Difficulty: Easy
Correct Answer: 1.0
Explanation:
Introduction / Context:
Steel connections often subject a single fastener to both shear and tension due to eccentricities or prying forces. Codes control this by an interaction check so that the combined utilization in shear and tension does not exceed the allowable envelope.
Given Data / Assumptions:
Concept / Approach:
A common interaction form is of the type (tau_vf,cal / tau_vf) + (sigma_tf,cal / sigma_tf) ≤ 1.0 or a similar squared-sum form bounded by 1.0. Regardless of the exact expression used, the non-dimensional combined utilization must not exceed unity for safety.
Step-by-Step Solution:
1) Compute the demand ratios in shear and in tension with respect to their allowable values.2) Substitute into the prescribed interaction formula.3) Check that the evaluated interaction ≤ 1.0.4) If > 1.0, redesign by increasing capacity or reducing demand.
Verification / Alternative check:
Design examples in steel manuals consistently use unity as the limiting combined utilization threshold.
Why Other Options Are Wrong:
1.2, 1.4, and 1.8 would permit overstress; interaction checks are capped at unity by definition.
Common Pitfalls:
Ignoring prying action in tension; double-counting or omitting load combinations; using nominal instead of design allowables.
Final Answer:
1.0
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