Difficulty: Medium
Correct Answer: 4.25 cm
Explanation:
Introduction:
To prevent tension anywhere on a compression member under eccentric loading, the load’s line of action must lie within the “kern.” For circular and annular (hollow circular) sections, the kern is a concentric circle of radius r_k. This question asks for the limiting eccentricity e for a hollow circular column so that no tensile stress develops.
Given Data / Assumptions:
Concept / Approach:
The kern radius is r_k = I / (A * c), where I is the second moment of area about any centroidal axis in the plane, A is area, and c is the distance from centroid to the extreme fiber in the load direction (for a ring, c = D/2). For a hollow circle: I = (pi/64) * (D^4 − d^4), A = (pi/4) * (D^2 − d^2). Simplifying yields a compact formula for the ring: r_k = (1/8) * (D^2 + d^2) / D.
Step-by-Step Solution:
Step 1: Write I = (pi/64) * (D^4 − d^4).Step 2: Write A = (pi/4) * (D^2 − d^2).Step 3: Take c = D/2.Step 4: Compute r_k = I / (A * c) = [(pi/64)(D^4 − d^4)] / {[(pi/4)(D^2 − d^2)] * (D/2)}.Step 5: Cancel pi and (D^2 − d^2); use D^4 − d^4 = (D^2 − d^2)(D^2 + d^2) to get r_k = (1/8) * (D^2 + d^2) / D.Step 6: Substitute D = 25 cm, d = 15 cm ⇒ r_k = (1/8) * (625 + 225) / 25 = (1/8) * 850 / 25 = (1/8) * 34 = 4.25 cm.
Verification / Alternative check:
A solid circle gives r_k = D/8; putting d = 0 in the ring formula reduces to the same result, confirming consistency.
Why Other Options Are Wrong:
2.75, 3.00, 3.50 cm — All are less than the derived kern radius; they are not the maximum permissible eccentricity.5.0 cm — Exceeds the kern; placing the load beyond this would induce tension.
Common Pitfalls:
Using the solid-circle kern radius D/8 for a hollow section; forgetting that P does not affect r_k; mixing diameter and radius in c.
Final Answer:
4.25 cm.
Discussion & Comments