Evaluating the B-bar method with simple examples

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$$ \newcommand{\B}{\text{B}} \newcommand{\F}{\text{F}} \newcommand{\I}{\mathbf I} \newcommand{\intD}[1]{\int_{\Omega_e}#1\mathrm{d}\Omega} $$

Evaluating the B-bar method with simple examples

This is a plane strain example demonstrating the mechanical behavior of a $1\text{m}\times1\text{m}$ domain under a constant pressure of 10 MPa.

We analyze two problems, referred to as the homogeneous and heterogeneous problems. The schematics of these problems are shown in the figures below:

Homogeneous problem	heterogeneous problem

As can be seen from the schematic figures, the settings of the two problems are identical except for the boundary conditions on the bottom edge. For the homogeneous problem, the bottom boundary has a roller support with the horizontal displacement fixed at the lower left corner. Under these boundary conditions, the strain and stress remain homogeneous throughout the domain. For the heterogeneous problem, all displacement components on the bottom are fixed, leading to heterogeneous strain, and stress within the domain.

Material compressibility is accounted for by using a Poisson’s ratio (nu) of 0.2 for the compressible material and 0.499 for the incompressible material.

We also consider the influence of element size on the accuracy of the solutions. Therefore, the tests are conducted using different meshes, generated by dividing the domain into a varying number of elements per edge: 2, 10, 15, 20, 25, and 30, respectively.

import os… (click to toggle)

import os
from pathlib import Path

import matplotlib.pyplot as plt
import numpy as np
import ogstools as ot

out_dir = Path(os.environ.get("OGS_TESTRUNNER_OUT_DIR", "_out"))
out_dir.mkdir(parents=True, exist_ok=True)


def run(bbar: bool, n: int, homo: bool, nu: float) -> ot.Mesh:
    name = f"{homo=}_{bbar=}_{nu=}".replace("=", "_").lower()
    model = ot.Project(
        input_file="simple_b_bar_test.prj", output_file=out_dir / "modified.prj"
    )
    model.replace_text(str(bbar).lower(), "./processes/process/use_b_bar")
    model.replace_text(name, "./time_loop/output/prefix")
    model.replace_text(2, "./process_variables/process_variable/order")
    if homo:
        model.replace_text("origin", ".//boundary_condition[1]/geometry")
    model.replace_text(str(nu), ".//parameter[name='nu']/value")
    model.replace_text(f"quad_edge_div_{n}.vtu", xpath="./mesh")
    model.write_input()
    model.run_model(logfile=out_dir / "out.txt", args=f"-o {out_dir} -m .")
    return ot.MeshSeries(out_dir / (name + ".pvd"))[-1]


def multi_run(bbar: bool, n: list[int], homo: bool, nu: float) -> list[ot.Mesh]:
    return [run(homo=homo, nu=nu, bbar=bbar, n=n_i) for n_i in n]


def contourplots(
    results: dict[bool, list[ot.Mesh]], var: ot.variables.Variable
) -> plt.Figure:
    fig, axs = plt.subplots(1, 2, figsize=[8, 3], sharey=True)
    for (bbar, meshes), ax in zip(results.items(), axs, strict=True):
        ot.plot.contourf(meshes[-1], var, fig=fig, ax=ax, show_edges=True)
        ax.set_title(f"{bbar=}")
    ot.plot.utils.update_font_sizes(fig.axes, 10)
    return fig


def center_data(meshes: list[ot.Mesh], var: ot.variables.Variable) -> np.ndarray:
    center_ids = [mesh.find_closest_point(mesh.center) for mesh in meshes]
    return np.asarray(
        [var.transform(mesh)[idx] for idx, mesh in zip(center_ids, meshes, strict=True)]
    )


