Beams
Stephen Krone, DSc, PE
University of Toledo
Shear and Moment
A beam is a structural member subjected to
transverse loads and negligible axial loads.
Internal Shear Causes Moment
These internal shear forces and bending
moments cause longitudinal axial stresses
and shear stresses in the cross-section.
Increasing Loads
Maximum bending
moment is reached
at extreme fibers
first, where the
stress reaches the
yield stress limit.
The Moment - Curvature (M-f) response With
Increasing Moment
My is the yield and Mp is the plastic moment
capacity of the cross-section.
The ratio Mp to My is the shape factor f.
For a rectangular section, f is equal to 1.5.
For a wide-flange section, f is equal to 1.1.
Developing the Plastic Moment
LRFD Raised the ASD Limit State
Columns’ Slenderness Ratio
Controls the Critical Buckling Stress
Lateral Torsional Buckling Behavior
Compression flange
begins to buckle out
of plane.
Bracing on the
compression flange
makes it more
difficult to fail.
Methods of Bracing Compression Flange
Beam and Column Similarities
• The design of the beam is usually based on
the compression flange.
• The braced length of support of the
compression flange, Lb, usually determines
the type of failure.
• There are three zones of failure: plastic,
inelastic buckling, and elastic buckling.
• Each failure zone has its own set of
equations.
Lateral Torsional Buckling:
Lb  Lp
AISC F2-1
AISC F2-2
AISC F2-3
Mn = Fcr Sx <= Mp
Plastic
Inelastic
Buckling
AISC F2-5
Lp = 1.76 ry E / Fy
Elastic
Buckling
Lr AISC F2-6 AISC 16.1-48
Plastic Buckling
•Most beams fall in the plastic zone.
•AISC Table 3-10
•Fully braced reaches full Mp. This is closely spaced lateral
bracing, where Lb < Lp
AISC F2-1
AISC F2-2
AISC
F2-3
M
n=F
cr Sx <= Mp
Plastic Section Modulus
y1
y2
sy A 1
F = s y A1 - s y A 2 = 0
\ A1 = A 2 = A / 2
A
´ ( y1 + y 2 )
2
Where , y1 = centroid of A1
\ M = sy
y 2 = centroid of A 2
Mp, Plastic Moment and
Plastic Section Modulus
W Shapes – Braced => Mp
Mp, Plastic Moment
Concrete floor slab provides lateral bracing.
Lb= 0.
Beam Example: Compact & Braced
Design a simply supported beam subjected to
uniformly distributed dead load of 450 lbs/ft.
and a uniformly distributed live load of 550
lbs/ft. The dead load does not include the selfweight of the beam.
Step I. Calculate the factored design loads
(without self-weight).
• wU = 1.2 wD + 1.6 wL = 1.42 kips / ft.
• MU = wuL2/8 = 1.42 x 302/8 = 159.75 kip-ft.
Beam Example (Cont.)
Step II. Select the lightest section from the
AISC Manual design tables.
• From the AISC manual, select W16 x 26
made from 50 ksi steel with fbMp = 166.0
kip-ft.
Step III. Add self-weight of designed section
and check design
• wsw = 26 lbs/ft
• Therefore, wD = 476 lbs/ft = 0.476 lbs/ft.
• wu = 1.2 x 0.476 + 1.6 x 0.55 = 1.4512 kips/ft.
• Therefore, Mu = 1.4512 x 302 / 8 = 163.26 kip-ft.
< fbMp of W16 x 26.
• OK!
Beam Example (Cont.)
Step IV. Check deflection at service loads.
• w = 0.45 + 0.026 + 0.55 kips/ft. = 1.026 kips/ft.
• D = 5 w L4 / (384 E Ix)
= 5 x(1.026/12) x (30 x 12)4/(384 x 29000 x 301)
• D = 2.142 in. > L/360
Step V. Redesign with service-load deflection
as design criteria
• L /360 = 1.0 in. > 5 w L4/(384 E Ix)
• Therefore, Ix > 644.8 in4
Example (Cont.)
