Geometric Non-Linear (GNL) Analysis using ANSYS

By

Santhosh M

Technical Support,

ANSYS India-Bangalore.

Webinar : 28 Sep 2011-

Way forward

This presentation does NOT discuss about the basics of non-linear finite element method, but about understanding ofnonlinear FEA and to answer the following questions:

What is it ?

When do we need it ?

How does it work ?

Importance of different controls/selections

Examples

Tips

2

No Less equationsBut mostly solid mechanics equation.

Contents

Motivation

Small and large deformation problems

Classifications of GNL problems

Large / Finite strain

Large displacements & Large rotation

Stress Stiffening

Spin Softening

Nonlinear strain & stress measures

Kinematic & Constitutive relations

Solution Schemes & Convergence

Meshing consideration

Result interpretation

3

How bad the result can go, if geometric nonlinearities not considered?

4

1m long 3mm thick 1m x 1 m steel plate with fixed edges

subjected to pressure of 0.1 Mpa

How bad the result can go, if geometric nonlinearities not considered?...

5

Load carried by bending

1 m long 3mm steel plate deflects by 782 mm !!!

783 mm

How bad the result can go, if geometric nonlinearities not considered?...

6

Load carried by in-plane tension (membrane effect)

Courtesy : Paul-Belcan

GNL Calculation

783 mm

16 mm

A simple hand calc: Analytical

7

L

d

F

E A

1

2

3

F

N12 N23

NNN 2312

Static :

The displacement U can be obtained

by equating the internal energy and

the external work.

External work done by the force F

UFW e2

1

Internal work for one bar

vvv

vAE

Nv

EvW d

d d

22

i 22

1

2

1

2

1

LAV

AN

E

i

AE

LNW

2

2

1

Internal work for the truss

)(cos

i

2

2

4 AE

LFW

)(cos A E 2

L F U W W

2ei

This problem can be solved using

classical structural mechanics.

)cos(2

FN (1)

(2)

F

U

F= K

U

We

Equate External & Internal work

Ref : KTH university

8

Ref: http://en.wikipedia.org/wiki/Direct_stiffness_method

2 1

2 1

cos

sin

x xC

L

y yS

L

L

1

2

(x1,y1)

(x2,y2)

T

L

d

F

E A

1

2

3

Truss Element matrix

A SIMPLE HAND CALC: FEM…

L

d

F

E A1

2

3

x

y

)(sin)sin( )cos(

)sin( )cos()(cos

2

2

L

AE12K

)(sin)sin( )cos(

)sin( )cos()(cos

2

2

L

AE32K

For the whole structure,

Assembling of the stiffness matrix:

00

02

2

2 F

V

U

L

AE

)(sin

)(cos

2

2

which gives

022 )(cos 2

2 V

AE

LFU

L = 1 m

d = 0.001m

5 kgsteel

1 cm

m 2

1 561

00100050102

819522112 .

..

.U

This result is obviously not correct.

Where is the mistake in our calculations ?

A SIMPLE HAND CALC: FEM…

9

Numerical Example

1 m long steel rod deflects by 1.56 m !!!

Element stiffness matrices;

Mistakes done using classical mechanics.

The equilibrium of the structure is

evaluated in the undeformed

configuration. But vertical

displacement U makes a large change

in the angle ’α’.

The expression of the external virtual

work, assumes that U is proportional

to F, which is not the case here. U is

also function of ’α’F

U

F= K

U

We

Mistakes done using FEM.

The transformation matrix T is

evaluated in the undeformed

configuration. Large change in the

angle ’α’ with respect in U is not

accounted in stiffness matrix.

For a single bar, only a displacement

in the bar direction gives a strain ε,

which is correct only for small

displacements.

Lu

v

L u = 0

v

L

u

0

WHAT IS GOING WRONG ?

10

Ld

F

L’U

α

The calculations have been done by assuming that the displacements are very

small compared to the geometry of the structure.

L

LL'EAEAN

L'

UdNF

2

)( UdL'L

AEF

112

222 dLUdL' )(

Ld

F

L’U

How to do it right ?

