Difficulty: Medium
Correct Answer: 9 N/mm^2
Explanation:
Introduction / Context:
When a pre-tensioned member is released, the steel force is partly transferred to the concrete by bond. Because steel and concrete then deform compatibly, the final steel and concrete stresses are related by the modular ratio. With tendons at the centroid (no eccentricity), the resulting concrete stress is uniform across the section.
Given Data / Assumptions:
Concept / Approach:
At transfer, compatibility enforces equal strain in steel and concrete: ε_s = ε_c. Thus f_s = m * f_c. Equilibrium requires that the compression developed in concrete equals the drop of force in steel: f_c * A_c = (f_si − f_s) * A_s. Solve for f_c using f_s = m f_c.
Step-by-Step Solution:
Write equilibrium: f_c * A_c = (f_si − m f_c) * A_s.Rearrange: f_c (A_c + m A_s) = f_si * A_s.Substitute values: A_c + m A_s = 50,000 + 10*500 = 55,000 mm^2.Compute numerator: f_si * A_s = 1000 * 500 = 500,000 N.Concrete stress: f_c = 500,000 / 55,000 ≈ 9.09 N/mm^2 ≈ 9 N/mm^2.
Verification / Alternative check:
Direct P/A using initial force P_i = f_si * A_s = 500 kN gives 10 N/mm^2, but compatibility reduces steel stress, transferring part of it back to steel; the modular-ratio solution (≈9 N/mm^2) is the correct compatible value at transfer.
Why Other Options Are Wrong:
11 N/mm^2 assumes no compatibility (simply P_i/A_c).7 and 5 N/mm^2 underestimate the compatible stress.13 N/mm^2 is not feasible with the given initial steel force and areas.
Common Pitfalls:
Final Answer:
9 N/mm^2
Discussion & Comments