MAE 323: Lecture 7. Modeling Topics: Part II. Modeling Topics Alex Grishin MAE 323 Lecture 7 FE Modeling Topics: Part 2

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1 Modeling Topics 1

2 Constraints 2

3 So far, we have seen how boundary conditions of the form u i =0, v i =0, θ i =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. 3

4 In general, constraints involve assignments to, or relations among DoFs which must not be violated during solution. Such a relation might look like this: 4u 8u = We might have a constraint that ties a rotational DoFof a beam to a solid, for example: u1 u2 = 2L sinθ z beam 1 2 L L Solid θz u 1 u 2 4

5 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 5

6 1. Direct Constraint Elimination Consider the common case where q=0: [ ] u 0 r Cr Cc = u c 0 Here u r and u c are,respectively, DoFs to be retained and DoFs to be eliminated. Because there are as many DoFs of u c as there are independent equations of constraint, matrix C c is square and non-singular. Solution for u c yields: Where uc = Crc ur C = C C 1 rc c r (2) (3) (4) 6

7 1. Direct Constraint Elimination We now combine the identity ur=ur and (4): u r = uc T u r where Now, if we write the constraintreduced equations as: K ' u ' = f ' I T = C rc Then the transformation, T (5) may be applied to the original partitioned system: K rr K rc ur fr K u = f = = K cr K cc uc fc (5) (6) 7

8 1. Direct Constraint Elimination where: T K ' = T K T T f ' = T f After (6) is solved for u r, Equation (3) yields u c. If q is not equal to zero in (1), additional terms appear in the RHS of (6). 8

9 1. Direct Constraint Elimination If Eq (2) simply sets certain DoFs (u c ) to zero, then C r =0 and C c =I. Hence C rc =0 is this case and Eq (6) is equivalent to discarding rows and columns associated with u c Otherwise, the choice of which DoFs to place in u c is not unique, so the choice of C c is not unique. One might then define C c 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 9

10 1. Direct Constraint Elimination Example 1 Let s take the three-spring system introduced in Chapter 2 and apply equal forces to all three nodes x y u 1,F 1 u 2,F 2 u 3,F k 1 k 2 k 3 Apply forces at nodes 1, 2, and 3 such that F 1 =F 2 =F 3 =F In addition, let k 1 =k 2 =k 2 Now, let s impose the constraint: u 2 =u 3 (7) From Chapter 2, we know the original governing equation is: u1 F1 k u = F u 3 F 3 10

11 1. Direct Constraint Elimination Example 1 Let uc = u 3 which results in the partition: u1 [ 0 1 1] u2 = C u u 3 so that: (8) (9) C C r c = [ 0 1] = 1 and: C = C C = 1 rc c r [ 0 1] (10) 11

12 1. Direct Constraint Elimination Example 1 The transformation, T is given by: 1 0 I T = 0 1 = Crc 0 1 At this point, we could either plug into the resultant form (see Chapter 7) explicitly, OR carry out the calculation: T K ' = T K T T F ' = T F Either way, we get the reduced system: (11) 2 1 K ' = k F ' = F 2 (12) 12

13 1. Direct Constraint Elimination Example 1 We can now solve for u r : ur = K ' f ' = k F and we can use (3) to solve for u c : F 3 F uc = u3 = Crc ur = [ 0 1] = 5 k 5 k and the original constraint condition u 2 =u 3 is immediately verified by inspection. 13

14 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 K u f = C u q = 0 0 (13) Now, use (13) to express the total potential energy of the constrained system: ( ) 0 T T T Π p = + = u K u u f λ C u q (14) 14

15 2. The Method of Lagrange Multipliers where λrepresents 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, λassociated 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, λ. This can be done by applying a generalized form of Castigliano s Second Theorem: Π p = u Π p = λ 0 0 (15) 15

16 2. The Method of Lagrange Multipliers Applying (15) to (14) yields: K C C 0 T u f = λ q (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* 16

17 2. The Method of Lagrange Multipliers 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: [ 0 1 1] u1 u2 = C u u 3 Thus, we have all the information we need to plug values directly into (16) : 17

