Strength of Materials – Radius of Gyration for a Rectangle A rectangular cross-section has depth D and width B. About the centroidal axis that is parallel to the width (i.e., the axis with bending across the depth), what is the correct expression for its radius of gyration?

Introduction / Context:
In strength of materials, the radius of gyration k of a section connects its moment of inertia I to its area A using k = √(I/A). For a rectangle, different centroidal axes produce different values of I; here we focus on the centroidal axis parallel to the width B (bending about the axis across the depth D).

Given Data / Assumptions:

• Section: rectangle with width B and depth D.
• Axis: centroidal axis parallel to B (i.e., the x-axis across the depth).
• Area: A = B * D.

Concept / Approach:
The moment of inertia of a rectangle about its centroidal axis parallel to the width is I = (B * D^3) / 12. Then k = √(I/A).

Step-by-Step Solution:
I = (B * D^3) / 12A = B * Dk = √(I / A) = √( (B * D^3 / 12) / (B * D) )k = √( D^2 / 12 )k = D / √12

Verification / Alternative check:
As the axis takes bending across depth, I scales with D^3; dividing by area BD yields D^2, giving k proportional to D. The √12 denominator is the standard centroidal result for rectangles about this axis.

Why Other Options Are Wrong:

• D / √3 and D / √8: incorrect constants; they do not match the rectangle's centroidal inertia form.
• D / 2 and D / (2√3): values are too large for the centroidal axis result.

Common Pitfalls:
Confusing the axis orientation (parallel to width vs parallel to depth) or using the wrong I formula (e.g., (D * B^3)/12) leads to incorrect k.

Final Answer:
D / √12