According to the Florida Building Code, what is the minimum clear width required for an accessible door opening?
Correct Answer
B) 32 inches
The Florida Building Code requires a minimum clear width of 32 inches for accessible door openings. This measurement is taken with the door open at 90 degrees, measured from the face of the door to the stop.
Why This Is the Correct Answer
The Florida Building Code mandates a minimum clear width of 32 inches for accessible door openings to ensure compliance with ADA requirements. This measurement is critical for wheelchair accessibility and must be measured with the door in the fully open position (90 degrees). The clear width is measured from the face of the door to the opposite door stop, not the rough opening or door frame width. This standard applies to all doors along accessible routes in commercial and public buildings.
Why the Other Options Are Wrong
Option A: 30 inches
30 inches is insufficient width for wheelchair accessibility and does not meet ADA or Florida Building Code requirements for accessible door openings.
Option C: 34 inches
34 inches exceeds the minimum requirement and while acceptable, it is not the minimum standard specified in the code.
Option D: 36 inches
36 inches exceeds the minimum requirement and while acceptable, it is not the minimum standard specified in the code.
Memory Technique
Think '32 and free' - 32 inches gives wheelchair users the freedom to pass through doorways safely and comfortably.
Reference Hint
Florida Building Code Chapter 11 - Accessibility, Section 1104 (Accessible Route) or Florida Accessibility Code for Building Construction
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