11

Need to be solved iteratively.A non-linear analysis is done by studyingthe structure in the deformedconfiguration, incrementally.

)(cos A E 2

L F U

SolutionLinear

2

50

F (N)

U (m)

0.014

Numerical application

1 m long steel cable deflecting 4 mm, makes sense !

11

Deflectio

n

10

20

1 2 3

I/P

30

0/P

What is nonlinearity ?

12

SystemI/P O/P

Force,Pressure,

Temp, etc,

Defl, Stress,Strain,etcK,M,C

How different nonlinear behavior is? From linear behavior

If a structure experiences large enough deformations, such that itsaltered configuration changes the stiffness of the structure, thenthis is referred as a geometric nonlinearity, which can cause thestructure to respond in a stiffened or a softened manner.

13

P=K.U

Example for Hardening| Softening

14

Example for Hardening| Softening…

15

F increment

U increment

Understanding about GNL…

Most engineers associate GNL problems with ‘large’ displacementseffects such as buckling.

However, GNL problems can also involve small displacements.

16GNL theory often predicts less displacement (realistic) than thecorresponding small deflection theory for hardening behaviour.

782 mm 15 mm

Linear analysis Non-Linear analysis

Why should a FE analyst worry about GNL?...

As the DOF(Deflections) are primary o/p in FEA, any error in deflection alsoaffects are derived results such as stress & strain.

Not always an easy to decide whether the obtained solution is a good or abad one.

If experimental or analytical results are available then easily possible toverify any finite element result.

In absence of reference results, Requires background about; Finite element method in general. Software used to solve the problem.

To be able to judge the appropriateness of the chosen elements andalgorithms from the ANSYS library.

17

Terminology : Small and large deformation problems

Compared to initial dimensions of the geometry

Small Infinitesimal / Negligible

• Say : 1 m long bar stretched by 1 mm

Finite Large but not infinite / not negligible

• Say : 10 mm long bar stretched by 1 mm

18Ref : NPTEL

Classifications of GNL problems

In order to classify GNL problems, it is best to focus on whether

strain is assumed small or large, as strain measure normalizes the

deformation with dimension of the geometry.

1. Small & Large (Finite) strain

2. Large displacements & Large rotation

3. Stress Stiffening

4. Spin Softening

19

Large rotation (rigid), no strain

Large strain

Large Deflection / Large Rotation but Small strain GNL problems

In this category, the change in the spatial orientation of theelements can be finite (large) but the induced strains remain small.

Example :

Shallow struts, shells and arches deflected by a transverse load.

Fishing rod bent under the weight of a heavy fish

Buckling of an imperfect column.

Snap- through buckling of deep arch.

20

Large Deflection / Large Rotation but Small strain GNL problem

Assumes that the rotations are large but the mechanical strains (those thatcause stresses) are evaluated using linearized expressions.

The structure is assumed not to change shape except for rigid bodymotions.

Example: VM 40

21

CERIG Vs MPC 184 in large rotation problems

22

MPC 184 does update the nodal coordinates based on deformed position of the structure

Different views

CERIG Vs MPC 184 in large rotation problem…

23

However, recognize that there are some situations in which coupling

or constraint equations can be valid in a nonlinear analysis. For

example:•You can couple together constrained DOFs at a rigid boundary.

•Constraint equations may be valid for large strain, small rotation

response.

Think carefully before using coupling or constraint equations in large rotation problems!

CERIG does not update the coefficients based on deformed position of the structure.

Ref : KTH university

Large (finite) strain GNL problems

In this category, as the structure deflects, the localized deformations are large enough that the strains are no longer infinitesimal.

Deflections and rotations may be arbitrarily large.

Shape changes (e.g. area, thickness, etc.) are also accounted for.

With large strains, it is also important to model material non-linearity such as plasticity, hyper elasticity, if required.

Example

Metal forming and manufacturing processes

Deformation of rubber-like materials is a typical analysis in which finite strains are experienced.

24

Classifications of GNL problems

1. Large / Finite strain

2. Large displacements / Large rotation

3. Stress Stiffening / Pre-stress

4. Spin Softening

25

Stress stiffening (SSTIF,ON)

Stress stiffening is the stiffening (or weakening) of a structure due to its stress state .

