Design of Steel Beams to AISC- LRFD

7/24/2019 Design of Steel Beams to AISC- LRFD

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A Lecture edited

By

Assistant Prof Dr Ehab B Matar

Design of steel beams LRFD-AISC



• The objectives of this lecture is to:-

1. Understanding the behavior of steel beams

under bending moments and shear

2. Identifying the different modes of failure for

laterally supported or laterally un-supported

steel beams

3. Practicing for the different code provisions

for the design of steel beams

Objectives



Reference

• AISC- Specification for structural steel

buildings- March, 2005

• “Steel Structures, Design and Behavior”, by,

Charles G. Salmon, John E. Johnson, Faris A.

Malhas- Pearson Prentice Hall, 5th edition,

2009



with codeignoranceResult of

Provisions









Load and Resistance Factor Design

(LRFD)

loadservicerelevanttheis

factor overloadtheis

capacitymemberorstrengthnominaltheis

factor reductionstrengththeis

:where

..

i

i

n

iin

Q

R

Q R



Resistance factor

• For tension members

=0.90 for yielding limit state

=0.75 for fracture limit state

• For beams =0.9 for shear and bending

• For compression members =0.85

•For fasteners =0.75



Load combinations for LRFD

Q i and i for various load combinations as follows:

• 1.4*D

• 1.2*D+1.6L+0.5(Lr or S or R)

• 1.2D+1.6(Lr or S or R)+(0.5L or 0.8W)• 1.2D+1.3W+0.5L+0.5(Lr or S or R)

• 1.2D±1.0E+0.5L+ 0.2S

• 0.9D±(1.3W or 1.0E)

Where, D: Dead load, L: Live load, Lr: roof live load,W: wind load, E: earthquake load, S: Snow load,R: Rain water load



Beams

Beams is a general word that can be applied to:

• Girders: which is the most important supporting elementfrequently spaced at wide distances

• Joists: the less important beams closely spaced with a truss

type webs• Purlins: roof beams spaning between trusses

• Stringers: longitudinal bridge beams spanning between X –girders

• Girts: horizontal wall beams supporting corrugated sheets

at side walls of factories• Lintels: members supporting a wall over a window or door



Examples for joists, lintels, purlins and

side girts



Modes of failure for beams

• Failure of beams subjected to major axis bendingcan take the following modes:-

1. Occurrence of local buckling for the

compression flange2. Occurrence of lateral torsional buckling

3. Occurrence of warping

4. Occurrence of shear buckling5. Exceeding the serviceability limits (deflection,

vibration,…etc.)



Bending &

local buckling



Bending: local buckling collapse

example from the JHU structures lab.



ELASTIC CRITICAL BUCKLING OF STIFFENED PLATEELASTIC CRITICAL BUCKLING OF STIFFENED PLATE

cr

2

2 2f = k

E

12(1- )(w / t)



FREQUENTLY USED k VALUESFREQUENTLY USED k VALUES



Section Class

– Class 1: Compact sections which achieve full plastic moment

capacity without local buckling

– Class 2: Non-compact sections which achieve yield momentcapacity without local buckling

– Class 3: Slender sections which cannot achieve yield moment

capacity without local buckling



Local buckling



Effect of local and dis-torsional buckling on

beams capacity

Pcrd

Py

local

distortional

Pcrd

Py

local

distortional



Section

class

Part Stress

Profile

Compact p Web Simple bending

Axial Comp.

Flange Uniform Comp. Rolled

Sec.&

B.U.S

Non-Compact

r

Web Simple bending

Axial Comp.

Flange Uniform Comp. Rolled

Sec.