18 2. The Method of Lagrange Multipliers Example 2 Plugging in values yields: 2k k 0 0 u1 F k 2k k 1 u F K C u f T 2 = = = 0 k k 1 u3 F C 0 λ q λ 0 (17) Solution of (17) yields: u1 3 F / k u 2 5 F / k = u3 5 F / k λ F (18) where the force (λ=-f) may be interpreted as the constraint force associated with a rigid link between nodes 2 and 3 18

19 3. Penalty-Based Methods 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 (18) where αis a diagonal matrix of penalty numbers with units of stiffness, then this expression may be substituted in the potential energy expression (14): ( ) T T T Π = u K u u f + ( α C u) C u q = 0 p (19) One may interpret this substitution as replacing the constraint force with an equivalent stiffness matrix based on the penalty parameter 19

20 3. Penalty-Based Methods Once again, taking the derivative with respect tou (Castigliano s Second Theorem) yields: T ( ) T K + C α C u = f + C α q (20) Thus, if α=0, there is effectively no constraint. The Lagrange Multiplier result is retrieved only in the limit as αapproaches infinity. Numerically, this is not possible and so the penalty method is always approximate. In practice, the value of αmust be such that C T α Cis greater than K, but not so great as to cause numerical difficulties 20

21 3. Penalty-Based Methods 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: [ 0 1 1] u1 u2 = C u u 3 In this case, let α 1 =α 2 =α 3. Upon plugging into (20), this yields: u1 F k α u = F u 3 F (21) 21

22 3. Penalty-Based Methods Example3 The solution to (21) is: u1 3 F / k u 2 5 F / k = u 3 5Fα + 6Fk k( α + k) Thus we see that when α=0, u 3 =6F/k (the same as if no constraint existed between node 2 and 3). As α, u 3 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 22

23 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 23

24 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 24

25 Some Practical Examples Direct Constraint Elimination is often used in cases where one wants to connect DoFsin 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 25

26 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. 26

27 Some Examples From ANSYS Workbench 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 lbfremote force scoped to the end of the beam. Remote force is located 20 inches from the fixed support. F=1 lbf 20 Moment Reaction 27

28 Some Examples From ANSYS Workbench 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). 28

29 Structural Nonlinearities 29

30 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 temperaturedependent 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 Kandf are considered independent of the primary variable u. In nonlinear problems, either Kor fis dependent on u 30

31 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 σ Yield Strength σ y Unloading Elastic Plastic ε 31

32 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) 32

33 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: K u = F F T A nr i i i Displacements are calculated incrementally at iteration, i according to: u = u + u + i 1 i i (22) 33

34 Solving Nonlinear Problems The Newton-Raphson Method In equation (22): K F T i nr i = Tangent Stiffness matrix at iteration i = Vector of restoring loads corresponding to element internal loads T nr Both K i and F i are evaluated based on the values given by u i. 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. 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 u i+1 5. Repeat steps 2 thru 4 until convergence is obtained 34

35 Solving Nonlinear Problems The Newton-Raphson Method 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 35

36 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 underconstrained models. 36

37 Nonlinear Solution Controls in ANSYS Workbench 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). 37

38 Nonlinear Solution Controls in ANSYS Workbench 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 38

39 Nonlinear Solution Controls in ANSYS Workbench 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 39

40 Contact Problems 40

41 Structural Contact Problems 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): e e T e ( ε ) Cδ ε dv = b wdv + FwdS e e V V S 1 where: δε ij = ( wi, j + wj, i ) 2 Also recall that terms involving δrefer 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 T e Π p = ( ε ) Cδε dv b δudv Fδu ds (23) e e e V V S 41

42 Structural Contact Problems Now, structural contact problems involve a new energy term which represents the interaction between two or more bodies. So (23) becomes: e e T e Π p = ( ε ) Cδε dv b δudv FδudS + "contact term" e e e V V S 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 42

43 Structural Contact Problems 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) K 1 K 1 K contact K 2 K 2 43

44 Structural Contact Problems 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: ( α δ + α δ ) "contact term"= ngn gn tgt gt da Lagrange Method: A ( λ δ g + λ δ g ) "contact term"= n n t t Augmented Lagrange Method: A da ([ λ + α ] δ + [ λ + α ] δ ) "contact term"= n ngn gn t tgt gt da A α n F g t αt g n *This description is taken from CADFEM online training documentation 44

