CE Practice Exam #52, why door thickness is added to calculate door width?
In this question, how do I understand the "2" door thickness, why it is added to calculate door width" in step 4 from the official answer?
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Okay, I got it now, and that kind of interpretation about clear width is wicked. Even I follow that interpretation, the door leaf width should be 1"(Door Stop) + 35" + 2"(Door Thickness) + 2"(Panic Bar) = 40". Unless, you say Panic Bar could be push in so it wouldn't be counted in.
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IBC Section 1010.1.1, Size of Doors.
"The required capacity of each door opening shall be sufficient for the occupant load thereof and shall provide a minimum clear opening width of 32 inches. The clear opening width of doorways with swinging doors shall be measured between the face of the door and the stop, with the door open 90 degrees."
So this leads me to believe that the NCARB answer is correct and that the panic hardware is not accounted for.
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Reference to the above IBC Sec # 1010.1.1 if the panic hardware is not accounted for, it means that the min leaf width = Clear Door Width (35") + 1" (Door Stop) on left side + 2" Door thickness on right side = 38"
Not sure why we count door thickness to figure out the leaf width ? It should be cleat door width + 2 * Door stops.
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So in effect the 38” door allows for a 2” thick panic bar/crash bar. I was probably making too many assumptions in my comments above - not to mention by using my 2” number for the panic bar the math wont work, but Haytham and Blane’s way of thinking is much more concise. Thanks for your comments everyone.
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