ASCE Windloads
ASCE Windloads
(OP)
I'm analyzing the wind loads on a 5/12 pitch roof truss for a post and frame structure (pole barn type). I've determined the winds loads along the surface of the truss, ASCE – 7, Fig. 6-10, footnote 1. I've got a partially enclosured structure, hence Method 2. To find the horizontal reactions on the truss, one would just breaks the ASCE wind loads into horizontal and vertical reactions. See attached jpeg.
My confusion lays in part w/ an example by NFBA's National Frame Builder Association example, where they just take the wind load from the surface an place it vertically.
My confusion lays in part w/ an example by NFBA's National Frame Builder Association example, where they just take the wind load from the surface an place it vertically.






RE: ASCE Windloads
RE: ASCE Windloads
Lets say you resolve the horizontal component as x=6plf*sin(22.6) =2.308 PLF. This load is acting along the surface of the top chord.
To find the equivalent of the load acting on the vertical projection of the truss, you have to find the rise (in the y direction) for every unit of length along the top chord.
thus for every foot along the top chord, you have a rise of
y=sin(q)\1 where q = roof slope.
y= sin(22.6)\1 = 0.38465
because sin(22.6) must equal to 0.38465/1.
Thus if you have a 2.308 PLF acting horizontally along the length of top chord, its equal to a load of 2.308 lbs acting over a vertical projection of 0.38465 ft.
If you have a load of 2.308 lbs over 0.38465 ft, you have a total load of 6.0 lbs acting over a 1.0 ft length in the vertical direction. (2.308/0.38465)
In other words, both diagrams are correct.
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C
RE: ASCE Windloads
EIT
RE: ASCE Windloads
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C
RE: ASCE Windloads
You have to consider both, plus the oblique...which is similar to the perpendicular sense.