Compression test modeling in Diana

COMPUTATIONAL MODELING OF STRUCTURES CASE STUDY Group 10

Maria Luisa Rosales Rob Smeekens

TABLE OF CONTENTS 1. INPUT....................................................................................................................................................... 3 Material Properties..................................................................................................................................... 3 Element Properties..................................................................................................................................... 3 Element Dimensions............................................................................................................................... 3 Finite Element Mesh................................................................................................................................ 3 Boundary Conditions.................................................................................................................................. 4 Constraints (at the bottom).................................................................................................................... 4 Constraints (at the top)........................................................................................................................... 4 Applied Loads......................................................................................................................................... 4 Prescribed Displacements....................................................................................................................... 5 2. ANALYSIS RESULTS................................................................................................................................. 6 2.1. Structural Linear Analysis............................................................................................................. 6 Reaction Forces....................................................................................................................................... 6 Displacements (Y direction, deformed shape)......................................................................................... 6 Displacements (Z direction, deformed shape)........................................................................................ 6 Stresses SZZ (outer face, deformed shape)............................................................................................ 6 Stresses SZZ (inner face, deformed shape)............................................................................................ 7 Strains EZZ (outer face, deformed shape).............................................................................................. 7 Strains EZZ (inner face, deformed shape).............................................................................................. 7 2.2. Non-Linear Analysis........................................................................................................................ 8 2.2.1. Physical Nonlinearity........................................................................................................................ 8 Displacements (Z direction, deformed shape)........................................................................................ 9 Displacements (Y direction, deformed shape)......................................................................................... 9 Stresses SZZ (outer face, Deformed Shape)........................................................................................... 9 Stresses SZZ (inner face, Deformed Shape).........................................................................................10 Strains EZZ (outer face, Deformed Shape)........................................................................................... 10 Strains EZZ (Inner face, Deformed Shape)............................................................................................10 2.2.2. Physical + Geometrical Nonlinearity.............................................................................................. 11 1

Displacements (Z direction, deformed shape)......................................................................................12 Displacements (Y direction, deformed shape).......................................................................................12 Stresses SZZ (outer face, Deformed Shape)......................................................................................... 12 Stresses SZZ (Inner face, Deformed Shape).........................................................................................13 Strains SZZ (outer face, Deformed Shape)...........................................................................................13 Strains SZZ (Inner face, Deformed Shape)............................................................................................ 13 2.2.3. Physical + Geometrical Nonlinearity (with point loads)..................................................................14 Displacements (Z direction, deformed shape)......................................................................................15 Displacements (Y direction, deformed shape).......................................................................................15 Stresses SZZ (outer face, Deformed Shape)......................................................................................... 16 Stresses SZZ (Inner face, Deformed Shape).........................................................................................16 Strains SZZ (outer face, Deformed Shape)...........................................................................................16 Strains SZZ (Inner face, Deformed Shape)............................................................................................ 17 3. COMMENTS AND CALCULATIONS....................................................................................................... 18 3.1. Hand Calculations............................................................................................................................. 18 3.2. Comments......................................................................................................................................... 21 3.3 Conclusions........................................................................................................................................ 24

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1. INPUT Dimension Assumed Stress Field

(Quadrilateral) 2D (Plane Stress)

Fig. 1.2 – CQ40S Element

ELEMENT DIMENSIONS As shown in the model Sketch.

Fig. 1.1 – Model Sketch The objective is to model a compression test that was done on an aluminium extrusion with an imperfection on its center.

MATERIAL PROPERTIES Name Type Elastic Modulus Poisson’s Ratio Model Type Initial Yield Stress

Aluminium Isotropic 70000 0.33 Von Mises 200

FINITE ELEMENT MESH In order to have the same mesh size for all the element, we chose a size of 5 mm.

N/mm2 N/mm2

ELEMENT PROPERTIES Name Data Name Type DIANA Element Type Nodes Degrees of Freedom Integration Scheme Type of Integration Type of Interpolation Shape Dimension Topological

ext ext Curved Shell (Hyperbolic) CQ40S 8 5x 8 Nodes = 40 D.O.F. (Displacements and Rotations) 2 x 2 x 3 = 12 Integration points Numerically Integrated 2th order (quadratical) 3D (Curved)

Fig. 1.3 – Finite Element Mesh – Size = 5 mm

2D 3

4

BOUNDARY CONDITIONS CONSTRAINTS (AT THE BOTTOM) For surfaces normal to the X Axis: UX, UZ, RX, RY, RZ for each node. For surfaces normal to the Y Axis: UY, UZ, RX, RY, RZ for each node.

