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Modeling Topics: Part IIMAE 323: Lecture 7

2011 Alex Grishin 1MAE 323 Lecture 7 FE Modeling Topics: Part 2

Modeling Topics



Constraints



•So far, we have seen how boundary conditions of the form ui=0, vi=0, θi=0 may

be applied to solve algebraic structural system equations.

•The setting of nodal DoFs to prescribed values in this manner falls under a

broader category of operations referred to generally as constraints

• A basic feature of FEM is that the system equations as generated do not

automatically contain such constraints within them*

•Frequently, it becomes necessary to specify more complicated constraints than

those we have already seen. So let’s start by seeing what these might look like

*In the mathematical derivations of FE solutions to differential equations, this difficulty is

overcome by specifying that the shape functions must be equal to boundary values at

boundaries.



•In general, constraints involve assignments to, or relations

among DoFs which must not be violated during solution.

•Such a relation might look like this:

2 34 8 8u u− =

•We might have a constraint that “ties” a rotational DoF of a

beam to a solid, for example:

beam

Solid

L L

1 2

1 2 2 sinz

u u L θ− = ⋅

u1 u2

θz



•All such relations have the same basic form:

⋅ =C u q (1)

•In building finite element models, such relations are extremely

useful. So, we’d like some general procedure of handling such

relations once we have assembled the global system equations:

⋅ =K u f

•There are three common methods of handling such relations

in a general way in FEM. They are:

1. Direct Constraint Elimination

2. Methods Based on Lagrange Multipliers

3. Penalty-Based Methods



[ ] r

r c

c

=

u 0C C

u 0

Here ur and uc are ,respectively, DoFs to be retained and DoFs

to be eliminated. Because there are as many DoFs of uc as

there are independent equations of constraint, matrix Cc is

square and non-singular. Solution for uc yields:

c rc r= ⋅u C u

Where

1

rc c r

−= − ⋅C C C

(2)

(3)

(4)

Consider the common case where q=0:




We now combine the identity ur=ur and (4):

r

r

c

= ⋅

uT u

uwhere

rc

=

IT

C(5)

Now, if we write the constraint-

reduced equations as:

' ' '⋅ =K u f

Then the transformation , T (5) may be applied

to the original partitioned system:

rr rc r r

cr cc c c

⋅ = = =

K K u fK u f

K K u f


(6)



where:

'

'

T

T

= ⋅ ⋅

= ⋅

K T K T

f T f

•After (6) is solved for ur, Equation (3) yields uc. If q is not equal

to zero in (1), additional terms appear in the RHS of (6).




•If Eq (2) simply sets certain DoFs (uc) to zero, then Cr=0 and

Cc=I. Hence Crc =0 is this case and Eq (6) is equivalent to

discarding rows and columns associated with uc

•Otherwise, the choice of which DoFs to place in uc is not

unique, so the choice of Cc is not unique. One might then

define Cc to be the last c linearly independent columns of C

•It is possible to apply constraints (1) serially without

partitioning and applying equation (6). All DoFs are retained

in such an approach, but the resulting system may not be

positive definite




Example 1

x

y

u2,F2

u3,F3

k2

k1

k3

1

2

3

Let’s take the three-spring system introduced in Chapter

2 and apply equal forces to all three nodes

u1,F1

•In addition, let k1=k2=k2

•Apply forces at nodes 1, 2, and 3

such that F1=F2=F3=F

1 1

2 2

3 3

2 1 0

1 2 1

0 1 1

u F

k u F

u F

−

− − = −

From Chapter 2, we know the original

governing equation is:

•Now, let’s impose the

constraint: u2=u3(7)




…so that:

Example 1

3c u=u

[ ]1

2

3

0 1 1

u

u

u

− = ⋅

C u

[ ]0 1

1

r

c

=

= −

C

C

…and:

[ ]1 0 1rc c r

−= − ⋅ =C C C

(8)

(9)

(10)


Let

…which results in the partition:



