8.1 2D plane strain plasticity - MIDAS...

1

midas NFX Benchmark Series 8-1

8.1 2D plane strain plasticity

REFERENCE NAFEMS [8-1]

KEYWORDS plane strain element

MODEL FILENAME MaterialNonlinearStatic01.nfxa

Figure 8.1.1 2D plane strain model

Material data

Young’s modulus

Poisson’s ratio

Perfect plasticity

Isotropic hardening

E = 250.0 103 N/mm

2

= 0.25

Y = 5.0 N/mm2

ET = 50.0 103 N/mm

2

L = 1.0 mm

A B

CD

uy

uxL = 1.0 mm

Figure 8.1.1 shows a 2D plane strain model undergoing elastic-plastic deformations. Two plasticity models are

considered; a perfect plasticity model and an isotropic hardening model. Plane strain elements are incorporated

to evaluate the nonlinear response. The loading conditions are summarized in table 8.1.1.

Material nonlinearity 8-2

(a) Perfect plasticity

(b) Isotropic hardening

Figure 8.1.2 Solutions for the stress variation per load step

-5.0

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7 8

Str

ess (N

/mm

2)

Step

σ_xx

σ_yy

σ_zz

σ_eff

Reference

-5.0

0.0

5.0

10.0

15.0

20.0

25.0

0 1 2 3 4 5 6 7 8

Str

ess (N

/mm

2)

Step

σ_xx

σ_yy

σ_zz

σ_eff

Reference

3




Figure 8.1.3 Solutions for the stress path

-10.0

-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

-10.0 -5.0 0.0 5.0 10.0

σy

y -σ

zz

σxx - σzz

Yield surface

midas NFX

Reference

-10.0

-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

-10.0 -5.0 0.0 5.0 10.0

σy

y -σ

zz

σxx - σzz

Yield surface

midas NFX

Reference


Table 8.1.1 Displacements prescribed in 8 increments (R = 2.5 10-5)

Table 8.1.2 Stress obtained at step 4, perfect plasticity

Table 8.1.3 Stress obtained at step 4, isotropic hardening

Step Disp. change x [mm] y [mm] Stress state

Step 1 Δux = R R 0.0 first yield

Step 2 Δux = R 2R 0.0 plastic flow

Step 3 Δuy = R 2R R elastic unloading

Step 4 Δuy = R 2R 2R plastic reloading

Step 5 Δux = -R R 2R plastic flow

Step 6 Δux = -R 0.0 2R plastic flow

Step 7 Δuy = -R 0.0 R elastic unloading

Step 8 Δuy = -R 0.0 0.0 plastic flow

Num. element xx [N/mm2] yy [N/mm

2] zz [N/mm

2] eff [N/mm

2]

Reference 16.31210 19.71433 13.97358 5.000000

QUAD-4 1 16.32229 19.71016 13.96855 5.000000

TRIA-3 2 16.32229 19.71016 13.96855 5.000000

QUAD-8 1 16.32229 19.71016 13.96855 5.000000

TRIA-6 2 16.32229 19.71016 13.96855 5.000000


2] zz [N/mm

2] eff [N/mm

2]

Reference 16.85673 20.24083 12.90243 6.361633

QUAD-4 1 16.86400 20.23665 12.89935 6.361184

TRIA-3 2 16.86400 20.23665 12.89935 6.361184

QUAD-8 1 16.86400 20.23665 12.89935 6.361184

TRIA-6 2 16.86400 20.23665 12.89935 6.361184

5


8.2 2D plane stress plasticity


KEYWORDS plane stress element


Figure 8.2.1 2D plane stress model

Material data

Young’s modulus

Poisson’s ratio

Perfect plasticity

Isotropic hardening

E = 250.0 103 N/mm

2

= 0.25

Y = 5.0 N/mm2

ET = 50.0 103 N/mm

2

L = 1.0 mm

A B

CD

uy

uxL = 1.0 mm

Figure 8.2.1 shows a 2D plane stress model undergoing elastic-plastic deformations. Two plasticity models are

considered; a perfect plastic model and an isotropic hardening model. Material nonlinear analyses are carried

out using plane stress elements in accordance with the loading condition summarized in table 8.2.1.





