LRFD-Steel Design
Chapter 6
6.1 INTRODUCTION
Most beams and columns are subjected to some degree of both
bending and axial load especially in statically indeterminate
structures.
Many columns can be treated as pure compression members with
negligible error.
For many structural members, there will be a significant amount of
both bending moment and axial load.
Such members are called beam-column.
Consider the rigid frame shown in the Figure:
For the given loading condition,
The horizontal member AB must not only support the vertical
uniform load but must also assist the vertical members in resisting
the concentrated load P1.
Member CD is a more critical case, because it must resist the load
P1 + P2 without any assistance from the vertical members.
The reason is that the bracing members, prevents sidesway in the
lower story. (in the direction of P, ED will be in tension and CF will be slack)
Member CD must transmit the load P1 + P2 from C to D.
The vertical members of this frame must also be treated as beam-
column.
In addition, at A and B, B.M. are transmitted from the horizontal
member through the rigid joints.
This is also occur at C and D and is true in any rigid frame.
Most columns in rigid frame are actually beam-columns, and the
effects of bending should not be ignored.
Another example of beam-column can sometimes be found in roof
trusses if purlins are placed between the joints of the top chord.
 6.2
INTRODUCTION FORMULAS
The inequality Equation may be written in the following form:
γ Q
i
φR n
i
load effects


 1.0
resistance
If both bending and axial compression are acting the interaction
pu
Mu
formula would be

 1.0
φ c Pn φ bMn
Where
Pu is the factored axial compressive load.
Фc Pn is the compressive design strength.
Mu is the factored bending moment.
Фc Mn is the flexural design strength.
For biaxial bending, there will be two bending ratios:
 Mux
Muy
pu


φ c Pn  φ bM nx φ bMny

  1.0

Two formulas are given in the specification:
One for small axial load and one for large axial load.
Pu
For
 0. 2
φ c Pn
Muy
pu
8  Mux
 

φ c Pn
9  φ b M nx
φ b Mny
Pu
For
 0. 2
φ c Pn

  1.0

 Mux
Muy
pu


2φ c Pn
φ b Mny
 φ b M nx

  1.0

 Example
6.1
Determine whether the member shown in the Figure satisfies the
appropriate AISC Specification interaction equation if the bending
is about the strong axis.
Solution:
From the column load tables:
Фc Pn = 382 kips.
Since bending is about the strong axis,
Фb Mn for Cb can be obtained from the beam
Design chart in Part 5 of the Manual.
For Lb = 17 ft, Фb Mn = 200 ft.kips. For the end condition and
loading of this problem, C = 1.32.
Фb Mn = Cb * 200 = 1.32 * 200 = 264.0 ft-kips.
This moment is larger than Фb Mp = 227 ft-kips (also from Manual)
So the design moment must be limited to Фb Mp. Therefore,
Фb Mn = 227 ft-kips.
Max. B.M. occurs at midheight, so Mu = 25*17/4 = 106.3 ft-kips.
Determine which interaction equation controls:
Pu
200

 05236  0.2
φ c Pn 382
Muy  200 8 106.3
pu
8  Mux

 

 
 0  0.94  1.0

φ c Pn 9  φ bM nx φ bMny  382 9  227

This member satisfies the AISC Specification.
 6.3
MOMENT AMPLIFICATION
The presence of the axial load produces secondary moment.
The total moment =
wL2
 P
8
The second term may be neglected if P is small.
Because the total deflection cannot be found
directly, this problem is nonlinear, and without
knowing the deflection, we cannot compute the
moment.
Ordinary structural analysis methods that do not take the displaced into
account are referred to as first-order methods.
Iterative numerical techniques, called second-order methods, can be
used to find the deflection and secondary moments.
These method are usually implemented with a computer program.
Most current design codes permit the use of either a second-order
analysis or the moment amplification method.
This method entails computing the maximum B.M. resulting from
flexural loading by a first-order analysis, then multiplying it by a
moment amplification factor to account for the secondary moment.
Mmax


1
 M0 

1  (Pu /Pe ) 
Where, M0 is the unamplified maximum moment.
Pe is the Euler buckling load =
π 2EI
(kL) 2
and Pu is factored load.
As we describe later, the exact form of the AISC moment
amplification factor can be slightly different.
Example 6.2. Compute the amplification factor for the beamcolumn of example 6.1.
π 2EI π 2EA
Pe 

