--- title: "Hollow versus solid shaft for power" subject: "Strength of Materials" chapter: 4 key: "semester_4/strength_of_materials_mechanical:hollow-versus-solid-shaft-power" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/hollow-versus-solid-shaft-power site: BCT Engineering --- # Hollow versus solid shaft for power Strength of Materials · Chapter 4: Torsion - Asked in: year not given, 8 marks - Video: [Example: Hollow vs Solid Torsional Shaft](https://www.youtube.com/watch?v=rjGtp4FROSQ) — Calvin Rans - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/hollow-versus-solid-shaft-power - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-4 ## Question A hollow circular shaft whose inside diameter is 0.6 times its outside diameter has to transmit 150 kW at 300 rpm. The allowable shear stress is 60 MPa. Determine the diameters of the hollow shaft, and compare it with a solid shaft designed for the same power and stress: find the diameter of the solid shaft and the percentage saving in weight. ## Short answer Torque = 4774.6 N m. Solid shaft diameter 74 mm (use 75 mm). Hollow shaft with d_i/d_o = 0.6: d_o = 77.5 mm (78 mm), d_i = 46.5 mm (47 mm). Weight saving ≈ 29.8%. ## Answer #### Torque to be transmitted $$ T = \frac{60P}{2\pi N} = \frac{60\times150\times10^3}{2\pi\times300} = 4774.6\ \text{N m} = 4.775\times10^6\ \text{N mm} $$ #### Solid shaft $$ T = \frac{\pi}{16}\tau d^3 \;\Rightarrow\; d^3 = \frac{16\times4.775\times10^6}{\pi\times60} = 405.3\times10^3 $$ $$ d = 74.0\ \text{mm} \;\rightarrow\; \text{provide } 75\ \text{mm} $$ #### Hollow shaft ($k = d_i/d_o = 0.6$) $$ T = \frac{\pi}{16}\tau\,\frac{d_o^4 - d_i^4}{d_o} = \frac{\pi}{16}\tau\,d_o^3(1-k^4) $$ $$ 1 - k^4 = 1 - 0.6^4 = 0.8704 $$ $$ d_o^3 = \frac{16\times4.775\times10^6}{\pi\times60\times0.8704} = 465.7\times10^3 \;\Rightarrow\; d_o = 77.5\ \text{mm} $$ $$ d_i = 0.6\times77.5 = 46.5\ \text{mm} $$ Provide $d_o = 78$ mm, $d_i = 47$ mm. #### Weight comparison For the same length and material, weight is proportional to cross-sectional area: $$ \frac{A_{hollow}}{A_{solid}} = \frac{d_o^2(1-k^2)}{d^2} = \frac{77.5^2(1-0.36)}{74.0^2} = \frac{3843.6}{5476} = 0.702 $$ $$ \text{Saving} = 1 - 0.702 = 29.8\ \% $$ The hollow shaft is more economical because the material near the axis, where shear stress is small, is removed. **Answer: hollow shaft $d_o = 77.5$ mm (78 mm), $d_i = 46.5$ mm (47 mm); solid shaft $d = 74.0$ mm (75 mm); weight saving $\approx 29.8\ \%$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the torque found from 150 kW at 300 rpm? A. T = 2 pi N P / 60 B. T = 60 P / (2 pi N) C. T = P N D. T = P / N Answer: B. P = 2 pi N T / 60, so T = 60P/(2 pi N) = 4774.6 N m. (Key point: Torque from power) 2. What equation is used to size the hollow shaft? A. T = (pi/16) tau d_o^3 (1 - k^4) B. T = (pi/16) tau d_o^3 (1 - k^2) C. T = (pi/32) tau d_o^3 D. T = (pi/16) tau d_o^2 (1 - k^4) Answer: A. Using J = pi(d_o^4 - d_i^4)/32 and r = d_o/2, T = (pi/16) tau d_o^3 (1 - k^4), with 1 - 0.6^4 = 0.8704. (Key point: Hollow shaft design equation) 3. Weight of the shafts compares as what? A. Ratio of polar moduli B. Ratio of cross-sectional areas, d_o^2(1-k^2) / d^2 C. Ratio of diameters D. Ratio of torques Answer: B. Same length and material means weight is proportional to area; 0.702 gives about 29.8 % saving. (Key point: Weight comparison) 4. Why is the hollow shaft more economical? A. Shear stress is greatest at the axis B. Material near the axis, where shear stress is small, is removed C. Hollow shafts have higher G D. It rotates faster Answer: B. Shear stress is zero at the axis, so that core carries little torque and can be removed. 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