nonlinear analysis using msc.nastran

7/21/2019 Nonlinear Analysis Using MSC.nastran

http://slidepdf.com/reader/full/nonlinear-analysis-using-mscnastran 1/783

MSC.Software Corporation2 MacArthur Place

Santa Ana, CA 92707, USATel: (714) 540-8900Fax: (714) 784-4056

Web: http://www.mscsoftware.com

United States

MSC.Patran Support

Tel: 1-800-732-7284

Fax: (714) 979-2990

Tokyo, Japan

Tel: 81-3-3505-0266

Fax: 81-3-3505-0914

Munich, Germany

Tel: (+49)-89-43 19 87 0

Fax: (+49)-89-43 61 716

Nonlinear Analysis Using MSC.Nastran

January 2004

NAS103 Course Notes

Part Number: NA*V2004*Z*Z*Z*SM-NAS103-NT1



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 2003 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 and MSC. are registered trademarks and service marks of MSC.Software Corporation. NASTRAN is a registered

trademark of the National Aeronautics and Space Administration. MSC.Nastran is an enhanced proprietary versiondeveloped and maintained by MSC.Software Corporation. MSC.Marc, MSC.Marc Mentat, MSC.Dytran, MSC.Patran,MSC.Fatigue, MSC.Laminate Modeler, and MSC.Mvision are all trademarks of MSC.Software Corporation.

All other trademarks are the property of their respective owners.

NAS103 Course Director:

[email protected]



DAY 1INTRODUCTION

NONLINEAR ANALYSIS STRATEGY

GEOMETRIC NONLINEAR ANALYSIS

WORKSHOPS

DAY 2BUCKLING ANALYSISMATERIAL NONLINEAR ANALYSIS

WORKSHOPS

COURSE OUTLINE

DAY 3NONLINEAR ELEMENTSNONLINEAR TRANSIENT ANALYSISWORKSHOPS

DAY 4NONLINEAR ANALYSIS WITH SUPERELEMENTS

SPECIAL TOPICSNONLINEAR ANALYSIS WITH SOL 600WORKSHOPS



S1-1NAS 103, Section 1, December 2003

SECTION 1

INTRODUCTION




TABLE OF CONTENTSPage

Purpose 1-4

Review Of Finite Element Analysis 1-5Linear Versus Nonlinear Structural Analysis 1-8

Nonlinear Analysis Capabilities 1-11

Basic Of A Nonlinear Solution Strategy 1-15

User Inter Face For Nonlinear Analysis 1-18Summary 1-20




PURPOSE To understand the following:

Differences between linear and nonlinear analysis.

Different types of nonlinearity. Nonlinear analysis capabilities available in MSC.NASTRAN.

Basics of a nonlinear solution strategy.

Basic user interface for nonlinear analysis.




REVIEW OF FINITE ELEMENT ANALYSIS A solution must satisfy:

1. Kinematics

eU =bg

T

beT T g U

g U = eg T eU Element

DeformationDisplacement

TransformationMatrix

GlobalDegrees of

Freedom




REVIEW OF FINITE ELEMENT ANALYSIS2. Element Compatibility and Constitute Relationships

a)

b)

ε = B eU

ElementStrains

Strain DeformationMatrix

ElementDeformations

σ = D ε Element

Stresses

Stress-Strain

Relationship

Element

Strains

V ∫ T B D B dV ee K =

Element

Stiffness

e F = ee K ∫=V

T

e dV BU σ

ElementForces

ElementStiffness

ElementDeformations




REVIEW OF FINITE ELEMENT ANALYSIS3. Equilibrium

4. Boundary Conditions

P = T

eg T Σ e F

External LoadVector

Force TransformationMatrix

ElementForces

α = g U Single and multipoint constraints




LINEAR VERSUS NONLINEAR STRUCTURAL

ANALYSIS Linear Analysis

Kinematic relationship is linear, and displacements are small.

Element compatibility and constitutive relationships are linear, and thestiffness matrix does not change. There is no yielding, and the strainsare small.

The equilibrium is satisfied in undeformed configuration.

Boundary conditions do not change.

The force transformation matrix is the transpose of the displacementtransformation matrix.

It follows that: Loads are independent of deformation.

Displacements are directly proportional to the loads. Results for different loads can be superimposed.





ANALYSIS Nonlinear Analysis

Geometric nonlinear analysis:

The kinematic relationship is nonlinear. The displacements and rotations arelarge. Equilibrium is satisfied in deformed configuration.

Follower forces: Loads are a function of displacements.

Large strain analysis: The element strains are nonlinear function of element deformations.

Material nonlinear analysis: Element constitutive relationship is nonlinear. Element may yield.

Element forces are no longer equal to stiffness times displacements (Kee •

Ue). Buckling analysis:

Force transformation matrix is not the transpose of displacementtransformation matrix. The equilibrium is satisfied in the perturbedconfiguration.





ANALYSIS Contact (interface) analysis:

Gap closure and opening, and relative sliding of different components.

Boundary conditions may change.

It follows that: Displacements are not directly proportional to the loads.

Results for different loads cannot be superimposed.




NONLINEAR ANALYSIS CAPABILITIES Geometric Nonlinearity

Large displacements and rotations, i.e., the displacement transformation

matrix is no longer constant. Both compatibility and equilibrium are satisfied in a deformed

configuration.

Effects of initial stress (geometric or differential stiffness) are included.

The follower force effect can be included

Examples: cable net, thin shells, tires, water hose, etc.

User interface: PARAM,LGDISPFollower Forces: FORCE1, FORCE2, MOMENT1,

MOMENT2, PLOAD, PLOAD2,

PLOAD4, PLOADX1, andRFORCE




NONLINEAR ANALYSIS CAPABILITIES Material Nonlinearity

Element stiffness matrix is not constant.

Two reasons for variable stiffness matrix:1. Stress-strain relationship is nonlinear (i.e., matrix D changes), but strains are

small (i.e., matrix B is linear). Example: Yielding structure (nonlinear elastic or plastic), creep

User Interface: MATS1 and CREEP Bulk Data entries

2. Strains are large (i.e., strain deformation matrix B is nonlinear). In general,stress-strain relationships and displacement transformation relationships arealso nonlinear.

Example: Rubber materialsUser Interface: MATHP, PLPLANE, and PLSOLID Bulk Data entries

Temperature-Dependent Material Properties Linear elastic materials (MATT1, MATT2, and MATT9).

Nonlinear elastic materials (MATS1, TABELS1, and TABLEST). Note: Nonlinear elastic composite materials cannot be temperature

dependent.




NONLINEAR ANALYSIS CAPABILITIES

Buckling Analysis Force transformation matrix is no longer a transpose of the

displacement transformation matrix. Equilibrium is satisfied in theperturbed configuration.

Example: Linear or nonlinear buckling User Interface: EIGB Bulk Data entry.

METHOD Case Control command.SOL 105 (linear buckling).PARAM, BUCKLE in SOL 106 (nonlinear buckling).

Contact (Interface) Analysis Treated by gap and 3-D slideline contact. Example: O-rings, rubber springs in the auto and aerospace

industry, auto or bicycle brakes, and rubber seals indisc brakes, etc.

User Interface: CGAP, PGAP, BCONP, BLSEG, BFRIC, BWIDTH,BOUTPUT Bulk Data entries.




NONLINEAR ANALYSIS CAPABILITIES

Boundary Changes User Interface: SPC, SPCD, and MPC Bulk Data entries and Case

Control commands.

Note: All different types of nonlinearities can be

combined together.




BASICS OF A NONLINEAR SOLUTION

STRATEGY A strategy is required to solve nonlinear problems.

A nonlinear strategy: Advances in increments (example: two load increments). Requires iterations for each increment (example: 5 iterations for the first

increment).

A solution is obtained when the convergence criteria is satisfied

(example: negligibly small unbalanced load).





STRATEGY Example:

Displacement, u

Unbalanced

Loads

P r e

d i c t o r

I t

e r a t i o n s

P2

P1

∆P

∆P

∆u1 ∆u2

R1

R2R3

R4

Load, PPredictor

Iterations





STRATEGY In MSC.NASTRAN

A number of different advancing schemes are available.

A number of different iteration schemes are available. A number of different convergence criteria are available.

User interface: NLPARM Solution strategy for nonlinear static analysis.

SPCD, SPC Displacement increments for nonlinear staticanalysis.

NLPCI Arc length increments for nonlinear static analysis.

TSTEPNL Solution strategy for nonlinear transient analysis.




USER INTERFACE FOR NONLINEAR

ANALYSIS Compatible with linear analysis

Analysis types Nonlinear static analysis: SOL 106 Quasi-static (creep) analysis: SOL 106

Linear buckling analysis: SOL 105

Nonlinear buckling analysis: SOL 106 (PARAM,BUCKLE)

Nonlinear transient response analysis: SOL 129

Subcase structure Allows changes in loads, boundary conditions, and methods.

Allows changes in output requests.




USER INTERFACE FOR NONLINEAR

ANALYSIS Bulk Data classification

Geometric data

Element data Material data

Boundary conditions

Loads and enforced motion Selectable in Subcases

Solution strategy




SUMMARY

In nonlinear analysis: Any one or more of the following relationship may be nonlinear:

Kinematics Element compatibility Constitutive relationship Equilibrium

Loads may be functions of displacements

Opening and closing of different components Boundary conditions may change

Nonlinear Solution Sequences: SOL 106: Nonlinear static analysis (geometric, material, large

strain, buckling, surface contact, and constraint

changes). SOL 129: Nonlinear transient analysis (geometric, material,

large strain, and surface contact). No constraintchanges are allowed.




SUMMARY

Basic User Interface: Solution strategy:

Solution strategy nonlinear static analysis. NLPARM Arc length increments for nonlinear static analysis. NLPCI

Solution strategy nonlinear transient analysis. TSTEPNL

Displacement-increment analysis. SPCD, SPC

Nonlinear materials: Nonlinear elastic and plastic. MATS1

Creep materials. CREEP

Hyper elastic (rubber-like) materials. MATHP

Temperature-dependent elastic materials. MATT1, MATT2, MATT9 Temperature-dependent

nonlinear elastic materials. TABLEST, TABLES1




SUMMARY

Geometric nonlinear: PARAM, LGDISP.

Follower forces: FORCE1, FORCE2, MOMENT1,MOMENT2, PLOAD, PLOAD2,

PLOADX1, and RFORCE. Nonlinear buckling analysis: PARAM, BUCKLE, in SOL 106.

Contact (interface): gap and 3-D slideline contact.

Boundary changes: SPC, SPCD, and MPC.




SECTION 2

NONLINEAR STATIC ANALYSISSTRATEGIES




TABLE OF CONTENTS

Page

Overview Of Nonlinear Analysis Methods 2-5

MSC.Nastran Nonlinear Static Analysis Flowchart (Simplified) 2-7

Classical (Standard) Newton-Raphson (NR) Method 2-8

Summary Of Basic Tasks In Nonlinear Analysis 2-13

Nonlinear Analysis Strategies In MSC.Nastran 2-14

Advancing Schemes In MSC.Nastran 2-15

Stiffness Update Schemes In MSC.Nastran 2-30

One-dimensional Example For Different Stiffness Update

Schemes 2-35

Displacement Prediction Schemes 2-38

Line Search 2-39Convergence Criteria 2-44




TABLE OF CONTENTS

Page

Special Logics 2-50

Restarts 2-53

Output For Solution Strategies 2-61

Result Output 2-66

Some Heuristic Observations 2-67

Hints And Recommendations 2-68

NLPARM Bulk Data Entry 2-69

Summary 2-70

Workshop Problems 2-73

Solution For Workshop Problem One 2-76

Solution For Workshop Problem Two 2-77




OVERVIEW OF NONLINEAR ANALYSIS

METHODS Concept

where Steps 1 through 5 = advancing (predicting) phase.Steps 6 through 9 = correcting (iterating) phase.

∆P1 need not equal ∆P2.

K0 need not equal K1.

2. K0 - Estimate of

Tangent Stiffness

6. K1 - Estimate

of TangentStiffness

5. R1 - Unbalanced

Load

4. F1 - Element

Force 8. F2 - ElementForce

9. R2 - Unbalanced

Load

Displacement, u

7. ∆U1 - Displacement

Correction

1. Load

Increment∆P1

∆P2

P2

P1

Load, P

3. ∆U0 - Displacement

Predictions




OVERVIEW OF NONLINEAR ANALYSIS

METHODS Algorithm

1. Determine an increment (e.g., load, displacement, or arc length) to move forwardon the equilibrium path.

2. Determine an estimate of a tangent stiffness matrix.3. Determine the displacement increment to move forward, generally by solving

equilibrium equations.4. Calculate the element resisting forces.5. Calculate the unbalanced load and check for convergence. If converged, go to

Step 1.

If not converged, continue as follows:6. Determine an estimate of tangent stiffness matrix.7. Determine the displacement increment due to the unbalanced load.8. Calculate the element resisting forces.9.

Calculate the unbalanced load and check for convergence. If converged, go toStep 1. If not converged, go to Step 6.

Steps 1 through 5 are called the advancing phase or predictingphase.

Steps 6 through 9 are called the correcting phase or iterating phase.




MSC.NASTRAN NONLINEAR STATIC

ANALYSIS FLOWCHART (SIMPLIFIED)• START with a converged solution u 0 and a corresponding load P0

Load step loop l = 1, l step

• Determine new load

• Start with converged deformation

Iteration loop i = 1, i max

• Calculate internal forces

• Calculate residual force:

• Update

• Solve equilibrium eqn:

• Update deformations

and satisfy boundary conditions

• Check convergence Converged

Continue iteration loop

Divergence occurred

• Reset load step counter

and try smaller load step

• or quit

Continue load step loop

• STOP

Yes

No

l l l P P P ∆+= −1

10 −= l l uu

)( 11 −− = i

l

i u F F

11 −− −= i

l

i F P R

1−i

K 111 −−− =∆ iii

Ru K

11 −− ∆+= ii

l

i

l uuu

1−= l l




CLASSICAL (STANDARD) NEWTON-RAPHSON

(NR) METHOD Advance forward by constant and positive load increments.

Tangent stiffness is formed at every iteration.

Displacement is predicted and corrected by solving equilibriumequations.





(NR) METHOD Mathematics

We want to solve:

Let u* be an approximation to the solution of R(u) = 0.

Taylor Series

where

K is called the tangent stiffness matrix.

K may not relate to an equilibrium state.

For loads independent of displacement:

R(u) = P(u) − F(u) = 0

Nonlinear Function of u

Kij

Ri

∂u j

-------- u*( )–=

Ki j

δ Fi

δu j

-------- u *( )=

*)(*)(*)(*)(*)(*)()( u K uuu Ruu

Ruuu Ru R T −−=∂−+= &





(NR) METHOD Algorithm

Solve:

Solve:

Solve:

until reaching convergence

Note: At each iteration, tangent K is computed from the currentelement state.

K u0

( )∆u0

P F0

∆R u0

( )==

u1 u0 ∆u0+=

K u1

( ) ∆ u1

R u1

( )=

u2

u1

∆u1

+=

K u 2( ) ∆ u 2 R u 2( )=

u3

u2

∆u2

+=

.

.

.





(NR) METHOD Weaknesses

1. Constant predetermined positive load increments cannot trace theunstable or post-buckling behavior.

Displacement

Load

Cannot traceequilibriumpath betweenA and B

A B





(NR) METHOD2. No convergence if total applied load is greater than the structure

strength.

3. Computation of tangent stiffness at each iteration is expensive and

unnecessary when the solution is close to convergence.4. Path-dependent state determination. Use of nonconverged reference

state may cause the inelastic material response to differ from the trueresponse.

5. Special logic is necessary if solution does not converge.

No Solution

Displacement

P

∆P4

∆P3

∆P2

∆P1




SUMMARY OF BASIC TASKS IN NONLINEAR

ANALYSIS1. Determination of an increment to advance forward on

the equilibrium path.

2. Determination of an estimate of tangent stiffness matrix.3. Prediction of the displacement for the increment.

4. Determination of the element state: deformation,

resisting forces, etc.5. Convergence check. Calculation of unbalanced forces

and satisfaction of convergence criteria.




NONLINEAR ANALYSIS STRATEGIES IN

MSC.NASTRAN Different schemes are available for advancing forward on

the equilibrium path.

Different schemes are available for estimating thetangent stiffness.

Different schemes are available for predicting thedisplacement increment.

Different convergence criteria are available. Note: Users can select different solution strategies based on

different combination of schemes selected for different tasks




ADVANCING SCHEMES IN MSC.NASTRAN

Constant load increments

Constant displacement increments

Arc-length increments





Constant Load Increment

Field ContentsID Identification number. (Integer > 0).NINC Number of increments. (0 < Integer < 1000).

Example:

SUBCASE = 10NLPARM = 10LOAD = 10BEGIN BULK

NLPARM,10,5FORCE,10,1,,100.,1.,0.,0.FORCE,10,3,,300.,0.,1.,0.MOMENT,10,6,,100.,0.,0.,1.

..

NINCIDNLPARM

10987654321

.

.





Displacement Increment Specify constant displacement for selected degrees of freedom.

Generally, specify displacement increment for one degree of freedom. May specify displacement increment for a set of degrees of freedom for

a rigid body movement.

Need to have some idea of the problem to avoid specifying aninconsistent displacement increment.

The value of displacement is a measure from the undeformed position. Displacement is processed incrementally in the subcase.

F

Displacement Increment

Subcase 1(Inc = 1)

Subcase 2

(Inc = 4)





May need tighter tolerances than the default for convergence criteria.

May be used in combination with load increment.

Cannot be used in combination with arc-length increments.

Specified in the Bulk Data entry SPCD or SPC.

If specified in Bulk Data entry SPCD: Selected by LOAD in Case Control.

SPCD cannot be combined in the Bulk Data LOAD.

The degree of freedom with the SPCD should be defined in theS-set (SPC).

Appropriate S-set should be selected in the subcase.





ID SLINE2U,V68TIME 300 $ FOR VAXSOL 106CEND$TITLE = SLINE2U: SYMMETRIC ELASTIC PUNCH WITH FRICTION$BOUTPUT = ALLDISP = ALLSUBCASE 1 $ VERTICAL LOADNLPARM = 420LOAD = 1$SUBCASE 2NLPARM = 120 $ DISPLACEMENT TO THE RIGHTLOAD = 2SPC = 20$BEGIN BULK

$PARAM,POST,0$$ GEOMETRY$GRID,100,,0.,0.,0.,,123456 $=,*1,,*(10.),== $=9 $..$$ LOAD FOR SUBCASE 2 : RIGHT HORIZONTAL DISPLACEMENT$FORCE,2,400,,-1000.,0.,1.,0.$FORCE,2,401,,-2000.,0.,1.,0.$FORCE,2,402,,-1000.,0.,1.,0.$SPCD,2,302,1,44.,301,1,44.0SPCD,2,300,1,44.0SPC1,20,1,300,301,302$$ NONLINEAR SOLUTION STRATEGY: AUTO METHOD WITH DEFAULTS$NLPARM,420,44,,AUTO,,,PW,YES,+NLP42 $+NLP42,,1.E-6,1.E-10 $ENDDATA

Displacement Increment Example

Note: May need tighter tolerances forconvergence criteria.

Displacement Increment entries

Note: May need tighter tolerancesfor convergence criteria

Case Control Commands for





Load increment for a specified arc length is larger for a stiff structurethan for a flexible structure. It is the opposite for displacementincrement.

P

Stiff

Flexible

∆µs = Load Increment for

Stiff Structure

∆µf = Load Increment for

Flexible Structure

∆l

∆l

∆ µ f ∆

µ s

µ





Contribution of load and displacement to arc length is unit dependent.

Use a scale factor w to control the contribution of the load term, i.e., arclength constraint becomes ∆l2 = ω2 ∆µ2 + ∆uT ∆u.

µ

u u

µ

Crisfield Method in Terms

of Combined Variables

Crisfield Method in Terms

of Displacements

w = 1(SCALE)

Circle

∆l

w = 0(SCALE)Cylinder

∆l





Based on numerical experience, Crisfield recommends that the loadterm not be included.

Becomes equivalent to displacement increment (Euclidian norm of

displacement increments), if the load term is not included. Local nonlinearities tend to get diluted for large degrees of freedom.

Need to solve the quadratic equation to enforce the arc lengthconstraint.

Riks method avoids the solution of the quadratic equation by enforcing anormal plane constraint.

µ1 2

3

∆µ

µ0

u





Modified Riks method continues to change the normal plane constraintwith every iteration.

∆µ

µ

u





Arc Length Increment - User Interface NLPCI combined with NLPARM

NLPCI Bulk Data entry

Example:

Field ContentsID Identification number of an associated NLPARM entry.

(Integer > 0).

TYPE Constraint type. (Character: "CRIS", "RIKS", or "MRIKS";Default = "CRIS").

MINALR Minimum allowable arc-length adjustment ratio betweenincrements for the adaptive arc-length method. (0.0 < Real <1.0; Default = 0.25).

MXINCDESITERSCALEMAXALRMINALRTYPEIDNLPCL

10987654321

101211CRIS10NLPCL

ADVANCING SCHEMES IN MSC NASTRAN





Field Contents

MAXALR Maximum allowable arc-length adjustment ratio betweenincrements for the adaptive arc-length method. (Real > 1.0;

Default = 4.0).SCALE Scale factor (w) for controlling the loading contribution in the

arc-length constraint. (Real > 0.0; Default = 0.0)

DESITER Desired number of iterations for convergence to be used forthe adaptive arc-length adjustment. (Integer > 0; Default =12).

MXINC Maximum number of controlled increment steps allowedwithin a subcase. (Integer > 0; Default = 20).






NLPARM Bulk Data Entry

Field Contents

MAXR Maximum ratio for the adjusted arc-length increment relativeto the initial value. (1.0 ≤ MAXR ≤ 40.0; Default = 20.0).

Example: NLPARM = 20

BEGIN BULKNLPARM,20,10NLPCI,20,CRIS,1.,1.,,,12,40ENDDATA

MAXR

IDNLPARM

10987654321






Option to specify either Crisfield, Riks, or modified Riksmethods.

Must be used in combination with a load increment, Initial arc length is based on the load increment specified

in NLPARM Bulk Data entry.

Can vary arc length based on the number of iterations.

Recommendation: Use constant arc length increments.

Disallowed with displacement increments (SPCD).

Line search* is not operational with arc lengthincrements.

Not allowed for creep analysis**Note: Will be discussed later on.

STIFFNESS UPDATE SCHEMES IN





MSC.NASTRAN At every iteration (NR method)

At every k-th iteration (modified NR method)

Based on the rate of convergence. Logic is hardwaredependent. For the same problem, the solution pathmay be different depending on the hardware.

On non-convergence or divergence

Quasi-Newton stiffness updates

MAXQN

MAXITERKSTEPKMETHODIDNLPARM

10987654321






MSC.NASTRANField Contents

KMETHOD Method for controlling stiffness updates. (Character ="AUTO", "ITER", or "SEMI"; Default = "AUTO").

KSTEP Number of iterations before the stiffness update for ITERmethod. (Integer > 1; Default = 5).

MAXITER Limit on number of iterations for each load increment.(Integer > 0; Default = 25).

MAXQN Maximum number of quasi-Newton correction vectors to besaved on the database. (Integer > 0; Default = MAXITER).





S SS U SC S

MSC.NASTRAN Quasi-Newton (QN) Stiffness Updates – Concept

Full Newton-Raphson is very expensive.

Modified Newton-Raphson converges slowly, if at all.

Hence we seek a simple but efficient way to update (rather than recompute) thestiffness, after each iteration.

Modified stiffness matrix should be a secant stiffness matrix for thedisplacements calculated in the previous iterations.

Modified stiffness should preserve symmetry and be positive definite.

Displacement increment using modified stiffness should be inexpensive tocalculate.





MSC.NASTRAN Consider a single degree of freedom

Km

(Modified)

F,P

K

Spring in the Direction ofUnbalanced Force

u i-1 ui

P

Displacement

K Kt

Ri = P – Fi

Ri–1 – Ri = Fi – Fi–1 = γi

P – Fi–1 = Ri–1

∆ui–1

Secant Stiffness Km

Ri 1– Ri–

∆ui 1–

------------------------- KRi

∆ui 1–

---------------- K Ks–=–= = =

Fi–1





MSC.NASTRAN Multi-Degrees of Freedom

Define

Equivalent to adding flexibility in the direction of unbalanced force.

Modification satisfies the secant stiffness criteria, i.e.,∆K ∆Ui-1 = Ri.

Modification preserves symmetry. Inverse of modified stiffness is inexpensive to calculate.

Km

KR

iR

i

T

Ri

T∆u

i 1–

------------------------; Ks

Ri

TR

i

Ri

T∆u

i 1–

------------------------ ; us

Ri

Ri

TR

i( )

1 2 ⁄ ---------------------------==–=

Direction of

UnbalancedForce

where = projection of Ri along Ri

= projection of ∆ui-1 along Ri (may be 0)

Ri

TR

i

RiT∆u

i 1–

ONE-DIMENSIONAL EXAMPLE FOR




DIFFERENT STIFFNESS UPDATE SCHEMES

F u( ) P=

u2

– 6u 8=+

u6

2---

6

2---

2

8–±=

u1 2=

u2 4=

1. 2. 3. 4.

U

Newton Method Illustrated

k = 2.5

P,F

P = 8

5

U0 U1 U2

k = 4

F1

F0

F2

Exact Solution





DIFFERENT STIFFNESS UPDATE SCHEMES Convergence Criteria: R < 0.01

Newton Method

Note: Quadratic rate of convergence

Modified Newton Method

Note: Linear rate of convergence.

Iteration Initial U Initial R K U Final U F Final R

1 1.0000 3.0000 4.0000 0.7500 1.7500 7.4375 0.5625

2 1.7500 0.5625 2.5000 0.2250 1.9750 7.9494 0.0506

3 1.9750 0.0506 2.0500 0.0247 1.9997 7.9994 0.0006


1 1.0000 3.0000 4.0000 0.7500 1.7500 7.4375 0.5625

2 1.7500 0.5625 4.0000 0.1406 1.8906 7.7692 0.2308

3 1.8906 0.2308 4.0000 0.0577 1.9483 7.8939 0.1061

4 1.9483 0.1061 4.0000 0.0265 1.9748 7.9490 0.0510

5 1.9748 0.0510 4.0000 0.0128 1.9876 7.9750 0.02506 1.9876 0.0250 4.0000 0.0063 1.9939 7.9878 0.0122

7 1.9939 0.0122 4.0000 0.0031 1.9970 7.9940 0.0060





DIFFERENT STIFFNESS UPDATE SCHEMES Modified Newton Method with QN Update

Ki = Ki-1 - Ri-1 /∆Ui-1


1 1.0000 3.0000 4.0000 0.75 1.75 7.4375 0.5625

2 1.7500 0.5625 3.2500 0.1731 1.9231 7.8403 0.1597

3 1.9231 0.1597 2.3274 0.0686 1.9917 7.9833 0.01674 1.9917 0.0167 2.0840 0.008 1.9997 7.9994 0.0006

DISPLACEMENT PREDICTION SCHEMES




DISPLACEMENT PREDICTION SCHEMES

Solution of equilibrium equation

Line search method

LINE SEARCH




LINE SEARCH

Concept Improves displacement increment calculated from the equilibrium

equation.

Displacement increment calculated from the equilibrium equation is notnecessarily the best estimate of the equilibrium state. Seek a multiple of displacement increment (a) that minimizes a measure

of work done by unbalanced forces. Applicable for each iteration.

Effective when the modified Newton method is used. Effective for contact problems. Phase 1: Seek upper and lower values of a that bound zero unbalance.

Calculate a measure of external work done by unbalanced loads for thebeginning of iteration (α0 = 0) and for the calculated displacement increment

(α1 = 1). If the unbalances at α0 and α1 are of opposite signs, the zero is bounded and

then go to phase 2. If the zero is not bounded, keep doubling ∆U until the zero is bounded or the

number of line searches allowed is performed.

LINE SEARCH




LINE SEARCH

Phase 2: Find a to minimize the unbalance. Let αk and αk – 1 be the scalar multiplies that bound the zero unbalance.

Based on the values of αk and αk – 1 , linearly interpolate to get a new value

of α. Evaluate the new unbalance at new a and keep interpolating between the

two a with opposite signs until the unbalance is less than the specifiedproportion of

or

the number of line searches allowed is performed.

αn( ) α0( )<

LINE SEARCH




LINE SEARCH

User Interface

Field Contents

MAXLS Maximum number of line searches allowed for each iteration.(Integer > 0; Default = 4)

LSTOL Line search tolerance. (0.01 ≤ Real ≤ 0.9; Default = 0.5)

LSTOLMAXLS

IDNLPARM

10987654321

LINE SEARCH




LINE SEARCH

Implementation Search for the local minimum point in

where feasible direction

Limit consecutive searches based on error:

where i = iteration counter

k = line search counter

Divergence if Ek

1 for α 1=>

Ek ∆u

i 1

Rk

i

∆ui 1–

Ri 1–

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

u

i

u

i 1

α∆u

i 1

+=

∆ui 1

K1R

i 1=

LINE SEARCH




S C

Linear interpolation if

−LSTOL

LSTOL

+1

−1

E

No LineSearch

Divergence

Doubling Scheme

Line Search

α

Ek

LSTOL<

αk 1+ αk

αk

αk 1–

Ek Ek 1––--------------------------Ek –=

CONVERGENCE CRITERIA




Criteria should: Be satisfied for the linear case at all times.

Be independent of structural units.

Be reliable and consistent; no cancellation errors.

Be independent of structural characteristics.

Be applicable to all loading cases

Have smooth transition after K updates and loading changes.

Be dimensionless.

Three criteria: Load (Ep) Work (Ew)

Displacement (Eu)

P = 0; ∆P = 0 (creep)





Load Criteria

Where

Note: If no loads are applied in more than two consecutivesubcases (creep) Pg

- i = 0, apply a dummy load.

R l

i 1

L--- ABS

l 1=

L

∑ ul

iR l

•( )=

pl

i 1

G---- ABS

g 1=

G

∑ ug

i pg

*•( )=

pg*

∆ p p +=

Increment for

(From Previous Subcase)

Moment or Load Sensitive

Nonmoment Load Sensitive

E p

i R l

pgi----------

R

P------ R u→ →=





Work Criteria

Where

Note: If no loads are applied in more than two consecutivesubcases (creep) Pg

- i = 0, apply a dummy load.

LineSearch

Ew

i

α Rl

i

∆ul

i 1–

•

pg

i-------------------------------------------------=

R l

i

∆Ul

i 1 –

• 1

L--- ABS

l 1=

L

∑ R l

i∆U

l

i 1 – •( )=





Displacement Criteria

Where

ITER,1 method Eui is not effective.

Eu

i qi

1 qi

–-------------

∆ul

i 1

ul

i---------------------=

ul

i 1 –

=

1

L--- ABS K ll u l

i

•( )l 1=

L

∑

qi = 2

3---

∆ul

i 1 –

∆ul

i 1 – ---------------------

1

3---qi 1 – +

qi = MAX qi .99;[ ] 1<

∆ul

i 1 – = α

L--- ABS K l l ∆u

l

i 1 – •( )

l 1=

L

∑





Convergence tolerances:

Loose tolerances cause inaccuracy and difficulties in subsequent steps. Tight tolerances cause a waste of computing resources.

Realistic Eu < 10 –3, Ep < 10 –3 and Ew < 10-7





User Interface Tested at every iteration after the line search

Field ContentsCONV Flags to select convergence criteria. (Character: “U”, “P”,

“W”, or any combination; Default = “PW”).EPSU Error tolerance for displacement (U) criterion. (Real > 0.0;

Default = 1.0 E-3).

EPSP Error tolerance for load (P) criterion. (Real > 0.0; Default =1.0E-3).EPSW Error tolerance for work (W) criterion. (Real > 0.0; Default =

1.0E-7).

EPSWEPSPEPSU

CONVIDNLPARM

10987654321

SPECIAL LOGICS




Bisection Algorithm Overcomes divergent problems due to nonlinearity.

Activated when divergence occurs.

Activated when MAXITER is reached. Activated when excessive ∆σ is detected.

Activated when an excessive rotation increment is detected.

Bisection continues until solution converges or MAXBIS is reached.

Activated with line search condition (see page 2-48).

SPECIAL LOGICS




If MAXBIS is reached, reiteration procedure is activatedto select the best attainable solution.

User Interface

Field ContentsFSTRESS Fraction of effective stress (σ) used to limit the sub-increment

size in the material routines. (0.0 < Real < 1.0; Default = 0.2).MAXBIS Maximum number of bisections allowed for each load

increment. (-10 ≤ MAXBIS ≤ 10; Default = 5).RTOLB Maximum value of incremental rotation (in degrees) allowed

per iteration to activate bisection. (Real > 2.0; Default = 20.0).

RTOLBMAXBIS

FSTRESS

IDNLPARM

10987654321

SPECIAL LOGICS




Time Expiration Criteria 5% of time reserved for data recovery.

Analysis is stopped to allow for data recovery.

Can restart.

RESTARTS




Need to save database.

Cold-starts are from a stress-free state with no

displacements or rotations. Must define database that stores all pertinent

information.

Changes in grid points, elements, or material properties

define a new problem. Can restart from any converged solution.

Can restart into: (a) nonlinear static, (b) nonlinear

transient, and (c) normal mode solution sequence.

RESTARTS




Restarting Into Nonlinear Static Analysis Requires two parameters:

PARAM,LOOPID,ι Converged solution to start from.

PARAM,SUBID,m Subcase to start into. Note: SUBID is not the same as SUBCASE ID. SUBID is the subcase

sequence number.

RESTARTS




Can restart into the same subcase or a new subcase. Must restart into a new subcase for follower forces and

temperature loads.

Follower loads are interpolated between A and B. Make a new subcasebetween A' and B.

Restart cases Next load step. (If follower forces are present, problems may result.) Next or new subcase (skip load steps). Data recovery (skip iteration).

A

BSC3

SC2

SC1

A'

RESTARTS




Example Cold start

Database Version 1

1. Restart into same subcase (next load step)LOOPID = 8SUBID = 2Same NLPARM specification

1 2 3 4 5 6 7 81 2 3 4 1 2 3 4 5

00 4.25 8.50 12.75 P1=161.0

201.50 P2=242.0

252.20 262.4 272.60 282.80 P3=293.0

Applying 16

1 2 3

LF = 1/4 LF = 1/8 LF = 1/5

INC

LOADLOADSTEP

Applying 16 + 8 Applying 24 + 5Subcase

Restart Here

RESTARTS




Database Version 2

2. Restart into new subcase before SUBID 3LOOPID = 8SUBID = 3NLPARM specification with 4 increments

INC

LOAD 0 16 20 22 24 29LOADSTEP 0.25 0.50 0.75 1.0 1.5 1.25 1.5 2.0 3.0

LOOPID 4 8 16 21

Restart

RESTARTS




Database Version 3

3. Restart into new subcase after SUBID 3

LOOPID = 8SUBID = 4NLPARM specification with 4 increments

INC

LOAD 0 16 20 22 24 29LOADSTEP 1.0 1.5 2.5 3.0 4.0

LOOPID 4 8 12 17

Restart

RESTARTS




Database Version 4

4. Restart for data recovery

PARAM,LOOPID,n (data recovery for LOOPID 1 through n)

PARAM,SUBID,m (m is the next subcase sequence number)

INC

LOAD 0 16 20 24LOADSTEP 1.0 1.5 3.5 4.0

LOOPID 4 8 12

Restart

RESTARTS




Restarting Into Nonlinear Transient Analysis Requires one parameter

PARAM,SLOOPID,LOOPID

See page 7-67 for more details

Restarting into Normal Mode Solution Sequences Requires one parameter

PARAM,NMLOOP,LOOPID

See page 9-2 for more details

Note: Results may not be accurate if the follower force effects wereincluded in the nonlinear static analysis.

OUTPUT FOR SOLUTION STRATEGIES




Standard Output EUI Normalized error in the displacement.

EPI Normalized error in the load vector.

EWI Normalized error in the energy. LAMBDA Rate of convergence is λi. Solution is diverging if

λi ≥ 1.0., λ1 = 0.1

DLMAG Absolute norm of the load error vector.

FACTOR Scale factor a for line search method.

E-First Initial error E1 before line search begins.

λi

1

2---

E p

i

E pi 1 – ------------- λi 1 –

*

+

= λi

*

min λi .7

λi

10------+ .99,,=

DLMAG R i=





E-FINAL Final error Ei after line search terminates.

N-QNV Number of quasi-Newton correction vectors to be used in thecurrent iteration.

N-LS Number of line searches performed. ENIC Expected number of iterations for convergence.

NDV Number of occurrences of probable divergence during theiteration.

MDV Number of occurrences of bisection conditions due to

excessive increments in stress and rotations.

Ni

EPSP E pi

⁄

λi

*

log

-------------------------log=





1 SLINE1U: UNSYMMETRIC RIGID PUNCH WITH FRICTION NOVEMBER 30, 1993 MSC/NASTRAN 11/29/93 PAGE 9

00 N O N - L I N E A R I T E R A T I O N M O D U L E O U T P U T STIFFNESS UPDATE TIME .89 SECONDS SUBCASE 1

ITERATION TIME .00 SECONDS LOAD FACTOR .250 - - - CONVERGENCE FACTORS - - - - - - LINE SEARCH DATA - - -0ITERATION EUI EPI EWI LAMBDA DLMAG FACTOR E-FIRST E-FINAL NQNV NLS ENIC NDV MDV 1 9.9000E+01 1.7374E-05 1.7374E-05 1.0000E-01 1.2127E-04 1.0000E+00 3.5268E-07 3.5268E-07 0 0 0 1

2 1.8484E-07 9.0935E-11 7.1947E-17 5.0003E-02 2.6099E-09 1.0000E+00 2.9490E-06 2.9490E-06 0 0 0 0 10*** USER INFORMATION MESSAGE 6186,

*** SOLUTION HAS CONVERGED *** SUBID 1 LOOPID 1 LOAD STEP .250 LOAD FACTOR .25000 ^^^ DMAP INFORMATION MESSAGE 9005 (NLSTATIC) - THE SOLUTION FOR LOOPID= 1 IS SAVED FOR RESTART

SubcaseSequence

Number

For RestartPurpose





Diagnostic Output DIAG 50 (NLPARM, NLPCI and SUBCASE data)

For every entry into NLITER module (AFTER STIFFNESS UPDATE)

N O N - L I N E A R I T E R A T I O N M O D U L E S O L U T I O N C O N T R O L D A T A LOOP CONTROLS :SUBCASE... 1 SUBCASE RECORD... 1 LARGE DISPLACEMENTS... NONLPARM DATA FOR SET : 110 MACHINE CHARACTERISTICS :NUMBER OF LOAD INCREMENTS ... 1 PRESENT OPEN CORE .... 1607856 WORDSINCREMENTAL TIME INTERVAL ... 0.0000E+00 MODULE’S WORK AREA ... 1601703 WORDSMATRIX UPDATE OPTION ........ ITER MAXIMUM G-SET SIZE ... 160170 TERMSMATRIX UPDATE INCREMENT ..... 1 ESTIMATION FOR NO SPILL ... 22 G-SET +

22 G-SET + 8 A-SET = 1601668MAXIMUM ITERATIONS .......... 25CONVERGENCE OPTIONS ......... PW

INTERMEDIATE OUTPUT ......... YES- DISPLACEMENT ... 1.0E-03 DMAP CONTROL PARAMETERS FROM PREVIOUSITERATION:

TOLERANCE - RESIDUAL FORCE . 1.0E-03- PLASTIC WORK ... 1.0E-07 CONVERGENCE ........ NO

DIVERGENCE LIMIT ............ 3 NEW SUBCASE ........ NOMAXIMUM QUASI-NEWTON VECTORS 0 NEW MATRIX ......... YESMAXIMUM LINE SEARCHES ....... 0 PREVIOUS ITERATIONS 1ERROR TOLERANCE IN YF ....... 2.0E-01 STIFFNESS UPDATES .. 1LINE SEARCH TOLERANCE ....... .500 DMAP LOOP NUMBER ... 1MAX. NUMBER OF BISECTIONS ... 5

LIMIT TO ADJUSTMENT FACTOR .. 20.000ROTATION LIMIT FOR BISECTION .200E+02CONTROLLED INCREMENTS OPTION CRISMINIMUM ARC FACTOR .......... 1.000MAXIMUM ARC FACTOR .......... 1.000SCALE FACTOR FOR LOAD FACTOR 0.000E+00DESIRED ITERATIONS .......... 12MAX. NUMBER OF C. I. STEPS .. 20

0*** USER INFORMATION MESSAGE 6188*** INITIAL ARC LENGTH IS 4.510799D-02

(Approximate)

NLPCI

N L P A R

M





DIAG 51 All the data needed to follow the solution process in detail

(displacement, nonlinear force, unbalanced load vector, etc.).

See Section 7.2.5 of the MSC.NASTRAN Handbook for Nonlinear Analysis for details.

Should not be used. It produces enormous output.

Used by developers when debugging.

DIAG 35 Penalty values (gap and friction) for each slave node.

Updated coordinates for slide line nodes.

Detail status for each slide line element (forces, gaps, connectivity, etc.).

RESULT OUTPUT




Results selected for output in the subcase. For example:DISP, FORCE, STRESS, etc. are printed at everyINTOUT load step.

Format:

Example:

Field ContentsINTOUT Intermediate output flag. (Character = “YES”, “NO”, or

“ALL”; Default = NO).

INTOUTNLPARM

10987654321

INTOUT515NLPARM

SOME HEURISTIC OBSERVATIONS




Loose tolerances for convergence test cause difficultiesin later stages.

Sometimes quasi-Newton updates seem to haveadverse effects in creep analysis.

SEMI is a good conservative method if AUTO does notwork. If desperate, use ITER with KSTEP=1 to get

started. A line search is comparable to an iteration in terms of

CPU. However, line searches may be required to getaround some difficulties in convergence.

HINTS AND RECOMMENDATIONS




Identify the type of nonlinearity; if unsure, perform linearanalysis.

Localize nonlinear region; use super-elements and linearelements for the linear region.

Nonlinear region usually needs a finer mesh.

Divide load history by subcases for convenience.

Loads should be subdivided, not to exceed 20 steps ineach subcase.

Select default values to start - NLPARM.

Choose GAP stiffness appropriately. Need to understand the basic theory to use the nonlinear

material.

NLPARM BULK DATA ENTRY




NLPARM with all its field is shown below

Parameters for Nonlinear Static Analysis Control

Defines a set of parameters for nonlinear static analysisiteration strategy.

Format:

Example:

RTOLBMAXRMAXBIS

LSTOLFSTRESSMAXLSMAXQNMAXDIVEPSWEPSPEPSU

INTOUTCONVMAXITERKSTEPKMETHODDTNINCIDNLPARM

10987654321

515NLPARM

SUMMARY




Five tasks in a nonlinear solution strategy Determine an increment to advance forward

Stiffness update

Displacement prediction Element state update

Unbalance force and convergence check

Advancing schemes Constant load increment

Displacement increment

Arc-length increment (Crisfield, Riks, and modified Riks)

SUMMARY




Stiffness update Every iteration (NR method)

Every k-th iteration

Based on the rate of convergence On nonconvergence or divergence

QN updates - Modify the stiffness matrix by two rank one additions

Displacement prediction Solution of equilibrium equations Line Search - Scale the calculated displacements to reduce unbalance

loads

State determination Update element state to calculate element forces

SUMMARY




Convergence criteria Displacement

Load

Energy

Special logics Divergence

Bisection

Time expiration criteria

User interface NLPARM (solution strategy)

SPCD and SPC (displacement increment)

NLPCI (arc-length increment)

WORKSHOP PROBLEMS




Purpose To demonstrate cold start and restart analysis procedures in SOL 106

Problem Description For the structure below:

Y

P = 29.E3

CROD

CELAS1

K = 1.E3

A = .01

E = 1.E7

L = 10.0

X

3

12

WORKSHOP PROBLEMS




1. Add Case Control commands and Bulk Data entries to:a) Perform geometric nonlinear analysis.

b) Apply a load of 16 × 103 lbs in the first subcase in four increments.

c) Apply a load of 24 × 103 lbs in the second subcase in eight increments.d) Apply a load of 29 × 103 lbs in the third subcase in five increments.

e) For the first subcase, print output at every load step.

f) For the second subcase, use only the work criteria for convergence,

and print output at every load step.g) For the third subcase, request output at the end of the subcase only.

2. Restart the analysis from a load of 20 × 103 lbs. Add anew subcase after the third subcase, and apply in it, a

load of 24 × 103 lbs, using 8 load steps. Also, printoutput at all load steps in this new subcase, and thenext (original subcase 3).

