# Re: Roof Truss Wind Pressure Coefficient

• To: seaint(--nospam--at)seaint.org
• Subject: Re: Roof Truss Wind Pressure Coefficient
• From: TARNO COLEMAN <TCOLEMAN(--nospam--at)OPENGOVT.OPEN.ORG>
• Date: Mon, 24 Jul 2000 15:25:14 -0700
```This is an awkward way that the code describes the change in pressure
coefficients.

I visualize it by plotting an x-y graph w/ Cq on the Y axis and Area
on the X axis.

Cq              Area

1.2                   0
1.2                 10
0.9                100   (where the reduction by 0.3 comes in)
0.8               1000   and upward (@ primary frame)

the linear interprolation between 100 sq. feet and 1000 sq. ft trib.
area gives the pressure coefficient that should be used but
is usually neglected.

Tarno Coleman, PE
Plan Check Engineer
503.566.3964
e-mail: tcoleman(--nospam--at)mail.open.org
Marion County Building Inspection
PO Box 14500, Salem, Oregon  97309

tcoleman(--nospam--at)mail.open.org

>>> Monty Hart <montyh(--nospam--at)gci.net> 07/24/00 02:37PM >>>
I have observed fabricators of pre-engineered wood roof
trusses routinely using the Primary Frames and Systems
Pressure Coefficients from Table 16-H of the '97 UBC,
instead of using the Elements and Components Coefficients,
to calculate wind uplift forces.  The difference is
significant (i.e. 0.7 vs. 1.3).  Foot note 2 of the table
says, "For tributary areas greater than 1,000 sf, use
primary frame values".  Under that criteria, a 24" oc wood
truss would have to span 500 ft, before it qualified for the
lower Primary Frames Coefficient.

I have now observed this practice in two high wind states
(Alaska and Hawaii), with calculations done by major truss
fabricators, using software supplied by two of the major
truss plate manufacturers.   I am wondering what other
engineers consider to be the appropriate pressure
coefficient for wood trusses, and if using the primary
frames coefficient is common in other areas.

Monty A. Hart

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