Damage, Failure and Characteristic Element Length

Damage is designed to model fracture and failure within a continuum material – the creation of new surface area and the associated localised deformation. When considering a uniaxial tension test there are two zones of deformation. The first is undergoing homogenous deformation, where the material law allows forces and displacements to be related to stresses and strains. The second deformation regime is in the region of the flaw which has developed. Here, there is no direct relationship between stresses and strains and forces and displacements due to the non-uniform stress and strain fields. The displacements produced within the region of uniform stress will be proportional to the length of the sample, so if we double the length of the sample the total displacement in this region will also double. However, the strains produced in the region of the flaw are not related to the length of the sample. The change in length in this region will not change whether the sample is 2, 4 or 8 times longer than the original.

To alleviate this problem, the traditional stress strain relationship is replaced with a stress displacement relationship to eliminate size effects post damage initiation. The characteristic length of the element is introduced into the calculation to create a stress displacement response. However, this approach can cause confusion when trying to calibrate a model or determine material data.

There are a number of ways in which the damage evolution can be defined within Abaqus, but for simplicity consider the case where the total plastic displacement (post initiation is defined) (pd), and for arguments sake is set to be 2.0. Consider the following cases.

Case One: Uniform Mesh

Simple Loading		Complex Loading
A simple uniaxial loading on a sample where each element is of a uniform size, 10 x 10 and carries an identical stress state. From the documentation, the strain that occurs in an element post damage initiation (DI) will be the plastic displacement divided by characteristic element length (cl). Therefore, for each element in this model the post DI strain will be 0.2 (from pd/cl). As the stress state is uniform, each element will damage at the same time – the same stress state. Therefore, in this case the total displacement of the sample post DI will be the number of elements  x the displacement to failure, 4 x 2.0 = 8.0 units.		Now consider that these elements are part of a larger body, or subject to some complex loading condition, such that failure will only occur in a single element. This single element must capture the localised deformation and cracking.Therefore, a single element will strain post DI by 0.2, corresponding to a displacement of 2.0.

Case Two: Single Element

Simple Loading		Complex Loading
The same uniaxial loading is applied to a single element, 10 x 40. For simplicity, consider that the characteristic element length is 40. Therefore the strain post DI will be 0.05 (from pd/cl).With a uniform stress state the displacement at failure will be 2.0. An apparent difference caused by different element sizes.		The single element is part of the same larger body or subject to the same complex loading scenario as above. Let’s assume that, whilst it is larger, it is sufficiently refined to capture the maximum stress as above. In this case the single element will fail, and must account for the required localised deformation. In this case the strain will be 0.05 – much lower than the refined mesh. However, the displacement to failure will be the same, 0.05 x 40.0 = 2.0

Under simple, uniform (and possibly artificial) loading the formulation appears to introduce a mesh dependence, borne out of the failure at the same time of all the elements, therefore the force displacement curve will not match when different meshes are used. In an experiment a single crack would occur causing the component to fail. But, the numerics of the stress state mean that a crack is being introduced in all four elements simultaneously leading to much larger displacement to failure for the component. However, under complex loading, the force deflection curve for the fine mesh and coarse mesh will match. The stress state is such that a crack or failure will initiate in only a single element. The plastic displacement at the point of failure will be the same, no matter what size the element.

The goal of the formulation and the introduction of the characteristic element length is to ensure that the plastic displacement at the point of failure (or the fracture energy dissipated during failure) match from one analysis to the next, no matter what the mesh density. This implies that as the mesh changes, the stress strain curve for those elements that are failing will also change (see the calculation for the complex loading scenario), but the load displacement curve for the component will remain constant.

Uniform stress distributions, such as those found in simple calibration analyses, can cause misleading results and need careful interpretation. Of course this never happens in the real world (usually things only break at one location). The reason is that physical materials have imperfections, meaning that there will be some regions of the material with slightly less strength than others. Under uniform loading, these weak regions will start to initiate damage first. In order to maintain stress equilibrium as the weaker regions undergo softening, the surrounding (stronger) regions will simply unload elastically without ever initiating damage. This leads to strain localization in the weaker regions and eventual fracture. The analysis is designed initiate a crack through the component. As this is a single crack it should only occur in a region that is one element wide. To reproduce this in the FE analysis where there is uniformity in the stress distribution, it is necessary to assign a slightly lower strength to one element (e.g. by reducing the damage initiation criterion slightly). In that case the deformation will localize in the weaker link, and the overall force displacement response should be independent of the mesh size. Note, that in many respects this is similar to perturbing the mesh when solving a buckling analysis; both problems lead to bifurcations in the solution.

In summary, once material properties for damage and failure have been determined from test, there is no need to modify the parameters based upon mesh density. One implication is that element aspect ratios should be as close to unity as possible. 

If you have any thoughts on damage and failure within Abaqus then please comment below and we'll answer your questions or take your points on board for future developments.