• Select the section from the moment of inertia
tables in the AISC Table 3-3. (p. 3-21)
Ix > 644.8 in4
• Select W21 x 44.
• W21 x 44 with Ix = 843 in4 and fbMp = 358
kip-ft. (50 ksi steel).
• Deflection at service load
D = 0.765 in. < L/360
- OK!
Beam Example (Cont.)
Step VI. Check shear for W16 x 26 .
wu = 1.2 wD + 1.6 wL = 1.42 kips / ft.
V = wl/2 = 1.42(30)/2 = 21.3 kips
Vn
= 0.6FyAwCv
= 0.6 (50ksi)(0.25x15.7)(1.0)
= 118 kips > 21.3 kips
Vn can be compared with Z Table (3-2) of
fvVnx = 106.0 kips > 21.3 kips
Non-plastic Failure
The development of a plastic stress distribution
over the cross-section can be hindered by
two different length effects:
(1) Local buckling of the individual plates
(flanges and webs) of the cross-section fail
before they develop the compressive yield
stress sy.
(2) Lateral-torsional buckling of the
unsupported length of the beam / member
fail before the cross-section develops the
plastic moment Mp.
Lateral-Torsional Buckling
• Lb of a beam-member can
undergo lateral-torsional
buckling due to the applied
flexural loading (bending
moment).
• Lateral-torsional buckling is
similar to the flexural
buckling or flexural-torsional
buckling of a column
subjected to axial loading.
• There is one very important
difference. For a column, the
axial load causing buckling
remains constant along the
length. But, for a beam,
usually the lateral-torsional
buckling causing bending
moment varies along the
unbraced length.
LTB: Lateral-Torsional Buckling
• bMn is a function of:
– Beam section properties
– (Zx, ry, x1, x2, Sx, G, J, A, Cw, Iy)
– Unbraced length, Lb
– Lb = distance between points which are
either braced against lateral displacement
of compression flange or braced against
twist of the cross section
– Lp : Limiting Lb for full plastic bending capacity
– Lr : Limiting Lb for inelastic LTB
• E : Modulus of elasticity, ksi
• Cb: Bending coefficient based on moment gradient.
Lateral Torsional Buckling:
Lp  Lb  Lr
• If the laterally unbraced length Lb is less than
or equal to a plastic length Lp then lateral
torsional buckling is not a problem and the
beam will develop its plastic strength Mp.
•
- for I members & Channels
(See
Lp = Pg.
1.7616.1-48,
ry E / FyF2-5)
• If Lb is greater than Lp then lateral torsional
buckling will occur and the moment capacity
of the beam is reduced below the plastic
strength Mp.
Lateral Torsional Buckling:
Lp L b  Lr
AISC F2-1
AISC F2-2
AISC F2-3
Mn = Fcr Sx <= Mp
AISC F2-5
Lp = 1.76 ry E / Fy
Lr AISC F2-6 AISC 16.1-48
Lateral Torsional Buckling:
Lp  L b  Lr
• When Lp  Lb  Lr, then the lateral-torsional
buckling Mn is given by:
Mn= Cb[Mp – (Mp -0.7Fy Sx )(Lb- Lp / Lr–Lp)]
•Linear interpolation between (Lp, Mp) and (Lr, Mr)
•AISC Table 3-2 - Lp, Lr, Mp, Mr
Mp
Mr
Lp
Lb
Lr
Lateral Torsional Buckling:
Lp  L b  Lr
• As the spacing of lateral bracing increases
but not all compression fibers will reach Fy.
• This is referred to as inelastic buckling.
• Easier to use BF:
– Mn= Mp – (BF )(Lb- Lp)
• Lr is the point where the transition is to
elastic buckling.
Lateral Torsional Buckling: Lb > Lr
•Elastic Buckling
•The buckling moment becomes smaller and smaller as
the unbraced length increases.