Doing a preliminary linear analysis with a random load we can interpolate the load required to get such displacement. From this analysis we get a distributed load qa=0.038 N/mm2 It is important to take into account that in this preliminary analysis we have to set a constraint UZ for the top part of the extrusion. This is because the prescribed displacements that are going to be applied in the analysis constraint the extrusion in this zone for this direction (putting a constraint in Z is the same as putting a displacement with a value of zero in the model, which is the initial step for the nonlinear analysis)

Fig. 1.4 – Constraints at the bottom of the element

CONSTRAINTS (AT THE TOP) For surfaces normal to the X Axis: UX, UZ, RX, RY, RZ for each node. For surfaces normal to the Y Axis: UY, UZ, RX, RY, RZ for each node.

Fig. 1.6 – Applied distributed load

Fig. 1.5 – Constraints at the top of the element

APPLIED LOADS To simulate an imperfection at the center of the element, we apply a distributed outward area load qa over an area of 40x40 mm in order to have a displacement of 0.1 mm.

5

Fig. 1.7 – Resulting Displacement for qa=0.038 N/mm2

6

PRESCRIBED DISPLACEMENTS

Fig. 1.8 – Prescribed Displacement

Using elastic and plastic theory we obtain an approximate value of 0.674 mm for the maximum elastic capacity (see calculations in part 3). In order to be able to observe the plastic behavior of the element after reaching these limit, we apply a prescribed displacement of 2 mm on each node all over the top of the element.

7

2. ANALYSIS RESULTS 2.1. STRUCTURAL LINEAR ANALYSIS Since there are no load steps on the linear analysis, we apply both the imperfection load qa and the prescribed displacement in the same Load Case.

REACTION FORCES Total Reaction Force on top= 121737.7

Fig. 2.1.1 – Reaction Forces FB(Y) ; N

DISPLACEMENTS (Y DIRECTION , DEFORMED SHAPE )

8

Fig. 2.1.2 – Displacements DyY(V) ; mm

Fig. 2.1.4 –Stresses SZZ Outer Face ; N/mm2 Max Stress SXX= 790 N/mm2 (>>> Yield Strenght)

STRESSES SZZ (INNER FACE, DEFORMED SHAPE ) DISPLACEMENTS (Z DIRECTION, DEFORMED SHAPE )

Fig. 2.1.5 –Stresses SZZ Inner Face ; N/mm2 Max Stress SXX= 696 N/mm2 (>>> Yield Strenght)

Fig. 2.1.3 – Displacements DyZ(V) ; mm

STRAINS EZZ ( OUTER FACE, DEFORMED SHAPE )

STRESSES SZZ (OUTER FACE, DEFORMED SHAPE )

Fig. 2.1.6 –Strains EZZ Outer Face ; N/mm2

9

Fig. 2.1.7 –Strains EZZ Outer Face ; N/mm2

STRAINS EZZ ( INNER FACE, DEFORMED SHAPE )

10

2.2. NON-LINEAR ANALYSIS 2.2.1. PHYSICAL NONLINEARITY In order to simulate the imperfection in the behavior of the extrusion, we set up two load cases and subsequently two execute blocks: one for the applied load qa and other for the prescribed displacement. In the non-linear analysis, we first apply the distributed load qa at once and then we apply the prescribed displacement using load steps.

60000 50000 40000 Reaction Force N

0 0 0.5 1 1.5 2 2.5 Displacement mm

Class: Steel Material Model: Von Mises and Tresca Plasticity Hardening Function: No hardening

Fig. 2.2.1.1 – Force vs. Displacement plot – Physical Nonlinearity

Iteration and Solution Properties (For applied load execute block)

Iteration and Solution Properties (For prescribed displacement execute block)

20000 10000

Material additional Properties for Non-linear Analysis

Equilibrium iteration method: NewtonRaphson Type: Regular First Tangent: Tangencial Convergence Norm: Energy Based Convergence Tolerance: 0.0001 Number of Steps: 1 Step Size: 1 Max. Number of Iterations: 10