At this point, we could either plug into the resultant

form (see Chapter 7) explicitly, OR carry out the

calculation:

Example 1

•The transformation, T is given by:

Either way, we get the reduced system:

(11)

(12)

1 0

0 1

0 1rc

= =

IT

C

'

'

T

T

= ⋅ ⋅

= ⋅

K T K T

F T F

2 1'

1 1

1'

2

k

F

− = −

=

K

F




Example 1

•We can now solve for ur:

13

' '5

r

F

k

− = ⋅ =

u K f

…and we can use (3) to solve for uc:

[ ]3

30 1 5

5c rc r

F Fu

k k

= = ⋅ = ⋅ =

u C u

…and the original constraint condition u2=u3 is

immediately verified by inspection.




2. The Method of Lagrange Multipliers

•The method of Lagrange multipliers is simple enough to state in

words: In this method, any constraints placed on a system (such

as (1)) must not contribute to the total potential energy of the

system

•To see how this statement is implemented, we start by writing

both the original governing equation AND the constraints

0

0

⋅ − =

⋅ − =

K u f

C u q

(13)

•Now, use (13) to express the total potential energy of the

constrained system:

( ) 0T T T

pΠ = ⋅ ⋅ − ⋅ + ⋅ ⋅ − =u K u u f λ C u q (14)




…where λλλλ represents unknown coefficients with units of force

required to enforce (14).

•Equation (14) may be interpreted as stating that the

requirement that constraints not change the total potential

energy (do no work) entails that there be forces, λλλλ associated

with such constraints . Furthermore, these forces must be solved

explicitly

•We must find a linear form the equations embodying the energy

expression (14) which also includes the variables, λλλλ. This can be

done by applying a generalized form of Castigliano’s Second

Theorem:

p

p

∂Π=

∂

∂Π=

∂

0u

0λ

(15)



•Applying (15) to (14) yields:

T =

u fK C

λ qC 0(16)

•It is apparent that applying (16) increases the matrix

system size by as many rows and columns as there are

constraints

•However, there are situations where application of (16) is

more advantageous than direct constraint elimination. One

such situation is when a single DoF is constrained to many

DoFs. Other situations involve the application of nonlinear

constraints*





Example 2

•In this example, let us re-use the model in example

1—applying the same loads, boundary conditions and

constraints– but this time, we will solve the problem

using the method of Lagrange Multipliers

•From equation (9), we have:

[ ]1

2

3

0 1 1

u

u

u

− = ⋅

C u

•Thus, we have all the information we need

to plug values directly into (16) :



…where the force (λ=-F) may be interpreted as the constraint

force associated with a rigid link between nodes 2 and 3


Example 2

1

2

3

2 0 0

2 1

0 1

0 1 1 0 0

T

k k u F

k k k u F

k k u F

λ

− − − = = = − − −

u fK C

λ qC 0

•Plugging in values yields:

(17)

•Solution of (17) yields:

1

2

3

3 /

5 /

5 /

u F k

u F k

u F k

Fλ

= −

(18)



•One may interpret this substitution as replacing the constraint force

with an equivalent stiffness matrix based on the penalty parameter


•A third method of handling constraints may be derived by an

approximation of the method of Lagrange Multipliers. Namely, if

one introduces the approximation:

= ⋅ ⋅λ α C u

where αααα is a diagonal matrix of “penalty” numbers with units of

stiffness, then this expression may be substituted in the potential

energy expression (14):

( )( ) 0T T T

pΠ = ⋅ ⋅ − ⋅ + ⋅ ⋅ ⋅ ⋅ − =u K u u f α C u C u q

(18)

(19)




•Once again, taking the derivative with respect to u (Castigliano’s

Second Theorem) yields:

( )T T+ ⋅ ⋅ ⋅ = + ⋅ ⋅K C α C u f C α q

•Thus, if α=0, there is effectively no constraint. The Lagrange

Multiplier result is retrieved only in the limit as α approaches

infinity. Numerically, this is not possible and so the penalty

method is always approximate. In practice, the value of α must

be such that CT⋅αααα⋅C is greater than K, but not so great as to cause

numerical difficulties

(20)




