Steel strucure lec # (19)

Prof. Dr. Zahid Ahmad Siddiqi
BEAM COLUMNS
Beam columns are structural members that are
subjected to a combination of bending and axial
stresses.
The structural behaviour resembles
simultaneously to that of a beam and a column.
Majority of the steel building frames have
columns that carry sizable bending moments in
addition to the usual compressive loads.

The sources of this bending moment are shown in
Figure 5.1 and explained below:
M=P´e
e
P
e
P
e
P
a) Out-Of-Plumb b) Initial Crookedness c) Eccentric Load
Figure 5.1. Sources of Eccentricity in Columns.

It is almost impossible to erect the columns
perfectly vertical and centre loads exactly on
columns.
Columns may be initially crooked or have other
flaws with the result that lateral bending is
produced.
In some cases, crane beams parallel to columns-
line and other perpendicular beams rest on brackets
projecting out from columns. This produces high
values of bending moments.

Wind and other lateral loads act within the
column height and produce bending.
The bending moments from the beams are
transferred to columns if the connections are
rigid.
CONTROLLING DESIGN FACTOR:
SECOND ORDER EFFECTS
The elastic analysis carried out to calculate
deflections and member forces for the given
loads is called 1st order and analysis.

The high axial load present in the column
combined with this elastic deflection produces
extra bending moment in the column, as is clear
from Figure 5.2.
The analysis of structure including this extra
moment is called 2nd order analysis.
Similarly, other higher order analysis may also be
performed.
In practice, usually 2nd order analysis is
sufficiently accurate with the high order results of
much lesser numerical value.

d
Maximum lateral
deflection due to
bending moment
(M)
P
M
M
P
Extra moment = P´d,
which produces more
deflections
Deflected shape
or elastic curve
due to applied
bending moment
(M)
Figure 5.2. Eccentricity Due to First Order Deflections.

The phenomenon in which the moments are
automatically increased in a column beyond the
usual analysis for loads is called moment
magnification or 2nd order effects.
The moment magnification depends on many
factors but, in some cases, it may be higher
enough to double the 1st order moments or even
more.
In majority of practical cases, this magnification
is appreciable and must always be considered for
a safe design.

1st order deflection produced within a member
(d) usually has a smaller 2nd order effect called P-
d effect, whereas magnification due to sides-way
(D) is much larger denoted by P-D effect (refer to
Figure 5.3).
P-Delta effect is defined as the secondary effect
of column axial loads and lateral deflections on
the moments in members.
The calculations for actual 2nd order analysis are
usually lengthy and can only be performed on
computers.

For manual calculations, empirical methods are
used to approximately cater for these effects in
design.
2nd order effects are more pronounced when loads
closer to buckling loads are applied and hence the
empirical moment magnification formula contains
a ratio of applied load to elastic buckling load.
The factored applied load should, in all cases, be
lesser than 75% of the elastic critical buckling load
but is usually kept much lesser than this limiting
value.

INTERACTION EQUATION AND
INTERACTION DIAGRAM
P D
P
Extra Moment
M = P´D
M
Figure 5.3.
A Deflected Beam-Column.
The combined stress at any
point in a member subjected to
bending and direct stress, as in
Figure 5.3, is obtained by the
formula:
f = ± ±
A
P
x
x
I
yM
y
y
I
xM

For a safe design, the maximum compressive
stress (f) must not exceed the allowable material
stress (Fall) as follows:
f = ± ± £ FallA
P
x
x
I
yM
y
y
I
xM
+ + £ 1
allAF
P
allx
x
FS
M
ally
y
FS
M
+ + £ 1
maxP
P
max,x
x
M
M
max,y
y
M
M
This equation is called interaction equation
showing interaction of axial force and bending
moment in an easy way.

If this equation is plotted against the various terms
selected on different axis, we get an interaction
curve or an interaction surface depending on
whether there are two or three terms in the
equation, respectively.
1.0
1.0
0,0
Figure 5.4. A Typical Interaction Curve.