-10.0

-5.0

0.0

5.0

10.0

0 1 2 3 4 5 6 7 8

Str

ess (N

/mm

2)

Step

σ_xx

σ_yy

σ_eff

Reference

-10.0

-5.0

0.0

5.0

10.0

0 1 2 3 4 5 6 7 8

Str

ess (N

/mm

2)

Step

σ_xx

σ_yy

σ_eff

Reference

7





-10.0

-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

-10.0 -8.0 -6.0 -4.0 -2.0 0.0 2.0 4.0 6.0 8.0 10.0

σy

y-σ

zz

σxx - σzz

Yield surface

midas NFX

Reference

-10.0

-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

-10.0 -8.0 -6.0 -4.0 -2.0 0.0 2.0 4.0 6.0 8.0 10.0

σy

y-σ

zz

σxx - σzz

Yield surface

midas NFX

Reference



Table 8.2.2 Stress obtained at step 4, perfect plasticity


Step Disp. change x [mm] y [mm] Stress state

Step 1 Δux = R R 0.0 first yield

Step 2 Δux = R 2R 0.0 plastic flow

Step 3 Δuy = R 2R R elastic unloading

Step 4 Δuy = R 2R 2R plastic reloading

Step 5 Δux = -R R 2R plastic flow

Step 6 Δux = -R 0.0 2R plastic flow

Step 7 Δuy = -R 0.0 R elastic unloading

Step 8 Δuy = -R 0.0 0.0 plastic flow


2] eff [N/mm

2]

Reference 3.308084 5.751942 5.000035

QUAD-4 1 3.375650 5.744146 5.000000

TRIA-3 2 3.375650 5.744146 5.000000

QUAD-8 1 3.375650 5.744146 5.000000

TRIA-6 2 3.375650 5.744146 5.000000


2] eff [N/mm

2]

Reference 5.685183 8.888501 7.797050

QUAD-4 1 5.790009 8.838760 7.776335

TRIA-3 2 5.790009 8.838760 7.776335

QUAD-8 1 5.790009 8.838760 7.776335

TRIA-6 2 5.790009 8.838760 7.776335

9


8.3 3D plasticity


KEYWORDS solid elements


Figure 8.3.1 3D solid model

Material data

Young’s modulus

Poisson’s ratio

Perfect plasticity

Isotropic hardening

E = 250.0 103 N/mm

2

= 0.25

Y = 5.0 N/mm2

ET = 50.0 103 N/mm

2

L = 1.0 mm

A B

CD

uz

ux

E F

GH

Z

X

Y

uy

L = 1.0 mm

L = 1.0 mm

Figure 8.3.1 shows a 3D cube model undergoing elastic-plastic deformations. Two plasticity models are

considered; a perfect plasticity and an isotropic hardening model. The boundary condition is depicted in figure

8.3.1 and the loading condition is summarized in table 8.3.1. 3D continuum elements are utilized to obtain

nonlinear responses.





-5.0

0.0

5.0

10.0

15.0

20.0

25.0

30.0

0 1 2 3 4 5 6 7 8 9 10 11 12

Str

ess (N

/mm

2)

Step

σ_xx

σ_yy

σ_zz

σ_eff

Reference

-5.0

0.0

5.0

10.0

15.0

20.0

25.0

30.0

0 1 2 3 4 5 6 7 8 9 10 11 12

Str

ess (N

/mm

2)

Step

σ_xx

σ_yy

σ_zz

σ_eff

Reference

11





-10.0

-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

-10.0 -8.0 -6.0 -4.0 -2.0 0.0 2.0 4.0 6.0 8.0 10.0

σy

y-σ

zz

σxx - σzz

Yield surface

midas NFX

Reference

-10.0

-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

-10.0 -8.0 -6.0 -4.0 -2.0 0.0 2.0 4.0 6.0 8.0 10.0

σy

y-σ

zz

σxx - σzz

Yield surface

midas NFX

Reference



Table 8.3.2 Stress obtained at of step 6, perfect plasticity


Step Disp. change x [mm] y [mm] z [mm]

Step 1 Δux = R R 0.0 0.0

Step 2 Δux = R 2R 0.0 0.0

Step 3 Δuy = R 2R R 0.0

Step 4 Δuy = R 2R 2R 0.0

Step 5 Δuz = R 2R 2R R

Step 6 Δuz = R 2R 2R 2R

Step 7 Δux = -R R 2R 2R

Step 8 Δux = -R 0.0 2R 2R

Step 9 Δuy = -R 0.0 R 2R

Step 10 Δuy = -R 0.0 0.0 2R

Step 11 Δuz = -R 0.0 0.0 R

Step 12 Δuz = -R 0.0 0.0 0.0


2] zz [N/mm

2] eff [N/mm

2]

Reference 22.27493 24.70006 28.02500 5.000002

HEXA-8 1 22.26431 24.71850 28.01719 5.000000

PENTA-6 2 22.26431 24.71850 28.01719 5.000000

HEXA-20 1 22.26431 24.71850 28.01719 5.000000

PENTA-15 2 22.26431 24.71850 28.01719 5.000000


2] zz [N/mm

2] eff [N/mm

2]

