOP #14 - MODES OF PRELOADEDSTRUCTURE (Cont.)

0.1 in

0.1 in

0.1 in

1.0 in

1.0 in

2.0 in

Length: 100 inHeight: 2 inWidth: 1 in

Thickness: 0.100 inArea: 0.38 in2

I1: 0.229 in4

I2: 0.017 in4


WORKSHOP # 14a

● Run soln14a.dat as a baseline (model without preload)


SOLUTION FILE FOR WORKSHOP #14a -MODES WITHOUT PRELOAD

SOL 103TIME 600CENDTITLE = Normal Modes ExampleSUBCASE 1

METHOD = 1SPC = 1VECTOR=ALL

BEGIN BULKPARAM WTMASS .00259PARAM COUPMASS 1EIGRL 1 3 0PBARL 1 1 I + A+ A 2. 1. 1. .1 .1 .1CBAR 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.CBAR 9 1 9 10 0. 1. 0.CBAR 10 1 10 11 0. 1. 0.MAT1 1 1.+7 .3 .101GRID 1 0. 0. 0. 345GRID 2 10. 0. 0. 345GRID 3 20. 0. 0. 345GRID 4 30. 0. 0. 345GRID 5 39.9999 0. 0. 345GRID 6 49.9999 0. 0. 345GRID 7 60. 0. 0. 345GRID 8 70. 0. 0. 345GRID 9 80. 0. 0. 345GRID 10 90. 0. 0. 345GRID 11 100. 0. 0. 345SPC1 1 1234 1SPC1 1 234 11ENDDATA


PARTIAL OUTPUT FILE FOR WORKSHOP#14a MODES WITHOUT PRELOAD

E I G E N V A L U E A N A L Y S I S S U M M A R Y (READ MODULE)

BLOCK SIZE USED ...................... 6

NUMBER OF DECOMPOSITIONS ............. 2

NUMBER OF ROOTS FOUND ................ 3

NUMBER OF SOLVES REQUIRED ............ 4

1 NORMAL MODES EXAMPLE MARCH 13, 2007 MD NASTRAN 3/ 1/07 PAGE 7

0 SUBCASE 1




WORKSHOP 14c

● Use soln14a.dat as a starting point as modify it to run inSOL 103 with preload


SOL 103DIAG 8CEND$TITLE = Normal Modes with Differential Stiffness$SPC = 1DISPLACEMENT=ALL$SUBCASE 1LOAD = 2$SUBCASE 2METHOD = 10STATSUB = 1$BEGIN BULKPARAM COUPMASS 1PARAM WTMASS .00259$EIGRL,10,,,3PBARL 1 1 I + B+ B 2. 1. 1. .1 .1 .1CBAR 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.CBAR 9 1 9 10 0. 1. 0.CBAR 10 1 10 11 0. 1. 0.$

$MAT1 1 1.+7 .3 .101GRID 1 0. 0. 0. 345GRID 2 10. 0. 0. 345GRID 3 20. 0. 0. 345GRID 4 30. 0. 0. 345GRID 5 39.9999 0. 0. 345GRID 6 49.9999 0. 0. 345GRID 7 60. 0. 0. 345GRID 8 70. 0. 0. 345GRID 9 80. 0. 0. 345GRID 10 90. 0. 0. 345GRID 11 100. 0. 0. 345LOAD 2 1. 1. 1SPC1 1 1234 1SPC1 1 234 11FORCE 1 11 0 500. 1. 0. 0.ENDDATA

SOLUTION FILE FOR WORKSHOP #14c -MODES WITH PRELOAD USING SOL 103


1 NORMAL MODES EXAMPLE MARCH 13, 2007 MD NASTRAN 3/ 1/07PAGE 11

0 SUBCASE 2



PARTIAL OUTPUT FILE FOR WORKSHOP #14cMODES WITH PRELOAD USING SOL 103


BLANK PAGE


SECTION 19

DYNAMIC DESIGN OPTIMIZATION


WHAT IS “DESIGN OPTIMIZATION”?

● Automated modifications of the analysis model parametersto achieve a desired objective while satisfying specifieddesign 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● Correlates analytical model with test results (see Chapter 20)● Others - (depends on designer’s creativity)


BASIC FEATURES IMPLEMENTED IN MDNASTRAN

● 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 analyses as the measure of efficiency


STRENGTHS OF MD NASTRAN STRUCTURALOPTIMIZATION

● 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 MD Nastran

analysis● High-level support as a part of MD Nastran● Access to the familiar analysis tools in MD Nastran


OPTIMIZATION CAPABILITIES SUPPORTED

● Multi-Disciplinary Optimization● Static Response Optimization● Buckling Response Optimization● Dynamic Response Optimization

● Direct Frequency● Modal Frequency● Modal Transient● Acoustics

● Superelement Optimization● Allows the design model to span superelement boundaries.

● Aeroelastic Optimization● Static Aeroelastic● Flutter


OPTIMIZATION CAPABILITIES SUPPORTED(Cont.)

● Shape Optimization● Four methods are available for Basis Vector Generation

● Manual Grid Variation● Direct Input of Shapes● Geometric Boundary Shapes● Analytic Boundary Shapes


BASIC OPTIMIZATION PROBLEMSTATEMENT

● Design variables

● Objective function: Minimize F(X)

● Subject to:● Inequality constraints:

● Equality constraints:

● Side constraints:

X X1 X2 Xn, , , =Find

Gj X 0 j = 1, 2, ..., M

Hk X 0= k = 1, 2, ..., L

XiL

Xi Xiu

j = 1, 2, ..., N


DESIGN OPTIMIZATION IN MD NASTRAN

● The ANALYSIS Case Control request allows you tospecify the type of optimization analysis discipline that youwant to perform for each subcase.