WORKSHOP PROBLEMS 1-2

I t Fil f M difi ti




Input File for Modification

ID CHAP2WS1, NAS103, Chap 2 $ Workshop 1

SOL 106 $ NONLINCENDTITLE=SIMPLE ROD SPRING - RESTART WORKSHOP

SUBTITLE=GEOMETRIC NONLINEARECHO=BOTH

DISP=ALLOLOAD=ALL$

SUBCASE 10 $LOAD=16.E03LABEL=APPLY LOAD P IN X DIRECTION = 16E+03

SUBCASE 20 $ LOAD=24.E03LABEL=APPLY LOAD P IN X DIRECTION = 24E+03SUBCASE 30 $ LOAD=29.E03

LABEL=APPLY LOAD P IN X DIRECTION = 29E+03BEGIN BULKPARAM,POST,0GRID 1 0 0.0 0.0 0.0 23456

GRID 3 0 0.0 10.0 0.0 123456CROD 3 3 3 1

CELAS1 2 2 1 1 0PROD 3 3 .01PELAS 2 1.0E3

MAT1 3 1.0E7 0.1 12.9-6FORCE 1 1 0 1.6E4 1.0

FORCE 2 1 0 2.4E4 1.0

FORCE 3 1 0 2.9E4 1.0ENDDATA

SOLUTION FOR WORKSHOP PROBLEM ONE

ID CHAP2WS1s NAS103 Chap 2 $ Workshop 1




BEGIN BULKPARAM,POST,0GRID 1 0 0.0 0.0 0.0 23456GRID 3 0 0.0 10.0 0.0 123456CROD 3 3 3 1CELAS1 2 2 1 1 0PROD 3 3 .01PELAS 2 1.0E3MAT1 3 1.0E7 0.1 12.9-6

FORCE 1 1 0 1.6E4 1.0FORCE 2 1 0 2.4E4 1.0FORCE 3 1 0 2.9E4 1.0PARAM, LGDISP, 1NLPARM, 10, 4, , SEMI, , , , YESNLPARM, 20, 8, , AUTO, , ,W, YESNLPARM, 30, 5, , AUTO, , ,W, NOENDDATA

ID CHAP2WS1s, NAS103, Chap 2 $ Workshop 1SOL 106 $ NONLINCENDTITLE=SIMPLE ROD SPRING - RESTART WORKSHOPSUBTITLE=GEOMETRIC NONLINEARECHO=BOTHDISP=ALLOLOAD=ALL$SUBCASE 10 $LOAD=16.E03

LABEL=APPLY LOAD P IN X DIRECTION = 16E+03LOAD=1NLPARM=10

SUBCASE 20 $ LOAD=24.E03LABEL=APPLY LOAD P IN X DIRECTION = 24E+03LOAD=2NLPARM=20

SUBCASE 30 $ LOAD=29.E03LABEL=APPLY LOAD P IN X DIRECTION = 29E+03LOAD=3NLPARM=30

SOLUTION FOR WORKSHOP PROBLEM TWO

RESTART, VERSION=1, KEEP




, ,

ASSIGN MASTER='chap2_ws_1s.MASTER'

ID CHAP2WS2s, NAS103, Chap 2 $ Workshop 2

SOL 106 $ NONLIN

CEND

TITLE=SIMPLE ROD SPRING - RESTART WORKSHOP

SUBTITLE=GEOMETRIC NONLINEAR

ECHO=BOTH

DISP=ALL

OLOAD=ALL

PARAM, LOOPID, 8

PARAM, SUBID, 3

SUBCASE 10 $LOAD=16.E03

LABEL=APPLY LOAD P IN X DIRECTION = 16E+03

LOAD=1

NLPARM=10

SUBCASE 20 $ LOAD=24.E03 LABEL=APPLY LOAD P IN X DIRECTION = 24E+03

LOAD=2

NLPARM=20

SUBCASE 21 $ LOAD=24.E03

LABEL=APPLY LOAD P IN X DIRECTION = 24E+03

LOAD=2

NLPARM=21

SUBCASE 30 $ LOAD=29.E03

LABEL=APPLY LOAD P IN X DIRECTION = 29E+03 LOAD=3

NLPARM=31

BEGIN BULK

NLPARM, 21, 8, , AUTO, , ,W, YES

NLPARM, 31, 10, , AUTO, , ,PW, YES

ENDDATA




SECTION 3





TABLE OF CONTENTS

Page




Page

Geometric Nonlinear Analysis 3-4

Simple Geometric Nonlinear Example 3-10

Treatment Of Large Rotations 3-14Follower Forces 3-20

Force1 Bulk Data Entry 3-21

Force2 Bulk Data Entry 3-22

Parameter K6ROT For QUAD4 And TRIA3 3-23

Example Problem One 3-25

Example Problem Two 3-29

Workshop Problem One 3-32

Workshop Problem Two 3-35


Solution For Workshop Problem Two 3-40


Large displacements and large rotations




Large displacements and large rotations Element deformations are a nonlinear function of the grid point

displacements (nonlinear displacement transformation matrix).

Large displacements Deflection of highly-loaded thin flat plates (geometric stiffening).

where u >> t

t

P

u


Large displacements and large rotations (Cont )




Large displacements and large rotations (Cont.) Large rotation.

P

Elastic


Follower forces




Follower forces Applied loads are functions of displacements.

Fluid pressure (changes in magnitude and direction).

Tire


Follower forces




Follower forces Centrifugal force (proportional to distance from spin axis).

Temperature loads (change in direction).

RFORCE

mr ω2

mr ω2


Large strains




g Element strains are nonlinear functions of element deformations.

Rubber Bearing (Hyper elastic Material)


User Interface




PARAM LGDISP 0 for no geometric nonlinearity (default).

1 for both nonlinear displacement transformation plus follower forces. 2 for nonlinear displacement transformation only.

Small or large strain depends on the element types.

SIMPLE GEOMETRIC NONLINEAR EXAMPLE

Truss bar with a spring




p g

Strain (ε) in truss, for small θ

z

wF

P

l ''

θ

l z'

l”2 /l 2 = (b2+(z+w)2 )/(b2+z 2 )= 1 +2(zw+w2 /2)/(b2+z 2 )

= 1 +2(zw+w2 /2)/ l 2 = (1+ (zw+w2 /2)/l 2 )2

... l”/l = 1+ (zw+w2 /2)/l 2

... l”/l – 1 = (zw+w2 /2)/ l 2 = ε

ε l '' l –

l -------------

zw

l

2-------

1

2---

w

l

2-------+≅=


Truss bar with a spring




g

Force in Truss (F)

z

wF

P

l ''

θ

l z'

F E A ε=E A

l 2

-------- z w1

2---w

2+

=


Equilibrium (deformed configuration)




Tangent stiffness

P F θs n K s w+=

F z w+( )

l ' '---------------------= K

s

w+

F z w+( )

l --------------------- K s w+≅

Linear Initial

Slope

Geometric

(Initial Stress or

Differential)

Spring+ + +

K tdP

dw-------

z w+

l -------------

d

d---

F

w----

F

l --- K s+ += =

EA

l --------=

z

l --

2

EA

l

3-------- 2zw w

2+( )

F

l --- K s+ ++

z

wF

P

l ''

θ

l z'


P, q

2

PEA = 10

7




11

x

y100.

14.

12.

10.

8.

6.

4.

2.

0

0

0.5 1.0 1.5 2.0 2.5 3.0

q

.235

1.765 2.16

18.

16.

1.Ks

Ks = 6.0

Ks = 3.0

TREATMENT OF LARGE ROTATIONS

Applicable to QUAD4, TRIA3, and BEAM.




Large rotations cannot be added vectorially.

Two approaches: Gimbal angle approach.

Default or selected by PARAM,LANGLE,1.

Rotation vector approach (recommended). Selected by PARAM,LANGLE,2.

User interface PARAM LANGLE.

Specified in Bulk Data Section (cannot specify in the Case ControlSection).

Cannot be changed between subcases or restart.


Gimbal Angle Approach - Concept




Rotation matrix is unique.

Several ways to go from one configuration to the other.

Orientation of a rigid body attached to the grid point is obtained by threesuccessive rotations. First, rotation of magnitude θz about the global z-axis.

Second, rotation of magnitude θY about the rotated y-axis.

Third, rotation of magnitude θX about the doubly rotated x-axis.

Note: Above definition produces elegant mathematics, but is difficult tovisualize.

Above definition is equivalent to saying: First, rotation of magnitude θX about the global x-axis.

Second, rotation of magnitude θY about the global y-axis. Third, rotation of magnitude θz about the global z-axis.

Note: With this definition, the mathematics is not elegant.


Gimbal Angle Approach - TheoryC id th fi it t ti (θ θ θ ) f t i th l b l




Consider the finite rotations (θX , θY , θz ) of a vector in the globalcoordinate system.

where

Rz, Ry, and Rx rotate z-axis by θz , rotated y-axis by θY , and doublyrotated x-axis by θX , respectively.

Ug( )rotated R z[ ] R y[ ] R x[ ] Ug[ ] R θg( )[ ] Ug = =

R x

1 0 0

0 θxcos θxsin

0 θxsin – θxcos

= R y

θycos 0 θxsin –

0 1 0

θxsin 0 θxcos

=

R z

θycos θxsin 0

θxsin – θycos 0

0 0 1

=


For small rotation (δθ)




Addition of gimbal angles

Where (δθ) = incremental rotations in the global system

∆θ = incremental gimbal angle

R δθ( )[ ] R z[ ]T

R y[ ]T

R x[ ]T

1 δθz – δθy

δθz 1 δθx –

δθy – δθx 1

≅=

R θ ∆θ+( )[ ] R δθg

( )[ ] R θg

( )[ ]=


Gimbal Angle Incrementsi




Mathematical singularity at θy = 90°.

This condition is usually caused by numerical ill-conditioning.

Use auxiliary angles to avoid singularity. Usual remedy is to use a smaller load increment.

∆θx δθy θz δθx θzcos+sin( ) θycos( )=

∆θy δθy θz δθx – cos θzs n=

∆θz δθz δθz θz δθx θzcos+s n( ) θycos( )[ ] θys n+=


Rotation Vector Approach (Version 67 plus)S l t d b PARAM LANGLE 2 i th B lk D t S ti




Selected by PARAM,LANGLE,2 in the Bulk Data Section.

The rotation components at a grid point are interpreted as components

of a rotation vector. Orientation of a rigid body attached to a grid point is obtained by rotating

the body by an amount of Φ about a principal axis of rotation p.

Consistent with enforced nonzero rotations.

Principal Axis

Magnitude

V

P

g3

g1

g2

V R ?asterisk14? V=

θx

θy

θz

φ

P1

P2

P3

=

FOLLOWER FORCES

Nodal forces change directions with displacements. Load vector(Pa) is a function of the displacement (Ua).




(Pa) is a function of the displacement (Ua).

Must specify PARAM,LGDISP,1.

Applicable to: PLOAD, PLOAD2, and PLOAD4 on QUAD4, TRIA3, HEXA, and PENTA.

PLOADX1 on QUADX and TRIAX hyper elastic elements.

FORCE1, FORCE2, MOMENT1, MOMENT2 (directions dependent upon GRIDlocations).

Temperature load. Centrifugal force.

Corrective loads are computed based on the updated geometry.

Total loads are computed based on the updated geometry.

Note: Tangential stiffness does not include the follower force effect.

FORCE1 BULK DATA ENTRY

FORCE1 Static Force, Alternate Form 1

Defines a static concentrated force by specification of a magnitude




Defines a static concentrated force by specification of a magnitudeand two grid points that determine the direction.

Format:

Example:

Field Contents

SID Load set identification number. (Integer > 0).

G Grid point identification number. (Integer > 0).

F Magnitude of the force. (Real).G1, G2 Grid point identification numbers. (Integer > 0; G1 and G2 may not

be coincident).

G2G1FGSIDFORCE1

10987654321

1316-2.93136FORCE1

FORCE2 BULK DATA ENTRY

FORCE2 Static Force, Alternate Form 2

Defines a static concentrated force by specification of a magnitude




Defines a static concentrated force by specification of a magnitudeand four grid points that determine the direction.

Format:

Example:

Field Contents

SID Load set identification number. (Integer > 0).

G Grid point identification number. (Integer > 0).

F Magnitude of the force. (Real).Gi Grid point identification numbers. (Integer > 0; G1 and G2 may not

be coincident; G3 and G4 cannot be coincident).

G4G3G2G1FGSIDFORCE2

10987654321

13171316-2.93136FORCE2

PARAMETER K6ROT FOR QUAD4 AND TRIA3

Stiffness of the normal rotation (θz) is not defined for theusual shell element on the flat plane.




usua s e e e e o e a p a e

This DOF cannot be constrained in the geometricnonlinear case.

K6ROT provides small stiffness to stabilize this DOF.

y

x

z

θz

PARAMETER K6ROT FOR QUAD4 AND TRIA3

Pseudo stiffness added to the relative rotation in theelement:




element:θz = rotation of a GRID from global displacement

; Rotation Measured in the Element

where G = Shear ModulusS = weighting factor

Pass the constant strain patch test.

No effect on the rigid-body rotation.

Insensitive to the mesh size. Default is K6ROT = 100. which is highly recommended.

Too large a value of K6ROT locks the varying strain by enforcingΩz ≅ θz

−=Ω

y

u

x

v z

δ

δ

δ

δ

2

1

ROT K S t G K z z 6****10)(for 6−=−Ω θ θ

EXAMPLE PROBLEM ONE

Purpose To illustrate geometric nonlinear analysis.




To illustrate geometric nonlinear analysis.

Problem Description Perform large deformation analysis of a hemisphere with a hole at the

top and loaded with four concentrated forces acting on the equator at90° intervals. The perimeter of the hole is constrained in z direction andthe equator is a free edge.

EXAMPLE PROBLEM ONE

Z

Y




X

Force ± 100.0Thickness t=0.04

Radius R=10.00

Hole

Youngs modulus E=6.825e+7Poisson’s ratio

018=θ

30.0=ν

EXAMPLE PROBLEM ONE

Solution Model (symmetric) 1/4 hemisphere with a mesh of 16X16 Quad4




( y ) p

Displacement in force direction: Node 1 = 2.587 (4.688*)

Node 289 = –3.791 (4.688*)

* Linear Solution

Y

X

Z

EXAMPLE PROBLEM ONE




Undeformed Shape Deformed Shape

EXAMPLE PROBLEM TWO

Purpose To illustrate geometric nonlinear analysis.




Problem Description Calculate the deformation of a corrugated sheet of paper coming out of

a copy machine. The paper deforms under its own weight.

Solution Perform a quasi-static analysis with a load of 2g to account for dynamic

effects. Model half of the paper taking advantage of symmetry.

EXAMPLE PROBLEM TWO

Model for a Corrugated Sheet of Paper




Thickness 0.0027 in

Radius 6.0000 in

Angle 40.0°Length 8.0000 in

EXAMPLE PROBLEM TWO

Undeformed and Deformed Sheet of Paper for a 2GGravity Load




WORKSHOP PROBLEM ONE

Purpose To demonstrate the use of geometric nonlinear analysis.




Problem Description Calculate the large deflection behavior of the cantilever beam for the

following four load cases:

1. P = 2000.

2. P = 4000.

3. P = 6000.4. P = 8000.

Compare the results with the linear analysis.


Properties

y P




L = 10A = 1.

I = 1.e –2

E = 10.e6

y P

x


Input File for ModificationBEGIN BULK

$ GEOMETRY




$ GEOMETRY

GRID,1,,0.,0.,0.,,345

=,*(1),=,*(1.),==$

=(9)

GRID,100,,0.,0.,1.,,123456

$ CONNECTIVITY

CBEAM,101,1,1,2,100

=,*(1),=,*(1),*(1),==$

=(8)

$ PROPERTIES

PBEAM,1,1,1.,1.-2,1.-2

MAT1,1,10.E6,,.0

$ CONSTRAINTS

SPC,1,1,123456

$ LOADING

FORCE,11,11,,1.E4,0.,1.,0.

LOAD,200,.2,1.,11

LOAD,400,.4,1.,11

LOAD,600,.6,1.,11

LOAD,800,.8,1.,11

$ PARAMETERS

PARAM,POST,0

$ SOLUTION STRATEGY

ENDDATA

ID CHAP3W1, NAS103W $ AR (12/03)

SOL 106

TIME 10

CEND

TITLE=TRACE LARGE DEFLECTION OF A CANTILEVERED BEAM

SUBTITLE=Ref: BISSHOPP & DRUCKER; QAM 3(1):272-275; 1945SPC=1

DISP=ALL

SPCF=ALL

$

SUBCASE 10

LOAD=200

$

SUBCASE 20

LOAD=400

$

SUBCASE 30

LOAD=600

$

SUBCASE 40

LOAD=800

WORKSHOP PROBLEM TWO

Purpose To demonstrate the use of geometric nonlinear analysis and arc length

increments




increments.

Problem Description Compute the load-deflection behavior of the three-rod structure shown

below.

Properties

3

21P

5 5

5

3

4

y

EA1 5.e5=

EA2 3.e6=

P 4.e5=

EA2

EA1

∆2

∆1

WORKSHOP PROBLEM TWO

Input File for ModificationID CHAP32W2, NAS103 Workshop 2 $ AR (12/03)SOL 106CEND




TITLE=GEOMETRIC NONLINEAR PROBLEMSUBTITLE=Ref: POWELL & SIMONS, IJNME, 17:1455-1467, 1981SPCF=ALL

SPC=12LOAD=10SUBCASE 10BEGIN BULK$ GEMOETRYGRID,1,,0.0,0.GRID,2,,5.,0.GRID,3,,10.,3.GRID,4,,10.,8.GRDSET,,,,,,,3456$ CONNECTIVITYCONROD,1,1,2,10,.5CONROD,2,2,3,11,1.CONROD,3,3,4,11,1.$ PROPERTIESMAT1,10,1.E6MAT1,11,3.E6$ CONSTRAINTSSPC1,12,1,3SPC1,12,2,1,2SPC1,12,12,4

$ LOADINGFORCE,10,1,,4.E5,1.,0.,0.$ PARAMETERSPARAM,POST,0$ SOLUTION STRATEGYENDDATA


ID CHAP3W1S, NAS103W $ AR (12/03)

SOL 106

TIME 10

CEND

TITLE=TRACE LARGE DEFLECTION OF A CANTILEVERED BEAM

SUBTITLE=Ref: BISSHOPP & DRUCKER; QAM 3(1):272-275; 1945

BEGIN BULK

$ GEOMETRY

GRID,1,,0.,0.,0.,,345

=,*(1),=,*(1.),==$

=(9)

GRID 100 0 0 1 123456




SUBTITLE Ref: BISSHOPP & DRUCKER; QAM 3(1):272 275; 1945

SPC=1

DISP=ALL

SPCF=ALLNLPARM=10

$

SUBCASE 10

LOAD=200

$

SUBCASE 20

LOAD=400

$

SUBCASE 30LOAD=600

$

SUBCASE 40

LOAD=800

GRID,100,,0.,0.,1.,,123456

$ CONNECTIVITY

CBEAM,101,1,1,2,100

=,*(1),=,*(1),*(1),==$=(8)

$ PROPERTIES

PBEAM,1,1,1.,1.-2,1.-2

MAT1,1,10.E6,,.0

$ CONSTRAINTS

SPC,1,1,123456

$ LOADING

FORCE,11,11,,1.E4,0.,1.,0.

LOAD,200,.2,1.,11LOAD,400,.4,1.,11

LOAD,600,.6,1.,11

LOAD,800,.8,1.,11

$ PARAMETERS

PARAM,POST,0

$ SOLUTION STRATEGY

NLPARM,10,10

PARAM,LGDISP,1

ENDDATA


P = 8000.

P = 6000.

P = 4000.




P = 2000.

PX

Y

L – ∆

δ

8

9

10

L ∆





00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1

2

3

4

5

6

7

Linear

Normalized Displacement

Normalized

Load,

L ∆ –

L-------------

δL---

$ CONNECTIVITY

CONROD,1,1,2,10,.5

CONROD,2,2,3,11,1.

CONROD,3,3,4,11,1.

$ PROPERTIES

MAT1 10 1 E6

ID CHAP32W2S, NAS103 Workshop 2 $ AR (12/03)

SOL 106

CEND

TITLE=GEOMETRIC NONLINEAR PROBLEM

SUBTITLE=Ref: POWELL & SIMONS, IJNME, 17:1455-1467, 1981

SPCF=ALL





MAT1,10,1.E6

MAT1,11,3.E6

$ CONSTRAINTS

SPC1,12,1,3SPC1,12,2,1,2

SPC1,12,12,4

$ LOADING

FORCE,10,1,,4.E5,1.,0.,0.

$ PARAMETERS

PARAM,POST,0

$ SOLUTION STRATEGY

NLPARM,10,40,,,,,,YES

NLPCI,10PARAM,LGDISP,1

ENDDATA

SPC=12

LOAD=10

SUBCASE 10NLPARM=10

BEGIN BULK

$ GEMOETRY

GRID,1,,0.0,0.

GRID,2,,5.,0.

GRID,3,,10.,3.

GRID,4,,10.,8.

GRDSET,,,,,,,3456





SECTION 4

NONLINEAR BUCKLING ANALYSIS




TABLE OF CONTENTS

Page

Instability Phenomena 4-4

Linear Versus Nonlinear Buckling 4-7




Nonlinear Buckling Analysis 4-9

Example Problem One 4-15Example Problem Two 4-20

Workshop Problems 4-29


Solution For Workshop Problem Two 4-34Solution For Workshop Problem Three 4-35

Solution For Workshop Problem Four 4-37

INSTABILITY PHENOMENA

Two Types:1 Snap-through (limit point): The loss of stability occurs at a stationary




1. Snap through (limit point): The loss of stability occurs at a stationary

point (relative maximum) in the load-deflection space. The critical loadis termed a limit point. For loads beyond the limit point, the structure“snaps-through” and assumes a completely different displacedconfiguration.

P P

Plimit


Snap-Through of Shallow Shells

P




Shallow arch is symmetric; Deep arch is anti-symmetric.

Question of stable and unstable path.

Arc-length increments are good for snap-through problems.


2. Bifurcation buckling: The loss of stability occurs when two or moreequilibrium paths intersect in the load-deflection space. The point ofintersection is termed a bifurcation point. For loads beyond thebifurcation point, the structure buckles.




p

Arc length increments may not pick a bifurcation buckling point.

P

Pcrit

P

∆

LINEAR VERSUS NONLINEAR BUCKLING

Linear Buckling

K λKd

+[ ] φ 0=




Kinematic relationship is linear.

Constitutive relationship is linear.

Equilibrium is satisfied in perturbed configuration.

Geometric stiffness is assumed proportional to the load.

Use SOL 105.

Nonlinear Buckling

withK n λ ∆K +[ ] φ 0 =

Incremental Stiffness

Actual Tangent Nonlinear∆K Kn

Kn 1–

= -

LINEAR VERSUS NONLINEAR BUCKLING

Kinematic relationship is nonlinear.

Constitutive relationship may be nonlinear.




Geometric stiffness is assumed proportional to displacement increment.

Equilibrium is satisfied in perturbed configuration.

Use SOL 106.


Two ways to predict the limit load: Arc length increments to trace the equilibrium path.

(may be expensive, and requires some idea of the limit load.)

PARAM BUCKLE th d




PARAM,BUCKLE method.

One way to predict bifurcation buckling: PARAM,BUCKLE method.


PARAM,BUCKLE Concept

Limit Point or

Bifurcation ∆ K

P

K




Note:The error in Ucr

may be large, but the corresponding error in Pcr

issmall.

Bifurcation

∆U

λ∆U

Un – 1 Un UUcr

Pcr

Pn

Pn – 1

α∆P∆P

Predicted by Analysis

UUn – 1 Un Ucr U'cr


PARAM,BUCKLE Theory Eigenvalue problem:

Kn

λ∆K+[ ] φ 0 =




with:

Kn and Kn–1 are evaluated at the known solution points in the vicinity ofinstability

n

Fcr

F un

( ) K u( ) udun

ucr

∫F

nK

o

λ

∫ λ( )∆u dλ+=+≅

Incremental Stiffness

Actual Tangent Nonlinear∆K Kn

Kn 1–

= -


Critical displacement:

with:

Ne w Assump tionProp ortional toDisplacementIncrement

ucr un λ ∆u +=

∆u un

un 1–

= -




Critical buckling load by matching virtual work∆uTFcr = ∆uTPcr

with:ResultF

crP

cr P

n α ∆P += =

∆P Pn

Pn 1–

=

α

λ ∆u T

Kn1

2--- λ∆K+ ∆u

∆u T ∆P ----------------------------------------------------------------------=

n n 1


Tangent stiffness is assumed to change linearly withdisplacement.

Internal loads are quadratic functions of displacement.




Run SOL 106 for static analysis until a negativedeterminant [K] is encountered. Make a restart run for buckling analysis.

Use PARAM,BUCKLE,1.

Include the restart parameters. Include PARAM,LGDISP,1. Provide two small loading steps below the buckling point. Specify KMETHOD = ITER or AUTO with KSTEP = 1 on the NLPARM

entry (if the number of iterations required to converge > 1).

Specify KMETHOD = ITER with KSTEP = 1 on the NLPARM entry (ifthe number of iterations required to converge = 1). Include EIGB via a METHOD command in the Case Control Section.


Make a restart run for buckling analysis. (Cont.) Provide mode shape PLOT commands if desired.

Sometimes, a negative determinant of [K] may be encountered due tonumerical reasons.




numerical reasons.

A good idea may be to perform at least two buckling analyses withdifferent restarting points and compare the calculated buckling (limit)load.

Look for sudden increase in displacement values to make sure that the

load is in the vicinity of a limit point.

EXAMPLE PROBLEM ONE

Purpose To illustrate the linear buckling capability




Problem Description Calculate the buckling load of a axially loaded thin cylinder

Radius 100 in

Length 800 in

Thickness 0.25 in.

EXAMPLE PROBLEM ONE

Solution Use SOL 105.

Only one half of the cylinder is modeled due to symmetry.

64 QUAD4 elements in the circumferential direction and 40 QUAD4




64 QUAD4 elements in the circumferential direction and 40 QUAD4

elements in the longitudinal direction.

Note:1. Could use cyclic symmetry to get buckling load.

2. Cannot plot buckling shape using cyclic symmetry for the full model.

EXAMPLE PROBLEM ONE

Model




EXAMPLE PROBLEM ONE

First Buckling Model Pexact = 41,700 pounds/in2

Reference: Flügge, W., Stresses in Shells, 2nd Ed.,Springer-Verlag New York, Heidelberg,Berlin,1973]




MSC.NASTRAN (Linear) = 0.999 * Pexact (Linear) MSC.NASTRAN (Nonlinear) = 0.984 * Pexact (Linear)

First Buckling Mode Shape



EXAMPLE PROBLEM TWO

Purpose To illustrate the nonlinear buckling capabilities.

Problem Description




Calculate the elastic-plastic buckling of a clamped spherical cap.

q

cR1

b

a

R

Geometry:

t 0.0251 in=

R 0.8251 in=

R 1 1.1506=

a 0.267 in=

γ 20o=

b 0.14328 in=

α 14.3065 o=

c 0.32908 in=

β 37.7612 o=

Z

Slope Et 1.1 106 psi×=

Kinematic

Strain Hardening:

Boundary Condition:

Periphery Clamped

Material:

7075 – T6 Aluminum

γ

α

β

θ

E 10.8 106 psi×=

ν 0.3=

sy 7.8 104 psi×=

EXAMPLE PROBLEM TWO

PLOAD2

Q

Grid 100

θ

φ




Grid 1000

z

y

x

Elastoplastic buckling of imperfect spherical shell, hydrostatic pressure applied,periphery clamped, undeformed shape.

R

Shell Model (QUAD4, TRIA)

3500 < scr < 3600

EXAMPLE PROBLEM TWO

Results Based on Version 2001

3.531.739410000.511

Principal StressElement 10 [104 psi]

Displacement -Uz

Grid 100 [10-3 In]No. Of

Iter.Load(psi)

LoopStep

LoopId

SubId




11.5613.60035722.715

11.0912.29635502.68810.6010.977635002.62512

10.049.573634002.511

9.658.613433002.37510

9.357.888532002.259

9.117.276431002.125838.96.741430027

8.525.840628001.86

8.255.174526001.65

8.044.608424001.44

7.884.142322001.2327.533.7124200012

1

23

4

EXAMPLE PROBLEM TWO

1. Upper fiber starts yielding2. Upper and lower fibers yield

3. After LOOPID 12, the negative factor diagonal occurs the




first time4. Last converged solution after several bisections

EXAMPLE PROBLEM TWO

Results Based on Version 2001

10.049.568434003.5114

Principal StressElement 10 [104 psi]

Displacement -Uz

Grid 100 [10-3 In]No. Of

Iter.Load(psi)

LoopStep

LoopId

SubId

Restart from LOOPID = 10




α .383 Pcr

Pn

= α∆P 3500 .383( ) 100( ) 3538.3=+=+=

10.5910.96853500412

Finite Difference (Reference)

MSC/NASTRAN - Shell Model

3546.8

5000

4000

3000

2000

1000

0 .002 .004 .006 .008 .010 .012 .014Central Deflection U100 (in)

p ( p s i )

Load Versus Central Deflection

P r e s s u r e

MSC/NASTRAN

+

++

++

+ + +

+

EXAMPLE PROBLEM TWO

Input File for Cold StartID SSBUK, NAS103 Example $ AR 12/03

SOL 106TIME 30

CENDTITLE=ELASTIC-PLASTIC BUCKLING OF IMPERFECT SPHERICAL SHELLSUBTITLE=HYDROSTATIC PRESSURE APPLIED, PERIPHERY CLAMPED

O 17 427 444 (1981)




LABEL=REF.: KAO; IJNME; 17:427-444 (1981)ECHO=UNSORT

DISP(SORT2)=ALLOLOAD=ALLSPCF=ALL

STRESS(SORT2)=ALLSPC=10

SUBCASE 1LOAD=10NLPARM=10

SUBCASE 2LOAD=20NLPARM=20

SUBCASE 3

LOAD=30NLPARM=30

BEGIN BULK

$ DEFINE SPHERICAL COORDINATE SYSTEMSCORD2S, 100, , 0., 0., 0., 0., 0., 1., +C2S1

+C2S1, 1., 0., 1.

CORD2S, 200, , 0., 0., -.32908,0., 0., 1., +C2S2+C2S2, 1., 0, 1.

$ GEOMETRYGRDSET, , , , , , 100, 345

EXAMPLE PROBLEM TWO

GRID, 100, 200, 1.1506, 0., 0., 0, 12456GRID, 101, 200, 1.1506, 0.715, -5.GRID, 102, 200, 1.1506, 0.715, 5.GRID, 103, 200, 1.1506, 1.43, -5.

GRID, 104, 200, 1.1506, 1.43, 5.GRID, 105, 200, 1.1506, 2.145, -5.

GRID, 106, 200, 1.1506, 2.145, 5.GRID, 107, 200, 1.1506, 2.86, -5.GRID, 108, 200, 1.1506, 2.86, 5.

GRID, 109, 200, 1.1506, 3.575, -5.

GRID 110 200 1 1506 3 575 5




GRID, 110, 200, 1.1506, 3.575, 5.GRID, 111, 200, 1.1506, 4.29, -5.GRID, 112, 200, 1.1506, 4.29, 5.

GRID, 113, 200, 1.1506, 5.005, -5.GRID, 114, 200, 1.1506, 5.005, 5.

GRID, 115, 200, 1.1506, 5.72, -5.GRID, 116, 200, 1.1506, 5.72, 5.GRID, 117, 200, 1.1506, 6.435, -5.

GRID, 118, 200, 1.1506, 6.435, 5.GRID, 119, 100, 0.8251, 10., -5.

GRID, 120, 100, 0.8251, 10., 5.GRID, 121, 100, 0.8251, 11.48, -5.GRID, 122, 100, 0.8251, 11.48, 5.

GRID, 123, 100, 0.8251, 12.96, -5.GRID, 124, 100, 0.8251, 12.96, 5.GRID, 125, 100, 0.8251, 14.44, -5.GRID, 126, 100, 0.8251, 14.44, 5.GRID, 127, 100, 0.8251, 15.92, -5.

GRID, 128, 100, 0.8251, 15.92, 5.GRID, 129, 100, 0.8251, 17.40, -5.

GRID, 130, 100, 0.8251, 17.40, 5.GRID, 131, 100, 0.8251, 18.8806, -5.

GRID, 132, 100, 0.8251, 18.8806, 5.



EXAMPLE PROBLEM TWO

Input File for Buckling AnalysisRESTART VERSION=last KEEP

ASSIGN MASTER = 'chap4_ex_2.MASTER'ID SSBUKR,NAS103 Example $ AR 12/03SOL 106 $TIME 30 $CENDTITLE=ELASTIC-PLASTIC BUCKLING OF IMPERFECT SPHERICAL SHELL

SUBTITLE=HYDROSTATIC PRESSURE APPLIED PERIPHERY CLAMPED1 2 (1981)




SUBTITLE=HYDROSTATIC PRESSURE APPLIED, PERIPHERY CLAMPEDLABEL=REF.: KAO; IJNME; 17:427-444 (1981)ECHO=UNSORTDISP(SORT2)=ALLOLOAD=ALLSPCF=ALLSTRESS(SORT2)=ALLSPC=10

METHOD=30PARAM BUCKLE 1PARAM SUBID 4PARAM LOOPID 10SUBCASE 1LOAD=10

NLPARM=10SUBCASE 2LOAD=20

NLPARM=20SUBCASE 3LOAD=30

NLPARM=30SUBCASE 4 $ ADDED FOR BUCKLING ANALYSISLOAD=40

NLPARM=40BEGIN BULKEIGB, 30, SINV, 0., 2., , 2, 2

NLPARM, 40, 2, , AUTO, 1, , , YES$ENDDATA

WORKSHOP PROBLEMS

Purpose To demonstrate use of (a) geometric nonlinear analysis, (b) linear and

nonlinear buckling analysis, and (c) arc length increments.

Problem DescriptionF th t t b l l l t




For the structure below, calculate:

100 = b

1 = zF

l

P, w EA = 107

Ks

WORKSHOP PROBLEMS

1. Linear Buckling load without spring.

2. Nonlinear Static with Large Deflection

3 N li B kli (PARAM BUCKLE 1)




3. Nonlinear Buckling (PARAM, BUCKLE, 1)4. Nonlinear Static with Arc Length Method

5. Repeat above with spring (for Ks = 3, and 6)

WORKSHOP PROBLEMS 1-4


SOL 105TIME 60CENDTITLE=SIMPLE ONE DOF GEOMETRIC NONLINEAR PROBLEM

SUBTITLE=SOLUTION SEQUENCE 105LABEL=Ref: STRICKLIN & HAISLER; COMP. & STRUC.; 7:125-136 (1977)




Qe : S C & S ; CO . & S UC.; : 5 36 ( 9 )ECHO=SORTDISP(SORT2)=ALL

BEGIN BULKPARAM, POST, 0$ GEOMETRYGRID, 1, , 0., 0., 0., , 123456GRID, 2, , 100., 1., 0., , 13456$ CONNECTIVITY

CROD, 10, 10, 1, 2CELAS1, 20, 20, 2, 2, 0$ PROPERTIESPROD, 10, 1, .1PELAS, 20, 3. MAT1, 1, 10.E7$ LOADSFORCE, 6, 2, , 6., 0. -1., 0.$

$ SOLUTION STRATEGY$ENDDATA


1.

PblbPzFK

l

z

l

EA K

z Pl F

elastic

222

2

θsinP

∗=

≅≅




For

z l

Pb

z

l

l

b

l

P

l

z

l

F K geometric 2=∗

∗=

∗=

9995.9)100)(005.100(

1102

7

2

2

2

2

2

2

=∗

=

=

=

lb

EAz

z l

b

l

z

l

EA

P cr


Linear Buckling Solution: Input FileSOL 105TIME 60CENDTITLE=SIMPLE ONE DOF GEOMETRIC NONLINEAR PROBLEM SUBTITLE=SOLUTION SEQUENCE 105LABEL=Ref: STRICKLIN & HAISLER; COMP. & STRUC.; 7:125-136 (1977)

ECHO=SORTDISP(SORT2)=ALL




DISP(SORT2)=ALLSUBCASE 10LOAD=6

SUBCASE 20 METHOD=30BEGIN BULKPARAM, POST, 0$ GEOMETRYGRID, 1, , 0., 0., 0., , 123456

GRID, 2, , 100., 1., 0., , 13456$ CONNECTIVITYCROD, 10, 10, 1, 2$CELAS1, 20, 20, 2, 2, 0$ PROPERTIESPROD, 10, 1, .1$PELAS, 20, 3. MAT1, 1, 10.E7$ LOADSFORCE, 6, 2, , 6., 0. -1., 0.

$$ SOLUTION STRATEGY$EIGB, 30, INV, 0., 3., 20, 2, 2ENDDATA


Nonlinear Solution for Problem 2: Input FileSOL 106TIME 60CENDTITLE=SIMPLE ONE DOF GEOMETRIC NONLINEAR PROBLEM LABEL=Ref: STRICKLIN & HAISLER; COMP. & STRUC.; 7:125-136(1977)ECHO=SORTDISP(SORT2)=ALL

SUBCASE 10




SUBCASE 10LOAD=6 NLPARM=20BEGIN BULKPARAM, POST, 0$ GEOMETRYGRID, 1, , 0., 0., 0., , 123456GRID, 2, , 100., 1., 0., , 13456$ CONNECTIVITY

CROD, 10, 10, 1, 2$CELAS1, 20, 20, 2, 2, 0$ PROPERTIESPROD, 10, 1, .1$PELAS, 20, 3. MAT1, 1, 10.E7$ LOADSFORCE, 6, 2, , 6., 0. -1., 0.$$ SOLUTION STRATEGY$PARAM, LGDISP, +1 NLPARM, 20, 10, , ITER, 5, 25, PW, ALL$ENDDATA

SOLUTION FOR WORKSHOP PROBLEM THREE

RESTART,VERSION=1,KEEP ASSIGN MASTER='chap4_ws_2s.MASTER'TIME 60 $SOL 106$CENDTITLE=SIMPLE ONE DOF GEOMETRIC NONLINEAR PROBLEM LABEL=Ref: STRICKLIN & HAISLER; COMP. & STRUC.; 7:125-136 (1977)ECHO=SORTDISP(SORT2)=ALL

PARAM,LOOPID,3PARAM,SUBID,2




, , METHOD,30SUBCASE 10LOAD=6 NLPARM=20SUBCASE 20LOAD=6 NLPARM=30

BEGIN BULKPARAM, BUCKLE, 1EIGB, 30, INV, 0., 3., 20, 2, 2 NLPARM, 30, 70, , ITER, 1, 25, PW, ALLENDDATA

RESULTS FOR WORKSHOP PROBLEMS 1 - 3

Linear Buckling Vs. Nonlinear Buckling Solution:

Pcr (Nonlinear)

PARAM, BUCKLE, 1

Pcr (Linear)

SOL 105

Ks




3.43616.9003

1.9249.9990

N/A25.6016

SOLUTION FOR WORKSHOP PROBLEM FOUR

Nonlinear Solution with Arc Length: Input File

SOL 106TIME 60CENDTITLE=SIMPLE ONE DOF GEOMETRIC NONLINEAR PROBLELABEL=Ref: STRICKLIN & HAISLER; COMP. & STRUC.;

ECHO=SORTDISP(SORT2)=ALLSUBCASE 10




SUBCASE 10LOAD=15 NLPARM=20BEGIN BULKPARAM, POST, 0$ GEOMETRYGRID, 1, , 0., 0., 0., , 123456GRID, 2, , 100., 1., 0., , 13456$ CONNECTIVITYCROD, 10, 10, 1, 2$CELAS1, 20, 20, 2, 2, 0$ PROPERTIESPROD, 10, 1, .1$PELAS, 20, 3. MAT1, 1, 10.E7$ LOADSFORCE, 15, 2, , 15., 0. -1., 0.$$ SOLUTION STRATEGY

$PARAM, LGDISP, +1 NLPARM, 20, 10, , ITER, 5, 25, PW, ALL NLPCI, 20, CRIS, 1., 1., , , , 40$ENDDATA





SECTION 5

MATERIAL NONLINEAR ANALYSIS



S5-1NAS103, Section 5, February 2004





Material Types in MSC.Nastran 5-5Nonlinear Elasticity 5-11

Workshop Problem 1: Nonlinear Elastic Material 5-27

Hyperelasticity 5-33

Workshop Problem 2: Hyperelastic Material 5-45




p ypElastic-Plastic Material 5-53

Workshop Problem 3: Elastic-Plastic Material 5-73

Creep (Viscoelastic) Material 5-81

Workshop Problem 4: Creep Material 5-96

Workshop Problem 5: Temperature Dependent Mat. 5-105

Workshop Problem 6: Elastic-Pefectly Plastic Mat. 5-111




MATERIALS TYPES IN MSC.NASTRAN

Time and temperature independent

Linear elastic Isotropic (MAT1)

Orthotropic (MAT3(axisym) or MAT8(shell)) ε

σ




Orthotropic (MAT3(axisym) or MAT8(shell))

Anisotropic (MAT2(shell) or MAT9(solid))

Nonlinear elastic Isotropic (MAT1 and MATS1)

ε

σ

ε

MATERIALS TYPES IN MSC.NASTRAN(Cont.)

Time and temperature independent (continued)

Hyperelastic




Isotropic (MATHP)

σ

λ1 2 3 4 5 6 (stretch)

Uniaxial Tension

σ

λ1 2 3 4 5 (stretch)

PureShear


Time and temperature independent (continued)

Elastic-plastic

σ σ




Isotropic (MAT1, MATS1) Anisotropic (MAT2(shell) or MAT9(solid), and MATS1)

ε

σ

Perfectly Plastic Linear Strain Hardening

ε

σ


Temperature dependent

Linear elastic Isotropic (MAT1, MATT1) Orthotropic (MAT3(axisym) and MATT3)

σ T2

T1




Orthotropic (MAT3(axisym) and MATT3)

Anisotropic (MAT2(shell) and MATT2, orMAT9(solid) and MATT9)

Nonlinear elastic

Isotropic (MAT1 and MATT1 and MATS1)

ε

σ

ε

T2

T1


Time dependent

Viscoelastic Isotropic (MAT1 and CREEP) Anisotropic (MAT2(shell) or MAT9(solid),

σ

σ0




Anisotropic (MAT2(shell) or MAT9(solid),and CREEP)

Slightly anisotropic onlyt (time)

t1

ε

t (time)

ε0

t1

Creep Recovery



NONLINEAR ELASTICITY

Applications Plastics

Metals

Example data




The data is from uniaxialtests

NONLINEAR ELASTICITY (Cont.)

Limitations in MSC.Nastran Small strain

Isotropic materials only

No time dependence – no creep Can use beam element, but not recommended to use offset




vectors, in solution sequences that use differential stiffness,because the vectors do not change angle(orientation)

User interface MAT1 used to specify E, G, ν, ρ

ΜΑΤΤ1 used to specify temperature dependence of E, G, ν, ρ

ΜΑΤS1 used, along with table TABLES1 or TABLEST(temperature dependence), to specify stress versus strain (fromuniaxial test)


MATS1 bulk data entry





MATS1 bulk data entry (continued)





MATS1 bulk data entry(continued)




TABLES1 bulk data entry





TABLES1 bulk data entry (continued)





TABLES1 bulk dataentry (continued)





TABLES1 bulk dataentry (continued)

1

E

σ

Loading




1

ε

Unloading

Must supply first &third quadrant data

Uniaxial stress versusstrain data curve

NONLINEAR ELASTICITY, RELATIONBETWEEN UNIAXIAL AND MULTIAXIAL

Following are remarks on stress and strain data from auniaxial test, and how it relates to the multiaxial stressand strain state as simulated by MSC.Nastran

As previously mentioned, the stress and strain dataused for the TABLES1 entry is from a uniaxial test(s)




The simulation by MSC.Nastran is not uniaxial, butmultiaxial

How does MSC.Nastran use the uniaxial data ?

A stress and strain, called equivalent stress (σ) andstrain (ε), is defined so that it is comparable to that of auniaxial stress and strain from tests

Calculate equivalent strain (ε) from a multiaxial stress(σ) and strain (ε) state by assuming that the work


(Cont.)




done by the corresponding stress states is equal

dεσdεσ ∫∫ ><=⋅

ε = function(ε)

Using the uniaxial stress versus strain data and theequivalent strain (ε), calculate the equivalent stress (σ)


(Cont.)

σ




Uniaxial stress versus

strain data curve

ε

σ

ε

Using the calculated equivalent stress (σ) and strain (ε),calculate the new multiaxial stress state (σ) and a newnonlinear constitutive tangential matrix ([Dne])


(Cont.)

[ ] εDEε

σσ currentenew =




where E is the elastic modulus from the MAT1 entry

Eε

)ε,],Dfunction([][D ene ε =



Include in model SOL 106

A subcase for loading, and a subcase for unloading

The bulk data entry NLPARM, and corresponding case control entry

Material entries for linear and nonlinear elastic properties

WORKSHOP PROBLEM 1: NONLINEARELASTIC MATERIAL PROPERTIES




Solution File (continued)




HYPERELASTICITY

Definition Materials that exhibit elastic behavior through large strains

Described using a scalar strain energy function

Derivative of strain energy function with respect to a strain componentdetermines the corresponding stress component

A li ti f d l t t l t i l




Application of model to actual materials Rubber material: O-rings, bushings, gaskets, seals, boots, tires

Plastic

Glass

Solid propellant

Other elastomers

HYPERELASTICITY (Cont.)

Sample of data





Limitations of MSC.Nastran Fully incompressible material (ν = 0.5) not currently implemented

Acceptable for nearly incompressible materials, e.g. ν = 0.4995





Comments on formulation of constitutive materialproperties

whereS d Pi l Ki hh ff t t t i t ith t

C

W2S

∂

∂=




S = second Piola-Kirchhoff stress tensor; symmetric stress with respectto undeformed state

C = right Cauchy-Green strain tensor W = elastic strain energy function; generalized Mooney-Rivlin model

∑∑ −−−+−−=

ND

i

2i

0vi

j

2

i

1

NA

ji, ij

))T(T1(JD3)(I3)(IAW α


The coefficients Aij and Di are determined usingexperimental data Aij = coefficients accounting for distortion

Di = coefficients accounting for volumetric change Tests performed are

Distortional deformation




Uniaxial tension/compression

Equibiaxial tension

Simple shear

Pure shear

Volumetric deformation Pure volumetric compaction


Number of experimental data points needed for desiredorder of W polynomial (accuracy of data fit) Distortional portion of W polynomial

2A10, A01 (Mooney-Rivlin)1

Minimum Number ofExperimental Points

Material ConstantsNA




20 Above & A50, A41, A32, A23, A14, A055

14 Above & A40, A31, A22, A13, A044

9 Above & A30, A21, A12, A0335 Above & A20, A11, A022

2 A10, A01 (Mooney Rivlin)1


Number of experimental data points needed for desiredorder of W polynomial (accuracy of data fit) (continued) Volumetric portion of W polynomial

1D11

Minimum Number ofExperimental Points

Material ConstantsND




5D1, D2, D3, D4, D55

4D1, D2, D3, D44

3D1, D2, D332D1, D22

1D11



User interface MATHP is used to supply or reference the material related data

Either the coefficients Aij and Di are to be specified, or data such asstress versus stretch is to be referenced

If experimental data is supplied, the value of the coefficients areestimated using least squares fitting of the data with polynomials

The coefficients are what is used by MSC.Nastran for the simulationof the constitutive properties





p p

Τhe experimental stress versus stretch, etc. data is supplied using

TABLES1 entries

MATHP bulk data entry





MATHP bulk data entry (continued)





MATHP bulk data entry(continued)





WORKSHOP PROBLEM 2:HYPERELASTIC MATERIAL PROPERTIES

The use of hyperelastic materials is demonstrated usinga model with a single hexahedral element. The elementis to deform with the interior angles remaining at 90

degrees, and with the deformation in only one direction.





The needed information for the creation of the model is 1x1x1 single hexahedral element

Constrain the eight GRIDs to allow enforced motion in only the X-direction as follows Constrain one GRID’s all six D-of-Fs, e.g. GRID, 1, … ,123456 Constrain the other three GRID’s, of the element face containing the GRID

just constrained and normal to the X-direction, from motion in the X-direction, but allowing motion due to the Poisson effect




Constrain the remaining four GRIDs like the first four GRIDs, but allowing

motion in the X-direction Using nine(9) MPCs constrain the hexahedral element to keep a

rectangular-box shape (all interior angles are 90 degrees), e.g.