Reference 21.84202 25.24927 27.90871 5.267201

HEXA-8 1 21.89467 25.21898 27.88635 5.199337

PENTA-6 2 21.89467 25.21898 27.88635 5.199337

HEXA-20 1 21.89467 25.21898 27.88635 5.199337

PENTA-15 2 21.89467 25.21898 27.88635 5.199337

13


8.4 Kinematic hardening plasticity


KEYWORDS plane strain elements


Figure 8.4.1 2D plane strain overlay model

Material data

Young’s modulus

Poisson’s ratio

Perfect plasticity

E = 250.0 103 N/mm

2

= 0.25

Y = 1000.0 N/mm2; w1 = 0.1724 (element 1)

Y = 5.0 N/mm2; w2 = 0.8276 (element 2)

L1 = 1 mm

A B

CD

ux

E

W1

X

Y

W2

A B

CD

L2 = 1 mmux

X

Y

Figure 8.4.1 shows a 2D plane strain overlay model which undergoes prescribed extension and compression

along x-axis as presented in table 8.4.1. One element has nearly elastic property and the other has perfectly

plastic property. The combination of plastic property produces kinematic hardening effect.


Figure 8.4.2 Solutions for the stress variation per load step (Kinematic hardening)

Figure 8.4.3 Solutions for the stress path (Kinematic hardening)

-20.0

-15.0

-10.0

-5.0

0.0

5.0

10.0

15.0

20.0

0 1 2 3 4 5 6 7 8

Str

ess (N

/mm

2)

Step

σ_xx

σ_yy

σ_zz

σ_eff

Reference

-10.0

-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

-10.0 -8.0 -6.0 -4.0 -2.0 0.0 2.0 4.0 6.0 8.0 10.0

σy

y-σ

zz

σxx - σzz

Yield surface

midas NFX

Reference

15



Table 8.4.2 Stress obtained at step 2, kinematic hardening

Step Disp. change x [mm]

Step 1 Δux = R R

Step 2 Δux = R 2R

Step 3 Δux = -R R

Step 4 Δux = -R 0.0

Step 5 Δux = -R -R

Step 6 Δux = -R -2R

Step 7 Δux = R -R

Step 8 Δux = R 0.0


2] zz [N/mm

2] eff [N/mm

2]

Reference 12.24140 6.379450 6.379450 5.862071

QUAD-4 1 12.24138 6.379310 6.379310 5.862069

TRIA-3 2 12.24138 6.379310 6.379310 5.862069

QUAD-8 1 12.24138 6.379310 6.379310 5.862069

TRIA-6 2 12.24138 6.379310 6.379310 5.862069


8.5 Two-bar assembly plasticity


KEYWORDS rod elements


Figure 8.5.1 Two-bar assembly model

Material data

Young’s modulus

Poisson’s ratio

Coefficient of thermal

expansion

Perfect plasticity

Kinematic hardening

E = 10.0 103 N/mm

2

= 0.0

= 1.0 10-5

/°C

Y = 10.0 N/mm2

ET = 1.0 103 N/mm

2

Section property Cross-sectional area A = 1 mm2

Bar 1 (ΔT1)

P = 15.0

Bar 2 (ΔT2)

Temperature

Half-cycleA = 1 mm2

A = 1 mm2

Figure 8.5.1 shows a pair of identical bars whose ends are constrained to move together axially. A constant

tensile force is applied so that the bars reach near-yielding point. And cyclic temperature variation is applied on

the bar assembly which changes the plastic state. Three load cases are tested; ratcheting (perfect plasticity with

1 100T and 2 0T ), shakedown (kinematic hardening with 1 100T and 2 0T ), and alternating

plasticity (kinematic hardening with 1 300T and 2 0T ).

17


(a) Perfect plasticity (Ratcheting)

(b) Kinematic hardening (Shakedown)

(c) Kinematic hardening (Alternating plasticity)

Figure 8.5.2 Solutions for the force variation per cycle

-5

0

5

10

15

20

25

0 5 10 15 20

Lo

ad

(N

)

Step

Bar 1 (midas NFX)Bar 2 (midas NFX)Bar 1 (Reference)Bar 2 (Reference)

-5

0

5

10

15

20

25

0 5 10 15 20

Lo

ad

(N

)

Step


-5

0

5

10

15

20

25

0 5 10 15 20

Lo

ad

(N

)

Step

Bar 1 (midas NFX) Bar 2 (midas NFX)

Bar 1 (Reference) Bar 2 (Reference)