Example3

•The model used previously will be re-used a third

time to demonstrate the use of the penalty method

of applying constraints

•As before, the constraint (9), C is re-used:

[ ]1

2

3

0 1 1

u

u

u

− = ⋅

C u

•In this case, let α1=α2=α3. Upon plugging into (20), this

yields:

1

2

3

2 1 0 0 0 0

1 2 1 0 1 1

0 1 1 0 1 1

u

k u

u

α

−

− − + − = − −

F

F

F

(21)




Example3

•The solution to (21) is:

1

2

3

3 /

5 /

5 6

( )

u F k

u F k

u F Fk

k k

α

α

= + +

•Thus we see that when α=0, u3=6F/k (the same as if no constraint

existed between node 2 and 3). As α→∞, u3→5F/k, as expected (the

same as previous solutions)

•Although the Penalty Method is approximate, it does not increase the

number of DoFs in the system of equations (although it may increase

matrix bandwidth), and if the penalty values are chosen carefully, the

resulting equations may have desirable convergence properties in

numerical solutions of nonlinear problems



Summary

•We have reviewed the three main methods of applying constraint

conditions to the numerical form of problems arising in structural

mechancs. We might summarize the pros and cons of each as follows:

1. Direct Constraint Elimination: This method splits a system of

equations into constrained and retained degrees of freedom.

Athough it may seem that the resulting system to be solved is

smaller, this is not true. In fact, direct constraint elimination

may be cumbersome and inefficient to implement in real

applications. This makes it suitable for only for linear problems

in which the total number of constrained DoFs is relatively

small. The advantage of the technique, however, lies in its

accuracy and ability to simply handle constraints of arbitrary

complexity



Summary

2. The Method of Lagrange Multipliers: This method is simpler to

implement numerically, but increases the number of DoFs to be

solved (one for each DoF constrained). It’s advantage over direct

constraint elimination lies in its ability to handle nonlinear problems

more easily. However, special nonlinear routines must be used in

such applications (such as constraint optimization routines) and

convergence may be difficult. In small linear problems, this method

may be generally preferable to direct constraint elimination

(commercial codes seem divided over this).

3. The Penalty Method: This method is approximate, and so is not the

best candidate for small linear problems where accuracy is

important. However, it’s straightfoward implementation in nonlinear

problems makes it a top choice for such applications



Some Practical Examples

•Direct Constraint Elimination is often used in cases where one wants to

connect DoFs in some fashion within a purely linear analysis (it becomes

very expensive for nonlinear analysis)

•A rather iconic example from structural analysis would be the modeling

of fasteners with beams connected to solid elements (see slide 4)

•Another example would be bonded contact between two surfaces, but

we will cover this later when we discuss contact



Some Examples From ANSYS Workbench

• Moment Loading :

– For solid bodies moments can be applied on a surface only.

– If multiple surfaces are selected, the moment load is evenly distributed.

– Vector or component method can be employed using the right hand rule.

– For surface bodies a moment can be applied to a vertex, edge or surface.

– Units of moment are in Force*length.




• Remote Force Loading :– Applies an offset force on a surface or edge of a body.

– The user supplies the origin of the force (geometry or coordinates).

– Can be defined using vector or component method.

– Applies an equivalent force and moment on the surface.

– Example: 10 inch beam with a 1 lbf remote force scoped to the end of the beam. Remote force is located 20 inches from the fixed support.

20”

F=1 lbf

Moment Reaction




• Bolt Pretension:– Applies a pretension load to a solid cylindrical section or beam using:

• Pretension load (force)

• OR

• Adjustment (length)

– For body loading a local coordinate system is required (preload in z direction).

– For sequenced loading additional options are available (see next page).



Structural

Nonlinearities



•In structural mechanics, a problem is nonlinear if the stiffness matrix

OR the load vector depend on the displacements. Nonlinearities can be

classified as material nonlinearity (associated with changes in material

properties, as in plasticity), or geometric nonlinearity (associated with

changes in configuration, as in large deflections of a slender elastic

beam).

