1<br />Mahmoud Shaaban<br />
Solution: for OS Beam<br />a/b = 5000 / 300 = 16.67<br />From the graph, assume   Є = 0,  kb = 24<br />2<br />Mahmoud Shaa...
σb = 24 π2x 205000  (3/300)2<br />12 (1 – 0.32)<br />σb = 444.6 Mpa<br />σ b &gt; Fsy; Design should be on Fsy in equation...
Since Y = 0, Frt & Frc = 0<br />Using eq. 2 for Nc<br />bc = bs = 2ts – 2 tsc<br />    = 150 – 2x3 – 0 = 144<br />α = 0.85...
Using eq 8  (y=0) to find Ns<br />Ns = 350 x 3 (2x300 + 150 – 2x55) – 215000<br />                       4 (3x350)<br />Ns...
WE BEAM<br />Y should equal d/2 or d/4<br />Y = d/2 = 300/2 = 150 mm<br />tse = 3 mm<br />Ast = Asc = 0<br />Using  Є = 0<...
Pb = Σo.x.Fb      where     Σo = 860 + 4Y = 1460 mm<br />     = 1460x2500x0.1       fb = 0.1<br />     = 365000 N         ...
Using eq 7. to find Ns<br />Ns = 350x3 (2x300+150-2x55)+2x150x3x350-365000<br />                           4 (3x350 + 3x35...
Since Y &gt; Ns then will use eq10 to find Mu<br />Mu = 3x350 (3002+300x150-2x74.052)<br />         + 3x350 (1502 – 2x74.0...
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Thin Walled Composite Filled Beam

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Thin Walled Composite Filled Beam (TWC); OS &amp; WE Sections

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Thin Walled Composite Filled Beam

  1. 1. 1<br />Mahmoud Shaaban<br />
  2. 2. Solution: for OS Beam<br />a/b = 5000 / 300 = 16.67<br />From the graph, assume Є = 0, kb = 24<br />2<br />Mahmoud Shaaban<br />
  3. 3. σb = 24 π2x 205000 (3/300)2<br />12 (1 – 0.32)<br />σb = 444.6 Mpa<br />σ b &gt; Fsy; Design should be on Fsy in equation 11 for the moment<br />For OS Beam:<br />Pb = Σo.x.Fb where Σo = 150+2x300+2x55 = 860<br /> = 860x2500x0.1 fb = 0.1<br /> = 215000 N x = 5000/2 = 2500 <br />3<br />Mahmoud Shaaban<br />
  4. 4. Since Y = 0, Frt & Frc = 0<br />Using eq. 2 for Nc<br />bc = bs = 2ts – 2 tsc<br /> = 150 – 2x3 – 0 = 144<br />α = 0.85 – 0.007 (30-28)<br /> = 0.836<br />Nc = 215000 <br /> 0.85x30x0.836x144<br />Nc = 70.03 mm<br />4<br />Mahmoud Shaaban<br />
  5. 5. Using eq 8 (y=0) to find Ns<br />Ns = 350 x 3 (2x300 + 150 – 2x55) – 215000<br /> 4 (3x350)<br />Ns = 108.8 mm<br />Using eq 11 to find Mu<br />Mu = 3x350 (3002 + 300x150 – 2x108.82) <br /> – 0.425 x 0.8362 x 70.032x150x30 <br />Mu = 110 KNm<br />W = 8 Mu/L2 = 8x 110 / 52 = 35.2 KN/m<br />5<br />Mahmoud Shaaban<br />
  6. 6. WE BEAM<br />Y should equal d/2 or d/4<br />Y = d/2 = 300/2 = 150 mm<br />tse = 3 mm<br />Ast = Asc = 0<br />Using Є = 0<br />a/b = 16.66 Kb = 24<br />σ b &gt; Fsy; Design should be on Fsy in equation 11 for the moment<br />6<br />Mahmoud Shaaban<br />
  7. 7. Pb = Σo.x.Fb where Σo = 860 + 4Y = 1460 mm<br /> = 1460x2500x0.1 fb = 0.1<br /> = 365000 N x = 5000/2 = 2500 <br />Using eq2. to find Nc<br />bc= 150-2x3-2x3 = 138<br />Nc = 365000<br /> 0.85x30x0.836x138<br />Nc = 124.07 mm<br />7<br />Mahmoud Shaaban<br />
  8. 8. Using eq 7. to find Ns<br />Ns = 350x3 (2x300+150-2x55)+2x150x3x350-365000<br /> 4 (3x350 + 3x350)<br /> Ns = 74.05 mm<br />Mahmoud Shaaban<br />8<br />
  9. 9. Since Y &gt; Ns then will use eq10 to find Mu<br />Mu = 3x350 (3002+300x150-2x74.052)<br /> + 3x350 (1502 – 2x74.052)<br /> - 0.425x0.8362x124.072x138x30<br />Mu = 123.4 kNm<br />W = 8Mu/L2 = 8x 123.4/52 = 39.48 kN/m<br />9<br />Mahmoud Shaaban<br />
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