def plot_center_per_refinement(results: dict[bool, list[ot.Mesh]]) -> plt.Figure:
    fig, axs = plt.subplots(1, 3, figsize=(9, 3))
    ax: plt.Axes
    for bbar, meshes in results.items():
        for ax, var in zip(axs, [u["y"], eps["yy"], sig["yy"]], strict=True):
            vals = center_data(meshes, var)
            ax.plot(n_range, vals, ls="-" if bbar else "--", label=f"{bbar=}")
            ax.set_xlabel("Number of elements per side")
            ax.set_ylabel(var.get_label())
            ax.legend()
    fig.tight_layout()
    return fig


def compare_data(results: dict[bool, list[ot.Mesh]], var: ot.variables.Variable):
    for bbar, meshes in results.items():
        vals = var.transform(meshes[-1], strip_unit=False)
        print(f"{var.output_name}, {bbar=}:", end="\t")
        print(f"mean={np.mean(vals):.3g}", end="\t")
        print(f"range of deviation={np.ptp(vals):.3g}")


def compare_with_analytical_solution(
    results: dict[bool, list[ot.Mesh]], nu: float, E: float = 1e10, p: float = -10e6
):
    def eps_ref(nu: float):
        lambd = nu * E / (1 + nu) / (1 - 2.0 * nu)
        G = 0.5 * E / (1 + nu)
        eps_11 = -0.25 * lambd * p / G / (lambd + G)
        eps_22 = eps_11 + 0.5 * p / G
        return [eps_11, eps_22, 0.0, 0.0]

    def u_ref(nu: float):
        eps_ref_val = eps_ref(nu)
        return [0.5 * eps_ref_val[0], 0.5 * eps_ref_val[1]]

    def sigma_ref(nu: float):
        return [0.0, p, nu * p, 0.0]

    def check(a, b, atol):
        np.testing.assert_allclose(a, np.tile(b, (len(a), 1)), 1e-10, atol)

    combinations = [[u, u_ref, 3e-5], [eps, eps_ref, 8e-16], [sig, sigma_ref, 4e-5]]
    for meshes in results.values():
        for var, ref, abstol in combinations:
            num_data = center_data(meshes, var.replace(data_unit=var.output_unit))
            check(num_data, ref(nu), abstol)


def test_center_difference_below(
    results: dict[bool, list[ot.Mesh]],
    var: ot.variables.Variable,
    atols: tuple[float, float],
):
    vals_a, vals_b = (center_data(results[True], var), center_data(results[False], var))
    max_diff = np.geomspace(atols[0], atols[1], len(n_range))
    np.testing.assert_array_less(np.abs(vals_a - vals_b), max_diff)


n_range = [2, 10, 15, 20, 25, 30, 40]
u = ot.variables.displacement.replace(output_unit="mm")
eps = ot.variables.strain
sig = ot.variables.stress

Homogeneous problem

The analytical solutions of the stress components on the plane are $\sigma_{11} = 0$, $\sigma_{22} = -10$ MPa. The analytical solutions of the remaining variables are

$$ \begin{align} \sigma_{33} &= \nu(\sigma_{11}+\sigma_{22})\\ \epsilon_{11} &= \frac{\lambda}{4G(\lambda+G)}\sigma_{22}\\ \epsilon_{22} &= \epsilon_{11} + \frac{\sigma_{22}}{2G} \\ u_{1} &= \epsilon_{11}x \\ u_{2} &= \epsilon_{11}y \end{align} $$

with $E$ the Young’s modulus, $G=E/2(1+\nu)$ the shear modulus, $\nu$ the Poisson ratio, and $\lambda=E\nu/((1+\nu)(1-2\nu))$ the Lame constant, $u$ the displacement, $\epsilon$ the strain, and $\sigma$ the stress.