•In thermal problems, nonlinearity can arise from temperature-

dependent conductivity, convection, or heat flux. An example of a

nonlinear heat flux would be that associated with radiation

⋅ =K u f

•To put more precisely: in linear problems, both K and f are

considered independent of the primary variable u. In nonlinear

problems, either K or f is dependent on u



Material Nonlinearity: Example—Plasticity

•When a ductile material is stressed beyond it’s elastic limit, it will

yield, acquiring permanent deformation

•The response of the material to loads beyond this limit is classified

as plastic

•Plasticity is brought about at the microstructural level by crystalline

planes slipping past one another. This behavior is mostly driven by

shear, so that volumetric strains are small if present at all

ε

σ

Yield Strength σy

Elastic Plastic

Unloading



Geometric Nonlinearity: Example—Large Deflection

•If a structure experiences large deformations, its changing geometric

configuration can cause the structure to respond nonlinearly. This happens

as a result of either changing load magnitudes and orientations (relative to

the deforming material), or changing stiffness, or both

•In the figure below*, the tip deflection of a fishing pole under a constant

weight is a nonlinear function of the rod’s flexure

*taken with permission from ANSYS 13.0 help documentation ( 2010 SAS IP, Inc)



Solving Nonlinear Problems – The Newton-Raphson Method

•Since the applied load or stiffness is no longer independent of the primary

(displacement) variable, nonlinear problems are usually solved incrementally

as progressively changing linear problems – each subsequent iteration

depending on the result of the last

•For static structural problems, the most common method of doing this is

the Newton-Raphson Procedure

•In this procedure, one equilibrium iteration may be written as:

T A nr

i i i⋅ ∆ = −K u F F

•Displacements are calculated incrementally at iteration, i

according to:

1i i i+ = + ∆u u u

(22)




In equation (22):T Tangent Stiffness matrix at iteration i

Vector of restoring loads corresponding to element internal loads

i

nr

i

=

=

K

F

•Both and are evaluated based on the values given by ui.

The RHS of (22) is called the residual or out-of-balance load vector

(the amount the system is out of equilibrium. A single solution

iteration is depicted graphically below for a one DOF model.

T

iK

nr

iF

•The steps performed in the Newton-Raphson procedure are

described below:

1. Assume u0. This is usually 0

2. Compute the updated tangent matrix and restoring load

from ui

3. Calculate ∆ui from (22)

4. Add ∆ui to ui in order to obtain next iteration ui+1

5. Repeat steps 2 thru 4 until convergence is obtained



•In the preceding, convergence (step 5) is determined based on a residual

criterion. In principle, this should be zero, but in practice, small positive

numerical values are used based on numerical considerations for efficiency

•Below left, the first iteration of a NR run is shown. Below right, after 2

iterations




Nonlinear Solution Controls in ANSYS Workbench

• The “Analysis Settings” details provide general control over the solution process:

• Step Controls:

– Manual and auto time stepping controls.

– Specify the number of steps in an analysis and an end “time” for each step.

– “Time” is a tracking mechanism in static analyses (discussed later).

• Solver Controls:

– Two solvers available (default program chosen):

• Direct solver (Sparse solver in ANSYS).

• Iterative solver (PCG solver in ANSYS).

– Weak springs:

• Mechanical tries to anticipate under-constrained models.




• Analysis Data Management:

– Solver Files Directory is the location where analysis files will be stored if a project has not yet been saved.

– Future Analysis: indicates whether a down stream analysis (e.g. pre-stressed modal) will use the solution. This is set automatically when coupled analyses are configured in the project schematic.

– Scratch Solver Files Directory: temporary directory used during solution.

– Save MAPDL db.

– Delete Unneeded Files: may choose to save all files for future use in Mechanical APDL.

– Solver Units: Active System or manual.

