VM29
VM29
Friction on a Support Block
Overview
Test Case
A support block is acted upon by forces Fx and Fy. For a given value of Fy determine the smallest value of Fx which will prevent the support block from sliding down the plane. Also determine the normal force Fn and sliding force Fs at the interface.
| Material Properties | Geometric Properties | Loading | |||
|---|---|---|---|---|---|
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Analysis Assumptions and Modeling Notes
A real problem of this nature would be solved in a trial-and-error, iterative fashion. In this case the theoretical answer is known, so the solution is verified by the "backward" process described below.
The normal stiffness of the sticking interface is arbitrarily selected at a high value. A value slightly greater than the calculated Fx value of 5.76728 lb is input in the first load step. A slightly lesser value is input in the second load step. The number of sub-steps is limited to one to prevent divergence due to the free motion of the block. The problem is first solved using CONTAC12 elements and then using CONTA178 elements. For CONTA178 elements the contact normals are defined using real constants NX, NY and NZ.
Results Comparison
The sliding force for CONTAC12, a 2D element, is an absolute value. The sliding force for CONTA178, a 3D element, will show the expected +/- sign.
| Status | Target | Mechanical APDL | Ratio |
|---|---|---|---|
| CONTAC12 elements | |||
| Fx = 5.7674 lb. | Sticking | Sticking | -- |
| Fn , lb | -95.942 | -95.942 | 1.000 |
| Fs , lb | 28.783 | 28.782 | 1.000 |
| Fx = 5.76720 lb. | Sliding | Sliding | -- |
| Fn , lb | -95.942 | -95.942 | 1.000 |
| Fs , lb | 28.783 | 28.783 | 1.000 |
| CONTA178 elements | |||
| Fx = 5.76724 lb. | Sticking | Sticking | -- |
| Fn , lb | -95.942 | -95.942 | 1.000 |
| Fs , lb | -28.783 | -28.782 | 1.000 |
| Fx = 5.76720 lb. | Sliding | Sliding | -- |
| Fn , lb | -95.942 | -95.942 | 1.000 |
| Fs , lb | -28.783 | -28.783 | 1.000 |