Let's solve the problem step by step! --- ### **Step 1: Given Data** - **Power (P)** = 50 kW = 50,000 W - **Speed (N)** = 1,750 rpm - **Material:** AISI C1045 as rolled - **Load:** Gradually applied - **Find:** Diameter of shaft (D) --- ### **Step 2: Material Properties** For **AISI C1045 (as rolled)**: - **Yield Strength (\( S_y \))** ≈ 370 MPa - **Allowable Shear Stress (\( \tau_{allow} \))** (for gradually applied load, use 30% of yield): \[ \tau_{allow} = 0.3 \times S_y = 0.3 \times 370 = 111~\text{MPa} \] (You may also use a factor of safety, typically 2, so \(\tau_{allow} \approx 185 \text{ MPa} / 2 = 92.5 \text{ MPa}\), but let's proceed with 111 MPa as per convention.) --- ### **Step 3: Torque Calculation** Power and speed relationship: \[ P = 2\pi N T / 60 \] \[ T = \frac{P \times 60}{2\pi N} \] Plug in the values: \[ T = \frac{50,000 \times 60}{2\pi \times 1750} \] \[ T = \frac{3,000,000}{10,995.6} \] \[ T \approx 273 \text{ Nm} = 273,000 \text{ Nmm} \] --- ### **Step 4: Shaft Diameter Formula** For a **solid circular shaft** subjected to torque: \[ T = \frac{\pi}{16} \tau_{allow} D^3 \] Rearrange to solve for \( D \): \[ D^3 = \frac{16T}{\pi \tau_{allow}} \] \[ D = \left( \frac{16T}{\pi \tau_{allow}} \right)^{1/3} \] --- ### **Step 5: Substitute the Values** - \( T = 273,000 \) Nmm - \( \tau_{allow} = 111 \) MPa \( = 111 \times 10^6 \) N/m² \( = 111 \) N/mm² \[ D = \left( \frac{16 \times 273,000}{\pi \times 111} \right)^{1/3} \] Calculate: \[ \text{Numerator: } 16 \times 273,000 = 4,368,000 \] \[ \text{Denominator: } \pi \times 111 \approx 348.0 \] \[ \frac{4,368,000}{348} \approx 12,552 \] \[ D = (12,552)^{1/3} \] \[ D \approx 23 \text{ mm} \] --- ## **Final Answer** ### **The required diameter of the pump shaft is approximately \(\boxed{23\, \text{mm}}\) if made of AISI C1045 (as rolled) and subjected to gradually applied load.** --- **Let me know if you'd like to see the steps for a hollow shaft or with a different factor of safety!**