A splined connection in a vehicle transmission consists of 10 splines cut into a 60 mm diameter shaft. The height of each spline is 5 mm, the splines in the hub are 70 mm long and the allowable pressure on the splines is 6 MPa.

Calculate the power that can be transmitted at 2 750 r/min.

1 answer

To calculate the power that can be transmitted through a splined connection, we first need to determine the contact area between the splines. Then, using the allowable pressure and the rotational speed, we can find the power transmission capability.

Step 1: Calculate the contact area

The total contact area \( A \) between the splines can be calculated using the formula:

\[ A = n \times L \times h \]

where:

  • \( n \) = number of splines = 10
  • \( L \) = length of splines = 70 mm = 0.070 m
  • \( h \) = height of each spline = 5 mm = 0.005 m

Substituting the values:

\[ A = 10 \times 0.070 , \text{m} \times 0.005 , \text{m} = 0.0035 , \text{m}^2 \]

Step 2: Calculate the allowable force

The allowable force \( F \) on the splines can be calculated using the formula:

\[ F = P \times A \]

where:

  • \( P \) = allowable pressure = 6 MPa = \( 6 \times 10^6 , \text{Pa} \)

Substituting the values:

\[ F = 6 \times 10^6 , \text{Pa} \times 0.0035 , \text{m}^2 = 21,000 , \text{N} \]

Step 3: Calculate the power

The power \( P \) transmitted can be calculated using the formula:

\[ P = \frac{F \times \text{RPM} \times \pi \times D}{60} \]

where:

  • \( F \) = force in Newtons = 21,000 N
  • \( D \) = diameter of the shaft = 60 mm = 0.060 m
  • \( \text{RPM} \) = rotational speed = 2,750 r/min

Substituting the values:

\[ P = \frac{21,000 , \text{N} \times 2,750 , \text{rev/min} \times \pi \times 0.060 , \text{m}}{60} \]

Calculating it step-by-step:

  1. Calculate \( \pi \times D \): \[ \pi \times 0.060 = 0.1884 , \text{m} \]

  2. Calculate the entire expression: \[ P = \frac{21,000 \times 2,750 \times 0.1884}{60} \]

  3. Calculate the numerator: \[ = 21,000 \times 2,750 \times 0.1884 = 108,267,000 \]

  4. Then divide by 60: \[ P = \frac{108,267,000}{60} = 1,804,450 , \text{W} \approx 1,804.45 , \text{kW} \]

  5. Finally convert to kW: \[ P \approx 1,804.45 , \text{W} \approx 1.80 , \text{kW} \]

Conclusion

The power that can be transmitted at 2,750 r/min is approximately 1.80 kW.