LRFD � Floor beam by ULJo77p

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									LRFD – Floor beam
      Unbraced top flange
      Lateral Torsion Buckling

 We have to check if there is plastic failure (yielding)
  or lateral-torsion buckling.
 This depends on the length between the lateral
  braces, related to the limiting lengths.
 Lp is the limiting length for plastic failure
 Lr is the limit length for torsional buckling.
 If Lb < Lp it is plastic failure
 If Lp < Lb < Lr we have a different failure criteria
 If Lb > Lr we use the lateral buckling stress criteria
               Plastic Failure


 If Lb < Lp
 Mn = Mp = sy Zx
       Zx is the plastic section modulus about the x axis
                  Lp < L b < L r


                                     L  L 
    Mn
              
                      
          Cb M p  M p  0.7sy Sx   
                                       L  L  M p
                                        b
                                        r
                                              p
                                                
                                                 
                                              p 
                                               




          Lb > L r


   Mn = scrSx ≤ Mp
The following definitions apply


                 Cb  E2
                                       Jc   Lb  2

           scr           2 1  0.078        
                 Lb               Sx ho  rts 
                       
                   rts 

                                           E
                           L p  1.76 ry
                                           sy

                 E         J c             0.7sy Sx h0 
                                                          2

Lr  1.95 rts                    1 1 6.76            
             0.7sy       Sx h0            EJc 
                       c


 For a doubly symmetric I-shape
   c=1

                      h0   Iy
 For a channel,   c
                      2    Cw

 Where h0 = distance between flange centroids
           
     Conservative simplifications


                                Cb  E
                                     2
                          scr 
                                Lb 
                                     
                                 rts 

                   E                              Iy Cw
     Lr  rts                              r 
                                             2
                                            ts
                 0.7s y                           Sx
       


 A beam of A992 steel with a span of 20 feet
  supports a stub pipe column with a factored load
  combination of 55 kips




   A992 Steel: structural steel, used in US for I-beams.
           Density = 7.85 g/cm3. Yield strength = 50 ksi.
   No flooring – no lateral bracing on top
                   flange



Find max moment.

Assume beam weighs 50 lbs/ft

From distributed load, Mmax = w L2/8

From point load, Mmax = P L / 4

Mmax = 55,000 * (20/4) + 50 * (20^2)/8 = 277.5
  kip-ft
                 Use trial method


•   Find a beam that has a fMp of at least 277.5 kip-ft
    •   Need to check if it will fail in plastic mode (Mp) or from
        flange rotation (Mr)
    •   Tables will show limiting unbraced lengths.
    •   Lp is full plastic capacity
    •   Lr is inelastic torsional buckling.
    •   If our length is less than Lp, use Mp. If greater than Lr,
        use Mr
Selected W Shape Properties –
          Grade 50

Prop           W18x35   W18x40   W21x50   W21x62
fMp (kip-ft)   249      294      416      540
Lp (ft)        4.31     4.49     4.59     6.25
Lr (ft)        11.5     12.0     12.5     16.7
fMr (kip-ft)   173      205      285      381

Sx (in3)       57.6     68.4     94.5     127
Iy (in4)       15.3     19.1     24.9     57.5
ho (in)        17.28    17.38    20.28    20.39
ry (in)        1.22     1.27     1.30     1.77
J (in4)        0.506    0.81     1.14     57.5
Cw             1140     1440     2560     5970
    W18 x 40 looks promising


294 > 277.5

But, Lp = 4.49. Our span is 20 feet.

And, Lr = 12.0 again, less than 20’

fMr = 205, which is too small.

W21x50 has Lr = 12.5, and fMr = 285.

That could work!
Nominal flexural design stress


 Mn = scr Sx

 The buckling stress, scr , is given as


               Cb  E
                    2
                                     Jc   Lb  2

         scr           2 1  0.078        
               Lb               Sx ho  rts 
                     
                 rts 




       Terms in the equation


 rts = effective radius of gyration

 h0 = distance between flange centroids

 J = torsional constant (torsional moment of
  inertia)

 Cw = warping constant

 c = 1.0 for doubly symmetric I-shape
     Effective radius of gyration


                        I y Cw
               r2
               ts   
                        Sx

              24.9 • 2560
      rts                        1.635 in
               94.5




           So the critical stress is




                                     1.14 • 1.0 20 • 12 
                                                         2
         1.0  2 29,000
scr                    2 1  0.078                       18.77 ksi
       20 x 12
                                 94.5 • 20.28  1.635 
                1.635
                     
  Then the nominal moment is


 Mn = scr Sx

 = 18.77 • 94.5 = 1,774 kip-in = 147.9 kip-ft

 We need 277.5!!
    If we had the AISC design manual, they show
     unbraced moment capabilities of beams.
    We would have selected W21x62, which turns
     out to handle 315.2 kip-ft unbraced.

								
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