– Solver Unit System: if the above setting is “manual”, you may choose 1 of 8 possible solver unit systems to insure consistency when data is shared with Mechanical APDL (does not affect results/load displays in the GUI).




• Step Controls:

– Multiple steps allow a series of static analyses to be set

up and solved sequentially.

– For a static analysis, the end time can be used as a

counter/tracker to identify the load steps and substeps.

– Results can be viewed step by step.

– Load values for each step can be entered in the “Tabular

Data” section provided.

The time and load value are

displayed in the graphics

window




• Results for each individual step can be viewed after the solution by selecting the desired step and RMB >“Retrieve This Result”.

Select desired step

and RMB to

retrieve result



Contact Problems



( )e e

ee T e

SV V

dV wdV wdSδ = +∫ ∫ ∫ε C ε b F

•First, recall Equation (5) from Chapter 5. This is the general

governing equation for an elastic continuum in weak form for

a single element (hence the superscript e):

( ), ,

1

2ij i j j i

w wδε = +where:

•Also recall that terms involving δ refer to the “variation” which we

are trying to minimize (in this case, by using a trial function for

displacement, w). In what follows, we’ll keep everything in terms of

δ to maintain consistency (we wont’ try to solve these equations—

just use their form to explain contact problems)

•Let’s move everything to the LHS to show that this equation

represents total potential energy, and sum over all elements:

( )e e

ee T e

p

e SV V

dV dV dSδ δ δ Π = − −

∑ ∫ ∫ ∫ε C ε b u F u (23)

Structural Contact Problems




•Now, structural contact problems involve a new energy term

which represents the interaction between two or more bodies. So

(23) becomes:

( ) "contact term"e e

ee T e

p

e SV V

dV dV dSδ δ δ Π = − − +

∑ ∫ ∫ ∫ε C ε b u F u

•This “contact term” is most easily expressed in terms of the

constraints we have already seen. However, such constraints are

typically nonlinear (this means that Direct Constraint Elimination is

usually not used).

•Most commercial codes give the user a choice (or perhaps choose

one for you) between the following three popular options:

•Penalty Methods

•Lagrange Methods

•Augmented Lagrange Methods



•To keep things simple, consider contact between two bodies

(labeled 1 and 2). The new contact term would produce a stiffness

matrix connecting the two bodies (because it involves a constraint

relating the DoFs between the two bodies)

K1

K2

K1

K2

Kcontact




•The Penalty and Lagrange Methods are the same as we’ve already

seen. Many commercial codes offer a hybrid combination between the

two, however, called the Augmented Lagrange Method (This is to

combine the best of both approaches). Below we summarize the form

of the contact term for each of the three methods*:


Penalty Method:

( )"contact term"= n n n t t t

A

g g dAα δ α δ+∫ g g

Lagrange Method:

( )"contact term"= n n t t

A

g dAλ δ λ δ+∫ g

Augmented Lagrange Method:

[ ] [ ]( )"contact term"= n n n n t t t t

A

g g dAλ α δ λ α δ+ + +∫ g g

*This description is taken from CADFEM online training documentation

F

gn

gt

αn αt



•Here, gn and gt are the normal and tangential contact “gaps” between

the two bodies. The tangential component is only needed for frictional

contact, and it should be noted that this is a tensor containing a static

and kinematic term (to include a stick and slip condition).

•Let us simplify the presentation further by focusing only on frictionless,

or normal contact




x

y

u1=0 k1

1

u4=0

k2

4

3u2,F2


Example 1: Penalty Method

2

u3,F3

g0

•To see how the Penalty Method

works in a problem, consider the 2-

spring system at left. The springs are

fixed at nodes 1 and 4. The following

summarizes the properties of the

system (the units are arbitrary):

1

4

2

3

0

1 2

0

0

5

0

1

2

u

u

F

F

g

k k k

=

=

=

=

=

= = =

•g0 is the “initial” geometric gap

between nodes 2 and 3





•We can use our knowledge of the matrix equation of a 1-dimensional

(canonical) spring element, together with the Direct Stiffness Method

of matrix addition (Chapter 2) to derive the basic matrix equation:

