Question:medium
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 5, 2026
Show Solution
Correct Answer: 2.5
Solution and Explanation
Step 1: Understanding the Concept:
We are asked to size a pinion module using the Lewis bending-strength equation for gear teeth.
The face width is fixed as a multiple of the module ($b = 10m$), so the unknowns reduce to a single variable, the module itself.
Step 2: Key Formula or Approach:
Instead of substituting numbers step by step, first combine the torque relation and the Lewis equation into a single design formula.
From torque: $T = F_t \cdot \dfrac{d}{2} = F_t \cdot \dfrac{zm}{2}$, so $F_t = \dfrac{2T}{zm}$.
From the Lewis equation: $F_t = \sigma (c m) Y m = \sigma Y c\, m^2$, where $b = c m$ and here $c = 10$.
Equating the two gives a single closed-form design equation:
\[ m^3 = \frac{2T}{\sigma Y c z} \]
Step 3: Detailed Explanation:
Substitute the known values directly into this single formula: $T = 95000$ N-mm, $\sigma = 180$ N/mm^2, $Y = 0.342$, $c = 10$, $z = 20$.
\[ m^3 = \frac{2 \times 95000}{180 \times 0.342 \times 10 \times 20} \]
\[ m^3 = \frac{190000}{12312} \]
\[ m^3 = 15.432 \]
Taking the cube root gives $m = 2.4897$ mm, which rounds to 2.5 mm.
Final Answer:
The pinion needs a module of 2.5 mm to safely carry the specified torque.
\[ \boxed{m = 2.5 \text{ mm}} \]
We are asked to size a pinion module using the Lewis bending-strength equation for gear teeth.
The face width is fixed as a multiple of the module ($b = 10m$), so the unknowns reduce to a single variable, the module itself.
Step 2: Key Formula or Approach:
Instead of substituting numbers step by step, first combine the torque relation and the Lewis equation into a single design formula.
From torque: $T = F_t \cdot \dfrac{d}{2} = F_t \cdot \dfrac{zm}{2}$, so $F_t = \dfrac{2T}{zm}$.
From the Lewis equation: $F_t = \sigma (c m) Y m = \sigma Y c\, m^2$, where $b = c m$ and here $c = 10$.
Equating the two gives a single closed-form design equation:
\[ m^3 = \frac{2T}{\sigma Y c z} \]
Step 3: Detailed Explanation:
Substitute the known values directly into this single formula: $T = 95000$ N-mm, $\sigma = 180$ N/mm^2, $Y = 0.342$, $c = 10$, $z = 20$.
\[ m^3 = \frac{2 \times 95000}{180 \times 0.342 \times 10 \times 20} \]
\[ m^3 = \frac{190000}{12312} \]
\[ m^3 = 15.432 \]
Taking the cube root gives $m = 2.4897$ mm, which rounds to 2.5 mm.
Final Answer:
The pinion needs a module of 2.5 mm to safely carry the specified torque.
\[ \boxed{m = 2.5 \text{ mm}} \]
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