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# section modulus for I or WF-beam?

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 zerts (Structural) 19 Feb 09 15:12
 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 (Structural) 19 Feb 09 15:19
 ps- I am looking for the Elastic, not Plastic modulus.
 IFRs (Petroleum) 19 Feb 09 16:48
 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 (Structural) 19 Feb 09 22:33
 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 (Petroleum) 20 Feb 09 11:15
 zerts (Structural) 20 Feb 09 12:15
 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 (Structural) 20 Feb 09 12:31
 Zerts:  Hokie's way works.  You just have to subtract the proper width.  Added section width = bf, subtracted section width = bf-tw.
 hokie66 (Structural) 20 Feb 09 16:49
 Thanks, nutte.
 RCinVA (Structural) 20 Feb 09 17:18
 S=I/cI = Iflanges + Iweb + Af(df)2c = centroidsimple as that RCAll that is necessary for the triumph of evil is that good men do nothing.    Edmund Burke
 frv (Structural) 20 Feb 09 21:19
 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 (Structural) 23 Feb 09 8:46
 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 (Structural) 23 Feb 09 13:20
 I can see that.

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