Anisotropic friction for Abaqus/Explicit

This problem contains basic test cases for one or more Abaqus elements and features.

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Elements tested
Features tested
Problem description
Results and discussion
Input files
Figures

ProductsAbaqus/Explicit

Elements tested

C3D8R S4R

Features tested

Anisotropic friction for general contact in Abaqus/Explicit.

Problem description

The models consist of a solid element pressed against a shell element, converted to a rigid element for convenience, and displaced along the shell surface in varying directions. At the beginning of the analysis both contact surfaces have a directional preference oriented with the major axis of the scaling ellipse pointing in the Y-direction (see the vector plot of CORIENT1 in Figure 1). The frictional directional preference factor, $ϵ,$ for both surfaces is equal to 0.5. The nominal or average friction coefficient, $μ_{i s o},$ is 0.2.

Model:

Model-1:


Step-1	The block is pressed on the shell element with a force of 25 N.
Step-2	The block is displaced with a boundary condition in the X-direction.
Step-3	Release displacement in the X-direction.
Step-4	The block is displaced with a boundary condition in the Y-direction.
Step-5	Release displacement in Y-direction.
Step-6	The block is displaced with a boundary condition in the X–Y direction (45°).

Model-2:


Step-1	The block is pressed on the shell element with a force of 25 N.
Step-2	The block is displaced with a boundary condition in the X-direction.
Step-3	Release displacement in X-direction and pressure.
Step-4	Rotate the shell element with a rotation of 90°.
Step-5	Apply the pressure again.
Step-6	The block is displaced again with a boundary condition in the X-direction.

After the shell is rotated, the two surfaces have orthogonal directional preferences and the equivalent scaling ellipse becomes a circle.

Material:


Density	$7.8 e^{- 9}$
Young's modulus	$7.0 e^{4}$
Poisson's ratio	$0.3$

Anisotropic behavior:

The scaling ellipses can be mathematically treated as tensors. The form in the principal coordinate system is

S = [\begin{matrix} 1 + ϵ & 0 \\ 0 & 1 - ϵ \end{matrix}] .

The form in a rotated coordinate system is

S = R S R^{T}

R = [\begin{matrix} \cos (θ) & - \sin (θ) \\ \sin (θ) & \cos (θ) \end{matrix}]

S = R [\begin{matrix} 1 + ϵ & 0 \\ 0 & 1 - ϵ \end{matrix}] R^{T}

S = [\begin{matrix} 1 + ϵ \cos (2 θ) & - ϵ \sin (2 θ) \\ - ϵ \sin (2 θ) & 1 - ϵ \cos (2 θ) \end{matrix}] .

Each ellipse is defined as

S_{s u r f 1} = S_{s u r f 2} = [\begin{matrix} 1 + 0.5 & 0 \\ 0 & 1 - 0.5 \end{matrix}] = [\begin{matrix} 1.5 & 0 \\ 0 & 0.5 \end{matrix}] .

The two surface scaling tensors (ellipses) are averaged to determine the contact scaling tensor (ellipse):

S_{c o n t} = ω_{1} S_{s u r f 1} + ω_{2} S_{s u r f 2} .

Initially the major axis of the two ellipses are aligned, and using a balanced weighting method, the contact scaling ellipse becomes

S_{c o n t} = 0.5 [\begin{matrix} 1.5 & 0 \\ 0 & 0.5 \end{matrix}] + 0.5 [\begin{matrix} 1.5 & 0 \\ 0 & 0.5 \end{matrix}] = [\begin{matrix} 1.5 & 0 \\ 0 & 0.5 \end{matrix}] .

The limit shear force in the principal direction is computed by multiplying $S$ by the average friction coefficient and the normal force,

F_{L i m i t} = F_{N} S_{c o n t} μ_{i s o} = 25 [\begin{matrix} 1.5 & 0 \\ 0 & 0.5 \end{matrix}] 0.2 = [\begin{matrix} 7.5 & 0 \\ 0 & 2.5 \end{matrix}] .

Results and discussion

The reaction forces are checked and compared with the analytical results.

Model-1: since the equivalent scaling ellipse is oriented as the Y-direction and the two surfaces are not rotating elatively, the expected reaction forces are equal to $F x$ = 2.5 N (Step-2) and $F y$ = 7.5 N (Step-4) during the sliding in the X- and Y-directions.(see Figure 3). In Step-6 the block is sliding in a diagonal direction in the X–Y plane (45°). By definition, the direction of the slip is orthogonal to the equivalent scaling ellipse (see Figure 2).
$F_{X} = A k_{x} y_{x}^{e l}, F_{y} = A k_{y} y_{y}^{e l},$
where
$\frac{k_{y}}{k_{x}} = {(\frac{1 + ε}{1 - ε})}^{2} = {(\frac{1.5}{0.5})}^{2} = 9.$
For $y_{x}^{e l} = y_{y}^{e l}$ (45° sliding),
$\frac{F_{y}}{F_{x}} = 9, θ = \tan^{- 1} (9) \approx 83.6 ° .$
The values of the forces $F_{x}$ and $F_{y}$ computed intersecting the scaling ellipse using the angle $θ = 83.6 °$ are
$F_{x} = 25 * 0.033 = 0.825$ and $F_{y} = 25 * 0.284 = 7.1$ (see Figure 3).
Model-2: during Step-2 the equivalent scaling ellipse is oriented as the Y-direction, so the expected reaction force is $F x$ = 2.5 N. After the rotation of the shell, the directional preferences of the two surfaces become orthogonal, generating an equivalent scaling ellipse equal to a circle for the isotropic case (see Figure 4).

Input files

xpl_anisFric_1.inp: Input file for Model-1.

xpl_anisFric_2.inp: Input file for Model-2.

Figures