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| 1. |
What is the velocity ratio for creep in the belt drive system for σ1 being the stress in tight side, σ2 being the stress on slack side and E is the Young’s modulus of elasticity of the belt material? |
| A. | \(\frac{{{N_1}}}{{{N_2}}} = \frac{{{d_1}}}{{{d_2}}} \times \frac{{E - \sqrt {{\sigma _2}} }}{{E - \sqrt {{\sigma _1}} }}\) |
| B. | \(\frac{{{N_1}}}{{{N_2}}} = \frac{{{d_1}}}{{{d_2}}} \times \frac{{E + \sqrt {{\sigma _2}} }}{{E - \sqrt {{\sigma _1}} }}\) |
| C. | \(\frac{{{N_2}}}{{{N_1}}} = \frac{{{d_1}}}{{{d_2}}} \times \frac{{E + \sqrt {{\sigma _2}} }}{{E + \sqrt {{\sigma _1}} }}\) |
| D. | \(\frac{{{N_1}}}{{{N_2}}} = \frac{{{d_1}}}{{{d_2}}} \times \frac{{E - \sqrt {{\sigma _2}} }}{{E + \sqrt {{\sigma _1}} }}\) |
| Answer» D. \(\frac{{{N_1}}}{{{N_2}}} = \frac{{{d_1}}}{{{d_2}}} \times \frac{{E - \sqrt {{\sigma _2}} }}{{E + \sqrt {{\sigma _1}} }}\) | |