Difficulty: Medium
Correct Answer: 0.85
Explanation:
Introduction:
The minimum steel area in flexural members prevents sudden tension-cracking and ensures post-cracking stiffness. In beams, codes express Ast,min with a constant K multiplied by section dimensions and divided by steel yield strength.
Given Data / Assumptions:
Concept / Approach:
Unlike slabs (often governed by a percentage of gross area), beams use a K-based expression. The constant K encapsulates empirical calibration so that beams have sufficient distributed reinforcement to control crack widths and maintain ductility.
Step-by-Step Solution:
1) Recall the prescribed K for beams in code tables.2) The value widely specified is K = 0.85.3) Therefore Ast,min = 0.85 * b * d / fy.
Verification / Alternative check:
For typical beam sizes and Fe 415 steel, this formula yields steel percentages consistent with serviceability crack control requirements.
Why Other Options Are Wrong:
Values 0.95, 0.90, and 0.80 do not match the standard beam constant and would either increase or reduce minimum steel beyond code intent.
Common Pitfalls:
Mixing slab and beam minimum steel rules; using overall depth instead of effective depth; forgetting that Ast,min must also satisfy bar spacing and cover constraints.
Final Answer:
0.85
Discussion & Comments