30000

Max Force ≈ 51900 N at δ=0.90 mm In order to show properly the behaviour of the extrusion in the different steps of the analysis, we show the output of it in the following steps:     

Step 3 (Elastic Behaviour) δ=0.40 mm Step 10 (Elastic Behaviour, just before yielding) δ=0.75 mm Step 25 (Yielding, Peak Load) δ=0.9 mm Step 40 (Yielding, Stable load) δ=1.05 mm Step 60 (Yielding, Stable load) δ=1.25 mm

Equilibrium iteration method: NewtonRaphson Type: Regular First Tangent: Tangencial Convergence Norm: Energy Based Convergence Tolerance: 0.0001 Number of Steps: 134 Step Size: 0.10 (3). 0.05(1), 0.005(130) Max. Number of Iterations: 10 Fig. 2.2.1.2 – Location of characteristic steps

DISPLACEMENTS (Z DIRECTION, DEFORMED SHAPE)

Fig. 2.2.1.3 – Displacements DyZ(V) for steps 3,10,25,40,60 ; mm

DISPLACEMENTS (Y DIRECTION , DEFORMED SHAPE)

Fig. 2.2.1.4 – Displacements DyY(V) for steps 3,10,25,40,60 ; mm

STRESSES SZZ (INNER FACE, DEFORMED SHAPE)

Fig. 2.2.1.5 –Stresses SZZ Outer Face for steps 3,10,25,40,60; N/mm 2

STRESSES SZZ (OUTER FACE, DEFORMED SHAPE)

Fig. 2.2.1.6 –Stresses SZZ Inner Face for steps 3,10,25,40,60; N/mm 2

STRAINS EZZ (INNER FACE, DEFORMED SHAPE)

Fig. 2.2.1.7 –Strains EZZ Outer Face for steps 3,10,25,40,60; N/mm 2

STRAINS EZZ (OUTER FACE, DEFORMED SHAPE)

Fig. 2.2.1.8 –Strains EZZ Inner Face for steps 3,10,25,40,60; N/mm 2

2.2.2. PHYSICAL + GEOMETRICAL NONLINEARITY

In order to model the buckling of the section that is shown in the experiment, we add geometrical nonlinear effects to the analysis.

50000 40000 30000

Material additional Properties for Non-linear Analysis

Reaction Force N 20000 10000

Class: Steel Material Model: Von Mises and Tresca Plasticity Hardening Function: No hardening

0 0 0.5 1 1.5 2 2.5

Iteration and Solution Properties (For applied load execute block) Equilibrium iteration method: NewtonRaphson Type: Regular First Tangent: Tangencial Convergence Norm: Energy Based Convergence Tolerance: 0.0001 Number of Steps: 1 Step Size: 1 Max. Number of Iterations: 10 Iteration and Solution Properties (For prescribed displacement execute block) Equilibrium iteration method: NewtonRaphson Type: Regular First Tangent: Tangencial Convergence Norm: Displacement Based Convergence Tolerance: 0.01 Number of Steps: 134 Step Size: 0.10 (3). 0.05(1), 0.005(130) Max. Number of Iterations: 100

Displacement mm

Fig. 2.2.2.1. – Force vs. Displacement plot – Physical + Geometrical Nonlinearity

Max Force = 43520 N at δ=0.73 mm (Step 8) In order to show properly the behaviour of the extrusion in the different steps of the analysis, we show the output of it in the following steps:  Step 2 (Elastic Behaviour) δ=0.20 mm  Step 4 (Elastic Behaviour, just before yielding) δ=0.60 mm  Step 8 (Yielding, Peak Load) δ=0.73 mm  Step 20 (Yielding, decreasing Load) δ=0.85 mm  Step 60 (Yielding, decreasing load) δ=1.25 mm

Geometrical Settings Type of geometrical nonlinearity: Total Lagrange

50000 40000 30000 Reaction Force N 20000 10000 0 0 0.5 1 1.5 2 2.5 Displacement mm

Fig. 2.2.2.2. – Location of characteristic steps




Fig. 2.2.2.4 – Displacements DyY(V) for steps 2,4,8,20,60; mm





STRAINS SZZ (INNER FACE, DEFORMED SHAPE)

Fig. 2.2.2.7 –Strains EZZ Outer Face for steps 2,4,8,20,60

STRAINS SZZ (OUTER FACE, DEFORMED SHAPE)

