workshop 09

MSC.Nastran 105 Exercise Workbook 9-1

Non-Linear Load Balancing

MPC MPC

1 2

3 4

56

1 2

3

1000# 1000#

WORKSHOP 9

Objectives:

Create 3 springs with different spring constants (K).

Implement different loadings in each subcase.

Simulate multiple boundary conditions with MPCs.

9-2 MSC.Nastran 105 Exercise Workbook

WORKSHOP 9 Non-Linear Load Balancing


IntroductionMany of the aerospace engineering challenges often involvefabrication and assembly of the components on the ground in 1-genvironment and subsequently launching these components in spacein 0-g environment. Assembly and Launch process introduces residualloads in the connecting components. The task at hand is madeinconvenient by the inherent nature of the process where loads,boundary conditions and the structure itself are changing during eachstep of the process. The finite element models associated with suchstructure are often large which further complicates the analysisprocess. Due to changing loads and boundary conditions, thestructural stiffness is changing between load cases. Moreover, thestrain energy stored in the structure in a load case is redistributed insubsequent load case. MSC.Nastran nonlinear capability allows thesetype of interaction between load cases where loads can be added andconstraints can be removed in a continuous manner allowing theenergy transfer to take place during process. This is done byappropriately selecting load sets, single-point-constraint set and multi-point constraint set for each load case. It is important to note that indeformed grid cannot be constrained in the deformed position in thesubsequent subcase. Constraining the deformed grid in later subcasewill force the deformation of the grid to zero. Multi-point-constraintsare linear equations connecting one degree of freedom displacementto one or more degree of freedom displacements.

Solution Method:MSC.Nastran nonlinear static solution sequence (SOL 106) will beused for the problem. It continuously, accounts for modification ofloads, boundary condition and multi-point constraint equationsbetween subcases. (i.e. converged solution at the end of a subcaseforms an initial condition for the next subcase). Multiple subcasesprovide tool for applying distinct loads and boundary conditions,which are part of a load history for the analysis under consideration.Within a subcase loading and boundary conditions can be subdividedinto number of increments to assist the convergence process. Thenumber of load increments and iteration methods are specified on theNLPARM bulkdata entry, which is selected within each subcase.Thus, subcases in nonlinear analysis provide partitioning of a loadhistory where each step of the loading history can be controlledindependently. For the present problem, response of the structure islinear for the given loading and incremental load steps are notrequired.


Demonstration of concept:It is invaluable to illustrate and validate the analysis underconsideration prior to attempting it on an actual problem. This is donehere using simple three-spring model, which is conceptually similar toproblem under investigation that has a close form solution. In thissample problem three co-linear springs are considered. Objective ofthe analysis is to fit the middle spring between the two outer springsand calculate the loads in the assembled structure. Two outer springsare constrained at the opposite ends and compressed to fit the middlespring. Subsequently, compression caused by the external forces fromthe two outer springs are released and load in the middle spring isobserved. The MSC.Nastran nonlinear solution result for the problemdescribed in figure 1 is compared to closed form solution obtained viaprinciple of minimum potential energy in the following table 1.Theoretical formulation is shown on next page and MSC.Nastraninput file is shown in appendix A.



Table 1: Comparison of a Testcase with MSC.Nastran nonlinear solution.

End of Subcase 1 End of Subcase 2

Nastran Theory Nastran Theory

Displacement A

B

-1.0 -1.0 -0.8 -0.8

1.0 1.0 0.8 0.8

Element forces 1

2

3

1000 1000 800 800

0.0 0.0 800 800

1000 1000 800 800

Figure 1

MPC MPC

1 2

3 4

56

Spring 1 Constant: 1000. Applied Force=-1000,0Spring 2 Constant: 1000. Applied Force=1000.0

Spring 3 Constant: 2000.

SPC = 1LOAD = 10$ Note MPCs are defined but not referenced.

1 2

3

Subcase 2:

SPC = 2 $ Spcs at 3 and 4 removed.MPC = 1 $ MPCs are active in this subcase.$ Load removed

Subcase 1:

1000# 1000#


Theoretical Formulation:

The potential energy of the system can be written as

Where,

x1, x2 = deformation of the outer spring prior to load release.x1, x2 = deformation at equilibriumk1, k2, k3 = spring stiffness of springs 1, 2, and 3.