-1.0171.016100MPC

-1.0171.013100MPC-1.0171.012100MPC

A2C2G2 A1C1G1SIDMPC

2

37

6

X1



The needed information for the creation of the model is(continued) Experimental data





The needed information for the creation of the model is(continued) SOL 106

PARAM, LGDISP, 1

The bulk data entry NLPARM, and corresponding case control entry






Solution File




ELASTIC-PLASTIC MATERIAL

Examples of elastic-plastic material The simple example shows that upon unloading

the path followed is not the path taken duringloading. The slopes are equal, but the paths are

offset from each other. ε

σ

Linear Strain Hardening




Another simple example

g

ε

σ Idealizationfor Nastran

Y

A

BC

ε

σData

Y

A

B C

Perfectly Plastic,Uniaxial Stress-Strain

ELASTIC-PLASTIC MATERIAL (Cont.)

Modeling of material yielding Yield criterion

Defines the initiation of inelastic response of the material

A descriptive statement that defines conditions under which yielding will

begin yield function f(σij,Y)

σij is multi-axial stress state

Y is yield strength in uniaxial tension/compression

Yield criterion satisfied when f(σij,Y) = 0




( ij )

If f(σij

,Y) < 0 the response is elastic

If f(σij,Y) > 0 the response is plastic

Example σe = function(σij) (effective stress, scalar)

f(σij,Y) = σe – Y

Note: there is a strain, εe, that corresponds to σe


Modeling of material yielding (continued) Hardening

The way the yield surface changes due to inelastic response Isotropic – yield surface expands uniformly

Kinematic – yield surface translates w/o distortion Combined – combination of isotropic and kinematic

Plastic flow direction Governs the plastic flow after yielding

Incremental stress versus incremental strain




Incremental stress versus incremental strain Plasticity is path dependent There is not a unique relationship between stress and strain

There is a unique relationship between infinitesimal increments of stressand strain


Yield criterion Tresca (maximun shear stress)

Used to model metals with crystals having slip planes (resistance toshear force is relatively small), such as brittle and some ductile metals

Yielding begins when the maximum shear stress at a point equals themaximum shear stress at yield in uniaxial tension/compression

Yσσ 32 ±=−

Yσσ 13 ±=−




where σi are the principal stresses

Yσσ 13 ±

Yσσ 21 ±=−


Yield criterion (continued) von Mises (distortional strain energy density)

Used to model metals with crystals having slip planes, such as ductilemetals

Yielding begins when the distortional strain energy density at a pointequals the distortional strain energy density at yield in uniaxialtension/compression

2222 Y1

])σ(σ)σ(σ)σ[(σ1

Y)f(σ ++




133221ij Y

3

])σ(σ)σ(σ)σ[(σ

6

Y),f(σ −−+−+−=


Yield criterion (continued) Mohr-Coulomb

Used to model cohesive materials such as rock or concrete

Generalization of the Tresca criterion that includes the influence of

hydrostatic stress

where c and φ are coefficients for the cohesion and angle of internalfriction, respectively

2ccosφ)sinφσ(σσσY),f(σ 3131ij −++−=




, p y


Yield criterion (continued) Drucker-Prager

Used to model cohesive materials such as sand or concrete

Generalization of von Mises criterion that includes the influence of

hydrostatic stress

where α and K are coefficients that are dependent on the cohesion andthe angle of internal friction

K JαIY),f(σ 21ij −+=




g


Hardening, work Isotropic hardening

Yield surface becomes larger by expanding uniformly about the origin instress space; it retains its shape.

Effective plastic strain is used as the measure of hardening The effective strain, εe, is used as a measure of the size of the yield

surface

The Bauschinger effect is not taken into account




ε2

ε3

Initial yield surface

Yield surface afterplastic deformation

O

Incremental loadingincreases size of surface

ε1


Hardening, work (continued) Kinematic hardening

Yield surface translates keeping its shape and size

Translate the yield surface in a direction normal to the yield surface

The location of the current center of the yield surface (o’) relative to theorigin of the principal strain coordinate system (o) is used as themeasure of hardening

The Bauschinger effect is accounted for

Physically reasonable results only for bilinear stress/strain relationship




ε2

ε3

Initial yield surface

Yield surface afterplastic deformation

O

ε1

O’


Hardening, work (continued) Combined hardening

Combination of isotropic and kinematic hardening

The Bauschinger effect is accounted for

σa

σ




σa

2σY

σa

σY

ε

Kinematic

Combined

Isotropic


Plastic flow direction MSC.Nastran uses associated flow rule to determine the plastic

flow direction

Uses plastic potential function equals the yield function

ε1 and σ1 axes coincide Plastic strain increment is calculated as follows

ij

P

ijσ

f λ dε

∂∂

=




Where

Plastic strain increment is in direction normal to yield surface

ij

P

ijε is plastic strain

λ is a scalar factor used to relate incremental strains to finite stress

f is the yield function; f(σij,Y) = 0 is the yield surface.


Incremental stress versus incremental strain Plasticity is path dependent

There is not a unique relationship between stress and strain

There is a unique relationship between infinitesimal increments of

stress and strain For incremental stress/strain relationship the strain increment is

divided into an elastic and plastic increment

σ




εP

ε

εEεP is the plastic strainεE is the elastic strain


Incremental stress versus incremental strain(continued) Relationship between differential increments of stress and plastic

strain

0dεε

f dσ

σ

f P

ijP

ij

ij

ij

=∂∂

+∂∂



MATS1 bulk data entry





MATS1 bulk data entry(continued)





WORKSHOP PROBLEM 3: ELASTIC-PLASTIC MATERIAL PROPERTIES

The use of elastic-plastic material is demonstrated usinga model of a rectangular plate loaded in tension in onedirection.

X

Y

WP




L

T = plate thickness


3 0 6E ( i)

0.1T (in.)

10.0W (in.)

50.0L (in.)

ValuePara-meter

Elastic-plastic model von Mises yield criterion

Isotropic hardening

950 0

1000.0

800.0

AppliedLoad, P

(lbf)




850.0σY (psi)

3.0e+4ET (psi)

0.25ν3.0e+6E (psi)

0.0

950.0



Two subcases for loading, and two subcases for unloading, e.g. SUBCASE 2

SUBTITLE = PLASTIC LOAD TO 1000 lbf LOAD = 2

NLPARM = 2

Use 1, 8, 5, and 2 increments for the NLPARM bulk data entries, for thefour subcases, e.g.

Case control




Case control SUBCASE 2

NLPARM = 2

Bulk data NLPARM, 2, 8,, AUTO,,,P




Input File for Modification (continued)





Solution File




CREEP (VISCOELASTIC) MATERIAL

This is for the class of materials that are viscoelastic(creeping). The application of a constant load causes adeformation that consists of an elastic and possiblyplastic part, and a viscous part. The elastic part mayoccur instantaneously (no mass), and the viscous partmay occur slowly over time.

Types of creep material behavior

Creep complianceC t t t ith t i i i ti




Creep, compliance Constant stress, σ0 , with strain, ε , increasing over time

O

A

B

C

time

ε

OA – instantaneous elastic response

AB – delayed elastic effect

BC – viscous flow over time

CREEP (VISCOELASTIC) MATERIAL(Cont.)

Types of creep material behavior (continued) Creep, compliance (continued)

Creep, three stage, with constant stress, σ0

Otime

ε

e

p

P i S d T ti

lr

r

er

sc

p




e – elastic deformation lr – load removed, then

p – plastic deformation different path

Primary – deformation rate decreases with time er – elastic recovery

Secondary – constant minimum creep rate r – ruptureTertiary – rapid increase of creep rate sc – secondary creep

Primary Secondary Tertiary


Types of creep material behavior (continued) Creep, compliance (continued)

Sample of materials Asphalt pavment

Solid propellant in rocket motors

High polymer plastics

Creep, relaxation Constant strain, ε0 , with stress, σ , decreasing over time

σ




O time

A

B

OA – instantaneous elastic response

AB – viscous flow over time


Types of creep material behavior (continued) Creep, relaxation (continued)

Sample of materials Prestress tendons in prestressed concrete

Prestress bolts at high temperatures that clamp rigid flanges of a machine





Creep analysis capability in MSC.Nastran Linear elastic isotropic, and elastic-plastic isotropic materials only

Anisotropic, nonlinear elastic, and hyperelastic materials cannot bemodeled

The creep law can be temperature dependent Primary and secondary creep modeling only; tertiary creep cannot be

modeled Primary creep model uses Kelvin model

Secondary creep model uses Maxwell model

kpc




For deviatoric stresses only

ke kscs

cp

σ

Kelvin MaxwellElastic

User interface MAT1 used to specify E, G, ν, ρ

CREEP is used to specify T0, Kp, Cp, Cs, TABLES1, etc.

TABLES1 is used to specify the creep

ΜΑΤS1 is used if there is plastic deformation





CREEP bulk data entry




CREEP bulk data entry (continued)





WORKSHOP PROBLEM 4: CREEPMATERIAL PROPERTIES

Calculate the creep strain in a cylindrical bar that issubjected to axial step loads/stresses.

Cross-sectionalarea = 1.0

E = 21.8e+6

ν = 0.32

CROD

All DOFfixed

Only DOF indirection of

force is free

Force

10.0 in





t g e f t r c )(]1)[( )( σ σ ε σ +−= −

σ σ 000208.0410476.3)( e f −∗=

094.25 )1000/(10991.3)( σ σ −∗=r

σ σ 000743.0111002.1)( e g −∗=

where

1.70e4

Force (lbf)




0 100 200 300 400 500

1.00e4

1.25e4

1.50e4

Time (hour)

Applied Force, Force



Five subcase pairs (total of 10 subcases), with the first subcase in apair for elastic loading and the second subcase in the pair for creeploading, e.g. SUBCASE 2O

SUBTITLE = ELASTIC

LOAD = 2 $ load = 1.25e4

NLPARM = 10 $ control of elastic solution

SUBCASE 21 SUBTITLE = CREEP




LOAD = 2 $ load = 1.25e4

NLPARM = 20 $ control of creep solution


Include in model (continued) For the bulk data it is necessary to specify the loading over time,

definition of the creep model, control of the elastic solution, and controlof the creep solution, e.g. FORCE, 2, 2, , 1.25E4, 1.0, 0.0, 0.0

CREEP, 1, … FORM = CRLAW

TYPE = 222

NLPARM, 10, 1

NLPARM, 20, 5, 20, , , , , YES Note: total creep time = NINC * DT





Input File for Modification (continued)





Solution File




WORKSHOP PROBLEM 5:TEMPERATURE DEPENDENT MATERIAL

PROPERTIES

Specify temperature dependent material properties, andsee that the strains in the output file (.f06) aremechanical strains (only).

The model is a single hexahedral element

The loading is uniaxial tension

Temperature change from 100 0F to 200 0F




The needed information for the creation of the model is 1x1x1 single hexahedral element; units are inch.

Constrain the eight GRIDs to allow enforced motion in only the X-direction as follows Constrain the four GRID’s, in the plane X=0 (Y-Z plane), all six D-of-Fs, e.g.

GRID, 1, … ,123456

Constrain the other four GRID’s, in the plane X=1, D-of-Fs 3456, e.g.GRID, 2, …, 3456

Elastic modulus, E, is 8.0e6, 100 0F ; and 4.0e6, 200 0F.

Poisson’s ratio, ν, is 0.3 at all temperatures Thermal expansion coefficient is 1.0e-5 at all temperatures

Applied load is 10 000 lbf


PROPERTIES




Applied load is 10,000 lbf

Initial and final temperature is 100 0F and 200 0F, respectively

The model has zero stress at the initial temperature


PROPERTIES

The needed information for the creation of the model is(continued) SOL 106

The case control section is used to select the initial temperature,

followed by two subcases, the latter of which is used to select the finaltemperature Case control

TEMP (INIT) = 10

SUBCASE 1

LOAD = 1 NLPARM = 1

SUBCASE 2




LOAD = 1

NLPARM = 2

TEMP (LOAD) = 20

The needed information for the creation of the model is(continued) The bulk data section is used to specify the initial and final temperature,

and the control of the nonlinear process Bulk data

$ INITIAL TEMPERATURE DISTRIBUTION

TEMPD, 10, 100.0

$ FINAL TEMPERATURE DISTRIBUTION

TEMPD, 20, 200.0

NLPARM, 1, 1

NLPARM, 2, 5


PROPERTIES





PROPERTIES

Nastran input file, with entries to be included





PROPERTIES

Nastran input file, complete




WORKSHOP PROBLEM 6: ELASTIC-PERFECTLY PLASTIC MATERIAL

PROPERTIES

Model is seven member truss, constrained from out ofplane motion

Use CROD elements

Material is elastic-perfectly plastic

Apply (enforced) displacements

Compute limit load for structure

2h

δ δ

σy1 σy1

σy2

σy2

σy22h = 10.0

A = 1.0

F = all 6 DOF fixed

ET

= 0.0

σy1 = 100.0

σy2 = 300.0




2h

σy2σy2

F F

E = 2.0e5y2

δ = 0.05


PROPERTIES


Case control NLPARM = 1

Bulk data Two entries for elastic-perfectly plastic properties

NLPARM, 1, 20, , , , , ,YES





PROPERTIES






PROPERTIES

Solution File




SECTION 6

NONLINEAR ELEMENTS




TABLE OF CONTENTS

Page

Types Of Nonlinear Elements 6-6Small Versus Large Strain 6-7

Small Strain Elements 6-10

Corotational Formulation 6-14

One-dimensional Small Strain Element Library 6-16Rod, Conrod, Tube (Small Strain) 6-17

Beam (Small Strain) 6-20

Two-dimensional Small Strain Element Library 6-30

Nonlinear Shell And Plate Elements 6-31

Output For Shell And Plate Elements 6-36

Three-dimensional Small Strain Element Library 6-38




Solid Elements 6-39

Large Strain Elements 6-43

Hyperelastic Elements 6-44

TABLE OF CONTENTS

Page

Total Lagrangian Formulation 6-47Volumetric Locking 6-49

Output For Hyperelastic Elements 6-51

Hyperelastic Element Limitations 6-54

Planar Hyperelastic Elements 6-55Solid Hyperelastic Elements 6-60

Contact (Interface) Elements 6-64

Gap Element 6-65

3-D Slideline Contact 6-80

BCONP Bulk Data Entry 6-89

BLSEG Bulk Data Entry 6-92

BFRIC B lk D t E t 6 95




BFRIC Bulk Data Entry 6-95

BWIDTH Bulk Data Entry 6-96

BOUTPUT Bulk Data Entry 6-99

TABLE OF CONTENTS

Page

BOUTPUT Case Control Command 6-100PARAM ADPCON 6-102

Summary 6-103

Large Strain (Hyperelastic) Physical Elements 6-109

Example Problem One 6-117Example Problem Two 6-122

Example Problem Three 6-124

Workshop Problem One 6-127

Solution To Workshop Problem One 6-130




TYPES OF NONLINEAR ELEMENTS

Physical elements Small strain (ROD, BEAM, QUAD4, TRIA3, HEXA, PENTA, TETRA). Large strain (QUAD4, QUAD8, QUAD, QUADX, TRIA3, TRIA6, TRIAX,

HEXA, PENTA, TETRA).

Contact (interface) elements GAP 3-D slideline.




SMALL VERSUS LARGE STRAIN

Longitudinal strain:

Shear strain:

Small strain does not include the quadratic terms in thesquare brackets

εx δuδx------ 1

2--- δu

δx------ 2 δv

δx------ 2 δw

δx------- 2+ ++=

εyδv

δy------

1

2---

δu

δy------

2 δv

δy------

2 δw

δy-------

2

+ ++=

+ [y

w

x

w

y

v

x

v

y

u

x

u

x

v

y

u xy

]




square brackets.


Large strain-displacement matrix (B) is nonlinear.

For large strains, different definitions of stress and strainare available.

Must use conjugate stress-strain definition.

All strain definitions give the same result for small strains(<10%).

Small (infinitesimal) strain:

Large (logarithmic) strain:

ε ∆ l

l 0

-----=

ε l d

l ---- ln

l

l 0

----=∫=




ε n 1 ε+( ) ε or ε 1≅= <<


At 10% stretch

Acceptable for engineering accuracy with a discrepancyof 0.47%.

At 100% stretchNot acceptable for engineering accuracy.

In metal forming problems, stretch could be more than100%, and large strain capability is required.

In most structural problems, small strain is adequate.

Large strain (> 10%) may be acceptable if it is highlylocalized, i.e., small compared to the total strain energy.

ε 0.1=( ) : ε ln 1.1 0.0953==

ε 1.0=( ) : ε ln 2 0.693==




SMALL STRAIN ELEMENTS

General May have large displacements and rotations (geometric nonlinear). May have nonlinear material constitutive relationship.

Elastic (isotropic, orthotropic, anisotropic)

Nonlinear elastic (isotropic, anisotropic)

Elastic-plastic (isotropic, anisotropic)

Temperature-dependent (Elastic: isotropic, orthotropic, anisotropic;Nonlinear elastic: isotropic)

Creep

Equilibrium is satisfied in deformed configuration.

Based on corotational formulation - A set of corotational axes thatcontinuously rotates with the element.

Use engineering strain, which is the strain in the element once theelement is rigidly rotated back to its original position




element is rigidly rotated back to its original position.


Displacement of an element is split into: Rigid body motion Element net deformation

Good for large global displacements and large globalrotations with small element strains.

Converges faster with fine meshes than in coarsemeshes.

Stiffness matrix is divided into material and geometric

parts. Geometric part is included by PARAM,LGDISP.

Nonlinear material is included by MATS1 MATTi (1 2




Nonlinear material is included by MATS1, MATTi (1, 2,

9), or CREEP.


Two types of OUTPUT FORMAT: NONLINEAR - STRESS in subcase LINEAR - FORCE in subcase

Strains in linear solution sequences are total strainsincluding thermal strains.

Strains in nonlinear solution sequences are themechanical strains, i.e., do not include thermal strains.

Output may be requested in SORT1 or SORT2.

SORT2 is applicable to linear format only.





Summary: Displacement transformation matrix may be nonlinear. Equilibrium is satisfied in deformed configuration.

Stress strain relationship may be nonlinear.

Strain displacement matrix is linear.

Small Strain Elements

PSOLIDCPENTA6-node Penta

PSOLIDCTETRA4-10 node Tetra

PSHELLCQUAD4, CQUAD84/8 node Shell

MATiMATS1

MATTi

CREEP

PSHELLCTRIA3, CTRIA63/6 node Shell

PBEAMCBEAM2-node Beam

PRODCROD2-node Rod

MaterialsPropertyConnectivityElement




PSOLIDCHEXA8-node Hexa

PSOLIDCPENTA6 node Penta

COROTATIONAL FORMULATION

Concept Applicable to small strain elements.

Consider a grid point Q.x X u+

Undeformed Element

Deformed Elemente1 Xe

e2

X0

b2

b1

Q'

O' d1

xe

u

x

x0

X

d2

wd

u

QO

u0




x X u+=

x0

X0

u0

+=

COROTATIONAL FORMULATION

Undeformed and deformed grid point position with respect to elementorigin.

Xe =X-X0

xe = x-x0

Total grid point displacement with respect to element origin

ue=xe-Xe

ue=u-u0

where X = undeformed position of grid point Q

x = deformed position of grid point Q

u = displacement of grid point Q

Xe = undeformed position of grid point Q w.r.t. element originxe = deformed position of grid point Q w.r.t. deformed elem. origin

ue = total displacement of grid point Q w.r.t. deformed elem. origin

Net deformation of grid point Q




g p

Ud( )

xe

d( ) Xe

e( )–=

DeformedSystem

DeformedSystem

UndeformedSystem

ONE-DIMENSIONAL SMALL STRAIN ELEMENTLIBRARY

ROD, CONROD, TUBE

BEAM

GA

GB




ROD, CONROD, TUBE (SMALL STRAIN)

Connected by two grid points.

Force components: axial force Ptorque T

Displacement components: uiθi

Straight, prismatic member.

Nonlinear capabilities: Geometric nonlinear

Only axial component may be material nonlinear

Small strain only

YesMATS1CONROD

MATiPRODCROD

Geometric

Nonlinearity

MaterialPropertyConnectivity




MATTIPTUBECTUBE


Nonlinear Output Format

NONLINEAR ELEMENT PROBLEM: NLELI65 FEBRUARY 20, 1986 MSC/NASTRAN 11/27/85 PAGE 33

INELASTIC LOADING

CHECK OUTPUT FORMATS FOR NONLINEAR ELEMENTS SUBCASE 2

LOAD STEP = 2.00000E+00

N O N L I N E A R S T R E S S E S I N R O D E L E M E N T S ( C R O D )

ELEMENT AXIAL STRESS EQUIVALENT TOTAL STRAIN EFF. STRAIN EFF. CREEP LIN.TORSIONAL

ID STRESS PLASTIC/NLELAST STRAIN STRESS

8900 4.500000E+04 4.500000E+04 3.000000E-03 1.500000E-03 0.0 0.0





Linear Output FormatNONLINEAR ELEMENT PROBLEM : NLELI65 FEBRUARY 20, 1986 MSC/NASTRAN 11/27/85 PAGE 52INELASTIC LOADING

CHECK OUTPUT FORMATS FOR NONLINEAR ELEMENTS NONLINEARSUBCASE 2

LOAD STEP = 2.00000E+00

S T R E S S E S I N R O D E L E M E N T S ( C R O D )

ELEMENT AXIAL SAFETY TORSIONAL SAFETY ELEMENT AXIAL SAFETY TORSIONAL

SAFETYID. STRESS MARGIN STRESS MARGIN ID. STRESS MARGIN STRESS

MARGIN

8900 4.500000E+04 0.0




BEAM (SMALL STRAIN)

Connected by two grid points

Element Coordinate System Orientation of cross-sectional bending properties are defined by the third

grid point or orientation vector v. Additional degrees of freedom must be defined for the warping variables

( ti l)

(0,0,0)

znb

Plane 2

Shear Center

Nonstructural MassCenter of Gravity

Neutral Axis

Grid PointGA

Plane 1

yelem

zna

yna

ymazelem

xelem

(xb ,0,0)

ynb

zmb

ymb

zma

Grid PointGB

I1 = Izz

I2 = Iyy

x x vwa Offset

wb Offset

v

Myy

z

Iyy

-------------

Mzz

y

Izz

-------------




(optional).

BEAM (SMALL STRAIN)

Force components: Axial force PTotal torque TWarping torque TwBending moments in planes 1 and 2 (Mi)Shears in planes 1 and 2 (Vi)

Displacement components: ui

θi / scalar point

Nonlinear capabilities Geometric nonlinear

Material nonlinear hinge at each end couples axial and bending components.

Small strain only

dθ dx( )i

MATSiPBCOMPY

MATiPBEAMCBEAM

Geometric NonlinearityMaterialPropertyConnectivity




CREEP

MATTiYes

BEAM (SMALL STRAIN)

Notes: 1. BAR is not a nonlinear element.2. Any kind of nonlinearity specified for BAR is ignored.

Plastic Hinges for the Beam Element Rationale:

If work-hardening is negligible, a plastic hinge appears in a frame at the pointwhere the bending moment is maximum.

The ratio of the collapse moment to the moment at first yield rangesfrom 1.0 to 2.0 for practical sections.

F i i b l i h d l d l h b di

P

Plastic Zone




For a prismatic beam element with end loads only, the bending moment

is maximum at one end.

BEAM (SMALL STRAIN)

MAT1 ’ MATS1 “Plastic”

Will only yield at grid point.

Plasticity is simulated by eight plastic rods that support extension andbending about two axes (y and z).

Taper is allowed.

l /8

Potential PlasticZones

A B

l

l /8




BEAM (SMALL STRAIN)

Arrangement of Equivalent Plastic Rods for Beam Ends

(radius of the gyration of the area)Ky

l zz

A-------- Kz;

I yy

A---------= =

Centroidof Section

y

y and z are principal axes.

Locations aredetermined by

I1 and I2.

z

0 2l Kz,( )

2 Ky 0,( )

Ky Kz–,( )

Ky– Kz,( ) Ky Kz,( )




( gy )

Matching moments of inertia (I1,I2) and the cross-sectional area.

BEAM (SMALL STRAIN)

Accuracy in Calculation of Ultimate Moment in Yielded

StateLetη

Calculated Ultimate Moment

Theoretical Ultimate Moment-----------------------------------------------------------------------=

Moment Axis

Any multiple of 450

from y-axis0.9481

η

Moment Axis

y or z 0.9856

η

t

z

y

z

y




BEAM (SMALL STRAIN)

htw wtf for y-axis for z- axis

0 0.8536 0.9856

0.5 0.8826 1.102

1.0 0.9031 1.207

2.0 0.9295 1.394

0.9856α α

h/w for y-axis for z- axis

0.0 0.8536 0.9856

0.5 0.9031 0.95431.0 0.9295 0.9295

2.0 0.9543 0.9031

0.9856 0.8536α

z

y

t

w

t << h,w

h

tf z

y

w

tw

tf , tw << h,w

h




BEAM (SMALL STRAIN)

Limitations of Nonlinear Beam Elements

The material is assumed to be elastic-perfectly plastic. Work hardeningcan cause errors since results depend on beam length.

Any material nonlinearity other than elastic-perfectly plastic will yieldincorrect answers.

Treatment for torsion, warping, and transverse shear is linear.

Pin-flags are not allowed for material nonlinear analysis, i.e., theycannot be used with MATS1, MATT1, or CREEP Bulk Data entries.However, pin-flags can be used for geometric nonlinear analysis.

Offsets are not allowed for geometric nonlinear analysis.

Linear or nonlinear buckling analysis with offsets may give wrongresults.

No distributed loads (PLOAD1) are allowed.




BEAM (SMALL STRAIN)

Nonlinear Output FormatN O N L I N E A R O U T P U T F O R B E A M

NONLINEAR ELEMENT PROBLEM: NLELI65 FEBRUARY 20, 1986 MSC/NASTRAN 11/27/85PAGE 30

INELASTIC LOADING

CHECK OUTPUT FORMATS FOR NONLINEAR ELEMENTSSUBCASE 2

LOAD STEP = 2.00000E+00

N O N L I N E A R S T R E S S E S I N B E A M E L E M E N T S ( C B E A M )

ELEMENT GRID POINT STRESS EQUIVALENT TOTAL STRAIN EFF. STRAIN EFF.

CREEPID ID STRESS PLASTIC/NLELAST

STRAIN9400 9401 C -4.973799E-14 0.0 -1.657933E-21 0.0 0.0

D 3.000000E+04 3.000000E+04 1.046283E-03 4.628333E-05 0.0E -4.973799E-14 0.0 -1.657933E-21 0.0 0.0F -3.000000E+04 3.000000E+04 -1.046283E-03 4.628333E-05 0.0

9402 C 7.105427E-15 0.0 2.368476E-22 0.0 0.0

D -1.153490E-12 0.0 -3.844965E-20 0.0 0.0E 7.105427E-15 0.0 2.368476E-22 0.0 0.0

F 1.167700E-12 0.0 3.892335E-20 0.0 0.0




BEAM (SMALL STRAIN)

Linear Output Format

NONLINEAR ELEMENT PROBLEM: NLELI65 FEBRUARY 20, 1986 MSC/NASTRAN

11/27/85 PAGE 48INELASTIC LOADING

CHECK OUTPUT FORMATS FOR NONLINEAR ELEMENTS NONLINEARSUBCASE 2

LOAD STEP = 2.00000E+00

S T R E S S E S I N B E A M E L E M E N T S ( C B E A M )

STAT DIST/ELEMENT-ID GRID LENGTH SXC SXD SXE SXF S-MAX S-MINM.S.-T M.S.-C

9400 9401 0.000 -4.973799E-14 3.000000E+04 -4.973799E-14 -3.000000E+04 3.000000E+04 -3.000000E+049402 1.000 7.105427E-15 -1.153490E-12 7.105427E-15 1.167700E-12 1.167700E-12 -1.153490E-12




TWO-DIMENSIONAL SMALL STRAIN ELEMENTLIBRARY

TRIA3 (3 nodes) TRIA6 (6 nodes)

QUAD4 (4 nodes) QUAD8 (8 nodes)

T3

Q4

T6

Q8




NONLINEAR SHELL AND PLATE ELEMENTS

QUAD4 and TRIA3

Isoparametric elements: QUAD4 and TRIA3. Membrane and plate bending applicable to nonlinear material.

Transverse shear (Mindlin) remains linear.

Simulate thick or thin curved shell.

QUAD4 is preferred. TRIA3 is too stiff in membrane. Each connecting node has 6 DOFs. Stiffness is not defined for rotation

about the normal to the plane. Therefore, use K6ROT.

Midplane offset may be used for geometric nonlinear only.

Pass constant stress patch test.

No shear locking, no spurious modes.

Poisson’s ratio locking exists, especially in plane strain.

Use of offsets will cause incorrect results in buckling analysis anddifferential stiffness.





QUAD4: Connected by four grid points. The orientation of the normal to

the surface is defined by the connectivity. TRIA3: Connected by three grid points. The orientation of the normal

to the surface is defined by the connectivity.

Force components: Membrane forces Fx, Fy, Fxy

Bending moments Mx, My, MxyTransverse shear forces Qx, Qy

Stress components: σx, σy, τxy (at center)


θx, θy (no rotation normal to element)





QUAD8 and TRIA6

Isoparametric elements: QUAD8 and TRIA6. Membrane and plate bending applicable to nonlinear material.

Transverse shear (Mindlin) remains linear.

Simulate thick or thin curved shell.

Each connecting node has 6 DOFs. Stiffness is not defined for rotationabout the normal to the plane. Therefore, use K6ROT.

Midplane offset may be used for geometric nonlinear only.

Pass constant stress patch test.

Use of offsets will cause incorrect results in buckling analysis and

differential stiffness.





QUAD8: Connected by eight grid points. The orientation of the normal to

the surface is defined by the connectivity. TRIA6: Connected by three grid points. The orientation of the normal

to the surface is defined by the connectivity.

Force components: Membrane forces Fx, Fy, Fxy

Bending moments Mx, My, MxyTransverse shear forces Qx, Qy

Stress components: σx, σy, τxy (at center)


θx, θy (no rotation normal to element)






Material nonlinear for membrane and bending components

CREEPCTRIA6

MATTiCQUAD8

MATS1CTRIA3

YesMATiPSHELLCQUAD4

Geometric

Nonlinearity





OUTPUT FOR SHELL AND PLATE ELEMENTS

Nonlinear Format

NONLINEAR ELEMENT PROBLEM: NLELI65 FEBRUARY 20, 1986 MSC/NASTRAN 11/27/85PAGE 34INELASTIC LOADING

CHECK OUTPUT FORMATS FOR NONLINEAR ELEMENTSSUBCASE 2LOAD STEP = 2.00000E+00

N O N L I N E A R S T R E S S E S I N T R I A N G U L A R E L E M E N T S ( T R I A 3 )

ELEMENT FIBRE STRESSES/ TOTAL STRAINS EQUIVALENT EFF. STRAIN EFF.CREEPID DISTANCE X Y Z XY STRESS PLASTIC/NLELAST

STRAIN

8800 -5.000000E-02 4.504111E+04 1.015688E+02 -6.266323E-14 4.499041E+04 1.499041E-03 0.02.993653E-03 -8.078549E-04 0.0

5.000000E-02 4.504111E+04 1.015688E+02 -6.266323E-14 4.499041E+04 1.499041E-03 0.02.993653E-03 -8.078549E-04 0.0

8801 -5.000000E-02 2.257134E+04 2.257134E+04 -2.246977E+04 4.499041E+04 1.499041E-03 0.01.092899E-03 1.092899E-03 -3.801508E-03

5.000000E-02 2.257134E+04 2.257134E+04 -2.246977E+04 4.499041E+04 1.499041E-03 0.0

1.092899E-03 1.092899E-03 -3.801508E-03




OUTPUT FOR SHELL AND PLATE ELEMENTS

Linear Format

NONLINEAR ELEMENT PROBLEM: NLELI65 FEBRUARY 20, 1986 MSC/NASTRAN11/27/85 PAGE 53

INELASTIC LOADING

CHECK OUTPUT FORMATS FOR NONLINEAR ELEMENTS NONLINEARSUBCASE 2LOAD STEP = 2.00000E+00

S T R E S S E S I N T R I A N G U L A R E L E M E N T S ( T R I A 3 )

ELEMENT FIBRE STRESSES IN ELEMENT COORD SYSTEM PRINCIPAL STRESSES (ZERO SHEAR)

ID. DISTANCE NORMAL-X NORMAL-Y SHEAR-XY ANGLE MAJOR MINORVON MISES8800 -5.000000E-02 4.504111E+04 1.015688E+02 -6.266323E-14 0.0000 4.504111E+04 1.015688E+02

4.499041E+045.000000E-02 4.504111E+04 1.015688E+02 -6.266323E-14 0.0000 4.504111E+04 1.015688E+02

4.499041E+048801 -5.000000E-02 2.257134E+04 2.257134E+04 -2.246977E+04 -45.0000 4.504111E+04 1.015688E+02

4.499041E+045.000000E-02 2.257134E+04 2.257134E+04 -2.246977E+04 -45.0000 4.504111E+04 1.015688E+02

4.499041E+04




THREE-DIMENSIONAL SMALL STRAINELEMENT LIBRARY

PENTA (6 nodes)

HEXA (8 nodes) TETRA (4 or 10 nodes)

Note:For HEXA & PENTA:All edge nodes must bedeleted for nonlinearanalysis.




SOLID ELEMENTS

HEXA: Connected by eight grid points.

PENTA: Connected by six grid points. TETRA: Connected by four (or ten) grid points.

Stress components: σx, σy, σz

τxy

, τyz

, τzx

(at center and corner points)



Material nonlinear

MATTiCTETRA

YesMATS1PSOLIDCPENTA

MATiCHEXA

Geometric NonlinearityMaterialPropertyConnectivity




CREEP

SOLID ELEMENTS

Note: 1. HEXA20 and PENTA15 are not nonlinear elements.

2. Any kind of nonlinearity specified for HEXA20 andPENTA15 is ignored.

Uses the strain function formulation that improves accuracy as

Poisson’s ratio approaches one-half. Internal degrees of freedom areintroduced to approximate quadratic shape function (for HEXA andPENTA only).



SOLID ELEMENTS

Nonlinear Output Format

1 ELASTIC-PLASTIC BUCKLING OF IMPERFECT SPHERICAL SHELL N10657 DECEMBER 2, 1993 MSC/NASTRAN 12/ 1/93 PAGE 177HYDROSTATIC PRESSURE APPLIED,PERIPHERY CLAMPED

0 SUBCASE 3LOAD STEP = 3.00000E+00

N O N L I N E A R S T R E S S E S I N H E X A H E D R O N S O L I D E L E M E N T S ( H E X A )

ELEMENT GRID/ POINT STRESSES/ TOTAL STRAINS EQUIVALENT EFF. STRAIN EFF. CREEPID GAUSS ID X Y Z XY YZ ZX STRESS PLAS/NLELAS STRAIN

0 11 GRID CENTER 5.8398E+04 4.4519E+04 -2.0017E+04 -3.0549E-08 3.1213E-08 3.7113E+02 7.8000E+04 2.3300E-02 .02.6695E-03 5.3108E-04 -1.3093E-04 -6.1377E-15 7.6010E-15 1.4537E-03

101 -1.6090E+05 -4.6212E+04 4.0530E+04 -1.8137E-07 7.5895E-08 -9.9750E+02 6.4259E+04 9.9663E-02 .0-9.1663E-02 -1.6827E-02 1.0228E-01 1.5330E-14 5.2356E-15 1.4537E-03

103 -1.3510E+05 -1.5007E+05 -3.2840E+04 -2.0672E-07 -1.3284E-08 -1.0080E+02 2.4847E+04 7.8079E-02 .0-5.1229E-02 -4.3337E-02 8.2691E-02 1.5330E-14 9.9665E-15 1.4537E-03

104 -1.3510E+05 -1.5007E+05 -3.2840E+04 -2.0672E-07 -1.3284E-08 -1.0080E+02 2.4847E+04 7.8079E-02 .0-5.1229E-02 -4.3337E-02 8.2691E-02 1.5330E-14 9.9665E-15 1.4537E-03

102 -1.6090E+05 -4.6212E+04 4.0530E+04 -1.8137E-07 7.5895E-08 -9.9750E+02 6.4259E+04 9.9663E-02 .0-9.1663E-02 -1.6827E-02 1.0228E-01 1.5330E-14 5.2356E-15 1.4537E-03

201 2.4837E+05 1.5368E+05 1.6060E+04 -2.5726E-08 8.0401E-08 -3.5225E+02 7.8000E+04 1.0156E-01 .01.0069E-01 1.7070E-02 -1.0227E-01 -2.7606E-14 5.2356E-15 1.4537E-03

203 1.9665E+05 1.8168E+05 2.0698E+04 -2.8129E-08 -1.7611E-08 -2.8327E+02 7.8000E+04 7.5058E-02 .05.2881E-02 4.5218E-02 -8.3228E-02 -2.7606E-14 9.9665E-15 1.4537E-03

204 1.9665E+05 1.8168E+05 2.0698E+04 -2.8129E-08 -1.7611E-08 -2.8327E+02 7.8000E+04 7.5058E-02 .05.2881E-02 4.5218E-02 -8.3228E-02 -2.7606E-14 9.9665E-15 1.4537E-03

202 2.4837E+05 1.5368E+05 1.6060E+04 -2.5726E-08 8.0401E-08 -3.5225E+02 7.8000E+04 1.0156E-01 .01.0069E-01 1.7070E-02 -1.0227E-01 -2.7606E-14 5.2356E-15 1.4537E-03




LARGE STRAIN ELEMENTS

General:

Can have large displacements and rotation. Only isotropic hyperelastic material is available (MATHP).

Strain to displacement matrix is nonlinear.

Equilibrium is satisfied in deformed configuration.




HYPERELASTIC ELEMENTS

Hyperelastic element characteristics:

All hyperelastic elements have hyperelastic materials only (MATHP BulkData entry). Hyperelastic material includes linear elastic material.

Total Lagrangian formulation with updated coordinates.

Green strain potential function.

Energy conjugate stress-strain pair: Cauchy stress and symmetric part

of the virtual displacement gradient. Deformation is split into volumetric and distortional components.

Mixed formulation: Separate interpolation for displacements andvolume ratio/pressure.




HYPERELASTIC ELEMENTS

Avoids volumetric locking for nearly incompressible material.

Stiffness matrix is divided into material and geometric parts. Geometric part is included by PARAM,LGDISP,1.

It is strongly recommended that PARAM,LGDISP,1 be used.

Temperature loads can be specified for all elements.

Follower pressure loads are available for all elements.

Missing grids (e.g., 5 node CQUAD) are not recommended.




HYPERELASTIC ELEMENTSElement Type Connectivity Property Hyperelastic

MaterialPlane Strain:

4-noded QUAD

5-9 noded QUAD

3-noded TRIA

6-noded TRIA

CQUAD4CQUAD8CQUAD

CQUAD8CQUAD

CTRIA3

CTRIA6

CTRIA6

PLPLANE

PLPLANE

PLPLANE

PLPLANE

MATHP

MATHP

MATHP

MATHP

Axisymmetric:

4-9 noded QUAD

3-6 noded TRIA

CQUADX

CTRIAX

PLPLANE

PLPLANE

MATHP

MATHP

Solid:

8-20 noded HEXA

6-15 noded PENTA

4-10 noded TETRA

CHEXA

CPENTA

C TETRA

PLSOLID

PLSOLID

PLSOLID

MATHP

MATHP

MATHP




TOTAL LAGRANGIAN FORMULATION

Concept x = X + u

The datum is always the initial state

Previous

Converged Solution

Initial State

Last Estimate

x

z

Basic Coordinate System

New Estimate

Vi

V0

ui( ) u

i 1+( )

Vi 1+

∆ui

uiu

0

X

ui 1+

ui

∆ui

+=

xi 1+

X ui 1+

+=

y




The datum is always the initial state.

TOTAL LAGRANGIAN FORMULATION Hyperelastic elements use the updated coordinates to form the

stiffness matrix.




VOLUMETRIC LOCKING

What Is Volumetric Locking?

Pressure

For nearly incompressible materials (Di α , J = 1) Stiffness matrix is ill-conditioned.

Spurious stresses.

Locking.

Volumetric Locking Avoidance Mixed formulation

Energy functional:

P( ) 2i J 1–( )2i 1–

Di

i 1=

ND

∑=

W u J p, ,( ) U I1 I2J,( ) p J J–( )+[ ] V0 Wext

u( )+d

B0

∫=




B0

VOLUMETRIC LOCKING

Virtual work of the internal forces:

Separate interpolations for displacements and volume ratio/pressure.

δWint

ST

δE V0 σT

∇S

δu( ) VdB∫

=dB0∫=




OUTPUT FOR HYPERELASTIC ELEMENTS

Cauchy stress σ

Defined from

where df = force in the deformed state

n = unit normal to the deformed areadA = deformed area

df σn dA=

x-y plane of basic Axisymmetricx-y plane of user-specified coordinate system; default=basicPlain Strain

BasicSolids

Output Coordinate SystemElements





Logarithmic strain

where λi = principal stretchesNi = unit vectors in the principal directions

ε ln λi N i N iT

i 1=

3

∑=





Note that in case of temperature strains, the total strains are output

Pressure (tension is positive)

Volumetric strain (volume increase is positive)

where J = det F Linear and nonlinear output format is available.

Output may be requested in SORT1 or SORT2.

SORT2 is applicable to linear format.

ε σ

E--- α∆T+=

p1

3--- tr σ=

εV

J 1= -




HYPERELASTIC ELEMENT LIMITATIONS

Fully incompressible material is not available yet; nearly

incompressible material is Poisson’s ratiov ≤ 0.4995 or D1 ≤ 1000. (A10 + A01)

Hyperelastic elements are only available in SOLs 106and 129 and are not available in SOL 66 or SOL 99.

SOL 101 does not produce a fatal error; however, itgives the wrong results.

Stress and strain output only in basic with no grid point

stress output, no center stress output and no user-defined coordinate system for output.




PLANAR HYPERELASTIC ELEMENTS

Plane strain: QUAD4, QUAD8, QUAD, TRIA3, TRIA6,

TRIA. Axisymmetric: QUADX, TRIAX.

Properties are specified by PLPLANE.

PLPLANE Bulk Data Entry Defines a finite deformation plane strain element.

Format:

Example:

CIDMIDPIDPLPLANE

10987654321

201204203PLPLANE




PLANAR HYPERELASTIC ELEMENTSField Contents

PID Element property identification number. (Integer > 0).

MID Identification number of MATHP entry. (Integer > 0).Identification number of a coordinate system defining theplane of deformation.

CID See Remarks 2 and 3. (Integer Š 0; Default = 0).

Remarks:1. PLPLANE can be referenced by a CQUAD, CQUAD4, CQUAD8, CQUADX,

CTRIA3, CTRIA6, or CTRIAX entry.

2. Plane strain hyperelastic elements must lie on the x-y plane of the CID coordinate

system. Stresses and strains are output in the CID coordinate system.3. Axisymmetric hyperelastic elements must lie on the x-y plane of the basic

coordinate system. CID may not be specified and stresses and strains are output inthe basic coordinate system.




PLANAR HYPERELASTIC ELEMENTS ID must be unique between PSHELL and PLPLANE; otherwise, User Fatal

Message 5410 is issued.

Fatal Error 6438 is issued if MATHP is not specified.

Output is in terms of Cauchy stress/log strains in the x-y plane of thereferred coordinate system at each Gauss point.





Nonlinear Output Format1 PURE SHEAR NOVEMBER 18, 1993 MSC/NASTRAN 11/17/93 PAGE 830

LOAD STEP = 1.00000E+00N O N L I N E A R S T R E S S E S I N H Y P E R E L A S T I C Q U A D R I L A T E R A L E L E M E N T S ( QUADFD )

ELEMENT GRID/ POINT CAUCHY STRESSES/ LOG STRAINS PRESSURE VOL. STRAINID GAUSS ID X Y Z XY

0 1 GAUS 1 1.678579E+02 -1.188351E-03 5.456201E+00 1.560323E-16 5.777097E+01 1.925699E-021.791759E+00 -1.772686E+00 .0 .0

2 1.678579E+02 -1.188351E-03 5.456201E+00 -1.191492E-15 5.777097E+01 1.925699E-021.791759E+00 -1.772686E+00 .0 .0

3 1.678579E+02 -1.188351E-03 5.456201E+00 -3.368891E-16 5.777097E+01 1.925699E-02

1.791759E+00 -1.772686E+00 .0 .04 1.678579E+02 -1.188351E-03 5.456201E+00 -7.083336E-16 5.777097E+01 1.925699E-02

1.791759E+00 -1.772686E+00 .0 .01 SIMPLE TENSION, AXISYMMETRIC ELEMENT SEPTEMBER 3, 1993 MSC/NASTRAN 9/ 2/93 PAGE 1030

LOAD STEP = 1.00000E+00NONLINEAR STRESSES IN H Y P E R E L A S T I C A X I S Y M M. Q U A D R I L A T E R A L ELEMENTS (QUADXFD)

ELEMENT GRID/ POINT CAUCHY STRESSES/ LOG STRAINS PRESSURE VOL. STRAINID GAUSS ID RAD YY THETA RY

0 1 GAUS 1 1.806839E-09 2.917973E+02 1.806839E-09 -1.456493E-15 9.726578E+01 3.242192E-02

-8.296252E-01 1.818534E+00 -3.250761E-01 -9.570014E-012 1.806910E-09 2.917973E+02 1.806910E-09 3.027552E-16 9.726578E+01 3.242192E-02

-8.296252E-01 1.818534E+00 -3.250761E-01 -9.570014E-013 1.807052E-09 2.917973E+02 1.807052E-09 -6.255834E-15 9.726578E+01 3.242192E-02

-8.296252E-01 1.818534E+00 -3.250761E-01 -9.570014E-01

4 1.807052E-09 2.917973E+02 1.807052E-09 -7.787191E-16 9.726578E+01 3.242192E-02-8.296252E-01 1.818534E+00 -3.250761E-01 -9.570014E-01





Linear Output Format

1 PURE SHEAR NOVEMBER 18, 1993 MSC/NASTRAN 11/17/93

PAGE 1250 NONLINEAR

LOAD STEP = 1.00000E+00

S T R E S S E S I N H Y P E R E L A S T I C Q U A D R I L A T E R A L E L E M E N T S ( QUADFD )ELEMENT GRID/ POINT ---------CAUCHY STRESSES-------- PRINCIPAL STRESSES (ZERO SHEAR)

ID GAUSS ID NORMAL-X NORMAL-Y SHEAR-XY ANGLE MAJOR MINOR0 1 GAUS 1 1.678579E+02 -1.188351E-03 1.560323E-16 .0000 1.678579E+02 -1.188351E-03

2 1.678579E+02 -1.188351E-03 -1.191492E-15 .0000 1.678579E+02 -1.188351E-033 1.678579E+02 -1.188351E-03 -3.368891E-16 .0000 1.678579E+02 -1.188351E-03

4 1.678579E+02 -1.188351E-03 -7.083336E-16 .0000 1.678579E+02 -1.188351E-031 SIMPLE TENSION, AXISYMMETRIC ELEMENT SEPTEMBER 3, 1993 MSC/NASTRAN 9/ 2/93PAGE 154

0 NONLINEARLOAD STEP = 1.00000E+00

S T R E S S E S I N H Y P E R E L A S T I C A X I S Y M M. Q U A D R I L A T E R A L E L E M E N T S (QUADXFD)ELEMENT GRID/ POINT STRESSES IN ELEMENT COORD SYSTEM PRINCIPAL STRESSES (ZERO SHEAR)

ID GAUSS ID RADIAL NORMAL-Y SHEAR-RY ANGLE MAJOR MINOR

0 1 GAUS 1 1.806839E-09 2.917973E+02 -1.456493E-15 -90.0000 2.917973E+02 1.806825E-092 1.806910E-09 2.917973E+02 3.027552E-16 90.0000 2.917973E+02 1.806939E-09

3 1.807052E-09 2.917973E+02 -6.255834E-15 -90.0000 2.917973E+02 1.807052E-094 1.807052E-09 2.917973E+02 -7.787191E-16 -90.0000 2.917973E+02 1.807052E-09




SOLID HYPERELASTIC ELEMENTS

HEXA, PENTA, and TETRA.