The out-of-plane stiffness of a structure can be significantly affected by the state of in-plane stress in that structure.

This coupling between in-plane stress and transverse stiffness, known as stress stiffening.

The effect of stress stiffening is accounted for by generating and then using anadditional stiffness matrix, hereinafter called the “stress stiffness matrix”.

26

Example•Guitar string•Cloth membrane

Spin softening

Spin softening / centrifugal softening accounts for the radial motion of a body'sstructural mass as it is subjected to an angular velocity. Hence it is a type of largedeflection but small rotation approximation.

27

F F

Initial stiffness

Spin softening

In large deflection analysis change in geometry is accounted no need to activate this effect explicitly.

Hierarchy of GNL problems

You may think of GNL in terms of a hierarchy:

1. Large strain

2. Large deformation & large rotation

3. Stress stiffening

4. Regular linear analysis

28

Spin softening

Small deflection & rotation

Stress stiffening

Large def & rot

Large strain

Terminology : Effective/Consistent Stiffness matrix

A fully consistent tangent stiffness matrix [Kenl] is made up of four

components:

[Kenl] = [Ke

ini]+ [Keinc] + [Ke

] + [Keu] + [Ke

a]

[Keini] => initial elastic stiffness matrix

[Keinc] => incremental tangent matrix.

[Ke] => stress-stiffening matrix.

[Keu] => initial displacement-rotation matrix, which includes the

effect of changing geometry in the stiffness relation.

[Kea] => pressure load stiffness matrix, which includes the effect of

changing pressure load orientation in the stiffness relation.

29

An element can experience large displacement, large rotation, large strain and non

linear material behavior.

The aim of the finite element solution is to evaluate the equilibrium position atdiscrete (load increments) time points 0, t, Δt, 2Δt,3Δt, …. Etc.

For a static analysis Δt corresponds to load increment between sub steps.

30

100 N

NSUBST,10

Load increment = 10 N

10 N

20 N30 N

100N

Kinematic relations in GNL

Kinematic relations in GNL

Kinematic relations

Change in geometry as the structure deforms is taken into account in setting up

• Strain-displacement relationship equations

• Equilibrium equations

Kinematic relations are the relations between displacements and the straincomponents

In linear analysis, strain components can be found from differentiating the shapefunction and multiplying with nodal displacements.

31

Example: axial strain component in x-direction.

Section displacementShape function

Node displacements

Constitutive relation

In physics, a constitutive equation is a relation between two physical quantities thatis specific to a material or substance, and approximates the response of thatmaterial to external forces.

As an example, in structural analysis, constitutive relations connect applied stressesto strains or deformations.

Constitutive relations are the relations between strain components and stresscomponents.

The stress-strain constitutive relation for linear materials commonly known asHooke's law.

Also called as

Stress-strain relation

Constitutive assumption

Equation of state

32

Constitutive relation in linear system

33

3D generalized Hook’s law

Plane stress

D matrix depends on problem dimensionality and material model

kinematic and constitutive relations

A nonlinear finite element equilibrium equation

contain the displacement and strain-displacement matrix plus the constitutive matrixof the material.

In order to use an element for a specific response prediction, it is necessary that boththe kinematic and constitutive descriptions to be appropriate.

34

][D

UBD

UB

]][[

][

From Constitutive law

From kinematic relation

UBDBUKv

T dv }]{[

Basic Solution Scheme

Numerical solutions of problems involving non-linearity usually attemptto replace the continuous non-linear displacement by a series oflinearised increments of displacements.

The challenge is to calculate the nonlinear displacement response using alinear set of equations.

ANSYS uses an iterative process called the Newton-Raphson method.

35

Nonlinear Response

Linear Response

Displacement

External Load

[K] {U} ={F}

Newton-Raphson algorithm

• Applies the load gradually, in increments.

• Performs iterations at each load increment to drive the incremental solution to equilibrium.

• Solves the equation [KT]{Du} = {F} - {Fnr}

[KT] = tangent stiffness matrix

{Du} = displacement increment

{F} = external load vector

{Fnr} = internal force vector

• Iterations continue until {F} - {Fnr} (difference between external and internal Loads) is within a tolerance.

