Question:
A custom pinion, with a width equal to 10 times the module, has 20 full depth teeth and a pressure angle of 20 degrees is being designed. It should transmit a torque of 95 N-m to a corresponding spur gear.
Considering the safe bending stress to be 180 MPa and form factor to be 0.342, the module of the pinion is ______ mm (rounded off to one decimal place).
Note: The Lewis equation gives the tangential force as \( F_t = \sigma b Y m \), where \( \sigma \) is the safe bending stress, \( b \) is the width, \( Y \) is the form factor, and \( m \) is the module.
A custom pinion, with a width equal to 10 times the module, has 20 full depth teeth and a pressure angle of 20 degrees is being designed. It should transmit a torque of 95 N-m to a corresponding spur gear.
Considering the safe bending stress to be 180 MPa and form factor to be 0.342, the module of the pinion is ______ mm (rounded off to one decimal place).
Note: The Lewis equation gives the tangential force as \( F_t = \sigma b Y m \), where \( \sigma \) is the safe bending stress, \( b \) is the width, \( Y \) is the form factor, and \( m \) is the module.
Show Hint
Use the Lewis equation \( F_t = \sigma b Y m \) with \( b = 10m \), and relate the tangential force to torque through \( F_t = 2T/d \) where \( d = zm \).
Updated On: Aug 3, 2026
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Verified By Collegedunia
Correct Answer: 2.5
Solution and Explanation
Step 1: Understanding the Concept:
This is a gear design problem based on the Lewis bending equation.
The pinion has 20 teeth, a face width of 10 times the module, and must safely carry the tangential load produced by a 95 N-m torque.
We need to find the module m that keeps the bending stress within the safe limit of 180 MPa.
Step 2: Key Formula or Approach:
The Lewis equation for the tangential (beam) strength of a gear tooth is given as:
\[ F_t = \sigma \, b \, Y \, m \]
The tangential force actually transmitted at the pitch circle from the torque is:
\[ F_t = \frac{2T}{d} \]
and the pitch circle diameter is \( d = z m \), where z is the number of teeth.
Step 3: Detailed Explanation:
Convert the torque to consistent units: \( T = 95 \text{ N-m} = 95000 \text{ N-mm} \).
The pitch circle diameter in terms of module: \( d = z m = 20m \).
So the tangential force from torque becomes:
\[ F_t = \frac{2 \times 95000}{20m} = \frac{9500}{m} \]
The Lewis equation with \( b = 10m \), \( \sigma = 180 \), \( Y = 0.342 \) gives:
\[ F_t = 180 \times 10m \times 0.342 \times m = 615.6\, m^2 \]
Equating both expressions for \( F_t \):
\[ \frac{9500}{m} = 615.6\, m^2 \]
\[ m^3 = \frac{9500}{615.6} = 15.432 \]
Taking the cube root:
\[ m = (15.432)^{1/3} \approx 2.4897 \text{ mm} \]
Final Answer:
Rounded off to one decimal place, the module works out to 2.5 mm, which lies safely within the official range.
\[ \boxed{m \approx 2.5 \text{ mm}} \]
This is a gear design problem based on the Lewis bending equation.
The pinion has 20 teeth, a face width of 10 times the module, and must safely carry the tangential load produced by a 95 N-m torque.
We need to find the module m that keeps the bending stress within the safe limit of 180 MPa.
Step 2: Key Formula or Approach:
The Lewis equation for the tangential (beam) strength of a gear tooth is given as:
\[ F_t = \sigma \, b \, Y \, m \]
The tangential force actually transmitted at the pitch circle from the torque is:
\[ F_t = \frac{2T}{d} \]
and the pitch circle diameter is \( d = z m \), where z is the number of teeth.
Step 3: Detailed Explanation:
Convert the torque to consistent units: \( T = 95 \text{ N-m} = 95000 \text{ N-mm} \).
The pitch circle diameter in terms of module: \( d = z m = 20m \).
So the tangential force from torque becomes:
\[ F_t = \frac{2 \times 95000}{20m} = \frac{9500}{m} \]
The Lewis equation with \( b = 10m \), \( \sigma = 180 \), \( Y = 0.342 \) gives:
\[ F_t = 180 \times 10m \times 0.342 \times m = 615.6\, m^2 \]
Equating both expressions for \( F_t \):
\[ \frac{9500}{m} = 615.6\, m^2 \]
\[ m^3 = \frac{9500}{615.6} = 15.432 \]
Taking the cube root:
\[ m = (15.432)^{1/3} \approx 2.4897 \text{ mm} \]
Final Answer:
Rounded off to one decimal place, the module works out to 2.5 mm, which lies safely within the official range.
\[ \boxed{m \approx 2.5 \text{ mm}} \]
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