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).

 

Hint:

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.

Answer 1

Correct Answer: 1.2

In both designs A and B, the maximum torque that a design can support is proportional to its polar moment of inertia \( J \). Since both designs are made from the same material with the same wall thickness, we will compare their polar moments of inertia. 
Step 1: Polar Moment of Inertia for Design A. For design A, the cross-sectional area is made of a semi-circular nose with a rectangular section. The polar moment of inertia for a hollow cylindrical section is given by: \[ J_A = \frac{1}{2} \times {thickness} \times {radius}^3 \] where the radius of the semi-circular section is \( 1 \, {cm} \) and the wall thickness is \( 1 \, {mm} \). \[ J_A = \frac{1}{2} \times 1 \times (1)^3 = 0.5 \, {cm}^4 \] Step 2: Polar Moment of Inertia for Design B. For design B, the cross-section has a similar semi-circular nose with the addition of a rectangular section. The polar moment of inertia for design B will be larger due to the more complex shape. For simplicity, assume that the polar moment of inertia is roughly proportional to the square of the dimensions of the cross-section. The calculation will be: \[ J_B \approx 1.25 J_A = 1.25 \times 0.5 = 0.625 \, {cm}^4 \] Step 3: Calculate the ratio of maximum torque for B to A. The ratio of the maximum torque that B can support to the maximum torque that A can support is simply the ratio of their polar moments of inertia: \[ \frac{T_B}{T_A} = \frac{J_B}{J_A} = \frac{0.625}{0.5} = 1.25 \] Thus, the ratio of maximum torque that B can support to the maximum torque that A can support is 1.25.