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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.

Answer Options
A
30 inches
B
32 inches
C
34 inches
D
36 inches

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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