1

2 2

3 3

4

1 1 0 0 0

1 1 0 0

0 0 1 1

0 0 1 1 0

u

u Fk

u F

u

−

− = − −

2 2

3 3

1 0

0 1

u Fk

u F

=

•We can eliminate the rows and columns associated with the fixed DoFs

(1 and 4). This leaves us with the unconstrained equation. Note,

because of the gap, DoFs u2 and u3 are not “connected”. These are, in

effect two different, simultaneous but unrelated spring problems





•Now, to apply the constraint, we first go back to the “contact

term”, which includes it – remembering to eliminate the frictional

term:"contact term"= n n n

A

g g dAα δ∫

•Here, the normal gap, gn is given by:

( ) [ ] 2

0 3 2 0

3

1 1n

ug g u u g

u

= − − = − −

•Since there is no contact area (the contact region is just a point),

we don’t’ have to perform the integral in (24).

•We see now that gn is the form C⋅u-q. The constraint we want to

enforce is actually gn≥0.

(24)





•Upon finding the first variation of (24), we have a stiffness term and a

residual (In other words, after taking the derivative with respect to δu.

Note that this yields an equation identical to equation (20)):

0

Stiffness Term =

Residual (force) Term = -

T

Tg

α

α

⋅ ⋅

⋅ ⋅

C C

C

•Here we are assuming a single scalar penalty stiffness, α.

These terms may now be added to the unconstrained

equations in a Newton-Raphson scheme:

( )0

T Tgα α+ ⋅ ⋅ ⋅ = − ⋅ ⋅K C C u f C

•The contact stiffness and residual terms only get added to the

system equations when: 0

ng ≤

(25)

(26)





•In fact, applying (25) when condition (26) is satisfied makes this a nonlinear

constraint. It ensures that gn≥0 within numerical bounds determined by the

numerical properties of the system equations with the inclusion of α

•Applying these constraints to our system equations results in

2 2

0

3 3

1 0

0 1

u Fk g

u F

α α α

α α α

− − + = −

−





•Let‘s see how this system behaves qualitatively. When gn >0

(when u3-u2<g0), the solution behaves the same as if no constraint

is present (α=0):

2 2

3 3

/

/

u F k

u F k

=

•But when gn≤0 (u3-u2≥g0), we have:

2 01

3 02

1

( 2 )

F gu k

F gu kk k

αα α

αα αα

++ = −++

•We can easily see that when α=0, (27) is recovered, but as α→∞,

(28) becomes singular. Well before this, the matrix becomes too ill-

conditioned to solve with finite precision on digital computers

(27)

(28)





•If we plot u2 vs. F2, we see that there is a “knee” in the curve at

F2=2 . This is when u1 equals the initial gap. At forces higher than

this value, the constraint is engaged and both springs resist F2. This

bilinear curve represents one the simplest kinds of nonlinearity we

can encounter in structural mechanics.





•Now, let’s see how the system behaves quantitatively by plugging

in the values from earlier:

1

4

2

3

0

1 2

0

0

5

0

1

2

100

u

u

F

F

g

k k k

α

=

=

=

=

=

= = =

=

•Since F2/k = 5/2 > g0, equation (28) is used.

In most commercial systems, this check

would be performed in the first iteration of

a nonlinear Newton-Raphson routine

•Plugging values at right into (28) yields:

2

3

1.7574

.74257

u

u

=

gn=-0.01485





•So, with a penalty value of 100 (50 times the spring stiffness), we

have a final gap of roughly -0.015 (a penetration of 1.5 percent the

original gap value)

•If we increase the value to 1000, we get:

2

3

1.7508

.74925

u

u

=

gn=-0.001499

•This behavior is characteristic of the penalty method. Namely, one needs

fairly large penalty values to enforce the constraint with high accuracy. The

inaccuracy is manifested by an unavoidable penetration. If one wants

increased accuracy, the Lagrange or Augmented Lagrange Method must be

used.