B.U.S

Note Kc=4/SQRT(dw/tw) but shall not be taken less than 0.35 nor greaterthan 0.76

FL=0.7Fy for minor axis bending , major axis bending of slender web

built up I shaped members and major axis bending of compact and

non-compact web built up I shaped members with Sxt/Sxc≥0.7 (where

Sxt and Sxc are the elastic section modulus of tension and

compression flanges in symmetrical I section

y yww F F E t d /1690/76.3/

NA

y y f F F E t C /170/38.0/

yww F E t d /7.5/

yww F E t d /49.1/

Local Buckling limits AISC- SI units (N,mm)

y f F E t C /0.1/

Lc f F E k t C /95.0/

Wid h / hi k i f i f



Width /thickness ratio r for non-compact sections for

beams with different steel yield strengthWeb

hw/twWelded sections

Rolled

sections,

un-stiffened

flange

bf /2tf

Fy (Mpa)

Flange

stiffened

box

section

bf /2tf

Flange un-

stiffened

welded

B.U.S.

bf /2tf

Kc=4/SQR

T(hw/tw)

h/tw

161.739.721.723.2

29.2

0.350.4

0.63

161.7100

40

27.7248

137.233.7

16.6

17.7

22.3

0.35

0.4

0.63

137.2

100

40

22.3345

130.832.1

15.4

16.5

20.8

0.35

0.4

0.63

130.8

100

40

21.0380

120.329.5

14.0

14.7

18.5

0.36

0.4

0.63

120.3

100

40

19.0448



Lateral torsional buckling



Warping of beams

St di t ib ti i b t



Stress distribution in beams at

different stages of loading• M

y: yield moment =S

x.F

y

• Mp: plastic moment= Zx.Fy

• Sx, Zx are the elastic and plastic section

modulus,• Shape factor =Zx/Sx is nearly 1.09-1.18



Restrained and Un-restrained

compression flange of steel beams



Overall lateral buckling of a whole system



Design of beams depend on:

• Section Class

•

Type of Steel

• Kind of stresses



Design of laterally supported steel

beams i.e. Luact=0 against Flexure

A- Case of compact sections

• Governing equation is:

.Mn≥Mu

=0.9

Mn=Z.Fy

Mu= ultimate moment due to ultimate loads

(N.mm)Z= plastic section modulus mm3

Fy= yield stress of steel (MPa i.e. N/mm2)

C ti



Continue

B- Non-compact sections

For sections exactly satisfies the limitations fornon-compact sections r for both of web and

flanges, Governing equation is

.Mn≥Mu

=0.9

Mn=Mr=S.(Fy-Fr)

Mu= ultimate moment due to ultimate loads (N.mm)S= elastic section modulus mm3

Fy= yield stress of steel (MPa i.e. N/mm2)

Fr= residual stress – for rolled sec. =68.95MPa

continue



continue

C- For partial compact sections

For partial compact sections with either flangeor web slenderness ratios lies in between

compact p and non-compact r sections

.Mn≥Mu

=0.9

sectionscompactnonorcompact

foror webflangesforlimitsrelativetheareor

c/tor/tdeitheris

*)(

r

f ww

p

pr

p

r p pn M M M M



Design of steel beams against shear

• Shear stress in I section is

given by

• Where =shear stress

• S= first moment of area

• I= second moment of area of

the whole section• b= width of section under

consideration

• Q= shear force

Ib

QS

continue



continue

• Governing designequation: .Vn≥Vu

=0.9

Vn=0.6.Fyw.Aw

Fyw= yield strength of web

(Mpa)Aw= web area (mm2)

Vu= ultimate shear force (N)

• The above governing

equation in condition thatthere is no shear bucklingin web i.e.d w /t w ≤ 1100/SQRT(F yw )



Serviceability limit states for deflection

• Governing condition is

DL+LL≤ L/360

L= beam span

• For continuous beams, anapproximate value can be

calculated as follows

Where

Ms= moment at mid span

Ma, Mb= moments at

interior supports

))(1.0(48

5 2

ba s M M M EI

L



Example 1• Given: the shown beam,

fully laterally supported

• Required: select thelightest section tosustain the shown loads

using steel A36

• Solution:

