User:Ayteebee/sandbox

$$ \text{depth of cut} = \frac{\text{diameter of rod} - \text{diameter of machined feature}}{2} $$

$$ \text{cutting velocity} = \frac{\text{cutting speed} \times \pi \times \text{diameter of rod}}{12} $$

$$ \text{cutting time} = \frac{\text{length of feature}}{\text{feed rate} \times \text{cutting speed}} $$

$$ \text{material removal rate} = \text{cutting speed} \times \text{feed rate} \times \text{depth of cut} \times \pi \times \frac{\text{diameter of rod} + \text{diameter of feature}}{2} $$

$$ \text{Line efficiency} = \frac{\text{Actual production}}{\text{Maximum possible production}} $$

$$ \text{Machine utilisation rate} = \frac{\text{Processing time}}{\text{Processing time + Idle time}} = \frac{\text{Processing time}}{\text{Total available time}} $$

$$ \text{Slenderness ratio } \lambda = \frac{\text{L}}{\text{k}} = \frac{\text{2.5}}{\frac{1}{2} \times 25\times10^-3} = 200 $$

Euler failure:

$$ \sigma_e = \frac{\pi^2 E}{\lambda^2} = \frac{\pi^2 \times 60\times10^9}{200^2} = 14.8 \text{ MPa} $$

Rankine-Gordon failure model:

$$ \sigma_{cr} = \frac{\sigma_y}{1 + \frac{\sigma_y \lambda^2}{\pi^2 E}} = \frac{24\times10^6}{1 + \frac{24\times10^6 \times 200^2}{\pi^2 \times 60\times10^9}} = 9.16 \text{ MPa} $$

Johnson's Parabola:

$$ \sigma_{cr} = \sigma_y - \left( \frac{\sigma_y^2}{4\pi^2 E} \right) \lambda^2 = 24\times10^6 - \left( \frac{\left(24\times10^6\right)^2 \times 200^2}{4 \times \pi^2 \times 60\times10^9} \right) = 14.3 \text{ MPa} $$

Perry-Robertson formula:

$$ \sigma_{cr} = \frac{\sigma_2}{2} - \sqrt{\frac{\sigma_2^2}{4} - \sigma_y \sigma_e } $$

$$ \sigma_2 = \sigma_y + \sigma_e \left( 1 + 0.003\lambda \right) $$

$$ \sigma_2 = 24\times10^6 + 14.8\times10^6 \left( 1 + \left(0.003 \times 200 \right) \right) = 47.7 \text{ MPa} $$

$$ \sigma_{cr} = \left( \frac{47.7\times10^6}{2} \right) - \sqrt{\frac{\left(47.7\times10^6\right)^2}{4} - \left( 24\times10^6 \times 14.8\times10^6 \right) } = 9.23 \text{ MPa} $$