According to the Florida Building Code, what is the minimum clear width required for an accessible doorway?
Correct Answer
B) 32 inches
The Florida Building Code requires a minimum clear width of 32 inches for accessible doorways when the door is open 90 degrees. This ensures wheelchair accessibility.
Why This Is the Correct Answer
The Florida Building Code, following ADA guidelines, requires a minimum clear width of 32 inches for accessible doorways when measured with the door open at 90 degrees. This measurement ensures that wheelchairs, which typically range from 24-27 inches wide, have adequate clearance to pass through comfortably. The 32-inch requirement provides the necessary maneuvering space for wheelchair users while accounting for the door frame and any potential obstructions. This standard is consistently applied across commercial and residential accessible entrances.
Why the Other Options Are Wrong
Option A: 30 inches
30 inches is insufficient width for wheelchair accessibility and does not meet the minimum requirements established by the Florida Building Code for accessible doorways.
Option C: 34 inches
34 inches exceeds the minimum requirement and while it would be acceptable, it is not the minimum standard required by the Florida Building Code.
Option D: 36 inches
36 inches exceeds the minimum requirement and while it would be acceptable, it is not the minimum standard required by the Florida Building Code.
Memory Technique
Think '32 for accessibility' - the number 32 sounds like 'thirty-two' which rhymes with 'through' - wheelchairs need to get 'through' the 32-inch opening.
Reference Hint
Florida Building Code Chapter 11 - Accessibility, Section 1104 (Accessible Route) or Florida Accessibility Code for Building Construction
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