Figure 8.5.3 Solutions for the strain variation per cycle

0

0.002

0.004

0.006

0.008

0.01

0.012

0 5 10 15 20

Str

ain

Step


0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0.004

0 5 10 15 20

Str

ain

Step


0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

0 5 10 15 20

Str

ain

Step


19





Figure 8.5.4 Solutions for the load-strain curve

-5

0

5

10

15

20

25

0 0.002 0.004 0.006 0.008 0.01 0.012

Lo

ad

(N

)

Strain


-5

0

5

10

15

20

25

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035

Lo

ad

(N

)

Strain


-5

0

5

10

15

20

25

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009

Lo

ad

(N

)

Strain



8.6 Rigid punch plasticity




Figure 8.6.1 Rigid punch on a deep plate

Material data

Young’s modulus

Poisson’s ratio

Perfect plasticity

Isotropic hardening

E = 1.0 103 N/mm

2

= 0.3

Y = 1.0 N/mm2

ET = 0.1 103 N/mm

2

A B

CD

Z

H = 160

W2 = 200

E

F

L = 20

W1 = 80

rigid

punchUnits: mm

Figure 8.6.1 shows a rigid punch pressed into a deep elastic foundation of finite width supported on the

frictionless plane. 2D plane strain condition is adopted for the foundation. The elastic perfectly plastic and

isotropic hardening material models are tested. Punch force is represented by the downward vertical

displacement, . Prescribed displacement is given in 6 steps: = 0.04, 0.12, 0.14, 0.16, 0.18, 0.24 mm.

21


Figure 8.6.2 Solution for the load-deflection curve

Table 8.6.1 Reaction force at deflection δ = 0.24 mm

0.0

25.0

50.0

75.0

100.0

125.0

150.0

0.00 0.05 0.10 0.15 0.20 0.25

Lo

ad

(N

)

Applied deflection of punch (mm)

Perfect plastic

Isotropic hardening

Reference

Perfect plastic Isotropic hardening

RF (N) RF (N)

Reference 110.530 123.230

QUAD-4 112.465 124.495

TRIA-3 113.060 124.854

QUAD-8 111.016 123.383

TRIA-6 110.420 123.044

HEXA-8 112.464 124.495

PENTA-6 119.525 129.139

HEXA-20 110.982 123.358

PENTA-15 111.481 123.607


8.7 Thermal ratcheting of uniform beam


KEYWORDS plane stress elements


Figure 8.7.1 Thermal ratcheting of uniform beam

Material data

Young’s modulus

Poisson’s ratio

Yield stress

Hardening coefficient

Coefficient of expansion

E = 200.0 103 N/mm

2

= 0.0

Y = 100.0 N/mm2

= 0.02

= 1.0 10-5

/°C

Temperature

distribution

T = -400 (10 half cycles)

Applied mechanical load (constant) 900 per unit width of beam

O

10

P P

T

Temperature

-400

1 100

Figure 8.7.1 shows a uniform beam with constant end load and cyclically varying linear through-thickness

temperature gradient. Plane sections are constrained to remain straight and parallel without curvature. The

material behavior is subjected to von Mises elasto-plastic model with linear isotropic hardening. Stress and

strain variations on the lower and upper surfaces are determined per load step. Each load step corresponds to a

half-cycle temperature variation.

23


(a) Lower surface

(b) Upper surface


-100

-50

0

50

100

150

200

0 1 2 3 4 5 6 7 8 9 10

Str

ess (N

/mm

2)

Step

midas NFX

Reference

-100

-50

0

50

100

150

200

0 1 2 3 4 5 6 7 8 9 10

Str

ess (N

/mm

2)

Step

midas NFX

Reference


(a) Lower surface

(b) Upper surface

Figure 8.7.3 Solutions for the strain variation per load step

-0.002

-0.001

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

0.011

0.012

0 1 2 3 4 5 6 7 8 9 10

Str

ain

Step

Total strain

Elastic strain

Plastic strain

Reference

-0.002

-0.001

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

0.011

0.012

0 1 2 3 4 5 6 7 8 9 10

Str

ain

Step

Total strain

Elastic strain

Plastic strain

Reference

25


8.8 An extended straight truss

REFERENCE Crisfield [8-3]



Figure 8.8.1 A straight truss model

Material data

Young’s modulus

Poisson’s ratio

Isotropic hardening

equation

E = 2.0 106 N/m

2

= 0.3

c = 22222κ + 200 (0 ≤ κ ≤ 0.0009)

c = 220 (0.0009 ≤ κ)

Section property Cross-sectional area A = 1 m2

Figure 8.8.2 Stress curve along axial displacement

X

Z

L = 1 m

prescribed

displacement 0.004 mcross-sectional area = 1 m2

0.0

50.0

100.0

150.0

200.0

250.0

0 0.5 1 1.5 2 2.5 3 3.5 4

σa

xia

l(N

/m2)

Axial displacement (mm)

midas NFX

Reference

Figure 8.8.1 shows a straight truss model fixed at one end in all directions. The other end is constrained to

move in axial direction and is extended to prescribed value using displacement controlled nonlinear analysis.