● The following analysis types are allowed in the ANALYSISCase Control request:● STATICS Statics● MODES Normal Modes● BUCK Buckling● DFREQ Direct Frequency● MFREQ Modal Frequency● MTRAN Modal Transient● SAERO Static Aeroelasticity● FLUTTER Flutter


DESIGN OPTIMIZATION IN MD NASTRAN(Cont.)

design constraint

design constraint

SOL 200cendspc = 100DESOBJ(MIN) = 15ANALYSIS = STATICSsubcase 1

subtitle=static load 1DESSUB = 10displacement = allstress = allload = 1

subcase 2subtitle=static load 1DESSUB = 20displacement = allstrain(fiber) = allload = 2

subcase 3subtitle=modal analysisANALYSIS = MODESDESSUB = 30method = 3

subcase 4subtitle=transient analysisANALYSIS = MTRANDESSUB = 40method = 4dload = 4TSTEP = N

begin bulk

● In the example below you optimize two static load cases insubcases 1 and 2, a modal response in subcase 3, and atransient response in subcase 4.


COMMONLY USED OPTIMIZATION BULKDATA ENTRIES

● DESVAR Defines a design variable.● DVPREL1 Defines the relation between an analysis model property

and design variables.● DLINK Relates one design variable to one or more other design variables.● DRESP1 Defines a set of direct structural responses that is used in the design either

as constraints (referenced by the DCONSTR Bulk Data entry) or as anobjective (referenced by the DESOBJ Case Control Command).

● DCONSTR Defines a design constraint (referenced by the DESSUB Case ControlCommand).

● DCONADD Defines the design constraints for a subcase as a union of DCONSTRentries.

● DRESP2 Defines a synthesized response that are used in the design. This responsecan be either a constraint or an objective.

● DEQATN Defines equation(s) for use in design sensitivity.● DVCREL1 Defines the relation between a connectivity property and design variables.● DVCREL2 Defines the relation between a connectivity property and design variables

using a user-supplied equation.● DVMREL1 Defines the relation between a material property and design variables.● DVMREL2 Defines the relation between a material property and design variables with a

user-supplied equation.


WORKSHOP #15

OPTIMIZATION USING NORMAL MODES


10”

10”

Node 1 Node 2 Node 3

Node 4

Area 1 Area 2 Area 1

Elastic Modulus: 10E6Poisson’s Ratio 0.33

Density: 0.1Wt.-Mass Conversion 0.00259

Area 1: 1.0Area 2: 2.0

WORKSHOP #15 - OPTIMIZATION USINGNORMAL MODES

Minimize the weight of the following three bar truss problem. The first modemust be between 1500-1550 Hz. The structure must remain symmetric.Below is a geometric representation of the truss. It also contains the loadsand boundary constraints.


OPTIMIZATION STATEMENT

● Design variables● The areas of the three rod elements (A1, A2, A3)

● Objective● Minimize the weight of the truss.

● Subject to the following constraints:● The first mode must be between 1500-1550 Hz.● A1 = A3 to impose symmetry.


WORKSHOP # 15

$$ wkshp15.dat$$ executive control, add : optimization solution sequence$$ case control, add : type of analysis$ appropriate case control callout$$ bulk data, add : design variable$ objective function$ constraints$ relate design variable to property$ID NAS102, WORKSHOP 15TIME 10SOL 200 $ OPTIMIZATIONCENDTITLE= SYMMETRIC THREE BAR TRUSS DESIGN OPTIMIZATION - VARIATION OF D200X1SUBTITLE= GOAL IS TO MIN WT WHILE KEEPING THE 1ST MODE BETWEEN 1500-1550 HZECHO= SORTSPC= 100DISP(PLOT) ALLDESOBJ(MIN)= 100 $ (DESIGN OBJECTIVE = DRESP ID)DESSUB= 200 $ DEFINE CONSTRAINT SET FOR BOTH SUBCASESSUBCASE 1

ANALYSIS= MODESMETHOD= 10

BEGIN BULK



WORKSHOP # 15$$---------------------------------------------------------------- ------$ ANALYSIS MODEL$---------------------------------------------------------------- ------$EIGRL, 10, , , 2PARAM, POST, -1PARAM, PATVER, 3.0$$ GRID DATA$ 2 3 4 5 6 7 8 9 10GRID, 1, , -10.0, 0.0, 0.0GRID, 2, , 0.0, 0.0, 0.0GRID, 3, , 10.0, 0.0, 0.0GRID, 4, , 0.0, -10.0, 0.0$ SUPPORT DATASPC, 100, 1, 123456, , 2, 123456SPC, 100, 3, 123456, , 4, 3456$ ELEMENT DATACROD, 1, 11, 1, 4CROD, 2, 12, 2, 4CROD, 3, 13, 3, 4$ PROPERTY DATAPROD, 11, 1, 1.0PROD, 12, 1, 2.0PROD, 13, 1, 1.0MAT1, 1, 1.0E+7, , 0.33, 0.1$PARAM, WTMASS, .00259

$$----------------------------------------------------------------------$ DESIGN MODEL$----------------------------------------------------------------------$$...DESIGN VARIABLE DEFINITION$$DESVAR,ID, LABEL, XINIT, XLB, XUB, DELXV(OPTIONAL)DESVAR, 1, A1, 1.0, 0.1, 100.0..

$$...IMPOSE X3=X1 (LEADS TO A3=A1)$$DLINK, ID, DDVID, CO, CMULT, IDV1, C1, IDV2, C2, +$+, IDV3, C3, ...DLINK, 1, 3, 0.0, 1.0, 1 1.00$$...DEFINITION OF DESIGN VARIABLE TO ANALYSIS MODEL

PARAMETER RELATIONS$$DVPREL1,ID, TYPE, PID, FID, PMIN, PMAX, CO, , +$+, DVID1, COEF1, DVID2, COEF2, ...DVPREL1, 10, PROD, 11, 4, , , , , +DP1+DP1, 1, 1.0

.

