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Material model 106

*MAT_ELASTIC_VISCOPLASTIC_THERMAL

$ MID RO E PR SIGY ALPHA LCSS

1 7.85E-6 1000

$ QR1 CR1 QR2 CR2 QX1 CX1 QX2 CX2

$ C P LCE LCPR LCSIGY LCR LCX LCALPH

1.0 1.0 1103 1104 1105

$ LCC LCP

1101 1102

1. table ID stress-strain curves for different temperatures

1

2. elastic constants as load curves vs. temperature

2

3. coefficient of thermal expansion as load curve vs. temperature

3

4. strain rate dependency parameters as load curves vs. temperature

4

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Material model 106

$----------------------

*DEFINE_TABLE

$ TBID1

$ VALUE

900.0

800.0

700.0600.0

{}

*DEFINE_CURVE$ LCID

2

0.00 150.0

0.02 158.0

... ...

1.00 250.0

{}

*DEFINE_CURVE$ LCID

3

{}

$----------------------

sY

eeff

T

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LSTC

Table look up

eff

T2

Y

Y ( ,T) = ?

T

T1

We have flow curves for the given T1and

T2in our table definition.

The desired temperature T is somewhere

between T1and T2: T1< T< T2

1) evaluate the yield stress and the plastichardening modul for the yield curves for T1

and T2

2) perform a linear interpolation over thetemperature to evaluate the actual yieldstress and hardening modul at thetemperature T

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Material model 106

$------------------------------------------------------------------------$ *MAT_ELASTIC_VISCOPLASTIC_THERMAL

$ MID RO E PR SIGY ALPHA LCSS

$ QR1 CR1 QR2 CR2 QX1 CX1 QX2 CX2

$ C P LCE LCPR LCSIGY LCR LCX LCALPH

$ LCC LCP{}

$------------------------------------------------------------------------$

p

C

1

0 1

e

ss

Cowper-Symonds Model for strainrate efffects

$----------------------

*DEFINE_CURVE$ LCID {}

*DEFINE_CURVE$ LCID {}

$----------------------

p =f(T)

C =f(T)

C

p

If a time scaling is

applied the parameter Cmust be scaled to avoid

an overestimation of the

rate sensitivity

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New 3D Table Option

*DEFINE_TABLE_3D...

20.0 101...

500.0 104...

800.0 108...

temp

erature

*DEFINE_TABL E ID 104

isothermal curvesvarying strain rates

*DEFINE_TABLE ID 108

isothermal curves

varying strain rates

enter all curves directly

must not fit Cowper-Symonds Parameters

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Identification of C and p

How to identify the parameters Cand p?

We have stress strain curves for different strain rates. So we canrewrite the equation above withe the known stresses on the left side.

p

C

1

0 1

e

ss

cppp

C

11

1

0

0

ee

s

ss

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Identification of C and p

When we logarithmize the equation

we get a simple linear equation of the form y = mx + b

It is simple to determine the intercept band the slope m. Can do this in Excel.

cpp11

0

0

es

ss

Cppln

1ln

1ln

0

0

e

s

ss

mp

1

pbeC

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Identification of C and p

Example: Numisheet Benchmark #3:

we have yield curves for each temperature for strain rate 0.1 s -1

the curves at rate 0.1 s-1 are declared as static and used for a table definition

we have yield curves at strain rate 1.0 s-1

for 650 C and 800 Cwe can determine C and p directly onlyfor 650 C and 800 C

only two stress values, one for 0.1 s-1and one for 1.0 s-1to determine C and p

here we can set C to 10 and choose p so that the error between the given yield

curve for 1.0 s-1and the calculated one is minimized

we get the following values:C p

650 C 10.0 2.2

800 C 10.0 1.33

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Identification of C and p

Example: Numisheet Benchmark #3:

deviation from given curve at 1.0 s-1is rather low, especially at higher plastic strains

-5,0%

-4,0%

-3,0%

-2,0%

-1,0%

0,0%

1,0%

2,0%

3,0%

4,0%

5,0%

650C

800C

650C -3,2% 0,8% 1,7% 1,3% 0,9%

800C -2,0% -1,0% 0,1% 0,4% 1,4%

0,05 0,1 0,15 0,2 0,25

Percentage error for strain rate 1.0 s-1vs. plastic strain

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Identification of C and p

Example: Numisheet Benchmark #3:

How to determine C and p for the other temperatures ?

The strain rate sensitivity of the boron steel nearly vanishes at 500 C.

If we let be C = 10.0 then we can choose for p a value where the yield stress atstrain rate 1.0 s-1is approximately equal to the given yield stress at 0.1 s-1.

with this pragmatic assumption we get a three sampling points and we can

interpolate p for all other temperatures with a convenient mathmatical function.

a simple exponential function might be a good choice

0096.10.1

110;10

1

p

CpC

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Identification of C and p

Example: Numisheet Benchmark #3:

With an exponential function with offset we get promising results. If you are not sure

which function to choose then try the great function finder at www.zunzun.com

ceaTp Tb )(

410314.1 a

2104623.1

b

22.1c

0,0

2,0

4,0

6,0

8,0

10,0

500 600 700 800 900

temperature

p

http://www.zunzun.com/http://www.zunzun.com/

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Identification of C and p

Example: Numisheet Benchmark #3:

we end up with the following Cowper-Symonds parameters:

T 500 550 600 650 700 800

p 10.0 5.44 3,25 2.2 1.69 1.33

C(T)= 10.0 = const.

p

C

1

0 1

e

ss