•AISC p. 16.1-47
AISC F2-1
AISC F2-2
AISC F2-3
Lateral Torsional Buckling: Lb > Lr
AISC F2-1
AISC F2-2
AISC F2-3
Mn = Fcr Sx <= Mp
Plastic
Inelastic
Buckling
AISC F2-5
Lp = 1.76 ry E / Fy
Elastic
Buckling
Lr AISC F2-6 AISC 16.1-48
Compact, Non-Compact, and Slender
• Slender sections cannot develop Mp due to elastic local
buckling. Non-compact sections can develop My but
not Mp before local buckling occurs.
• Only compact sections can develop the plastic moment
Mp.
• Applies to major and minor axis bending
Local Buckling of Flange Due to
Compressive Stress
• Equations for local
buckling of steel plates
have limiting slenderness
ratios for the individual
plate elements of the
cross-sections.
• AISC B4 (page 16.14),
Table B4.1 (16.1-13) and
Page 16.1-223
M
• Steel sections are
classified as compact,
non-compact, or slender
depending upon the
slenderness (l) ratio of
the individual plates of
the cross-section.
M
Local Buckling
Limit States for Local Buckling
Table B4.1 (p 16.1-16)
Limit States for Local Buckling
Table B4.1 (p 16.1-16)
Strength Limit State for Local Buckling
•  = Slenderness parameter; must be
calculated for flange and web buckling
(actually no web slender for W, M, C)
• p= limiting slenderness parameter for
compact element
• r = limiting slenderness parameter for
non-compact element
•   p : Section capable of developing fully
plastic stress distribution
Strength Limit State for Local Buckling
• p    r : Section capable of developing
yield stress before local buckling occurs; will
buckle before fully plastic stress distribution
can be achieved.
•   r : Slender compression elements; will
buckle elastically before yield stress is
achieved.
Compact, Non Compact, or Slender
• For inelastic buckling because of unbraced
length:
Mn= Mp – (Mp -0.7Fy Sx )(Lb- Lp / Lr–Lp)]
AISC F2-2
• For Flange Local Buckling because flange in
not compact:
Mn= Mp – (Mp -0.7Fy Sx )( - p / r - p)
AISC F3-1
The Bending Coefficient, Cb
• Worse case scenario is for compression
flange under constant uniform moment
• Moments usually vary over unbraced
length, so AISC uses Cb to modify for only
segment that undergoes LTB under the
highest moment.
• This factor is similar to the effective length K
that we modified in column buckling.
– (Old formula like G/G)
– Cb =1.75 + 1.05(M1/M2)+0.3(M1/M2 )
The Bending Coefficient, Cb
• Cb can always be taken as 1.
Cb =12.5 Mmax/(2.5Mmax+3Ma +4Mb + 3Mc)
• Mmax = maximum value in braced section
• Ma = value at quarter point
• Mb = value at center point
• Mc = value at three quarter point
The AISC Specification says that:
• Cb is a multiplier.
• Cb is always greater than 1.0 for nonuniform bending moment.
– Cb is equal to 1.0 for uniform bending moment.
– If you cannot calculate or figure out Cb, then it
can be conservatively assumed as 1.0.
Moment Capacity versus Lb
(for non-uniform moment case)
• Bending
Coefficient,
Cb
• AISC
Table 3-1
Example
• Design the beam shown below. The
unfactored uniformly distributed live load is
equal to 3 kips/ft. There is no dead load.
Lateral support is provided at the end
reactions.
Step I. Calculate the factored loads assuming a
reasonable self-weight.
• Assume self-weight = wsw = 100 lbs/ft.
• Dead load = wD = 0 + 0.1 = 0.1 kips/ft.
• Live load = wL = 3.0 kips/ft.
• Ultimate load =
– wu = 1.2wD + 1.6wL = 4.92 kips/ft.
• Factored ultimate moment =
– Mu = wu L2/8 = 354.24 kip-ft.
Step II. Determine unsupported length Lb & Cb
• One unsupported span with Lb = 24 ft.
