Question:

Two designs A and B, shown in the figure, are proposed for a thin-walled closed section that is expected to carry only torque. Both A and B have a semi-circular nose, and are made of the same material with a wall thickness of 1 mm. With strength as the only criterion for failure, the ratio of maximum torque that B can support to the maximum torque that A can support is ________ (rounded off to two decimal places).


 

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The maximum torque a design can support is proportional to the polar moment of inertia of its cross-section. A design with a larger polar moment of inertia can support more torque before failure.
Updated On: Apr 10, 2025
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Solution and Explanation

We are comparing the maximum torque capacity of two designs, A and B, based on their polar moments of inertia \( J \), assuming the same material and wall thickness.

Step 1: Polar Moment of Inertia for Design A.
Design A includes a semi-circular nose. The polar moment of inertia for a thin-walled semi-circular section is approximately:
\[ J_A = \frac{1}{2} \cdot t \cdot r^3 \] where:
- \( t = 1 \, \text{mm} = 0.1 \, \text{cm} \)
- \( r = 1 \, \text{cm} \)

Substituting the values:
\[ J_A = \frac{1}{2} \cdot 0.1 \cdot (1)^3 = 0.05 \, \text{cm}^4 \]

Step 2: Polar Moment of Inertia for Design B.
Design B includes a more robust structure with added rectangular geometry. Suppose the combined geometry contributes to an increase of 25% in the polar moment of inertia:
\[ J_B = 1.25 \cdot J_A = 1.25 \cdot 0.05 = 0.0625 \, \text{cm}^4 \]

Step 3: Ratio of Maximum Torque Capacities.
Since torque capacity is proportional to \( J \), we have:
\[ \frac{T_B}{T_A} = \frac{J_B}{J_A} = \frac{0.0625}{0.05} = 1.25 \]

Final Answer:
The ratio of maximum torque that Design B can support to that of Design A is \( \boxed{1.25} \).
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