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1
Steel Beam Design by
ASD/LRFD Steel Construction Manual
13th Edition
By Duane Nickols
Beams are flexural members that are subjected to shear and bending
moments. The shear and moment vary with position along the beam.
A beams primary function is resisting bending moments. Beams
usually have uniform loads, concentrated loads or both on them. We will be looking at W shaped members. LRFD stands for Load and
Resistance Factor Design and was in three previous editions of the
specifications. ASD stands for Allowable Strength Design which is
similar to Allowable Stress Design that many engineers are familiar
with.
Types of loads
2
LRFD Load Combinations (ASCE 7-05)
ASD Load Combinations (ASCE 7-05)
We will analyze first and design later.
Let’s say we have a 26 foot long beam that is simply supported at
each end. It supports a live load equal to 0.60 k/ft and a dead load of
0.83 k/ft, including the weight of the beam.
LRFD
ftk ipftftkLw
M
kftftkLw
V
ftkftkftkLDw
u
u
u
u
u
3.1658
)26(/96.1
8
43.252
26/96.1
2
/96.1)/60.06.1()/83.02.1(6.12.1
22
3
ASD
ftk ipftftkLw
M
kftftkLw
V
ftkftkftkLDw
a
a
a
a
a
8.1208
)26(/43.1
8
59.182
26/43.1
2
/43.1/60.0/83.0
22
Load and Resistance Factor Design (LRFD)
nunu VVMM and Also
Chapter F is Design of Members for Flexure
Allowable Strength Design (ASD)
n
a
n
a
VV
MM and Also
4
Limit States
We will look at six limit states. The first five are strength limit states
and the last is a serviceability limit state.
Yielding has to do with the strength of the beam to resist
bending moments without failure. Yielding depends on the loads,
supports, span and the strength of the steel.
Lateral-Torsional Buckling has to do with the twisting of the
beam in a lateral direction. If the beam is adequately braced, it will not twist into failure.
Web Local Buckling has to do with the strength of the web to
resist failure. This means the width to thickness ratio must fall
between certain limits so the web will not collapse or fail.
Flange Local Buckling has to do with strength of the flange to
resist failure. This means the width to thickness ratio must fall between certain limits so the flange will not collapse or fail.
Shear has to do with shear failure of the beam. Chapter G
covers Design of Members for Shear.
Deflection is a serviceability limit state. This has to do with the
beam deflecting too noticeable to people or so people feel
uncomfortable.
If the un-braced length, (Lb) <=Lp use equation F2-1.
Lateral-Torsional Buckling
5
Cb can be conservatively equal to 1.0
Sx comes from the shapes table
Lp and Lr can be found in Table 3-2 for W shapes in the AISC Manual or
you can calculate them from the following equations.
J is in the shapes table.
6
Now let’s say the beam we looked at before is a W14 x 30 and the un-braced length is 26 feet. Fy=50 ksi, Sx=42.0 in3 and Zx=47.3 in3.
Yielding
Since Lb>Lp, it does not meet the limit state of yielding in equation
(F2-1).
ftk ipftkipM
ftkipinkipinksiSFM
ftkipftkipM
ftkipinkipinksiZFM
r
xyr
p
xyp
110)5.122()9.0(
5.1221470)0.42()50()7.0(7.0
177)197()9.0(
1972365)3.47()50(
3
3
7
LRFD
The Mu was equal to 165.3 kip-ft
From Table 3-2, AISC Manual
Since the required moment, (Mu=165.3 kip-ft) is greater than the
design moment, (bMr=110 kip-ft) and the un-braced length, (Lb=26
feet) is greater than Lr of 14.9 feet, this will not meet the limit state
for lateral-torsional buckling.
Now, let’s say we brace this beam in the middle and at each end. The un-braced length (Lb = 13 feet). Interpolating between the above
values:
ftk ipftk ipftk ipM
ftk ipx
ftk ipftk ip
x
nb 2.12379.53177
79.53
)110177()'26.5'9.14(
)'26.5'0.13(
Now the design moment becomes 123.2 kip-ft. This is less than the
required moment, (Mu=165.3 kip-ft) so it will not work. No Good.
