Let's solve the problem step by step. --- # Q1. Hollow Shaft Design ## **Given Data** - Power, \( P = 600 \) kW \( = 600 \times 10^3 \) W - Speed, \( N = 200 \) rpm - Max shear stress, \( \tau_{max} = 60 \) N/mm² \( = 60 \times 10^6 \) N/m² - Shear modulus, \( G = 80 \) GPa \( = 80 \times 10^9 \) N/m² - Shaft length, \( L = 2 \) m Let: - Outer diameter = \( D_o \) - Inner diameter = \( D_i \) - Let \( D_i = k D_o \) (commonly, \( k = 0.6 \) for hollow shafts, unless otherwise specified) --- ## **1. Finding External and Internal Diameters** ### **Step 1: Calculate Torque (\( T \))** Power transmitted by a shaft: \[ P = 2\pi N T / 60 \] \[ T = \frac{P \times 60}{2\pi N} \] Plug in values: \[ T = \frac{600,000 \times 60}{2\pi \times 200} \] \[ T = \frac{36,000,000}{1,256.64} \] \[ T \approx 28,651 \text{ Nm} \] --- ### **Step 2: Torsion Formula for a Hollow Shaft** \[ \tau_{max} = \frac{T \cdot R_o}{J} \] Where - \( R_o = D_o/2 \) - \( J = \frac{\pi}{32} (D_o^4 - D_i^4) \) (polar moment of inertia) So, \[ \tau_{max} = \frac{T \cdot (D_o/2)}{\frac{\pi}{32} (D_o^4 - D_i^4)} \] \[ \tau_{max} = \frac{16T D_o}{\pi (D_o^4 - D_i^4)} \] Given \( D_i = k D_o \): \[ D_o^4 - D_i^4 = D_o^4 (1 - k^4) \] Rewriting, \[ \tau_{max} = \frac{16T}{\pi D_o^3 (1 - k^4)} \] \[ D_o^3 = \frac{16T}{\pi \tau_{max} (1 - k^4)} \] Let \( k = 0.6 \): - \( 1 - k^4 = 1 - (0.6)^4 = 1 - 0.1296 = 0.8704 \) Plug in the values: - \( T = 28,651 \) Nm \( = 28,651 \times 10^3 \) Nmm - \( \tau_{max} = 60 \) N/mm² \[ D_o^3 = \frac{16 \times 28,651 \times 10^3}{\pi \times 60 \times 0.8704} \] Calculate the denominator: - \( \pi \times 60 \times 0.8704 \approx 3.1416 \times 60 \times 0.8704 \approx 164.213 \) Now: \[ D_o^3 = \frac{458,416,000}{164.213} \] \[ D_o^3 \approx 2,791,424 \text{ mm}^3 \] \[ D_o = \sqrt[3]{2,791,424} \approx 140.3 \text{ mm} \] Now, \[ D_i = k D_o = 0.6 \times 140.3 \approx 84.2 \text{ mm} \] --- ### **Final Answers for 1:** - **External diameter \( D_o \):** **140.3 mm** - **Internal diameter \( D_i \):** **84.2 mm** --- ## **2. Angle of Twist Over 2 m** Angle of twist, \[ \theta = \frac{T L}{G J} \] - \( L = 2 \) m \( = 2000 \) mm - \( G = 80 \) GPa \( = 80 \times 10^3 \) N/mm² - \( T = 28,651 \times 10^3 \) Nmm - \( J = \frac{\pi}{32} (D_o^4 - D_i^4) \) Calculate \( D_o^4 \) and \( D_i^4 \): \[ D_o = 140.3 \text{ mm},\ D_i = 84.2 \text{ mm} \] \[ D_o^4 = (140.3)^4 \approx 388,175,000 \] \[ D_i^4 = (84.2)^4 \approx 50,252,000 \] \[ D_o^4 - D_i^4 = 388,175,000 - 50,252,000 = 337,923,000 \] So, \[ J = \frac{\pi}{32} \times 337,923,000 \approx 0.09817 \times 337,923,000 \approx 33,186,000 \text{ mm}^4 \] Now, \[ \theta = \frac{T L}{G J} = \frac{28,651,000 \times 2000}{80,000 \times 33,186,000} \] Calculate denominator: - \( 80,000 \times 33,186,000 = 2,654,880,000,000 \) Numerator: - \( 28,651,000 \times 2000 = 57,302,000,000 \) So, \[ \theta = \frac{57,302,000,000}{2,654,880,000,000} \] \[ \theta \approx 0.0216 \text{ radians} \] Convert to degrees: - \( 0.0216 \times \frac{180}{\pi} \approx 1.24^\circ \) --- ### **Final Answer for 2:** - **Angle of twist over 2 m = \( 0.0216 \) radians (or \( 1.24^\circ \))** --- ## **3. Advantage of a Hollow Shaft Over a Solid Shaft** - **Weight Reduction:** Hollow shafts are lighter than solid shafts for the same torque capacity because material is distributed away from the center, where it is less effective. - **Material Efficiency:** For the same weight, a hollow shaft can transmit more torque than a solid shaft. - **Less Stress Concentration:** Hollow shafts can better withstand stress concentrations, especially in applications with bending and torsional loads. - **Cost Savings:** Less material usage can lower cost, especially in high-strength materials. - **Torsional Stiffness:** By optimizing the outer and inner diameters, hollow shafts can be designed for high torsional rigidity. --- ## **Summary Table** | Part | Answer | |----------|------------------------------------------------------| | 1 | \( D_o = 140.3 \) mm, \( D_i = 84.2 \) mm | | 2 | Angle of twist = \( 0.0216 \) radians \( (1.24^\circ) \) | | 3 | Hollow shafts are lighter, more material-efficient, and can handle the same torque as solid shafts with less weight and material. | If you need further clarifications or derivations, let me know!