U12---k1x1

2 12---k2x2

2 12---k3 x1 x1( ) x2 x2'( )+[ ]

2+ +=

dUdx2---------- 0 k2x2 k3 1( ) x1 x1' x2 x2'+[ ]+ + 0= =

k2x2' k3 x1 x2 x2'+[ ] k3x1'=

k2x2' k3 x1 x2+[ ] k3x2'k3k3

k1 k3+------------------ x1 x2 x2'+[ ]=

Substituting x1

k2x2 k3k3k3

k1 k3+------------------ x1 x2+[ ] k3x2'

k3k3k1 k3+------------------x2'+=

dUdx1'---------- k1x1' 0 k3 1( ) x1 x1' x2 x2'+[ ]+ + 0= =

k1x1' k3 x1 x2 x2'+[ ] k3x1'=

x1'k3

k1 k3+------------------ x1 x2 x2'+[ ]=



ResultsSubcase 1: Max = 1

On Subcase 2: Max = 0.8

Verification

k2 k+ 3

k32

k1 k3+------------------ x2' k3

k32

k1 k3+------------------ x1 x2+[ ]=

k2k1k3

k1 k3+------------------+ x2'

k1k3k1 k3+------------------ x1 x2+[ ]=

x2'k1k3

k1k2 k2k3 k1k3+ +------------------------------------------------ x1 x2+[ ]=

x2k1k3

k1k2 k2k3 k1k3+ +------------------------------------------------ x1 x2+[ ]= k1 k2 1000= =

k3 2000=

2 1 1+( )1 2 2+ +--------------------- 4

5--- 0.8= = =


SOL 106CENDOLOAD = ALLDISP = ALLSPCFORCE = ALLMPCFORCE = ALLESE = ALL$SUBCASE 1 NLPARM = 1 LOAD = 1 SPC = 1$SUBCASE 2$$ - UNLOAD$ - CHANGE SPC TO INCLUDE THE MIDDLE SPRING$ - CONNECT OUTER SPRINGS TO THE MIDDLE SPRING$ NLPARM = 2 SPC = 2 MPC = 1BEGIN BULK$NLPARM,1,1NLPARM,2,1$GRID,1,,0.GRID,2,,10.GRID,3,,10.GRID,4,,14.GRID,5,,14.GRID,6,,24.$CELAS1,1,1,1,1,2,1CELAS1,2,2,3,1,4,1CELAS1,3,1,5,1,6,1$PELAS,1,1000.PELAS,2,2000.$FORCE,1,2,,1000.,-1.FORCE,1,5,,1000.,1.$$ SUBCASE1$ FIXED BOTH ENDS$ GRID 2 & 6 ARE FREE IN x$SPC1,1,1,1,6SPC1,1,23456,1,2,5,6$SPC1,1,123456,3,4 $ - REMOVE FROM SOLUTION (NOT CONNECTED)$$ SUBCASE 2$ FIXED BOTH ENDS$ ALL OTHER GRIDS ARE FREE IN x$SPC1,2,1,1,6SPC1,2,23456,1,2,3,4,5,6$MPC,1,4,1,1.,5,1,-1.MPC,1,3,1,1.,2,1,-1.ENDDATA



Exercise Procedure:

1. Create a new file.

Click on the Node Size icon.

2. Create the finite elements.

File/New...

New Database Name load_balance

OK

Finite Elements

Action: Create

Object: Node

Method: Edit

Associate with Geometry

Node Location List: [0 0 0]

Apply

Node ID List: 2


Apply

Node ID List: 3


Apply

Node ID List: 4


Apply

Node ID List: 5

Node Size


Create the Properties.


