Moment of Inertia - 18-inch deep 2x4 Modified Warren Flat Truss

[DCEngr1]

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.

 

[KootK]

I typically use the 100% chord value, as I believe you have done, and reduce it by 15% to account for shear deformation. I don't know of anywhere where this is tabulated.

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.

 

[DCEngr1]

KK . . . Thank you for responding. I am attempting to verify some deflection calculations from a plate mfgr. software program. Do you know if they reduce their I in their calculations?

 

[KootK]

They wouldn't have an "I" per se. Rather, they would use true truss theory to determine displacements, similar to how we might do in SAP, RISA, etc. Or, goodness forbid, by hand.

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.

 

[DCEngr1]

Using the 1312.79 I came up with a smaller deflection, using TPI formulas, than the deflection that shows up on the individual truss print. 0.72 inches by my calc. - 1.19 inches on the print

 

[KootK]

I would expect their value to be higher than yours as their value includes the shear deflection that will occur via elongation of the truss webbing. In this case, my 15% estimate would have been considerably low.

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.

 

[BAretired]

When I was just starting in the engineering profession, my boss told me to increase deflection by 10% when using moment of inertia determined from the chords only. He was referring to open web steel joists, but it should be about the same for wood trusses 18" deep.

For 2x4 chords and depth of 18", moment of inertia cannot be as high as you have calculated.

 

[BAretired]

I believe I = 552 in⁴. That is about 42% of your calculated value. So your deflection should have been .72/0.42 = 1.71". Increase that by 10% = 1.88" which is 58% more than shown on the print. Something's wrong somewhere.

 

[hokie66]

Agree with BA. Your arithmetic is obviously wrong.

 

[DCEngr1]

Guys . . . so what equasion should I be using to figure for a wood truss 18 inches deep with 2x4 chords & 2x4 webs - 4 inch dimension horizontal? Obviously, I am using the wrong equasion.

 

[hokie66]

You quoted the AISC "equal rectangle" equation. That is correct. bd^3/12, and deduct bd1^3/12. If the 2 x 4 members are flat, your I will be larger than BA's answer, which I think assumed the members vertical. I get 716, based on 1.5 x 3.5 members. I don't know exactly how big a 2 x 4 is in the US these days.

 

[BAretired]

You should use the following:
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

 

[BAretired]

I don't know what formula the TPI recommends. It might be the difference between two rectangular areas, the outer one being 1.5 x 18 and the inner one being 1.5 x 11 (the clear distance between chords).

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.

 

[hokie66]

BA, in his last post, he said the members are horizontal, so the two rectangles would be 3.5 x 18 and 3.5 x 15.

 

[BAretired]

Okay hokie, I missed that. In that case I = 2*5.25*8.25² = 714.6 in⁴

Or, using the other formula, I = 3.5(18³ - 15³)/12 = 716.6 which is only a little more because I_chord is small.

 

[DCEngr1]

BAretired & hokie66 - Thanks for your patience & diligence. I had the right equasion, just somehow used the wrong d1.

 

[BAretired]

Using the 1312.79 I came up with a smaller deflection, using TPI formulas, than the deflection that shows up on the individual truss print. 0.72 inches by my calc. - 1.19 inches on the print

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.

 

[SteelPE]

A little off topic, but I too was always told to take a 15% reduction for shear deformation of the web members when analyzing a truss based upon the moment of inertia of the chords. This was from one of my professors in school.... and I believe my old boss as well. I still use this number today, however, I could never prove that it is correct or incorrect if required. Good to know I can at least do one thing right .

 

[manstrom]

Just subtract the I of a 3.5x15" from a 3.5x18" rectangle. Easier math.

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

 

[BAretired]

Easier than what?