Pr = required axial compressive strength
(Pu in LRFD)
Pc = available axial compressive
strength
= fcPn, fc = 0.90 (LRFD)
= Pn / Wc, Wc = 1.67 (ASD)
Mr= required flexural strength (Mu in
LRFD)
Mc = available flexural strength
= fbMn, fb = 0.90 (LRFD)
= Mn / Wb, Wb = 1.67 (ASD)

AISC INTERACTION EQUATIONS
The following interaction equations are
applicable for doubly and singly symmetric
members:
If ³ 0.2, axial load is considerable, and
following equation is to be satisfied:
c
r
P
P
£ 1.0
÷
÷
ø
ö
ç
ç
è
æ
++
cy
ry
cx
rx
c
r
M
M
M
M
P
P
9
8

If < 0.2, axial load is lesser, beam
action is dominant, and the applicable
equation is:
c
r
P
P
£ 1.0÷
÷
ø
ö
ç
ç
è
æ
++
cy
ry
cx
rx
c
r
M
M
M
M
P
P
2
MOMENT ADJUSTMENT FACTOR
(Cmx or Cmy)
Moment adjustment factor (Cm) is based on
the rotational restraint at the member ends
and on the moment gradient in the members.
It is only defined for no-sway cases.

1. For restrained compression members in
frames braced against joint translation (no
sidesway) and not subjected to transverse loading
between their supports in the plane of bending:
Cm = 0.6 – 0.4
2
1
M
M
where M1 is the smaller end moment and M2 is
the larger end moment.
is positive when member is bent in
reverse curvature and it it is negative when
member is bent in single curvature (Figure 5.5b).
21 / MM

P
P
M2
M1
a) Reverse Curvature
P
P
M2
M1
b) Single Curvature
Figure 5.5. Columns Bent in Reverse and Single Curvatures.
When transverse load is applied between the
supports but or sway is prevented,
for members with restrained ends Cm = 0.85
for members with unrestrained ends Cm = 1.0

K-VALUES FOR FRAME BEAM-COLUMNS
K-values for frame columns with partially fixed
ends should be evaluated using alignment charts
given in Reference-1.
However, if details of adjoining members are not
given, following approximate estimate may be
used:
K = 1.2 – 1.5 if sidesway is permitted with
partially fixed ends
K = 1 if sidesway is prevented but end
conditions are not mentioned

MOMENT MAGNIFICATION FACTORS
Moment magnification factors (B1 and B2) are
used to empirically estimate the magnification
produced in the column moments due to 2nd order
effects.
These are separately calculated for sway or lateral
translation case (lt-case) and for no-sway or no
translation case (nt-case).
Accordingly, the frame is to be separately
analysed for loads producing sway and not
producing sway.

Mlt = moment due to lateral loads producing
appreciable lateral translation.
B2 = moment magnification factor to take
care of PuD effects for sway and
deflections due to lateral loads.
Mnt = the moment resulting from gravity
loads, not producing appreciable lateral
translation.
B1 = moment magnification factor to take
care of Pud effects for no translation
loads.

Mr = required magnified flexural strength
for second order effects
= B1 Mnt + B2 Mlt
Pr = required magnified axial strength
= Pnt + B2 Plt
No-Sway Magnification
B1 = ³ 1.0
11 er
m
PP
C
a-

where
a = 1.0 (LRFD) and 1.60 (ASD)
Pe1 = Euler buckling strength for
braced frame
= p2 EI / (K1 L)2
K1= effective length factor in the
plane of bending for no lateral
translation, equal to 1.0 or a
smaller value by detailed analysis

Sway Magnification
The sway magnification factor, B2, can be
determined from one of the following formulas:
B2 =
2
1
1
e
nt
P
P
å
å
-
a
where,
a = 1.0 (LRFD) and 1.60 (ASD)
SPnt = total vertical load supported by
the story, kN, including gravity loads

SPe2 = elastic critical buckling
resistance for the story
determined by sidesway
buckling analysis
= Sp2 EI / (K2 L)2
where I and K2 is calculated in the plane of
bending for the unbraced conditions

SELECTION OF TRIAL BEAM-
COLUMN SECTION
The only way by which interaction of axial
compression and bending moment can be
considered, is to satisfy the interaction equation.
However, in order to satisfy these equations, a
trial section is needed.
For this trial section, maximum axial compressive
strength and bending strengths may be
determined.

The difficulty in selection of a trial section for a
beam column is that whether it is selected based
on area of cross-section or the section modulus.
No direct method is available to calculate the
required values of the area and the section
modulus in such cases.
For selection of trial section, the beam-column
is temporarily changed into a pure column by
approximately converting the effect of bending
moments into an equivalent axial load.

