asce-7 02 external pressure coefficien
 

asce-7 '02 external pressure coefficient
im making a program for wind calculation..
in figure 6-6(con't), in getting the value of roof pressure coefficient cp, where do you get the value of cp when roof angle is than 10 and h/l is between 0.5 and 1.0?
since the table shows only values of h/l <=0.5 and >=1.0
help much appreciated!

you would use a linear interpolation between the values provided for less than or equal to h/l of 0.5 and greater than or equal to h/l of 1.0. for instance if you have an h/l of 0.75 then the cp for distances from 0 to h/2 from the edge would be midway between -0.9 and -1.3 (-1.1).
i agree with willisv, to a point.
you would have to do four linear interpolations,
1) for dist from windward edge 0 to h/2 cp = -1.1 or -.18
2) for dist from windward edge h/2 to h cp = -0.8 or -.18
3) for dist from windward edge h to 2h cp = -0.6 or -.18
4) for dist from windward edte > 2h cp = -0.4 or -.18
also, note that the first one can have a reduction factor applied to the interpolation number for the high side (h/l = 1.0; cp = -1.3) based on effective area; this reduction does not apply to the interpolation number on the low side (h/l = 0.5; cp = -0.9)
regards,
chichuck
thanks!
thats what i did for now.. interpolate for the upper, interpolate for the lower.. then interpolate again..