For a circular column section, the core (kern) is the region where a compressive load must act to avoid tension anywhere. What is the radius of this core expressed as a fraction of the column radius?

Introduction / Context:The kern (core) concept ensures that under axial compression with small eccentricity, no part of the section goes into tension. For different shapes, the kern size varies; for circles, it is a fixed fraction of the radius.

Given Data / Assumptions:

• Circular cross-section with radius r.
• Linear elastic stress distribution.
• Load resultant is within kern to avoid tensile stress.

Concept / Approach:For a circular section, the radius of the kern is r/4. This is derived from the eccentricity limit where the compressive stress at the farthest fiber just reaches zero under combined axial load and bending.

Step-by-Step Solution:Stress at extreme fiber: σ = P/A ± M*c/I.Zero tension condition at the farthest fiber gives the limiting eccentricity e = I / (A * c).For a circle, I = π r^4 / 4, A = π r^2, and c = r → e = (π r^4 / 4) / (π r^2 * r) = r/4.Thus the core radius is r/4.

Verification / Alternative check:Textbook kern tables list r/4 for circles and b/6 for rectangles, confirming the calculation.

Why Other Options Are Wrong:

• One-half, one-third, one-fifth, one-sixth: do not match the derived eccentricity limit for a circle.

Common Pitfalls:

• Mixing the circular kern (r/4) with the rectangular middle-third rule (± b/6).

Final Answer:one-quarter of the radius