Understanding the Tubular Shaft Problem

This problem involves calculating the rotational speed (rpm) required for a hollow tubular shaft to transmit a specific amount of power without exceeding a maximum allowable shear stress. We need to use the principles of torsional mechanics and power transmission.

Key Information Provided:

• Shaft Type: Tubular (Hollow)
• Inner Diameter ($d_i$): 30 mm = 0.030 m
• Outer Diameter ($d_o$): 40 mm = 0.040 m
• Power Transmitted ($P$): 80 kW = $80 \times 10^3$ W
• Maximum Allowable Shear Stress ($\tau_{max}$): 50 MPa = $50 \times 10^6$ Pa

Calculating Required Shaft Parameters

To find the speed, we need to relate power, torque, speed, and stress. The steps are as follows:

1. Calculate the Polar Moment of Inertia ($J$)

For a hollow circular shaft, the polar moment of inertia is calculated using the formula:

$$J = \frac{\pi}{32} (d_o^4 - d_i^4)$$

Substituting the given values:

$$J = \frac{\pi}{32} ((0.040 \text{ m})^4 - (0.030 \text{ m})^4)$$ $$J = \frac{\pi}{32} (2.56 \times 10^{-6} \text{ m}^4 - 0.81 \times 10^{-6} \text{ m}^4)$$ $$J = \frac{\pi}{32} (1.75 \times 10^{-6} \text{ m}^4)$$ $$J \approx \frac{3.14159}{32} \times 1.75 \times 10^{-6} \text{ m}^4$$ $$J \approx 0.098175 \times 1.75 \times 10^{-6} \text{ m}^4$$ $$J \approx 1.7179 \times 10^{-7} \text{ m}^4$$

2. Determine the Maximum Torque ($T$)

The relationship between maximum shear stress ($\tau_{max}$), torque ($T$), outer radius ($r_o$), and polar moment of inertia ($J$) is given by:

$$\tau_{max} = \frac{T \times r_o}{J}$$

Where the outer radius $r_o = d_o / 2 = 0.040 \text{ m} / 2 = 0.020 \text{ m}$.

Rearranging the formula to solve for torque ($T$):

$$T = \frac{\tau_{max} \times J}{r_o}$$

Substituting the values:

$$T = \frac{(50 \times 10^6 \text{ Pa}) \times (1.7179 \times 10^{-7} \text{ m}^4)}{0.020 \text{ m}}$$ $$T = \frac{8.5895 \text{ N} \cdot \text{m}}{0.020 \text{ m}}$$ $$T = 429.475 \text{ N} \cdot \text{m}$$

Let's re-evaluate the torque calculation with more precision, as the intermediate calculation seemed off previously. Let's use the formula directly relating Power, Torque, and Speed.

3. Relate Power, Torque, and Speed

The power ($P$) transmitted by a shaft is related to the torque ($T$) and the angular velocity ($\omega$) in radians per second by the formula:

$$P = T \times \omega$$

We know the power ($P$) and we can calculate the torque ($T$) using the stress condition. First, let's use the stress formula to find T and then use the power formula.

Let's recalculate Torque using the Stress equation directly:

$$T = \frac{\tau_{max} \cdot J}{r_o}$$ $$T = \frac{(50 \times 10^6 \, \text{N/m}^2) \cdot (\frac{\pi}{32} (0.040^4 - 0.030^4) \, \text{m}^4)}{0.020 \, \text{m}}$$ $$T = \frac{(50 \times 10^6) \cdot (\frac{\pi}{32} (1.75 \times 10^{-6}))}{0.020}$$ $$T = \frac{(50 \times 10^6) \cdot (1.7179 \times 10^{-7})}{0.020}$$ $$T = \frac{8.5895}{0.020} = 429.475 \, \text{N} \cdot \text{m}$$

Okay, the torque calculation is correct. Let's re-examine the power formula relation.

The angular velocity ($\omega$) is related to the speed of rotation ($N$) in revolutions per minute (rpm) by:

$$\omega = \frac{2 \pi N}{60}$$

Substituting this into the power equation:

$$P = T \times \frac{2 \pi N}{60}$$

Now, we can rearrange this formula to solve for the speed ($N$):

$$N = \frac{60 \times P}{2 \pi \times T}$$

Substituting the known values of Power ($P$) and the calculated Torque ($T$):

$$N = \frac{60 \times (80 \times 10^3 \text{ W})}{2 \pi \times (429.475 \text{ N} \cdot \text{m})}$$ $$N = \frac{4800000}{2 \pi \times 429.475}$$ $$N = \frac{4800000}{2 \times 3.14159 \times 429.475}$$ $$N = \frac{4800000}{2698.55}$$ $$N \approx 1778.77 \text{ rpm}$$

Final Result

The calculated speed of rotation is approximately 1778.77 rpm. Comparing this with the given options, the closest value is 1778.7 rpm.