Peq = equivalent or effective axial load
= Pr + Mrx mx + Mry my
mx (for first trial) = 8.5 - 0.7K1xLx
my (for first trial) = 17 - 1.4K1yLy
mx = 10 - 14(d / 1000)2 - 0.7K1xLx
my = 20 - 28(d / 1000)2 - 1.4K1yLy

The above equation is evaluated for Peq and a
column section is selected from the
concentrically loaded column tables for that
load.
The equation for Peq is solved again using a
revised value of m.
Another section is selected and checks are then
applied for this trial section.

WEB LOCAL STABILITY
For stiffened webs in combined flexural and axial
compression:
If £ 0.125 lp =
yb
u
P
P
f ÷
÷
ø
ö
ç
ç
è
æ
-
yb
u
y P
P
F
E
f
75.2
176.3
For A36 steel, lp = ÷
÷
ø
ö
ç
ç
è
æ
-
yb
u
P
P
f
75.2
17.106
If > 0.125 lp =
yb
u
P
P
f yyb
u
y F
E
P
P
F
E
49.133.212.1 ³
÷
÷
ø
ö
ç
ç
è
æ
-
f
3.4233.28.31 ³
÷
÷
ø
ö
ç
ç
è
æ
-
yb
u
P
P
f
For A36 steel, lp =
where l = h / tw and Py = Fy Ag

FLOW CHART FOR DESIGN OF
BEAM-COLUMNS
Known Data: Pu, Mntx, Mltx , Mnty, Mlty, KxLx, KyLy
Mr = Mu = Mnt + Mlt for the first trial
Calculate Mr both in the x and y directions
Peq = Pr + Mrx(mx) + Mry(my)
Assume an approximate magnification of
15% for the moments only.

Select section as a simple column depending
upon the following criteria:
1. Asel » Areq
2. Minimum weight
3. Connecting leg width b > bmin
4. Depth of W-section £ 360 mm
my (for first trial) = 17 - 1.4K1yLy
mx = 10 - 14(d / 1000)2 - 0.7K1xLx
my = 20 - 28(d / 1000)2 - 1.4K1yLy

The column selection tables may also be
employed to select the section using the values of
Peq and KyLy.
See rx/ry from column selection table for selected
section
Calculate (KyLy)eq =
yx
xx
rr
LK
Re-enter the table for greater of KyLy and (KyLy)eq
and revise to obtain suitable section for the load
Peq.

Find new values of m for subsequent trials.
Select a new section and repeat until values
of load capacities, Peq, and m are stabilized.
Peq = Pr + Mrx (mx) + Mry (my)
Select a new section and repeat until values of
load capacities, Peq and m are stabilized.

Calculate Cmx and Cmy for no sway conditions
Calculate , , and
x
xx
r
LK1
y
yy
r
LK1
R = maximum of the above values
Check for maximum slenderness ratio: R £ 200
x
xx
r
LK2
y
yy
r
LK2

Axial strength of trial section:
Calculate fcFcr corresponding to the R-value or
directly read it from the table in Reference-1 and
evaluate the compression capacity by multiplying
with the area of cross-section.
Pc = fcPn = fcFcr Ag / 1000
Calculate Euler buckling strength (Pe1)x, (Pe1)y
, (Pe2)x and (Pe2)y for both lt and nt cases.
Pe1 = p2 EI / (K1 L)2 / 1000 (kN)

Calculate no-sway moment magnification factors
B1x = ³ 1.0 : B1y = ³ 1.0
Note: Pr in the above formulas is the actual
factored axial load and not Peq.
Calculate B2x and B2y.
B2 =
( )xe
r
mx
P
P
C
1
1 a-
( )ye
r
my
P
P
C
1
1 a-
where a = 1.0 for LRFD procedure.
2
1
1
e
nt
P
P
å
å
-
a

Calculate design moments
Mrx = Mux = B1x Mntx + B2x Mltx
Mry = Muy = B1y Mnty + B2y Mlty
Bending strength of the trial section:
fbMny = fb Fy Zy / 106 (kN-m)
There are no chances of lateral buckling because
the lateral direction for y-axis bending is the
stronger direction.