Fig. 2.2.2.8 –Strains EZZ Inner Face for 2,4,8,20,60

2.2.3. PHYSICAL + GEOMETRICAL NONLINEARITY (WITH

POINT LOADS)

In order to find the effects of the distribution of load that simulate the imperfection, and also to have a better possibility to check the experiment and the modeling by hand calculation, we apply a point load instead of distrubted load. To simulate an imperfection at the center of the element, we apply a point load in the center of each flange (instead of applying it directly in the web) in order to have a displacement of 0.1 mm. Doing a preliminary linear analysis with a random load we can interpolate the load required to get such displacement. From this analysis we get a point load of 260 N on each flange.

Fig. 2.2.3.2. – Resulting displacement for point load of 260 N on each flange

The non linear analysis is done using the same properties for the Material, Iteration and Solution and Geometrical Effects as in the previous analysis 50000 40000 30000 Reaction Force N

20000 10000

Fig. 2.2.3.1. – Applied point load of 260 N on each flange

0 0 0.5 1 1.5 2 2.5 Displacement mm

Fig. 2.2.3.3. – Force vs. Displacement plot – Physical Nonlinearity + Geometrical Nonlinearity with point loads

Max Force = 43825 at δ=0.73 mm (Step 8)

In order to show properly the behaviour of the extrusion in the different steps of the analysis, we show the output of it in the following steps:     

Step Step Step Step Step

2 (Elastic Behaviour) δ=0.20 mm 4 (Elastic Behaviour, just before yielding) δ=0.60 mm 8 (Yielding, Peak Load) δ=0.73 mm 20 (Yielding, decreasing Load) δ=0.85 mm 60 (Yielding, decreasing load) δ=1.25 mm


20000 10000 0 0 0.5 1 1.5 2 2.5 Displacement mm

Fig. 2.2.3.4. – Location of characteristic steps




Fig. 2.2.3.6 – Displacements DyY(V) for steps 2,4,8,20,60; mm





STRAINS SZZ (INNER FACE, DEFORMED SHAPE)

Fig. 2.2.3.8 –Strains EZZ Outer Face for steps 2,4,8,20,60

STRAINS SZZ (OUTER FACE, DEFORMED SHAPE)

Fig. 2.2.3.9 –Strains EZZ Inner Face for steps 2,4,8,20,60

3. COMMENTS AND CALCULATIONS 3.1. HAND CALCULATIONS

In order to perform the respective hand We idealize this situation as a column with length L=300 clamped on top and bottom, with an applied horizontal load on its center and subsequent prescribed displacements. From the geometrical properties of the section we can calculate the Elastic Moment and the Plastic moment. The elastic moment is the moment due to the applied force that causes a displacement of 0.1 mm, which can be calculated with elastic theory. After that we calculate the elastic moment using equilibrium.

F=

δ∗192∗E∗I 3 L M el=

FL 8

Elastic cross section b h L t E Atot fy;d

80 25 300 2 70000 252 200

mm mm mm mm N/mm2 mm2 N/mm2

z0 zb h

5,56 19,44 25,00

mm mm + mm

I1 I2 I

3216 10020 13236

mm4 mm4 + mm4

M0;el σb;1 σb;2

24707 -36,28 10,39

Nmm N/mm2 N/mm2

u

0,1

mm

Fhor

659

N

1/8FL M*zb/I M*zo/I

u*192EI/ L3

Table 3.1.1. – Elastic cross section calculations

On the other hand, the plastic moment can be calculated using plastic theory. We can also calculate the applied force necessary to reach this plastic moment, as this happens when all the fibers of the section reach the yield strenght of the material.

F pl =f y ∗A M pl=f y∗W pl

Plastic cross section (Full) z0 zb h S1 A1 z1 S2 A2 A1 + A2 z2 S3

1,58 23,43 25 99 92 11,93 1097 34 126 0,21 7

mm mm mm mm3 mm mm mm3 mm mm mm mm3

Wpl;u;d

1204

mm3

Mpl;u;d Fpl;u;d

24071 0 50400

Nmm N

½*A/b h-zo zo+zb ½*b*zo2

S1+S2+S 3

Wpl;u;d * fu;d A * fu;d

Table 3.1.2. – Plastic cross section calculations

This is the situation that is modeled in the first analysis (just physical nonlinearity). However, due to the results of the modeling compared to the ones of the experiment (see section 3.2.), we also need to include geometrical nonlinearity effects. In order to do this, we gradually increase the elastic moment and the supports by increasing it with the effect of the load eccentricity due to the displacement in the center of the extrusion, until it reaches the yield strenght (elastic limit). After that, we find a force F vert that generates a moment at the supports that is equal to the plastic moment of a reduced cross section. Such cross section is the cross section of the extrusion, excluding the area bearing the axial force of the force F vert and the area bearing the shear force due to the horizontal load applied in the center of the extrusion.

