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 < Lb < Lr
Mn

 Cb M p  M p  0.7sy Sx



L  L 
b
p

L  L 
 M p
 r
p 


Lb > Lr

Mn = scrSx ≤ Mp
The following definitions apply
2


Cb  E
Jc Lb
scr 
 
2 1  0.078
Lb 
Sx ho rts 


r
 ts
2
Lp 1.76 ry

E
Lr 1.95 rts
0.7
s
y

E
sy
2


0.7sy Sx h0
J c
1 1 6.76

Sx h0
 EJc 
c
 For a doubly symmetric I-shape
 c=1
 For a channel,
h0 Iy
c
2 Cw
 Where h0 = distance between flange centroids


Conservative simplifications
Cb  E
scr 
Lb 


 rts
2
Lr  rts

E
0.7s y
r 
2
ts
I y Cw
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
2


Cb  E
Jc Lb
scr 
 
2 1  0.078
Lb 
Sx ho rts 


r
 ts
2

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
2
ts
r
rts 



I y Cw
Sx
24.9 • 2560
 1.635 in
94.5
So the critical stress is
2
1.0  2 29,000
1.14 • 1.0 20 • 12
scr 

  18.77 ksi
2 1  0.078
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.