The material behavior is subjected to von Mises elasto-plastic model with linear isotropic hardening combined

with perfectly plastic model.


8.9 Square plate under uniformly distributed load


KEYWORDS shell elements


Figure 8.9.1 Square plate model

Material data

Young’s modulus

Poisson’s ratio

Perfect plasticity

E = 30000.0 N/mm2

= 0.3

Y = 30.0 N/mm2

Section property Thickness t = 0.4 mm

Table 8.9.1 Load per unit area for the plastic state

40

40

0.4Units: mm

Simply supported Clamped edge

limP [N/mm2] limP [N/mm

2]

Reference 0.01877 0.03852

QUAD-4 0.01880 0.03850

TRIA-3 0.01880 0.03850

QUAD-8 0.01880 0.03850

TRIA-6 0.01861 0.03850

Figure 8.9.1 shows a square plate subjected to uniformly distributed load. The load is applied up to the yielding

point. The plate material follows the von Mises elastic perfectly plastic material model without hardening. The

shell section is integrated in the thickness direction using 13-point Simpson integration rule. Two boundary

edge conditions are tested: simply supported and clamped edge.

27


(a) Simply supported boundary condition

(b) Clamped edge boundary condition

Figure 8.9.2 Pressure-center deflection curve

0.000

0.005

0.010

0.015

0.020

0.0 1.0 2.0 3.0 4.0 5.0 6.0

Pre

ssure

(N

/mm

2)

Deflection at center of plate (mm)

midas NFX

Layered formulation (reference)

Gross yield formulation

P_Limit

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035

0.040

0.0 1.0 2.0 3.0 4.0 5.0

Pre

ssure

(N

/mm

2)

Deflection at center of plate (mm)

midas NFX

Layered formulation (reference)

Gross yield formulation

P_Limit


8.10 Uniformly loaded circular plate

REFERENCE Owen et al. [8-4]

KEYWORDS axisymmetric elements


Figure 8.10.1 Circular plate model

Material data

Young’s modulus

Poisson’s ratio

Yield stress

E = 1.0 107 psi

= 0.24

Y = 16000.0 psi

PY

X

Z

PY

R = 10

h = 1

Units: in

C

Figure 8.10.1 shows a circular plate model simply supported around its outer rim under uniform pressure. The

circular plate undergoes material nonlinear deformations. The material follows elastic, perfectly plastic von

Mises model. the limit pressure is determined using various types of axisymmetric elements.

29


Figure 8.10.2 Pressure-center deflection curve

Table 8.10.1 Limit pressure obtained using axisymmetric elements

*

2

2

6 . 5 2,

4

y y

lim y

M hP M

R

0.0

50.0

100.0

150.0

200.0

250.0

300.0

0.0 0.5 1.0 1.5 2.0

Pre

ssure

(p

si)

Deflection at center of plate (in)

midas NFX

P_Limit

limP [psi]

Reference 260.8

QUAD-8 261.2

TRIA-6 261.6


8.11 Two coaxial tubes

REFERENCE Crandall et al. [8-5]



Figure 8.11.1 Two coaxial tubes

Material data (Inner

tube: steel 1020-RC)

Young’s modulus

Poisson’s ratio

Yield stress

Esteel = 1.84375 105 N/mm

2

steel = 0.3

/Y steel = 590 N/mm2

Material data (Outer

tube: aluminum alloy

2024-T4)

Young’s modulus

Poisson’s ratio

Yield stress

Ealloy = 7.6 104 N/m

2

alloy = 0.3

/Y alloy = 380 N/mm2

R1 = 4.069718

R2 = 2.478169

t = 0.5

Units: mm

δ = 0.1

L = 10

Figure 8.11.1 shows two coaxial tubes made of steel (1020-CR) and aluminum alloy (2024-T4) compressed

between two rigid flat ends. Both tubes are assumed to follow the von Mises elastic perfectly plastic material

model with associated flow rule.