.$DOPTPRM, DESMAX, 30$$.......2.......3.......4.......5.......6.......7.......8.......9.......0ENDDATA



ID NAS102, WORKSHOP 15TIME 10SOL 200 $ OPTIMIZATIONCENDTITLE= SYMMETRIC THREE BAR TRUSS DESIGN OPTIMIZATION - VARIATION OF D200X1SUBTITLE= GOAL IS TO MIN WT WHILE KEEPING THE 1ST MODE BETWEEN 1500-1550 HZECHO= SORTSPC= 100DISP(PLOT) ALLDESOBJ(MIN)= 100 $ (DESIGN OBJECTIVE = DRESP ID)DESSUB= 200 $ DEFINE CONSTRAINT SET FOR BOTH SUBCASESSUBCASE 1ANALYSIS= MODESMETHOD= 10

BEGIN BULK$$----------------------------------------------------------------------$ ANALYSIS MODEL$----------------------------------------------------------------------$EIGRL, 10, , , 2PARAM, POST, -1$$ GRID DATA$ 2 3 4 5 6 7 8 9 10GRID, 1, , -10.0, 0.0, 0.0GRID, 2, , 0.0, 0.0, 0.0GRID, 3, , 10.0, 0.0, 0.0GRID, 4, , 0.0, -10.0, 0.0$ SUPPORT DATASPC, 100, 1, 123456, , 2, 123456SPC, 100, 3, 123456, , 4, 3456$ ELEMENT DATACROD, 1, 11, 1, 4CROD, 2, 12, 2, 4CROD, 3, 13, 3, 4$ PROPERTY DATAPROD, 11, 1, 1.0PROD, 12, 1, 2.0PROD, 13, 1, 1.0MAT1, 1, 1.0E+7, , 0.33, 0.1$PARAM, WTMASS, .00259


$$----------------------------------------------------------------------$ DESIGN MODEL$----------------------------------------------------------------------$$...DESIGN VARIABLE DEFINITION$$DESVAR,ID, LABEL, XINIT, XLB, XUB, DELXV(OPTIONAL)DESVAR, 1, A1, 1.0, 0.1, 100.0DESVAR, 2, A2, 2.0, 0.1, 100.0DESVAR, 3, A3, 1.0, 0.1, 100.0$$...IMPOSE X3=X1 (LEADS TO A3=A1)$$DLINK, ID, DDVID, CO, CMULT, IDV1, C1, IDV2, C2, +$+, IDV3, C3, ...DLINK, 1, 3, 0.0, 1.0, 1 1.00$$...DEFINITION OF DESIGN VARIABLE TO ANALYSIS MODEL PARAMETER RELATIONS$$DVPREL1,ID, TYPE, PID, FID, PMIN, PMAX, CO, , +$+, DVID1, COEF1, DVID2, COEF2, ...DVPREL1, 10, PROD, 11, 4, , , , , +DP1+DP1, 1, 1.0DVPREL1, 20, PROD, 12, 4, , , , , +DP2+DP2, 2, 1.0DVPREL1, 30, PROD, 13, 4, , , , , +DP3+DP3, 3, 1.0$$...STRUCTURAL RESPONSE INDENTIFICATION$$DRESP1 ID LABEL RTYPE PTYPE REGION ATTA ATTB ATT1 +$+ ATT2 ...DRESP1 100 W WEIGHTDRESP1 210 MODE1 EIGN 1$$...CONSTRAINTS$$DCONSTR,DCID, RID, LALLOW, UALLOWDCONSTR, 200, 210, 8.883E7, 9.485E7$$...OPTIMIZATION CONTROL$DOPTPRM, DESMAX, 30$ENDDATA

SOLUTION FOR WORKSHOP #15 (Cont.)


DIRECT SPECIFICATION OF COMMONLYUSED FUNCTIONS

● Direct Support of Commonly-used Functions on theDRESP1 entry (e.g, SUM, RSS, AVG, etc.) for transientand frequency response analysis.

● Example:

● DRESP1,100,RSSCAL,FRDISP,,,3,RSS,100

● The above DRESP1 performs a root sum square of thedisplacement at grid point 100, component 3, across the wholeforcing frequency range.


OPTIMIZATION EXAMPLE USINGFREQUENCY RESPONSE

The automobile model shown below has the front left wheel out of balance. The amountof mass is .3 units and the radius from the center of the wheel to the mass is 10 inches.The frequency range of interest is from 0.5-50 Hz. The tire must not displace by morethan 0.5 inches at any frequency. The driver’s seat must not displace by more than0.25 inches between 0.5 and 25 hz. The goal is to minimize the SRSS response of thedriver’s seat across the whole forcing frequency range. Use the modal method.