36

Displacement

F

[KT]

1

23

4

equilibrium

iterations

Fnr

Du

Newton-Raphson algorithm

37

Internal forces(1)

Residual(1)

[Kt]

Known equilibrium configuration

Residual(2)

Internal forces(2)

100 N

50 N

100 N

NSUBST,2Level of external forces – constant

D

Fap

The stiffness matrix is updated at every equilibrium iteration.

{R}={Fa} - {Fi} ~ 0

Tolerance & Convergence criteria

The iteration process described in the previous section continues until convergence is achieved.

{R}={Fa} - {Fi} Residual (out-of- balance) force

||R|| < εr.Rref Criterion

|| Δ u | | < εu.Uref

ε R and εu are tolerances (TOLER on the CNVTOL command)

ε R = Defaults to 0.005 (0.05 % ) for force and moment,

εu = Defaults 0.05 (5%) for displacement

38

F

u

Criterion

The solution is converged when Residual < Criterion

Graphical Solution Tracking

39

Convergence criteria guidelines:

Default convergence criteria work well most of the time.

• You should rarely need to change the criteria.

Do not use a “loose” criterion to eliminate convergencedifficulties.

• This simply allows the solution to “converge” to anincorrect result!

Tightening the criterion requires more equilibriumiterations, but more close to actual equilibrium solution.

40

Meshing consideration

Use appropriate element types.

Not all elements support geometric nonlinearities!

Some have no geometric nonlinear capability.

For example, CONTAC52 and PRETS179.

Others have only limited geometric nonlinear capability.

For example, SHELL63 does not support large strain.

Workbench take care of element selection automatically.

Anticipate mesh distortion.

ANSYS shape-checking examines the quality of the mesh prior to the firstiteration.

In a large strain analysis, the mesh can become significantly distorted afterthe first iteration.

Poor element shapes are undesirable in every iteration.

Prevent poor shapes from developing by modifying the original mesh.

Rezoning feature shall be looked at.

41

Meshing consideration…

42

Replace quad elements with two triangles to prevent 180° corner angles.

Output Interpretation

When post processing, realize that:

• ANSYS reports true stress & true strain.

• Calculated nodal displacements are reported in the original directions, becausenodal coordinate system orientations are not updated for large deflections.

• Stress and strain components rotate with most elements, because most elementcoordinate systems follow the element.

43

Y

X

Other learning resources

44

When to do Large deflection analysis ?

Initial guide lines, not always applicable

•In Plates, transverse deflection in a linear analysis is more than about half thethickness of the plate.

•In Beams, if the moment arm is expected to change, amount of stretch orcompression is more or load is expected to change orientation.

•In Solids, most of the time bulky metal structures does not under go largedeformation. However exception are 3D structures made up of Rubber , Plastics,etc.

45

If you are not sure to include or exclude the geometric nonlinear

effects in the simulation, nonlinear calculation is for sure always the

better choice.

However, if a linear calculation has already been performed and the

resulting strains are small for example below 2% the error in the

analysis will be small and acceptable. If more than 5% strain is

indicated a geometric nonlinear analysis should be performed to obtain

better results.

How to Activate GNL analysis ?

ANSYS doesn’t know whether to the problem is linear or nonlinear.

User need to NLGEOM,ON Command to instruct ANSYS to do nonlinearanalysis.

46

Summary Geometric nonlinearities

•Geometric nonlinearities refer to the nonlinearities in the structure due to thechanging geometry as it deflects.

•That is, the stiffness [K] is a function of the displacements {u}.

•GNL is accounted by NLGEOM,ON command.

•GNL problem are solved iteratively using Newton-Rapshon technique.

•Combination of right kinematics, right constitutive relation, right mesh andright solution control parameters need to be used to over come divergenceissues.

47

Topic : Webinar : Linear & NonLinear Buckling Analysis using ANSYS

Date : November 9, 2011

Time : 12:30 am, Eastern Standard Time (New York, GMT-05:00)

Contact : Mahek.Khubchandani@ansys.com

Next Webinar on this series