Properties are specified by PLSOLID. PLSOLID Bulk Data Entry

Defines a finite deformation solid element.

Format:

Example:

MIDPIDPLSOLID

10987654321

2120PLSOLID




SOLID HYPERELASTIC ELEMENTSField Contents


MID Identification number of a MATHP entry. (Integer > 0).

Remarks:1. PLSOLID can be referenced by a CHEXA, CPENTA or CTETRA entry.

2. Stress and strain are output in the basic coordinate system.

IDs must be unique between PSOLID and PLSOLID; otherwise, User FatalMessage 5410 is issued.

Fatal Error 6438 is issued if MATHP is not specified. Output is in terms of Cauchy stress/log strain in the basic coordinate

system at each Gauss point.



CONTACT (INTERFACE) ELEMENTS GAP

3-D slideline




GAP ELEMENT

Connects two grid points with the orientation (gapdirection).

Opening or closing (contact) is determined in the gapdirection.

Uses hard surface contact, i.e., no penetration of grid

points is allowed in the gap direction. Can specify friction between the two points.

Uses the penalty method for both contact and friction.

Can have a large opening between the two points. No large relative slipping between the two points is

permitted.

No large rotation for the two points (relative or rigid).




GAP ELEMENT

CGAP Bulk Data Entry Defines a gap or frictional element for nonlinear analysis.

Format:

Example:

Alternate Format and Example:

NoPGAPCGAP

Geometric

Nonlinearity


CIDX3X2X1GBGAPIDEIDCGAP

10987654321

-6.10.35.2112110217CGAP

CIDGOGAGAPIDEIDCGAP

13112110217CGAP




S6 67NAS 103 S ti 6 D b 2003

GAP ELEMENT CID identifies the element coordinate system.

T1, T2, and T3 of CID are the element x-, y-, and z-axis, respectively.

For noncoincident grid points GA and GB if CID is not defined GA - GB defines the x-axis.

Orientation vector is given by x1, x2, and x3, (like beam element) or GA -GO defines the x-y plane.

For coincident grid points GA and GB, If CID is blank, the job is terminated with a fatal message.




S6 68NAS 103 Section 6 December 2003

GAP ELEMENT

CGAP Element Coordinate System.

GA

GB

KA − KBKB

Note: KA and KB in thisfigure are from thePGAP entry.

v

zelem

yelem

xelem




S6 69NAS 103 Section 6 December 2003

GAP ELEMENT

PGAP Bulk Data Entry Defines the properties of the gap element (CGAP entry).

Format:

Example:

0.250.251.0E+61.0E+62.50.0252PGAP

TRMINMARTMAX

MU2MU1KTKBKAF0U0PIDPGAP

10987654321




S6-70NAS 103 Section 6 December 2003

GAP ELEMENT

Shear Force for GAP Element.

F0

Fx (Compression)

Slope = KB

Slope = KA

(Tension)

Slope KA is used when UA – U

B ≥ UO

(Compression)UA – UBUO

Nonlinear Shear

Unloading

Slope = KT

MU1 ?asterisk14? Fx

MU2 ?asterisk14? Fx

∆V or ∆W

GAP Element Force-Deflection Curve for Nonlinear Analysis.





GAP ELEMENT There are two kinds of GAP element:

New and adaptive (TMAX >=0., preferred choice). New GAP can forcebisection and stiffness updates.

Old and non-adaptive (TMAX = –1.0).

New GAP element is recommended.

Old GAP element will not be covered.

Initial GAP opening is defined by U0, not by the distance.

Preload is defined by F0 (not recommended). Closed stiffness Ka is used when U A – UB ≥ U0.

The default for open stiffness Kb = 10 –14 Ka.



S6 71NAS 103, Section 6, December 2003


GAP ELEMENT

The transverse shear stiffness KT becomes active uponcontact. (The default = µ

1

* Ka

).

The continuation line is applicable for adaptive features

of the new GAP element only. Adaptive features are specified by TMAX > 0.

Penalty values are adjusted based on the penetration.

If the penetration is greater than TMAX, the penaltyvalue is increased by a magnitude.

New Default

1 Static Friction 0.0

2 Kinetic Friction 1



, ,


GAP ELEMENT

If the penetration is less than TRMIN * TMAX, thepenalty value is decreased by a magnitude.

MAR defines the lower and upper bounds for the penaltyvalue adjustment ratio.

Proper Estimation of Gap Stiffness

The stiffness of the beam at points A and B

The stiffness of the beam at points A and B

A

B

KA

3EI

L3

--------- 1= = KB

48EI

L3

------------ 16= =

KA

1000 * MAX K AK

B,( ) 16 10

3×=≥

KB

103–

* MIN KAK

B,( ) 0.001=≤




GAP ELEMENT The recommended stiffness acts rigid when closed and acts free when

open with an error of 0.1%.

Factors (103 or 10 –3) may be reduced to facilitate convergence at theexpense of accuracy.

Recommended stiffnesses are based on the decoupled stiffnesses.

Friction Features Friction effect is turned off with Kt = 0. Static and kinetic frictions are allowed.

Frictional gap problem is path dependent.

Sticking with elastic stiffness Kt before slipping.

Slipping is similar to plasticity. Sub-incremental process similar to plasticity is used for the new gap.

No sub-incremental process for the old gap.

Accuracy deteriorates if the increment produces large changes in thedisplacements with friction.




GAP ELEMENT

The slip locus is generalized by an ellipse:.

Fy2

Fz2

µsFx( )2

≤+ Closed and Sticking

Fy

2F

z

2µ

k F

x( )

2>+ Closed and Slipping




GAP ELEMENT

Caution for Using GAP Element Large rotation capability is not implemented.

When used for linear analysis, GAP stays linear with the initial stiffness.

The penalty values (Ka and Kt) should be as small as possible forsolution efficiency, but large enough for acceptable accuracy.

Penalty values are constants while the structural stiffness in the

adjacent structure changes continuously during loading. Avoid friction unless its effect is significant.

Use smaller increments if friction is involved.

Avoid complications by using isotropic friction (for old gap).

Typical coefficients of friction: Steel on steel (dry) 0.4 to 0.6

Steel on steel (greasy) 0.05 to 0.1

Brake lining on cast iron 0.3 to 0.4

Tire on pavement (dry) 0.8 to 0.9




GAP ELEMENT

Output Format

1 NONLINEAR STATIC CONTACT OF A SPHERE ON A RIGID NG6603 JANUARY 26, 1993 MSC/NASTRAN 1/25/93 PAGE 353

FLAT PLANE. LOAD IN THE -Z IS 60. WITH FRICTION0 RESULTS ARE FOR A HEXA MODEL SUBCASE 3

LOAD STEP = 3.00000E+00S T R E S S E S ( F O R C E S ) I N G A P E L E M E N T S ( C G A P )

ELEMENT - F O R C E S I N E L E M S Y S T - - D I S P L A C E M E N T S I N E L E M S Y S T -

ID COMP-X SHEAR-Y SHEAR-Z AXIAL-U TOTAL-V TOTAL-W SLIP-V SLIP-W STATUS2001 5.41107E+00 -3.66852E-01 .0 2.20054E-02 -1.81889E-03 .0 -1.81522E-03 .0 STICK2002 1.03702E+01 -2.07404E+00 -1.66861E-14 1.97010E-01 -1.78610E-02 -6.96423E-19 -1.78403E-02 -5.29562E-19 SLIP2003 -1.21279E-09 .0 .0 4.20721E-01 -1.52148E-02 .0 -1.52148E-02 .0 OPEN

2004 5.23067E+00 .0 .0 2.20193E-02 -1.80472E-02 .0 -1.80472E-02 .0 SLIDE




GAP ELEMENT

STRESS output request in the Case Control Section.

Output quantities are in the element coordinates. Output shows GAP status: open, slide, stick, slip.

Positive Fx is a compression force.

Total displacement is from the original position.

Slip displacement for the sticking or slipping condition isthe slip from the current contact position or slip center.

Slip-V

Force

Displacement




GAP ELEMENT

Slip displacement for the open or sliding condition is thesame as the total displacement.

If open , ( µ = 0 or µ ≠ 0 ), Total-V = Slip-V.

If sliding, Total-V = Slip-V ≠ 0 for new gap.

If sticking, Slip-V ≠ Total-V ≠ 0.

If slipped, Total-V ≠ Slip-V = Vs from slip center.




3-D SIDELINE CONTACT A master/slave segment is the line joining two consecutive nodes.

Master/slave nodes are the grid points in the contact region.

The slideline plane is the plane in which the master and slave nodesmust lie.

The master and slave nodes can have large relative motion within theslideline plane.

Relative motions outside the slideline plane are ignored. Therefore,they must be small.

Contact is determined between the slave nodes and the master line(very important).




3-D SIDELINE CONTACT 3-D Slideline Element

Consists of three nodes: slave, master node 1, and master node 2.

where S, m1, m2 = slave, master node 1 and master node 2, respectivelya, a0 = current and previous surface coordinate

gn = penetration of slave node into the master segment

gt = sliding of the slave node on the master segment

n = normal direction for the master segment

x2 x 1–6

5

1

4 3

n

St

m1

gn

gt

x2 x 1–

x2 x1–

a

a0

m2

2




3-D SIDELINE CONTACT Note that the master nodes to which a slave node connects change

continually.

The only way an internal element can be identified is by the external gridnumber of the slave node.




3-D SIDELINE CONTACT Output

1 NONLINEAR STATIC CONTACT OF A SPHERE ON A RIGID PLANE WITHOUT FRIC JANUARY 26, 1993 MSC/NASTRAN 1/25/93 PAGE53

0 SUBCASE 3LOAD STEP = 3.00000E+00

R E S U L T S F O R S L I D E L I N E E L E M E N T S (IN ELEMENT SYSTEM)

SLAVE CONTAC MASTER SURFACE NORMAL SHEAR NORMAL SHEAR NORMAL SLIP SLIP SLIP

GRID ID GRID1 GRID2 CORDINATE FORCE FORCE STRESS STRESS GAP RATIO CODE110 1 315 313 3.5261E-01 1.1142E+01 .0 1.8074E+01 .0 6.6289E-03 -4.7546E-02 .0 SLIDE

108 1 315 313 2.2509E-01 9.9720E+00 .0 1.6162E+01 .0 5.8047E-03 -3.5960E-02 .0 SLIDE105 1 315 313 1.0759E-01 5.3024E+00 .0 1.7893E+01 .0 1.9600E-03 -1.8563E-02 .0 SLIDE208 2 315 314 1.0000E+00 .0 .0 .0 .0 -1.1209E+00 .0 .0 OPEN

176 2 315 314 4.7187E-01 2.7194E+00 .0 4.5926E+00 .0 1.5293E-03 -3.9368E-02 .0 SLIDE170 2 315 314 1.0759E-01 5.2982E+00 .0 1.7879E+01 .0 1.9684E-03 -1.8563E-02 .0 SLIDE




3-D SIDELINE CONTACT General Features

Can have as many slideline contact regions as desired.

Contact is determined only for slave nodes and the master line. May specify symmetric penetration, i.e., contact is determined for both

slave and master nodes into master and slave line, respectively.

Initial penetration of slave nodes into master line is not allowed.

User Warning Message 6315 is issued, if the initial penetration is lessthan 10% of the master segment length.

Coordinates of the slave node are changed internally to precludepenetration.

User Fatal Message 6314 is issued, if initial penetration for any slavenode is greater then 10% of the master segment length.

The master and slave nodes must be in the slideline plane in the initialgeometry; otherwise, Fatal Message 6312 is issued.




3-D SIDELINE CONTACT During the analysis, no check is made to ensure that the master and

slave nodes are in the slideline plane.

The slave or master nodes need not be attached to the physical element(model rigid surface).

Ensure that the contact region is properly defined so that there are noerroneous overhangs.

Forces/stresses are associated with slave nodes.

Output can be requested in SORT1 or SORT2. There is only one output format.




3-D SIDELINE CONTACT User Interface

Bulk Data entries: BCONP Defines the parameters for a contact region and its

properties.

BLSEG Defines the grid points on the master/slave line.

BFRIC Defines the frictional properties.

BWIDTH Defines the width/thickness associated with each slave

node. BOUTPUT Defines the output requests for slave nodes in a slideline

contact region.

Case Control command: BOUTPUT Selects contact region for output

DMAP parameter: ADPCON Adjusts penalty values on restart.




BCONP BULK DATA ENTRY Description:

Defines the parameters for a contact region and its properties

Format:

Example:

Field ContentsID Contact region identification number (Integer > 0)SLAVE Slave region identification number (Integer > 0).

MASTER Master region identification number (Integer > 0)

SFAC Stiffness scaling factor. This factor is used to scale thepenalty values automatically calculated by the program. (Real> 0 or blank)

1331151095BCOMP

CIDPTYPEFRICIDSFACMASTERSLAVEIDBCONP

10987654321




BCONP BULK DATA ENTRYField Contents

FRICID Contact friction identification number (Integer > 0 or blank)

PTYPE Penetration type (Integer = 1 or 2; Default =1).1: unsymmetrical (slave penetration only) (default)

2: symmetrical

CID Coordinate system ID to define the slide line plane vector andthe slide line plane of contact. (Integer > 0 or blank; Default =0 which means the basic coordinate system)




BCONP BULK DATA ENTRY

Can have as many contact regions as desired. Penalty values are automatically selected based on the diagonal terms

of grid points. In symmetrical penetration, both the slave and master nodes are

checked for penetration into the master and slave surface, respectively. The t3 direction of CID is the z-direction of all the 3-D slideline elements

(one corresponding to each slave node and also to each master nodefor symmetric penetration) of the contact region.

kSlave Line

Slideline Plane Vector Direction

Master Line

x

y

z

k-1th Slave Segment

1-th Master Segment

k − 1k + 1

l + 1

l − 1




BLSEG BULK DATA ENTRY Description:

Defines a curve which consists of a number of line segments via grid

numbers that may come in contact with other body. A line segment isdefined between every two consecutive grid points. Thus, number of linesegments defined is equal to the number of grid points specified minus1. A corresponding BWlDTH Bulk data entry may be required to definethe width/thickness of each line segment. If the corresponding BWlDTH

is not present, the width/thickness for each line segment is assumedunity

Format:

G12G11

G10BYG9THRUG8

G7G6G5G4G3G2G1IDBLSEG

10987654321




BLSEG BULK DATA ENTRY Examples:

Field Contents

ID Line segments identification number (Integer > 0)Gi Grid numbers on a curve in a continuous topological order so

that the normal to the segment points towards other curve.

Grid points must be specified in topological order.

Normals (z × t) of the master segments must face toward the slave linefor unsymmetric penetration.

Normals of master and slave segments must face each other forsymmetric penetration

44THRU35

33323027

14BY21THRU515BLSEG




BLSEG BULK DATA ENTRY These conditions are accomplished by traversing counterclockwise or

clockwise from the master line to the slave line depending on whetherthe slideline vector forms the right-hand rule or the left-hand rule.

The master line must have at least two grid points.

The slave line may have only one grid point for unsymmetricalpenetration.

Two grid points in a line cannot be the same or coincident except for the

first point and the last point, which signify a close region.




BFRIC BULK DATA ENTRY Description:

Defines frictional properties between two bodies in contact.

Format:

Example:

Field ContentsFID Friction identification number (Integer > 0)FSTIF Frictional stiffness in stick (Real > 0.0). Default =

automatically selected by the program.MU1 Coefficient of static friction (Real > 0.0).

(Note that no distinction is made between static and kinetic friction.)

MU1FSTIFFIDBFRIC

10987654321

0.333BFRIC




BWIDTH BULK DATA ENTRY Description

Defines width/thickness for line segments in 3-D/2-D slideline contact

defined in the corresponding BLSEG BULK Data entry. This entry maybe omitted if the width/thickness of each segment defined in the BLSEGentry is unity. Number of thicknesses to be specified is equal to thenumber of segments defined in the corresponding BLSEG entry. If there is no corresponding BLSEG entry, the width/thickness specified in

the entry are not used by the program. Format:

W12W11

W10BYW9THRUW8

W7W6W5W4W3W2W1IDBWIDTH

10987654321




BWIDTH BULK DATA ENTRY Examples:

Field Contents

ID Width/thickness set identification number (Real > 0.0).Wi Width/Thickness values for the corresponding line segments

defined in the BLSEG entry. (Real > 0.0).

44THRU35

2222

1BY5THRU215BWIDTH




BWIDTH BULK DATA ENTRY ID is the same as the slave line (BLSEG) ID.

Widths/thicknesses are specified for slave nodes only. Default = unity.

Widths/thicknesses are used for calculating contact stresses. Each slave node is assigned a contributory area.

The number of widths to be specified is equal to the number of slavenodes -1.

For only one slave node, specify the area in W1 field.




BOUTPUT BULK DATA ENTRY Description

Defines the slave nodes at which the output is requested.

Format:

Example:

Field ContentsID Boundary identification number for which output is desired

(Integer > 0.0).Gi Slave node numbers for which output is desired. Note: The ID is the same as the corresponding BCONP ID. This entry

can selectively specify the slave grid points for which OUTPUT isdesired.

B10BYG9THRUG8

G8G7G6G5G4G3G2G1

ALLIDBPOUTPUT

10987654321

ALL15BOUTPUT




BOUTPUT CASE CONTROL COMMAND Description:

Selects slave nodes specified in the Bulk Data entry BOUTPUT for

history output. Format:

Example: BOUTPUT = ALL BOUTPUT = 5

Field ContentsSORT1 Output is presented as a tabular listing of slave nodes for

each load or time depending on the solution sequence.SORT2 Output is presented as a tabular listing of load or time for

each slave node.

BOUTPUT

SORT1, PRINT

SORT2, PUNCH

PLOT ALL

n

None

=




BOUTPUT CASE CONTROL COMMANDField Contents

PRINT The print file (Fortran I/O unit 6) is the output media.

PUNCH The punch file is the output media.PLOT Generate slave node results history but do not print.

ALL Histories of all the slave nodes listed in all the BOUTPUTbulk data entries are output. If no BOUTPUT bulk data entriesare specified, histories of all the lave nodes in all the contact

regions are output.n Set identification of previously appearing set command. Only

contact regions whose identification numbers.

none Result histories for no slave nodes are output.

Note: This command selects the contact region for which output isdesired.




PARAM ADPCON User interface

PARAM,ADPCON,(real value)

On restart, ADPCON can be used to increase ordecrease the penalty values for all the line contactregions.

A negative value of ADPCON implies that penalty values

are calculated at the beginning of a subcase only. Thisis useful for contact between elastic bodies.

Penalty values for a line contact region are given by

where ks = number calculated automatically for a slave node by theprogram

SFAC = scale factor specified in BCONP

| ADPCON |* SFAC * k s




SUMMARY Small Strain Physical Elements




QUAD8

TRIA6




SUMMARY Small Strain Physical Elements (Cont.)

One-dimensional stress-strain curves use MAT1.

All other elements may be used for nonlinear analysis as long as theyremain linear.




TETRA10




LARGE STRAIN (HYPERELASTIC) PHYSICAL

ELEMENTS





ELEMENTS





ELEMENTS Contact Interface Elements




EXAMPLE PROBLEM ONE

Purpose To illustrate the use of hyperelastic elements

Problem Description Determine the force versus displacement curve for the rubber bushing

unit.

Assumptions Rubber material is perfectly bonded to frame and shaft. Frame and shaft are rigid.

Rubber

Frame

Shaft

15mm

30mm

Rubber material is theMooney-Rivlin type with:

Rubber Bushing

A10 0.177 N mm2

⁄ =

A01 0.045 N mm2

⁄ =

D1 333 N m m2

⁄ =

∆




EXAMPLE PROBLEM ONE

Solution Model one-half of rubber bushing taking advantage of symmetry.

Fully constraint the grid points at the outer boundary (between rubberand frame).

Constraint the horizontal degree of freedom for grid points at the innerboundary (between rubber and shaft) and tie the vertical motiontogether with MPC.

Force-DisplacementCurve of a Rubber

Bushing.




EXAMPLE PROBLEM ONE




EXAMPLE PROBLEM 1: .DAT File

ID, chap6E1, NAS103 Chap 6, EX 1 $ AR (12/03)SOL 106

CENDTITLE = Rubber Bushing, NAS103 chapter 6 Ex 1MPC = 13SUBCASE 1

NLPARM = 1SPC = 2LOAD = 1DISPLACEMENT=ALL

BEGIN BULKPARAM POST 0PARAM AUTOSPC NOPARAM LGDISP 2PARAM PRTMAXIM YESNLPARM, 1, 10, , AUTO, 1, 25, PW, YES

MATHP, 1, .177, .045, 333.PLPLANE, 1, 1CQUAD4, 1, 1, 1, 2, 9, 8=, *1, =, *7, *7, *7, *7=10CQUAD4, 13, 1, 2, 3, 10, 9=, *1, =, *7, *7, *7, *7=10CQUAD4, 25, 1, 3, 4, 11, 10=, *1, =, *7, *7, *7, *7=10

CQUAD4, 37, 1, 4, 5, 12, 11

=, *1, =, *7, *7, *7, *7

=10

CQUAD4, 49, 1, 5, 6, 13, 12

=, *1, =, *7, *7, *7, *7

=10

CQUAD4, 61, 1, 6, 7, 14, 13

=, *1, =, *7, *7, *7, *7

=10

GRID, 1, 1, 30., 0., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 8, 1, 30., 15., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 15, 1, 30., 30., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 22, 1, 30., 45., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 29, 1, 30., 60., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 36, 1, 30., 75., 0.

=, *1, =, *(-2.5), =, =

=5




EXAMPLE PROBLEM 1: .DAT File (Cont.)

GRID, 43, 1, 30., 90., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 50, 1, 30., 105., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 57, 1, 30., 120., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 64, 1, 30., 135., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 71, 1, 30., 150., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 78, 1, 30., 165., 0.

=, *1, =, *(-2.5), =, =

=5

GRID, 85, 1, 30., 180., 0.

=, *1, =, *(-2.5), =, =

=5

MPCADD, 13, 1, 2, 3, 4, 5, 6, 7,

, 8, 9, 10, 11, 12

MPC, 1, 14, 2, -1., 7, 2, 1.

MPC, 2, 21, 2, -1., 7, 2, 1.

MPC, 3, 28, 2, -1., 7, 2, 1.

MPC, 4, 35, 2, -1., 7, 2, 1.

MPC, 5, 42, 2, -1., 7, 2, 1.

MPC, 6, 49, 2, -1., 7, 2, 1.

MPC, 7, 56, 2, -1., 7, 2, 1.

MPC, 8, 63, 2, -1., 7, 2, 1.

MPC, 9, 70, 2, -1., 7, 2, 1.

MPC, 10, 77, 2, -1., 7, 2, 1.

MPC, 11, 84, 2, -1., 7, 2, 1.

MPC, 12, 91, 2, -1., 7, 2, 1.

SPCADD, 2, 1, 3, 4

SPC1, 1, 12, 1, 8, 15, 22, 29, 36,

, 43, 50, 57, 64, 71, 78, 85

SPC1, 3, 1, 7, 14, 21, 28, 35, 42,

, 49, 56, 63, 70, 77, 84, 91

SPC1, 4, 1, 1, THRU, 7SPC1 4, 1, 85, THRU, 91

FORCE, 1, 7, , 1200., 0., -1., 0.

CORD2C, 1, , 0., 0., 0., 0., 0., 1.,

, 0., -1., 0.

ENDDATA




EXAMPLE PROBLEM TWO

Purpose To illustrate the use of axisymmetric hyperelastic elements and follower

forces.

Problem Description A circular plate is 15 inches in diameter and 0.5 inches thick. It is simply

supported along the edge and is subjected to a uniform pressure of 45psi. Plot the deformed shape at various pressures.

Rubber material properties:

Solution The problem is solved in two ways:

Model a ten-degree wedge using HEXA8 and PENTA6 elements withaxisymmetric boundary conditions.

Model it using axisymmetric QUAD4 elements.

A10

80 psi, A01

20 psi==

D1

5 104

psi×=




EXAMPLE PROBLEM TWO

Deformed Shapes for the Wedge Model.Deformed Shapes for the AxisymmetricModel.




EXAMPLE PROBLEM THREE

Purpose To illustrate the 3-D slideline contact capability.

Problem Description Determine the deformed shape for a pipe being pushed in and out of a

clip.

Pipe Diameter = 10.1 mm

E = 2.1 × 105n = 0.3

Clip Diameter = 10.0 mm

E = 2.1 × 103n = 0.3





Solution Reduce the problem to a two-dimensional model.

Undeformed Shape Deformed Shape,∆Pipe = 5.0 mm





Deformed Shape, Deformed Shape

∆Pipe = 10.30 mm ∆Pipe = 7.5 mm





Purpose To demonstrate the use of 3-D slideline contact.

Problem Description An elastic punch is punched into an elastic foundation and then movedhorizontally to the right by 30 inches. The details of the model are asshown below.

Modify the input file to define a symmetric contact region. Use the displacement increment to push the punch horizontally to the

right by a total of 10 inches. Use increment of one inch per load step. Plot the deformed shapes at the end of subcases one and two.






ID CHAP6WS1,NAS103, Chap 6, Workshop 1 $ AR (12/0

TIME 300

SOL 106

CEND

$

TITLE = SYMMETRIC ELASTIC PUNCH WITH FRICTION

$

DISP = ALL

SUBCASE 1 $ VERTICAL LOAD

LOAD = 1

NLPARM = 410

SUBCASE 2 $ DISPLACEMENT TO THE RIGHT

LOAD = 1

$

BEGIN BULK

PARAM,POST,0

$

$ GEOMETRY

GRID,100,,0.,0.,0.,,123456

=,*1,,*(10.),===9

GRID,200,,0.,,20.,,2456

=,*1,,*(10.),==

=9

GRID,300,,45.,,20.,,2456

GRID,301,,55.,,20.,,2456

GRID,302,,65.,,20.,,2456

GRID,400,,45.,,25.,,2456

GRID,401,,55.,,25.,,2456GRID,402,,65.,,25.,,2456

$

$ ELEMENTS

CQUAD4,100,1,100,101,201,200

=,*1,=,*1,*1,*1,*1

=8

CQUAD4,200,2,300,301,401,400

=,*1,=,*1,*1,*1,*1

PSHELL,1,1,1.,-1PSHELL,2,2,1.,-1

MAT1,1,1.E5,,0.0

MAT1,2,1.E5,,0.0





$

$ PUNCH LOAD: VERTICAL LOAD

FORCE,1,400,,-1000.,0.,0.,1.

FORCE,1,401,,-2000.,0.,0.,1.

FORCE,1,402,,-1000.,0.,0.,1.

$

$ LOAD FOR SUBCASE 2 : RIGHT HORIZONTAL DISPLACEMENT

$$ SLIDELINE CONTACT

$

$ NONLINEAR SOLUTION STRATEGY: AUTO METHOD WITH DEFAULTS

NLPARM, 410, 1 , ,AUTO, , ,PW, YES, +NLP41

+NLP41, ,1.E-6, 1.E-10

$

ENDDATA

Input File for Modification (Cont’d)




SOLUTION TO WORKSHOP PROBLEM ONEID CHAP6WS1S,NAS103, Chap 6, Workshop 1 $ AR(12/03)

TIME 300

SOL 106

CEND

$

TITLE = SYMMETRIC ELASTIC PUNCH WITH FRICTION$

DISP = ALL

SUBCASE 1 $ VERTICAL LOAD

LOAD = 1

NLPARM = 410

SUBCASE 2 $ DISPLACEMENT TO THE RIGHT

LOAD = 1

NLPARM=420

SPC=2

$

BEGIN BULK

PARAM,POST,0

$

$ GEOMETRY

GRID,100,,0.,0.,0.,,123456

=,*1,,*(10.),==

=9

GRID,200,,0.,,20.,,2456

=,*1,,*(10.),==

=9

GRID,300,,45.,,20.,,2456

GRID,301,,55.,,20.,,2456

GRID,302,,65.,,20.,,2456

GRID,400,,45.,,25.,,2456

GRID,401,,55.,,25.,,2456

GRID,402,,65.,,25.,,2456

$$ ELEMENTS

CQUAD4,100,1,100,101,201,200

=,*1,=,*1,*1,*1,*1

=8

CQUAD4,200,2,300,301,401,400

=,*1,=,*1,*1,*1,*1

PSHELL,1,1,1.,-1PSHELL,2,2,1.,-1

MAT1,1,1.E5,,0.0

MAT1,2,1.E5,,0.0




SOLUTION TO WORKSHOP PROBLEM ONE

$

$ PUNCH LOAD: VERTICAL LOAD

FORCE,1,400,,-1000.,0.,0.,1.

FORCE,1,401,,-2000.,0.,0.,1.FORCE,1,402,,-1000.,0.,0.,1.

$

$ LOAD FOR SUBCASE 2 : RIGHT HORIZONTAL DISPLACEMENT

SPC, 2, 300, 1, 10.

SPC, 2, 302, 1, 10., 301, 1, 10.

$

$ SLIDELINE CONTACT

BCONP, 10, 10, 20, , 10., 10, 2, 10

BFRIC, 10, 1, , 0.1

BLSEG, 10, 302, 301, 300

BLSEG, 20, 200, 201, 202, 203, 204, 205, 206,

, 207, 208, 209, 210

CORD2R, 10, , 0., 0., 0., 0., -1., 0.

, 1., 0., 0.

$

$ NONLINEAR SOLUTION STRATEGY: AUTO METHOD WITH DEFAULTS

NLPARM, 410, 1 , ,AUTO, , ,PW, YES, +NLP41

+NLP41, ,1.E-6, 1.E-10

NLPARM, 420, 10, ,AUTO, , ,PW, YES, +NLP42

+NLP42, ,1.E-6, 1.E-10

$

ENDDATA




SOLUTION TO WORKSHOP PROBLEM ONE



S7-1NAS 103, Section 7, December, 2003

SECTION 7

NONLINEAR TRANSIENT ANALYSIS




TABLE OF CONTENTS

Page

Review Of Transient Analysis 7-5

User Interface 7-11

Example Input For Sol 129 7-15

General Features 7-16

General Limitations 7-17

Integration Schemes 7-18

Nonlinear Transient Solution Strategy 7-21

Mass Specification 7-30

Damping 7-31

Damping Specification 7-34

Load Specification 7-37Dynamic Loads 7-38

Dynamic Loads Example 7-47




TABLE OF CONTENTS

PageStatic Loads In Transient Analysis 7-49LSEQ Entry 7-50

Example: Static Loads In Transient Analysis 7-53Nonlinear Loads 7-55Example: Nonlinear Loads 7-63Initial Conditions 7-66

Restarts For Nonlinear Transient Analysis 7-67Hints And Recommendations For Sol 129 7-68Example Problem One 7-69Example Problem Two 7-74Workshop Problems One Through Three 7-76

Workshop Problem Four 7-80Solution For Workshop Problem One 7-84Solution For Workshop Problem Two 7-89Solution For Workshop Problem Three 7-90

Solution To Workshop Problem Four 7-91




REVIEW OF TRANSIENT ANALYSIS

Static analysis: Compute a solution U that satisfies the equilibrium equation:

F(U) = P

Transient analysis: Compute a solution U that satisfies the equilibrium equation:

U)P(t, t)F(U, t),U D( )t ,U ( I =++ &&&

InertiaForces DampingForces ElementForces ExternalLoad





For a linear system

For a general nonlinear system Mass of the system may change

Damping may change

Stiffness may change

Load may be function of system response In MSC.NASTRAN mass and damping cannot change. Therefore, the

equilibrium equation is

P(t)KU U BU M =++ &&&

U)P(t,F(U(t))t U B )t ( U M =++ &&&





Nonlinear Transient Analysis Nonlinear transient analysis proceeds by dividing the time into a

number of small time steps.

Beginning of k-th Time Step

t = total time

End of k-th Time Step

Note: Time steps may not be equal.

∆t1

∆tk ∆tn





The solution at the end of a time step provides the initial conditions forthe next time step.

For each time step, a relationship is assumed between displacement,

velocity, and acceleration (integration scheme).

un Displacement at time tn approximated by dn.

un Velocity at time tn approximated by vn.

un Acceleration at time tn approximated by an.

.

..

t

F(d)

Fn

dn

Fn + 1

dn + 1d

a, v, u, danvn

dn

d(t)

u(t)

∆ttn tn + 1

dn + 1

an + 1

vn + 1

∆d





There are a number of different integration schemes available in theliterature. Implicit integration: dn + 1 is obtained by using the equilibrium conditions at

time tn + 1. Explicit integration: dn + 1 is obtained by using the equilibrium conditions at

time tn.

Use of the integration scheme reduces the transient equilibriumequation to a static equilibrium equation form.

Effective dynamic stiffness and load vector depend on the integrationscheme used.

For example, for the average acceleration scheme, also called thetrapezoidal rule or Newmark scheme (γ = 1/2, β = 1/4),

K * ( M, B, K, ∆t ) ∆U = P * ( ∆t, Ů, Ü, M, B, ∆P )

Effective Dynamics Effective DynamicStiffness Load Vector

.

U





The equilibrium is satisfied at the beginning and at the end of a time step. The equilibrium is not satisfied within the time step. Therefore, the selection of ∆t

is important. A large value of ∆t reduces accuracy. A small value of ∆t increases computing cost. A strategy is needed that automatically adjusts the time step value to achieve an

optimum value in terms of accuracy and computing cost. Adjustment of time step value requires the reformation and decomposition of the

dynamic stiffness.

K Bt

M t

K +∆

+∆

=24

2

*

)(2)](4

)(2[)(* t U Bt U t

t U M t P P &&&& +∆

++∆=




USER INTERFACE

Solution sequences SOL 129 or SOL 99.

Solution strategy TSTEPNL Bulk Data entry. TSTEPNL Case Control command (always required).

SEALL or equivalent Case Control command is required

for SOL 99 Mass specification RHO field in MATi Bulk Data entries. CMASSi Bulk Data entries for scalar mass elements. CONMi Bulk Data entries for concentrated mass elements. PARAM,COUPMASS, to specify the generation of coupled rather than

lumped mass matrices for elements with coupled mass capability. PARAM,WTMASS.




USER INTERFACE

Damping specification CVISC Bulk Data entry for the viscous damper element.

Field GE in MATi Bulk Data entries for nonlinear element damping PARAM, G for overall structural damping.

PARAM, W3 to convert structural damping to equivalent viscousdamping.

PARAM, W4 to convert element damping to equivalent viscous

damping. PARAM, NDAMP to specify numerical damping.




USER INTERFACE

Load Specification

Selected by DLOAD, LOADSET, and NONLINEAR Case Controlcommands.

Nonlinear transient load as a negative variable raised to a power.NOLIN4Nonlinear transient load as a positive variable raised to a power.NOLIN3

Nonlinear transient load as the product of two variables.NOLIN2

Nonlinear transient load as a tabular function.NOLIN1

Generate transient load history for static loads.LSEQ

Transient load scale factors.DAREATransient load as defined by analytical functions.TLOAD2

Transient load as ordered time, force pairs.TLOAD1




USER INTERFACE

Initial conditions specification TIC Bulk Data entry

IC Case Control command Additional entries for nonlinear analysis

Similar to nonlinear static analysis

Material nonlinear only

MATS1 Geometric nonlinear only

PARAM,LGDISP,+1

Contact (interface) only CGAP/PGAP

BCONP, BLSEG, BWIDTH, BFRIC, BOUTPUT

Combined material and geometric nonlinear MATS1

PARAM,LGDISP,+1




EXAMPLE INPUT FOR SOL 129

ID MSC, NL129TIME 30SOL 129DIAG 50 $ Print nonlinear iteration informationCENDTITLE = MATERIAL NONLINEAR TRANSIENT ANALYSIS

SPC = 123DISP = ALL

STRESS = ALLIC = 50

SUBCASE 10DLOAD = 100TSTEPNL = 10

SUBCASE 20DLOAD = 200TSTEPNL = 20

BEGIN BULK..

(Usual entries for model definition)..MAT1,10,30.+6,,.3, 0.1, , , 1.E-4MATS1,10,,PLASTIC,0.,,,30.+3$PARAM,LGDISP,1$ LOAD ENTRIESTLOAD2,100,10,,0,0.0,10.0,1.0TLOAD2,110,20,,0,10.0,20.0,1.0DLOAD,200,1.0,1.0,100,1.0,110

DAREA,10,15,1,10.0DAREA,20,18,1,5.0$ INITIAL CONDITIONSTIC,50,5,1,1.0,-2.0TIC,50,6,2,-2.0,4.0$ SOLUTION STRATEGY ENTRIESTSTEPNL,10,10,.01,1,AUTO,,10,PTSTEPNL,20,20,.01,1,AUTO,,10,P$ENDDATA

Initial Conditions

Load Selection

Solution Strategy




GENERAL FEATURES

Transient material nonlinear, geometric nonlinear,combined geometric and material nonlinear, and contact

problems can be solved using this solution sequence. Linear superelements can be combined with nonlinear

elements.

Modal reduction (SEQSET,EIGR) and generalizeddynamic reduction (DYNRED) are available for the linearsuperelements.

Parameter-controlled restarts from the end of any SOL

129 subcase or from SOL 106.




GENERAL LIMITATIONS

No constraint changes after first subcase - including restart.

No thermal loads or enforced displacements.

Reduction (GDR, Guyan reduction) only for superelements. PARAM “G” damping only applies to linear elements.

Nonlinear element damping provided by GE on MAT Bulk Dataentries (PARAM “W4” must also be used) only for initial K.

Damping remains constant. No element force output for nonlinear elements.

Upstream loads are ignored in the superelement data recovery.

No grid point stresses for nonlinear elements.

Mass cannot change.




INTEGRATION SCHEMES

Two-Point Integration Scheme Use the following equilibrium equation:

Assume that the acceleration for a time step is equal to the average ofthe beginning and end of the step.

Velocity and displacement are obtained by integration.

111n1n UU ++++ =++ nn P F B M &&&

2(t)U 1++= nn U U

&&&&

&&

211

11

4

2

t U U

t U U U

t U U

U U

nnnnn

nnnn

∆

++∆+=

∆

++=

++

++

&&&&&

&&&&

&&

INTEGRATION SCHEMES




INTEGRATION SCHEMES

Rearrange the equilibrium equation in terms of incremental values.

Calculate velocity as follows:

Note that the acceleration need not be calculated since it does notappear in the incremental equilibrium equation.

For postprocessing purposes, acceleration is calculated as:

][24

424

1212 nnnnnnT U U C t

M t

U M t

F P P U K Bt

M t

−

∆

+∆

−∆

+−+=∆

+∆

+∆ ++

&

Dynamic Stiffness Dynamic Load Factor

nnnn U t U U U &&

−∆−= ++

2

)( 11

∆∆

−

∆∆

−∆

∆+

∆∆

∆+∆= −

+

+

+−

++

1

11

1

11

11

n

n

nn

n

n

n

nn

n

n

nn

n U t

t U

t

t

t

t U

t

t

T t U &&&&&&&

INTEGRATION SCHEMES




INTEGRATION SCHEMES

Two-point integration scheme is the same as thetrapezoidal rule or average acceleration method except

for the calculation of acceleration in postprocessing. For linear problems, this scheme is second-order

accurate, is unconditionally stable, and has no numericaldamping.

Easy starting, restarting, ending. Residual error carried over effectively.

Equilibrium is satisfied without the need of calculating 0 U &&

NONLINEAR TRANSIENT SOLUTION





STRATEGY Specified by TSTEPNL Bulk Data entry

Selected by TSTEPNL Case Control command

TSTEPNL Bulk Data Entry Description: Defines parametric controls and data for nonlinear

transient analysis

Format:

Examples:

RTOLBUTOLMAXRRBMSTEP ADJUSTMAXBIS

FSTRESSMAXLSMAXQNMAXDIVEPSWEPSPEPSU

CONVMAXITIERKSTEPNODTNDTIDTSTEPNL

10987654321

0.1160.75055

0.0221021.00E-061.00E-03

PW-1021250TSTEPNL






STRATEGYField Contents

ID Identification number. (Integer > 0).

NDT Number of time steps of value DT. (Integer > 4).

DT Time increment. (Real > 0.0).

NO Time step interval for output. Every NO-th step will be savedfor output. (Integer > 0; Default = 1).

KSTEP If METHOD = “TSTEP”, then KSTEP is the time step interval

for stiffness Updates. If METHOD = “ADAPT”, then KSTEPis the number of converged bisection solutions betweenstiffness updates. (Integer > 0; Default = 2)

MAXITER Limit on number of iterations for each time step. (Integer ≠ 0;Default = 10)






STRATEGYField Contents (Cont.)

CONV Flags to select convergence criteria. (Character: “U”, “P”,“W”, or any combination; Default = “PW”)

EPSU Error tolerance for displacement (U) criterion. (Real > 0.0;Default = 1 .0E-2)

EPSP Error tolerance for load (P) criterion. (Real > 0.0; Default =1.0E-3)

EPSW Error tolerance for work (W) criterion. (Real > 0.0;Default = 1 .0E-6)

MAXDIV Limit on the number of diverging conditions for a time stepbefore the solution is assumed to diverge. (Integer > 0;Default = 2)

MAXQN Maximum number of quasi-Newton correction vectors to besaved on the database. (Integer ≥ 0; Default = 10)

MAXLS Maximum number of line searches allowed per iteration.(Integer ≥ 0; Default = 2)





O S SO U O


FSTRESS Fraction of effective stress (s) used to limit the subincrementsize in the material routines. (0.0 < Real < 1.0;

Default = 0.2)MAXBIS Maximum number of bisections allowed for each time step.

(- 9 ≤ Integer ≤ 9; Default = 5)

ADJUST Time step skip factor for automatic time step adjustment.(Integer ≥ 0; Default = 5)

MSTEP Number of steps to obtain the dominant period response.(10 ≤ Integer ≤ 200; Default = variable between 20 and 40)

RB Define bounds for maintaining the same time step for thestepping function if METHOD = “ADAPT”. (0.1 ≤ Real ≤ 1.0;

Default = 0.75)MAXR Maximum ratio for the adjusted incremental time relative to

DT allowed for time step adjustment. (1.0 ≤ Real ≤ 32.0;Default = 16.0)






UTOL Tolerance on displacement increment beneath which there isno time step adjustment. (0.001 > Real ≤ 1.0; Default = 0.1)

RTOLB Maximum value of incremental rotation (in degrees) allowedper iteration to activate bisection. (Real > 2.0;Default = 20.0)





STRATEGY Automatic Time Step Adjustment (Adaptive Method)

Two-Point Integration Scheme

Time step is automatically adjusted (Use ADJUST = 0, to deactivate)

Stiffness is automatically updated to improve convergence

(KSTEP = # of converged bisection solutions between stiffness updates)

Accurate, efficient, and user-friendly

Based on the dominant frequency in the incremental deformationpattern:

Number of steps (MSTEP) for a period is adaptive, based on thestiffness ratio:

n

T

n

nn

T

n

n

T

n

n

T

nn

U M U

F F U

U M U

U K U

∆∆−∆

=∆∆∆∆

= − )( 12ω

nnn

n

t MSTEP t

t r

∆=

∆∆

= + 1211

ω

π





STRATEGY Thrashing is prevented by the stepping function:

With f = 0.25for r < 0.5 * RBf = 0.5 for 0.5 < RB ≤ r < RBf = 1.0 for RB ≤ r < 2.0

f = 2.0 for 2.0 ≤ r < 3.0/RBf = 4.0 for r ≥ 3.0/RB

Bounds for ∆t adjustment:

Undesirable effects due to GAP, plasticity, large mass, massless points,etc., are filtered out.

nn t r f t ∆=∆ + )(1

MAXR DT t AXR DT n *<∆<





STRATEGY Stepping Function for Time Step Adjustment with

Rb = 0.75f(r)

4.0

3.0

2.0

1.0

.5 Rb 1 2 3 4 5Rb

0.5

0.25r





STRATEGY Bisection Algorithm

To overcome divergent problems due to nonlinearity.

Activated when divergence occurs.

Activated when MAXITER is reached.

Activated when excessive ∆σ is detected.

Decomposition at every bisection.

Update [K] at every KSTEP-th converged bisection.

Bisection continues until solution converges or MAXBIS is reached.

If MAXBIS is reached, the reiteration procedure is activated to select thebest attainable solution.

MASS SPECIFICATION




MASS SPECIFICATION

Similar to linear transient analysis.

CMASS1 and CMASS2 define scalar mass elements.

CMASS3 and CMASS4 define scalar mass elements connectedonly to scalar points.

CONM1 defines a 6 x 6 mass matrix for a grid point.

CONM2 defines a diagonal mass matrix for translational degrees offreedom and a 3 x 3 full matrix for rotational degrees of freedom at agrid point.

Element mass density is defined on the RHO field of the MATi BulkData entry.

PARAM,COUPMASS,1 specifies the coupled mass matrix forelements with coupled mass capability (BAR, BEAM, ROD, HEXA,PENTA, TRIA, and TUBE elements).

DAMPING




DAMPING

Damping represents energy dissipation observed instructures.

Difficult to accurately model since damping results frommany mechanisms: Viscous effects (dashpot, shock absorber)

External friction (slippage in structural joints)

Internal friction (characteristic of material type) Structural nonlinearities (plasticity)

Analytical conveniences are used to model damping.

Viscous damping force proportional to velocity

ub f v &=

pkuubum =++ &&&

DAMPING




DAMPING

Structural damping force proportional to displacement

Viscous and structural damping are equivalent at

frequency ω3.

with

1 −== iuGk i f s

tcoefficiendampingstructuralG )1( ==++ pkuiGum &&

G bω3

k ----------

2ξω3

ωn

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

ξ c

2mωn

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

DAMPING




DAMPING

Damping Structural Damping, f s = iGKu

Equivalent

Viscous b = Gk/ω3

ω3 ω

f v b u·=

DAMPING SPECIFICATION





Similar to linear transient analysis.