Plastic cross section (Reduced) Fvert Vd AVd Atot/2 AF/2

47570 329 2,85 126,00 118,93

Anet

5,65

h1 a1 S1 h2 a2 S2 Wpl;u;d Mpl;u;d

0,07 1,54 9 1,41 22,72 128 137 27401

N N mm2 mm2 mm2 mm2/sid e mm mm mm3 mm mm mm3 mm3 Nmm

½*Fhor Vd*31/2/fy;d ½*F/fy;d Atot/2-AF/2-Avd/2

Anet*a1

Anet*a2 Sl+S2 Wpl;u;d * fy;d

Table 3.1.3. – Reduced Plastic cross section calculations

A 1= A 2

Fig. 3.1.1. – Reduced plastic cross section

We can calculate the buckling load using euler’s formula with a buckling length L buc=0.5L=150 mm (under the assumption of double clamping).

Fbuc =

π 2 EI Lbuc2

We calculate the horizontal displacement in the center of the extrusion with:

uend =

n u n−1 0

With n=Fbuc/Fvert.

Elastic

Plastic

N, 40641 40641 40641 40641 4757 4757 4755 4755 4755 Lbuc = 150 7 7 7 7 0 0 9 9 9 mm 4756 4756 4755 4753 4719 20000 30000 39631 47570 N 7 3 9 7 5 20,32 13,55 10,26 8,54

Fbuc

40641 7

406417

Fvert

0

10000

n

1

40,64

n/(n1)

1

1,03

uo

0

0,100

0,100 0,100 0,100 0,100 0,100 0,100 0,100 0,100 0,100

mm

uend

0

0,103

0,105 0,108 0,111 0,113 0,115 0,117 0,119 0,130 0,300

mm

Mo

24707

24707

24707 24707 24707 24707

2470 2470 2470 2470 2470 7 7 7 7 7

Nmm

Mend 24707

25732

26811 27946 29098 27401

2744 2749 2753 2779 3178 2 0 7 7 6

Nmm

σFvert

0

-40

1,05

-79

1,08

-119

1,11

-157

π2EI/Lbuc2

Fbuc/Fvert

1,13

N/mm2

n/(n-1)*uo

Mo+½*uend*F vert

Fvert/Atot

σM;end

-36

-38

-39

-41

-43

σmax

-36

-77

-119

-160

-200

wvert;top

0

0,170

N/mm2 + -200 -200 -200 -200 -200 -200

Mend*zb/I

N/mm2

0,340 0,510 0,674 0,809

mm

Fvert*L/(EA)

Table 3.1.3. – Geometrical nonlinearity calculations

Load/displacement top 60000 47570 39631 30000 20000 10000

40000 Load Fvert [N]

20000

0 0 0.000 0.200 0.400 0.600 0.800 1.000 wvert [mm]

Fig. 3.1.2. – Load vs Displacement on top of the extrusion

Geometric non-linearity 60000 50000

47195

40000 Load Fvert [N]

30000 20000 10000 0 0.1000.1500.2000.2500.3000.350 uhor mid section [mm]

Fig. 3.1.3. – Load vs horizontal Displacement of the extrusion

With this calculations we find the limit for the elastic capacity at F=39631 N with a prescribed displacement of 0.674 mm, and a limit for the plastic capacity at F=47570 N with a prescribed displacement of 0.809 mm.

3.2. COMMENTS First we are going to compare the results of the linear analysis and the first two nonlinear analysis, against the hand calculations for the Geometrical nonlinearity.