2
(kL)
(kL/r) 2
π * 29000 * 14.4
Pe 
 1874 kips
2
(1.0 * (17 * 12)/4.35)
2
1
Amplificat ion factor 
1  (Pu /Pe )
1
Amp. factor 
 1.12
1  (200/1874)
This represents a 12 % increase in B.M.
Mmax = 1.12 * 106.3 = 119 ft-kips
 6.4
WEB LOCAL BUCKLING IN BEAM-COLUMNS
The determination of the design moment requires that the cross
section be checked for compactness.
The web is compact for all tabulated shapes if there is no axial
load.
If λ ≤ λp, the shape is compact
If λp ≤ λ ≤ λr, the shape is noncompact; and
If λr ≤ λ, the shape is slender
AISC B5, Table 5.1, prescribes the following limits:
Pu
For
 0.125,
φ bPy
E
λ p  3.76
Fy
 2.75Pu 
1 

φ bPy 

Pu
Pu 
E 
E
For
 0.125, λ p  1.12
2.33 
  1.49
φ bPy
Fy 
φ bPy 
Fy
Pu
Pu 
E 
For any value of
, λ r  5.70
1.0  0.74

φ bPy
Fy 
φ bPy 
Where Py = Ag Fy
Because Py is variable, compactness of the web cannot be
checked and tabulated.
Some rolled shapes satisfy the worst case limit of
1.49 E / Fy
Shapes listed in the column load tables in Part 4 of the Manual
that do not satisfy this criterion are marked, and these shapes
need to be checked for compactness of the web.
Shapes whose flanges are not compact are also marked
Example 6.3
A W12 x 58 of A992 steel is subjected to a bending moment
and a factored axial load of 300 kips. Check the web for
compactness.
Pu
Pu
300


 0.3922  0.125
φ bPy φ b A g Fy 0.90 *17.0 * 50
E
λ p  1.12
Fy

Pu 
29000
2.33  0.3922  52.27
2.33 
  1.12
φ bPy 
50

The upper limit is
E
29000
1.49
 1.49
 35.88  52.27
Fy
50
  p  52.27
From the dimension and properties tables λ = h/tw = 27.0 < λp
The web is therefore compact
 6.5
BRACED VERSUS UNBRACED FRAMES
Two amplification factors are used in LRFD
One to account for amplification resulting from the member
deflection and the other to account for the effect of sway when the
member is part of unbraced frame. The following Figure illustrates
these two components.
In Figure a, the member is restrained against
sidesway, and the max. secondary moment is Pδ.
If the frame is unbraced, there is an additional
.
component of the secondary moment, shown in
Figure b, and the max value of it is PΔ, which
represents an amplification of the end moment.
The amplified moment to be used in design is:
Mu = B1 Mnt + B2 Mlt
Where
Mnt = maximum moment assuming that no sidesway occurs.
Mlt = maximum moment caused by sidesway, = 0.0 in the actuall
braced frame.
B1 = amplification factor for the moment occurring in the member
when it is braced against sidesway.
B2 = amplification factor for the moment resulting from sidesway.
The following sections explain the evaluation of B1 and B2 .
 6.6
MEMBERS IN BRACED FRAMES
The amplification factor given in section 6.3 was derived for a
member braced against sidesway.
The following Figure shows a member of this type
Maximum moment amplification occurs at the
center, where the deflection is largest.
For equal end moment, the moment is constant
throughout the length of the member,
So the maximum primary moment also occurs at the center.
If the end moments are not equal, the maximum primary and
secondary moments will occur near each other.
If the end moments produce reverse-curvature bending as shown.
Here the max. primary moment is at one of the ends, and the max.
amplification occurs between the ends.
Therefore, the max. moment in a beam-column depends on the
distribution of bending moment within the member.
This distribution is accounted for by a factor, Cm, applied to the
amplification factor given in section 6.3.
The amplification factor given in section 6.3 was derived for the
worst case, so Cm will never be greater than 1.0.
The final form of the amplification factor is
Cm
B1 
1
1  (Pu /Pe1 )
 EAg
2
where
Pe1 
( KL / r )
2
Note:
When computing pe1, use KL/r for the axis of bending and an effective length
factor K less than or equal to 1.0 (braced condition)
Evaluation of Cm
The factor Cm applies only to the braced condition.
There are two categories of members:
1.
Members with transverse loads applied between it’s ends.
2.
Members with no transverse loads.
If there are no transverse loads acting on the member,
 M1 
C m  0.6  0.4

