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

Difficulty: Easy

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

Explanation:


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:

Recall: I_x(solid circle) = (pi / 64) * D^4.By symmetry: I_y = I_x.Compute J: J = I_x + I_y = 2 * (pi / 64) * D^4 = (pi / 32) * D^4.Use in torsion: tau_max = T * R / J, theta = T * L / (G * J).


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

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