Physical Nonlinearity

30000

Phyisical + Geometrical Nonlinearity

20000

Linear

10000

Hand Calculations

0 0 1 2 Displacement mm

Fig. 3.2.1. – Comparison between analysis

Fig. 3.2.2. – Experiment Results

From this we can see that geometrical nonlinearity must be used in order to obtain an output for the extrusion that is closer to its real behaviour. Without the geometrical nonlinearity the extrusion can keep deforming undefinitely over a constant load, which is not the case of the experiment since the extrusion clearly reaches a failure after which the load decreases under a increasing displacement. Taking into account the geometrical effects, we get a curve that is very close to the one obtained in the experiment, which confirms that the selection of the boundary conditions was appropiate for this model. The maximum load is basically the same to the one obtained in the experiment, but the softening part of the curve is slightly lower than the one from the experiment (in the experiment the load tends to stabilize at 30 KN while in the model it keeps going down). It must be taken into account as well that the deformed shape of the model is very similar to the final shape of the extrusion after the experiment, which also confirms that the model output is very close to the real behaviour of the extrusion.

From the hand calculations we obtain a plastic limit that is about 4 KN higher (47,57 vs. 43,83 kN = ~ 9 %) then the one derived from the experiment and from the model. This can be declared with the fact that the n/(n-1) method is a elastic method to calculate the second order displacements according to buckling and bending. In the quasi-linear part of the experiment, the strains (and displacements) are not linear any more with the loads and the load-displacement line will become curved and more flat. To evaluate the effects of the imperfection in the behaviour of the extrusion, we compare the results between the model with an distributed load (applied in the web) that generates a displacement of 0.1 mm, and a point load (applied in the flanges) that generates the same displacement. It must be taken into account that with the area load we get a 0.1 mm displacement just in the mid point of the web while with the point load we get a displacement of 0.1 in a line over the center of the web (see output above). Hence the total load applied in the first case (0.038x40x40=608 N) is much lower than the one applied in the second case (260x2=520 N). 50000 40000 30000 Reaction Force N

20000

Area Loading Point Loading

10000 0 0 0.5 1 1.5 2 Displacement mm

Fig. 3.2.2. – Area Loading vs. Point loading comparison

From this graph we conclude that, even though both applied loads are different, the behaviour of the extrusion tends to be similar. The slight difference between the maximum load can be explained with the difference between the values of the two applied loads. But apart from that, the imperfection does not seem to have a big influence in the capacity and behaviour of the extrusion. This can be confirmed by running a new analysis without applying any loads to simulate imperfections. 50000 40000 30000 Reaction Force N

Area Loading

20000

Point Loading No imperfection

10000 0 0

1

2

Displacement mm

Fig. 3.2.2. – Comparison, with and without imperfection

3.3 CONCLUSIONS -This is a satisfactory modeling of the experiment because the maximum load and the prescribed displacement corresponding to that load are very similar between the model and the experiment. There is a slight difference in the behaviour of the model after reaching that peak (softening) compared to the experiment, which may be due to other material and element properties that cannot be taken into account for the modeling. But in terms of getting the actual capacity of the extrusion, the experiment was properly modeled. -To perform the hand calculations, it is correct to idealize this experiment as a double clamped column. However, it is not possible to take into account other factors that greatly affect the behaviour of the element (such as local stiffness) which is why the results vary between the calculations and the model (for example, the load needed to get a 0.1 mm dislacement varies from 659 N for the hand calculations to 520 N for the model). But doing hand calculations is still a good approach to understand the behaviour of the extrusion and to detect possible mistakes done in the modeling. -From Figure 3.2.2. we can conclude that, even though both applied loads are different, the behaviour of the extrusion tends to be similar. The slight difference between the maximum load can be explained with the difference between the values of the two loads. But apart from that, the imperfection does not seem to have a big influence in the behaviour of the extrusion. This can be confirmed by running a new analysis without applying any loads to simulate imperfections. We can say that due to the short length of the extrusion and its geometrical properties, the geometrical nonlinear effects due to the imperfection are not relevant in the whole behavior of the extrusion. However, geometrical nonlinear effects due to other factors (i.e. local buckling of the web and flanges) must be taken into account in order to be able to model the experiment properly, even though these effects cannot be hand calculated. -In order for the nonlinear analysis to converge (especially when taking into account geometrical nonlinear effects) we have to select the right analysis and iteration parameters. For example, using smaller load steps and increasing the maximum number of iterations can increase significantly the computation time, but must be done due to the complex behavior of the extrusion in the nonlinear range.

Compression test modeling in Diana

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