Example 2: Lagrange Method

•We will demonstrate the Lagrange Method on the same

spring system as before. Again, we ignore friction:

"contact term"= n n

A

g dAλ δ∫

( ) [ ] 2

0 3 2 0

3

1 1 0n

ug g u u g

u

= − − = − − ≥

•Once again, we want to apply the constraint:

•We re-write this as:

[ ] 2

0

3

1 1u

gu

− ≥ −

(29)





•This time, we’ll save a little effort and note that we have

everything we need to plug into equation (17) (just as we could

have plugged directly into equation (20) in the previous example)

2 2

3 3

0

0 1

0 1

1 1 0

Tk u F

k u F

gλ

− = = = − −

u fK C

λ qC 0(30)

•However, we should note a subtlety. Equation (29) involves an

inequality, which we have simply ignored in (30). In real

implementations of the Lagrange Method, we can’t do this. And we

can’t just add in the terms C and CT when the gap reaches a threshold

with many solvers. This difficulty reflects the fact that this is really a

constraint optimization problem, and some more sophisticated

techniques are required to enforce (29). In any case, we will ignore this

for now and pretend that we already know the gap is closed





•Solving (30) yields:

2

3

1.75

0.75

3.5

u

u

λ

=

•The λ variable gives us the constraint force, which we know

must act opposite the applied load. This reveals a benefit of

the Lagrange Method: It automatically provides some useful

reaction force information. For example, we immediately

know that that there is a 3.5 lbf tensile load and 1.5 lbf (5-3.5)

compressive load acting on nodes 2 and 3:



ANSYS Workbench Implementation

• When importing assemblies of solid parts, contact regions are automatically created between the solid bodies.

– Contact allows non-matching meshes at boundaries between solid parts

– Tolerance controls under “Contact” branch allows the user to specify distance of auto contact detection via slider bar




• In Mechanical, the concept of contact and target surfaces are used for each contact region:

– One side of a contact region is referred to as a contact surface, the other side is referred to as a

target surface.

– The contact surfaces are restricted from penetrating through the target surface.

• When one side is designated the contact and the other side the target, this is called asymmetric contact.

• If both sides are made to be contact & target this is called symmetric contact.

• By default, Mechanical uses symmetric

contact for solid assemblies.

• For ANSYS Professional licenses and

above, the user may change to

asymmetric contact, as desired.

Symmetric

ContactAsymmetric

Contact




• Five contact types are available:

– Bonded and No Separation contact are linear and require only 1

iteration.

– Frictionless, Rough and Frictional contact are nonlinear and

require multiple iterations.

• Nonlinear contact types allow an “interface treatment”

option:

• “Add Offset”: input zero or non-zero value for initial adjustment

• “Adjusted to Touch”: ANSYS closes any gap to a just touching position

(ANSYS Professional and above)

Contact Type Iterations Normal Behavior (Separation) Tangential Behavior (Sliding)

Bonded 1 No Gaps No Sliding

No Separation 1 No Gaps Sliding Allowed

Frictionless Multiple Gaps Allowed Sliding Allowed

Rough Multiple Gaps Allowed No Sliding

Frictional Multiple Gaps Allowed Sliding Allowed




• Interface treatment options:

Add offset: contact surface is numerically

offset a given amount in positive or negative

direction (offset can be ramped on).

Adjusted to touch: offsets contact surface

to provide initial contact with target

regardless of actual gap/penetration.

TCC T




• Advanced options (see chapter 3 for additional details on the pinball region):– Pin Ball Region:

• Inside pinball = near-field contact

• Outside pinball = far-field contact

• Allows the solver to more efficiently process contact calculations.

• For ANSYS Professional licenses and above, mixed assemblies of shells and solids are supported as well as more contact options.

In this case, the gap between

the two parts is bigger than the

pinball region, so no automatic

gap closure will be performed.

lecture7 fe modeling topics part ii€¦ · mae 323: lecture 7 modeling topics: part ii 2011 alex...

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