31


Figure 8.11.2 Reaction force-applied displacement curve

Table 8.11.1 Reaction force at the bottom surface with the whole model in plasticity state

7.0484.E+03

8.6900.E+03 8.6900.E+03

0

1,000

2,000

3,000

4,000

5,000

6,000

7,000

8,000

9,000

10,000

0.00 0.02 0.04 0.06 0.08 0.10

Sum

of r

eactio

n fo

rce (N

)

Applied displacement (mm)

midas NFX

Reference yield points

RF [N]

Reference 8690.0

HEXA-8 8681.1

PENTA-6 8683.7

HEXA-20 8681.1

PENTA-15 8683.7


8.12 Residual stress problem

REFERENCE Crandall et al. [8-5]



Figure 8.12.1 3 member frame model

Material data

Young’s modulus

Poisson’s ratio

Perfect plasticity

E = 30.0 106 psi

= 0.3

Y = 30000.0 psi

Section property Cross-sectional area A = 1.0 in2

Table 8.12.1 Residual stress results of center member after loading-unloading

A = 1.0 in2 A = 1.0 in2 A = 1.0 in2

P

elastic [in] res [psi]

Reference 0.1000 -5650

midas NFX 0.0998 -5650

Figure 8.12.1 shows a chain hoist attached to the ceiling through three tie rods which is made of cold-rolled

steel with yield strength y . The three-rod structure is loaded in the vertical direction and then unloaded.

While the structure is loaded, the center rod member yields firstly while side members remain elastic. When

the frame is fully loaded, all three rods become fully plastic.

33


Figure 8.12.2 Stress-deflection curve of center member

-5.650.E+03

-1.0E+04

0.0E+00

1.0E+04

2.0E+04

3.0E+04

4.0E+04

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14

Axia

l str

ess o

f cen

ter ro

d (p

si)

Deflection at point A (in)

midas NFX

Reference


8.13 Nonlinear equation solution tests




Figure 8.13.1 3 overlay elements mesh model

Material data

(element 1)

Young’s modulus

Poisson’s ratio

Perfect plasticity

E = 100.0 103 N/mm

2

= 0.25

Y = 3.0 N/mm2

Material data

(element 2)

Young’s modulus

Poisson’s ratio

Perfect plasticity

E = 60.0 103 N/mm

2

= 0.25

Y = 6.0 N/mm2

Material data

(element 3)

Young’s modulus

Poisson’s ratio

Perfect plasticity

E = 40.0 103 N/mm

2

= 0.25

Y = 8.0 N/mm2

P/2

P/2

P = 1.0

Figure 8.13.1 shows a mesh with three overlaid elements, each with yield stress. This problem is tailored to test

the performance of nonlinear equation solver. Three elements are welded together at the nodes with identical

element connectivity. Three different stiffness update schemes are tested. Namely, initial stiffness, modified

Newton-Raphson and full Newton-Raphson methods are tested for number of iterations to reach convergence.

Nonlinear analyses are carried out with 6 increments and 0.01% internal force norm is used as convergence

criteria.

35


Figure 8.13.2 Target equilibrium curve obtained by initial stiffness matrix method

* Values and bracketed values are numbers of iteration required to reach convergence using midas NFX and the reference

number, respectively.

Figure 8.13.3 Target equilibrium curve obtained by modified Newton-Raphson method



1(1)

1(1)

13(10)

13(10)

40(27)

40(27)

0.0

3.0

6.0

9.0

12.0

15.0

18.0

0.00E+00 6.00E-05 1.20E-04 1.80E-04 2.40E-04

σx(N

/mm

2)

δx (mm)

midas NFX

Reference

P_limit

1(1)

1(1)

2(3)

2(3)

6(3)

2(3)

0.0

3.0

6.0

9.0

12.0

15.0

18.0

0.00E+00 6.00E-05 1.20E-04 1.80E-04 2.40E-04

σx(N

/mm

2)

δx (mm)

midas NFX

Reference

P_limit


Figure 8.13.4 Target equilibrium curve obtained by full Newton-Raphson method



1(1)

1(1)

2(3)

2(3)

3(2)

2(2)

0.0

3.0

6.0

9.0

12.0

15.0

18.0

0.00E+00 6.00E-05 1.20E-04 1.80E-04 2.40E-04

σx(N

/mm

2)

δx (mm)

midas NFX

Reference

P_limit

37


8.14 Stiffened cylindrical panel


KEYWORDS shell elements


Figure 8.14.1 Stiffened cylindrical panel under compressive end load

Material data

Young’s modulus

Poisson’s ratio

Perfect plasticity

E = 2.1 105 N/mm

2

= 0.3

Y =350 N/mm2

Section property Thickness t = 1.0 mm

End A

Y

Z

9o

L = 400d = 10

X, R

y

9o

θ

x

z

R = 400

t = 1.0

Load = 18649

Symmetry

Definition of nodal forces to simulate

evenly distributed load

Total load

= 12.875P P4P

P4P

2P

0.07955P0.31818P0.15909P0.31818P

D, solution point

Figure 8.14.1 shows a stiffened cylindrical panel under compressive end load. The underlying material is

elastic perfectly plastic von Mises material model. Automatic incrementation of the applied load using arc

length procedure is adopted until the global Z displacement at point D exceeded 0.2. There is an initial

imperfection in both the cylindrical panel and the stiffener.

