Section properties – polar moment: For a solid circular shaft of diameter D, what is the polar moment of inertia J about its own axis?
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AJ = (pi / 32) * D^4
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BJ = (pi / 64) * D^4
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CJ = (pi / 16) * D^4
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DJ = (pi / 8) * D^4
Answer
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