Damping matrix B comprised of several matrices:

Where B1 = damping elements (VISC,DAMP)

G = overall structural damping coefficient (PARAM,G)

W3 = frequency of interest - rad/sec (PARAM,W3)K1 = global stiffness matrix

Ge = element structural damping coefficient (GE on the MATientry)

W4 = frequency of interest - rad/sec (PARAM,W4

Ke = element stiffness matrix

∑++=e

ee K GW

K W G B B

4

1

3

1 1





G S C C O

Default values for W3 and W4 are 0.0. In this case, the associated dampingterms are ignored.

Nonlinear element damping provided with PARAM,W4 and field GE in the

MATi entry using initial K. Damping matrix is not rotated.

Caution for large rotation.





DAMPING PARAMETERS PARAM,G, factor (default = 0.0)

Overall structural damping coefficient to multiply stiffness matrix for linear

elements. 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 set to greater than zero or PARAM,G will be ignored.

LOAD SPECIFICATION




Three ways: Dynamic loads

Static loads

Nonlinear loads

DYNAMIC LOADS




Dynamic loads require both temporal and spatialdistribution.

A user needs to follow four steps to specify dynamicloads.

The four steps are:1. Define the load as a function of time (TLOADi).

2. Define the spatial distribution of the load (DAREA).3. Combine the TLOADi entries via DLOAD entry.

4. Select the loads via the DLOAD Case Control command.

DYNAMIC LOADS




TLOAD1 Bulk Data Entry Description: Defines a time-dependent dynamic load or enforced

motion of the form

for use in transient response analysis.

Format:

Example:

)t ( F * A )t ( P τ −=

TIDTYPEDELAYDAREASIDTLOAD1

10987654321

1375TLOAD1

DYNAMIC LOADS




Field Contents

SID Set identification number. (Integer > 0).

DAREA Identification number of DAREA entry set or a thermal load

set (in heat transfer analysis) which defines A. (Integer > 0).DELAY Identification number of DELAY entry set that defines t.

(Integer ≥ 0, or blank).

TYPE Defines the nature of the dynamic excitation. (Integer 0, 1, 2,3, or blank).

TID Identification number of TABLEDi entry that gives F(t-t).(Integer > 0).

Enforced Acceleration3

Enforced Velocity2

Enforced Displacement1

Force or Moment0 or blank

Excitation FunctionInteger

DYNAMIC LOADS




DAREA Bulk Data Entry Description: Defines scale (area) factors for dynamic loads. DAREA is

used in conjunction with RLOADi and TLOADi entries.

Format:

Example:

Field ContentsSID Identification number. (Integer > 0).Pi Grid, extra, or scalar point identification number.(Integer > 0).Ci Component number. (Integer 1 through 6 for grid point;

blank or 0 for extra or scalar point). Ai Scale (area) factor. (Real).

A2C2P2 A1C1P1SIDDARIA

10987654321

10.11158.2263DARIA

DYNAMIC LOADS




TLOAD2 Bulk Data Entry Description: Defines a time-dependent dynamic load or enforced

motion of the form

for use in a transient response problem where = t - T1 - t.

Format:

Example:

+≤≤++

+>+<=

)(T2t)(T1, )~

2cos(~

)(T2or t)(T1t, 0)( ~

τ τ π

τ τ

P t F et At P

t C

t ~

BC

PFT2T1TYPEDELAYDAREASIDTLOAD2

10987654321

2

124.72.1104TLOAD2

DYNAMIC LOADS




Field Contents

SID Set identification number. (Integer > 0).

DAREA Identification number of DAREA entry set or a thermal load

set (in heat transfer analysis) that defines A. (Integer > 0).DELAY Identification number of DELAY entry set that defines t.

(Integer ≥ 0, or blank).

TYPE Defines the nature of the dynamic excitation. (Integer 0, 1, 2,3 or blank).

T1 Time constant. (Real ≥ 0.0).

T2 Time constant. (Real; T2 > T1).

F Frequency in cycles per unit time. (Real ≥ 0.0; Default =0.0).

P Phase angle in degrees. (Real; Default = 0.0).C Exponential coefficient. (Real; Default = 0.0).

B Growth coefficient. (Real; Default = 0.0).

DYNAMIC LOADS




For a constant load, leave fields F, P, C, and B blank.

For a cosine wave, specify F = 1.0, and leave fields P, C,

and B blank. For a sine wave, specify F = 1.0, P = - 90° and leave

fields C and B blank.

DYNAMIC LOADS




DLOAD Bulk Data Entry Description: Defines a dynamic loading condition for frequency

response or transient response problems as a linear combination of

load sets defined via RLOAD1 or RLOAD2 entries for frequencyresponse or TLOAD1 or TLOAD2 entries for transient response.

Format:

Example:

L4S4

L3S3L2S2L1S1SSIDDLOAD

10987654321

9-2

827-262117DLOAD

DYNAMIC LOADS




Field Contents

Sid Load set identification number. (Integer > 0).

S Scale factor. (Real).

Si Scale Factors. (Real).Li Load set identification numbers of RLOAD1, RLOAD2,

TLOAD1, and TLOAD2 entries. (Integer > 0).

DYNAMIC LOADS EXAMPLE




Step 1 P1 : TLOAD1,101,1,0,0,1

P2 : TLOAD1,102,2,0,0,2

or TLOAD2,102,2,0,0.0,10.0

P3 : TLOAD2,103,3,,0,0.0,10.0,1.0,-90.0

Constant

P1 P3

P210 11 12

7 8 9

4 5 6

1 2 3

10.0

TimeP1 = 1.0

Load

Sine Wave

10.0

10.010.0

P2 = 2.0

P3 = 10.0

DYNAMIC LOADS EXAMPLE




Step 2 DAREA,1,10,2,-1.0

DAREA,2,12,1,-2.0

DAREA,3,11,2,-10.0

Step 3 DLOAD,10,1.0,1.0,101,1.0,102,1.0,103

Step 4 DLOAD=10 in Case Control

STATIC LOADS IN TRANSIENT ANALYSIS




A user needs to follow five steps to specify static loads intransient analysis.

The five steps are:1. Define the static loads using FORCEi, GRAV, MOMENTi, etc., that arereferenced by the LOAD Case Control command.

2. Define a LSEQ Bulk Data entry to point to a TLOADi entry and to a loadset that is referenced by a LOAD Case Control command.

3. Define a TLOAD1 or TLOAD2 entry to define a constant function withtime.

4. Combine all the TLOADi entries through the DLOAD Bulk Data entry.

5. Select the DLOAD entry through the DLOAD Case Control command

and the LSEQ entry through the LOADSET Case Control command.

LSEQ ENTRY




Defines static loads that will be applied dynamically.

Relationship to other commands and entries:

DLOAD LOADSET

DLOAD LSEQ

Case Control:

Bulk Data:

Dynamic

Load

DAREA Static

Load

LSEQ ENTRY




LSEQ Bulk Data Entry Description: Defines a sequence of static load sets.

Format:

Example:

Field Contents

SID Set identification of the set of LSEQ entries. (Integer > 0).

DAREA The DAREA set identification assigned to this static loadvector. (Integer > 0).

TIDLIDDAREASIDLSEQ

10987654321

10011000200100LSEQ

LSEQ ENTRY




Field Contents

LID Load set identification number of a set of static load entriessuch as those Referenced by the LOAD Case Control

command. (Integer > 0 or blank).TID Temperature set identification of a set of thermal load entriessuch as those referenced by the TEMP(LOAD) Case Controlcommand. (Integer > 0 or blank).

EXAMPLE: STATIC LOADS IN TRANSIENTANALYSIS




Aim: to specify gravity load in transient analysis.

Solution:

Case Control Section Step 5: DLOAD = 50011

LOADSET = 5000

Bulk Data Set

Step 4: DLOAD, 50011, 1.0, 1.0, 5001, 1.0, 4444,….

Step 3: TLOAD2, 5001, 5002, , 0, 0.0, 99999., 0., 0.

to LSEQ

Normal Dynamic Loads

EXAMPLE: STATIC LOADS IN TRANSIENTANALYSIS




Step 3: TLOAD2, 5001, 5002, , 0, 0.0, 99999., 0., 0.

Step 2: LSEQ, 5000, 5002, 5555

Step 1: GRAV, 5555, , 380., 0., 0., 1.0

Defines a function = cos (0) = 1.0

LOADSET DAREA

NONLINEAR LOADS




Allows for the specification of load at a particular degreeof freedom to be the function of displacement andvelocity at another degree of freedom.

Example:

Load at grid point 1, displacement component 2 as afunction of the displacement component 1 at grid point 3.

P(t)

65 4 3 2 1

P(t) = f (u3)

NONLINEAR LOADS




Useful for specifying nonlinear springs and nonlineardamping.

Nonlinear loads are specified using NOLINi entries. Four NOLINi entries (NOLIN1, NOLIN2, NOLIN3, and

NOLIN4) to specify mechanical loads.

Nonlinear loads are selected via the NONLINEAR Case

Control command. Nonlinear loads cannot be selected via the DLOAD Case

Control command.

All degrees of freedom referenced on NOLINi entry mustbe members of the solution set.

NONLINEAR LOADS




Velocity for an independent degree of freedom (for thepurpose of loads) is calculated as

Note: This may be different from that calculated in theintegration scheme. But it is acceptable.

In all NOLINi entries a degree of freedom is specified bythe grid number and its component number.

All loads generated with NOLINi entries lag behind by one

time step ∆t.

t

t t t

t

U U U

∆

∆−−=&

NONLINEAR LOADS




NOLIN1 Bulk Data Entry Description: Defines nonlinear transient forcing functions of the form.

Function of displacement: Pi(t) = S * T(u j(t)) (1)

Function of velocity: Pi(t) = S * T(u j(t)) (2)where u j(t) and u j(t) are the displacement and velocity at point GJ in thedirection of CJ.

Format:

Example:

.

.

TIDCJGJSC1G1SIDNOLIN1

10987654321

61032.14321NOLIN1

NONLINEAR LOADS




Field Contents

SID Nonlinear load set identification number. (Integer > 0).

GI Grid, scalar, or extra point identification number at which

nonlinear load is to be applied. (Integer > 0).CI Component number for GI. (0 < Integer ≤ 6; blank or zero if

GI is a scalar or extra point).

S Scale factor. (Real).

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

CJ Component number for GJ according to the following table:

TID Identification number of a TABLEDi entry. (Integer > 0).

Integer = 10Blank or ZeroExtra

Integer = 10Blank or ZeroScalar

11 < Integer < 161 < Integer < 6Grid

VelocityDisplacementType of point

NONLINEAR LOADS





where and can be either displacement or velocity at points GJ and GKin the directions of CJ and CK.

Format:

Example:

Pi(t) = S * X j(t) * Xk(t)

CKGKCJGJSC1G1SIDNOLIN2

10987654321

2122.91214NOLIN2

NONLINEAR LOADS





where may be a displacement or a velocity at point GJ in the directionof CJ.

Format:

Example:

≤

>= 0)(, 0

0)(,)]([*)( t X

t X t X S t P

j

j

A

j

i

ACJGJ5C1G1SIDNOLIN3

10987654321

2152-6.11024NOLIN3

NONLINEAR LOADS





where may be a displacement or a velocity at point GJ in the directionof CJ.

Format:

Example:

≥<−−= 0)(, 0 0)(,)]([*)(

t X t X t X S t P

j

j

A

ji

ACJGJSC1G1SIDNOLIN4

10987654321

16.31012642NOLIN4

EXAMPLE: NONLINEAR LOADS




Use of NOLINi Entries

k

c

g(x)

mf(t)

x

Nonlinear Spring

x

g(x)

g=0

x2

m x·· cx· kx g x( ) f t( )=+ + +

mx·· cx· kx f t( ) g x( ) – =+ +





Assume: X100 represents the displacement of the moving mass (X100 = X).

How to define g(x)? Define two scalar points, for example, 200 and 300 with k=1.

Use a NOLIN1 entry to define a force acting at scalar points 200 and300

1

1

Table 3333X200

X300

X if X 0≥

0 if X 0<

=⇒

NOLIN1,SID,200,1,1.0,100,1,3333

NOLIN1,SID,300,1,1.0,100,1,3333

Table ID

Table ID





Use a NOLIN2 entry to define a force acting at GRID100:

NOLIN2,SID,100,1,-1.000,200,1,300,1

⇒ We define a force acting at the mass (GRID 100) equal

to

Note: This approach is more accurate than using just one NOLIN1 to

define -g(x), where Table 3333 would be a square function rather than alinear function.

≤

>−

0Xif 0

0X2 if X

INITIAL CONDITIONS




May impose initial displacements and/or velocities with aTIC Bulk Data entry.

IC Case Control command selects TIC entries in theBulk Data Section.

Warning: Initial conditions for unspecified degrees offreedom are set to zero.

Initial conditions may be specified only for A-set degreesof freedom.

RESTARTS FOR NONLINEAR TRANSIENTANALYSIS

St ti f i t i t l i




Starting from a previous transient analysis Restarts are allowed only from the end of subcases. Set parameters:

PARAM,LOOPID,I I = loop number on printoutPARAM,STIME,To To = starting value of time To should be the last printed value for subcase I. The database will be modified starting from LOOPID+1, T = To.

Starting from a previous nonlinear static analysis Set parameter:PARAM,SLOOPID,I I = loop number on SOL 106 run

Initial transient load should be identical to static loads at restart state.(SPC, etc., may change)

Caution:The database will be completely overwritten.Transient analysis will destroy the static analysisdatabase.

HINTS AND RECOMMENDATIONS FOR SOL129

Id tif th t f li it




Identify the type of nonlinearity. Localize nonlinear region. Divide time history by subcases for convenience. Each subcase should not have more than 200 time

steps. Select default values to start - TSTEPNL.

Pick time step size for highest frequency of interest.Twelve or more steps per cycle and frequent content ofinput.

Some damping is desirable for numerical stability.

Avoid massless degrees of freedom. Choose GAP stiffness carefully. Increase MAXITER if convergency is poor.

EXAMPLE PROBLEM 1

D i ti T i t A l i f Si l S t d




Description: Transient Analysis of a Simply SupportedBeam with a Restrained Motion

0.02 in

990

25 in

50 in

0.02 in

Fnz

x

Stopper P(t)

Fn(U10010)

(U10010)0.011 sec

t

P

47.2

Forcing Function NOLIN1 Representing GAP

20 Beam Elements50 in

A = 0.314 in2

I = 0.157 in4

ρ = 0.3 lb./in3

EXAMPLE PROBLEM 1 (Contd.)

Di l t t th L di P i t (DT 0 0002)




Displacement at the Loading Point (DT=0.0002)



EXAMPLE PROBLEM 1: .dat File

ID, chap7e1, NAS103, chap 7, example 1 $ (AR12/28/03)




SOL, 129CENDTITLE=SS Beam with a Restrained Motion (NOLIN1)SUBTITLE=Direct Transient Response, NonlinearForceLABEL= NOLIN in SOL 129

SEALL = ALL

ECHO=SORTEDSPC=1002SET 1 = 10005SET 2 = 10010SET 3 = 10005,10010DISP=3VELO=3OLOAD=1NLLOAD=2

SUBCASE 1

DLOAD=30TSTEPNL=20NONLINEAR=13 $ Select Nonlinear Force

$BEGIN BULKPARAM, POST, 0PARAM, GRDPNT, 10010PARAM, WTMASS, 0.002588$CBAR, 101, 100, 10000, 10001, 0.0, 0.0, 1.

CBAR, 102, 100, 10001, 10002, 0.0, 0.0, 1.CBAR, 103, 100, 10002, 10003, 0.0, 0.0, 1.CBAR, 104, 100, 10003, 10004, 0.0, 0.0, 1.CBAR, 105, 100, 10004, 10005, 0.0, 0.0, 1.

CBAR, 106, 100, 10005, 10006, 0.0, 0.0, 1.CBAR, 107, 100, 10006, 10007, 0.0, 0.0, 1.CBAR, 108, 100, 10007, 10008, 0.0, 0.0, 1.CBAR, 109, 100, 10008, 10009, 0.0, 0.0, 1.CBAR, 110, 100, 10009, 10010, 0.0, 0.0, 1.CBAR, 111, 100, 10010, 10011, 0.0, 0.0, 1.CBAR, 112, 100, 10011, 10012, 0.0, 0.0, 1.CBAR, 113, 100, 10012, 10013, 0.0, 0.0, 1.CBAR, 114, 100, 10013, 10014, 0.0, 0.0, 1.CBAR, 115, 100, 10014, 10015, 0.0, 0.0, 1.CBAR, 116, 100, 10015, 10016, 0.0, 0.0, 1.CBAR, 117, 100, 10016, 10017, 0.0, 0.0, 1.CBAR, 118, 100, 10017, 10018, 0.0, 0.0, 1.CBAR, 119, 100, 10018, 10019, 0.0, 0.0, 1.CBAR, 120, 100, 10019, 10020, 0.0, 0.0, 1.$CONM2, 12, 10010, , .1$GRID, 10, , 50., 0.,-1.GRID, 10000, , 0., 0., 0., , 1246GRID, 10001, , 5., 0., 0., , 1246GRID, 10002, , 10., 0., 0., , 1246GRID, 10003, , 15., 0., 0., , 1246GRID, 10004, , 20., 0., 0., , 1246GRID, 10005, , 25., 0., 0., , 1246GRID, 10006, , 30., 0., 0., , 1246GRID, 10007, , 35., 0., 0., , 1246GRID, 10008, , 40., 0., 0., , 1246GRID, 10009, , 45., 0., 0., , 1246GRID, 10010, , 50., 0., 0., , 1246GRID, 10011, , 55., 0., 0., , 1246GRID, 10012, , 60., 0., 0., , 1246

EXAMPLE PROBLEM 1: .dat File (Contd.)

GRID, 10013, , 65., 0., 0., , 1246




GRID, 10014, , 70., 0., 0., , 1246GRID, 10015, , 75., 0., 0., , 1246GRID, 10016, , 80., 0., 0., , 1246GRID, 10017, , 85., 0., 0., , 1246GRID, 10018, , 90., 0., 0., , 1246GRID, 10019, , 95., 0., 0., , 1246GRID, 10020, ,100., 0., 0., , 1246

$MAT1, 1000, 3.E7, , 0.3, 0.3$PBAR, 100, 1000, 0.31416, 0.15708, 1., 0.$SPC, 1002, 10, 123456SPC, 1002, 10020, 3, , 10000, 3$ Modeling Information for Center SpringCROD, 10, 10, 10, 10010

MAT1, 10, 10., , 0.PROD, 10, 10, 1.$MATS1, 10, , PLASTIC, 0., 1, 1, 3.E8$ Loading and Solution InformationTLOAD2, 30, 33, , , 0., 0.011, 90.91, -90.DAREA, 33, 10005, 3, 47.2TSTEPNL, 20, 200, 0.0002, 1, ADAPT$ Modeling Information for Nonlinear SpringNOLIN1, 13, 10010, 3, 1., 10010, 3, 13

TABLED1, 13,, -2.5E-2, 4.95, -2.0E-2, 0., 0., 0., ENDTENDDATA

EXAMPLE PROBLEM TWO

Purpose




Purpose To illustrate the use of slideline contact and nonlinear transient analysis

in bumper crash applications.

Problem Description A rigid barrier moving at 5 mph. impacts a bumber fixed at the bumper

brackets. Plot the deformed shape of the bumper after 20 msec ofcontact.

EXAMPLE PROBLEM TWO

Solution




Solution Five separate contact regions are defined with the barrier as the master

and the bumper as the slave.

Each master region consists of two master nodes. Each slave region consists of 23 slave nodes.

DEFORMED BUMPER

WORKSHOP PROBLEMS ONE THROUGHTHREE

Purpose




Purpose To demonstrate the use of cold start and restart procedures for

nonlinear transient analysis (SOL 129).

Problem Description For the massless rod given below, calculate and plot (a) the rod stress

time history and (b) displacement, velocity, and acceleration time historyfor the mass. Request the output every tenth time step.


σ




P(t)

Max30000 lbs

t

10000 lbs

240 in

P(t)Massless Rod

σy = 67895 psi

ε

E = 30.E 6

A = .6672

g = 386 in/sec2

= 0.00259071

g---


1 Modify the input file to perform the analysis in one




1. Modify the input file to perform the analysis in onesubcase for a total duration of 0.3 seconds with an initialtime increment of 0.0025 seconds.

2. Modify the input file to perform the analysis in threesubcases. The duration for the first, second, and thirdsubcase is 0.125, 0.100, and 0.075 seconds,

respectively.3. Restart the analysis from the end of subcase two.






Input File for ModificationID CHAP7WS1, NAS103 Workshop $ AR 12/15/03SOL 129CENDTITLE=ELASTO PLASTIC VIBRATION PROBLEM NAS103 Chapter 6SUBTITLE=NONLINEAR TRANSIENT ANALYSISECHO=BOTHSET 1 = 1SET 2 = 2DISP=1ACCE=1VELO=1STRESS=2SUBCASE 1OUTPUT(XYPLOT)

XTITLE = TIME IN SECSXGRID LINES = YESYGRID LINES = YES

YTITLE = DISPLACEMENT GRID 1XYPLOT DISP RESP/1(T2)YTITLE = VELOCITY GRID 1XYPLOT VELO RESP/1(T2)YTITLE = ACCELERATION GRID 1XYPLOT ACCE RESP/1(T2)YTITLE = STRESS IN RODXYPLOT STRESS RESP /2(2)

BEGIN BULK$ GEOMETRY AND CONNECTIVITYGRID, 1, , 0., 0., 0., , 13456GRID, 2, , 0., 240., 0., , 123456

CROD, 2, 2, 2, 1CMASS2, 1, 10000., 1, 2$ PROPERTIESPROD, 2 2 .6672MAT1, 2, 30.E06MATS1, 2, , PLASTIC, 0., 1, 1, 67895.68$ SOLUTION STRATEGY$ LOADINGPARAM, POST, 0ENDDATA

WORKSHOP PROBLEM FOUR

Purpose




p To demonstrate the use of (a) GAP element and (b) material damping

and initial condition in nonlinear transient analysis.

Problem Description Modify the input file to specify (a) a gap element between the rod and

rigid body, (b) damping for the rod element, and (c) initial conditions.


Rod length, L = 100.m




Area of rod, A = 1.m2

Young’s modulus, E = 103 N/m2

Poisson’s ration, ν = 0.3 Mass density, ρ = 1.0 kg/m3

Mass of rod, m = ρ = AL = 10. kg

Mass of rigid body, M = 20.kg

Velocity of impact for Vo = 0.1 m/sec

Damping = 0.1% at first mode

570796 .1El L2

TT )1n2 ( =

−= ρ ω






pID CHAP7WS4, NAS103 Workshop $ AR 12/15/03

SOL 129CENDTITLE = TRANSIENT RESPONSE OF SHOCK WAVE IN BAR -- IMPACT

SUBTITLE = BAR STRUCK BY A MOVING MASS AT THE FREE ENDECHO = UNSORTSET 1 = 21,99SET 2 = 101,120,899DISP = 1VELOCITY = 1

STRESS = 2SUBCASE 1 $ UP TO 6 SECONDSTSTEPNL = 200

OUTPUT(XYPLOT)

CSCALE = 1.5XAXIS = YESYAXIS = YESXGRID LINES = YES

YGRID LINES = YESXTITLE = TIMEYTITLE = FORCETCURVE = FORCE IN THE GAP (ELEMENT 899)XYPLOT STRESS /899(2)

YTITLE = DISPLACEMENT

TCURVE = DISP. (T1) AT MASS PT. (GP99), FREE END (GP21)XYPLOT DISP /99(T1),21(T1)YTITLE = STRESS

TCURVE = STRESS AT FREE END (ELEMENT 120)XYPLOT STRESS /120(2)

TCURVE = STRESS AT FIXED END (ELEMENT 101)XYPLOT STRESS /101(2)


Input File for Modification (cont.)




BEGIN BULK

$ GEOMETRY

GRDSET, , , , , , , 23456

GRID, 1, ,0., 0., 0., , 123456

GRID, 2, ,5., 0., 0.

=,*1,=,*5.,== $

=18

GRID, 99, , 100., 0., 0.

$ ELEMENT CONNECTIVITY

CONROD, 101, 1, 2, 100, 1.

=,*1,*1,*1,== $

=18

$ MATERIAL PROPERTIES

CONM2, 999, 99, , 20.

$ GAP ELEMENT CONNEVTIVITY

$ GAP ELEMENT PROPERTIES

$ INITIAL CONDITIONS

$ PARAMETERS

param, post, 0

PARAM, COUPMASS, 1

$ SOLUTION STRATEGY

TSTEPNL 200 600 .01 1

$

ENDDATA


ID CHAP7WS1S, NAS103 Workshop $ AR 12/15/03SOL 129 BEGIN BULK




SOL 129CENDTITLE=ELASTO PLASTIC VIBRATION PROBLEM NAS103Chapter 7SUBTITLE=NONLINEAR TRANSIENT ANALYSISECHO=BOTH

SET 1 = 1SET 2 = 2DISP=1ACCE=1VELO=1STRESS=2SUBCASE 1

DLOAD=100TSTEPNL=100

OUTPUT(XYPLOT)

XTITLE = TIME IN SECSXGRID LINES = YESYGRID LINES = YESYTITLE = DISPLACEMENT GRID 1XYPLOT DISP RESP/1(T2)YTITLE = VELOCITY GRID 1XYPLOT VELO RESP/1(T2)YTITLE = ACCELERATION GRID 1XYPLOT ACCE RESP/1(T2)YTITLE = STRESS IN ROD

XYPLOT STRESS RESP /2(2)

BEGIN BULK$ GEOMETRY AND CONNECTIVITYGRID, 1, , 0., 0., 0., , 13456GRID, 2, , 0., 240., 0., , 123456CROD, 2, 2, 2, 1CMASS2, 1, 10000., 1, 2

$ PROPERTIESPROD, 2 2 .6672MAT1, 2, 30.E06MATS1, 2, , PLASTIC, 0., 1, 1, 67895.68$ SOLUTION STRATEGYTSTEPNL, 100, 120, .0025, 1$ LOADINGDAREA, 100, 1, 2, 30000.TLOAD1, 100, 100, , 0, 100TABLED1,100, , , , , , , , +TAB

+TAB, 0., 1., 10., 1., ENDT$ PARAMETERSPARAM, POST, 0PARAM, WTMASS, .0025907ENDDATA






ID CHAP7WS2S, NAS103 Workshop $ AR 12/15/03SOL 129CENDTITLE=ELASTO PLASTIC VIBRATION PROBLEM NAS103Chapter 7

BEGIN BULK$ GEOMETRY AND CONNECTIVITY




Chapter 7SUBTITLE=NONLINEAR TRANSIENT ANALYSISECHO=BOTHSET 1 = 1SET 2 = 2DISP=1ACCE=1VELO=1STRESS=2SUBCASE 1DLOAD=100TSTEPNL=100

SUBCASE 2DLOAD=100TSTEPNL=200


OUTPUT(XYPLOT)XTITLE = TIME IN SECSXGRID LINES = YESYGRID LINES = YESYTITLE = DISPLACEMENT GRID 1XYPLOT DISP RESP/1(T2)YTITLE = VELOCITY GRID 1XYPLOT VELO RESP/1(T2)YTITLE = ACCELERATION GRID 1XYPLOT ACCE RESP/1(T2)YTITLE = STRESS IN RODXYPLOT STRESS RESP /2(2)

GRID, 1, , 0., 0., 0., , 13456GRID, 2, , 0., 240., 0., , 123456CROD, 2, 2, 2, 1CMASS2, 1, 10000., 1, 2$ PROPERTIES

PROD, 2 2 .6672MAT1, 2, 30.E06MATS1, 2, , PLASTIC, 0., 1, 1, 67895.68$ SOLUTION STRATEGYTSTEPNL, 100, 50, .0025, 1TSTEPNL, 200, 40, .0025, 1TSTEPNL, 300, 30, .0025, 1$ LOADINGDAREA, 100, 1, 2, 30000.TLOAD1, 100, 100, , 0, 100

TABLED1,100, , , , , , , , +TAB+TAB, 0., 1., 10., 1., ENDT$ PARAMETERSPARAM, POST, 0PARAM, WTMASS, .0025907ENDDATA

SOLUTION FOR WORKSHOP PROBLEMTHREE

RESTART,VERSION=1,KEEPASSIGN MASTER='chap7_ws_2s.MASTER'ID CHAP7_WS_3S, NAS103 Workshop $ AR 12/15/03SOL 129CEND




CENDTITLE=ELASTO PLASTIC VIBRATION PROBLEM NAS103 Chapter7SUBTITLE=NONLINEAR TRANSIENT ANALYSISECHO=BOTH$ INITIAL STATE FOR RESTART

PARAM,LOOPID,2PARAM,STIME,0.225$SET 1 = 1SET 2 = 2DISP=1ACCE=1VELO=1STRESS=2SUBCASE 1DLOAD=100TSTEPNL=100



OUTPUT(XYPLOT)XTITLE = TIME IN SECSXGRID LINES = YESYGRID LINES = YESYTITLE = DISPLACEMENT GRID 1XYPLOT DISP RESP/1(T2)

YTITLE = VELOCITY GRID 1XYPLOT VELO RESP/1(T2)YTITLE = ACCELERATION GRID 1XYPLOT ACCE RESP/1(T2)YTITLE = STRESS IN RODXYPLOT STRESS RESP /2(2)

BEGIN BULKENDDATA


ID CHAP7WS4S, NAS103 Workshop $ AR 12/15/03SOL 129

CEND

TITLE = TRANSIENT RESPONSE OF SHOCK WAVE IN BAR -- IMPACT

SUBTITLE = BAR STRUCK BY A MOVING MASS AT THE FREE END




ECHO = UNSORT

SET 1 = 21,99

SET 2 = 101,120,899

DISP = 1

VELOCITY = 1

STRESS = 2SUBCASE 1 $ UP TO 6 SECONDS

IC = 1

TSTEPNL = 200

OUTPUT(XYPLOT)

CSCALE = 1.5

XAXIS = YES

YAXIS = YES

XGRID LINES = YES

YGRID LINES = YES

XTITLE = TIMEYTITLE = FORCE

TCURVE = FORCE IN THE GAP (ELEMENT 899)

XYPLOT STRESS /899(2)

YTITLE = DISPLACEMENT

TCURVE = DISP. (T1) AT MASS PT. (GP99), FREE END (GP21)

XYPLOT DISP /99(T1),21(T1)

YTITLE = STRESS

TCURVE = STRESS AT FREE END (ELEMENT 120)


TCURVE = STRESS AT FIXED END (ELEMENT 101)





Force in the GAP Element





Displacement for the Free End and Rigid Body





Stress in the Rod at the Free End





Stress in Rod at the Fixed End




SECTION 8

NONLINEAR ANALYSIS WITHSUPERELEMENTS




TABLE OF CONTENTS

Page Advantage Of Superelement Analysis 8-4Typical Aircraft Superelement Arrangement 8-6




Typical Aircraft Superelement Arrangement 8-6How Are Superelements Defined In MSC.Nastran? 8-7

Grid Point Partitioning 8-8Interior Versus Exterior 8-10Element Partitioning 8-11Solution Terminology 8-12Super Command 8-14Superelement Example Input 8-16Nonlinear Analysis Features 8-17Hierarchy Of Load Data 8-20Example Of Case Control With Upstream Loads 8-21

Example Of Bulk Data To Specify Upstream Loads 8-22Workshop Problem 1 8-23Solution For Workshop Problem 1 8-28

ADVANTAGE OF SUPERELEMENT ANALYSIS

Large problems (i.e., allows solving problems thatexceed your hardware capabilities).




Less CPU or wall clock time per run (reduced risk since

each superelement may be processed individually). Partial redesign requires only partial solution (cost). Allows more control of resource usage. Partitioned input desirable.

Organization Repeated components

Partitioned output desirable. Organization

Comprehension Components may be modeled by subcontractors.

ADVANTAGE OF SUPERELEMENT ANALYSIS

Multi-step reduction for dynamic analysis.

Zooming (or global-local analysis).




g ( g y )

Allows for efficient configuration studies (“What if...”).

TYPICAL AIRCRAFT SUPERELEMENTARRANGEMENT

1




2

5

63

4

1 2 3 4 5 6

0

1 2 3 4 5 6

0

56

Single-Level Tree Multilevel Tree

Small

Big

Body Tail Wing

123

HOW ARE SUPERELEMENTS DEFINED INMSC.NASTRAN?

Superelements are identified using numbers (SEID). Each superelement (SEID > 0) is defined with its own set

f id l t t i t l d t




of grids, elements, constraints, loads, etc. Interior grid points are assigned (partitioned) to a superelement by the

user. Exterior grid points, elements, loads, and constraints are automatically

partitioned by the program based on interior grid point assignments.

The residual structure is a superelement that contains

grid points, elements, etc., which are not assigned to anyother superelement. Last superelement (SEID = 0) to be processed. Superelement on which the assembly analysis (nonlinear, transient

response, frequency response, buckling, system modes, etc.) isperformed.

A superelement may also be defined as an image of asuperelement or obtained from outside MSC.NASTRAN.

GRID POINT PARTITIONING

Bulk Data Entries

SEIDETC.GIDGRID

10987654321




Only interior points need to be defined.

247GRID

57THRU470SESET

G2“THRU”G1SEIDSESET

10987654321

Superelements are identifiedby an integer

GRID POINT PARTITIONING

SESET takes precedence over GRID. For the example shown above, Grid Point 47 will belong to the residual

structure (SEID=0)




structure (SEID=0).

Elements, constraints, loads, etc., are automaticallypartitioned.

Points not assigned belong to the residual structure bydefault. A model with no grid point assignments is

defined as a residual structure-only model.

INTERIOR VERSUS EXTERIOR

A grid point assigned to a superelement by the user isinterior to that superelement.

Th i d d fi t i t f id i t




The processing order defines exterior sets of grid points

for each superelement. A grid point that is connected to a superelement and is

interior to a downstream superelement is exterior to theupstream superelement.

Scalar points are interior only to the residual structurebut may be exterior to any number of superelements.

ELEMENT PARTITIONING

Automatically performed by the program. All element identification numbers must be unique.

An element that is connected entirel b the interior




An element that is connected entirely by the interior

points of a superelement is assigned to thatsuperelement. Branch element - An element that is connected to the

interior points of more than one superelement - is

assigned to the most upstream superelement. Boundary element - An element that is connected by all

exterior points of one or more superelements - is sentdownstream (SEELT can be used to assign it upstream).

Concentrated mass element (CONMi) is assigned asinterior to the superelement that contains the attachmentGRID point.

SOLUTION TERMINOLOGY

Superelement matrix generation “SEMG” Generate structural matrices (KGG, KJJ), MJJ, BJJ.

Enforced displacements rigid elements MPCs check singularities




Enforced displacements, rigid elements, MPCs, check singularities.

Superelement load generation “SELG” (statics only) Generate load matrices (PG, PJ).

Superelement stiffness (K) reduction “SEKR” (stiffnessonly)

Superelement mass (and damping) reduction “SEMR” Assemble upstreams.

Reduce to (boundary exterior) points.

Superelement load reduction “SELR” Loads, mass or damping.

Float downstream to assemble, reduce.

SOLUTION TERMINOLOGY

Superelement data recovery “SEDR” Expand boundary displacements.

Compute internal loads stresses element strain energy etc




Compute internal loads, stresses, element strain energy, etc.

SUPER COMMAND

Partitions (assigns) a subcase to a superelement(s).

Associates a superelement(s) with requests forparameters loads constraints and output




parameters, loads, constraints, and output.

Subcase is required for each superelement and for eachload condition.

If the Case Control Section does not contain a SUPER

command, then loads, constraints, and output requestsare applied to the residual structure only.

The SUPER command may reference a superelement ora SET of superelements. Note: The SET ID must be unique with respect to any superelement

IDs.

SUPER COMMAND

Examples:SUPER = ALL

or but

SET 1 = 10, 20, 0 SET 10 = 10, 20, 0




Form of SUPER command

SUPER = i,jwhere i = superelement ID or set of superelements j= load sequence number (a counter on loading conditions)

The load sequence number for a superelement cannot be greaterthan the number of loading conditions for the residual structure (seethe MSC.NASTRAN Quick Reference Guide).

The appropriate SE_ _ = n commands must also appear above thesubcase level.

SUPER = 1 (SET) SUPER = 10 (SEID)

or

Defaults differently than other entries.

SUPER = 10

SUPERELEMENT EXAMPLE INPUT

P

10 302515 2016 19 21 24 26 29




Case Control:SEALL = ALL

SUPER = ALL Bulk Data:

SESET, 1, 16, THRU, 19



SEID = 1 SEID = 2 SEID = 3

NONLINEAR ANALYSIS FEATURES

Linear assumptions - only the residual structure isallowed to be nonlinear (material or geometric).

Nonlinear superelement analysis can be restarted from




p ylinear analysis (databases from SOLs 101 and 109).

Restarts - No recalculations are required for upstreamsuperelements if there is no change in superelements.

For Unstructured Solution Sequences 66 and 99, specify

for every superelement unless SEALL=ALL is used. SELG, SELR for changes in loads. SEMG, SEKR, SEMR for changes in elements. Always do SEALL on residual superelements.

Recommendation: Read the MSC.NASTRANSuperelement Analysis User’s Guide or MSC.NASTRANSuperelement Seminar Notes.


Load vectors for the upstream elements must begenerated before the nonlinear solutions.

Case Control command SUPER is used to partition the




pproper subcase to a superelement.

All the subcases should include the SUPER command(default, SUPER=0) except when SUPER=ALL isspecified above the subcases.

Case Control command LOADSET selects LSEQ loads. Only one LOADSET may appear in Case Control and

must be above all the subcases. Bulk Data CLOAD entry is designed to apply static loads

to upstream superelements by combining loads definedin LSEQ.


Case Control command CLOAD must be specified in theresidual solution subcases to have loads on thesuperelements.




p

The Case Control command CLOAD must be specifiedin all the subcases to have data recovery forsuperelements.

Usual static load entries (LOAD, FORCE, etc.) applied tothe upstream superelements cannot be directlyreferenced by a Case Control command LOAD.

Any loads which are referenced by a CLOAD entry

should not be again referenced by a LOAD entry,otherwise, the load will be doubled, e.g., GRAV, TEMP.

HIERARCHY OF LOAD DATA

LOAD CLOAD LOADSET

SE 0 Upstream Superelement




Bulk Data CLOAD

DAREA2DAREA1

LSEQ

Static Loads

EXAMPLE OF CASE CONTROL WITHUPSTREAM LOADS

.

.

.

SEALL = ALL

LOADSET = 1000$ Selects LSEQ 1000 for

This command

processes upstream

Points to




LOADSET 1000

SUPER = ALLDISP = ALL

ETC.

.

.

.

SUBCASE 10

CLOAD = 1001

NLPARM = 12

.

.

.

SUBCASE 20

CLOAD = 1002NLPARM = 22

LOAD = 10

$ Identify superelementsSets Up

$ Refers to CLOAD Bulk Data

$ Convergence control

Nonlinear Solutions

for Residual

Points to

LSEQ

Residual $ Residual superelement forces

EXAMPLE OF BULK DATA TO SPECIFYUPSTREAM LOADS

$ LSEQ selected by LOADSET/DAREA may be referenced by RLOAD, TLOAD

LOADSET = SID

$ (LOADSET) (DAREA) (P-ID) (TID)

Load Column (lowest to highest)




$ (LOADSET) (DAREA) (P-ID) (TID)

LSEQ 1000 101 1LSEQ 1000 102 2LSEQ 1000 103 27

$ Usual LOAD entries

FORCE 1 etc.PLOAD 2 etc.

GRAV 27 etc.

$ CLOAD combines LSEQ loads for upstream superelements$ CID S S1 DAREA S2 DAREA

CLOAD 1001 1.0 386 103 1.0 101CLOAD 1002 1.0 386. 103 1.0 102

CLOAD = CID

WORKSHOP PROBLEM 1

Purpose To demonstrate one possible way to specify upstream and residual

loads.

P bl D i ti




Problem Description The model shown below consists of three superelements:

superelement 100, superelement 200, and superelement 0.

SE 100 SE 0 SE 200

101Q4 1003BM 201Q4

1002BM

102 202

101 201

2 4

31

1001Q4

z

yx

WORKSHOP PROBLEM 1 (Contd.)

Perform the analysis for the following loads:

Subcase Upstream Load Combination Residual Load

1 No Load 1 0 (PLOAD2 1000)




1 No Load -1.0 (PLOAD2 1000)

2 -1.0 (PLOAD2 112) No Load

3 -1.0 (PLOAD2 113) -1.0 (PLOAD2 1000)

4 0.2 (PLOAD2 113)

+.5 (PLOAD2 114)

-1.0 (PLOAD2 1000)

5 1.4 (PLOAD2 113)+1.0 (PLOAD2 114)+.5 (PLOAD2 115)

-1.0 (PLOAD2 1000)

SID


Input File for ModificationID CHAP8WS1,NAS103, Chap 8 Workshop 1 $ AR (12/28/03)SOL 106CENDTITLE=SUPERELEMENT LOAD COMBINATION TESTSUBTITLE=TWO TIPS PLUS A RESIDUAL




SUBTITLE TWO TIPS PLUS A RESIDUALECHO = BOTH

DISP=ALLSTRESS=ALLSPC = 20SUPER = ALL

SUBCASE 1LABEL=1000 PSI RESIDUALNLPARM=10SUBCASE 2LABEL=MINUS 1500 PSI SE 100NLPARM=10SUBCASE 3

LABEL=1000 PSI RESIDUAL MINUS 1500 PSI SE 100NLPARM=10SUBCASE 4LABEL=1000 PSI RESIDUAL PLUS 1300 PSI SE 100 PLUS 750 PSI SE 200NLPARM=10SUBCASE 5LABEL=1000 PSI RESIDUAL PLUS 5600 PSI SE 100 PLUS 3000 PSI SE 200NLPARM=20BEGIN BULK$ PARAMETERSPARAM, POST, 0

NLPARM, 10, 2, , AUTO, 10, , PW, NONLPARM, 20, 2, , AUTO, 10, , PW, YES$ PROPERTIESMAT1, 1, 29.E6, , 0.3, .001, 6.5E-4MAT1, 10, 29.E6, , 0.3, .001, 6.5E-4MATS1, 10, , PLASTIC, 2.9E6, 2, 2, 33.E3PSHELL, 100, 1, 0.5, 1PSHELL, 1000, 10, 0.5, 10


Input File for Modification (Cont.)$ LINEAR ELEMENTS IN RESIDUALCBEAM, 1002, 10, 1, 4, 2CBEAM, 1003, 10, 2, 3, 1PBEAM, 10, 1, 0.2, 8.333E-5, 8.333E-3

, -0.5, -0.1, , , 0.5, 0.1




$ BOUNDARY CONDITIONS

SPC1, 20, 12, 1, 2SPC1, 20, 13, 1, 3$$ LOADING CONDITIONS$ ASSIGNS LOAD VECTORS TO THE SUPERELEMENTS AND LABELS THEM$ APPLIED LOADSPLOAD2, 111 , 0., 101PLOAD2, 112 , 1.5E3, 101PLOAD2, 113 , 1.5E3, 101PLOAD2, 114 , 2.0E3, 101PLOAD2, 115 , 3.0E3, 101PLOAD2, 114 , 1.5E3, 201PLOAD2, 115 , 3.0E3, 201PLOAD2, 1000, -1.0E3, 1001$ COMBINE LOADS$ GEOMETRYGRID, 1, , , -1.0, 0., , 4, 0GRID, 2, , , -1.0, 1., , 4, 0GRID, 3, , , 1.0, 0., , 4, 0GRID, 4, , , 1.0, 1., , 4, 0GRID, 101, , , -2.0, 0., , 4, 100GRID, 102, , , -2.0, 1., , 4, 100GRID, 201, , , 3.0, 0., , 4, 200

GRID, 202, , , 3.0, 1., , 4, 200CQUAD4, 101, 100, 1, 2, 102, 101CQUAD4, 201, 100, 4, 3, 201, 202CQUAD4, 1001, 1000, 1, 3, 4, 2ENDDATA


Hints Bulk Data changes:

Define a dummy load PLOAD2, 111 for superelement 100.

Define CLOADs 1010, 1020, 1030, 1040, and 1050 to apply upstream loads




Define CLOADs 1010, 1020, 1030, 1040, and 1050 to apply upstream loads

in subcases 1, 2, 3, 4, and 5, respectively. Define LOAD 10 to apply residual load.

Define LSEQ,100 entries to select PLOAD2 entries with ID = 111 through115.

Case Control changes: Define SUPER = ALL above subcase level. Define LOADSET = 100 above subcase level.

Select LOAD and CLOAD entries for each subcase.

SOLUTION FOR WORKSHOP PROBLEM 1

ID CHAP8WS1s,NAS103, Chap 8 Workshop 1 $ AR (12/28/03)SOL 106CENDTITLE=SUPERELEMENT LOAD COMBINATION TESTSUBTITLE=TWO TIPS PLUS A RESIDUALECHO = BOTHDISP=ALLSTRESS=ALLSPC = 20LOADSET = 100 $ REFERING TO LSEQ FOR UPSTREAM LOADSSUPER = ALL

SUBCASE 1




SUBCASE 1LABEL=1000 PSI RESIDUAL

NLPARM=10LOAD = 10CLOAD = 1010SUBCASE 2LABEL=MINUS 1500 PSI SE 100NLPARM=10CLOAD = 1020SUBCASE 3LABEL=1000 PSI RESIDUAL MINUS 1500 PSI SE 100NLPARM=10LOAD = 10CLOAD = 1030SUBCASE 4

LABEL=1000 PSI RESIDUAL PLUS 1300 PSI SE 100 PLUS 750 PSI SE 200NLPARM=10LOAD = 10CLOAD = 1040SUBCASE 5LABEL=1000 PSI RESIDUAL PLUS 5600 PSI SE 100 PLUS 3000 PSI SE 200NLPARM=20LOAD = 10CLOAD = 1050BEGIN BULK$ PARAMETERSPARAM, POST, 0NLPARM, 10, 2, , AUTO, 10, , PW, NO

NLPARM, 20, 2, , AUTO, 10, , PW, YES$ PROPERTIESMAT1, 1, 29.E6, , 0.3, .001, 6.5E-4MAT1, 10, 29.E6, , 0.3, .001, 6.5E-4MATS1, 10, , PLASTIC, 2.9E6, 2, 2, 33.E3PSHELL, 100, 1, 0.5, 1PSHELL, 1000, 10, 0.5, 10

SOLUTION FOR WORKSHOP PROBLEM 1(Contd.)