Check conditions of compact section as a beam.
Find Lp and Lr from column table and check
against Lbx.
Calculate fbMnx as for a beam using Lbx, Lp, Lr
and beam selection tables. Use Cb = 1.0 in the
expressions.
Calculate to see which interaction
equation is applicable.
c
r
P
P

Check interaction equations:
£ 1.0For ³ 0.2
£ 1.0For < 0.2
Get the value of Left Side of equation (LS) up to
2nd decimal place, truncating the 3rd decimal digit,
which should not be more than 1.00.
This means that LS can be as high as 1.0099 but
not 1.01.
c
r
P
P
c
r
P
P
÷
÷
ø
ö
ç
ç
è
æ
++
cy
ry
cx
rx
c
r
M
M
M
M
P
P
9
8
÷
÷
ø
ö
ç
ç
è
æ
++
cy
ry
cx
rx
c
r
M
M
M
M
P
P
2

Values of LS between 0.9 and 1.0 ® Very economical design
Values of LS between 0.8 and 0.9 ® Economical design
Values of LS between 0.7 and 0.8 ® May be acceptable,
but better to try
an economical section
Values of LS lesser than 0.7 ® Revise by selecting
a lighter section
Values of LS greater than 1.0099 ® Select a stronger section
Check shear strength, which may usually be
omitted in hot rolled W sections because of very
high available strength.
Write the final solution using standard designation.

Example 5.1: Design the columns in a single-
bay multi-storey unbraced frame shown in Figure
5.6, where P is the load from the top stories.
Ratio of moment of inertia of beams with respect
to columns may be assumed as shown in the
figure. Approximate analyses results are also
provided in Figures 5.7 and 5.8. Assume that
sway is not allowed in the y-direction.
Solution:
Total Factored Loads
1. Load Combination 1, Gravity Load
Combination (1.2 D + 1.6 L)

8.5 m
II
PP
wI I
H
6.0 m
6.0 m
1.4 I
Figure 5.6. Frame And Loading For Example 5.1.

Pu = 1.2(1025) + 1.6(410) = 1886 kN
wu = 1.2(7.3) + 1.6(22.0) = 43.96 kN/m
P = 1025 kN dead load
= 410 kN live load
w = 7.3 kN/m dead load
= 22.0 kN/m live load
H = 345 kN wind load

1886kN1886kN
43.96kN/m
75.8 kN-m
37.9 kN-m
227 kN-m
Figure 5.7. Partial Gravity Load Analysis Results.

2. Load Combination 2, Wind Load
Combination (1.2D+0.5L+1.3W)
Pu = 1.2(1025) + 0.5(410) = 1435 kN
Hu = 1.3(345) = 448.5 kN
wu = 1.2(7.3)+0.5(22.0) = 19.76 kN/m
Value of Kx
Gtop = = = 2.02
( )
( ) beamsforLI
columnsforLI
å
å ( )
5.84.1
62
I
I
Gbotton = 1.0 for sway columns

1435kN1435kN
19.76kN/m
51.2 kN-m
25.6 kN-m
( No Sway Part )
93.7 kN-m
138 kN
138 kN
448.5kN
586 kN-m
759 kN-m
Doh
( Sway Part )
586 kN-m
759 kN-m
Figure 5.8. Partial Lateral Load Analysis Results.

Kx = 1.0 for braced frame
Kx = 1.45 for unbraced conditions
Value of Ky:
No data of connected elements is given for y-
direction and hence the approximate value may
conservatively be assumed for no sway in this
direction.
Ky = 1.0

Here, design is made for the wind combination and
check is then made for the gravity combination.
Design for Combination 2:
Pr = 1435 + 138 + 19.76 ´ 8.5/2
= 1656.7 kN
According to AISC, max. moments for different
types of loading (nt or lt case), acting at different
locations or of different signs, are to be added
magnitude-wise in any combination.
The Right column is critical for the axial load.

Mntx = 51.2 kN-m
Mltx = 759 kN-m
Mnty = Mlty = 0
K2xLx= 1.45 ´ 6 = 8.7 m for lt-case
K1xLx = 1.00 ´ 6 = 6.00 m for nt-case
K1yLy = 1.00 ´ 6 = 6.00 m
Peq = Pu + 1.15 Mux (m)
= 1656.7 + 1.15 (51.2 + 759.0) (4.3)
= 5663 kN
= 8.5 - 0.7 ´ 6 = 4.3
Assume 15% magnification of moments.

