Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
(OP)
I have calculated the moment of inertia of a 2x4, 18-inch deep modified metal-plated flat truss as 1312.79 from standard "equal rectangle" formula in the AISC manual. Can someone tell me where to obtain wood floor truss section properties listed? I believe this looks correct. I am not into wood trusses, but I need to check some deflection figures. Hmmm.
Thank you for your assistance.
Thank you for your assistance.






RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
For 2x4 chords and depth of 18", moment of inertia cannot be as high as you have calculated.
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
I = 2*A*y2
A is the area of one 2x4 = 1.5*3.5 = 5.25in2
y is the distance from the centroid of the truss to the centroid of 2x4 = 7.25"
so I = 2*5.25*(7.25)2 = 551.9
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
That would give you I = 1.5 (183 - 113)/12 = 562.6, a little more than found before. The difference is the moment of inertia of two 2x4 which I believe should be ignored.
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
Or, using the other formula, I = 3.5(183 - 153)/12 = 716.6 which is only a little more because Ichord is small.
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
There is still a problem, however. Using 714.6 in4, you should calculate a deflection of 0.72*1312.79/714.6 = 1.32" which is 11% more than the value on the print; and if you increase deflection to allow for web member distortion, you would have 1.32*1.15 = 1.52" according to KootK or 1.32*1.10 = 1.45 according to my old boss, neither agreeing with the published value within acceptable tolerance.
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
When I am working on a problem, I never think about beauty but when I have finished, if the solution is not beautiful, I know it is wrong.
-R. Buckminster Fuller
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
RE: Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss
If you make a little truss in RISA, you get something close to the 85% number. Could have just been dumb luck on my part. I can never come close to matching the Mitek software deflections.