OPTIMIZATION STATEMENT

● Design Variables:● The springs and dampers of the car

● Objective function:● Minimize the SRSS response at the driver’s seat

● Subject to the following constraints:● The maximum vertical displacement of the tire must be less than

0.5 inches● The maximum vertical displacement of the driver seat must be

less than 0.25 inches


DYNAMIC LOAD INPUT

● Use DAREA to describe magnitude of Fx and Fy (mr)

● Use one RLOAD1 entry for each loading

● Use a DPHASE to describe the phase relationship

● Use a TABLED4 to describe the frequency relationship of the loading (2)

● Use a DLOAD entry to combine the loadings (RLOAD1s)

Direction of rotation

F = mr2

= tFx = F cos(t)Fy = F sin( t)


TABLED4 ENTRY FOR THIS SAMPLE

● Set X1 = 0. X2 = 1. X3 = 0. X4 = 1000. (More than highest freq of interest)

● Since X is in cycles per unit time, it must be multiplied by 2to get radians perunit time - enter 2

● The input force is mr2 - we will enter mr using DAREA entries, but need the2 term

● We want to use the second (or the 2) term from the TABLED4, This can beaccomplished by setting A1=0.0 and A2=(2)2

Y Ai

X X1–

X2---------------------- i

i 0=

N

=


DYNAMIC RESPONSE SENSITIVITY

● Allows calculation of sensitivities of selected dynamicresponses with respect to changes in design variablesdri/dxj.

● These quantities provide insight on which areas are mosteffective in driving the design.

● Sensitivity coefficients can be requested with the DSAPRTcase control command

● Analysis disciplines:● Direct frequency response● Modal frequency response● Modal transient response


DESCRIPTION OF THE DESIGN MODEL

● Identify structural properties to be related to designvariables (DVPREL1, DVPREL2).

● Identify response quantities of interest (DRESP1,DRESP2).

● Specify bounds on responses and (optionally) screeningcriteria (DCONSTR, DSCREEN).

● Select output frequencies or time steps in Case Control(OFREQ or OTIME sets).


INPUT FOR OPTIMIZATION MODEL

BEGIN BULK..include ’car.blk’include ’springs.blk’include ’optim1.blk’..ENDDATA

Executive and Case Control

SOL 200CENDTITLE = Sample dynamic analysis modelset 999 = 358,471DISP(phase) = 999SUBCASE 1

ANALYSIS = MFREQDESSUB = 100 $ constraintsDESOBJ(min) = 300 $ design objective - minimize driver’s responseDLOAD = 1METHOD = 1FREQ = 1

Bulk Data Section


PARTIAL LISTING OF CAR.BLK

EIGRL,1,-1.0,100.DLOAD,1,1.,1.,11,1.,12RLOAD1,11,20,,,,111RLOAD1,12,30,,,40,111DPHASE 40 358 2 90.DAREA 20 358 1 3.DAREA 30 358 2 3.TABLED4,111,0.,1.,0.,1000.,0.,0.,39.478,ENDT$$ PLUS THE REST OF THE MODEL DESCRIPTION$

Input for Dynamic Loading


PARTIAL LISTING OF SPRINGS.BLK

CROD 1002 1002 402 1402CROD 1012 1001 825 1825CROD 1022 1001 358 1358CROD 1032 1002 869 1869$PROD 1001 1000 1000.PROD 1002 1000 800.$ select material so that value of PROD$ is spring stiffness,$ therefore, E = l = 10.MAT1 1000 10.

Input for Springs

Input for Shock Absorbers $

$ add dampers for shock absorbers$$ frontcvisc 2011 2001 825 1825cvisc 2021 2001 358 1358$ backcvisc 2001 2002 402 1402cvisc 2031 2002 869 1869$ damper propertiespvisc 2001 10. 0.pvisc 2002 5. 0.$


PARTIAL LISTING OF OPTIM.BLKDefine Design Variables

desvar,1,frntdamp,1.,.1,10.desvar,2,reardamp,1.,.2,20.desvar,3,frntstif,1.,.4,2.desvar,4,rearstif,1.,.5,2.5

Link Design Variables to Properties$$ relation between properties and variables$dvprel1,101,pvisc,2001,3,1.,,,,+dv101+dv101,1,10.dvprel1,102,pvisc,2002,3,1.,,,,+dv102+dv102,2,5.dvprel1,103,prod,1001,4,4.,,,,+dv103+dv103,3,10.dvprel1,104,prod,1002,4,4.,,,,+dv104+dv104,4,8.$

Define Design Constraints$ require that maximum tire displacement be .5 inches$dconstr,101,200,-.5,.5$$ require that maximum driver displacement be .25 inches$dconstr,102,201,-.25,.25,0.5,25.$ combine constraints into set 100$dconadd,100,101,102


PARTIAL LISTING OF OPTIM.BLK (Cont.)Select Response Quantities

$ select displacement Y at driver seat and mount point as$ response quantities$$ mount point$dresp1,200,disp,frdisp,,,2,,358$$ define driver’s seat disp as a response$dresp1,201,driver,frdisp,,,2,,471

Define Objective Function - SRSS of Driver Response

dresp1,300,srss,frdisp,,,2,RSS,471


WORKSHOP #16

OPTIMIZATION USING FREQUENCYRESPONSE


SOLUTION TO WORKSHOP #16$$ input file to optimize (minimize) the response of a car to a$ rotating imbalance - V68 - June, 1994$ use modal approach - up to 100 hz$SOL 200diag 8CENDTITLE = Sample dynamic analysis modelSUBTITLE = Rotating force due to tire out of balanceLABEL = perform optimization to minimize driver responseset 999 = 358,471DISP(phase) = 999SUBCASE 1

ANALYSIS = MFREQDESSUB = 100 $ constraintsDESOBJ(min) = 300 $ design objective - minimize driver responseDLOAD = 1METHOD = 1FREQ = 1

BEGIN BULKeigrl,1,0.,100.doptprm,desmax,25include ’car.blk’include ’springs.blk’include ’optim1.blk’param,post,0$

$$ DATA RELATED TO FREQUENCY RESPONSE$DLOAD 1 1. 1. 11 1. 12RLOAD1 11 20 111RLOAD1 12 30 40 111DPHASE 40 358 2 90.DAREA 20 358 1 3.DAREA 30 358 2 3.TABLED4 111 0. 1. 0. 100.

