Friction models in Abaqus/Standard

The first two steps of the analysis establish contact between each rod and the rigid surface and set up an equilibrium solution in which each beam element is compressed by a force of 300. The temperature of the secondary node is specified as 20° and that of the rigid surface, as 0°; therefore, the average surface temperature is 10° when contact is established. In Step 3 the normal force is increased to 400, and a shear force is applied to the first rod such that ${\tau }_{eq}<{\tau }_{crit}$ and the rod remains sticking. The shear force is removed in Step 4. In Step 5 the friction model for rod 1 is modified. The normal force is increased to 550, and a shear force is applied such that ${\tau }_{eq}<{\tau }_{crit}$ and the rod still remains sticking. The shear forces are removed in Step 6. In Step 7 the original friction model is specified with the friction properties reset to their original values. The pressure on rod 1 is increased to 850, and a slip is applied. In Step 8 a slip velocity–dependent friction model is introduced for rod 2. In Step 9 a slip is applied to rod 2 in which the slip rate is varied by prescribing the displacement with an amplitude curve during the static step.

Surface interaction for rod 1:

Step 1

${\mu }_1=$ 0.005$\overline{\theta }$ + 2.5 × 10⁻⁴($p$ − 100).

${\mu }_2=$ 0.005$\overline{\theta }$ + 3.3 × 10⁻⁴($p$ − 100). (Anisotropic model only for case with 2 local tangent directions.)

Step 5

$\mu =$ 0.002$\overline{\theta }$ + 3.3 × 10⁻⁴$p$ for 100 $<p\le$ 500.

$\mu =$ 0.1650 + 0.002$\overline{\theta }$ + 5.5 × 10⁻⁴($p$ − 500) for 500 $<p\le$ 900.

Step 7

Same as specified in Step 1.

Step 8

$\mu =$ 0.0 for the elastic slip formulation.

Rough friction model for the Lagrange multiplier formulation.

Surface interaction for rod 2:

Step 1

$\mu =$ 0.0 for the elastic slip formulation.

Rough friction model for the Lagrange multiplier formulation.

Step 8

$\mu ={\mu }_s$ for ${\dot{\gamma }}_{eq}=$ 0;

$\mu ={\mu }_s-({\mu }_s-{\mu }_k){\dot{\gamma }}_{eq}/$2.0 for 0 $<{\dot{\gamma }}_{eq}\le$ 2.0;

$\mu ={\mu }_k$ for ${\dot{\gamma }}_{eq}>$ 2.0, where

${\mu }_s=$ 0.2 and ${\mu }_k=$ 0.0 for $p=$ 100.0 and

${\mu }_s=$ 0.4 and ${\mu }_k=$ 0.2 for $p=$ 500.0.

The first two steps of the analysis establish contact between each rod and the rigid surface and set up an equilibrium solution in which each beam element is compressed by a force of 300. The pressure is kept constant throughout the analysis. In Step 3 a shear force is applied to rod 1 such that ${\tau }_{eq}<{\tau }_{crit}$ and the rod remains sticking. The shear force is removed in Step 4. In Step 5 the friction model for rod 1 is modified by providing test data. A shear force is applied such that ${\tau }_{eq}<{\tau }_{crit}$ and the rod remains sticking. The shear forces are removed in Step 6. In Step 7 the original friction model is specified with the friction properties reset to their original values. A slip is applied to rod 1. In Step 8 a new friction model is introduced for rod 2. In Step 9 a slip is applied to rod 2 in which the slip rate is varied by prescribing the displacement with an amplitude curve during the static step.

Surface interaction for rod 1:

Step 1

${\mu }_s=$ 0.3;

${\mu }_k=$ 0.1;

$d_c=$ 4.

Step 5

Test data input:

${\mu }_1=$ 0.5, ${\dot{\gamma }}_1=$ 0.0;

${\mu }_2=$ 0.3, ${\dot{\gamma }}_2=$ 0.2;

${\mu }_3=$ 0.2, ${\dot{\gamma }}_3=\mathrm{\infty }$.

Step 7

Same as specified in Step 1.

Step 8

$\mu =$ 0.0 for the elastic slip formulation.

Rough friction model for the Lagrange multiplier formulation.

Surface interaction for rod 2:

Step 1

$\mu =$ 0.0 for the elastic slip formulation.

Step 8

Test data input:

${\mu }_1=$ 0.3, ${\dot{\gamma }}_1=$ 0.0;

${\mu }_2=$ 0.1, ${\dot{\gamma }}_2=$ 0.2.

It is assumed that $\frac{\left({\mu }_2-{\mu }_{\mathrm{\infty }}\right)}{\left({\mu }_1-{\mu }_{\mathrm{\infty }}\right)}=$ 0.05.