# 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 –

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

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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