0. 39.478 ENDTFREQ1 1 .5 .5 100$ENDDATA


SOLUTION TO WORKSHOP #16 (Cont.)$$ springs.blk$CONM2 3001 1402 1.E8CONM2 3002 1825 1.E8CONM2 3003 1358 1.E8CONM2 3004 1869 1.E8$GRID 1402 159.870 11.0000 -14.3750 13456GRID 1358 20.6116 11.0000 -15.1000 13456GRID 1825 20.6116 11.0000 -54.9000 13456GRID 1869 159.870 11.0000 -55.6250 13456$CELAS2 1001 10000. 402 1 1402 1CROD 1002 1002 402 1402CELAS2 1003 10000. 402 3 1402 3CELAS2 1011 10000. 825 1 1825 1CROD 1012 1001 825 1825CELAS2 1013 10000. 825 3 1825 3CELAS2 1021 10000. 358 1 1358 1CROD 1022 1001 358 1358CELAS2 1023 10000. 358 3 1358 3CELAS2 1031 10000. 869 1 1869 1CROD 1032 1002 869 1869CELAS2 1033 10000. 869 3 1869 3$$ properties for springs$PROD 1001 1000 1000.PROD 1002 1000 800.$ select material so that value of PROD is spring stiffness,$ therefore, E = l = 10.MAT1 1000 10.

$$ add dampers for shock absorbers$ frontcvisc 2011 2001 825 1825cvisc 2021 2001 358 1358$ backcvisc 2001 2002 402 1402cvisc 2031 2002 869 1869$ damper propertiespvisc 2001 10. 0.pvisc 2002 5. 0.$$ end of springs.blk$


SOLUTION TO WORKSHOP #16 (Cont.)$$ file - car.blk$$ MODEL COURTESY LAPCAD ENGINEERING$ CHULA VISTA, CALIFORNIA$GRID 1 79.0000 56.0000-2.00000GRID 2 157.000 40.1000-1.30000..CQUAD4 29 2 22 19 12 21 0.00E+0 0.00E+0CQUAD4 30 2 21 20 11 22 0.00E+0 0.00E+0..PSHELL 1 1 0.025 1PSHELL 2 1 0.200 1..MAT1 11.000E+73.759E+63.300E-12.600E-41.370E-57.000E+1.3$$ end of car.blk$


SOLUTION TO WORKSHOP #16 (Cont.)$$ beginning of optim1.blk$$ data for design sensitivity$$ define design variables$desvar,1,frntdamp,1.,.1,10.desvar,2,reardamp,1.,.2,20.desvar,3,frntstif,1.,.4,2.desvar,4,rearstif,1.,.5,2.5$$ relation between properties and variables$dvprel1,101,pvisc,2001,3,1.,,,,+dv101+dv101,1,10.dvprel1,102,pvisc,2002,3,1.,,,,+dv102+dv102,2,5.dvprel1,103,prod,1001,4,4.,,,,+dv103+dv103,3,10.dvprel1,104,prod,1002,4,4.,,,,+dv104+dv104,4,8.$

$$ select displacement Y at tire and driver seat as$ response quantities$$ maximum tire displacement be < +/-0.5 inches$dconstr,101,200,-.5,.5dresp1,200,disp,frdisp,,,2,,358$$ define driver's seat disp as a response$$$ require that maximum driver displacement be < +/- 0.25 inches$ between .5 and 25 hz$dconstr,102,201,-.25,.25,0.5,25.0dresp1,201,driver,frdisp,,,2,,471$$ combine constraints into set 100$dconadd,100,101,102$$ define objective = minimize srss of response$dresp1,300,rss,frdisp,,,2,rss,471$$ end of optimization input$$ end of optim1.blk$


SOLUTION TO WORKSHOP #16 (Cont.)


RESULTS OF OPTIMIZATION

● Design criteria satisfied in 10 cycles.● Results follow:


OBJECTIVE FUNCTION HISTORY


DESIGN VARIABLES HISTORY


WHEEL RESPONSE

Before OptimizationBefore Optimization

After OptimizationAfter Optimization


DRIVER’S SEAT RESPONSE

Before OptimizationBefore Optimization

After OptimizationAfter Optimization


CONCLUSION

● With minimum effort, the design is modified to satisfy constraints and minimizeselected responses.

● SOL 200 is a valuable design tool for dynamic analysis.


BLANK PAGE


SECTION 20

TEST-ANALYSIS CORRELATION


INTRODUCTION TO TEST-ANALYSISCORRELATION

● MD Nastran results and test data may not match due tomodeling and testing uncertainties.

● Common saying: “No one believes the analytical results(except for the modeler), and everyone believes the testdata (except for the experimentalist).”

● Sources of modeling uncertainties:● Physics of the problem being simulated● Boundary conditions● Material properties● Joint flexibility● “As-built” versus “as-designed”● Damping


INTRODUCTION TO TEST-ANALYSISCORRELATION (CONT.)

