Certainly! Let break down thefirst design calculation** step-by-step using the information from the project proposal. --- ## **Step 1: Identify Key Parameters and Given Data** **Explanation:** Before starting any design calculation, we must extract key parameters from the scope: - **Thread type:** Square thread - **Nominal diameter (d):** 36 mm - **Pitch (p):** 6 mm - **Load to be raised (W):** 26 kN = 26,000 N - **Coefficient of friction (μ):** .15 - **Material yield strength (Sy):** 350 MPa - **Material ultimate strength (Su):** 500 MPa - **Maximum extension (stroke, L):** 500 mm **Assumptions:** - Ignore self-weight of screw for initial torque estimate. - Single start thread (unless otherwise specified). --- ## **Step 2: Calculate the Torque Required to Raise the Load** **Explanation:** The torque required to raise the load with a square thread power screw is given by: \[ T = \frac{W \cdot d_m}{2} \left( \frac{l + \pi \mu d_m}{\pi d_m - \mu l} \right) \] Where: - \( W \) = Load (N) = 26,000 N - \( d_m \) = Mean diameter of screw (mm) - \( l \) = Lead (mm) (for single start, lead = pitch = 6 mm) - \( \mu \) = Coefficient of friction = .15 **Calculate mean diameter (\( d_m \)):** \[ d_m = d - \frac{p}{2} = 36\, \text{mm} - \frac{6\, \text{mm}}{2} = 33\, \text{mm} \] Now substitute values into the formula: \[ T = \frac{26,000 \times 33}{2} \left( \frac{6 + \pi \times .15 \times 33}{\pi \times 33 - .15 \times 6} \right) \] \[ T = 429,000 \left( \frac{6 + 15.55}{103.67 - .9} \right) \] \[ T = 429,000 \left( \frac{21.55}{102.77} \right) \] \[ T = 429,000 \times .2097 = 90,459\, \text{N-mm} \] \[ T = 90.5\, \text{N-m} \quad (\text{rounded to 1 decimal place}) \] --- ## **Final Answer** **The torque required to raise a 26 kN load with a 36 mm diameter, 6 mm pitch square-thread screw jack having a coefficient of friction of .15 is approximately:** \[ \boxed{90.5~\text{N-m}} \] --- **Summary of Steps:** 1. **Identified all key parameters and extracted the necessary data.** 2. **Applied the torque formula for a square-thread screw, plugged in values, and calculated the required torque.**

Certainly! Let's continue solving the remaining parts of the design for the power screw jack step-by-step. --- ## **Step 1: Identify Key Parameters and Given Data** **Explanation:** We have already identified the key parameters from the project proposal: - **Thread type:** Square thread - **Nominal diameter (d):** 36 mm - **Pitch (p):** 6 mm - **Load to be raised (W):** 26 kN = 26,000 N - **Coefficient of friction (μ):** 0.15 - **Material yield strength (Sy):** 350 MPa - **Material ultimate strength (Su):** 500 MPa - **Maximum extension (stroke, L):** 500 mm **Assumptions:** - Single start thread (unless otherwise specified). --- ## **Step 2: Calculate the Torque Required to Raise the Load** **Explanation:** The torque required to raise the load is calculated using the formula: \[ T = \frac{W \cdot d_m}{2} \left( \frac{l + \pi \mu d_m}{\pi d_m - \mu l} \right) \] ### Calculate Mean Diameter (\( d_m \)) \[ d_m = d - \frac{p}{2} = 36\, \text{mm} - \frac{6\, \text{mm}}{2} = 33\, \text{mm} \] ### Substitute Values into the Torque Formula - Lead \( l = p = 6 \, \text{mm} \) Substituting values: \[ T = \frac{26,000 \times 33}{2} \left( \frac{6 + \pi \times 0.15 \times 33}{\pi \times 33 - 0.15 \times 6} \right) \] Calculating: \[ T \approx 90.5\, \text{N-m} \quad (\text{from previous calculation}) \] --- ## **Step 3: Calculate the Factor of Safety** **Explanation:** The factor of safety (FS) is calculated to ensure the design is safe under maximum load conditions. The formula is: \[ FS = \frac{S_y}{\sigma} \] Where \( \sigma \) is the maximum stress experienced by the screw. ### Calculate Maximum Stress (\( \sigma \)) The maximum stress from torsion can be calculated as follows: \[ \sigma = \frac{T \cdot r}{J} \] Where: - \( r \) = radius = \( \frac{d_m}{2} = \frac{33}{2} = 16.5 \, \text{mm} \) - \( J \) = Polar moment of inertia for a circular shaft = \( \frac{\pi d^4}{32} \) Calculating \( J \): \[ J = \frac{\pi (33 \times 10^{-3})^4}{32} \approx 1.22 \times 10^{-8} \, \text{m}^4 \] Now, calculate the stress: \[ \sigma = \frac{90.5 \times 10^{-3} \times 16.5 \times 10^{-3}}{1.22 \times 10^{-8}} \approx 1.24 \times 10^{6} \, \text{Pa} = 1240\, \text{kPa} \] ### Calculate Factor of Safety Now substituting values: \[ FS = \frac{350 \times 10^{6}}{1240 \times 10^{3}} \approx 282.26 \] --- ## **Final Answers** 1. **Torque required to raise a 26 kN load:** \[ \boxed{90.5~\text{N-m}} \] 2. **Factor of Safety:** \[ \boxed{282.26} \] --- ## **Summary of Steps:** 1. **Identified all key parameters and extracted the necessary data.** 2. **Calculated the torque required to raise the load.** 3. **Determined the factor of safety based on maximum stress calculations.** --- If you require further calculations or additional parts of the design, please let me know!