$ LINEAR ELEMENTS IN RESIDUALCBEAM, 1002, 10, 1, 4, 2CBEAM, 1003, 10, 2, 3, 1PBEAM, 10, 1, 0.2, 8.333E-5, 8.333E-3

, -0.5, -0.1, , , 0.5, 0.1$ BOUNDARY CONDITIONSSPC1, 20, 12, 1, 2SPC1, 20, 13, 1, 3$ LOADING CONDITIONSLSEQ, 100, 11, 111LSEQ, 100, 12, 112LSEQ 100 13 113




LSEQ, 100, 13, 113

LSEQ, 100, 14, 114LSEQ, 100, 15, 115$ ASSIGNS LOAD VECTORS TO THE SUPERELEMENTS AND LABELS THEM$ APPLIED LOADSPLOAD2, 111 , 0., 101PLOAD2, 112 , 1.5E3, 101PLOAD2, 113 , 1.5E3, 101PLOAD2, 114 , 2.0E3, 101PLOAD2, 115 , 3.0E3, 101PLOAD2, 114 , 1.5E3, 201PLOAD2, 115 , 3.0E3, 201PLOAD2, 1000, -1.0E3, 1001$ COMBINE LOADSLOAD, 10, 1.0, -1.0, 1000CLOAD, 1010, 1.0, -1.0, 11CLOAD, 1020, 1.0, -1.0, 12CLOAD, 1030, 1.0, -1.0, 13CLOAD, 1040, 1.0, 0.2, 13, 0.5, 14CLOAD, 1050, 1.0, 1.4, 13, 1.0, 14, 0.5, 15$ GEOMETRYGRID, 1, , , -1.0, 0., , 4, 0GRID, 2, , , -1.0, 1., , 4, 0GRID, 3, , , 1.0, 0., , 4, 0GRID, 4, , , 1.0, 1., , 4, 0GRID, 101, , , -2.0, 0., , 4, 100GRID, 102, , , -2.0, 1., , 4, 100GRID, 201, , , 3.0, 0., , 4, 200

GRID, 202, , , 3.0, 1., , 4, 200CQUAD4, 101, 100, 1, 2, 102, 101CQUAD4, 201, 100, 4, 3, 201, 202CQUAD4, 1001, 1000, 1, 3, 4, 2ENDDATA




SECTION 9

SPECIAL TOPICS




TABLE OF CONTENTS

PageSpecial Topics 9-5Normal Modes of Deformed Structure 9-6Normal Modes of Prestressed Structure 9-7Normal Modes With Differential Stiffness 9-8




Example Problem 1 –Modes of Preloaded Structure 9-9Input File For Problem 1A – Modes Without Preload 9-11Partial Output File For Problem #1A Modes Without Preload 9-12Input File: Problem #1B – Modes With Preload Using SOL 106 9-13Partial Output File For Problem #1B – Modes With Preload Using

SOL 106 9-14Input File: Problem #1C – Modes With Preload Using SOL 103 9-15Partial Output File For Problem #1C – Modes With Preload Using

SOL 103 9-16

Composite Elements 9-17Features Of Nonlinear Composite Beam 9-18Input Data Entry PBCOMP Beam Property Alternate From For

PBEAM 9-19


Beam Cross-Sectional Area Lumping Scheme For VariousSections 9-21

Smeared Cross-Sectional Properties (I1,I2,I12 Ignored on ParentEntry) 9-23

Features Of Composite Plates 9-25




Laminate And Ply Orientation (PCOMP) 9-262-D Orthotropic Material 9-27Composite Material Specification 9-28Composite Element Specification 9-30

PCOMP – Mat Relationship 9-33 Anisotropic Material In Mat2 9-34Example Problem 2: Composite Cantilever Beam 9-36Failure Theory For Composites 9-40Output For Composite Element 9-42

Smeared Material Properties In PSHELL And MAT2 9-43Layer Stresses In Composite Elements 9-44Failure Index Table 9-45

SPECIAL TOPICS

Nonlinear modal analysis Composite analysis




Normal Modes of Deformed Structure




Large Geometry Changes

u1

Nonlinear Material

k1

F

k0

K0 = k1

Normal Modes of Prestressed Structure

Procedures for obtaining frequencies of a preloadedstructure.

Method 1 (Nonlinear Solution for Preload)




Use SOL 106. Include linear or nonlinear material properties as required by modeling.

If material is linear, then only linear material properties are referenced.

Include PARAM,LGDISP,1 in the Bulk Data Section.

Only the residual structure (SEID=0) may contain nonlinear elements. All upstream superelements must be linear.

A METHOD = X Case Control Command in subcase calls out theappropriate EIGRL entry.

Include PARAM,NMLOOP,Y where Y is the loopid that you want to

calculate the normal modes at.

Normal Modes WithDifferential Stiffness (Cont.)

Procedures for obtaining frequencies of a preloadedstructure.

Method 2 (Linear Solution for Preload)




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 calculates the modes with a METHOD = X

case control command, where X is the appropriate EIGRL ID.

The second subcase must also contain a STATSUB = Y command,where Y is subcase ID of the first subcase.

EXAMPLE PROBLEM 1 -Modes of PreloadedStructure

Consider the simply supported beam as shown below.Calculate the first bending frequency: Case A: Without preload

Case B: With preload using SOL106




Case C: With preload using SOL 103

P

EXAMPLE PROBLEM 1 -Modes of PreloadedStructure (Cont.)

0.1 in

0.1 in

1.0 in

2.0 in




0.1 in

1.0 in

Length: 100 inHeight: 2 in

Width: 1 in

Thickness: 0.100 in

Area: 0.38 in2

I1: 0.229 in4

I2: 0.017 in4

Input File For Problem 1A - Modes WithoutPreload

SOL 103

DIAG 8

CEND

TITLE = Normal Modes, Unloaded

$

SUBCASE 1

METHOD = 10

SPC 1

$

MAT1, 1, 1.+7, , .3, .101

GRID, 1, , 0., 0., 0., ,345

GRID, 2, , 10., 0., 0., ,345

GRID 3 20 0 0 345




SPC = 1VECTOR=ALL

$

BEGIN BULK

PARAM, COUPMASS, 1

PARAM, WTMASS, .00259

$

EIGRL,10,,,3

PBARL, 1, 1, , I, , , , ,+PB+PB, 2., 1., 1., .1, .1, .1

CBAR, 1, 1, 1, 2, 0., 1., 0.

CBAR, 2, 1, 2, 3, 0., 1., 0.

CBAR, 3, 1, 3, 4, 0., 1., 0.

CBAR, 4, 1, 4, 5, 0., 1., 0.

CBAR, 5, 1, 5, 6, 0., 1., 0.

CBAR, 6, 1, 6, 7, 0., 1., 0.

CBAR, 7, 1, 7, 8, 0., 1., 0.

CBAR, 8, 1, 8, 9, 0., 1., 0.

CBAR, 9, 1, 9, 10, 0., 1., 0.

CBAR, 10, 1, 10, 11, 0., 1., 0.

GRID, 3, , 20., 0., 0., ,345GRID, 4, , 30., 0., 0., ,345

GRID, 5, , 40., 0., 0., ,345

GRID, 6, , 50., 0., 0., ,345

GRID, 7, , 60., 0., 0., ,345

GRID, 8, , 70., 0., 0., ,345

GRID, 9, , 80., 0., 0., ,345

GRID, 10, , 90., 0., 0., ,345

GRID, 11, , 100., 0., 0., ,345SPC1, 1, 1234, 1

SPC1, 1, 234, 11

FORCE, 1, 11, 0, 500., 1., 0., 0.

ENDDATA

Partial Output File For Problem #1A ModesWithout Preload

0

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 ...................... 7

NUMBER OF DECOMPOSITIONS ............. 2

NUMBER OF ROOTS FOUND ................ 3




NUMBER OF SOLVES REQUIRED ............ 4

1 NORMAL MODES EXAMPLE APRIL 8, 1998 MSC.Nastran 4/ 6/98 PAGE 5

0 SUBCASE 1

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

MODE EXTRACTION EIGENVALUE RADIANS CYCLES GENERALIZED GENERALIZED

NO. ORDER MASS STIFFNESS

1 1 2.239398E+04 1.496462E+02 2.381693E+01 1.000000E+00 2.239398E+04

2 2 3.549898E+05 5.958102E+02 9.482614E+01 1.000000E+00 3.549898E+053 3 1.771818E+06 1.331096E+03 2.118506E+02 1.000000E+00 1.771818E+06

Input File: Problem #1B - Modes With Preload Using SOL 106

SOL 106

TIME 600CEND

TITLE = Normal Modes, Prestressed (nonlinear)

METHOD = 10

SUBCASE 1

NLPARM = 1

SPC = 1

LOAD = 1

DISPLACEMENT=ALL

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 3 101




DISPLACEMENT=ALL$

BEGIN BULK

PARAM, COUPMASS, 1


$

PARAM, LGDISP, 1

NLPARM, 1, 5, , AUTO, 5, 25, PW, NO

+NLP, .001, 1.-7

PARAM,NMLOOP,5

$

EIGRL,10,,,3

PBARL, 1, 1, , I, , , , ,+PB

+PB, 2., 1., 1., .1, .1, .1

CBAR, 1, 1, 1, 2, 0., 1., 0.

CBAR, 2, 1, 2, 3, 0., 1., 0.CBAR, 3, 1, 3, 4, 0., 1., 0.

CBAR, 4, 1, 4, 5, 0., 1., 0.

CBAR, 5, 1, 5, 6, 0., 1., 0.

MAT1, 1, 1.+7, , .3, .101

GRID, 1, , 0., 0., 0., ,345

GRID, 2, , 10., 0., 0., ,345

GRID, 3, , 20., 0., 0., ,345

GRID, 4, , 30., 0., 0., ,345

GRID, 5, , 40., 0., 0., ,345

GRID, 6, , 50., 0., 0., ,345

GRID, 7, , 60., 0., 0., ,345GRID, 8, , 70., 0., 0., ,345

GRID, 9, , 80., 0., 0., ,345

GRID, 10, , 90., 0., 0., ,345

GRID, 11, , 100., 0., 0., ,345

SPC1, 1, 1234, 1

SPC1, 1, 234, 11

FORCE, 1, 11, 0, 500., 1., 0., 0.

ENDDATA

Partial Output File For Problem #1B Modes With PreloadUsing SOL 106



NO. ORDER MASS STIFFNESS

1 1 2.735837E+04 1.654037E+02 2.632481E+01 1.000000E+00 2.735837E+04

2 2 3.748482E+05 6.122484E+02 9.744236E+01 1.000000E+00 3.748482E+05

3 3 1.816508E+06 1.347779E+03 2.145057E+02 1.000000E+00 1.816508E+06




Input File: Problem #1C - Modes With Preload Using SOL 103

SOL 103

DIAG 8

CEND

TITLE = Normal Modes, preloaded (linear)

SPC = 1

DISPLACEMENT=ALL

$

SUBCASE 1

LOAD = 1

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 3 101




LOAD 1SUBCASE 2

METHOD = 10

STATSUB = 1

$

BEGIN BULK

PARAM, COUPMASS, 1


$EIGRL,10,,,3

PBARL, 1, 1, , I, , , , ,+PB

+PB, 2., 1., 1., .1, .1, .1

CBAR, 1, 1, 1, 2, 0., 1., 0.

CBAR, 2, 1, 2, 3, 0., 1., 0.

CBAR, 3, 1, 3, 4, 0., 1., 0.

CBAR, 4, 1, 4, 5, 0., 1., 0.

CBAR, 5, 1, 5, 6, 0., 1., 0.

MAT1, 1, 1.+7, , .3, .101

GRID, 1, , 0., 0., 0., ,345

GRID, 2, , 10., 0., 0., ,345

GRID, 3, , 20., 0., 0., ,345

GRID, 4, , 30., 0., 0., ,345

GRID, 5, , 40., 0., 0., ,345

GRID, 6, , 50., 0., 0., ,345

GRID, 7, , 60., 0., 0., ,345GRID, 8, , 70., 0., 0., ,345

GRID, 9, , 80., 0., 0., ,345

GRID, 10, , 90., 0., 0., ,345

GRID, 11, , 100., 0., 0., ,345

SPC1, 1, 1234, 1

SPC1, 1, 234, 11

FORCE, 1, 11, 0, 500., 1., 0., 0.

ENDDATA

Partial Output File For Problem #1C Modes With PreloadUsing SOL 103

1 NORMAL MODES WITH DIFFERENTIAL STIFFNESS APRIL 9, 1998 MSC.Nastran 4/ 6/98

PAGE 9

0 SUBCASE 2



NO. ORDER MASS STIFFNESS1 1 2 735837E+04 1 654037E+02 2 632481E+01 1 000000E+00 2 735837E+04




1 1 2.735837E+04 1.654037E+02 2.632481E+01 1.000000E+00 2.735837E+04

2 2 3.748482E+05 6.122484E+02 9.744236E+01 1.000000E+00 3.748482E+05

3 3 1.816508E+06 1.347779E+03 2.145057E+02 1.000000E+00 1.816508E+06

COMPOSITE ELEMENTS

Beam with PBCOMP QUAD4 and TRIA3 with PCOMP and MAT8




FEATURES OF NONLINEAR COMPOSITEBEAM

BEAM properties in PBCOMP. May be used for geometric and material nonlinear

problems.

Distribution of lumped areas of the BEAM cross section




Distribution of lumped areas of the BEAM cross sectionin arbitrary configuration.

Different material for each of the lumped areas allowed.

Maximum of 20 lumped areas may be input. The BEAM is assumed to be uniform (non-tapered).

Warping effects are ignored.

INPUT DATA ENTRY PBCOMP BEAMPROPERTY ALTERNATE FORM FOR PBEAM

Can replace PBEAM for linear or nonlinear analysisWill be ignored if 2nd to 21stcontinuation entry is present

MATS1

1230002.9639PBCOMP

abcNSMJI12I2I1 AMIDPIDPBCOMP




234000001123

bcdSECTIONN2N1M2M1K2K1

Shear Stiffness KAG

Symmetry Option

4562500.90.245

def SOUT2MID2C2Z2Y2cde

INPUT DATA ENTRY PBCOMP BEAMPROPERTY ALTERNATE FORM FOR PBEAM

345NO1801.2-0.534

cdeSOUT1MID1C1Z1Y1bcd

If blank, use parent entry.

For heat transfer, use only MAT4 and/or MAT 5.




Continue 18 more times for a total of 21 continuationentries.

Need E, ν or E, G on MAT1 entry for parent entry.

p y

BEAM CROSS-SECTIONAL AREA LUMPINGSCHEME FOR VARIOUS SECTIONS

Zre f 1

2

3

4

5

6

7

8

Yref

0 2 K z,( )

Ky Kz,( )

2 K y 0,( )

Zre f Zre f

Yre f Yre f

1

2

3

4

5

6

8

7

1 2

3

4

5 6

8

7




SECTION=0 ( default)Symmetric about y and z

SECTION=1(with continuation entry)Symmetric about y and z

SECTION=2Symmetric about y

SECTION=3

Symmetric about z

SECTION=4

Symmetric about y=z=0

SECTION=5

No symmetry

Izz - Moment of inertia about z-axis

Iyy - Moment of inertia about y-axis

Ky

Izz

A------ Kz

Iyy

A------ C1

1

8---=,=,=

Y1 Y3 Y5– Y– 7= = =

Z1 Z3– Z5 Z7 etc.,–= = =

Y1 Y5=

Z1 Z– 5 etc.,=

Zre f Zre f Zr ef

Yref Yref Yre f

1

2

3

4

5

6

7

8

12

3

4

8

75 6

1 2 3 4

5

6

7

8

Y1 Y5 Z1 Z5 etc.,=,= Y1 Y5 Z1 Z5 etc.,=,=

BEAM CROSS-SECTIONAL AREA LUMPINGSCHEME FOR VARIOUS SECTIONS

Notes: 1. Integration points (lumped areas) are numbered 1-8, to be referenced

by stress output request (SO field).

2. User-specified points are denoted by •, and the program default

points are denoted by ¤.




points are denoted by . 3. Underlined words refer to fields on the PBCOMP entry (Section 5 of

the MSC/NASTRAN Quick Reference Guide).

4. Use 1/2 areas on the symmetric boundary.

SMEARED CROSS-SECTIONAL PROPERTIES(I1,I2,I12 Ignored on Parent Entry)

Offset of neutral axis

yNA

yi C

i E

i

i 1=

n

∑

Ci E i

n

∑

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

PBCOMP Field 4 of each continuation

SEAL2 Field 1

MAT1




Effective cross-sectional area

i 1=∑ PBCOMP Field 4 of each continuation

line greater than 1.

zNA

zi C

i E

i

i 1=

n

∑

Ci E

i

i 1=

n

∑

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

A A= zi

Ci E

iEo

---------------

i 1=

n

∑

PBCOMP Field 4

SMEARED CROSS-SECTIONAL PROPERTIES(I1,I2,I12 Ignored on Parent Entry)

Effective moment of inertia

I1 AC

i E

iy

iy

NA–( )

2

Eo

------------------------------------------

i 1=

n

∑=

MID Parent Entry

C E z z( )2n




PBCOMPField 8

Override I1, I2, I12of Parent Entry

I2 AC

i E

iz

iz

NA–( )

Eo

------------------------------------------

i 1=

∑=

I12 A

Ci E

iy

iy

NA–( ) z

izNA

–( )

Eo

------------------------------------------------------------------

i 1=

n

∑=

J JG

i

nGo

----------

i 1=

n

∑=

FEATURES OF COMPOSITE PLATES

Classical lamination theory is used. Equations for laminate (aggregate) are derived from those of laminae. Each individual lamina is in plane stress. The laminate is presumed to consist of perfectly bonded laminae.

Bond is presumed to be very thin and nonshear deformable. No lamina can slip relative to each other; laminate acts as a single layer




p y No lamina can slip relative to each other; laminate acts as a single layer.

Plate elements (QUAD4, QUAD8, TRIA3, TRIA6) areavailable for modeling composites.

Limited to the linear material. User interface: PCOMP and MAT8. Pre-(IFP6) and post-(SDRCOMP) processing of PCOMP

and MAT8. Stress output for user-specified plies available. Failure indices for elements can be requested.

LAMINATE AND PLY ORIENTATION (PCOMP)

Z

Y

n




T

θ

X

Y

X

21

12

Z0

Y

2-D ORTHOTROPIC MATERIAL

Orthotropic material in plane stress requires:

Material constants in terms of E1, E2, ν12, AND G12

σ3

0 τ13

0 and τ23

0=,=,=

E1 E1ν21

0




where E2 ν12 = E1 ν21

Transverse shear effects included (G1z, G2z) (Mindlin orReisner plate).

σ1

σ2

τ

12

1

1 ν12ν21–--------------------------

1 21

1 ν12ν21–-------------------------- 0

E2ν

12

1 ν12

ν21

–--------------------------

E2

1 ν12

ν21

–-------------------------- 0

0 0 G12

ε1

ε2

γ

12

=

COMPOSITE MATERIAL SPECIFICATION

SYcYtXcXtTREF A2 A1

00.0561.5+63.+62.+60.31.+630.+6100MAT8

RHOG2zG1zG12n12E2E1MIDMAT8

10987654321




Field Contents

MAT8 Input data for each ply. May be replaced by MAT2

E1, E2 Moduli in principal directions

n12 e2/e1 uniaxial loading in 1-direction

01.+38.+22.+21.5+41.+41551.5-628.-60

1.-30

F12GE

COMPOSITE MATERIAL SPECIFICATION

Field ContentsG12 In-plane shear modulus

G1z, G2z Transverse shear moduli in 1-z and 2-z planes

RHO Mass density

TREF Reference temperature for calculation of thermalloads or a temperature dependent thermal




p pexpansion coefficient

A1, A2 Thermal expansion coefficients in the 1- and 2-directions

Xt, Xc, Yt Tension and compression allowable stresses in 1-and 2- directions

S Allowable stress in plane shear for failure index

GE Structural damping

F12 Interaction term in Tsai-Wu failure theory

Example:

COMPOSITE ELEMENT SPECIFICATION

Etc.SOUT3THETA3T3MID3

SOUT2THETA2T2MID2SOUT1THETA1T1MID1

LAMGETREFFTSBNSMZ0PIDPCOMP10987654321




Example:

Field Contents

PID Property identification number. (0 < Integer < 106)

Z0 Distance from the reference plane to the bottomsurface. See Remark 10. (Real; Default = -1/2 thethickness of the element)

90-45

45YES00.056171

HOFF1000.07.45-0.224181PCOMP


Field ContentsNSM Nonstructural mass per unit area. (Real)SB Allowable shear stress of the bonding material (allowable

interlaminar shear stress). Required if failure index is desired.(Real > 0.0)

FT Failure theory. The following theories are allowed (Characteror blank. If blank, then no failure calculation will be




o b a b a , t e o a u e ca cu at o beperformed):

“HILL” for the Hill theory“HOFF” for the Hoffman theory

“TSAI” for the Tsai-Wu theory“STRN” for the maximum strain theory

TREF Reference temperature. See Remark 3. (Real).GE Damping coefficient. See Remark 4g. (Real; Default =0.0).

LAM Symmetric lamination option. If LAM = “SYM”, only


Field ContentsMIDi Material ID of the various plies. The plies are identified by

serially numbering them from 1 at the bottom layer. The MIDsmust refer to MAT1, MAT2, or MAT8 Bulk Data entries. SeeRemark 1. (Integer > 0 or blank except MID1 must be

specified).Ti Thickness of the various plies. See Remark 1. (Real or blank




except T1 must be specified).THETAi Orientation angle of the longitudinal direction of each ply with

the material axis of the element. (If the material angle on the

element connection entry is 0.0, the material axis and side102 of the element coincide). The plies are to be numberedserially starting with 1 at the bottom layer. The bottom layer isdefined as the surface with the largest –Z value in theelement coordinate system. (Real; Default = 0.0).

SOUTi Stress or strain output request. See Remarks 5 and 6.

(Character: “YES” or “NO”; Default = “NO”).

PCOMP - MAT RELATIONSHIP

CQUAD4

PBCOMP

MAT1 MAT3MAT2




EQUIV PSHELL*

MID1MAT2

MID4MAT4

MID3MAT3

MID2MAT2

ANISOTROPIC MATERIAL IN MAT2

+M225.1+36.2+36.2+3100MAT2

RHOG33G23G22G13G12G11MIDMAT210987654321

+M2320.+50.0025006.5-66.5-6+M22

SSSCSTGETREF A12 A2 A1




Field Contents

Gii Material constants

RHO Mass density A1, A2, A12 Thermal expansion coefficients

1003+M29

MCSID

ANISOTROPIC MATERIAL IN MAT2

Field ContentsTREF Reference temperature for the calculation of thermal loads

GE Structural damping

ST, SC, SS Stress limit in tension, compression and shear for

computing margin of safetyMCSID Material coordinate system ID




σ1

σ2

τ12

G11 G12 G13

G22 G23

SYM G33

ε1

ε2

γ12

T T0–( )

A1

A2

A12

–=

EXAMPLE PROBLEM 2: COMPOSITECANTILEVER BEAM

Y

A0.1 in

360 in




X360 in360 in

5 105

× lb 5 105 lb×

E1 1 107 psi×= E2 1 10

6×=

ν12 0.3= G12 4 106 psi×=

θ 45o for Layers 1 and 3=

θ 00

for Layer 2=

Input File For Example 2 – CompositeCantilever Beam

ID, chap9ex2b, NAS103, chap 8 Ex 2 (Nonlinear)SOL 106

CEND

TITLE = Cantilever Composite Beam of NAS103 chapter 9

ECHO=SORT

SUBCASE 1

SPC = 1

LOAD = 1

NLPARM = 10

DISPLACEMENT(SORT1,REAL)=ALL

SPCFORCES(SORT1,REAL)=ALL

$ ElementsCQUAD4, 1, 1, 1, 2, 15, 14

CQUAD4, 2, 1, 2, 3, 16, 15

CQUAD4, 3, 1, 3, 4, 17, 16

CQUAD4, 4, 1, 4, 5, 18, 17

CQUAD4, 5, 1, 5, 6, 19, 18

CQUAD4, 6, 1, 6, 7, 20, 19

CQUAD4, 7, 1, 7, 8, 21, 20

CQUAD4, 8, 1, 8, 9, 22, 21

CQUAD4, 9, 1, 9, 10, 23, 22

CQUAD4, 10, 1, 10, 11, 24, 23




SPCFORCES(SORT1,REAL) ALL

STRESS(SORT1,REAL,VONMISES,BILIN)=ALL

BEGIN BULK

PARAM, POST, 0

PARAM, AUTOSPC, YES

PARAM, PRTMAXIM, YES

PARAM, LGDISP, 2

$ NONLINEAR SOLUTION STRATEGY

NLPARM, 10, 1, , AUTO, , , PW, YES, +NLP1

+NLP1, , 1.E-2, 1.E-3

$ Composite Material

PCOMP, 1,

, 1, .02, 45., YES, 1, .06, 0., YES

, 1, .02, 45., YES

CQUAD4, 10, 1, 10, 11, 24, 23

CQUAD4, 11, 1, 11, 12, 25, 24

CQUAD4, 12, 1, 12, 13, 26, 25

….

….

CQUAD4, 61, 1, 66, 67, 80, 79CQUAD4, 62, 1, 67, 68, 81, 80

CQUAD4, 63, 1, 68, 69, 82, 81

CQUAD4, 64, 1, 69, 70, 83, 82

CQUAD4, 65, 1, 70, 71, 84, 83

CQUAD4, 66, 1, 71, 72, 85, 84

CQUAD4, 67, 1, 72, 73, 86, 85

CQUAD4, 68, 1, 73, 74, 87, 86

CQUAD4, 69, 1, 74, 75, 88, 87CQUAD4, 70, 1, 75, 76, 89, 88

CQUAD4, 71, 1, 76, 77, 90, 89

CQUAD4, 72, 1, 77, 78, 91, 90

Input File For Problem 1A - Modes WithoutPreload

$ Material Properties

MAT8, 1, 1.E7, 1.E6, .3, 4.E6

$ Nodes

GRID, 1, , 0., 0., 0.

GRID, 2, , 60., 0., 0.

GRID, 3, , 120., 0., 0.

GRID, 4, , 180., 0., 0.

GRID, 5, , 240., 0., 0.

GRID, 6, , 300., 0., 0.

GRID, 79, , 0., 360., 0.

GRID, 80, , 60., 360., 0.

GRID, 81, , 120., 360., 0.

GRID, 82, , 180., 360., 0.

GRID, 83, , 240., 360., 0.

GRID, 84, , 300., 360., 0.

GRID, 85, , 360., 360., 0.

GRID, 86, , 420., 360., 0.

GRID, 87, , 480., 360., 0.

GRID 88 540 360 0




GRID, 6, , 300., 0., 0.

GRID, 7, , 360., 0., 0.

GRID, 8, , 420., 0., 0.

GRID, 9, , 480., 0., 0.

GRID, 10, , 540., 0., 0.

GRID, 11, , 600., 0., 0.GRID, 12, , 660., 0., 0.

GRID, 13, , 720., 0., 0.

….

….

GRID, 88, , 540., 360., 0.

GRID, 89, , 600., 360., 0.

GRID, 90, , 660., 360., 0.

GRID, 91, , 720., 360., 0.

$

SPC1, 1, 123456, 1, 14, 27, 40, 53, 66, 79

$

FORCE, 1, 7, 0, 5.E5, 0., -1., 0.

FORCE, 1, 13, 0, 5.E5, 0., -1., 0.

$

ENDDATA

EXAMPLE PROBLEM 2: COMPOSITE

CANTILEVER BEAM (Contd.)

Vertical Displacement at Point A

(with a 12X6 Mesh of Quad4)




LAYER THICKNESSES DISPLACEMENT

Linear (SOL 101) Nonlinear (SOL 106)

0.02/0.06/0.02 28.95 28.49

0.03/0.04/0.03 33.48 32.90

0.04/0.02/0.04 41.94 41.08

FAILURE THEORY FOR COMPOSITES

Allowable stresses are direction dependent. Failure envelope is defined in the stress space.

Failure index is a measure whether the stress state in

the worst stressed lamina is within or outside theenvelope.




Inter-laminar shear stress is checked against theallowable bonding stress (Sb).

Failure index for the laminate is the larger of the two.

FAILURE THEORY FOR COMPOSITES

Laminate has failed if the failure index is greater than 1. Failure envelope is defined by:

Hills’s theory (ellipsoidal).

Hoffman’s theory. Accounts for differing tension and compression.

Tensor polynomial theory (Tsai-Wu): closed envelope.

Maximum strain theory.




Maximum strain theory.

OUTPUT FOR COMPOSITE ELEMENT

Smeared material properties for equivalent PSHELL andMAT2 data. Requires ECHO = SORT.

Smeared stresses in linear stress output format. Usualstress output request.

Stresses in individual lamina (including inter-laminarh t ) R i YES SOUT fi ld




shear stresses). Requires YES on SOUT field onPCOMP.

Failure index table Requires ELFORCE and ELSTRESS requests.

Allowable stresses must be provided on MAT8.

SMEARED MATERIAL PROPERTIES INPSHELL AND MAT2

For ECHO = SORTCOMPOSITE ELEMENTS SOL 106 MARCH 15, 1993 MSC/NASTRAN 3/12/93 PAGE 5*** USER INFORMATION MESSAGE 4379, THE USER SUPPLIED PCOMP BULK DATA CARDS ARE REPLACED BY THE FOLLOWING PSHELL AND MAT2CARDS.

PSHELL 100 100000100 2.0000E-01 200000100 1.0000E+00 300000100 1.0000E+00 0.0000E+00-1.0000E-01 1.0000E-01 0

MAT2 100000100 1.4286E+07 4.2857E+06 2.5241E-08 1.4286E+07 1.5806E-09 5.0000E+06 0.0000E+000.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00

0MAT2 200000100 1.4286E+07 4.2857E+06 2.0446E-08 1.4286E+07 1.2803E-09 5.0000E+06 0.0000E+00

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+000

MAT2 300000100 4 1667E+06 0 0000E+00 0 0000E+00 4 1667E+06 0 0000E+00 0 0000E+00 0 0000E+00




MAT2 300000100 4.1667E+06 0.0000E+00 0.0000E+00 4.1667E+06 0.0000E+00 0.0000E+00 0.0000E+00

0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+000

LAYER STRESSES IN COMPOSITE ELEMENTS

COMPOSITE ELEMENTS SOL 106 MARCH 15, 1993 MSC/NASTRAN 3/12/93 PAGE 103

SUBCASE 1

S T R E S S E S I N L A Y E R E D C O M P O S I T E E L E M E N T S ( Q U A D 4 )ELEMENT PLY STRESSES IN FIBRE AND MATRIX DIRECTIONS INTER-LAMINAR STRESSES PRINCIPAL STRESSES (ZERO SHEAR) MAXID ID NORMAL-1 NORMAL-2 SHEAR-12 SHEAR-1Z SHEAR-2Z ANGLE MAJOR MINOR SHEAR101 1 -4.07141E+05 -1.35714E+06 -1.69456E+02 7.84304E+02 2.02505E-14 -.01 -4.07141E+05 -1.35714E+06 4.75001E+05101 3 4.07141E+05 1.35714E+06 1.69456E+02 5.49630E-12 1.41913E-28 89.99 1.35714E+06 4.07141E+05 4.75001E+05102 1 -4.07139E+05 -1.35714E+06 -8.32131E+02 5.71212E+02 4.69811E-12 -.05 -4.07138E+05 -1.35714E+06 4.75002E+05102 3 4.07139E+05 1.35714E+06 8.32131E+02 4.00298E-12 3.29237E-26 89.95 1.35714E+06 4.07138E+05 4.75002E+05103 1 -4.07137E+05 -1.35714E+06 -1.19496E+03 3.82783E+02 2.43006E-13 -.07 -4.07136E+05 -1.35714E+06 4.75003E+05

103 3 4.07137E+05 1.35714E+06 1.19496E+03 2.68250E-12 1.70295E-27 89.93 1.35714E+06 4.07136E+05 4.75003E+05104 1 -4.07137E+05 -1.35714E+06 -9.55886E+02 2.18618E+02 -8.70770E-13 -.06 -4.07136E+05 -1.35714E+06 4.75003E+05104 3 4.07137E+05 1.35714E+06 9.55886E+02 1.53205E-12 -6.10225E-27 89.94 1.35714E+06 4.07136E+05 4.75003E+05105 1 -4.07136E+05 -1.35714E+06 -4.99785E+02 7.09528E+01 2.78697E-12 -.03 -4.07136E+05 -1.35714E+06 4.75003E+05105 3 4.07136E+05 1.35714E+06 4.99785E+02 4.97228E-13 1.95307E-26 89.97 1.35714E+06 4.07136E+05 4.75003E+05




FAILURE INDEX TABLE

1 COMPOSITE ELEMENTS SOL 106 MARCH 15, 1993 MSC/NASTRAN 3/12/93 PAGE 104

0 SUBCASE 1 $

F A I L U R E I N D I C E S F O R L A Y E R E D C O M P O S I T E E L E M E N T S ( Q U A D 4 )ELEMENT FAILURE PLY FP=FAILURE INDEX FOR PLY FB=FAILURE INDEX FOR BONDING FAILURE INDEX FOR ELEMENT FLAGID THEORY ID (DIRECT STRESSES/STRAINS) (INTER-LAMINAR STRESSES) MAX OF FP,FB FOR ALL PLIES101 HILL 1 582.0204

.78432 .0000

.78433 582.0204

582.0204 ***102 HILL 1 582.0205

.57122 .0000

.57123 582 0206




3 582.0206582.0206 ***

103 HILL 1 582.0208.3828

2 .0000.3828

3 582.0208 582.0208 ***104 HILL 1 582.0205

.21862 .0000

.21863 582.0205

582.0205 ***105 HILL 1 582.0202

.07102 .0000

.07103 582.0203582.0203 ***




SECTION 10



10-1NAS 103, SOL 600, December 2003

SOL 600




TABLE OF CONTENTS

PageMSC.Nastran Implicit Nonlinear(SOL 600) Analysis 10-7

Overview Of NonLinear Analysis Using MSC.Nastran SOL 600 10-8

MSC.Nastran SOL 600 Overview 10-9

What Is MSC.Nastran SOL 600? 10-10Nonlinear Capabilities In MSC.Nastran 10-11

What Is MSC Marc? 10-12




What Is MSC.Marc? 10 12

What Is MSC.Marc? – Summary 10-15

What Is MSC.Nastran SOL 600 10-16

What Is MSC.Patran? 10-17

MSC.Patran Is A Pre- & Post-Processor 10-19

MSC.Nastran SOL 600 Is Open Architecture 10-20

What Is MSC.Nastran SOL 600? 10-21

Why Should I Use MSC.Nastran SOL 600? 10-22

How Does MSC.Nastran SOL 600 Work? 10-24

TABLE OF CONTENTS

PageFeatures And Capabilities 10-28

Summary of MSC.Nastran SOL 600 Nonlinear Analysis Capabilities 10-29

Non-Linear Capabilities In MSC.Nastrain SOL 600 10-30

Geometric Non-Linearities 10-31Geometric Non-Linearities - Finite Deformation 10-32

Geometric Nonlinearity – Follower Forces 10-33




y

Geometric Nonlinearity – Updated Lagrange 10-34

MSC.Nastran SOL 600 Materials 10-35

MSC.Nastran SOL 600 Plasticity Models 10-36

SOL 600 Hyperelasitic Models 10-37

Nonlinear Material Models 10-38

Boundary Condition Non-Linearity 10-39

Rigid & Deformable Bodies 10-41

MSC.Nastran SOL 600 Contact 10-42

TABLE OF CONTENTS

PageBoundary Condition Non-Linearity 10-44Example –Analysis Of A Rubber Boot 10-46MSC.Nastran SOL 600 Features 10-47Features – Matrix Solver Options 10-48Features – Distributed Memory Parallel 10-49Parallel Processing: Distributed Memory Parallel Method 10-50F t Di t ib t d M P ll l M th d 10 51




Features – Distributed Memory Parallel Method 10-51Features – Advanced Element Technology 10-53

Features – Conclusion 10-54Summary of MSC.Nastran SOL 600 Nonlinear Analysis Capabilities 10-55When To Use SOL 600 VS 106/129 10-56Future Capabilities 10-57Enhancements Planned 10-58Future Capabilities – User- Subroutines 10-59Future Capabilities –Thermal And Coupled Analysis 10-60

TABLE OF CONTENTS

PageFuture Capabilities – Adaptive Global Remeshing 10-61

More Info On MSC.Nastran SOL 600 Features 10-62

To Learn More – MSC.Nastran SOL 600 Documentation 10-63

MSC.Nastran SOL 600 Is Easy To Learn 10-64Learn Through On-Line Example Problems 10-65

Learn Through On-Line Example Problems 10-66




Learn Through On Line Example Problems 10 66

MSC Client Support 10-67

Nonlinear Summary 10-68Conclusion 10-69

MSC.NASTRAN IMPLICIT

NONLINEAR (SOL 600) ANALYSIS

Nonlinear

Analysis CapabilitiesFor 3D Contact and




Highly Nonlinear Problems

OVERVIEW OF NONLINEAR ANALYSIS USINGMSC.NASTRAN SOL 600

OVERVIEW WHAT IS MSC.NASTRAN SOL 600

WHO SHOULD USE MSC.NASTRAN SOL 600

MSC.NASTRAN, MSC.MARC AND MSC.PATRAN

SUMMARY OF MSC.NASTRAN SOL 600 NONLINEAR ANALYSISCAPABILITIES

MSC.NASTRAN SOL 600 FEATURES




SUMMARY AND CONCLUSIONS

MSC.NASTRAN SOL600 OVERVIEW

MSC.Nastran SOL 600 =

MSC.MARC ALGORITHMS +MSC.NASTRAN INTERFACE

What is MSC.Nastran SOL600?

Powerful General Purpose

Robust




User Friendly

Features and Capabilities: Contact

Geometric and Material Nonlinear

Added value: Adaptive Re-meshing and

DDM Conclusions

WHAT IS MSC.NASTRAN SOL 600?

Integrated Package: MSC.Nastran Interface

MSC.Marc Algorithms

MSC.Patran GUI

Access to most MSC.Marc Capabilities Access to all MSC.Marc Structural / Thermal



10-10NAS 103, SOL 600, December 2003

/ Coupled Analysis Capabilities

Easy to Use Intuitive

Powerful

Makes nonlinear finite element analysisEASY!!!

NONLINEAR CAPABILITIES IN MSC.NASTRAN

MSC.Nastran AdvancedNonlinear – SOL600:

Provides FEA capability forthe analysis of 3D contactand highly nonlinearproblems.



10-11NAS 103, SOL 600, December 2003

Combines the world’s most

advanced nonlinear finiteelement technology with theworld’s most widely usedfinite element code,

MSC.Nastran

WHAT IS MSC.MARC?



10-12NAS 103, SOL 600, December 2003

WHAT IS MSC.MARC? First commercially available, general-purpose, non-

linear, FE code - used in industry for over 30 years Parallel-processing on multiple platforms

Coupled-thermal structural analysis

User-subroutines to create new material models, applynew boundary conditions

Particularly powerful for highly nonlinear problems



10-13NAS 103, SOL 600, December 2003

y p g y p

Rigid-Deformable and Deformable-Deformable

Contact Analytic or Discrete Rigid Contact Surfaces with Velocity,

Force/Moment, or Displacement Control

Glued, Stick-Slip or Continuous Friction Models

Elastic, Plastic, Hyper-elastic, Creep and Visco-elasticMaterial Models Ample Library of Built in Material Models

Composite Damping and Failure MaterialsLarge Displacements Buckling

WHAT IS MSC.MARC?



10-14NAS 103, SOL 600, December 2003

Composite, Damping and Failure Materials

Large Element Library 0,1,2 and 3-D Elements may be Combined

User Control on Integration Methods

Advanced Solution and Modeling Features User Subroutines

Global and Local Adaptive Re-meshing Parallel Processing using Domain Decomposition - Manual or

Automatic Sub-division of the Model

X

Y

Z

Contact Resolution

Non-Linear Material

(Hyper-elastic rubberin this example)

WHAT IS MSC.MARC ? - SUMMARY

MSC.MARC is a general-purpose,non-linear FEA code.

It has been used extensively for thelast 3 decades in various types of

industries MSC.Marc - DDM is a completely

ll li d fi it l t



10-15NAS 103, SOL 600, December 2003

parallelized finite element process

Based on MSC.Marc’s nonlinearalgorithms, Nastran SOL 600 is avery powerful tool for solving largeand complex highly nonlinear

problems

Typical Application:Rubber Boot with Hyper-elastic material and self-

contact

WHAT IS MSC.NASTRAN SOL600

MSC.Nastran SOL600 is thenonlinear capabilities ofMSC.MARC delivered in anMSC.Nastran user interface

MSC.PATRAN provides: un-paralleled geometry integration

capabilities (who else can integrate



10-16NAS 103, SOL 600, December 2003

p ( gwith Catia as strongly as we do?)

robust automated meshingalgorithms (the new parasolidgeometry editing features trulyexpand your meshing options)

feature-rich, mature pre- and post-

processing capabilities

WHAT IS MSC.PATRAN?



10-17NAS 103, SOL 600, December 2003

WHAT IS MSC.PATRAN?

MSC.PATRAN is a finite element pre- and post-processor, which has been integrated withseveral nonlinear analysis solvers includingMSC.MARC, MSC.NASTRAN, and ABAQUS/Standard for implicit solutions; and

MSC.DYTRAN and LS-DYNA3D for explicitsolutions.

All model definition, analysis submittal and



10-18NAS 103, SOL 600, December 2003

results evaluation can be done through

MSC.PATRAN and driven via the graphical userinterface.

MSC.PATRAN on-line help facility includesdocumentation for all GUI forms and topics as

well as help on MSC.MARC.

MSC.Patran

Geometric Representationof Model

MSC.Patran

P r e - P r o c e

s s i n g

P o s t - P r o c e s s i n g

Results Visualization

MSC.PATRAN IS A PRE- & POST-PROCESSOR



10-19NAS 103, SOL 600, December 2003

MSC.Nastran

MSC.Marc

MSC.Marc

MSC.Fea

Abaqus

ANSYS

MSC.Dytran

MSC.Thermal

MSC.Structural Opt.

MSC.Fatigue

LS-DYNA3D

SUPPORTED SOLVERS

Patran

CAD:UG

ProE

CATIA

Euclid

Ideas

Other:Parasolid

Acis

Iges

Step

Express

MSC.NASTRAN SOL 600 IS OPENARCHITECTURE

Strengths of MSC.Nastran SOL600 …

Open Architecture – Interfaces to Any CAD or Analysis Program

MSC.Nastran SOL 600 hasinterfaces to all major CAD and Analysis Codes – includes inputdeck readers for all most analysis



10-20NAS 103, SOL 600, December 2003

Analysis:Nastran

Marc

Abaqus

Ansys

IdeasLS-Dyna

Sinda

Other:

Neutral

Iges

Step

Fatigue

New preference

Mapping feature

In Patran 2002

Provides Complete

Model Conversion

ycodes. Provides “customizable”

hooks for importing and exportingmodel information. Allows you to bring model data to

anywhere/ from anywhere …

WHAT IS MSC.NASTRAN SOL600 ?

Allows Nastran users to perform: nonlinear structural thermal * coupled thermo-structural analysis *

Includes contact, large deflection,

rotation, and strain analysis capabilitiesnever before available in Nastran Can use input decks from the many

thousands of existing MSC.Nastran



10-21NAS 103, SOL 600, December 2003

gmodels.

Provides solutions for simple to complexengineering problems including multi-body contact and advanced elastomeric(rubber) material models

* Note: Starred capabilities (on any page) may

not be in first release

WHY SHOULD I USE MSC.NASTRAN SOL600 ?

Allows Companies to Use aSingle Model Format (BDF)

Single Input Format Allows: Common Model for All Analysis

Needs Elimination of Model Re-creation

Effort

Reduced Time to Market

$ NASTRAN input file created by MSC.Nastran input file

$ Direct Text Input for File Management Section$ Nonlinear II Analysis

SOLMARC 600 EXEMARC PATH=3$ Direct Text Input for Executive ControlCENDSEALL = ALL

SUPER = ALLTITLE = MSC.Nastran job created12-Oct-01 at 09:38:33ECHO = NONE$ Direct Text Input for Global Case Control Data

BCONTACT = ALLSUBCASE 1$ Subcase name : Default

SUBTITLE=Default

NLPARM = 1BCONTACT = 1

SPC = 2LOAD = 2



10-22NAS 103, SOL 600, December 2003

Reduced Time-to-Market

Increased Efficiency Further value to FEA simulation

Allows Re-use of Thousandsof Existing Models That Cost

Millions to Create

DISPLACEMENT(SORT1,REAL)=ALL


STRESS(SORT1,REAL,VONMISES,BILIN)=ALLBEGIN BULKPARAM POST 0PARAM AUTOSPC NO

PARAM LGDISP 1PARAM,NOCOMPS,-1PARAM PRTMAXIM YESNLPARM 1 10 AUTO 5 25

$ Direct Text Input for Bulk Data$ Elements and Element Properties for region : shell

PSHELL 1 1 .25 1 1$ Pset: "shell" will be imported as: "pshell.1"CQUAD4 1 1 1 2 13 12

CQUAD4 2 1 2 3 14 13

WHY SHOULD I USE MSC.NASTRAN SOL600 ?(Cont.)

Brings the Following toMSC.Nastran: Contact

Large Deformation and Rotation

Large Strain Advance Nonlinear Materials:

Plasticity for Polymers and Metals

Hyper-elastic for Elastomers



10-23NAS 103, SOL 600, December 2003

yp

Gaskets for Engine Blocks

Brings Powerful, Mature, RobustNonlinear Technology to theMSC.Nastran Community

HOW DOES MSC.NASTRAN SOL600 WORK ?

