Question
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.
Calculate the power that can be transmitted at 2 750 r/min.
Answers
GPT-4o mini
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**.
### 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**.