Compressible material (nu=0.2)

homogeneous_compressible = {… (click to toggle)

homogeneous_compressible = {
    bbar: multi_run(bbar, n_range, homo=True, nu=0.2) for bbar in [True, False]
}

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The displacement field is linear and the same in both cases:

fig = contourplots(homogeneous_compressible, u["y"])

png

The vertical stress is constant everywhere aside from random noise.

compare_data(homogeneous_compressible, sig["yy"])

stress_yy, bbar=True:	mean=-10 megapascal	range of deviation=6.8e-13 megapascal
stress_yy, bbar=False:	mean=-10 megapascal	range of deviation=1.24e-12 megapascal

For this homogeneous problem, the solutions obtained using the B-bar method have almost the same accuracy as those obtained with the standard approach. This is confirmed by the following comparisons with the analytical solutions:

compare_with_analytical_solution(homogeneous_compressible, nu=0.2)

Incompressible material (nu=0.499)

homogeneous_incompressible = {… (click to toggle)

homogeneous_incompressible = {
    bbar: multi_run(bbar, n_range, homo=True, nu=0.499) for bbar in [True, False]
}

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Again, the displacement field is linear and the same in both cases:

fig = contourplots(homogeneous_incompressible, u["y"])

png

Also the vertical stress is again constant everywhere aside from random noise.

compare_data(homogeneous_incompressible, sig["yy"])

stress_yy, bbar=True:	mean=-10 megapascal	range of deviation=1.03e-10 megapascal
stress_yy, bbar=False:	mean=-10 megapascal	range of deviation=1.88e-10 megapascal

The following figure shows the vertical components of displacement, strain and stress at the center over the different discretizations. It shows, that this model, with or without the B-bar method, delivers accurate solutions and exhibits excellent mesh size independence. This reflects the fact that this test has no volumetric locking though the material is incompressible.

fig = plot_center_per_refinement(homogeneous_incompressible)

png

compare_with_analytical_solution(homogeneous_incompressible, nu=0.499)

Heterogeneous problem

Compressible material (nu=0.2)

heterogeneous_compressible = {… (click to toggle)

heterogeneous_compressible = {
    bbar: multi_run(bbar, n_range, homo=False, nu=0.2) for bbar in [True, False]
}

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In the following figures we see, that with and without the B-bar method, the variable distribution patterns look quite similar. In the bottom corners the stresses deviate from each other.

fig = contourplots(heterogeneous_compressible, u["y"])
fig = contourplots(heterogeneous_compressible, sig["yy"])

png

test_center_difference_below(heterogeneous_compressible, u["y"], (3e-2, 1e-4))… (click to toggle)

test_center_difference_below(heterogeneous_compressible, u["y"], (3e-2, 1e-4))
test_center_difference_below(heterogeneous_compressible, eps["yy"], (1e-2, 1e-5))
test_center_difference_below(heterogeneous_compressible, sig["yy"], (0.7, 1e-3))

The following figure shows, that the differences in the center between the results with B-Bar and without B-bar tend to become tiny as the mesh refinement level increases.

fig = plot_center_per_refinement(heterogeneous_compressible)

png

Incompressible material (nu=0.499)

heterogeneous_incompressible = {… (click to toggle)

heterogeneous_incompressible = {
    bbar: multi_run(bbar, n_range, homo=False, nu=0.499) for bbar in [True, False]
}

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In the following figures we see, that the bottom corners exhibit even stronger stress concentrations, which is slightly reduced by the B-bar method.

fig = contourplots(heterogeneous_incompressible, u["y"])
fig = contourplots(heterogeneous_incompressible, sig["yy"])

png

The following figure shows, that the differences in the center between the results with B-Bar and without B-bar tend to become smaller as the mesh refinement level increases, but not as much, as in the compressible model.

fig = plot_center_per_refinement(heterogeneous_incompressible)

png

test_center_difference_below(heterogeneous_incompressible, u["y"], (0.2, 3e-3))… (click to toggle)

test_center_difference_below(heterogeneous_incompressible, u["y"], (0.2, 3e-3))
test_center_difference_below(heterogeneous_incompressible, eps["yy"], (2e-2, 3e-4))
test_center_difference_below(heterogeneous_incompressible, sig["yy"], (14, 0.3))

Conclusions

In the FEM analysis of small deformation problems,

The B-bar method can significantly alleviate volumetric locking in incompressible materials with a high-quality mesh.
When applied to problems with compressible materials, the B-bar method can still give accurate solutions with a fine mesh.
The B-bar method may better find the correct solution when the material is highly incompressible and there is volumetric locking.