Cylindrical panel (cylindrical system, RZ )

2

0.569sin where -9 9, 0 2007200

ZR Z

Stiffener (local cartesian system, xyz)

0.3sin sin where 0 200, 0 10400 20

x yz x y


Table 8.14.1 Limit point solution

Figure 8.14.2 Load-displacement curve

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

Ap

plied lo

ad (N

)x 1

0000

Z-displacement (mm)

midas NFX

Reference

Reference midas NFX

Limit point 1 Load [N] 23742 24347

Z-disp [mm] 0.1679 0.1725

Limit point 2 Load [N] 15456 16279

Z-disp [mm] 0.1377 0.1405

39


8.15 Necking of a circular bar

REFERENCE Simo et al. [8-7]

KEYWORDS axisymmetric elements, solid elements


Figure 8.15.1 A circular bar model with geometric imperfection and its deformed shape

Material data

Shear modulus

Bulk modulus

Initial flow stress

Residual flow stress

Linear hardening

coefficient

Saturation exponent

= 80.1938 103 N/mm

2

= 164.206 103 N/mm

2

Y = 450 N/mm2

Y = 715 N/mm2

K = 129.24 N/mm2

= 16.93

X

Y

ZGeometric

Imperfection

R = 6.413

L/2 = 26.667

Units: mm

Figure 8.15.1 shows a circular bar with geometrical imperfection (0.982 % of radius) to be imposed at

symmetric plane to trigger strain localization. The model is considered in the context of large strain problem

and is subjected to a uniform extension of 7 mm. The linear hardening law is replaced by a more general

nonlinear hardening law of the form

0( ) : ( )[1 exp( )], 0yk K K K .


Figure 8.15.2 Stress-strain curve

Figure 8.15.3 Necking ratio versus elongation ratio

Table 8.15.1 Radius ratios of final step

0

100

200

300

400

500

600

700

800

900

1000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Pla

stic s

tress

(N

/mm

2)

Plastic strain

0

0.25

0.5

0.75

1

0 0.1 0.2

Rad

ius /

Rad

ius

0

ΔL/L0

midas NFXReferenceExperimental:2499R2515ST2501R2502R

R/R0 Element type R/R0

QUADX-4 0.4384 HEXA8 0.4292

TRIAX-3 0.7087 PENTA6 0.6962

QUADX-8 0.4281 HEXA20 0.4302

TRIAX-6 0.4269 PENTA15 0.4267

41


8.16 Pressurized rubber disc

REFERENCE Oden [8-8]



Figure 8.16.1 Pressurized rubber disc

Material data

Deviatoric strain energy

coefficients

Poisson’s ratio

C10 = 80 lb/in2

C01 = 20 lb/in2

= 0.4995

t = 0.5 in

Pressure = 50 psi

Symmetry

R = 7.5 in

Figure 8.16.1 shows a rubber disc pinned around its outside edge. The disc is subjected to pressure so that it

bulges into a spherical shape. For obvious axisymmetric consideration, a pie of disc is modeled with solid

elements. The Mooney-Rivlin material model is used to represent nearly incompressible rubber elasticity. The

pressure load is non-conservative. Additional constraining equations are required to illustrate that the edge is

constrained to remain straight.


Figure 8.16.2 Thickness strain vs. central displacement

Figure 8.16.3 Pressure vs. deflection curve

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5 3

Th

ickn

ess

/ Ori

gin

al t

hic

kness

Uz / R0

midas NFX

Reference

0

10

20

30

40

50

60

0 2 4 6 8 10 12 14 16 18 20

Pre

ssure

(p

si)

Uz (in)

midas NFX

Reference

43


8.17 Inflation of a spherical rubber balloon

REFERENCE Souza et al. [8-9]



Figure 8.17.1 Inflation of a spherical rubber balloon

Material data


coefficients

Poisson’s ratio

1 = 1.3

1 = 6.3 N/mm2

2 = 5.0

2 = 0.012 N/mm2

3 = -2.0

3 = -0.1 N/mm2

= 0.4995

Figure 8.17.1 shows a spherical rubber membrane-like structure inflated under internal pressure. The three-

term Ogden material model is used to represent nearly incompressible rubber elasticity. The solid mesh

discretizes one octant of the sphere with symmetry boundary conditions imposed along the edge. The pressure

load is non-conservative. Arc-length method is employed to allow equilibrium to be found beyond the

instability point.


