Let's solve each part step-by-step. --- ## Q1. Given Data - Power transmitted, \( P = 600\, \text{kW} = 600,000\, \text{W} \) - Speed, \( N = 200\, \text{rpm} \) - Max shear stress, \( \tau_{max} = 60\, \text{N/mm}^2 = 60 \times 10^6\, \text{N/m}^2 \) - Shear modulus, \( G = 80\, \text{GPa} = 80 \times 10^9\, \text{N/m}^2 \) - Length, \( L = 2\, \text{m} \) Let: - Outside diameter = \( D_o \) - Inside diameter = \( D_i \) - \( D_i = k D_o \), usually \( k \) is assumed (let's use \( k = .6 \) for typical design unless specified) --- ## 1. Find External and Internal Diameters ### **Step 1: Find Torque (\( T \))** \[ P = 2\pi N T / 60 \] \[ T = \frac{P \times 60}{2\pi N} \] \[ T = \frac{600,000 \times 60}{2 \pi \times 200} \] \[ T = \frac{36,000,000}{1,257} \] \[ T \approx 28,646\, \text{Nm} \] --- ### **Step 2: Torsion Formula for Hollow Shaft** \[ T = \frac{\pi}{16} \tau_{max} \frac{D_o^4 - D_i^4}{D_o} \] Let \( D_i = k D_o \) So \( D_i^4 = k^4 D_o^4 \) \[ T = \frac{\pi}{16} \tau_{max} D_o^3 (1 - k^4) \] Plug in values: - \( T = 28,646\, \text{Nm} = 28,646 \times 10^3\, \text{Nmm} \) - \( \tau_{max} = 60\, \text{N/mm}^2 \) \[ 28,646 \times 10^3 = \frac{\pi}{16} \times 60 \times D_o^3 (1 - k^4) \] \[ D_o^3 = \frac{28,646 \times 10^3 \times 16}{\pi \times 60 \times (1 - k^4)} \] Assume \( k = .6 \): \[ 1 - k^4 = 1 - (.6)^4 = 1 - .1296 = .8704 \] Calculate denominator: \[ \pi \times 60 \times .8704 \approx 3.1416 \times 60 \times .8704 \approx 164.16 \] Calculate numerator: \[ 28,646 \times 10^3 \times 16 = 458,336,000 \] So, \[ D_o^3 = \frac{458,336,000}{164.16} = 2,791,225 \] \[ D_o = \sqrt[3]{2,791,225} \approx 140.1\, \text{mm} \] \[ D_i = .6 \times 140.1 = 84.1\, \text{mm} \] ### **Final Answers for (1):** - **External diameter, \( D_o \) = 140 mm** - **Internal diameter, \( D_i \) = 84 mm** --- ## 2. Angle of Twist (\( \theta \)) Over 2 m \[ \theta = \frac{T L}{J G} \] Where: - \( J \) is the polar moment of inertia for a hollow shaft: \[ J = \frac{\pi}{32} (D_o^4 - D_i^4) \] Plug in values: \[ D_o = 140\, \text{mm}, \quad D_i = 84\, \text{mm} \] \[ D_o^4 = (140)^4 = 384,160,000 \] \[ D_i^4 = (84)^4 = 497,871 \] \[ D_o^4 - D_i^4 = 384,160,000 - 497,871 = 383,662,129\, \text{mm}^4 \] \[ J = \frac{\pi}{32} \times 383,662,129 = 37,669,652\, \text{mm}^4 \] Convert \( J \) to \( \text{m}^4 \): \[ J = 37,669,652 \times 10^{-12} = 3.77 \times 10^{-5} \text{ m}^4 \] Plug in values: - \( T = 28,646\, \text{Nm} \) - \( L = 2\, \text{m} \) - \( G = 80 \times 10^9\, \text{N/m}^2 \) \[ \theta = \frac{28,646 \times 2}{3.77 \times 10^{-5} \times 80 \times 10^9} \] \[ = \frac{57,292}{3.016 \times 10^6} \] \[ \approx .019\, \text{radians} \] \[ = .019 \times \frac{180}{\pi} \approx 1.09^\circ \] ### **Final Answer for (2):** - **Angle of twist = \( .019 \) radians (or \( 1.09^\circ \))** --- ## 3. **Advantages of Hollow Shaft Over Solid Shaft** - **Weight Saving:** Hollow shafts are lighter for the same torque capacity, reducing overall system weight. - **Material Efficiency:** Most torque is transmitted at the outer fibers; removing inner material doesn’t much affect strength. - **Cost Saving:** Less material is used, reducing cost. - **Better Torsional Strength-to-Weight Ratio:** For a given weight, hollow shafts can transmit more torque. - **Application:** Used in automotive, aerospace, and marine applications where weight is critical. --- ## **Summary Table** | Part | Answer | |------|--------| | 1 | \( D_o = 140\, \text{mm},\, D_i = 84\, \text{mm} \) | | 2 | Angle of twist \( \theta = .019 \) rad (\( 1.09^\circ \)) | | 3 | Hollow shafts are lighter, more efficient, & cost-effective for same torque | --- If you want a more detailed explanation of any step, let me know!