The formula to calculate the Resolved Shear Stress (τR) is:
\[ \tau_R = \sigma \cos(\phi) \cos(\lambda) \]
Where:
Resolved Shear Stress is the shear component that exists at all but parallel or perpendicular alignments to the stress direction.
Applied Stress is the external force applied to a material.
Slip Plane Angle is the angle between the normal to the slip plane and the applied stress direction.
Slip Direction Angle is the angle between the slip and stress directions.
Let's assume the following values:
Using the formula:
\[ \tau_R = 93.3 \cos(0.5236) \cos(0.5236) \]
Evaluating:
\[ \tau_R \approx 69.975 \, \text{Pa} \]
The Resolved Shear Stress is approximately 69.975 Pa.
| Applied Stress (σ) (Pa) | Slip Plane Angle (φ) (radians) | Slip Direction Angle (λ) (radians) | Resolved Shear Stress (τR) (Pa) |
|---|---|---|---|
| 90 | 0.5 | 0.5 | 69.3136 |
| 90 | 0.5 | 0.5236 | 68.4007 |
| 90 | 0.5 | 0.55 | 67.3345 |
| 90 | 0.5236 | 0.5 | 68.4007 |
| 90 | 0.5236 | 0.5236 | 67.4999 |
| 90 | 0.5236 | 0.55 | 66.4477 |
| 90 | 0.55 | 0.5 | 67.3345 |
| 90 | 0.55 | 0.5236 | 66.4477 |
| 90 | 0.55 | 0.55 | 65.4118 |
| 93.3 | 0.5 | 0.5 | 71.8551 |
| 93.3 | 0.5 | 0.5236 | 70.9088 |
| 93.3 | 0.5 | 0.55 | 69.8034 |
| 93.3 | 0.5236 | 0.5 | 70.9088 |
| 93.3 | 0.5236 | 0.5236 | 69.9749 |
| 93.3 | 0.5236 | 0.55 | 68.8841 |
| 93.3 | 0.55 | 0.5 | 69.8034 |
| 93.3 | 0.55 | 0.5236 | 68.8841 |
| 93.3 | 0.55 | 0.55 | 67.8103 |
| 95 | 0.5 | 0.5 | 73.1644 |
| 95 | 0.5 | 0.5236 | 72.2008 |
| 95 | 0.5 | 0.55 | 71.0753 |
| 95 | 0.5236 | 0.5 | 72.2008 |
| 95 | 0.5236 | 0.5236 | 71.2499 |
| 95 | 0.5236 | 0.55 | 70.1392 |
| 95 | 0.55 | 0.5 | 71.0753 |
| 95 | 0.55 | 0.5236 | 70.1392 |
| 95 | 0.55 | 0.55 | 69.0458 |