DESIGN OF STEEL BEAMS
DR. BASHAR BEHNAM
GENERAL
• When a beam bends, the compression region
above the neutral axis is analogous to a column
and it will buckle if the member is slender enough.
• The compression portion/region of the cross section
is restrained by the tension portion.
• The outer deflection (flexural buckling) is
accompanied by twisting (torsion)
LATERAL-TORSIONAL BUCKLING
• This form of instability is called lateral-torsional
buckling (LTB).
• Lateral-Torsional buckling can be prevented by
bracing the beam against twisting at sufficiently
close intervals. This can be accomplished two
different types of stability bracing.
STABILITY BRACING
• There are two types of stability bracing
1.
Lateral bracing:
• X-bracing
• Concrete deck
2.
Torsional bracing:
• Cross frame
• Diaphragm
LATERAL BRACING
X-Bracing
Concrete Deck
TORSIONAL BRACING
Cross-Frame
Diaphragm
FAILURE MODES
• There are four failure modes
1.
2.
3.
4.
The cross section of the beam becomes fully plastic.
Lateral-torsional buckling (LTB), either elastically or
inelastically.
Flange local buckling (FLB), either elastically or
inelastically.
Web local buckling (WLB), either elastically or inelastically.
CLASSIFICATION OF SHAPES
• AISC Manual (American institute of Steel
Construction) classifies cross-sectional shapes as:
• Compact
• Noncompact
• Slender
THE UNBRACED LENGTH
• The unbraced length Lb is the distance between
points of lateral support, or bracing. The moment
strength of compact shapes is function of the
unbraced length, Lb.
NOMINAL FLEXURAL STRENGTH
• If the unbraced length, no longer than Lp, the beam
is considered to have full lateral support and the
nominal flexural strength can be calculated
M n  M p  Z Fy
NOMINAL FLEXURAL STRENGTH
• If Lb is greater than Lp but less than or equal to the
parameter Lr, then the strength is based on inelastic
LTB.

 Lb  L p 
  M p
M n  Cb  M p  M p  0.7 Fy S x 
 L  L 

p 
 r
NOMINAL FLEXURAL STRENGTH
• If Lb is greater than Lr, the strength is based on
elastic LTB.
M n  Fcr S x  M p
Fcr 
Cb  E
2
 Lb 
 
 rts 
2
Jc  Lb 
 
1  0.078
S x ho  rts 
2