Using column tables of Reference 1 for this Peq
and KyLy = 6.0 m;
Trial Section = W360 ´ 262
Peq = Pu + 1.15 Mux (m)
= 1656.7 + 1.15 (51.2 + 759.0) (3.99)
= 5374 kN
Revised mx = 10 - 14(d / 1000)2 - 0.7K1xLx
= 10 - 14 ´ 0.362 - 0.7 ´ 6
= 3.99

Trial Section-1: W360 ´ 237
A = 30,100 mm2
rx = 162 mm, ry = 102 mm
rx/ry = 1.60
Ix = 79,100 ´ 104 mm4
M1 / M2 is positive because of reverse curvature
Cmx = 0.6 – 0.4
2
1
M
M
= 0.6 – 0.4 = 0.4÷
ø
ö
ç
è
æ
2.51
6.25
= = 37.04 ( for nt case)
x
xx
r
LK1
162
10000.6 ´
yx
xx
rr
LK
/
2
60.1
7.8
(KyLy)eq = = = 5.44 m (not critical)

= = 53.70 (for lt-case)
x
xx
r
LK2
162
10007.8 ´
= = 58.82
y
yy
r
LK1
102
10000.6 ´
R » 59 < 200 OK
fcFcr = 187.09 MPa
Pc = fcFcr Ag = = 5,631 kN
1000
100,3009.187 ´
for nt-case
Pe1x = p2 EI / (K1x L)2
=
10006000
10100,79000,200
2
42
´
´´´p
= 43,371 kN

B1x = =
= 0.42 B1x = 1.0
Pe2x = p2 EI / (K2x L)2
=
10008700
10100,79000,200
2
42
´
´´´p
= 20,628 kN for lt-case
xer
mx
PP
C
,11 a- 371,437.165611
4.0
´-
SPnt = 1435 ´ 2 + 19.75 ´ 8.5 = 3038 kN
SPe2,x = 2 ´ 20,628 = 41,256 kN

B2x = = = 1.08
Mrx = B1x Mntx + B2x Mltx
= 1.0 (51.20) + 1.08(759.00)
= 870.9 kN-m
From column selection table:
Lp = 5.06 m, Lr = 25.43 m
xe
nt
P
P
,2
1
1
å
å
-
a
256,41
30380.1
1
1
´
-
Pr = Pnt + B2 Plt
= 1518.98 + 1.08(138)
= 1668.02kN

Check conditions of compact section:
= 6.5 < lp = 10.8 OK
f
f
t
b
2
= = 0.245
yb
u
P
P
f ( ) 100,301000/2509.0
7.1656
´´
For web, lp = 3.4233.28.31 ³
÷
÷
ø
ö
ç
ç
è
æ
-
yb
u
P
P
f
= 66.3 for A36 steel
= 15.3 < lp OK
wt
h

Lb = 6.00m > Lp, bending strength is to be
calculated using the inelastic LTB formula.
Mp = 250 ´ 4700 ´ 103 / 106 = 1175.0 kN-m
fbMp = 0.9 ´ 1175 = 1057.5 kN-m
Mr = 0.7 ´ Fy ´ Sx / 106
= 0.7 ´ 250 ´ 4160 ´ 103 / 106
= 728.0 kN-m
BF = = 21.94 kN÷
ø
ö
ç
è
æ
-
-
=
-
-
06.543.25
7281175
pr
rp
LL
MM

Mcx = fb[Mp - BF(Lb - Lp)]
= 0.9 [1175 - 21.94(6.0 - 5.06)]
= 1038.9 kN-m
Check Interaction Equation:
c
r
P
P
631,5
02.1668
= = 0.296 > 0.2
÷÷
ø
ö
çç
è
æ
+
cx
rx
c
r
M
M
P
P
9
8
= 0.296 + ÷
ø
ö
ç
è
æ
9.1038
92.870
9
8
= 1.041 > 1.00 NG

Trial Section-2: W360 ´ 262
A = 33,400 mm2
rx = 163 mm, ry = 102 mm
rx/ry = 1.60
Ix = 89,100 ´ 104 mm4
(KyLy)eq =
yx
xx
rr
LK
/
2
=
60.1
7.8
= 5.44 m (not critical)