Figure 8.17.2 Normalized pressure vs. expansion ratio

* Normalized internal pressure and expansion ratio are defined as 0 02p pr t and 0r r , where r and 0r are the

current and initial radii of the balloon respectively, t0 is the initial thickness of the rubber membrane and p is the current

internal pressure.

0

1

2

3

0 1 2 3 4 5 6 7 8

No

rmalize

d p

ressu

re, p

*

Expansion ratio, λ

midas NFX

Reference (Ogden, 1972)

45


8.18 Inflation of an axisymmetric ellipsoidal balloon

REFERENCE Souza et al. [8-9]



Figure 8.18.1 Inflation of an axisymmetric ellipsoidal balloon

Material data


coefficients

Poisson’s ratio

1 = 1.3

1 = 6.3 N/mm2

2 = 5.0

2 = 0.012 N/mm2

3 = -2.0

3 = -0.1 N/mm2

= 0.4995

r2 = 145

r1 = 95

t = 3Units: mm

Figure 8.18.1 shows an axisymmetric ellipsoidal balloon to be inflated. The initial geometry of the balloon

corresponds to the shape of an official rugby ball. This ball is assumed to be made of the three-term Ogden

material model. Solid elements are used to discretize one octant of the sphere with symmetry boundary

conditions imposed along the edge. The pressure load is non-conservative. Arc-length method is employed to

allow equilibrium to be found beyond the instability point.


Figure 8.18.2 Pressure-expansion diagram for rugby ball

0

0.05

0.1

0.15

0.2

0 40 80 120 160

pre

ssure

, p (N

/mm

2)

radii r1, r2 (mm)

midas NFX

Reference

r2

r1

47


8.19 Contact between a rigid body and a hyperelastic body

REFERENCE Zhi-Qiang Feng et al. [8-10]



Figure 8.19.1 A rigid cylinder into a hyperelastic body model

Material data Shear modulus G = 10.0 N/mm2

R = 50

b = 80

a = 280

F

δ = 50

rigid

bottom side is fixed

Figure 8.19.1 shows the indentation of a rigid cylinder into a hyperelastic body. The Blatz-Ko hyperelastic

material model is used to represent sponge-type material behavior which requires only one material constant,

shear modulus. The results are obtained by carrying out nonlinear analysis with displacement control. Friction

coefficients of 0.4 and 0.0 are tested.


Figure 8.19.2 Load-displacement curve

Figure 8.19.3 Deformed shape of the sponge type body (μ = 0.4)

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

0 10 20 30 40 50

Fo

rce (N

)

Displacement of the cylinder (mm)

midas NFX

Reference

μ= 0.4

μ= 0.0

49


References

[8-1] NAFEMS, Background to material Non-Linear Benchmarks, Ref . R0049, NAFEMS, Glasgow, 1998

[8-2] NAFEMS, Selected Benchmarks for Material Non-Linearity, Ref . R0026, NAFEMS, Glasgow, 1993

[8-3] M. A. Crisfield, Non-linear Finite Element Analysis of Solids and Structures, England, John Wiley &

Sons Ltd., 1994

[8-4] D.R.J. Owen and E. Hinton, Finite elements in plasticity – Theory and Practice, Pineridge Press Limited,

Swansea, U.K., 1980

[8-5] S. H. Crandall and N. C. Dahl, An Introduction to the Mechanics of Solids, McGraw-Hill Book Co., Inc.,

New York, NY, 1959

[8-6] NAFEMS, A Review of Benchmark Problems for Geometric Non-linear Behaviour of 3-D Beams and

Shells, Ref . R0024, NAFEMS, Glasgow, 1993

[8-7] J. C. Simo, T. J. R. Hughes, Computational inelasticity, Springer-Verlag, New York, pp. 326-335 1998

[8-8] J. T. Oden, Finite elements of nonlinear continua, McGraw-Hill, 1972

[8-9] E. A. de Souza Neto, D. Peric and D. R. J. Owen, Computational Methods for plasticity, Wiley, New

York, 2008

[8-10] Zhi-Qiang Feng, Francois Peyraut, Nadia Labed, “Solution of large deformation contact problems with

friction between Blatz-Ko hyperelastic bodies,” International Journal of Engineering Science, Vol. 41, pp.

2213-2225, 2003

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