Section properties – polar moment: For a solid circular shaft of diameter D, what is the polar moment of inertia J about its own axis?

Introduction / Context:The polar moment of inertia J quantifies a shaft’s resistance to torsional deformation. For circular shafts, J is directly used in torsion equations to compute shear stress and angle of twist, which are critical in power transmission and drive-train design.

Given Data / Assumptions:

• Circular cross-section, solid, diameter D.
• Homogeneous, isotropic material.
• Axis of interest is the shaft’s centroidal axis.

Concept / Approach:For circular sections, J equals the sum of the second moments of area about two orthogonal centroidal axes: J = I_x + I_y. For a solid circle, I_x = I_y = (pi / 64) * D^4. Therefore, J = 2 * (pi / 64) * D^4 = (pi / 32) * D^4.

Step-by-Step Solution:

Verification / Alternative check:Dimensionally, J has units of length^4, matching D^4. Values appear in design tables for quick reference and align with standard formulae.

Why Other Options Are Wrong:

• (pi / 64) * D^4: this is I_x or I_y for a solid circle, not J.
• (pi / 16) * D^4 and (pi / 8) * D^4: overestimate J by factors of 2 and 4, respectively.

Common Pitfalls:Confusing polar moment J with area moment I. Remember: torsion uses J, bending uses I.

Final Answer:J = (pi / 32) * D^4