Cmx = 0.4 (as before)
x
xx
r
LK1
163
10000.6 ´
= = 36.81 ( for nt case)
x
xx
r
LK2
163
10007.8 ´
= = 53.37 ( for lt case)
y
yy
r
LK1
102
10000.6 ´
= = 58.82
R » 59 < 200 OK
fcFcr = 187.09 MPa

Pc = fcPn = fcFcr Ag
=
1000
400,3309.187 ´
= 6,248 kN
Pe1x = p2 EI / (K1x L)2
=
10006000
10100,89000,200
2
42
´
´´´p
= 48,854 kN for nt-case
Pe2x = p2 EI / (K2x L)2
=
10008700
10100,89000,200
2
42
´
´´´p
= 23,236 kN for lt-case

xer
mx
PP
C
,11 a-
854,487.165611
4.0
´-
B1x =
= = 0.41 B1x = 1.0
SPnt = 1435 ´ 2 + 19.75 ´ 8.5
= 3038 kN
SPe2,x = 2 ´ 23,236 = 46,472 kN
B2x =
xe
nt
P
P
,2
1
1
å
å
-
a =
472,46
30380.1
1
1
´
-
= 1.07

Mrx = B1x Mntx + B2x Mltx
= 1.0 (51.20) + 1.07(759.00)
= 863.33 kN-m
Pr = Pnt + B2 Plt
= 1518.98 + 1.07(138)
= 1666.64 kN
From column selection table:
Lp = 5.08 m, Lr = 30.44 m

Check conditions of compact section:
f
f
t
b
2
= 6.0 < lp = 10.8 OK
yb
u
P
P
f ( ) 400,331000/2509.0
7.1656
´´
= = 0.220
For web,
lp = 3.4233.28.31 ³
÷
÷
ø
ö
ç
ç
è
æ
-
yb
u
P
P
f
for A36 steel
= 67.1
wt
h
= 13.7 < lp OK

Lb = 6.00m > Lp, bending strength is to be
calculated using the inelastic LTB formula.
Mp = 250 ´ 5240 ´ 103 / 106
= 1310.0 kN-m
fbMp = 0.9 ´ 1310 = 1179 kN-m
Mr = 0.7 ´ Fy ´ Sx / 106
= 0.7 ´ 250 ´ 4600 ´ 103 / 106
= 805.0 kN-m

BF = ÷
ø
ö
ç
è
æ
-
-
=
-
-
08.544.30
8051310
pr
rp
LL
MM
= 19.91 kN
Mcx = fb[Mp - BF(Lb - Lp)]
= 0.9 [1310 - 19.91(6.0 - 5.08)]
= 1162.5 kN-m
c
r
P
P
248,6
64.1666
= = 0.267 > 0.2

÷÷
ø
ö
çç
è
æ
+
cx
rx
c
r
M
M
P
P
9
8
÷
ø
ö
ç
è
æ
5.1162
33.863
9
8
= 0.267 +
= 0.927 < 1.00 OK
Section Selected For Wind
Combination: W360 ´ 262
Check for Combination 1:
Pr = Pu = 1886 + 43.96 ´ 8.5/2
= 2073 kN
Mntx = 75.8 kN-m

Cmx = 0.4 same as before
xer
mx
PP
C
,11 a- 854,48207311
4.0
´-
B1x = =
= 0.42 B1x = 1.0
Mrx = B1x ´ Mntx = 75.8 kN-m
c
r
P
P
248,6
2073
= = 0.332 > 0.2

÷÷
ø
ö
çç
è
æ
+
cx
rx
c
r
M
M
P
P
9
8
÷
ø
ö
ç
è
æ
5.1162
8.75
9
8
= 0.332 +
= 0.39 < 1.00 OK
Final Selection: W360 ´ 262

Example 5.2: Design the column for the
following data:
1.Braced frame
2.Pu = 1750 kN
3.Mntx = 330 kN-m
4.Mltx = 0
5.Mnty = 105 kN-m
6.K1xLx = K1yLy = 7.3 m
7.Lb = 7.3 m
8.Cm = 0.85
9.Fy = 250 MPa

