According to the Florida Building Code, what is the minimum width required for an accessible door opening?
Correct Answer
B) 32 inches
The Florida Building Code requires accessible door openings to provide a minimum clear width of 32 inches when the door is open 90 degrees. This typically requires a 36-inch door to achieve the 32-inch clear opening.
Why This Is the Correct Answer
The Florida Building Code, which adopts the International Building Code (IBC) accessibility standards, requires accessible door openings to have a minimum clear width of 32 inches when measured with the door open at 90 degrees. This clear width measurement is taken from the face of the door to the stop on the latch side when the door is fully opened. The 32-inch requirement ensures adequate passage width for wheelchairs and mobility devices. This standard is consistent with ADA guidelines and is critical for compliance in commercial and accessible residential construction.
Why the Other Options Are Wrong
Option A: 30 inches
30 inches is insufficient and does not meet the minimum accessibility requirements established by the Florida Building Code and ADA standards.
Option C: 34 inches
34 inches exceeds the minimum requirement, and while it would be compliant, it is not the minimum width required by code.
Option D: 36 inches
36 inches refers to the typical door width needed to achieve the 32-inch clear opening, not the clear opening width itself.
Memory Technique
Remember '32 Clear' - the clear opening must be 32 inches, achieved by a 36-inch door (36-4=32, accounting for door thickness and hardware).
Reference Hint
Florida Building Code Chapter 11 (Accessibility) or International Building Code Chapter 11, Section 1010.1.1 for door opening requirements
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