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.