MSC.Nastran Look and feel: Input a standard BDF Read byNastran IFP

Runs Marc “Under the Hood” Results Read back to Nastran

database via Toolkit Standard Output from Nastran

New Nastran text input: Executive Command:

$ NASTRAN input file created by MSC.Nastran input file

$ Direct Text Input for File Management Section$ Advanced Nonlinear AnalysisSOL 600, NLSTATIC$ Direct Text Input for Executive Control

CENDSEALL = ALL

SUPER = ALLTITLE = MSC.Nastran job created12-Feb-03 atECHO = NONE

$ Direct Text Input for Global Case Control Data

BCONTACT = ALLSUBCASE 1

$ Subcase name : DefaultSUBTITLE=Default

NLPARM = 1BCONTACT = 1SPC = 2LOAD = 2



10-24NAS 103, SOL 600, December 2003

Executive Command:SOL600, NLSTATIC

New Case Control Command for 3Dcontact

New Bulk Data Entries for 3Dcontact

New Bulk Data Entry for gasketmaterial

LOAD = 2DISPLACEMENT(SORT1,REAL)=ALL


STRESS(SORT1,REAL,VONMISES,BILIN)=ALLBEGIN BULKPARAM POST 0

PARAM AUTOSPC NOPARAM LGDISP 1PARAM,NOCOMPS,-1

PARAM PRTMAXIM YESNLPARM 1 10 AUTO 5 25

$ Direct Text Input for Bulk Data$ Elements and Element Properties for region : shell

PSHELL 1 1 .25 1 1$ Pset: "shell" will be imported as: "pshell.1"CQUAD4 1 1 1 2 13 12

CQUAD4 2 1 2 3 14 13


Nastran-Marc Translator: Start Nastran, read the Nastran

input file

Generate a Marc input file andrun Marc in the background

Marc run-time error messagespiped to .f06

Nastran deletes intermediatefiles

Nastran InputFile

Nastran IFP

Nas-MarcTranslator

Spawn MarcRun

Nastran .f06File



10-25NAS 103, SOL 600, December 2003

files

Needs a Marc and a Nastranexecutable (both will be includedon the Nastran CD)

Future version will eliminateseparate Marc input and run

Nastran ResultsDatabase

File

Marc .t16File

Nastran .xdbFile

Nastran .op2File

MSC.Nastran SOL600 Runs

MSC.Marc as a BackgroundProcess Version 2004: Two Executables

Marc Files: jobname.marc.xxx

Version 2005: Single Executable

Version 2004 gives users asmuch (next page) or as littlecontrol of MSC.Marc run asth d i




10-26NAS 103, SOL 600, December 2003

they desire: Input File May be Edited Job Submittal

License Usage

Output File Format

Job Messages can beconsolidated in .f06 file

Marc files can be automaticallyremoved


MSC.Nastran SOL600 isthe nonlinear capabilities ofMSC.MARC delivered in anMSC.Nastran user interface

MSC.PATRAN provides: un-paralleled geometry

integration capabilities (who elsecan integrate with Catia as

MSC.Nastran Input Deck

Use std Nast output req -

deck echo and

Write jobname.marc.dat

IFP Processes Input Deck

SuccessfulTranslation?

Submit MarcAnalysis?

Marc writes .out,.t16,.t19

Post-processingDMAP in place?

Submit Marc job -see note

.t16/19 results to Nast db

Nastran .f06,

.f04, .log files

error messages

Yes

Yes

Yes

generate std xdb,op2,f06

sts etc (these will be

No

No



10-27NAS 103, SOL 600, December 2003

can integrate with Catia as

strongly as we do?) robust automated meshing

algorithms (the new parasolidgeometry editing features trulyexpand your meshing options)

feature-rich, mature pre- andpost-processing capabilities

Stop

Is marccpy= 1or 2?

Append runtime error

Yes

.sts,etc (these will bedeleted later by Nastranif marccpy = 1 or 3) -.sts

messages to .f06 and .log

and .log may be used byMSC.Patran to monitorthe progress of the jobwhile it is running

Note - every attempt will be

made to have the Nastran InputFile Processor (IFP) catch allinput format errors. However,this may not be possible

in early releases. It maysometimes be necessary for theuser to debug the Marc analyisis.

See Chapter 16 on “TroubleShooting Analysis Runs” for

debugging suggestions if thisoccurs.

No

FEATURES AND CAPABILITIES

Supports the following StructuralCapabilities: Contact

Nonlinear Materials :

Elastic - Plastic

Hyper-elastic

Creep and Visco-elastic



10-28NAS 103, SOL 600, December 2003

Composite

Large Deformation – Large Strain

Patran Preference for SOL600

capabilities identical to Marc Preference

1

23

1

23

80.

UNIAXIAL TENSION

SUMMARY OF MSC.NASTRAN SOL600NONLINEAR ANALYSIS CAPABILITIES

The following Analysis Solutions are supported withMSC.NASTRAN SOL600 (more detail on these in Chapter 2)

Linear Static Analysis Nonlinear Static Analysis

Geometric Nonlinearity

Material Nonlinearity Contact Nonlinearity Example: Rubber (Hyperelastic Material)



10-29NAS 103, SOL 600, December 2003

1

23

1

23

1

23

1

23

0. 2. 4. 6.

NOMINAL STRAIN

0.

20.

40.

60.

BIAXIAL TENSION

PLANAR TENSION

Treloar’s Experimental Data

N O M I N A L S T R E S S ( K G F / C M * * 2 )

NON-LINEAR CAPABILITIES IN MSC.NASTRANSOL600

Materially Non-linearModels

Geometric Non-linearity's

Boundary Condition Non-linearity's (Contact)

All Non-linear BehaviorsCan be Combined …



10-30NAS 103, SOL 600, December 2003

Example Application: BallJoint

Axi-symmetric model

of ball-joint assembly

GEOMETRIC NON-LINEARITIES

Large Displacement andRotations

Large Strain Analyses

Buckling of Structures

Post-buckling behavior

Axially Loading Critical Mode



10-31NAS 103, SOL 600, December 2003

GEOMETRIC NONLINEARITY - FINITEDEFORMATION

Finite Deformation Large Deflection, Rotation andStrain: Large Deformation and Rotation of RBE’S Large (Finite) Strain With Choice of Strain

Definitions Finite Strain Plasticity

Robust and User-FriendlyAdaptive Load Incrementation



10-32NAS 103, SOL 600, December 2003

Adaptive Load Incrementation Total and Updated Lagrange

Procedures Choice of Solvers Including

Iterative and MSC.Nastran’s FastSparse

Distributed loads are taken into account bymeans of equivalent nodal loads; changes indirection and area can be taken into accountusing the MSC.Marc parameter optionFOLLOW FOR

Where on MSC.Patran?

GEOMETRIC NONLINEARITY - FOLLOWERFORCES



10-33NAS 103, SOL 600, December 2003

The pressure stays normal to the deformedshape thus changes direction, in turn alsoproducing a change in the reaction forcesand moments.

Updated Lagrange is especially useful for

beam and shell structures with large rotationsand for large strain plasticity problems;activated using the UPDATE parameter option

0

1

2

Total Lagrange

GEOMETRIC NONLINEARITY - UPDATEDLAGRANGE



10-34NAS 103, SOL 600, December 2003

3

0

1

23

Updated Lagrange

Isotropic 2-D and 3-D Orthotropic 2-D and 3-D Anisotropic Laminated and 3D

Composites, Gaskets forEngine Blocks Material properties can be

temperature dependent e.g.

MSC.NASTRAN SOL600 MATERIALS



10-35NAS 103, SOL 600, December 2003

Young’s Modulus Poisson Ratio Coefficient of Thermal Expansion Specific Heat Thermal Conductivity And more…

MSC.NASTRAN SOL600 PLASTICITY MODELS

Perfectly Plastic and Rigid Plastic

Elastic Plastic with Hardening orSoftening

Plastic - Hardening Laws: Isotropic

Kinematic

Combined and others

Plastic - Yielding: With a dependence of the yield stress



10-36NAS 103, SOL 600, December 2003

p y

on strain rate Von Mises and Drucker-Prager

Linear and Parabolic Mohr-Coulomb

Various Oak Ridge National Laboratorymodels and others

Hyper-elastic Including Graphical Feedback on

Experimental Data Fitting

Large strain for elastic materials(rubber) using; Neo-Hookean

Mooney-Rivlin

Ogden

Gent

SOL600 HYPERELASITIC MODELS



10-37NAS 103, SOL 600, December 2003

Arruda-Boyce Jamus-Green-Simpson models

Large strain, elastic analysis ofcompressible foams

Nonlinear Material Models Creep

Behavior for materials where, for aconstant stress state, strainincreases with time

Relaxation where stress decreases

with time at constant deformation Visco-elastic

Behavior for elastic materials thatrelax and dissipate energy undertransient loadings

NONLINEAR MATERIAL MODELS



10-38NAS 103, SOL 600, December 2003

Failure Hill, Hoffman, Tsai-Wu, MaximumStress or Strain

Damping Using Mass or Stiffness Matrix, or

Numerical Multipliers - can be used

together

BOUNDARY CONDITION NON-LINEARITY

Contact Developed in Marc in late ’80s

Automatic detection of contact

2D and 3D contact

Finds widespread use in areas likeManufacturing Simulations for sheetmetal forming, deep drawing, mountingseals and other process simulations,bio- medical simulations and more



10-39NAS 103, SOL 600, December 2003


Contact Automatic Re-meshing during contact

Friction models Stick-slip model

Coloumb model

Shear Friction for rolling



10-40NAS 103, SOL 600, December 2003

RIGID & DEFORMABLE BODIES

Analytic or Discrete Rigid Contact Surfaces with Velocity,Force/Moment, or Displacement Control

Glued, Stick-Slip or Continuous Friction Models



10-41NAS 103, SOL 600, December 2003

MSC.NASTRAN SOL600 CONTACT

Contact Capabilities Brings Advanced Contact

Capabilities to MSC.Nastran: Easy to Use Multi-Body Capability

2-D and Full 3-D Contact Supports Rigid-Deformable Contact

Position, Velocity or Load ControlledRigid Bodies

Rigid Geometry Defined Via NURBS DeformableContact stress

Contact area



10-42NAS 103, SOL 600, December 2003

g y

Discrete or Analytical DefinitionDeformableStructure (including friction)

Calculated

MSC.NASTRAN SOL600 CONTACT

Contact Capabilities Include Deformable-Deformable Contact With:

Initial Interference Fit

Stress – Free Initial Mesh Adjustment Single or Double – Sided Contact

Detection Force or Stress–Based Separation

Multiple Friction Models



10-43NAS 103, SOL 600, December 2003

Glued Contact Automatic or User – Defined

Contact Tolerance Distance

(CTD) Bias on CTD


Multi-Body Contact Very Easy to Set-Up

Automatic detection of contact surfaces

2D and 3D contact

Finds widespread use in areaslike: Manufacturing Simulationsfor sheet metal forming, deepdrawing, mounting seals and



10-44NAS 103, SOL 600, December 2003

other process simulations, bio-medical simulations and more

Try setting this up with contactpair contact …


Contact Capabilities: Rigid and Deformable

Automatic Re-meshing duringcontact

Reports Interface Results

Surface Interactions Contact Distance Tol

Bias on Distance Tol

Quadratic Elements



10-45NAS 103, SOL 600, December 2003

Friction models Glued Contact

Separation Force

EXAMPLE - ANALYSIS OF A RUBBER BOOT

All Non-Linearities Can Appear Together

Constant Velocity Rubber Boot Elastomeric material model (Mooney, non-linear elastic)

Large rotations and strains

Multi-body and Self-contact (default) Local buckling



10-46NAS 103, SOL 600, December 2003

MSC.NASTRAN SOL 600 FEATURES

MSC.NASTRAN SOL 600 Features: Structural, Thermal and Coupled

Analysis (thermal and coupled in2004)

Material, Geometric and Contact

Non-linearity Parallel Processing (DMP)

available (in 2003)

Experimental Data fitting for elastomers (in 2003)



10-47NAS 103, SOL 600, December 2003

User Defined Subroutines (2004)

Global Re-meshing (2004)

FEATURES - MATRIX SOLVER OPTIONS

Matrix Solution Methods (more on this later) Direct Solver

Direct Sparse

Iterative Solver (Conjugate Gradient)

Sparse Iterative

New BCS Solver (Multi-frontal solver)



10-48NAS 103, SOL 600, December 2003

FEATURES – DISTRIBUTED MEMORYPARALLEL

MSC.Nastran DMP Parallel Processing

Automatic Subdivision based on Metis

Manual Decomposition based on MSC.PatranGroups

Nastran SOL 600 allows DMP using a singleinput file



10-49NAS 103, SOL 600, December 2003

1 CPU 4 CPUs

PARALLEL PROCESSING: DISTRIBUTEDMEMORY PARALLEL METHOD

Mesh is broken up into several domains, each submittedto a different CPU



10-50NAS 103, SOL 600, December 2003

Linearly-Scalable Distributed

Memory Parrallel MSC.Nastran SOL600 DMP often

gives what is called “Super-Linear”scalability – meaning the you getbetter than 1/# cpu performance

increase. This occurs because the% of in-core solution time goes wayup …

In a recent comparisonMSC.Nastran SOL600’s DMP

bili d l

PARALLEL PROCESSING: DISTRIBUTEDMEMORY PARALLEL METHOD



10-51NAS 103, SOL 600, December 2003

capability was used to solve anengine block problem in 2.5 hoursthat took our competitor 7 days tosolve using a single cpu solution.

Who wouldn’t want to cut their

solution times down by an order ofmagnitude …

Domain 1 Domain 2

Domain 3

Domain 4

Entire Engine ModelGenerally linear scaling!

FEATURES - DISTRIBUTED MEMORYPARALLEL

Example 448,361 Elements

1.8 Million DOFs

10 Increment -

Transient Thermal

Analysis 78 minute on single

processor

12 minutes on 8 CPUs



10-52NAS 103, SOL 600, December 2003

Domain 5Domain 6

Domain 7Domain 8

FEATURES - ADVANCED ELEMENTTECHNOLOGY

Advanced ElementTechnology Linear and Quadratic

Herrmann Formulation forIncompressible Materials

Assumed Strain – CapturesStress Distribution in Bending

Global and Local Adaptive Re-meshing

P ll l P i i

Linear

Quadratic

Mid-body (Hermann formulation)



10-53NAS 103, SOL 600, December 2003

Parallel Processing usingDISTRIBUTED MEMORYPARALLEL - Manual or AutomaticSub-division of the Model

User Subroutines

Mid-body (Hermann formulation)

FEATURES - CONCLUSION

Take advantage of the MSC.Marcfeatures through an MSC.Nastraninterface:

1. Easy to Set-up Multi-body Contact

2. Global adaptive re-meshing

3. Experimental data fitting with graphical userfeedback

4. Linearly scalable DDM 300150 450 600 750 900Strain

S t r e s s

15

30

45

60

75



10-54NAS 103, SOL 600, December 2003

Domain 1 Domain 2

Domain 3

Domain 4

448,361 Elements1.8 Million DOFs10 Increment -TransientThermal Analysis78 min. on single

processor 12 minutes on 8CPUs

1.0A B

SUMMARY OF MSC.NASTRAN SOL 600NONLINEAR ANALYSIS CAPABILITIES

General Solution Features No fixed problem size limits, DISTRIBUTED MEMORY PARALLEL avail.

for parallel solution

Automated procedures for load step, convergence control, andequilibrium/stability control in nonlinear analysis

Reliable Newton-Raphson algorithm Arc length control for static collapse problems



10-55NAS 103, SOL 600, December 2003

Load, P

Displacement

Proportional loading with unstable response.

WHEN TO USE SOL600 VS 106/129

Most Common Reasons to Use MSC.Nastran SOL 600:Capability SOL 106/129 SOL 600

2D Def-Def Contact Slidelines Multi-Body

2D Rigid-Def Contact No Multi-Body

3D Def-Def Contact Slidelines Multi-Body

3D Rigid-Def Contact No Multi-Body

Beam Contact No Multi-Body

Elastic-Perfectly Plastic via Bi-Linear Yes

Bi-linear Elastic Plastic Yes via Multi-Linear

Multi-linear Elastic Plastic Yes Yes

Temp-Dependent Elastic-Plastic No Yes

Multi-linear Elastic Yes No

Mooney-Rivlin for 1D (beam) elements No Yes

Mooney-Rivlin for 2D elements Yes YesMooney-Rivlin for 3D elements Yes Yes

Need to Model 3D or Multi-BodyContact

Strain Level > 10-15%

RBE’s/MPC’s need large

rotation capability Elastic-Plastic or Hyper-Elastic

Material Properties areTemperature Dependent

Need to Model 3D SolidComposites



10-56NAS 103, SOL 600, December 2003

Mooney-Rivlin for 3D elements Yes Yes

Other hyperelastic (Ogden,Gent…) for all

element types No Yes

Temp-Dependent Hyperelastic No Yes

Composite Beams Yes Yes

Composite Shells Yes Yes

Continuum (2D Solid & 3D) Composites No Yes

Composites User Defined Subroutines

Need Global Adaptive Re-meshing

FUTURE CAPABILITIES



10-57NAS 103, SOL 600, December 2003

ENHANCEMENTS PLANNED

Product enhancements for MSC.Nastran SOL 600 are planned.

Major enhancements for v 2005. Examples of MSC.Nastran Version 2003 enhancements: 3D contact,

increased robustness, new rubber models and element technology,improvements in rigid-plastic flow and structural-acoustic analyses,general contact post-processing including area, force and stresscalculation between deformable bodies (surface-to-surface)

Deformable

Contact area



10-58NAS 103, SOL 600, December 2003

DeformableStructure

Contact stress(including friction)

Calculated

FUTURE CAPABILITIES - USER-SUBROUTINES

User-subroutines are apowerful way to input newcapabilities by the user forspecific needs

User-subroutines can beused to create: Material Models

Work-hardening varying as afunction of temperature

Damage models etc. Shape memory alloy material



10-59NAS 103, SOL 600, December 2003

Shape memory alloy materialmodels

Boundary Conditions Heat flux varying spatially or

with other BCs Friction varying as a function of

temperature

FUTURE CAPABILITIES - THERMAL ANDCOUPLED ANALYSIS

Steady state and Transient Analysis

Conductivity and radiation across interfaces can bemodeled

Temperature-dependent material properties can be used

Latent heat exchange during phase changes can bemodeled



10-60NAS 103, SOL 600, December 2003

FUTURE CAPABILITIES – ADAPTIVE GLOBALREMESHING

Global Adaptive Re-Meshing (2005 ?) MSC.Nastran SOL600 will have

a wide variety of methodsavailable for specifying re-

meshing criteria. Specifying thearea to be re-meshed is as easyas setting up a contact body (infact that is what you do).

Global adaptive re-meshingis the “silver bullet” for



10-61NAS 103, SOL 600, December 2003

is the silver bullet forsolving mesh distortionproblems …

MORE INFO ON MSC.NASTRAN SOL 600FEATURES

For More Information: See the MSC.Nastran SOL 600

Product Spec Sheet

Get the Power-point presentationon the Nastran SOL 600 Webinar or

the New MSC.Nastran Preference(down-load from:http://www.pm.macsch.com/nastran/presentations/naspref2003.ppt)

On-line documentation for

MSC.Patran, MSC.Nastran SOL600 d th MSC N t P f



10-62NAS 103, SOL 600, December 2003

,600 , and the MSC.Nastran Pref.Guide

TO LEARN MORE - MSC.NASTRAN SOL 600DOCUMENTATION

MSC.Nastran SOL 600User’s Guide

MSC.Marc OnlineDocumentation

MSC.Nastran PreferenceGuide

MSC.Nastran UserManuals: Quick Reference Guide



10-63NAS 103, SOL 600, December 2003

Quick Reference Guide

Reference Manual

MSC.Patran User’s Guide

MSC.NASTRAN SOL 600 IS EASY TO LEARN

MSC.Nastran SOL 600 STUDENT

VERSION: There is a node/element/entity

limited version available foreducational use.

Has all capabilities of the fullproduct except fornode/element/entity limits

For More Information or to Order:

Go to the MSC Engineering e-comSoftware Mart



10-64NAS 103, SOL 600, December 2003

Software Mart

Get the Power-point presentation onthe New MSC.Marc Preference (down-load from: http://www.engineering-e.com/software)

LEARN THROUGH ON-LINE EXAMPLEPROBLEMS

Truly “general purpose” FEA capability. MSC.NastranSOL 600 is fully modular. All capabilities can be mixedand used together.

MSC.PATRAN incorporates the most commonly used

features of the MSC.MARC analysis code to produce anintegrated interface as MSC.Nastran SOL 600: Code specific translator.

Analysis model set-up and submission of MARC jobs supported through

MSC.PATRAN Customer support provided for setting up analyses for all Nastran SOL



10-65NAS 103, SOL 600, December 2003

Customer support provided for setting up analyses for all Nastran SOL600 procedures.

Nastran SOL 600’S general modeling and PCL customizationcapabilities, along with direct text input, help support advanced

modeling capabilities.

LEARN THROUGH ON-LINE EXAMPLEPROBLEMS

Http://www.mscsoftware.com/support/online_ex/Patran

MSC.Nastran SOL 600 Example Problems



10-66NAS 103, SOL 600, December 2003

With corporate headquarters in Santa Ana, California, MSC.Software

maintains regional sales and support offices worldwide. MSC Technical Support Hotline 1-800-732-7284 (USA/Canada).

Staffed Monday through Friday 7:00 a.m. to 3:00 p.m. Pacific StandardTime (10:00 a.m. to 6:00 p.m. Eastern Standard Time)

E-mail support (USA/Canada) at

[email protected], MSC.Marc Mentat, MSC.Patran Marc Preferencesupport

[email protected] Patran other than Marc Preference support)

MSC CLIENT SUPPORT



10-67NAS 103, SOL 600, December 2003

MSC.Patran –other than Marc Preference- support)

Support (USA/Canada) Fax 714-979-2900 Internet support http://www.mscsoftware.com

NONLINEAR SUMMARY

Non-linearity results fromcontact, geometric and/ormaterial response. All non-linearities can appear

together in any analysis

General-purpose, mature,robust FE capability used inindustry for over 30 years

Parallel-processing for large

models using DMP Coupled-thermal structural



10-68NAS 103, SOL 600, December 2003

panalysis (V 2005?)

User-subroutines to create newmaterial models, apply new

boundary conditions (V 2005?)

CONCLUSION

MSC.Nastran SOL 600 is a powerful, easy to use tool forsimulating manufacturing processes and componentdesigns



10-69NAS 103, SOL 600, December 2003

CONCLUSION

Combine the World’s Most Advanced

Contact and Nonlinear Finite ElementTechnology With the World’s Leading Analysis Code and You GetMSC.Nastran Implicit Nonlinear –

SOL600 This Powerful Combination Will Lead To:

Common Analysis Model Format

Increased Efficiency

Reductions in: Need for Physical Prototypes Model Re-creation Effort



10-70NAS 103, SOL 600, December 2003

Model Re creation Effort

Product Development Time

Increased Value of FEA Simulation – an Already Indispensable Tool !!

SECTION 11

APPENDIX A

(NONLINEAR DATA)



A-1NAS 103, Appendix A, December 2003

( )




TABLE OF CONTENTS

Page

Summary Of Nonlinear Case Control Data A-5Summary Of Nonlinear Bulk Data A-7

Summary Of Parameters In Nonlinear Analysis A-11

Description Of Specific Nonlinear Bulk Data A-13

BCONP A-14BFRIC A-19

BLSEG A-21

BOUTPUT A-24

BWIDTH A-26CGAP A-29




CREEP A-34

MATHP A-44

MATS1 A-49NLPARM A-58

TABLE OF CONTENTS

Page

NLPCI A-70PBCOMP A-76

PGAP A-85

PLPLANE A-92

PLSOLID A-94TABLES1 A-95

TABLEST A-98

TSTEPNL A-100




SUMMARY OF NONLINEAR CASE CONTROLDATA

Requests output for 3-D slideline contactBOUTPUT

Output Request

Selects iteration methods for nonlinear transient analysisTSTEPNL

Selects iteration methods for nonlinear static analysisNLPARM

Selects methods for eigenvalue analysisMETHOD

Solution MethodSelection

Selects initial conditions for transient responseIC

Selects nonlinear loading (NOLINi) for transient responseNONLINEAR

Selects static load sets defined on the Bulk Data entry LSEQLOADSET

Selects dynamic loading conditionsDLOAD

Selects static load combination for superelementsCLOADSelects static loading conditionLOAD

Load Selection




Requests output for acceleration of physical points ACCELERATION

Requests output for velocities of physical pointsVELOCITY

Requests output for displacements of physical pointsDISPLACEMENT

q p

SUMMARY OF NONLINEAR CASE CONTROLDATA

Specifies the superelement identification numbers for which the staticSELR

Specifies the superelement identification numbers for which loadvectors are generated

SELG

Specifies the superelement identification numbers for which stiffnessmatrices are assembled and reduced

SEKR

Combines the functions of SEMG, SELG, SEKR, SEMR, and SELRSEALL

Specifies the superelement identification number and the loadsequence number

SUPER

Superelement Control

Requests the beginning of the plotter outputOUTPUT (Plot)

Requests output for NOLINi in transient responseNNLOAD

Requests output for constraint forces of SPC pointsSPCFORCES

Requests output for element stressesSTRESS

Requests output for element forcesELFORCE

(Cont.)Output Request




Specifies the superelement identification numbers for which the massand damping matrices are assembled and reduced

SEMR

Specifies the superelement identification numbers for which stiffness,mass, and damping matrices are generated

SEMG

Specifies the superelement identification numbers for which the staticload matrices are assembled and reduced

SELR

SUMMARY OF NONLINEAR BULK DATA

Defines properties for CBEAMPBEAM

Defines properties for composite CBEAMPBCOMP

Element Properties

Defines connection for a tubeCTUBE

Defines connection for triangular element with bending and membrane stiffnessCTRIA3

Defines connection for four-sided solid elementCTETRA

Defines connection for rod with axial and torsional stiffnessCROD

Defines connection for quadrilateral element with bending and membranestiffness

CQUAD4

Defines connection for five-sided solid elementCPENTA

Defines connection and properties for rodCONROD

Defines connection for six-sided solid elementCHEXA

Defines connection for beam elementCBEAM

Element Connectivity




Defines properties for CRODPROD

Defines properties for large strain CQUAD4 and CTRIA3PLPLANE

Defines properties for large strain CHEXA, CPENTA, and CTETRAPLSOLID

Defines properties for composite material laminatePCOMP


Constraints

Combines many TABLES1 entries for temperature- dependent material propertiesTABLEST

Defines a function for stress-dependent material propertiesTABLES1

Defines properties for plastic and nonlinear elastic materialsMATS1

Defines properties for hyperelastic materialMATHP

Defines anisotropic material properties for solid elementsMAT9

Defines orthotropic material properties for shell elementsMAT8

Defines anisotropic material properties for shell elementsMAT2

Defines creep material propertiesCREEP

Material Properties

Defines properties for CTUBEPTUBE

Defines properties for CHEXA, CPENTA, and CTETRAPSOLID

Defines properties for CTRIA3 and CQUAD4PSHELL

(Cont.)Element Properties




Defines a linear relationship for two or more degrees of freedomMPC

Defines single-point constraintsSPC1

Defines single-point constraints and enforced displacementsSPC

Constraints


Defines temperature field for line elementsTEMPRB

Defines temperature field for surface elementsTEMPPi

Defines temperature at grid pointsTEMP

Defines load due to centrifugal force fieldRFORCE

Defines pressure loads on surfaces of HEXA, PENTA, TETRA, TRIA3, andQUAD4 elements

PLOAD4

Defines pressure loads on shell elements, QUAD4, and TRIA3PLOAD2Defines pressure loads on QUAD4 and TRIA3\PLOAD

Defines nonlinear transient loadNONLINi

Defines moment at a grid pointMOMENTi

Defines static load sets for dynamic analysisLSEQ

Defines concentrated load at grid pointFORCEi

Defines a static load combination for superelement loadsCLOAD

Loads




Defines loads as a function of timeTLOADi

Specifies initial values for displacement and velocityTIC

p


Defines properties for CGAPPGAP

Defines connection for gap or frictional elementCGAP

Defines the width/thickness for line segments in 3-D/2-D slideline contact defined

in the corresponding BLSEG Bulk Data entry

BWIDTH

Defines slave nodes at which output is requestedBOUTPUT

Defines a curve consisting of a number of line segments via grid numbers thatmay come in contact with other bodies

BLSEG

Defines frictional properties between two bodies in contactBFRIC

Defines the parameters for contact between two bodiesBCONP

Contact

Defines eigenvalue extraction method for buckling analysisEIGB

Specifies integration and iteration methods for nonlinear transient analysisTSTEPNL

Defines arc-length methods for nonlinear static analysisNLPCI

Defines iteration methods for nonlinear static analysisNLPARM

Solution Methods




Defines properties for CGAPPGAP

SUMMARY OF PARAMETERS IN NONLINEARANALYSIS

Specifies number of integration points through thickness forQUAD4 and TRIA3

5BBNLAYERSSpecifies numerical damping in ADAPT method0.025ENDAMP

Maximum number of iterations for internal loop5EMAXLP

Specifies LOOPID in the database for restarts0EELOOPID

Selects large displacement effects-1EELGDISP

Specifies large rotation approach1BBLANGLE Assigns stiffness to normal rotation of QUAD4, TRIA30.0EEK6ROT

Selects nonlinear buckling analysis for restarts-1EBUCKLE

Specifies automatic SPC for residual structureNOBE AUTOPSPCR

Scale factor to adjust automatic calculated penalty valuesfor slideline elements

1.0E ADPCON

DescriptionDefault

129106

Parameter Name SolutionSequence




Specifies LOOPID for nonlinear normal mode analysis0ENMLOOP

Sets Defaults for the CONV, EPSU, EPSP, and EPSWfields of NLPARM Bulk Data Entry

2ENLTOL

SUMMARY OF PARAMETERS IN NONLINEARANALYSIS

Selects frequency for conversion of element damping0.0BW4

Selects the frequency for the conversion of structuraldamping

0.0BW3

Tests for negative terms on factor diagonal-2 (N), 1(A)ETESTNEG

Specifies subcase ID for restarts0ESUBID

Specifies LOOPID from SOL 106 database for restarts0BSLOOPID

DescriptionDefault

129106

Parameter Name Solution

Sequence




Note: B=usable in the Bulk Data Section onlyE=usable in either the Bulk Data or Case Control Section

DESCRIPTION OF SPECIFIC NONLINEAR

BULK DATA




BCONP

Description: Defines the parameters for a contact region and its properties

Format:

Example:

Field Contents

ID Contact region identification number (Integer > 0)SLAVE Slave region identification number (Integer > 0).


1331151095BCOMP

CIDPTYPEFRICIDSFACMASTERSLAVEIDBCONP

10987654321





SFAC Stiffness scaling factor. This factor is used to scale thepenalty values automatically calculated by the program. (Real

> 0 or blank)

BCONP

Field Contents

FRICID Contact friction identification number (Integer > 0 or blank)PTYPE Penetration type (Integer = 1 or 2; Default =1).

1: unsymmetrical (slave penetration only) (default)

2: symmetrical

CID Coordinate system ID to define the slide line plane vector andthe slide line plane of contact. (Integer > 0 or blank; Default =0 which means the basic coordinate system)

Remarks1.

ID field must be unique with respect to all other BCONP identificationnumbers.

2. The referenced SLAVE is the identification number in the BLSEG Bulk




Data entry. This is the slave line. The width of each slave segmentmust also be defined to get proper contact stresses. See BWIDTH Bulk

Data entry for the details of specifying widths.

BCONP

Remarks (Cont.)3. The referenced MASTER is the identification number in the BLSEG Bulk

Data entry. This is the master line. For symmetrical penetration, thewidth of each master segment must also be defined. See BWlDTH BulkData entry for the details of specifying widths.

4. SFAC may be used to scale the penalty values automatically calculated

by the program. The program calculates the penalty value as a functionof the diagonal stiffness matrix coefficients that are in the contactregion. In addition to SFAC, penalty values calculated by the programmay be further scaled by the ADPCON parameter (see description of

ADPCON parameter for more details). The penalty value is then equalto k * SFAC * |ADPCON|, where k is a function of the local stiffness. It

should be noted that the value in SFAC applies to only one contactregion, whereas the ADPCON parameter applies to all the contactregions in the model.




regions in the model.5. The referenced FRlClD is the identification number of the BFRlC Bulk

Data entry. The BFRlC defines the frictional properties for the contact

region.

BCONP

Remarks (Cont.)

6. In an unsymmetrical contact algorithm only slave nodes are checked forpenetration into master segments. This may result in master nodes penetratingthe slave line. However, the error involve depends only on the meshdiscretization. In symmetric penetration both slave and master nodes arechecked for penetration. Thus, no distinction is made between slave and master.Symmetric penetration may be up to thirty percent more expensive than the

unsymmetric penetration.7. In Figure 1, the unit vector in the Z-axis of the coordinate system defines theslideline plane vector. Slideline plane vector is normal to the slideline plane.Relative motions outside the slideline plane are ignored, therefore must be smallcompared to a typical master segment. For a master segment the direction frommaster node 1 to master node 2 gives the tangential direction (t). The normal

direction for a master segment is obtained by cross product of the slideline planevector with the unit tangent vector (i.e., n = z x t). The definition of the coordinatesystem should be such that the normal direction must point toward the slaveregion. For symmetric penetration the normals of master segments and slave




region. For symmetric penetration the normals of master segments and slavesegments must face each other. This is generally accomplished by traversingfrom master line to slave line in a counter-clockwise or clockwise fashiondepending on whether the slideline plane vector forms right hand or left handcoordinate system with the slideline plane.

BCONP

• X-Y plane is the slide line plane. Unit normal in the Z-directionis the slide line plane vector.

• Arrows show positive direction for ordering nodes. Counter-

k-th Slave Segment

l -th Master Segment

k k - 1Slave Line

Master Line

k + 1

Slideline Plane Vector Direction

Y

X

Z

l − 1l + 1




p gclockwise from master line to slave line.

• Slave and master segment normals must face each other.

BFRIC

Description: Defines frictional properties between two bodies in contact.

Format:

Example:

Field Contents

FID Friction identification number (Integer > 0)

MU1FSTIFFIDBFRIC

10987654321

0.333BFRIC




FSTIF Frictional stiffness in stick (Real > 0.0). Default =automatically selected by the program.

MU1 Coefficient of static friction (Real > 0.0).

BFRIC

Remarks:1. This identification number must be unique with respect to all otherfriction identification numbers. This is used in the FRlClD field of

BCONP Bulk Data entry.2. The value of frictional stiffness requires care. A method of choosing its

value is to divide the expected frictional strength (MU1 × the expected

normal force) by a reasonable value of the relative displacement whichmay be allowed before slip occurs. The relative value of displacementbefore slip occurs must be small compared to expected relativedisplacements during slip. A large stiffness value may cause poorconvergence, while too small value may cause poor accuracy.Frictional stiffness specified by the user is selected as the initial value.If convergence difficulties are encountered during the analysis, thefrictional stiffness may be reduced automatically to improveconvergence.




g3. The stiffness matrix for frictional slip is unsymmetric. However, the

program does not use the true unsymmetric matrix. Instead the program

uses only the symmetric terms. This is to avoid using the unsymmetricsolver to reduce CPU time.

BLSEG

Description: Defines a curve which consists of a number of line segments via grid

numbers that may come in contact with other body. A line segment isdefined between every two consecutive grid points. Thus, number of linesegments defined is equal to the number of grid points specified minus1. A corresponding BWlDTH Bulk data entry may be required to define

the width/thickness of each line segment. If the corresponding BWlDTHis not present, the width/thickness for each line segment is assumedunity

Format:

G10BYG9THRUG8

G7G6G5G4G3G2G1IDBLSEG

10987654321




G12G11

G10BYG9THRUG8

BLSEG

Examples:

Field ContentsID Line segments identification number (Integer > 0)Gi Grid numbers on a curve in a continuous topological order so

that the normal to the segment points towards other curve.

Remarks1. ID must be unique with respect to all other BLSEG entries. Each line

segment has a width in 3-D sideline and a thickness in a 2-D slidelinecontact to calculate contact stresses. The width/thickness of each line

t i d fi d i BWIDTH B lk D t t Th ID i BLSEG

44THRU35

33323027

14BY21THRU515BLSEG




segment is defined via BWIDTH Bulk Data entry. The ID in BLSEGmust be same as the ID specified in the BWlDTH. That is, there must

be one to one correspondence between BLSEG and BWlDTH.BWlDTH Bulk Data entry may be omitted only if the width/thickness ofeach segment is unity.

BLSEG

Remarks (Cont.)2. Gi may be automatically generated using the THRU and BY keywords.

For first line, THRU and BY can only be specified in the fourth and thesixth fields, respectively. For continuation lines, THRU and BY can onlybe specified in the third and the fifth fields, respectively. For automaticgeneration of grid numbers the default value for increment is 1 if grid

numbers are increasing or -1 if grid numbers are decreasing (i.e., theuser need not specify BY and the increment value).

The normal to the segment is determined by the cross product of theslideline plane vector (i.e., the Z direction of the coordinate systemdefined in the ‘ClD’ field of BCONP Bulk Data entry) and the tangentialdirection of the segment. The tangential direction is the direction fromnode 1 to node 2 of the line segment




node 1 to node 2 of the line segment.

A curve may be closed or open. A closed curve is specified by havingthe last grid number same as the first grid number.

BOUTPUT

Description Defines the slave nodes at which the output is requested.

Format:

Example:

Field Contents

ID B d id ifi i b f hi h i d i d

B10BYG9THRUG8

G8G7G6G5G4G3G2G1

ALLIDBPOUTPUT

10987654321

ALL15BOUTPUT




ID Boundary identification number for which output is desired(Integer > 0.0).

Gi Slave node numbers for which output is desired.

BOUTPUT

Remark:1. The grid numbers may be automatically generated using the THRU and

BY keywords. For first line, THRU and BY can only be specified in thefourth and the sixth fields, respectively. For continuation lines, THRUand BY can only be specified in the third and the fifth fields,respectively. If output is desired for all the slave nodes, specify the

word ALL in the third field of the first line or just include the contactregion ID in the Case Control command BOUTPUT.




BWIDTH

Description Defines width/thickness for line segments in 3-D/2-D slideline contact

defined in the corresponding BLSEG BULK Data entry. This entry maybe omitted if the width/thickness of each segment defined in the BLSEGentry is unity. Number of thicknesses to be specified is equal to thenumber of segments defined in the corresponding BLSEG entry. If

there is no corresponding BLSEG entry, the width/thickness specified inthe entry are not used by the program.

Format:

W12W11

W10BYW9THRUW8

W7W6W5W4W3W2W1IDBWIDTH

10987654321




W12W11

BWIDTH

Examples:

Field ContentsID Width/thickness set identification number (Real > 0.0).

Wi Width/Thickness values for the corresponding line segmentsdefined in the BLSEG entry. (Real > 0.0).

Remarks:1 The ID field must be unique with respect to all other BWlDTH entries It

44THRU35

2222

1BY5THRU215BWIDTH




1. The ID field must be unique with respect to all other BWlDTH entries. Itmust be the same as the ID field in the corresponding BLSEG entry.

BWIDTH

Remarks: (Cont.)2. The widths may be automatically generated using the THRU and BY

keywords. For first line, THRU and BY can only be specified in thefourth and the sixth fields, respectively. For continuation lines, THRUand BY can only be specified in the third and the fifth fields,respectively. For automatic generation of the width values the default

value for increment is 1.0 if the width is increasing or -1.0 if the width isdecreasing. That is the user need not specify BY and the incrementvalue. If the number of width specified are less than the number ofsegments defined in the corresponding BLSEG entry, the width for theremaining segments is assumed to be equal to the last width specified.

3. If there is only one grid point in the corresponding BLSEG entry, there isno contributory area associated with the grid point. To compute correctcontact stresses an area may be associated with the single grid point by




specifying the area in field W1.

CGAP

CGAP Bulk Data Entry Defines a gap or frictional element for nonlinear analysis.

Format:

Example:

Alternate Format and Example:

CIDX3X2X1GBGAPIDEIDCGAP

10987654321

-6.10.35.2112110217CGAP

CIDGOGAGAPIDEIDCGAP

13112110217CGAP




CGAP

Field Contents

EID Element identification number. (Integer > 0).PID Property identification number of a PGAP entry. (Integer > 0;

Default = EID).

GA, GB Connected grid points at ends A and B. (Integers > 0; GA ≠GB).

X1, X2, X3 Components of the orientation vector , from GA, in thedisplacement coordinate system at GA. (Real).

G0 Alternate method to supply the orientation vector using gridpoint G0. Direction of is from GA to G0. (Integer).

CID Element coordinate system identification number. SeeRemark 3. (Integer ≥ 0 or blank).

v

v




CGAP

Remarks:

1. The CGAP element is intended for the nonlinear solution sequences66, 99, 106, 129, 153 and 159. However, it will produce a linearstiffness matrix for the other solutions, but remains linear with the initialstiffness. The stiffness used depends on the value for the initial gapopening (U0 field in the PGAP entry).

2.

If the grid points GA and GB are coincident (distance from A to B < 10-4

)and the CID field is blank, the job will be terminated with a fatal errormessage.

3. The gap element coordinate system is defined by one of two followingmethods:

a) If the coordinate system (CID field) is specified, the element coordinatesystem is established using that coordinate system, in which the element x-axis is in the T1 direction and the y-axis in the T2 direction. The orientationvector will be ignored in this case.

b) If the CID field is blank and the grid points GA and GB are not coincident




) g p(distance from A to B ≥ 10-4), then the line AB is the element x-axis and theorientation vector lies in the x-y plane (like the CBEAM element).

CGAP

Remarks:4. The element coordinate system does not rotate as a result of

deflections.

5. Initial gap openings are defined on the PGAP entry and not by theseparation distance between GA and GB.

6. Forces, which are requested with the STRESS Case Control command,are output in the element coordinate system. Fx is positive for compression.




CGAP

GA

GB

KA − KBKB

Note: KA and KB in thisfigure are from thePGAP entry.

v

zelem

yelem

xelem

Figure 2. CGAP Element Coordinate System.




CREEP

Creep Bulk Data Entry Defines creep characteristics based on experimental data or known

empirical creep law. This entry will be activated if a MAT1, MAT2, orMAT9 entry with the same MID is used and the NLPARM entry isprepared for creep analysis.

Format:

Example:

Gf edcbaTYPE

Gf edcbaTYPE

THRESHTIDCSTIDCPTIDKPFORMEXPTOMIDCREEP

10987654321




+CR1.-5CRLAW10-911008CREEP

CREEP

Field ContentsMID Material identification number of a MAT1, MAT2, or MAT9

entry. (Integer > 0).T0 Reference temperature at which creep characteristics are

defined. See Remark 2. (Real; Default = 0.0).EXP Temperature-dependent term, e(-∆H/R(R*T0)), in the creep rate

expression. See Remark 2. (0.0 < Real ≤ 1.0;Default = 1.0E-9).

FORM Form of the input data defining creep characteristics.(Character: “CRLAW” for empirical creep law, or “TABLE” fortabular input data of creep model parameters).

TIDKP Identification number of a TABLES1 entry which defines theTIDCP creep model parameters Kp(σ), Cp(σ), and Cs(σ),TIDCS respectively. See Remarks 3 through 5. (Integer > 0).THRESH Threshold limit for creep process. Threshold stress under

which creep does not occur is computed as THRESH




which creep does not occur is computed as THRESHmultiplied by Young’s modulus. (0.0 < Real < 1.0E-3;

Default = 1.0E-5).

CREEP

Field Contents (Cont.)

TYPE Identification number of the empirical creep law type. SeeRemark 1. (Integer: 111, 112, 121, 122, 211, 212, 221, 222,or 300).

a through g Coefficients of the empirical creep law specified in TYPE.Continuation should not be specified if FORM = “TABLE”.

See Remark 1. (Real). Remarks:

1. Two classes of empirical creep law are available. Creep Law Class 1

The first creep law class is expressed as

Parameters A(σ) R(σ) and K(σ) are specified in the following form as

εc

σ t,( ) A σ( ) 1 eR σ( )t –

– [ ] K σ( ) t+= (1)




Parameters A(σ), R(σ), and K(σ) are specified in the following form, asrecommended by Oak Ridge National Laboratory

CREEP

Remarks: (Cont.)

TYPE = ijk where i, j, and k are digits equal to 1 or 2 according to the desiredfunction in the table above. For example, TYPE = 122 defines A(σ) = aσb,R(σ) = cσd, and K(σ) = eef σ

Creep Law Class 2

The second creep law class (TYPE = 300) is expressed as:

where the values of b and d must be defined as follows:

k =2eef σk = 1e*[sinh (f σ)]gK(s)

j = 2cσd j = 1cedσR(s)

i = 2aebσi = 1aσb A(s)

DigitFunction 2DigitFunction 1Parameter

εc

σ t,( ) aσ b

td

= (2)




1.0 < b < 8.0

and0.2 < d < 2.0

CREEP

Remarks: (Cont.) The coefficient g should be blank if TYPE = 112, 122, 222, or 212 and c, e, f,and g should be blank if TYPE = 300. The coefficients a through g are

dependent on the structural units; caution must be exercised to make theseunits consistent with the rest of the input data.

2. Creep law coefficients a through g are usually determined by least

squares fit of experimental data, obtained under a constant temperature.This reference temperature at which creep behavior is characterizedmust be specified in the T0 field if the temperature of the structure isdifferent from this reference temperature. The conversion of thetemperature input (°F or °C) to °K (degrees Kelvin) must be specified in

the PARAM,TABS entry as follows:

PARAM,TABS,273.16 (If Celsius is used)




PARAM,TABS,459.69 (If Fahrenheit is used)

CREEP

Remarks: (Cont.)

When the correction for the temperature effect is required, thetemperature distribution must be defined in the Bulk Data entries(TEMP, TEMPP1 and/or TEMPRB), which are selected by the CaseControl command TEMP(LOAD) = SID within the subcase.

From the thermodynamic consideration, the creep rate is expressed as:

where ∆H = energy activation

R = gas constant (= 1.98 cal/mole ° K)T = absolute temperature (°K)

= strain/sec per activation

ε·c

ε·A e ∆H RT –

( )= (3)

aε&




If the creep characteristics are defined at temperature T0, the creep rateat temperature T is corrected by a factor.

CREEP

Remarks: (Cont.)

Where = Corrected creep rate

= creep rate at T0

Exp(T0/T-1)

= correction factor 3. If the creep model parameters Kp, Cp, Cs and are to be specified with

FORM = “TABLE” then TABLES1 entries (whose IDs appear in TIDXXfields) must be provided in the Bulk Data Section. In this case, thecontinuation should not be specified.

4. Creep model parameters Kp, Cp, and Cs represent parameters of theuniaxial rheological model as shown in the following figure.

Tabular values (Xi, Yi) in the TABLES1 entry correspond to (σi, Kpi), (σi,Cpi), and (σi, Csi) for the input of Kp, Cp, and Cs, respectively. For linear

ε·c

ε·o

c----- EXP T 0 T ⁄ 1 – ( )= (4)

c

oε&

cε&




Cpi), and (σi, Csi) for the input of Kp, Cp, and Cs, respectively. For linearviscoelastic materials, parameters K

p, C

p, and C

s, are constant and two

values of si must be specified for the same value of Kpi, Cpi, and Csi

CREEP

Remarks: (Cont.)

Creep model parameters, as shown in the figures below, must havepositive values. If the table look-up results in a negative value, thevalue will be reset to zero and a warning message (TABLE LOOK-UPRESULTS IN NEGATIVE VALUE OF CREEP MODEL PARAMETER IN

SecondaryCreep

PrimaryCreepElastic

Ke Cs(σ)Cp(σ)

Kp(σ)

σ(t)

Figure 1. CREEP Parameter Idealization




ELEMENT ID = ****) will be issued.