M
 2
M1/M2 is the ratio of the bending moments at the ends of the
member. (M1 is the smallest value and M2 is the largest one).
The ratio is positive for member bent in reverse curvature and
negative for single curvature bending as shown in the Figure.
For transversely loaded members, Cm can be taken as 0.85 if the
ends are restrained against rotation (fixed) and 1.0 if the ends are
unrestrained against rotation (pinned).
End restrained will usually result from the stiffness of members
connected to the beam-column.
Although the actual end condition may lie between full fixity and a
frictionless pin, use of one of the two values given here will give
satisfactory results.
Example 6.5
The horizontal beam-column shows in the Figure is subjected to
the service live loads shown. This member is laterally braced at its
ends, and bending is about the x-axis. Check for compliance with
the AISC Specification.
Solution
The factored load = Pu =1.6*28=44.8 kips, wu=12*0035=0.042 kips
The maximum moment is
44.8 * 10 0042 * (10) 2
Mnt 

 112.5 ft  kips
4
8
The member is braced against end translation, so Mlt =0.0
Compute the moment amplification factor
The member braced against sidesway, transversely loaded, and
with unrestrained ends, Cm can be taken as 1.0.
Pe1 
π 2EA g
(Kl/r) 2
π 2 * 29000 * 10.3

 2522 kips
2
(34.19)
The amplification factor is:
Cm
1
B1 

 1.018  1
1- (Pu /Pe1 ) 1- (44.8/2522 )
For the axis of bending
Mu  B1Mnt  B 2MLt  1.018 * 112.5  0  114.5
ft - kips
Maximum
KL
K yL

r
λc 
ry
KL
Fy
rπ
E
λ c  1.5
Fcr  (0.658


1.0(10 * 12)
 59.11
2.03
29.11
50
π
29000
 0.7813
inelastic
λc2
)Fy  (0.658)
(0.7813) 2
(50)  38.73 ksi
φ c Pn  φ c A gFcr  0.85 * 10.3 * 38.73  339.1 kips
For the flexural strength, first check for compactnes s of the flange
b
λ  f  8.10,
2t
f
λ p  0.38
E
Fy
since λ  λ p , the shape is compact
 0.38
29000
50
 9.152
Lateral torsional buckling.
L b  10 ft
E
29000
L  1.76 r
 1.76 * 2.03
 86 .04 in  7.17 ft
p
y F
50
y
r X
y 1
L 
1  1  X ( F  10 ) 2
2 y
r ( F  10 )
y
2.03 * 3610
L 
1  1  (763 * 10  6 )( 50  10 ) 2  289 .1 in  24 .09 ft
r
(50  10 )
Since L
L L
p
b
r
L  L 
b
p

Mn  Mp  (Mp  Mr )
L L 
p
 r
Mp  Fy Z  50 * 34.7  1735 in  kips
Mr  (Fy  10)S  (50  10) * 31.2  1248 in  kips

 10  7.17  
Mn  1735  (1735  1248)
 1473 in  kips


 24.09  10  

φ bMn  0.9 * 1473  122.8 ft  kips
Because the beam weight is very small in relation to the
concentrated live load, Cb may be taken from Figure 5.15c as
1.32. This value results in a design moment of
φbMn  1.32 * 122.8  162.1ft  kips
This moment is greater than the plastic moment = 0.90*1735/12
=130.1 ft-kips, so the design strength must be limited to this value.
Check the interaction formula:
Pu
44.8

 0.1321  0.20
φ c Pn
339.1
 Mux
Muy 
Pu



2φ c Pn
φ c Mny 

 φ c Mnx

44.8
114.1


 0   0.943  1.0
2(339.1) 130.1

So a W8 x 35 is adequate
(OK)
Please read the remaining examples, 6.4 and 6.6 from the text
book.
 6.7
MEMBERS IN UNBRACED FRAMES
The amplification factor given in section 6.3 was derived for a
member braced against sidesway.
The following Figure shows a member of this type