ArmstrongFrederick

Armstrong-Frederick Steel Model

References

Theory

Implementation can be found in a separate document.

A von Mises type yield function is used. The associated plasticity is assumed. Both isotropic and kinematic hardening rules are employed.

Although the plastic flow is associative, the hardening rules are not. As the result, the consistent tangent modulus is not symmetric.

Isotropic Hardening

An exponential function is added to the linear hardening law.

k=σy+ks(1emp)+klp,k=\sigma_y+k_s(1-e^{-mp})+k_lp,

where σy\sigma_y is the initial elastic limit (yielding stress), ksk_s is the saturation stress, klk_l is the linear hardening modulus, mm is a constant that controls the speed of hardening, dp=23dεp:dεp\mathrm{d}p=\sqrt{\dfrac{2}{3}\mathrm{d}\varepsilon^p:\mathrm{d}\varepsilon^p} is the rate of accumulated plastic strain pp.

Kinematic Hardening

The Armstrong-Frederick type rule is used. Multiple back stresses are defined,

β=βi\beta=\sum\beta^i

in which

dβi=23ai dεpbiβ dp,\mathrm{d}\beta^i=\sqrt{\dfrac{2}{3}}a^i~\mathrm{d}\varepsilon^p-b^i\beta~\mathrm{d}p,

where aia^i and bib^i are material constants. Note here a slightly different definition is adopted as in the original literature 23\dfrac{2}{3} is used instead of 23\sqrt{\dfrac{2}{3}}. This is purely for a slightly more tidy derivation and does not affect anything.

Syntax

The following applies to v3.6 and later. Check the older syntax in the older version of the documentation.

material ArmstrongFrederick (1) (2) (3) (4) (5) (6) (7) [(8) (9)...] [10]
# (1) int, unique material tag
# (2) double, elastic modulus
# (3) double, poissons ratio
# (4) double, yield stress
# (5) double, linear hardening modulus
# (6) double, saturation stress
# (7) double, m, saturation rate
# (8) double, a, kinematic hardening parameter
# (9) double, b, kinematic hardening parameter
# [10] double, density, default: 0.0

History Layout

location
parameter

initial_history(0)

accumulated plastic strain

initial_history(1-6)

back stress for the first pair of a1a^1 and b1b^1

initial_history(7-12)

back stress for the second pair of a2a^2 and b2b^2 if defined

...

more back stresses

Example

Here a few examples are shown.

Isotropic Hardening Only

There is no difference between the classic J2 plasticity model and this AF steel model if only isotropic hardening is defined.

material ArmstrongFrederick 1 2E2 .2 .1 1. .05 1000.

Kinematic Hardening Only

If a0a\neq0 and b=0b=0, there is no difference between the classic J2 plasticity model and this AF steel model. A linear kinematic hardening rule is implied. Normally at least one set of a1a^1 and b1b^1 is defined.

material ArmstrongFrederick 1 2E2 .2 .1 0. 0. 0. 50. 500.

By definition, if one set of a1a^1 and b1b^1 is defined, then the maximum stress can be computed as

σmax=σy+32a1b1.\sigma_{max}=\sigma_y+\sqrt{\dfrac{3}{2}}\dfrac{a^1}{b^1}.

In this case, it is

σmax=0.1 GPa+3250500 GPa=222.47 MPa.\sigma_{max}=0.1~\mathrm{GPa}+\sqrt{\dfrac{3}{2}}\dfrac{50}{500}~\mathrm{GPa}=222.47~\mathrm{MPa}.

The cyclic response is shown as follows.

Of course, multiple sets of aia^i and bib^i can be defined.

material ArmstrongFrederick 1 2E2 .2 .1 0. 0. 0. 50. 500. 100. 600.

Accordingly, the maximum stress is

σmax=σy+32aibi=0.1+32(50500+100600)=426.60 MPa.\sigma_{max}=\sigma_y+\sqrt{\dfrac{3}{2}}\sum\dfrac{a^i}{b^i}=0.1+\sqrt{\dfrac{3}{2}}\left( \dfrac{50}{500}+\dfrac{100}{600}\right)=426.60~\mathrm{MPa}.

The cyclic response is shown as follows.

Zero Elastic Range

It is possible to define a zero plastic range response, although the initial stiffness cannot be explicitly assigned.

material ArmstrongFrederick 1 2E2 .2 0. 0. 0. 0. 40.82482305 500.

The corresponding maximum stress is 100 MPa100~\mathrm{MPa}.

With some linear isotropic hardening,

material ArmstrongFrederick 1 2E2 .2 0. .1 0. 0. 40.82482305 500.

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