A hollow shaft of 150 mm inside diameter and 50 mm thickness transmits power at 220 rev/min. If the maximum shear stress is 70 MN/m², find: a) the power transmitted, b) the percent... A hollow shaft of 150 mm inside diameter and 50 mm thickness transmits power at 220 rev/min. If the maximum shear stress is 70 MN/m², find: a) the power transmitted, b) the percentage by which the shaft would be strengthened if made solid instead of hollow, for the same external diameter. Hint: Compare the torque, c) Sketch the stress distribution diagrams for both shafts.
Understand the Problem
The question describes a hollow shaft transmitting power and asks us to calculate: a) the power transmitted, b) the percentage increase in strength if the shaft were solid instead of hollow (for the same outer diameter), and c) to sketch the stress distribution diagrams for both hollow and solid shafts. This involves concepts of shear stress, torque, power transmission, and shaft design.
Answer
a) $4.31 \text{ MW}$ b) $14.9 \%$ c) Stress distribution diagrams: Linear variation of shear stress from the inner radius to the outer radius for the hollow shaft and from the center to the outer radius for the solid shaft.
Answer for screen readers
a) The power transmitted is approximately $4.31 \text{ MW}$. b) The percentage increase in strength is approximately $14.9%$. c) Stress Distribution Diagrams: - Hollow Shaft: Linear variation of shear stress from the inner radius ($r_i$) to the outer radius ($r_o$), with $\tau = 0$ at $r < r_i$, and $\tau_{max}$ at $r=r_o = 0.125m$. - Solid Shaft: Linear variation of shear stress from the center ($r = 0$) to the outer radius ($r_o$), with $\tau = 0$ at $r=0$ and $\tau_{max}$ at $r=r_o = 0.125m$.
Steps to Solve
- Calculate the outer diameter of the hollow shaft
Given the inside diameter ($d_i$) is 150 mm and the thickness ($t$) is 50 mm, the outer diameter ($d_o$) is:
$d_o = d_i + 2t = 150 + 2(50) = 250 \text{ mm} = 0.25 \text{ m}$
- Calculate the polar moment of inertia ($J$) for the hollow shaft
$J = \frac{\pi}{32}(d_o^4 - d_i^4) = \frac{\pi}{32}(0.25^4 - 0.15^4) = \frac{\pi}{32}(0.00390625 - 0.00050625) = 3.34 \times 10^{-4} \text{ m}^4$
- Calculate the torque ($T$) transmitted by the hollow shaft using the maximum shear stress ($\tau_{max}$)
$\frac{T}{J} = \frac{\tau_{max}}{r_o}$
Where $r_o = d_o / 2 = 0.25 / 2 = 0.125 \text{ m}$
$T = \frac{\tau_{max} \cdot J}{r_o} = \frac{(70 \times 10^6) \cdot (3.34 \times 10^{-4})}{0.125} = 186848 \text{ Nm}$
- Calculate the power ($P$) transmitted by the hollow shaft
Convert the rotational speed from rev/min to rad/s: $N = 220 \text{ rev/min} = \frac{220}{60} \text{ rev/s} = \frac{220}{60} \cdot 2\pi \text{ rad/s} = 23.04 \text{ rad/s}$
$P = T \cdot N = 186848 \cdot 23.04 = 4305488.19 \text{ W} \approx 4.31 \text{ MW}$
- Calculate the polar moment of inertia ($J_s$) if the shaft were solid
$J_s = \frac{\pi}{32} d_o^4 = \frac{\pi}{32}(0.25)^4 = 3.835 \times 10^{-4} \text{ m}^4$
- Calculate the torque ($T_s$) that the solid shaft could transmit
$T_s = \frac{\tau_{max} \cdot J_s}{r_o} = \frac{(70 \times 10^6) \cdot (3.835 \times 10^{-4})}{0.125} = 214760 \text{ Nm}$
- Calculate the percentage increase in strength
Percentage increase $= \frac{T_s - T}{T} \times 100 = \frac{214760 - 186848}{186848} \times 100 = \frac{27912}{186848} \times 100 = 14.94% \approx 14.9 %$
- Stress distribution diagram (Conceptual)
For both hollow and solid shafts, the shear stress is zero at the center and increases linearly to the maximum value at the outer surface.
- Hollow shaft: Stress varies from a non-zero value at the inner radius to the maximum value at the outer radius.
- Solid shaft: Stress varies from zero at the center to the maximum value at the outer radius.
a) The power transmitted is approximately $4.31 \text{ MW}$. b) The percentage increase in strength is approximately $14.9%$. c) Stress Distribution Diagrams: - Hollow Shaft: Linear variation of shear stress from the inner radius ($r_i$) to the outer radius ($r_o$), with $\tau = 0$ at $r < r_i$, and $\tau_{max}$ at $r=r_o = 0.125m$. - Solid Shaft: Linear variation of shear stress from the center ($r = 0$) to the outer radius ($r_o$), with $\tau = 0$ at $r=0$ and $\tau_{max}$ at $r=r_o = 0.125m$.
More Information
The power transmitted by a shaft is directly related to the torque it can withstand and its rotational speed. A solid shaft, for the same outer diameter, provides a higher polar moment of inertia, thus increasing its torque-carrying capacity and strength compared to a hollow shaft.
Tips
A common mistake is forgetting to convert units to be consistent (e.g., mm to m). Another mistake is using diameter instead of radius in the shear stress formula. Also, students may incorrectly calculate the polar moment of inertia (J) for hollow and solid shafts. Watch for the order of operations.
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