● Goals of test-analysis correlation:● Assess the degree of correlation between MD Nastran results and

test data.● Refine the MD Nastran model to match test data.

● Person doing the correlation must understand the testdata, the MD Nastran results, and the uncertainties inboth.


INTEGRATED TEST-ANALYSIS PROCEDURE

● Four phases of an integrated test-analysis procedure:● 1. Pre-test planning (test simulation)● 2. Data acquisition. (Capture raw data, such as,

accelerations.)● 3. Data reduction and analysis. (Process raw data into

quantities of interest, such as, mode shapes.)● 4. Post-test evaluation. (Assess goodness of fit between

test data and MD Nastran results; refine theMD Nastran model to match the test data.)

● The analyst is involved in phases 1 and 4, and theexperimentalist is involved in phases 2 and 3.


PRE-TEST PLANNING

● Create a baseline MD Nastran model to determine optimal excitationand measurement locations. There are two methods for doing this:● Simulation and inspection● Cross-orthogonality check

● Simulation and inspection. Use MD Nastran to simulate the test, andchoose input and output locations that give maximum response.

● Cross-orthogonality check:● From a proposed set of measurement locations create an A-set, use

Guyan reduction, and compute the mode shapes normalizing to a unitmodal mass. Call this set of vectors (as in “test” modes). Output themodes that span the frequency range of interest and the A-set massmatrix.

● Then, remove the A-set, repeat the modal calculation computingresponses for the A-set DOFs for the full model, and output thesemodes (a, as in “analysis” modes).


● In a third MD Nastran run, read both sets of results andcompute:

● If the proposed measurement locations (A-set) areadequate, then the resulting matrix has 1 on the diagonaland 0 as off-diagonal terms. If the off-diagonal terms arenot 0, the proposed measurement locations are notadequate and a new set must be formulated. (In actuality,off-diagonal terms less than 0.05 are acceptable.)

● DMAP alter for pre-test planning, premaca.vxx is on theMD Nastran delivery.

PRE-TEST PLANNING (Cont.)

tT

Maa


● Modal effective mass1 and modal kinetic energy2 calculations can also bemade in MD Nastran to ensure that the test specimen is well understoodbefore testing.

● Test and analytical locations need to align fully (location and coordinatedirection) to facilitate pre-test planning and post-test evaluation. RBARs,MPCs, and alternate output coordinate systems can be used to “line up” thelocations.

● Once the excitation and measurement locations are verified, then the testshould be simulated to ensure that the test specimen is not overstressedduring testing.

1. For the i-th mode, effective mass =where M = mass matrix and Dm = rigid-body vector.

2. For the i-th mode, kinetic energy = . This shows the energydistribution within a mode.

PRE-TEST PLANNING (Cont.)

iT

MDm 2

Mi2


POST-TEST EVALUATION

● Compare MD Nastran results to test data by inspection orcross-orthogonality checks.

● Inspection—graphics● XY plots:

● Plot test data and MD Nastran results together and assess thedegree of correlation in magnitude and frequency content.

● Make sure that the curves plotted together represent the samespatial location and direction.

● Structure plots:● Plot test and analytical mode shapes together and assess the degree

of correlation. (Plots can also be animated.)

● Inspection can also be applied to non-graphical data suchas, comparison of measured and computed resonantfrequencies.


POST-TEST EVALUATION (Cont.)

tTMaaa COR=

● Cross-orthogonality checkQuantitative check on the degree of test-analysis correlation is the cross-orthogonality check defined by

where t = test modesMaa = A-set mass matrixa = analysis modes computed for the A-set

Off-diagonal terms should be less than 10% of the diagonal terms in order tohave a reliable match between test and analysis.Modes may be “switched” which is reflected by large off-diagonal terms inCOR, the correlation matrix.

● If there are discrepancies between test data and analytical results, it may bedesirable to refine the MD Nastran model to get a better match.



y

x1 2 3 4 5 6 7 8 9 10 11

= grid point

= A-set point

● COR example:Consider the 2-D beam model shown below. Assume thataccelerometers are placed on the beam at every other gridpoint. The translational accelerations are measured in thex and y directions.


Grid Point Mode 1 Mode 2 Mode 3 Mode 4 Mode 5 Mode 62x 0 0.15 0 0.45 1 02y 0 0 0.15 0 0 0.34x 0 0.45 0 1 0.95 04y 0.15 0 0.75 0 0 16x 0 0.7 0 0.7 -1 06y 0.4 0 1 0 0 0.058x 0 0.9 0 -0.15 -0.95 08y 0.7 0 0.45 0 0 -0.85

10x 0 1 0 -0.85 1 010y 1 0 -0.75 0 0 0.25

Frequency(Hz) 14.2 50.5 85.2 144.1 245.3 257.3


Test Frequencies and Mode Shapes



MD Nastran Frequencies and Mode Shapes

Test and MD Nastran frequencies are close

Grid Point Mode 1 Mode 2 Mode 3 Mode 4 Mode 5 Mode 62x 0 0.254 0 0.773 0 1.3162y 0.038 0 0.207 0 -0.526 04x 0 0.736 0 1.681 0 1.3164y 0.309 0 1.183 0 -1.759 06x 0 1.147 0 1.203 0 -1.3166y 0.77 0 1.62 0 -0.078 08x 0 1.445 0 -0.266 0 -1.3168y 1.34 0 0.747 0 1.524 0

10x 0 1.602 0 -1.516 0 1.31610y 1.957 0 -1.145 0 -0.474 0

Frequency(Hz)

13.91 49.45 86.38 150.59 244.03 256.5


Post-Test Evaluation (Cont.)