• Cb = 1.14 for the parabolic bending moment
diagram, See values of Cb shown in Table 3-1.
Step III. Select a wide-flange shape
• The moment capacity of the selected section
fb Mn > Mu (Note fb = 0.9)
• fbMn = moment capacity = Cb x (fbMn for the case
with uniform moment)  fbMn
• Table 3-10 in the AISC-LRFD manual shows plots
of fbMn - Lb for the case of uniform bending
moment (C =1.0)
Therefore, in order to select a section, calculate
Mu/ Cb and use it with Lb to find the first
section with a solid line.
• Mu/ Cb = 354.24/1.14 = 310.74 kip-ft.
• Select W16 x 67 (50 ksi steel) fbMn =357 kip-ft. for
Lb = 24 ft. and Cb =1.0
• For the case with Cb = 1.14,
• fbMn = 1.14 x 357 = 406.7 kip-ft., which must be 
fbMn = 491 kip-ft.
OK!
• Thus, W16 x 67 made from 50 ksi steel with
moment capacity equal to 406.7 kip-ft. for an
unsupported length of 24 ft. is the designed section.
Step IV. Check for local buckling. Also
found in footnotes of sections.
• l = bf / 2tf = 7.7; lp = 0.38 (E/Fy)0.5 = 9.192
Therefore, l < lp - compact flange
• l = h/tw = 34.4; lp = 3.76 (E/Fy)0.5 = 90.5
Therefore, l < lp - compact web
Step V. Check shear. (W16 x 67 made from 50
ksi steel)
wu = 1.2wD + 1.6wL = 4.92 kips/ft.
V = wl/2 = 4.92(24)/2 = 59 kips
Vn = 0.6FyAwCv
= 0.6 (50ksi)(0.395x16.3)(1.0)
= 193 kips
OK
Step VI. Check deflection. (W16 x 67 made
from 50 ksi steel)
D = Wl3/384 EI
D = (4.92 klf x 24ft)(24 x 12) 3/384(29,000 ksi)(954 in4)
D = 0.27in
Dallow = L /360 = 24 x 12/ 360 = 0.8 in.
OK
Summary LRFD Beam Design
1. Determine the location and magnitude of
the loads and draw a load diagram.
2. Determine if the shape is compact by
calculating  slenderness parameter
  p
Full plastic moment, Mp
p    r Mn= Mp – (Mp 0.7Fy Sx )( - p / r - p)
 > r
Mn= Fcr Sx
Summary LRFD Beam Design
Check Shape: If the shape is non compact
because of the flange, (calculate 
slenderness parameter)
  p
Full plastic moment, Mp
p    r Mn= Mp – (Mp 0.7Fy Sx )( - p / r - p)
 > r
Mn= Fcr Sx
Summary LRFD Beam Design
3. Find Moment - Lateral Torsional Buckling
AISC F2-1
AISC F2-2
AISC F2-3
Mn = Fcr Sx <= Mp
AISC F2-5
Lp = 1.76 ry E / Fy
Lr AISC F2-6 AISC 16.1-48
Summary LRFD Beam Design
4. Check shear in the web
h/tw = 2.24
AISC G2-1
Most cases with
W shapes
fv = 1 and Cv = 1,
fv = 0.9
AISC G2-3
E / Fy
AISC G2-3 and AISC G2-3
Summary LRFD Beam Design
5. Check deflection.
Summary
• Local buckling is not an issue for most
beams for A992 beams
• Effective bracing is provided at ends to
restrain rotation about the longitudinal axis.
• Effective bracing reduces Lb and prevents
twist of the cross section and/or lateral
movement of the compression flange.
• Cb is important in economical design when
plastic moment is not developed
Works Cited
• Segui, William T. 2007. Steel Design. 4th
ed. Thompson Engineering.
• Steel Construction Manual, 13th ed. AISC,
2005.
• CE 405: Design of Steel Structures –Dr. A.
Varma