• Loads and strainingactions

Wu=1.2Wd.L+1.6WL.L=1.2*2.9+1.6*11.67=22.16KN/m’

Mu=Wu*L2

/8=99.698KN.m



Continue• Selecting a trial section

For steel A36, Fy=250MPaMn≥Mu

Assuming compact section, then

*Zx*Fy ≥Mu Zx ≥Mu/(*Fy)

Zx ≥99.698E6/(0.9*250) ≥443102mm3

Try HEA 220, bf =220mm, tf =11mm, dw=152mm, tw=7mm,

Zx=Wpl.y=568.5cm3, own weight= 50.5 kg/m’

• Checking cross section class

bf /2tf =220/(2*11)=10<10.8 flange is compact

dw/tw=152/7=21.7<107 web is compact

Whole section is compact



Continue• Checking section capacity

Mn= plastic section moment capacity as the section iscompact= Zx*Fy

*Zx*Fy=0.9*568.5E3*250=127.91E6N.mm=127.91KN.

m

Considering the own weight of steel beam, then

Mu=(1.2*(2.9+0.5)+1.6*11.67)*62/8=102.384KN.m

Then , Mn>Mu section is safe for flexure



Example 2• Given: shown simply

supported beam, fulllaterally supported

• Required: select the lightestsection, consideringdeflection using A572 Gr 50

• Solution:

• Loads and straining actions:

a- service loadsW=7.3+14.59=21.89KN/m’

b- ultimate loadsWu=1.2*7.3+1.6*14.59=32.104

KN/m’

Mu=32.104*12.62/8=637.1KN.

m



Continue• Selecting a trial section

For steel A572 Gr 50, Fy=350MPa

Mn≥Mu

Assuming compact section, then

*Zx*Fy ≥Mu Zx ≥Mu/(*Fy)

Zx ≥637.1E6/(0.9*350) ≥2022552mm3

Deflection can be only limited to L/360, if the inertia of the beam:

max=5/384*W*L4/(EI)≤L/360 5/384*21.89*126004/(2E5*I)≤12600/360 I≥1,026,286,655mm4

Try HEA 550, bf =300mm, tf =24mm, dw=438mm, tw=12.5mm,Zx=Wpl.y=4622cm3, I=111900cm4, own weight= 166 kg/m’

• Checking cross section class

bf /2tf =300/(2*24)=6.25<9.2 flange is compactdw/tw=438/12.5=35.04<90.5 web is compact

Whole section is compact



Continue• Checking section capacity

Mn= plastic section moment capacity as the section iscompact= Zx*Fy

*Zx*Fy=0.9*4622E3*350=1455.9E6N.mm=1455.9KN.m

Considering the own weight of steel beam, thenMu=(1.2*(7.3+1.66)+1.6*14.59)*12.62/8=676.635KN.m

Then , Mn>Mu section is safe for flexure

it should be noted that the cross section is constrainedby serviceability requirements not by ultimate limitstates



Design of laterally un-supported steel

beams

• The elastic lateral critical torsional buckling moment under

action of constant moment is given by

• For non-uniform moment, the value

of Mcr is multiplied by the moment

gradient factor Cb

)..(3

1constanttorsionalJ

bendingaxisminoraboutinertiaof momentI

/4.hI/2.hIconstanttorsionalwarpingC

75800MPa))E/(2(1modulusshearG

modulussYoung'E

.....

3

y

2

y

2

f w

2

ii

y ywcr

t b

J G I E I C L

E

L M



Zone 1: when Lu<Lpd then plastic moment capacity is

reached with large plastic rotation capacity- plastic

analysis is permitted- section should be compact

flangencompressioof gyrationof radius

Z.Fcapacitymoment plasticM

curvature)reversewhen(segmentunbraced

laterallytheof endsat themomentsmaller

/1520024800

y p

1

1

ry

M

r F

M M L y

y

p

pd

9.0

.