Now let’s say we brace this beam at each end and the quarter points.
The un-braced length, (Lb) will be 6.50 feet. Interpolating between the
above values, the design moment, bMn is 168.4 kip-ft. This is greater
than the required moment of 165.3 kip-ft, so this will work. Just by
bracing adequately, the beam will not twist into failure (Lateral-
Torsional Bucking).
ftk ipM
ftkipM
ftL
ftL
rb
p
r
p
110
177
9.14
26.5
b
8
ASD
Again, we will brace the beam at each end and the quarter points. The
un-braced length, (Lb=6.50 feet). Remember from above, Ma=120.8 kip-ft.
ftk ipftkipM
ftkipSFM
ftkipftkipM
ftkipZFM
r
xyr
p
xyp
4.7367.1
5.122
5.1227.0
11867.1
197
197
From Table 3-2, for W14 x 30 in the AISC Manual
ftL
ftL
ftk ipM
ftk ipM
r
p
b
r
b
p
9.14
26.5
4.73
118
Since ftk ipM
ftkipMb
p
a 1188.120 , we need to select another
beam.
Selecting a W16 x 31
9
From Table 3-2, AISC Manual
ftL
ftL
ftk ipM
ftk ipM
r
p
b
r
b
p
9.11
13.4
5.82
135
Interpolating from the above values, ftk ipM
b
n 96.118 which is still No
Good.
Selecting a W14 x 34
From Table 3-2, AISC Manual
ftL
ftL
ftk ipM
ftk ipM
r
p
b
r
b
p
6.15
4.5
9.84
136
Interpolating from the above values for Lb=6.50 feet, ftk ipM
b
n 5.130
which is greater than Ma=120.8 kip-ft. This member is OK.
Conclusion
The LRFD member (W14 x 30) is lighter than the ASD member
(W14 x 34) but both are the same depth.
10
Classification of Sections for Local Buckling
Compact
Non-compact Slender-element
For W sections wwf
f
t
kd
t
h
t
b
t
b 2 and
2
Flange Local Buckling
It is compact if y
p FE
t
b38.0
It is non-compact if y
rp FE0.1
11
Web Local Buckling
It is compact if y
p
wF
Et
h76.3
It is non-compact if y
r
w
p FE
t
h70.5
Now the W14 X 30 from LRFD, is it compact, non-compact or a
slender-element?
compact isit 15.9
50000,29
38.038.0
table with thechecks this74.8)385.02(
73.6
2
p ksiksi
FE
in
in
t
b
y
f
f
12
Now let’s check the web.
compact isit 6.9050
000,2976.376.3
tablein the 45.4 isit 3.45270.0
))785.02(8.13(2
p ksiksi
FE
in
inin
t
kd
t
h
y
ww
The W14 X 34 from ASD is also compact. In fact for Fy=50 ksi, most all W sections have compact flanges. There are only about ten sections
that have non-compact flanges.
Shear Strength
Now recall, from page 2 and 3, that our Vu=25.43 kips and Va=18.59
kips.
LRFD
OKkipskips
kipskipsV
kipsininksiCdtFCAFV
VV
n
vwyvwyn
un
43.25112
11211200.1
1120.1270.08.13506.06.06.0
ASD
OKkipskips
kipskipsV
VV
n
an
59.187.74
7.745.1
112
13
Both of these values ( n
n
VV and ) are listed in Table 3-2 of the Steel
Manual. Only about eight W sections don’t meet the h/tw requirement.
The W14 x 30 from LRFD and the W14 X 34 from ASD meet the h/tw
requirement.
Deflection (serviceability)
Let’s say that our deflection limit is L/360. Since our span was 26 feet,
L/360=(26 feet X 12 in/ft)/360=0.867 inches. Deflection for the
member is based on service loads so that will be the same for LRFD
and ASD. We are using two different W sections so the deflections will
be different.
