Difficulty: Easy
Correct Answer: 1.25 times imposed load
Explanation:
Introduction / Context:Load tests verify that constructed members satisfy service performance by limiting residual deflection and damage under a sustained test load. Codes specify the magnitude and duration of such tests to provide a uniform basis for acceptance.
Given Data / Assumptions:
Concept / Approach:
The test load must be high enough to exercise the structure while remaining within safe limits. A commonly adopted magnitude for the live component is 1.25 times the imposed load, applied in combination with full dead load, to provide a meaningful performance check of stiffness and serviceability.
Step-by-Step Solution:
1) Identify required combination: full dead load + k * imposed load.2) Select k = 1.25 as the widely taught factor.3) Maintain the load for 24 hours; record deflections and inspect for cracks.Verification / Alternative check:
Many acceptance criteria require that residual deflection after unloading not exceed a small fraction of the maximum test deflection, confirming elastic recovery at the 1.25 live-load level.
Why Other Options Are Wrong:
Common Pitfalls:
Not maintaining the load for the full duration; measuring deflection from an unstable datum; neglecting temperature effects during long tests.
Final Answer:
1.25 times imposed load
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