Apply

Node ID List: 6


Apply

Action: Create

Object: Element

Method: Edit

Shape: Bar

Auto Execute

Element ID List: 1

Node 1 = node 1

Node 2 = node 2

Apply

Element ID List: 2

Node 1 = node 3

Node 2 = node 4

Apply

Element ID List: 3

Node 1 = node 5

Node 2 = node 6

Apply

Properties

Action: Create

Object: 1D

Method: Spring

Property Set Name: spring_1



Click on the Beam Element icon.

Select Elements 1 and 3.

Select Element 2.

Create loads and boundary conditions.

Input Properties

Spring Constant: 1000

DOF at Node 1: UX

DOF at Node 2: UX

OK

Select Members

Add

Apply

Property Set Name: spring_2

Input Properties

Spring Constant: 2000

DOF at Node 1: UX

DOF at Node 2: UX

OK

Select Members

Add

Apply

Loads/BC

Action: Create

Beam Element


Object: Force

Type: nodal

New Set Name: force_1

Input Data

Force :

OK

Select Application Region

Geometry Filter: FEM

Select Nodes: node 2

Add

OK

Apply

New Set Name: force_2

Input Data

Force :

OK



Select Nodes: node 5

Add

OK

Apply

Action: Create

Object: Displacement

Type: nodal

New Set Name: outside

Input Data

Translations :

Rotations :



OK



Select Nodes: node 1 6

Add

OK

Apply

New Set Name: inside

Input Data

Translations :

Rotations :

OK



Select Nodes: node 3 4

Add

OK

Apply

Finite Elements

Action: Create

Object: MPC

Type: Explicit

MPC ID: 1

Define Terms

Auto Execute

Create Dependent


Create Load Cases.

Select from menu.

Node List: node 3

DOFs: UX

Apply

Create Independent

Coefficient 1.0

Node List: node 2

DOFs: UX

Apply

Apply

MPC ID: 2

Create Dependent

Node List: node 4

DOFs: UX

Apply

Create Independent

Coefficient 1.0

Node List: node 5

DOFs: UX

Apply

Apply

Load Cases

Action: Create

Load Case Name: case_1

Assign/Prioritize Loads/BCs

Select Individual Loads/BCs: Displ_inside

Displ_outside



Select from menu.

Do the Analysis.

Under Subcases For Solution Sequence, click on case 1 and case 2and they should come up under Subcases Selected. Then deselectDefault under Subcases Selected by clicking on it.

Force_force_1

Force_force_2

OK

Apply

Load Case Name: case_2

Assign/Prioritize Loads/BCs

Select Individual Loads/BCs: Displ_outside

OK

Apply

Analysis

Action: Analyze

Object: Entire Model

Method: Analysis Deck

Solution Type

Nonlinear Static

OK

Subcase Select

OK

Apply


Now we need to modify the input decks.

1) Delete MPC = 3 in Case Control section

2) Add MPC = 3 under Subcase 2



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

POINT ID. TYPE T1 T2 T3 R1 R2 R3

1 G 0.0 0.0 0.0 0.0 0.0 0.0

2 G -1.000000E+00 0.0 0.0 0.0 0.0 0.0

3 G 0.0 0.0 0.0 0.0 0.0 0.0

4 G 0.0 0.0 0.0 0.0 0.0 0.0

5 G 1.000000E+00 0.0 0.0 0.0 0.0 0.0

6 G 0.0 0.0 0.0 0.0 0.0 0.0

1 JUNE 9, 1999 MSC/NASTRAN 6/25/98 PAGE 12

0 SUBCASE 2

LOAD STEP = 2.00000E+00

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

POINT ID. TYPE T1 T2 T3 R1 R2 R3

1 G 0.0 0.0 0.0 0.0 0.0 0.0

2 G -8.000000E-01 0.0 0.0 0.0 0.0 0.0

3 G 2.000000E-01 0.0 0.0 0.0 0.0 0.0

4 G -2.000000E-01 0.0 0.0 0.0 0.0 0.0

5 G 8.000000E-01 0.0 0.0 0.0 0.0 0.0

6 G 0.0 0.0 0.0 0.0 0.0 0.0

1 JUNE 9, 1999 MSC/NASTRAN 6/25/98 PAGE 13

0 SUBCASE 1

LOAD STEP = 1.00000E+00

workshop 09

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