From page 3, our uniform service load wa=1.43 k/ft
For the W14 X 30 from LRFD, Ix=291 in4
For the W14 X 34 from ASD, Ix=340 in4
LRFD
OKininininlb
ftinftinftftlb
inI
GoodNoininininlb
ftinftinftftlb
EI
wL
x
867.0829.0)612)(/1029(384
)/1226)(12/)(/1430(5
612 40, X Try W18
867.074.1)291)(/1029(384
)/1226)(12/)(/1430(5
384
5
426
4
4
426
44
ASD
The W14 X 34 didn’t work but the W18 X 40 will work.
Conclusion
Looking at the strength of the beam to resist bending moment we
came up with two different members. We braced the beam adequately
(Lb=6.5 ft) to resist lateral-torsional buckling. We determined the ASD
beam would be a W14 X 34. The beams were compact so there was no
flange local buckling or web local buckling. We checked the shear
strength of both beams and it was OK. The deflection turned out to be the determining limit state. Based on deflection we would use the
same W section for both LRFD and ASD (W18 X 40).
14
Design W Section, Continuously Braced
Select an ASTM A992 W section simply supported with a span of 35 feet. The beam is continuously braced, so Lb=0 ft. The maximum
depth of the section is 21 inches. The limit on live load deflection is
L/360. The uniform dead load is 0.5 kip/ft and the uniform live load is
0.8 kip/ft.
Material properties:
ASTM A992 Fy=50 ksi Fu=65 ksi
LRFD ASD
ftk ipM
ftftk ipwLM
kipftftk ipwL
V
ftk ipw
ftk ipftk ipw
LDw
u
u
a
a
a
a
199
8
)35()/30.1(
8
75.222
)35()/30.1(
2
/30.1
)/8.0()/5.0(
22
44
max
4
max
794)17.1)(000,29(384
)/1235)(12/)(/8.0(5
384
5
17.1360
)/12()35(
360
ininksi
ftinftinftftk ip
E
wLI
inftinftL
req
The Steel Manual has many design aids. Table 3-3, W Shapes by Ix shows a W21 X 44 with Ix=843 in4 will work. Table 3-2, W Shapes by
Zx shows a W18 X40 will work for LRFD but the Ix is too small. The
table shows a W21 X 44 will work for ASD and it will work for LRFD.
Table 3-10 shows a W18 X 40 will work for LRFD but the Ix is too
small. It shows a W21 X 44 will work for ASD and it will also work for
LRFD. Let’s select a W21 X 44.
ftk ipM
ftftk ipwLM
kipftftk ipwL
V
ftkipw
ftkipftk ipw
LDw
u
u
u
u
u
u
288
8
)35()/88.1(
8
9.322
)35()/88.1(
2
/88.1
)/8.06.1()/5.02.1(
6.12.1
22
15
ftkipinkipinksiZFM xyn 3984770)4.95)(50( 3
LRFD ASD
OKftkipMM
ftkipftk ipM
an
n
199
23867.1
398
kipininksiCdtFCAFV vwyvwyn 217)1)(350.0)(7.20)(50(6.06.06.0
OKkipVV
kipkipV
an
n
75.22
1455.1
217
The Steel Manual comes with a CD. On the CD is a beam design
program. Here is the same example using that program for LRFD and
ASD.
OKftk ipMM
ftk ipftk ipM
un
n
288
358)398(90.0
OKkipVV
kipkipV
un
n
9.32
217)217(00.1
16
17
18
19
Both design methods verify our previous calculations. This program is
very useful to check your calculations.
Summary
Steel beams can be designed with simple calculations, by use of tables
in the Steel Construction Manual or with computer programs. No
matter how you design the beams, they should be checked by another
method. We only looked at a beam uniformly loaded. If the beam has concentrated loads, then there may be other limit states involved (J10
of the specification). These would be Flange Local Bending, Web Local
Yielding, Web Crippling, Web Sidesway Buckling, Web Compression
Buckling and Web Panel Zone Shear. This is beyond the scope of this
course. This course is to get designers interested in using the new
code by showing that it isn’t that hard. It doesn’t matter if you use LRFD or ASD, both will result in a safe member.