Solution:
Peq = Pu + Mux mx + Muy my
For first trial: mx = 8.5 - 0.7 K1xLx
= 8.5 - 0.7 ´ 7.3 = 3.39
my = 17 - 1.4 K1yLy
= 17 - 1.4 ´ 7.3 = 6.78
Assume 15% magnification.
Peq = 1750 + 1.15 ´ 330 ´ 3.39 + 1.15
´ 105 ´ 6.78
= 3855 kN

KyLy = 7.3 m
From column load table, the trial section is:
W360 ´ 196
mx = 10 - 14 (d/1000)2 - 0.7 K1xLx
= 10 - 14 (0.36)2 - 0.7 ´ 7.3 = 3.08
my = 20 - 28 (d/1000)2 - 1.4 K1yLy
= 20 - 28 (0.36)2 - 1.4 ´ 7.3 = 6.15
Peq = 1750 + 1.15 ´ 330 ´ 3.08
+ 1.15 ´ 105 ´ 6.15
= 3661 kN

W360 ´ 179
rx/ry = 1.67
K1xLx / 1.67 = 4.37 < KyLy
KyLy is critical
Trial Section No. 1: W360 ´ 179

Ag = 22,800 mm2
rx = 158 mm
ry = 95.0 mm
Lp = 4.73 m
Lr = 21.20 m
Ix = 57,400 ´ 104 mm4
Iy = 20,600 ´ 104 mm4
Zx = 3,474 ´ 103 mm3

Zy = 1,671 ´ 103 mm3
Sx = 3,110 ´ 103 mm3
Mp = 868.5 kN-m
Mr = 544.25 kN-m
BF = 19.69 kN
Mcx = fbMnx = 736.11 kN-m
Cm = 0.85 (given)

158
10003.71 ´
=
x
xx
r
LK
0.95
10003.71 ´
=
y
yy
r
LK
= 46.20
R » 77 < 200 OK
fcFcr = 164.32 MPa
= 76.84
Pc = fcFcrAg =
1000
32.164
´ 22,800 = 4108 kN

10007300
10400,57000,200
2
42
´
´´´p
10007300
10600,20000,200
2
42
´
´´´p
Pe1,x =
= 21,262 kN
Pe1,y =
= 7,630 kN
B1x =
262,21
17501
1
85.0
1
,1
´
-
=
-
xe
nt
m
P
P
C
a
= 0.93

B1x = 1.0
B1y =
630,7
17501
1
85.0
´
-
= 1.10
Pr is not magnified as Plt = 0.
Mux = B1x ´ Mntx = 330 kN-m
Muy = B1y ´ Mnty = 1.10 ´ 105 = 115.5 kN-m

6
3
10
10671,12509.0 ´´´
c
r
P
P
3746
1750
Mcy = fbMpy
=
= 375.98 kN-m
bf / 2tf = 7.8 < 10.8 OK
h / tw = 19.3 < 42.3 (worst case) OK
= = 0.467 > 0.2

÷
÷
ø
ö
ç
ç
è
æ
++
cy
ry
cx
rx
c
r
M
M
M
M
P
P
9
8
÷
ø
ö
ç
è
æ
+
98.375
5.115
11.736
330
9
8
Interaction Equation:
= 0.467 +
= 1.139 > 1.0 NG
Ag = 25,000 mm2
rx = 160 mm
ry = 95.5 mm

Ix = 63,700 ´ 104 mm4
Iy = 22,800 ´ 104 mm4
Zx = 3,835 ´ 103 mm3
Zy = 1,852 ´ 103 mm3
Sx = 3,420 ´ 103 mm3
Mp = 958.75 kN-m
Mr = 598.50 kN-m
BF = 19.74 kN

Pc = 4108 kN
Pe1,x = 23,595 kN
Pe1,y = 8,445 kN
B1x = 1.0
B1y = 1.07
Mux = 330 kN-m
Muy = 112.5 kN-m
Mcy = fbMpy = 416.7 kN-m

c
r
P
P
4108
1750
= = 0.426 > 0.2
÷
÷
ø
ö
ç
ç
è
æ
++
cy
ry
cx
rx
c
r
M
M
M
M
P
P
9
8
÷
ø
ö
ç
è
æ
+
7.416
5.112
57.817
330
9
8
= 0.426 +
= 1.025 > 1.0 NG

Ag = 27,500 mm2
rx = 161 mm
ry = 101 mm
Lp = 5.03 m
Lr = 25.43 m
Ix = 71,200 ´ 104 mm4
Iy = 28,200 ´ 104 mm4
Zx = 4,260 ´ 103 mm3

