Mises creep
Elements tested
Problem description
Material:
- Elasticity
Young's modulus, E = 20.0E6
Poisson's ratio,
= 0.3
- Creep
Time-hardening/strain-hardening power law
A = 2.5E−27
n = 5.0
m = −0.2
Hyperbolic-sine law
A = 2.5E−27
B = 4.4E−4
n = 5.0
= 0.0
R = 8.314
Anand law
= 1 .6
= 2.5E-14
= 1.0E4
= -600.0
= 0.0
= 0.0
= 0.0
= 10000.0
= -500.0
= 0.0
= 1.0E-4
= 5.0
= 0.5
= 1.0
= 1.0
= 0.0
Darveaux law
= 1.0E6
= 2.25E-12
= 2.0
= 5.0
= 1.0E-5
= 2.0E-4
= 0.0
Double power law
= 1.5E-10
= 2.5E-11
= 1.0
= 4.0
= 2.0
= 5.0
= 10000.0
= 0.0
(The units are not important.)
Results and discussion
The tests in this section are set up as cases of homogeneous deformation of
a single element. Consequently, the results are identical for all integration
points within the element. The elements have unit dimensions except in the
loading direction in which they have a length of 10. The constitutive path is
integrated with the quasi-static procedure using automatic incrementation.
Therefore, the number of increments varies from test to test. The results are
reported at a convenient increment near the halfway point of the response and
at the end of the test.
Hill creep
Elements tested
Problem description
Material:
- Elasticity
Young's modulus, E = 20.0E6
Poisson's ratio,
= 0.3
- Creep
A = 2.5E−27
n = 5.0
m = −0.2
Anisotropic creep ratios: 1.5, 1.2, 1.0, 1.0, 1.0, 1.0
(The units are not important.)
Results and discussion
The constitutive path is integrated with the quasi-static procedure using
automatic incrementation. Therefore, the number of increments varies from test
to test.
Mises creep and plasticity
Elements tested
B32 C3D8 C3D8R CPS4 S4 S4R T3D2
Problem description
Material:
- Elasticity
Young's modulus, E = 20.0E6
Poisson's ratio,
= 0.3
- Plasticity
Hardening curve:
Yield stress
| Plastic strain
|
---|
10.0E3
| 0.00
| 50.0E3
| 0.02
|
- Creep
A = 1.0E−24
n = 5.0
m = −0.2
- Swelling
Volumetric swelling rate = 2.0E−6
(The units are not important.)
Results and discussion
The tests in this section verify the coupled Mises creep and plasticity
model for problems involving uniaxial tension, shear, bending, and torsion. The
test cases consider stress spaces with 1, 2, or 3 direct components. Both time
and strain creep laws, as well as volumetric swelling, are considered with the
constitutive path integrated by the quasi-static procedure using automatic
incrementation. Explicit and implicit time integration are employed, with
automatic switching to the implicit scheme once a material point goes plastic.
The solution's accuracy is verified by comparing it to test cases employing
extremely fine time integration.
Drucker-Prager creep and plasticity
Elements tested
Problem description
Material:
- Elasticity
Young's modulus, E = 300.0E3
Poisson's ratio,
= 0.3
- Plasticity
Angle of friction,
= 40.0
Dilation angle,
= 40.0
Third invariant ratio, K = 1.0
Hardening curve:
Yield stress
| Plastic strain
|
---|
6.0E3
| 0.00
| 9.0E3
| 0.02
| 11.0E3
| 0.063333
| 12.0E3
| 0.11
| 12.0E3
| 1.0
|
- Creep
For the time and strain creep laws:
A = 0.5E−7
n = 1.1
m = −0.2
The Singh-Mitchell creep law parameters are varied. For example:
A = 0.002
= 1.0E−6
m = −1.0
= 1.0
(The units are not important.)
Results and discussion
The tests in this section verify the coupled Drucker-Prager creep and
plasticity model. The tests are set up as cases of homogeneous deformation of a
single solid element of unit dimension subjected to uniaxial tension and
compression, shear, and hydrostatic tension. The Singh-Mitchell, time, and
strain hardening creep laws are considered with the constitutive path
integrated by the quasi-static procedure. Explicit and implicit time
integration are employed, with automatic switching to the implicit scheme once
a material point goes plastic.
Cap creep and plasticity
Elements tested
Problem description
Material:
- Elasticity
Young's modulus, E = 300.0E4
Poisson's ratio,
= 0.3
- Cap
plasticity
Material cohesion, d = 2.0E4
Material angle of friction,
= 40.0
Cap eccentricity, R = 0.3
Initial cap yield surface,
= 0.5
Transition surface radius,
= 0.0
Third invariant ratio, K = 1.0
Hardening curve:
Hydrostatic pressure yield stress
| Volumetric plastic strain
|
---|
6.01E3
| 0.0
| 6.04E3
| 0.4
| 1.432E4
| 0.5
| 3.5E4
| 0.7
| 8.7E4
| 1.0
|
- Creep (for both
cohesion and consolidation)
Time-hardening power law:
A = 1.0E−24
n = 5
m = 0.0
Strain-hardening power law:
A = 7.0E−26
n = 5
m = 0.0
Singh-Mitchell type law:
A = 0.002
= 1.6E−4
m = 0.0
= 1.0
User-specified creep law:
The user subroutine for the time creep law specified earlier is implemented.
(The units are not important.)
Results and discussion
The tests in this section verify the cap creep and plasticity model. The
tests are set up as cases of homogeneous deformation of a single solid element
of unit dimension. To validate the model, the element is subjected to various
stress paths including uniaxial tension and compression, shear, hydrostatic
tension and compression, and triaxial compression. The Singh-Mitchell creep
law, the time and strain hardening creep laws, and a user-defined creep model
are considered with the constitutive path integrated by the quasi-static
procedure. Explicit and implicit time integration are employed, with automatic
switching to the implicit scheme once a material point goes plastic.
Additional verification problems for the coupled creep and plasticity
capability
Elements tested
Problem description
Additional verification problems were obtained by adding creep to the
plasticity model of
Limit load calculations with granular materials
and
Finite deformation of an elastic-plastic granular material.
For these cases a small creep strain rate was selected to verify the plasticity
component of the coupled creep and plasticity models. Thus, the results should
be comparable to the equivalent problem without creep, although they are
separate
Abaqus
material models. These verification problems test both the Drucker-Prager creep
and the Drucker-Prager/Cap creep models.
Further verification problems for Mises creep and plasticity were obtained
by adding plasticity to the problems described in
Creep of a thick cylinder under internal pressure
and
Ct-integral evaluation.
For the example described in
Creep of a thick cylinder under internal pressure,
the initial application of the pressure plastifies the cylinder during the
first step of the analysis; and the creep response is then developed in the
second step. For the example described in
Ct-integral evaluation,
the plastic deformation is very small and localized. Plastification occurs only
during the preloading static step. As a result, the -integrals
calculated by
Abaqus
in the early stages of the quasi-static step are expected to differ somewhat
from the ones calculated in the creep-only case and are not path independent.
Later on, when larger scale creep dominates the stress fields, the
-integrals
calculated should converge toward the same values as obtained in the creep-only
case and become path independent.
Results and discussion
The results obtained show good agreement with the corresponding example
problems. The addition of creep in the first two problems has little effect on
the plastic results, and the addition of plasticity in the second two problems
has little effect on the creep results.
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