Question

The polar section modulus for a circular shaft of diameter \(d\) is given by which formula?

• A. \( \dfrac{\pi d^3}{16} \)
• B. \( \dfrac{\pi d^3}{32} \)
• C. \( \dfrac{\pi d^4}{32} \)
• D. \( \dfrac{\pi d^4}{16} \)

Hint:

For solid circular shaft:
\(J = \frac{\pi d^4}{32}\), \quad \(Z_p = \frac{\pi d^3}{16}\)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the mathematical formula for the polar section modulus, denoted as \(Z_p\), for a solid circular shaft of diameter \(d\). This property is crucial in analyzing shafts under torsion.
Step 2: Key Formula or Approach:
The polar section modulus (\(Z_p\)) is defined as the ratio of the polar moment of inertia (\(J\)) to the distance from the center to the outermost fiber (\(R\)), which is the radius.
\[ Z_p = \frac{J}{R} \] For a solid circular shaft of diameter \(d\), the polar moment of inertia is:
\[ J = \frac{\pi d^4}{32} \] And the radius is:
\[ R = \frac{d}{2} \] Step 3: Detailed Explanation:
We substitute the formulas for \(J\) and \(R\) into the definition of \(Z_p\):
\[ Z_p = \frac{\left(\frac{\pi d^4}{32}\right)}{\left(\frac{d}{2}\right)} \] To simplify this expression, we multiply the numerator by the reciprocal of the denominator:
\[ Z_p = \frac{\pi d^4}{32} \times \frac{2}{d} \] \[ Z_p = \frac{2\pi d^4}{32d} = \frac{\pi d^3}{16} \] Step 4: Final Answer:
The correct formula for the polar section modulus of a solid circular shaft is \( \dfrac{\pi d^3}{16} \). Therefore, option (A) is correct.