36 / FRAME BUILDER - VOL7 NO3
engaged during high wind events.
Rh = horizontal reaction @ head
Rj = horizontal reaction @ jamb
S = wind suction pressure
w = door width
H = door height
X = vertical location of jamb reaction
Summing moment about head S x w x H2/2 = Rj x X
Summing forces in the X direction (S x w x H) + Rh = Rj
½ Rh and ½ Rj are applied to each jamb column under 
the suction wind case. Half of door weight is included 
on each jamb under dead load.
Mono-fold Door — Operation Loads
When a mono-fold door is stationary, the entire 
assembly consisting of the frame, door leaf and 
hydraulic cylinders can be assumed to be a rigid body. 
The door weight can be assumed to be 6 psf for unlined 
doors and 8 psf for lined doors.
When the door is closed the door hangs from 
the hinges. The door dead load is transferred via 
diaphragm action of the door leaf to the exterior hinge 
points and then to the vertical frames that are lagged 
to the jamb column. Again, it is common practice to 
ignore dead load redistribution via diaphragm action 
and apply the entire dead weight of the door as a 
uniform line load to the head framing system.
As the door opens the weight of the door begins 
to move away from the building wall. This causes 
eccentricity in the rigid body door assembly. This 
eccentricity causes the door to want to pull out of the 
door opening. A force couple develops to counter 
this rotational force. The eccentric dead load and the 
resisting force couple are at maximum when the door 
is fully open.
The cylinder is a pin ended axial link. Hence the 
resultant of W1 and F must align with the cylinder axis. 
The cylinder force can be calculated directly if system 
geometry is well known. However, it is not common 
in actual practice to know the precise orientation 
RESEARCH & TECHNOLOGY
continued from page: 35
THE OFFICIAL NFBA MAGAZINE
Mono-Fold Hydraulic Door

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