Difficulty: Easy
Correct Answer: replacing larger bars by a greater number of smaller bars
Explanation:
Introduction / Context:
For a given member size and external load, detailing choices affect stress distribution, crack width control, and bond performance. Using many smaller bars instead of a few large bars spreads the reinforcement more uniformly and improves bond, which reduces peak steel stress and crack widths for the same total steel area.
Given Data / Assumptions:
Concept / Approach:
Stress in steel ≈ internal force / steel area. With more, smaller-diameter bars, the steel is distributed closer to the tensile face, increasing the effective lever arm slightly and providing better crack control through more uniform strain distribution and greater perimeter for bond. This combination reduces peak stress demand in individual bars and improves serviceability.
Step-by-Step Reasoning:
Keep total As approximately constant but distribute it among more bars.Better distribution → improved crack control and reduced peak bar stress.Often a small increase in effective lever arm results due to placement flexibility.
Verification / Alternative check:
Design guides for crack control favor smaller bars at closer spacing because maximum crack width is proportional to bar spacing and steel strain. Thus, swapping a few large bars for many small bars is beneficial for reducing stress concentrations and crack width.
Why Other Options Are Wrong:
Decreasing lever arm increases internal force demand in steel for equilibrium.Increasing perimeter alone (option b) is a partial explanation; option c operationalizes it effectively.Using larger bars (option d) concentrates force and worsens crack control.
Common Pitfalls:
Final Answer:
replacing larger bars by a greater number of smaller bars
Discussion & Comments