To solve the problem, we need to approach it systematically, beginning with the given parameters and requirements.
Given Data
- Diameter of the shaft, \( D = 60 , \text{mm} = 0.06 , \text{m} \)
- Number of splines, \( N = 12 \)
- Maximum speed, \( n = 50 , \text{rev/s} \)
- Height of each spline, \( h = 5 , \text{mm} = 0.005 , \text{m} \)
- Length of each spline, \( L = 50 , \text{mm} = 0.05 , \text{m} \)
- Allowable pressure on the splines, \( p = 5 , \text{MPa} = 5 \times 10^6 , \text{Pa} \)
2.1.1 Calculate the Power that can be Transmitted
To calculate the maximum power that can be transmitted through the splined connection, we first determine the total area of the splines that is carrying the load.
Step 1: Determine the Total Area of the Splines
Each spline has a rectangular cross-section. The area \( A \) for one spline is given by:
\[ A_{\text{spline}} = \text{height} \times \text{length} = h \times L = 0.005 , \text{m} \times 0.05 , \text{m} = 0.00025 , \text{m}^2 \]
The total area for all 12 splines is:
\[ A_{\text{total}} = N \times A_{\text{spline}} = 12 \times 0.00025 , \text{m}^2 = 0.003 , \text{m}^2 \]
Step 2: Calculate the Maximum Load
The maximum load \( F_{\text{max}} \) that can be transmitted through the splines is related to the allowable pressure:
\[ F_{\text{max}} = p \times A_{\text{total}} = 5 \times 10^6 , \text{Pa} \times 0.003 , \text{m}^2 = 15000 , \text{N} \]
Step 3: Calculate the Maximum Torque
The maximum torque \( T \) transmitted can be calculated using the formula \( T = F \cdot r \), where \( r \) is the radius of the shaft.
\[ r = \frac{D}{2} = \frac{0.06 , \text{m}}{2} = 0.03 , \text{m} \]
Thus, the maximum torque is:
\[ T = F_{\text{max}} \times r = 15000 , \text{N} \times 0.03 , \text{m} = 450 , \text{Nm} \]
Step 4: Calculate the Power
Power \( P \) can be calculated using the formula \( P = T \cdot \omega \), where \( \omega \) is the angular speed in radians per second.
First, convert the revolutions per second to radians per second:
\[ \omega = 2\pi n = 2\pi \times 50 , \text{rev/s} \approx 314.16 , \text{rad/s} \]
Then, calculate the power:
\[ P = T \cdot \omega = 450 , \text{Nm} \times 314.16 , \text{rad/s} \approx 141372 , \text{W} \approx 141.37 , \text{kW} \]
2.1.2 Calculate the Coefficient of Friction
The coefficient of friction \( \mu \) can be determined from the relationship between the torque, load, and friction:
\[ T = F_{\text{max}} \cdot r \cdot \mu \]
Rearranging for \( \mu \):
\[ \mu = \frac{T}{F_{\text{max}} \cdot r} \]
We already have \( T = 450 , \text{Nm} \), \( F_{\text{max}} = 15000 , \text{N} \), and \( r = 0.03 , \text{m} \):
\[ \mu = \frac{450 , \text{Nm}}{15000 , \text{N} \times 0.03 , \text{m}} = \frac{450}{450} = 1.0 \]
Summary of Results
- Power that can be transmitted: \( \approx 141.37 , \text{kW} \)
- Coefficient of friction: \( \mu = 1.0 \)
These results suggest that under the given conditions, the splined connection can transmit a significant amount of power while maintaining adequate friction characteristics.