Zy = 2,180 ´ 103 mm3
Sx = 3,800 ´ 103 mm3
Mp = 1065 kN-m
Mr = 665 kN-m
BF = 19.61 kN
c
r
P
P
4665
1750
= = 0.375 > 0.2

÷
÷
ø
ö
ç
ç
è
æ
++
cy
ry
cx
rx
c
r
M
M
M
M
P
P
9
8
÷
ø
ö
ç
è
æ
+
5.490
3.107
44.918
330
9
8
= 0.375 +
= 0.889 > 1.0 OK

Example 5.3: Design
the beam column
shown in Figure 5.9, if
sidesway is allowed
along weak axis but is
prevented along strong
axis. The moments
shown are factored
and are due to lateral
loads. The column
ends are partially fixed.
Pu =290 kN
Pu =290 kN
5.2m
220
kN-m
320 kN-m
320 kN-m
Figure 5.9. Column of Example 5.3.

Solution:
Although lateral load is present, sway is not
allowed along strong axis. Hence, the
moments may be considered to be of nt-case.
Pu = 290 kN; Mntx = 320 kN-m; Mltx = 0 kN-m
Due to unavailability of the connection data, use
approximate values of Kx and Ky as follows:
K1x = 1.0 ; K1y = 1.0 ; K2y = 1.2
K1xLx = 5.2 m ; K1yLy = 5.2 m
K2yLy = 6.24 m

For first trial: mx = 8.5 - 0.7 K1xLx
= 8.5 - 0.7 ´ 5.2 = 4.86
Assume 15% magnification.
Peq = Pu + 1.15 Mux (mx)
= 290 + 1.15 ´ 320(4.86)
= 2079 kN
W360 ´ 110 is uneconomical.
From column selection table, the trial section
is: W310 ´ 97

mx = 10 - 14 (d/1000)2 - 0.7 K1xLx
= 10 - 14 (0.31)2 - 0.7 ´ 5.2 = 5.01
Peq = 290 + 1.15 ´ 320 ´ 5.01
= 2134 kN
W310 ´ 97
rx/ry = 1.75
K1xLx / 1.75 = 2.97 < KyLy
KyLy is critical

Trial Section No.1: W310 ´ 97
Ag = 12,300 mm2
rx = 134 mm
ry = 76.7 mm
Lp = 3.82 m
Lr = 13.90 m
Ix = 22,700 ´ 104 mm4
Iy = 7,240 ´ 104 mm4

Zx = 1,586 ´ 103 mm3 ; Zy = 723 ´ 103 mm3
Sx = 1,440 ´ 103 mm3 ; Mp= 396.5 kN-m
Mr= 252 kN-m ; BF= 14.34 kN
Mcx = fbMnx= 339.04 kN-m
Check for local stability:
Þ Compact Section
bf/2tf = 9.9 < lp
h/tw = 24.9 < lp 42.3 for the worst case

Cmx = 1.0 (Consider member with
unrestrained ends to be on conservative side.)
K1xLx/rx = 5200/134 = 38.81
K1yLy/ry = 5200/76.7 = 67.80
K2yLy/ry = 6240/76.7 = 81.36
R = 82 < 200 OK
fcFcr = 157.54 MPa
Pc = fcFcrAg / 1000
= 157.54 ´ 12,300/1000 = 1938 kN

10005200
10200,22000,200
2
42
´
´´´p
xe
nt
mx
P
P
C
,1
1
a
-
206,16
2901
1
1
´
-
Pe1,x =
= 16,206 kN
B1x = ³ 1.0 =
= 1.018
Mrx = B1x Mntx
= 1.018 ´ 320 = 325.76 kN-m

c
r
P
P
1938
290
= = 0.150 < 0.2
÷÷
ø
ö
çç
è
æ
+
cx
rx
c
r
M
M
P
P
2 04.339
76.325
2
150.0
+=
= 1.036 > 1.0 NG
Trial Section No.2: W310 ´ 107

÷÷
ø
ö
çç
è
æ
+
cx
rx
c
r
M
M
P
P
2
77.379
12.325
2
134.0
+=
= 0.923 < 1.0 OK

Steel strucure lec # (19)

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