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

Concept: Polar section modulus is used in torsion problems to determine the strength of a shaft. It is defined as: \[ Z_p = \frac{J}{R} \] where \(J\) is the polar moment of inertia and \(R\) is the outer radius.
Step 1: Write the polar moment of inertia.
For a solid circular shaft: \[ J = \frac{\pi d^4}{32} \]
Step 2: Substitute radius \(R = \frac{d}{2}\).
\[ Z_p = \frac{J}{R} = \frac{\frac{\pi d^4}{32}}{\frac{d}{2}} \]
Step 3: Simplify the expression.
\[ Z_p = \frac{\pi d^4}{32} \times \frac{2}{d} = \frac{\pi d^3}{16} \]
Conclusion:
Thus, the correct formula is \( \boxed{\dfrac{\pi d^3}{16}} \).