Let's break down the problem step by step: --- ## **Given Data** - **Shaft Length (ABC)** 1.2 m - **Power Transmitted (P = 15 kW = 15,000 W - **Rotational Speed (N)** = 600 rpm - **Radial Force B (F_r)** = 100 N - **Axial Force at B (F_a)** = 650 N - **Yield Strength (\(\sigma_{y}\))** = 310 MN/m² = \(310 \times 10^6\) N/m² - **Factor of Safety (FOS)** = 3 --- ## **Step 1: Calculate the Torque Transmitted** \[ P = 2\pi N T / 60 \] \[ T = \frac{P \times 60}{2\pi N} \] \[ T = \frac{15000 \times 60}{2\pi \times 600} \] \[ T = \frac{900000}{3769.9} \] \[ T = 238.7 \text{ Nm} = 238,700 \text{ Nmm} \] --- ## **Step 2: Find Maximum Bending Moment (M)** Since the gear is at the midpoint (B), reactions at A and C are equal and each force acts as a point load at midspan. ### **Vertical Plane (Radial Force, 100 N)** - Each bearing reaction (vertical) = \(100/2 = 500\) N - Maximum bending moment at B (center): \[ M_{max,V} = 500 \text{ N} \times .6 \text{ m} = 300 \text{ Nm} = 300,000 \text{ Nmm} \] ### **Horizontal Plane (Axial Force, 650 N)** - Each bearing reaction (horizontal) = \(650/2 = 325\) N - Maximum bending moment at B (center): \[ M_{max,H} = 325 \text{ N} \times .6 \text{ m} = 195 \text{ Nm} = 195,000 \text{ Nmm} \] ### **Resultant Maximum Bending Moment at B:** \[ M_{max} = \sqrt{M_{max,V}^2 + M_{max,H}^2} \] \[ M_{max} = \sqrt{(300,000)^2 + (195,000)^2} \] \[ M_{max} = \sqrt{90 \times 10^{9} + 38.025 \times 10^{9}} \] \[ M_{max} = \sqrt{128.025 \times 10^{9}} \] \[ M_{max} \approx 357,850 \text{ Nmm} \] --- ## **Step 3: Equivalent Twisting Moment (Combined Bending + Torsion)** \[ T_{eq} = \sqrt{M_{max}^2 + T^2} \] \[ T_{eq} = \sqrt{(357,850)^2 + (238,700)^2} \] \[ = \sqrt{128.05 \times 10^9 + 57.02 \times 10^9} \] \[ = \sqrt{185.07 \times 10^9} \] \[ = 430,117 \text{ Nmm} \] --- ## **Step 4: Allowable Stress** \[ \text{Allowable Stress} = \frac{\sigma_{y}}{\text{FOS}} = \frac{310 \times 10^6}{3} = 103.33 \times 10^6 \text{ N/m}^2 = 103.33 \text{ N/mm}^2 \] --- ## **Step 5: Shaft Diameter Calculation** For a solid shaft: \[ \tau_{max} = \frac{16 T_{eq}}{\pi d^3} \] \[ d^3 = \frac{16 T_{eq}}{\pi \tau_{allowable}} \] \[ d^3 = \frac{16 \times 430,117}{\pi \times 103.33} \] \[ d^3 = \frac{6,481,872}{324.61} \] \[ d^3 = 19,972.6 \] \[ d = \sqrt[3]{19,972.6} \] \[ d \approx 27.2 \text{ mm} \] --- ## **Step 6: Final Shaft Diameter (with rounding)** Always round **up** to the next standard shaft size for safety. So, **shaft diameter = 28 mm (Standard Size)**. --- # **Final Answer** ### **The recommended shaft diameter is:** \[ \boxed{28\ \text{mm}} \] --- ### **Summary of Steps:** 1. **Calculated Torque** from power and speed. 2. **Calculated Reactions and Max Bending Moments** (vertical & horizontal). 3. **Found Resultant Bending Moment** at gear location. 4. **Got Equivalent Twisting Moment** combining bending and torsion. 5. **Checked Allowable Shear Stress** using yield and FOS. 6. **Solved for Minimum Shaft Diameter** and rounded up. --- **If you want to see calculations for a hollow shaft or further explanation on any step, let me know!**