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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 opening width of 32 inches for accessible doors when the door is open 90 degrees. This is measured from the face of the door to the stop on the latch side.

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

Why This Is the Correct Answer

The Florida Building Code requires a minimum clear opening width of 32 inches for accessible doors to comply with ADA accessibility standards. This measurement is taken when the door is fully open at 90 degrees, measured from the face of the door to the door stop on the latch side. This 32-inch clear width ensures adequate space for wheelchair users and individuals with mobility devices to pass through comfortably. The requirement is consistent with federal ADA guidelines that Florida has adopted into its building code.

Why the Other Options Are Wrong

Option A: 30 inches

30 inches is insufficient to meet ADA accessibility requirements and would not provide adequate clearance for wheelchair users or individuals with mobility devices to pass through safely and comfortably.

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 for accessible door openings.

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 for accessible door openings.

Memory Technique

Think '32 for accessibility' - the number 32 sounds like 'thirty-two' which rhymes with 'ADA through' to remember that 32 inches is what ADA users need to get through doors.

Reference Hint

Florida Building Code, Chapter 11 - Accessibility, or Florida Accessibility Code for Building Construction (FACBC)

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