section modulus for I or WF-beam? 1

[zerts]

Please help, I don't have the AISC Manual, and Googling this brought up conflicting formulae.

All I need is the formula for Section Modulus, expressed in AISC terms or any generic terms for an I or WF-section. I don't need the expression for that small area included in the fillet radius (flange to web)

Thanks in advance!

 

[zerts]

ps- I am looking for the Elastic, not Plastic modulus.

 

[IFRs]

The strong axis section modulus for a symmetrical I-beam can be approximated as Tw / 12 * ( D - 2 x Tf ) ^ 3 + 2 x ( B x Tf ^ 3 / 12 + B x Tf x ( ( D - Tf ) / 2 ) ^2 ) all divided by D / 2.

This assumes square cornered web and flanges. D is the overall height. B is the flange width. Tf is the flange thickness. Tw is the web thickness.

Please be sure to verify this agains a known shape.

 

[hokie66]

You just take the modulus of inertia of the overall rectangle, subtract the modulus of inertia of the part left out, and divide by half the depth. I=bd^3/12 if you don't know that.

 

[IFRs]

See this !!

http://www.eng-tips.com/viewthread.cfm?qid=238029&page=1

 

[zerts]

hokie66,
I think that only works if I am looking for the Section Modulus of the flanges only (as if the flanges were effective, but the web was left out of the calculation) - it is pretty close, a bit conservative.

 

[nutte]

Zerts: Hokie's way works. You just have to subtract the proper width. Added section width = bf, subtracted section width = bf-tw.

 

[hokie66]

Thanks, nutte.

 

[PEinVA]

S=I/c

I = Iflanges + Iweb + Af(df)2
c = centroid

simple as that

RC
All that is necessary for the triumph of evil is that good men do nothing.
Edmund Burke

 

[frv]

RCraine has it right.

If you don't know the exact moment of inertia of the wide flange, you should be able to approximate it as three rectangles, conservatively neglecting fillets.

btw.. that's not an AISC provision, it's a mechanics of solids concept.. back to basics!

 

[WillisV]

Hokie's way is actually more straight forward because it doesn't involve the parallel axis theorem (everything is still about the elastic neutral axis - in the same terms as IFR:

S = [(B/12)x D^3-((B-Tw)/12)x(D-2xTf)^3]/(D/2)

 

[frv]

I can see that.