The locationsof the asetpoints are inthe samegeometriclocation as inthe test article

ID PRETEST, DYNOTESSOL, 103compile xread $ALTER ' read.*mxx.*phix'$MATPRN MXX,,,,//$MATPCH MXX,PHIX,,,// $CENDTITLE= Cantilever Beam for Normal ModesSUBTITLE=MODES CASE CONTROLDISP=ALLSPC=1METHOD=100BEGIN BULKPARAM, AUTOSPC, YESGRDSET, , , , , , ,345BAROR, , , , ,0., 1., 0.GRID, 1, , 0.0, 0.0, 0.0GRID, 2, , 1.0, 0.0, 0.0GRID, 3, , 2.0, 0.0, 0.0GRID, 4, , 3.0, 0.0, 0.0GRID, 5, , 4.0, 0.0, 0.0GRID, 6, , 5.0, 0.0, 0.0GRID, 7, , 6.0, 0.0, 0.0GRID, 8, , 7.0, 0.0, 0.0GRID, 9, , 8.0, 0.0, 0.0GRID,10, , 9.0, 0.0, 0.0GRID,11, ,10.0, 0.0, 0.0CBAR, 1, 1, 1, 2CBAR, 2, 1, 2, 3CBAR, 3, 1, 3, 4CBAR, 4, 1, 4, 5CBAR, 5, 1, 5, 6CBAR, 6, 1, 6, 7CBAR, 7, 1, 7, 8CBAR, 8, 1, 8, 9CBAR, 9, 1, 9,10CBAR,10, 1,10,11$Use Master Set DOF, to reduce to Test DOFASET1, 12, 2, 4, 6, 8, 10SPC1, 1, 123456, 1PBAR, 1, 1, 0.01, 0.016, 0.016MAT1, 1, 3.E7, , 0.3, 7.7EIGRL, 100, , 10.E3, 6ENDDATA


POST-TEST EVALUATION (Cont.)ID COR, DYN.NOTESTIME 30SOL 100COMPILE USERDMAP, SOUIN=MSCSOU, NOLIST, NOREFALTER 2 $$----------------------------------------------------------$ DMAP TO COMPUTE CROSS-ORTHOGONALITY$ INPUTS FROM MSC.Nastran RUN: MXX (A-SET MASS)$ PHIX (A-SET MODE SHAPES)$ (PREVIOUS M/N RUN USED MATPCH TO PUNCH DMI ENTRIES)$ INPUT FROM TEST: PHITEST (A-SET MODE SHAPES)$ OUTPUTS: UNITCHK (UNIT MASS CHECK)$ COR (CROSS-ORTHOGONALITY MATRIX)$----------------------------------------------------------$ READ DMI INPUTDMIIN DMI,DMINDX/PHIX,PHITEST,MXX,,,,,,,/ $$ VERIFY INPUT MATRICESMATPRN PHIX,PHITEST,MXX,,// $$ MULTIPLY PHIX(TRANS)*MXX = PHITMASSMPYAD PHIX,MXX,/ PHITMASS /1///$$ MULTIPLY PHITMASS*PHIX = UNITCHKMPYAD PHITMASS,PHIX,/ UNITCHK // $$ PRINT TITLE AND UNITCHKMATPRN UNITCHK,,,,// $MESSAGE // ’CHECK ON UNIT MASS’/ $$ MULTIPLY PHITMASS*PHITEST = CORMPYAD PHITMASS,PHITEST,/ COR // $$ PRINT TITLE AND CORMATPRN COR,,,,// $MESSAGE // ’CROSS-ORTHOGONALITY MATRIX’/ $ENDALTERCENDTITLE = CROSS-ORTHOGONALITY CHECKBEGIN BULKDMI,PHITEST,0,2,1,0,,10,6DMI,PHITEST,1,1,0.,.0,0.,.15,0.,,.40,0.,.70,0.,1.0

DMI,PHITEST,2,1,.15,0.,.45,0.,.70,,0.,.90,0.,1.0,0.

.

. (rest of PHITEST)

.


DMI MXX 0 6 1 0 10 10DMI* MXX 1 1 9.62499976E-02* 3 1.92499999E-02DMI* MXX 2 2 1.38867334E-01* 4 2.67470982E-02 6 -1.37760025E-02* 8 4.32037748E-03 10 3.94638191E-05.. (rest of MXX).DMI PHIX 0 2 1 0 10 6DMI* PHIX 1 1 -1.02694275E-17* 3.79799381E-02 -3.21330419E-17 3.09179097E-01 1.62494801E-17* 7.69559503E-01 -1.60461922E-17 1.34014440E+00 -4.83748987E-17* 1.95740056E+00DMI* PHIX 2 1 2.53673702E-01* -3.71932167E-18 7.36189783E-01 -2.11487186E-17 1.14664245E+00* -1.32509834E-17 1.44485378E+00 -5.04831637E-19 1.60163271E+00* 1.19194476E-17.. (rest of PHIX).ENDDATA




MATRIX COR (GINO NAME 101 ) IS A DB PREC 6 COLUMN X 6 ROW SQUARE MATRIX.COLUMN 1 ROWS 1 THRU 6 --------------------------------------------------ROW

1) 5.1380D-01 6.7188D-19 1.8822D-03 -1.6038D-17 8.3091D-03 -1.5467D-16COLUMN 2 ROWS 1 THRU 6 --------------------------------------------------ROW

1) -1.0247D-17 6.1972D-01 3.1542D-18 -5.7035D-03 -1.0390D-17 -8.8287D-09COLUMN 3 ROWS 1 THRU 6 --------------------------------------------------ROW

