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Course C-2020

Steel Beam Design by ASD/LRFD SteelConstruction Man ual

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1

Steel Beam Design byASD/ 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. Beamsusually have uniform loads, concentrated loads or both on them. We

will be looking at W shaped members. LRFD stands for Load andResistance Factor Design and was in three previous editions of the

specifications. ASD stands for Allowable Strength Design which issimilar to Allowable Stress Design that many engineers are familiarwith.

Types of loads

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LRFD Load Combinations (ASCE 7-05)

ASD Load Combinations (ASCE 7-05)

We will analyze first and design later.

Lets say we have a 26 foot long beam that is simply supported ateach 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

ftkipftftkLw

M

k

ftftkLw

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

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3

ASD

ftkipftftkLw

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 andAlso

Chapter F is Design of Members for Flexure

Allow able Strength Design (ASD)

na

n

a

VV

MM andAlso

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4

Limit States

We will look at six limit states. The first five are strength limit statesand the last is a serviceability limit state.

Yielding has to do with the strength of the beam to resistbending 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 thebeam 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 fallbetween certain limits so the web will not collapse or fail.

Flange Local Buckling has to do with strength of the flange toresist failure. This means the width to thickness ratio must fallbetween certain limits so the flange will not collapse or fail.

Shear has to do with shear failure of the beam. Chapter Gcovers Design of Members for Shear.

Deflection is a serviceability limit state. This has to do with thebeam deflecting too noticeable to people or so people feel

uncomfortable.

If the un-braced length, (Lb)

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5

Cb can be conservatively equal to 1.0Sx comes from the shapes table

Lp and Lr can be found in Table 3-2 for W shapes in the AISC Manual oryou can calculate them from the following equations.

J is in the shapes table.

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6

Now lets say the beam we looked at before is a W14 x 30 and theun-braced length is 26 feet. Fy=50 ksi, Sx=42.0 in

3 and Zx=47.3 in3.

Yielding

Since Lb>Lp, it does not meet the limit state of yielding in equation

(F2-1).

ftkipftkipM

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

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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 thedesign 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, lets say we brace this beam in the middle and at each end. The

un-braced length (Lb = 13 feet). Interpolating between the abovevalues:

ftkipftkipftkipM

ftkipx

ftkipftkip

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 lets say we brace this beam at each end and the quarter points.

The un-braced length, (Lb) will be 6.50 feet. Interpolating between theabove 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 bybracing adequately, the beam will not twist into failure (Lateral-

Torsional Bucking).

ftkipM

ftkipM

ftL

ftL

rb

p

r

p

=

=

=

=

110

177

9.14

26.5

b

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8

ASD

Again, we will brace the beam at each end and the quarter points. Theun-braced length, (Lb=6.50 feet). Remember from above, Ma=120.8

kip-ft.

ftkipftkipM

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

ftkipM

ftkipM

r

p

b

r

b

p

9.14

26.5

4.73

118

=

=

=

=

Since ftkipM

ftkipMb

p

a

=

>= 1188.120 , we need to select another

beam.

Selecting a W16 x 31

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9

From Table 3-2, AISC Manual

ftL

ftL

ftkipM

ftkipM

r

p

b

r

b

p

9.11

13.4

5.82

135

=

=

=

=

Interpolating from the above values, ftkipM

b

n =

96.118 which is still No

Good.

Selecting a W14 x 34

From Table 3-2, AISC Manual

ftL

ftL

ftkipM

ftkipM

r

p

b

r

b

p

6.15

4.5

9.84

136

=

=

=

=

Interpolating from the above values for Lb=6.50 feet, ftkipM

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.

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10

Classification of Sections for Local Buck ling

Compact Non-compact Slender-element

For W sectionswwf

f

t

kd

t

h

t

b

t

b 2and

2

===

Flange Local Buckling

It is compact ify

p FE

t

b38.0==

It is non-compact ify

rp FE0.1=

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Web Local Buckling

It is compact ify

p

wF

Et

h76.3=

It is non-compact ify

r

w

p FE

t

h70.5=

Now the W14 X 30 from LRFD, is it compact, non-compact or a

slender-element?

compactisit15.950

000,2938.038.0

tablewith thechecksthis74.8)385.02(

73.6

2

p ===

=

==

ksiksi

FE

in

in

t

b

y

f

f

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12

Now lets check the web.

compactisit6.9050000,2976.376.3

tablein the45.4isit3.45270.0

))785.02(8.13(2

p ===

=

=

=

ksiksiFE

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 sectionsthat have non-compact flanges.

Shear Strength

Now recall, from page 2 and 3, that our Vu=25.43 kips and V

a=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

a

n

59.187.74

7.745.1

112

==

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13

Both of these values (

n

n

VV and ) are listed in Table 3-2 of the Steel

Manual. Only about eight W sections dont 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)

Lets 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 themember 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 willbe different.

From page 3, our uniform service load wa=1.43 k/ftFor 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

61240,XTry W18

867.074.1)291)(/1029(384

)/1226)(12/)(/1430(5

384

5

426

4

4

426

44

=

==

ASD

The W14 X 34 didnt work but the W18 X 40 will work.

Conclusion

Looking at the strength of the beam to resist bending moment wecame up with two different members. We braced the beam adequately

(Lb=6.5 ft) to resist lateral-torsional buckling. We determined the ASDbeam would be a W14 X 34. The beams were compact so there was noflange local buckling or web local buckling. We checked the shear

strength of both beams and it was OK. The deflection turned out to bethe determining limit state. Based on deflection we would use the

same W section for both LRFD and ASD (W18 X 40).

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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 isL/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

ftkipM

ftftkipwLM

kipftftkipwL

V

ftkipw

ftkipftkipw

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

ftinftinftftkip

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. Thetable 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 toosmall. It shows a W21 X 44 will work for ASD and it will also work for

LRFD. Lets select a W21 X 44.

ftkipM

ftftkipwLM

kipftftkipwL

V

ftkipw

ftkipftkipw

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

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15

ftkipinkipinksiZFM xyn ==== 3984770)4.95)(50(3

LRFD ASD

OKftkipMM

ftkipftkipM

a

n

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 andASD.

OKftkipMM

ftkipftkipM

un

n

=

===

288

358)398(90.0

OKkipVV

kipkipV

un

n

9.32

217)217(00.1

=

==

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Both design methods verify our previous calculations. This program isvery 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. Nomatter how you design the beams, they should be checked by anothermethod. We only looked at a beam uniformly loaded. If the beam has

concentrated loads, then there may be other limit states involved (J10of the specification). These would be Flange Local Bending, Web Local

Yielding, Web Crippling, Web Sidesway Buckling, Web CompressionBuckling and Web Panel Zone Shear. This is beyond the scope of this

course. This course is to get designers interested in using the newcode by showing that it isnt that hard. It doesnt matter if you use

LRFD or ASD, both will result in a safe member.