Load Testing of Structural Components – 24-Hour Test Level When load-testing an individual part of a structure for 24 hours, the applied load equals the full dead load plus what multiple of the imposed (live) load?

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:

  • Test duration: 24 hours.
  • Loading composition: full dead load plus a multiple of imposed load.
  • Acceptance intent: check deflection recovery and absence of distress.


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:

  • 1.0 imposed load: Less demanding; may not reveal stiffness deficiencies.
  • 1.15 imposed load: Still conservative but below the commonly prescribed level.
  • 1.3 imposed load: More severe than customary 24-hour test requirements in many building codes.


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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