1) -1.5467D-02 -6.3725D-18 6.3014D-01 -2.7673D-17 -4.6145D-03 9.5503D-17COLUMN 4 ROWS 1 THRU 6 --------------------------------------------------ROW

1) 2.5192D-18 1.1390D-02 8.1214D-17 5.8007D-01 -1.0636D-16 7.5993D-03COLUMN 5 ROWS 1 THRU 6 --------------------------------------------------ROW

1) -1.0452D-17 5.3899D-03 -2.1283D-17 -1.3447D-02 1.2668D-14 7.4474D-01COLUMN 6 ROWS 1 THRU 6 --------------------------------------------------ROW

1) -3.2927D-03 -2.5409D-18 8.7221D-03 -2.3220D-16 -5.6143D-01 9.6348D-15THE NUMBER OF NON-ZERO TERMS IN THE DENSEST COLUMN = 6THE DENSITY OF THIS MATRIX IS 100.00 PERCENT.^^^CROSS-ORTHOGONALITY MATRIX


SUMMARY

1. COR matrix shows good agreement in mode shapes. (Off diagonalare small compared to the diagonal terms.)

2. Mode 5 (test) is Mode 6 (analysis) and vice versa. This is shown bylarge off-diagonal terms in 5,6 and 6,5.

3. The magnitude of diagonal terms are not 1.0 because the test modesare not normalized to unit modal mass. (They are normalized to amaximum component of 1.0.)

4. Some effort is required to reformat test data for DMI input.5. For a large number of modes, additional DMAP may be written to

simplify the cross-correlation output.6. DMAP alter for test-analysis cross-orthogonality, postmaca.vxx, is on

the MD Nastran delivery.


MODEL REFINEMENT

● There are three ways to update an MD Nastran model to match testdata:● 1. Using brute force● 2. Using the sensitivity matrix● 3. Using design optimization

● All methods update MD Nastran model parameters such as I and A forBARs and t for QUADs. Base flexibility can also be a parameter if itwas explicitly modeled (such as with ELASs).

● Brute force:● Make model changes based on inspection of the results and make an

educated “guess” as to the type of changes necessary.● After the model is updated, another analysis is made and the MD

Nastran results are again compared to test data to see if the match isbetter.

● If the match is not better, then make more changes and repeat.


● Sensitivity matrix:● Sensitivity is the gradient of response with respect to model parameter. “Responses”

are MD Nastran-computed results, and “parameters” are property values.

● Each term in the sensitivity matrix S is given by

● where Ri is the i-th response and Pj is the j-th parameter.● The greater the magnitude Sij, the greater the response sensitivity.● SOL 200 in MD Nastran computes the sensitivity matrix.● S can be used by inspection to indicate which parameters need to be changed to

change the response.

MODEL REFINEMENT (Cont.)

Response

Po

Parameter

S P P0RP-------= =

S ij Ri P j=



● S can also be output and used in a least-squares sense tominimize the difference between test and analysis asfollows:

Pn = Po + (STS)-1ST(Rt - Ra)where Pn = updated parameters for test-analysis match

Po = parameters from original modelS = sensitivity matrixRt = test responseRa = analytical results


● Design optimization:● Implemented in SOL 200● Want to minimize the difference between test and analysis but with a model that changes least

from the baseline model● Use DEQATN to write an objective function that is the weighted difference between test data

and analytical results.

● where RTi = test data● PF = final parameters● RAi = analysis results● POj = original parameters● WRi = test weighting● WPj = parameter weighting● wt = weighting for test as a whole● wp = weighting for model parameters as a whole

● SOL 200 minimizes the objective (E) subject to design constraints.


E wt WR iRT i RA i– 2

i 1=

test data

= wp WPjPFj POj– 2

j 1=+

model params



● Example:● Disk drive enclosure with 1406 grid points, 1354 plate elements, four

design variables (plate thicknesses), and four measured flexible modes● Minimize the difference between computed and measured resonant

frequencies.



Test Baseline Final346 234 3461307 892 12951567 1165 14601678 1267 1695

Frequencies (Hz)

Baseline Lower Bound Upper Bound Final.08 .05 .125 .125.12 .05 .20 .075.10 .05 .15 .084.20 .10 .30 .227

Plate Thicknesses (in.)

● Example (Cont.)● Test data weighted 100 times more than the baseline model parameters.● Put upper and lower bounds on allowable plate thicknesses.● Five iterations to converge● Results:



Plate Thickness versus Design Cycle

Error versus Design Cycle



● Can also use SOL 200 to match frequency response testdata.

● Can match static results as well.● For best results, use as much test data as feasible

(including the weight of the structure).● Also, limit design variables to only the parameters that are

truly uncertain.


REFERENCES

Blakely, K. and Dobbs, M., “Integrated System Identification: The Union of Testing and Analysis,”Proceedings First International Modal Analysis Conference, November 1982.

Kientzy, D., Richardson, M. and Blakely, K., “Using Finite Element Data to Set Up Modal Tests,”Sound and Vibration, June 1989.

Blakely, K., “Updating MSC.Nastran Models to Match Test Data,” Proceedings MSC World Users’Conference, March 1991.

Blakely, K. and Rose, T., “Cross-Orthogonality Calculations for Pre-Test Planning and ModelVerification,” Proceedings MSC World Users’ Conference, May 1993.

Blakely, K., “Matching Frequency Response Test Data with MSC.Nastran,” Proceedings MSC WorldUsers’ Conference, June 1994.

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