If a section has moment of inertia I about a given axis and effective area A, the radius of gyration r about that axis is r = sqrt(I / A).

Difficulty: Easy

Correct Answer: sqrt(I / A)

Explanation:


Introduction / Context:
The radius of gyration links a section's second moment of area with its area to provide a convenient measure of how area is distributed about an axis. It is heavily used in column buckling to compute the slenderness ratio and to identify the critical (least) axis.


Given Data / Assumptions:

  • Moment of inertia about an axis: I.
  • Effective cross-sectional area: A.
  • Section is prismatic and properties refer to the same axis.


Concept / Approach:
By definition, I = A * r^2. Rearranging gives r = sqrt(I / A). This applies to any axis about which I and A are taken, whether principal or non-principal.


Step-by-Step Solution:
Start from definition: I = A * r^2Rearrange: r^2 = I / ATake square root: r = sqrt(I / A)


Verification / Alternative check:
Dimensional check: I has units of length^4; A has length^2. I / A has length^2, and sqrt(I / A) has length, which is consistent for a radius measure.


Why Other Options Are Wrong:

  • I / A: has units of length^2, not length.
  • A / I: has units of 1/length^2, not a length.
  • sqrt(A / I): has units of 1/length, incorrect dimensionally.


Common Pitfalls:

  • Forgetting to use the least radius of gyration when assessing buckling.
  • Mixing axes (using I about one axis and r about another).


Final Answer:
sqrt(I / A)

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