45 Structural Contact Problems Here, g n and g t 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 45

46 Structural Contact Problems Example 1: Penalty Method x y u 1 =0 k 1 1 u 2,F 2 u 3,F 3 u 4 =0 2 3 k 2 4 g 0 is the initial geometric gap between nodes 2 and 3 g 0 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): u u 1 F F g = = = = = 1 k = k = k =

47 Structural Contact Problems Example 1: Penalty Method 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: u u 2 F 2 k = u F u4 0 k 1 0 u2 F2 0 1 = u F 3 3 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 u 2 and u 3 are not connected. These are, in effect two different, simultaneous but unrelated spring problems 47

48 Structural Contact Problems Example 1: Penalty Method Now, to apply the constraint, we first go back to the contact term, which includes it remembering to eliminate the frictional term: "contact term"= α ng δ n gnda (24) Here, the normal gap, g n is given by: A ( ) [ 1 1] g g u u g u 2 n = = 0 u3 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 g n is the form C u-q. The constraint we want to enforce is actually g n 0. 48

49 Structural Contact Problems Example 1: Penalty Method 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)): Stiffness Term = C Residual (force) Term = -C T α C T α g 0 (25) 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: T T ( K + C α C) u = f C α g0 The contact stiffness and residual terms only get added to the system equations when: gn 0 (26) 49

50 Structural Contact Problems Example 1: Penalty Method In fact, applying (25) when condition (26) is satisfied makes this a nonlinear constraint. It ensures that g n 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 1 0 α α u F α 2 2 k g = α α u3 F3 α 50

51 Structural Contact Problems Example 1: Penalty Method Let s see how this system behaves qualitatively. When g n >0 (when u 3 -u 2 <g 0 ), the solution behaves the same as if no constraint is present (α=0): u 2 F2 / k = (27) u3 F3 / k But when g n 0 (u 3 -u 2 g 0 ), we have: u1 1 k + α α F2 + α g0 = u k( k + 2 α) α k α + F α g (28) We can easily see that when α=0, (27) is recovered, but as α, (28) becomes singular. Well before this, the matrix becomes too illconditioned to solve with finite precision on digital computers 51

52 Structural Contact Problems Example 1: Penalty Method If we plot u 2 vs. F 2, we see that there is a knee in the curve at F 2 =2. This is when u 1 equals the initial gap. At forces higher than this value, the constraint is engaged and both springs resist F 2. This bilinear curve represents one the simplest kinds of nonlinearity we can encounter in structural mechanics. 52

53 Structural Contact Problems Example 1: Penalty Method Now, let s see how the system behaves quantitatively by plugging in the values from earlier: Since F 2 /k = 5/2 > g 0, 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: u = u g n = u u 1 F F g = 0 = 0 = 5 = 0 = 1 k = k = k = 1 2 α =

54 Structural Contact Problems Example 1: Penalty Method So, with a penalty value of 100 (50 times the spring stiffness), we have a final gap of roughly (a penetration of 1.5 percent the original gap value) If we increase the value to 1000, we get: u = u g n = 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. 54

55 Structural Contact Problems Example 2: Lagrange Method We will demonstrate the Lagrange Method on the same spring system as before. Again, we ignore friction: n "contact term"= λ δ g da n n Once again, we want to apply the constraint: u g = g u u = g We re-write this as: A ( ) [ ] [ ] u g0 u u3 (29) 55

56 Structural Contact Problems Example 2: Lagrange Method 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) k 0 1 u2 F2 T 0 k 1 K C u f u3 = F3 = = λ g C 0 λ q 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 Cand C T 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 56

57 Structural Contact Problems Example 2: Lagrange Method Solving (30) yields: u u3 = 0.75 λ 3.5 The λvariable 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 lbftensile load and 1.5 lbf(5-3.5) compressive load acting on nodes 2 and 3: 57

58 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 58

59 ANSYS Workbench Implementation 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 Contact Asymmetric Contact 59

60 ANSYS Workbench Implementation Five contact types are available: 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 Bondedand No Separation contact are linear and require only 1 iteration. Frictionless,Rough and Frictionalcontact 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) 60

61 ANSYS Workbench Implementation Interface treatment options: C T C T 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. 61

62 ANSYS Workbench Implementation 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 Professionallicenses 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. 62