CREEP

Remarks: (Cont.)

0 5 10 15 20 25 30

5000

4000

3000

2000

1000

σ(ksi)

Kp(Kips/in2)

Figure 2. Kp Versus σ Example for CREEP

250 x 106

200 x 106

150 x 106

100 x 106

50 x 106

05 10 15 20 25 30

σ(ksi)

C pK ps- oursin

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

Figure 3. Kp Versus σ Example for CREEP

40,000 x 106

30,000 x 106

20,000 x 106

10,000 x 106

05 10 15 20 25 30

50,000 x 106

Cs( )Kips-hoursin

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




σ(ksi)

Figure 3. Cs Versus σ Example for CREEP

CREEP

Remarks: (Cont.)

5. Creep analysis requires an initial static solution at t = 0, which can beobtained by specifying a subcase which requests an NLPARM entrywith DT = 0.0.




MATHP

Specifies material properties for use in nonlinear analysis

of rubber-like materials (elastomers). Format:

TABDTAB4TAB3TAB2TAB1D5 A05 A14 A23 A32 A41 A50

D4 A04 A13 A22 A31 A40

D3 A03 A12 A21 A30

D2 A02 A11 A20

NDNA

GETREF AVRHOD1 A01 A10MIDMATHP

10987654321




MATHP

Field Contents

MID Identification number of a MATHP entry. (Integer > 0; No default)

Aij Material constants related to distortional deformation. (Real; Default= 0.0)

Di Material constants related to volumetric deformation. (Real ≥;Default for D1 is 103 *(A10 + A01); Default for D2 through D5is 0.0)

RHO Mass density in original configuration. (Real; Default = 0.0) AV Coefficient of volumetric thermal expansion. (Real; Default = 0.0)

TREF Reference temperature. See MAT1 entry. (Real; Default = 0.0)

GE Structural damping element coefficient. (Real; Default = 0.0)

NA Order of the distortional strain energy polynomial function. (0 <Integer < 5; Default = 1)

ND Order of the volumetric strain energy polynomial function. (0 <Integer < 5; Default = 1)

TAB1 Table identification number of a TABLES1 entry that contains simple




tension/compression data to be used in the estimation of the materialconstants Aij. See Section 15.3.3 of the MSC.NASTRAN ReferenceManual. (Integer > 0 or blank)

MATHP


TAB2 Table identification number of a TABLES1 entry that contains

equibiaxial tension data to be used in the estimation of the materialconstants Aij. See Section 15.3.3 of the MSC.NASTRAN ReferenceManual. (Integer > 0 or blank).

TAB3 Table identification number of a TABLES1 entry that contains simpleshear data to be used in the estimation of the material constants Aij.

See Section 15.3.3 of the MSC.NASTRAN Reference Manual.(Integer > 0 or blank)

TAB4 Table identification number of a TABLES1 entry that contains pureshear data to be used in the estimation of the material constants Aij.See Section 15.3.3 of the MSC.NASTRAN Reference Manual.(Integer > 0 or blank)

TABD Table identification number of a TABLES1 entry that contains purevolumetric compression data to be used in the estimation of thematerial constant Di. See Section 15.3.3 of the MSC.NASTRANReference Manual. (Integer > 0 or blank)




MATHP

Remarks:

The generalized Mooney-Rivlin strain energy function may beexpressed as follows:

where I1 and I2 are the first and second distortional strain invariants,

respectively; J = det F is the determinant of the deformation gradient;and 2D1 = k and 2(A10 + A01) = G at small strains, in which K is thebulk modulus and G is the shear modulus. The model reduces to aMooney-Rivlin material if NA = 1 and to a Neo-Hookean material if NA =1 and A01 = 0.0. (See Remark 2). For Neo-Hookean or Mooney-Rivlinmaterials no continuation command is needed. T is the currenttemperature and T0 is the initial temperature.

Conventional Mooney-Rivlin and Neo-Hookean materials are fullyincompressible. Full incompressibility is not presently available but maybe simulated with a large enough value of D1. A value of D1 higherthan 103 * (A10 + A01) is however not recommended

U J I1 I2, ,( ) Ai j I1 3 – ( )i

I2 3 – ( ) j

Di J 1 AV T T0 – ( ) – – ( )2i

A00 0=,

i 1=

D

∑+

i j, 0≥

A

∑=




than 103 (A10 + A01) is, however, not recommended.

MATHP

Remarks: (Cont.)

3. Di are obtained from least squares fitting of experimental data. One ormore of 4 experiments (TAB1 to TAB4) may be used to obtain Aij. Dimay be obtained from pure volumetric compression data (TABD). IfTABD is blank, the program expects Di to be manually input. If all TAB1through TAB4 are blank, the program expects Aij to be manually input.

Parameter estimation, specified through any of the TABLES1 entries,supersedes the manual input of the parameters.

4. IF ND = 1 and a nonzero value of D1 is provided or is obtained fromexperimental data in TABD, then the parameter estimation of thematerial constants Aij takes compressibility into account in the cases of

simple tension/compression, equibiaxial tension, and general biaxialdeformation. Otherwise, full incompressibility is assumed in estimatingthe material constants.




MATS1

Material Stress Dependence

Specifies stress-dependent material properties for use in applicationsinvolving nonlinear materials. This entry is used if a MAT1, MAT2, orMAT9 entry is specified with the same MID in a nonlinear solutionsequence (SOLs 66, 99, 106, and 129).

Format:

Example:

Field Contents

MID Identification number of a MAT1, MAT2, or MAT9 entry.(Integer > 0)

LIMIT2LIMIT1HRYFHTYPETIDMIDMATS1

10987654321

2. +4110.0PLASTIC2817MATS1




(Integer > 0).

MATS1

Field Contents (Cont.)TID Identification number of a TABLES1 or TABLEST entry. If H

is given, then this field must be blank. See Remark 3.(Integer ≥ 0 or blank).

TYPE Type of material nonlinearity. See Remarks. (Character:“NLELAST” for nonlinear elastic or “PLASTIC” forelastoplastic).

H Work hardening slope (slope of stress vs. plastic strain) inunits of stress. For elastic-perfectly plastic cases, H = 0.0.For more than a single slope in the plastic range, the stress-strain data must be supplied on a TABLES1 entry referencedby TID, and this field must be blank. See Remark 2. (Real).

YF Yield function criterion, selected by one of the followingvalues (Integer):1 = von Mises (Default)2 = Tresca3 = Mohr-Coulomb




4 = Drucker-Prager

MATS1


HR Hardening Rule, selected by one of the following values(Integer):

1 = Isotropic (Default)2 = Kinematic3 = Combined isotropic and kinematic hardening

LIMIT1 Initial yield point. See . (Real).LIMIT2 Internal friction angle for the Mohr-Coulomb and Drucker-

Prager yield criteria. See Table 1. (0.0 ≤ Real < 45.0°).

Angle of InternalFriction φ (in Degrees)

2*Cohesion, 2c (in unitsof stress)

Mohr-Coulomb (3) orDrucker-Prager (4)

Not UsedInitial Yield Stress InTension, Y1

Von Mises (1) or Tresca (2)

LIMIT2LIMIT1 Yield Function (YF)

Table 1. Yield Functions Versus LIMIT1 and LIMIT2.




MATS1

Remarks:

1. If TYPE = “NLELAST”, then MID may refer to a MAT1 entry only. Also,the stress-strain data given in the TABLES1 entry will be used todetermine the stress for a given value of strain. The values H, YF, HR,LIMIT1, and LIMIT2 will not be used in this case.

Thermoelastic analysis with temperature-dependent material properties

is available for linear and nonlinear elastic isotropic materials (TYPE =“NLELAST”) and linear elastic anisotropic materials. Four options ofconstitutive relations exist. The relations appear in Table 2 along withthe required Bulk Data entries.




MATS1

Remarks: (Cont.)

In Table 2 σ and ε are the stress and strain vectors, [Ge] the elasticitymatrix, the effective elasticity modulus, and E the reference elasticitymodulus.

MAT1, MATT1, MATS1, TABLEST, and TABLES1

MAT1, MATS1, TABLEST, and TABLES1

MAT1, MATT1, MATS1, and TABLES1

MATi and MATTi where i = 1, 2, or 9

Required Bulk Data EntriesRelation

ε(T)][Gσ e=

ε(T)][G

ε),(

σ e E

E

σ

=

ε][Gε),,(

σ e E

T E

σ =

ε(t)][Gε),,(

σ e

T E

σ =

Table 2. Constituative Relations and Required Material Property Entries




MATS1

Remarks: (Cont.)

2. If TYPE = “PLASTIC”, either the table identification TID or the workhardening slope H may be specified, but not both. If the TID is omitted,the work hardening slope H must be specified unless the material isperfectly plastic. The plasticity modulus (H) is related to the tangentialmodulus (ET) by.

where E is the elastic modulus and ET = dY/dε is the slope of the

uniaxial stress-strain curve in the plastic region. See Figure 1.

Hr ET

1ET

E------- –

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




MATS1

Remarks: (Cont.)

e

E

0

ET

Y1

Y or s( )

Figure 1. Stress-Strain Curve Definition When H is Specified in Field 5.




MATS1

Remarks: (Cont.)

3. If TID is given, TABLES1 entries (Xi, Yi) of stress-strain data (εk, Yk)must conform to the following rules (see Figure 2):a. If TYPE = “PLASTIC”, the curve must be defined in the first quadrant. The

first point must be at the origin (X1 = 0, Y2 = 0) and the second point (X2,Y2) must be at the initial yield point (Y1 or 2c) specified on the MATS1 entry.The slope of the line joining the origin to the yield stress must be equal to thevalue of E. Also, TID may not reference a TABLEST entry.

b. If TYPE = “NLELAST”, the full stress-strain curve (-• < x < •) may be definedin the first and the third quadrant to accommodate different uniaxialcompression data. If the curve is defined only in the first quadrant, then thecurve must start at the origin (X1 = 0.0, Y1 = 0.0) and the compression

properties will be assumed identical to tension properties




MATS1

Remarks: (Cont.)

0

Y or s( )

ε2

p

Y3

Y2

Y1

H2H1

E

k = 1

k = 2

k = 3

εε3ε3

pε2ε1

H3

If TYPE = PLASTIC

εk

pEffective Plastic Strain=

Hk

Yk 1+ Yk –

εk 1+

pεk

p –

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

Figure 2 Stress-Strain Curve Definition When TID is Specified in Field 3




Figure 2. Stress Strain Curve Definition When TID is Specified in Field 3



NLPARM


DT Incremental time interval for creep analysis. See Remark 3.(Real ≥ 0.0; Default = 0.0 for no creep).

KMETHOD Method for controlling stiffness updates. (Character ="AUTO", "ITER", or "SEMI"; Default = "AUTO").

KSTEP Number of iterations before the stiffness update for ITER

method. (Integer > 1; Default = 5).MAXITER Limit on number of iterations for each load increment.

(Integer > 0; Default = 25).

CONV Flags to select convergence criteria. (Character: “U”, “P”,“W”, or any combination; Default = “PW”).

INTOUT Intermediate output flag. See Remark 8. (Character =“YES”, “NO”, or “ALL”; Default = NO).

EPSU Error tolerance for displacement (U) criterion. (Real > 0.0;Default = 1.0 E-2).




NLPARM


EPSP Error tolerance for load (P) criterion. (Real > 0.0; Default =1.0E-2).

EPSW Error tolerance for work (W) criterion. (Real > 0.0; Default =1.0E-2).

MAXDIV Limit on probable divergence conditions per iteration before

the solution is assumed to diverge. See Remark 9.(Integer ≠ 0; Default = 3

MAXQN Maximum number of quasi-Newton correction vectors to besaved on the database. (Integer > 0; Default = MAXITER).

MAXLS Maximum number of line searches allowed for each iteration.

(Integer > 0; Default = 4)FSTRESS Fraction of effective stress (σ) used to limit the sub-increment

size in the material routines. (0.0 < Real < 1.0; Default = 0.2).

LSTOL Line search tolerance. (0.01 ≤ Real ≤ 0.9; Default = 0.5)




NLPARM


MAXBIS Maximum number of bisections allowed for each loadincrement. (-10 ≤ MAXBIS ≤ 10; Default = 5).

MAXR Maximum ratio for the adjusted arc-length increment relativeto the initial value. See Remark 14. (1.0 ≤ MAXR ≤ 40.0;Default = 20.0

RTOLB Maximum value of incremental rotation (in degrees) allowedper iteration to activate bisection. (Real > 2.0; Default = 20.0).

Remarks:1. The NLPARM entry is selected by the Case Control command NLPARM

= ID. Each solution subcase requires an NLPARM command.




NLPARM

Remarks: (Cont.)

2. In cases of static analysis (DT = 0.0) using Newton methods, NINC isthe number of equal subdivisions of the load change defined for thesubcase. Applied loads, gravity loads, temperature sets, enforceddisplacements, etc., define the new loading conditions. The differencesfrom the previous case are divided by NINC to define the incremental

values. In cases of static analysis (DT = 0.0) using arc-length methods,NINC is used to determine the initial arc-length for the subcase, and thenumber of load subdivisions will not be equal to NINC. In cases ofcreep analysis (DT > 0.0), NINC is the number of time step increments.

3. The unit of DT must be consistent with the unit used on the CREEP

entry that defines the creep characteristics. Total creep time for thesubcase is DT multiplied by the value in the field NINC; i.e., DT * NINC.




NLPARM

Remarks: (Cont.)

4. The stiffness update strategy is selected in the KMETHOD field.a. If the AUTO option is selected, the program automatically selects the most

efficient strategy based on convergence rates. At each step the number ofiterations required to converge is estimated. Stiffness is updated, if (i)estimated number of iterations to converge exceeds MAXITER, (ii) estimatedtime required for convergence with current stiffness exceeds the estimated

time required for convergence with updated stiffness, and (iii) solutiondiverges. See Remarks 9 and 13 for diverging solutions.

b. If the SEMI option is selected, the program for each load increment (i)performs a single iteration based upon the new load, (ii) updates thestiffness matrix, and (iii) resumes the normal AUTO option.

c. If the ITER option is selected, the program updates the stiffness matrix atevery KSTEP iterations and on convergence if KSTEP ≤ MAXITER.However, if KSTEP > MAXITER, stiffness matrix is never updated. Note thatthe Newton-Raphson iteration strategy is obtained by selecting the ITERoption and KSTEP = 1, while the Modified Newton-Raphson iterationstrategy is obtained by selecting the ITER option and KSTEP = MAXITER




NLPARM

Remarks: (Cont.)

5. For AUTO and SEMI options, the stiffness matrix is updated onconvergence if KSTEP is less than the number of iterations that wererequired for convergence with the current stiffness.

6. The number of iterations for a load increment is limited to MAXITER. Ifthe solution does not converge in MAXITER iterations, the load

increment is bisected and the analysis is repeated. If the load incrementcannot be bisected (i.e., MAXBIS is attained or MAXBIS = 0) andMAXDIV is positive, the best attainable solution is computed and theanalysis is continued to the next load increment. If MAXDIV is negative,the analysis is terminated.

7. The test flags (U = displacement error, P = load equilibrium error, and W= work error) and the tolerances (EPSU, EPSP and EPSW) define theconvergence criteria. All the requested criteria (combination of U, Pand/or W) are satisfied upon convergence. See the MSC/NASTRAN

Handbook for Nonlinear Analysis for more details on convergence




criteria.

NLPARM

Remarks: (Cont.)

8. INTOUT controls the output requests for displacements, element forcesand stresses, etc. YES or ALL must be specified in order to be able toperform a subsequent restart from the middle of a subcase.

a. For the Newton family of iteration methods (i.e., when no NLPCI command isspecified), the option ALL is equivalent to option YES since the computed

load increment is always equal to the user-specified load increment.b. For arc-length methods (i.e., when the NLPCI command is specified) the

computed load increment in general is not going to be equal to the user-specified load increment, and is not known in advance. The option ALLallows the user to obtain solutions at the desired intermediate loadincrements.

For every computed and user-specified load increment. ALL

For the last load of the subcaseNO

For every computed load incrementYES

Output ProcessedINTOUT




NLPARM

Remarks: (Cont.)

9. The ratio of energy errors before and after the iteration is defined asdivergence rate Ei, i.e.

Depending on the divergence rate, the number of diverging iteration(NDIV) is incremented as follows:

The solution is assumed to diverge when NDIV w |MAXDIV|. If thesolution diverges and the load increment cannot be further bisected (i.e.,MAXBIS is attained or MAXBIS is zero), the stiffness is updated basedon the previous iteration and the analysis is continued

Ei ∆u

i R

i

∆ui

T

R i 1 –

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

If E 1≥ or E 1012

– < then, NDIV NDIV 2+=

If 1012

E1

1 – < < – then, NDIV NDIV 1+=




NLPARM

Remarks: (Cont.)

If the solution diverges again in the same load increment while MAXDIVis positive, the best solution is computed and the analysis is continuedto the next load increment. If MAXDIV is negative, the analysis isterminated on the second divergence.

10. The BFGS update is performed if MAXQN > 0. As many as MAXQN

quasi-Newton vectors can be accumulated. The BFGS update withthese QN vectors provides a secant modulus in the search direction. IfMAXQN is reached, no additional ON vectors will be accumulated.

Accumulated QN vectors are purged when the stiffness is updated andthe accumulation is resumed.

11. The line search is performed as required, if MAXLS > 0. In the linesearch, the displacement increment is scaled to minimize the energyerror. The line search is not performed if the absolute value of therelative energy error is less than the value specified in LSTOL.




NLPARM

Remarks: (Cont.)12.

The number of subincrements in the material routines (elastoplastic andcreep) is determined so that the subincrement size is approximatelyFSTRESS (equivalent stress). FSTRESS is also used to establish atolerance for error correction in the elastoplastic material; i.e.,

Error in yield function < FSTRESS *σ

If the limit is exceeded at the converging state, the program will exit witha fatal error message. Otherwise, the stress state is adjusted to thecurrent yield surface.

The number of bisections for a load increment/arc-length is limited to|MAXBIS|. Different actions are taken when the solution diverges

depending on the sign of MAXBIS. If MAXBIS is positive, the stiffness isupdated on the first divergence, and the load is bisected on the seconddivergence. If MAXBIS is negative, the load is bisected every time thesolution diverges until the limit on bisection is reached. If the solutiondoes not converge after |MAXBIS| bisections, the analysis is continued

or terminated depending on the sign of MAXDIV See Remark 9




or terminated depending on the sign of MAXDIV. See Remark 9.

NLPARM

Remarks: (Cont.)14.

MAXR is used in the adaptive load increment/arc-length method todefine the overall upper and lower bounds on the load increment/arc-length in the subcase; i.e.,

where ∆ln is the arc-length at step n and ∆l0 is the original arc-length.The arc-length method for load increments is selected by an NLPCIBulk Data entry. This entry must have the same ID as the NLPARMBulk Data entry.

15. The bisection is activated if the incremental rotation for any degree offreedom (∆θx, ∆θy, and ∆θz exceeds the value specified by RTOLB.This bisection strategy is based on the incremental rotation andcontrolled by MAXBIS.

1

AXR ------------------

∆ n

∆lo

-------- MA XR ≤ ≤




NLPCI

Parameters for Arc-Length Methods in Nonlinear Static

Analysis Defines a set of parameters for the arc-length incremental solutionstrategies in nonlinear static analysis (SOLs 66 and 106). This entry willbe used if a subcase contains an NLPARM command (NLPARM = ID).

Format:

Example:

MXINCDESITERSCALEMAXALRMINALRTYPEIDNLPCI

10987654321

101211CRIS10NLPCI




NLPCI

Field ContentsID Identification number of an associated NLPARM entry.

(Integer > 0).TYPE Constraint type. (Character: "CRIS", "RIKS", or "MRIKS";

Default = "CRIS").MINALR Minimum allowable arc-length adjustment ratio between

increments for the adaptive arc-length method. (0.0 < Real <

1.0; Default = 0.25).MAXALR Maximum allowable arc-length adjustment ratio between

increments for the adaptive arc-length method. (Real > 1.0;Default = 4.0).

SCALE Scale factor (w) for controlling the loading contribution in the

arc-length constraint. (Real > 0.0; Default = 0.0)DESITER Desired number of iterations for convergence to be used for

the adaptive arc-length adjustment. (Integer > 0; Default =12).

MXINC Maximum number of controlled increment steps allowed

within a subcase (Integer > 0; Default = 20)




within a subcase. (Integer > 0; Default = 20).

NLPCI

Remarks:1. The NLPCI entry is selected by the Case Control command NLPARM =

ID. There must also be an NLPARM entry with the same ID. However,for creep analysis (DT 0 0.0 in NLPARM entry), the arc-length methodscannot be activated, and the NLPCI entry is ignored if specified. TheNLPCI entry is not recommended for heat transfer analysis in SOL 153.

2. The available constraint types are as follows:TYPE = “CRIS”

TYPE = “RIKS”:

TYPE = “MRIKS”

uni

un0

– uni

un0

– w2

µi

µ0

– ( ) ∆l n2

=+

uni

uni 1 –

– un1

un0

– w2∆µ

i0=+

uni

uni 1 –

– uni 1 –

un0

– w2∆µ

iµ

i 1 – µ

0 – ( ) 0=+




NLPCI

Remarks: (Cont.)Where w = the user specified scaling factor (SCALE)

µ = the load factor

∆l = the arc length

The constraint equation has a disparity in the dimension by mixing thedisplacements with the load factor. The scaling factor (w) is introducedas user input so that the user can make constraint equation unit-dependent by a proper scaling of the load factor m. As the value of w isincreased, the constraint equation is gradually dominated by the loadterm. In the limiting case of infinite w, the arc-length method isdegenerated to the conventional Newton’s method.




NLPCI

Remarks: (Cont.)3. The MINALR and MAXALR fields are used to limit the adjustment of the

arc-length from one load increment to the next by

The arc-length adjustment is based on the convergence rate (i.e.,number of iterations required for convergence) and the change instiffness. For constant arc-length during analysis, use MINALR =MAXALR = 1.

4.

The arc-length ∆l for the variable arc-length strategy is adjusted basedon the number of iterations that were required for convergence in theprevious load increment (Imax) and the number of iterations desired forconvergence in the current load increment (DESITER) as follows:

INALR ∆ new

∆l old

---------------- MAXALR ≤ ≤




NLPCI

Remarks: (Cont.)

5. The MXINC field is used to limit the number of controlled incrementsteps in case the solution never reaches the specified load. This field is

useful in limiting the number of increments computed for a collapseanalysis

max I l DESITERl old

new∆⋅=∆




PBCOMP

Beam Property (Alternate form of PBEAM) Alternate form of the PBEAM entry to define properties of a uniform

cross-sectional beam referenced by a CBEAM entry. This entry is alsoused to specify lumped areas of the beam cross section for nonlinearanalysis and/or composite analysis.

Format:

-etc.-

MID2C2Z2Y2

NID1C1Z1Y1

SECTIONN2N1N1M2M1K2K1

NSMJI12I2I1 AMIDPIDPCOMP

10987654321




PBCOMP

Example:

Field Contents

PID Property identification number. See Remark 1. (Integer > 0)

MID Material identification number. See Remarks 2 and 5.(Integer > 0)

A Area of beam cross section. (Real > 0.0)

I1 Area moment of inertia in plane 1 about the neutral axis. SeeRemark 6. (Real > 0.0)

I2 Area moment of inertia in plane 2 about the neutral axis. See

Remark 6. (Real > 0.0).

0.150.90.2

180.11.2-0.5

1

2.9639PBCOMP




( )

PBCOMP


I12 Area product of inertia. See Remark 6. (Real; Default =

I1 ∗ 12(I12)\J Torsional stiffness parameter. See Remark 6. (Real > 0.0;

Default = 0.0).

NSM Nonstructural mass per unit length. (Real > 0.0;

Default = 0.0)K1, K2 Shear stiffness factor K in K ∗ A ∗ G for plane 1 and plane 2.See Remark 4. (Real > 0.0; Default = 1.0)

M1, M2 (y,z) coordinates of center of gravity of nonstructural mass.See the figure in the CBEAM entry description. (Real;

Default = 0.0)N1, N2 (y,z) coordinates of neutral axis. See the figure in the

CBEAM entry description. (Real; Default = 0.0)

SECTION Symmetry option to input lumped areas for the beam crosssection. See Figure 1 below and Remark 7. (0 ≤ Integer ≤ 5;Default = 0)




PBCOMP


Yi, Zi (y,z) coordinates of the lumped areas in the element

coordinate system. See Remark 1. (Real)Ci Fraction of the total area for the i-th lumped area. (Real >

0.0; Default = 0.0)

MIDi Material identification number for the i-th integration point.

See Remark 5. (Integer > 0) Remarks:

1. The PID number must be unique with respect to other PBCOMP entriesas well as PBEAM entries. The second continuation entry may berepeated 18 more times. A maximum of 21 continuation entries isallowed; i.e., a maximum of 20 lumped areas may be input if SECTION= 5. If SECTION = 1 through 4, the total number of areas input plus thetotal number generated by symmetry must not exceed 20. If these arenot specified, the program defaults, as usual, to the ellipticallydistributed 8 nonlinear rods. See Figure 1




PBCOMP

Remarks:

SECTION=0 ( default)Symmetric about y and z

SECTION=1(with continuation entry)Symmetric about y and z

SECTION=2Symmetric about y

SECTION=3Symmetric about z

SECTION=4Symmetric about y=z=0

SECTION=5No symmetry

Izz - Moment of inertia about z-axis

Iyy - Moment of inertia about y-axis

Zr ef

1

2

3

4

5

6

7

8

Yref

0 2 K z,( )

Ky Kz,( )

2 K y 0,( )

Zre f

Zr ef

Yre f Yref

1

2

3

4

5

6

8

7

Ky

Izz

A------ Kz

Iyy

A------ C1

1

8---=,=,=

1 2

3

4

5 6

8

7

Y1 Y3 Y5– Y– 7= = =

Z1 Z3– Z5 Z7 etc.,–= = =

Y1 Y5=

Z1 Z– 5 etc.,=

Zr ef

Zr ef

Zre f

Yref

Yref

Yre f

1

2

3

4

5

6

7

8

12

3

4

8

75 6

1 2 3 4

5

6

7

8

Y1 Y5 Z1 Z5 etc.,=,= Y1 Y5 Z1 Z5 etc.,=,=

Figure 1. PBCOMP Entry SECTION Types




PBCOMP

Remarks: (Cont.) Figure Notes:

Integration points (lumped area) are numbered 1 through 8. User-specified points are denoted by l and the program default point denoted by

m.

2. For structural problems, MID and MIDi must reference a MAT1 materialentry. For material nonlinear analysis, the material should be perfectlyplastic since the plastic hinge formulation is not valid for strainhardening. For heat transfer problems, MID and MIDi must reference aMAT4 or MAT5 material entry.

3. For the case where the user specifies I1, I2 and I12 on the parent entry,he may specify the stress-output location on continuation entries. The(y,z) coordinates specified on these entries will serve as stress output

locations with the corresponding Ci’s set to 0. Stress output is providedat the first four lumped area locations only. If one of the symmetryoptions is used and fewer than four lumped areas are input explicitly,the sequence of output locations in the imaged quadrants is shown inFigure 1. For one specific example in the model shown in Remark 7

(Figure 2), output can be obtained at points 1 and 2 and in the imagepoints 3 and 4




PBCOMP

Remarks: (Cont.)4. Blank fields for K1 and K2 are defaulted to 1.0. If a value of 0.0 is used

for K1 and K2, the transverse shear stiffness becomes rigid and thetransverse shear flexibilities are set to 0.0.

5. The values E0 and G0 are computed based on the value of MID on theparent entry. MIDi will follow the same symmetry rules as Ci depending

on the value of SECTION. If the MIDi field on a continuation entry isblank, the value will be that of MID on the parent entry. MIDi valuesmay be input on continuations without the corresponding Yi, Zi, and Civalues to allow different stress-strain laws.

6. If the lumped cross-sectional areas are specified, fields I1, I2, and I12

will be ignored. These and other modified values will be calculatedbased on the input data (Yi, Zi, Ci, MIDi) as follows:




PBCOMP

Remarks: (Cont.)

where n is the number of lumped cross-sectional areas specified

y NA

Yi Ci Ei

i 1=

n

∑

Ci Ei

i 1=

n

∑--------------------------------=

z NA

Zi Ci Ei

i 1=

n

∑

Ci Ei

i 1=

n

∑-------------------------------=

A ACi Ei

Eo

-------------

i 1=

n

∑=

I1 ACi E i Yi y NA – ( )2

Eo

-------------------------------------------

i 1=

n

∑=

I2 A Ci E i Zi z NA – ( )

2

Eo--------------------------------------------

i 1=

n

∑=

J JCi G i

Go

-----------------

i 1=

n

∑=

I12 ACi Ei Yi y NA – ( ) Zi z NA – ( )

Eo

---------------------------------------------------------------------

i 1=

n

∑=




PBCOMP

Remarks: (Cont.) As can be seen from Figure 1, if the user chooses to leave the SECTION field

blank, the program defaults to the elliptically distributed 8 nonlinear rods, similarto the PBEAM entry. For this particular case it is illegal to supply Ci and MIDivalues. For a doubly symmetric section (SECTION = 1), if the lumped areas arespecified on either axis, the symmetry option will double the areas. For example,for the section shown in Figure 2, points 2 and 4 are coincident and so are points6 and 8. In such cases, it is recommended that users input the value of area ashalf of the actual value at point 2 to obtain the desired effect.

For SECTION = 5, at least one Yi and one Zi must be nonzero

1

2

4

37

8

6

5

Yre f

Zref

Figure 2. Doubly Symmetric PBCOMP Section




PGAP

Gap Element Property Defines the properties of the gap element (CGAP entry).

Format:

Example

Field Contents

PID Property identification number. (Integer > 0).U0 Initial gap opening. See Figure 2. (Real; Default = 0.0).

F0 Preload. See Figure 2. (Real ≥ 0.0; Default = 0.0)

KA Axial stiffness for the closed gap; i.e., Ua − Ub > U0. See

Figure 2. (Real > 0.0)

0.250.251.0E+61.0E+62.50.0252PGAP

TRMINMARTMAX

MU2MU1KTKBKAF0U0PIDPGAP

10987654321




PGAPField Contents (Cont.)

KB Axial stiffness for the open gap; i.e., Ua − Ub > U0. See

Figure 2. See Remark 2. (Real ≥ 0.0; Default = 10-14 ∗ KA)KT Transverse stiffness when the gap is closed. See Figure 3.

It is recommended that KT ≥ (0.1 ∗ KA) (Real ≥ 0.0; Default =MU1 ∗ KA).

MU1 Coefficient of static friction (µs) for the adaptive gap elementor coefficient of friction in the y transverse direction (µy) forthe nonadaptive gap element. See Figure 3. (Real ≥ 0.0;Default = 0.0)

MU2 Coefficient of kinetic friction (µk) for the adaptive gap element

or coefficient of friction in the z transverse direction (µz) forthe nonadaptive gap element. See Figure 3. (Real ≥ 0.0 forthe adaptive gap element, MU2 ≤ MU1; Default = MU1).

TMAX Maximum allowable penetration used in the adjustment ofpenalty values. The positive value activates the penalty

value adjustment. See Remark 4. (Real; Default = 0.0).




PGAPField Contents (Cont.)

MAR Maximum allowable adjustment ratio for adaptive penalty

values KA and KT. See Remark 5. (1.0 < Real < 106;Default = 100.0).

TRMIN Fraction of TMAX defining the lower bound for the allowablepenetration. See Remark 6. (0.0 ≤ Real ≤ 1.0;Default = 0.001)

Remarks:1. Figures 1 through 3 show the gap element and the force-displacement

curves used in the stiffness and force computations for the element.

2. For most contact problems, KA (penalty value) should be chosen to bethree orders of magnitude higher than the stiffness of the neighboringgrid points. A much larger KA value may slow convergence or causedivergence, while a much smaller KA value may result in inaccurateresults. The value is adjusted as necessary if TMAX > 0.0.




PGAP

Remarks: (Cont.)3. When the gap is open, there is no transverse stiffness. When the gap is

closed and there is friction, the gap has the elastic stiffness (KT) in thetransverse direction until the friction force is exceeded and slippagestarts to occur.

4. There are two kinds of gap elements: adaptive gap and nonadaptive

gap. If TMAX ≥ 0.0, the adaptive gap element is selected by theprogram. When TMAX = 0.0, penalty values will not be adjusted, butother adaptive features will be active (i.e., the gap-induced stiffnessupdate, gap-induced bisection, and subincremental process). The valueof TMAX = -1.0 selects the nonadaptive (old) gap element. The

recommended allowable penetration TMAX is about 10% of the elementthickness for plates or the equivalent thickness for other elements whichare connected to the gap




A 89NAS 103 A di A D b 2003

PGAP

Remarks: (Cont.)5. The maximum adjustment ratio MAR is used only for the adaptive gap

element. Upper and lower bounds of the adjusted penalty are definedby

where Kinit is either KA or KT.

6. TRMIN is used only for the penalty value adjustment in the adaptive

gap element. The lower bound for the allowable penetration iscomputed by TRMIN * TMAX. The penalty values are decreased if thepenetration is below the lower bound

MAR K K AR

K init init

⋅≤≤




A 90NAS 103 Appendix A December 2003

PGAP

Remarks: (Cont.)

y

V A

U A

G AW A

z

VB

UB xGBWB

Slope KA is used when

U A − UB ≥ U0

F0

Slope = KB

Slope = KA

(compression)U A - UBU0(tension)

Fx (compression)

Figure 1. The CGAP ElementCoordinate System

Figure 2. CGAP Element Force-Deflection Curve for Nonlinear Analysis




A 91NAS 103 Appendix A December 2003

PGAP

Remarks: (Cont.)

Nonlinear Shear

Unloading

Slope = KT

∆V or ∆W

MU1 ∗ Fx

MU2 ∗ Fx

Figure 3. Shear Force for CGAP Element




A-92NAS 103 Appendix A December 2003

PLPLANE

Properties of Fully Nonlinear Plane Strain Elements Defines the properties of a finite deformation, hyperelastic plane strain

or axisymmetric element.

Format:

Example:

Field Contents


MID Identification number of MATHP entry. (Integer > 0).

CID Identification number of a coordinate system defining theplane of deformation. See Remarks 1and 2 (Integer ≥ 0;

Default = 0)

CIDMIDPIDPLPLANE

10987654321

201204203PLPLANE





PLPLANE

Remarks:1. PLPLANE can be referenced by a CQUAD, CQUAD4, CQUAD8,

CQUADX, CTRIA3, CTRIA6, or CTRIAX entry.2. Plane strain hyperelastic elements must lie on the x-y plane of the CID

coordinate system. Stresses and strains are output in the CIDcoordinate system.

3. Axisymmetric hyperelastic elements must lie on the x-y plane of thebasic coordinate system. CID may not be specified and stresses andstrains are output in the basic coordinate system.



A 93NAS 103, Appendix A, December 2003


PLSOLID

Finite Deformations Solid Element Properties Defines a finite deformation hyperelastic solid element.

Format:

Example:

Field Contents


MID Identification number of a MATHP entry. (Integer > 0)

Remarks:1. PLSOLID can be referenced by a CHEXA, CPENTA or CTETRA entry.

2. Stress and strain are output in the basic coordinate system.

MIDPIDPLSOLID

10987654321

2120PLPLANE



, pp ,


TABLES1

Material Property Table, Form 1 Defines a tabular function for stress-dependent material properties such

as the stress-strain curve and creep parameters

Format:

Example:

Field Contents

TID Table identification number. (Integer > 0)

xi, yi Tabular values. (Real)

-etc.-y3x3y2x2y1x1

TIDTABLES1

10987654321

ENDT15000..0210000..010.00.0

32TABLES1



pp


TABLES1

Remarks:1. xi must be in either ascending or descending order, but not both.

2. Discontinuities may be specified between any two points except the twostarting points or two end points. For example, in Figure 1discontinuities are allowed only between points x2 through x7. Also, if yis evaluated at a discontinuity, then the average value of y is used. InFigure 1 the value of y at x = x3 is y = (y3 + y4)/2.

3. At least one continuation entry must be present.

4. Any xi-yi pair may be ignored by placing “SKIP” in either of the two fieldsused for that entry.

5. The end of the table is indicated by the existence of “ENDT” in either of

the two fields following the last entry. An error is detected if anycontinuations follow the entry containing the end-of table flag ENDT.

6. TABLES1 is used to input a curve in the form of

y = yT(x)




TABLES1

Remarks: (Cont.)where x is input to the table and y is returned. The table look-up is

performed using linear interpolation within the table and linear extrapolation outside the table using the two starting or end points. SeeFigure 1. No warning messages are issued if table data is inputincorrectly.

y

x

DiscontinuityDiscontinuity

Linear

Extrapolationof Segment

x

x1 x2 x3, x5 x6 x7,, x4 x8

x valueRange of




TABLEST Material Property Temperature-Dependence Table

Specifies the material property tables for nonlinear elastic temperature-

dependent materials. Format:

Example:

Field ContentsTID Table identification number. (Integer > 0)Ti Temperature values. (Real)TIDi Table identification numbers of TABLES1 entries.

(Integer > 0)

-etc.-T3Tid2T2Tid1T1

TIDTABLEST

10987654321

ENDT20175.010150.0

101TABLEST




TABLEST

Remarks:1. TIDi must be unique with respect to all TABLES1 and TABLEST table

identification numbers.2. Temperature values must be listed in ascending order.

3. The end of the table is indicated by the existence of ENDT in either ofthe two fields following the last entry. An error is detected if any

continuations follow the entry containing the end-of-table flag ENDT.4. This table is referenced only by MATS1 entries that define nonlinear

elastic (TYPE = “NLELAST”) materials.




TSTEPNL

Parameters for Nonlinear Transient Analysis Defines parametric controls and data for nonlinear transient structural or

heat transfer analysis. TSTEPNL is intended for SOLs 129, 159, and99.

Format:

Example:

RTOLBUTOLMAXRRBMSTEP ADJUSTMAXBIS

FSTRESSMAXLSMAXQNMAXDIVEPSWEPSPEPSUCONVMAXITIERKSTEPNODTNDTIDTSTEPNL

10987654321

200.1160.75055

0.0221021.00E-061.00E-03

PW-1021250TSTEPNL




TSTEPNLField Contents

ID Identification number. (Integer > 0).

NDT Number of time steps of value DT. (Integer > 4).DT Time increment. (Real > 0.0).

NO Time step interval for output. Every NO-th step will be savedfor output. (Integer > 0; Default = 1).

KSTEP Number of converged bisection solutions betweenstiffness updates. (Integer > 0; Default = 2)

MAXITER Limit on number of iterations for each time step. (Integer ≠ 0;Default = 10)

CONV Flags to select convergence criteria. (Character: “U”, “P”,

“W”, or any combination; Default = “PW”)




TSTEPNLField Contents (Cont.)

EPSU Error tolerance for displacement (U) criterion. (Real > 0.0;

Default = 1 .0E-2)EPSP Error tolerance for load (P) criterion. (Real > 0.0; Default =

1.0E-3)

EPSW Error tolerance for work (W) criterion. (Real > 0.0;Default = 1 .0E-6)

MAXDIV Limit on the number of diverging conditions for a time stepbefore the solution is assumed to diverge. (Integer > 0;Default = 2)

MAXQN Maximum number of quasi-Newton correction vectors to be

saved on the database. (Integer ≥ 0; Default = 10)MAXLS Maximum number of line searches allowed per iteration.(Integer ≥ 0; Default = 2)

FSTRESS Fraction of effective stress (s) used to limit the subincrementsize in the material routines. (0.0 < Real < 1.0;

Default = 0.2)





MAXBIS* Maximum number of bisections allowed for each time step.

(- 9 ≤ Integer ≤ 9; Default = 5) ADJUST* Time step skip factor for automatic time step adjustment.

(Integer ≥ 0; Default = 5)

MSTEP* Number of steps to obtain the dominant period response.(10 ≤ Integer ≤ 200; Default = variable between 20 and 40)

RB* Define bounds for maintaining the same time step for thestepping function if METHOD = “ADAPT”. (0.1 ≤ Real ≤ 1.0;Default = 0.75)

MAXR* Maximum ratio for the adjusted incremental time relative to

DT allowed for time step adjustment. (1.0 ≤ Real ≤ 32.0;Default = 16.0)

UTOL* Tolerance on displacement increment beneath which there isno time step adjustment. (0.001 > Real ≤ 1.0; Default = 0.1)





RTOLB Maximum value of incremental rotation (in degrees) allowed

per iteration to activate bisection. (Real > 2.0;Default = 20.0)

*These fields are only valid for METHOD = “ADAPT”

Remarks:1. The TSTEPNL Bulk Data entry is selected by the Case Control

command TSTEPNL = ID. Each subcase (residual superelementsolutions only) requires a TSTEPNL entry and either applied loads via

TLOADi data or initial values from a previous subcase. Multiplesubcases are assumed to occur sequentially in time with the initialvalues of time and displacement conditions of each subcase defined bythe end conditions of the previous subcase.




TSTEPNL Remarks: (Cont.)

2. IF METHOD = “ADAPT”, NDT is used to define the total duration for

analysis, which is NDT * DT. (Since DT is adjusted during the analysisfor METHOD = “ADAPT”, the actual number of time steps, in general,will not be equal to NDT). Also, DT is used only as an initial value forthe time increment.

3. For printing and plotting the solution, data recovery is performed at timesteps 0, NO, 2 * NO, ..., and the last converged step. The Case Controlcommand OTIME may also be used to control the output times.

4. The stiffness update strategy as well as the direct time integrationmethod is selected in the METHOD field.

a. METHOD = “AUTO”: The stiffness matrix is automatically updated toimprove convergence. The KSTEP value is ignored.

b. METHOD = “TSTEP”: The stiffness matrix is updated every KSTEPthincrement of time.





c. METHOD = “ADAPT”: The program automatically adjusts the incremental

time and uses bisection. During the bisection process, the stiffness matrix isupdated every KSTEPth converged bisection solution in order to reducecomputing cost.

In all methods the stiffness matrix is always updated for a new subcaseor restart. The ADAPT method allows linear transient analysis, but

AUTO or TSTEP will abort the run if the model does not have any datarepresenting nonlinearity.

5. The number of iterations for a time step is limited to MAXITER. IfMAXITER is negative, the analysis is terminated when the divergencecondition is encountered twice during the same time step or the solutiondiverges for five consecutive time steps. If MAXITER is positive, theprogram computes the best solution and continues the analysis untildivergence occurs again. If the solution does not converge in MAXITERiterations, the process is treated as a divergent process. See Remark 7.





6. The convergence test flags (U = displacement error test, P = load

equilibrium error test, W = work error test) and the error tolerances(EPSU, EPSP, and EPSW) define the convergence criteria. Allrequested criteria (combination of U, P, and/or W) are satisfied uponconvergence. Note that at least two iterations are necessary to checkthe displacement convergence criterion.

7. MAXDIV provides control over diverging solutions. Depending on therate of divergence, the number of diverging solutions (NDIV) isincremented by 1 or 2. The solution is assumed to diverge when NDIVreaches MAXDIV during the iteration. If the bisection option is used(allowed MAXBIS times) with the ADAPT method, the time step isbisected upon divergence. Otherwise, the solution for the time step is

repeated with a new stiffness based on the converged state at thebeginning of the time step. If NDIV reaches MAXDIV again within thesame time step, the analysis is terminated.

8. Nonzero values of MAXQN and MAXLS will activate the quasi-Newtonupdate and the line search process, respectively.





9. The number of subincrements in the material routines is determined

such that the subincrement size is approximately FSTRESS * .FSTRESS is also used to establish a tolerance for error correction inelastoplastic material, i.e.,

error in yield function < FSTRESS * yield stress

If the limit is exceeded at the converging state, the program will EXITwith a fatal error message. Otherwise, the stress state is adjusted to thecurrent yield surface, resulting in δ = 0.

10. The bisection process is activated when divergence occurs andMAXBIS ≠ 0. The number of bisections for a time increment is limited to

|MAXBIS|. If MAXBIS is positive and the solution does not convergeafter MAXBIS bisections, the best solution is computed and the analysisis continued to the next time step. If MAXBIS is negative and thesolution does not converge in |MAXBIS| bisection, the analysis isterminated.

σ





11. ADJUST controls the automatic time stepping for METHOD = ADAPT.

Since the automatic time step adjustment is based on the mode of response and not on the loading pattern, it may be necessary to limit theadjustable step size when the period of the forcing function is muchshorter than the period of dominant response frequency of the structure.It is the user’s responsibility to ensure that the loading history is properly

traced with the ADJUST option. The ADJUST option should besuppressed for the duration of short pulse loading. If unsure, start witha value for DT that is much smaller than the pulse duration in order toproperly represent the loading pattern.

a. If ADJUST = 0, then the automatic adjustment is deactivated. This is

recommended when the loading consists of short duration pulses.b. If ADJUST > 0, the time increment is continually adjusted for the first few

steps until a good value of ∆ is obtained. After this initial adjustment, the timeincrement is adjusted every ADJUST-th time step only.

c. If ADJUST is one order greater than NDT, then automatic adjustment is

deactivated after the initial adjustment.





12. MSTEP and RB are used to adjust the time increment during analysis

for METHOD = ADAPT. The recommended value of MSTEP for nearlylinear problems is 20. A larger value (e.g., 40) is required for highlynonlinear problems. By default, the program automatically computesthe value of MSTEP based on the changes in the stiffness.

The time increment adjustment is based on the number of time steps

desired to capture the dominant frequency response accurately. Thetime increment is adjusted as follows:

Where:

n1n f(r)∆∆t t =+

r 1

MSTEP--------------------

2πωn

-------

1

∆tn

--------

=





With f = 0.25 for r < 0.5 * RB

f = 0.5 for 0.5 * RB ≤ r < RBf = 1.0 for RB ≤ r < 2.0

f = 2.0 for 2.0 ≤ r < 3.0/RB

f = 4.0 for r > 3.0/RB

13. MAXR is used to define the upper and lower bounds for adjusted timestep size, i.e.,,

DT MAXRt MAXR

DT MAXBIS

DT ⋅≤∆≤ ,min 2





14. UTOL is a tolerance used to filter undesirable time step adjustments;

i.e.,

Under this condition no time step adjustment is performed in a structuralanalysis (SOLs 99 and 129). In a heat transfer analysis (SOL 159) thetime step is doubled.

15. The bisection is activated if the incremental rotation for any degree offreedom (∆θx, ∆θy, ∆θz) exceeds the value specified by RTOLB. This

bisection strategy is based on the incremental rotation and controlled byMAXBIS

Un

U·

max

-------------------- UTOL<

nonlinear analysis using msc.nastran

Documents