.

yn

un

F Z M

M M



Case 2: Lu≤Lp plastic moment is reached (Mn=Mp) with

little rotation capacity- section should be compact

y

y

p r F

L790

9.0

.

.

yn

un

F Z M

M M

Case 3: when Lp< Lu≤Lr- lateral torsional buckling



Case 3: when Lp< Lu≤Lr lateral torsional buckling

of compact sections may occur in the inelastic

range (Mp>Mn≥Mr)

bendingaxismajoraboutmodulussectionelasticS

68.95MPastressresidualF

flangencompressioof strengthyieldF

axisminoraboutinertiaof momentI

flangencompressioof gyrationof radiusr

modulusshearG

200000MPamodulussYoung'E

sectiontheof areasectionalcrossA

constanttorsionalJ

constantwarping

4

2

)(11)(

.

x

r

yf

y

y

2

2

1

2

2

1

w

x

y

w

x

r yf

r yf

y

r

C

GJ S

I C X

EGJA

S X

F F X F F

X r L



Continue case 3

lengthunbracedof length3/4andhalf quarter,atmoment bendingM,M,M

lengthunbracedhinmoment wit bendingmax.

3435.25.12

)

CBA

max

max

max

M

M M M M M C

M L L

L L M M M C M

M M

C B A

b

p

pr

pu

r p pbn

un

Case 4 L <L ≤L General limit state for any section



Case 4: Lp<Lu≤Lr General limit state for any section

where nominal moment strength Mn occurs in the

inelastic range

• When Lp<Lu≤Lr or p< <

r whether for flange or

web, then

pr

pu

r p pbn

pr

p

r p pn

L L

L L M M M C M

M M M M

bucklingtorsionallateralforstateLimit

bucklinglocalof stateslimitM.M un

Case 5: Lu>Lr general limit state where nominal



Case 5: u r ge e a t state e e o a

moment strength Mn equals the elastic buckling

strength Mcr

J G I E I C L

E

LC M Mn

M M

y yw

uu

bcr

un

....

.

2



Example 3

• Given: the shown beam,laterally un-supported

• Required: select the

lightest section to sustain

the shown loads using

steel A36- considering

D.L=20% of W

• Solution:

• Loads and straining

actions



Continue Ex. 3

• Wu=14.6(1.2*0.2+1.6*0.8)=22.2KN/m’

•

Mu=22.2*15

^2

/8=624.375KN.m- Assuming compact sec.,

- Zx≥Mu /(*F y ) ≥624.4E6/(0.9*250)=2775E3mm3

• Try HEA 450, Zx=3216cm3, Sx=2896cm3,Iy=9465cm4, ry=7.29cm, bf =300mm, tf =21mm,dw=344mm, tw=11.5mm, A=178cm2, G=140Kg/m’, J=It=243.8cm4, Cw=Iw=4.148cm6

•Checking sec. class: flange; bf /2tf =7.14<10.8

flange is compact

• dw/tw=29.91<107 web is compact, wholesec. is compact



continue

• Checking max. lateral un-supported length

r uact p

x

y

w

x

L

l

y

r

uact y

y

p

L L L

mm E Lr

E E

E

E

E

GJ

S

I

C X

E E E E

EGJAS

X

F X F

X r L

mm Lmmr F

L

11286)95.68250(*11305.411)95.68250(

19820*9.72

11305.448.243*75800

32896

49465

6148.4*44

198202*)3.01(2

2178*48.243*)52(328962

.11.

750036429.72*250

790.790

2

22

2

2

1

2

2

1



continue

• Calculating moment gradient factor

3.1

.65.629

.72.503

.84.293

.625.6718/15*)4.1*2.1(4.624

3435.25.12

2

max

max

max

b

C

B

A

C B A

b

C

m KN M

m KN M

m KN M

m KN M

M M M M M C



continue

• Moment capacity for the section:-

•Mn=Cb[Mp-(Mp-Mr){(Luact-Lp)/(Lr-Lp)}]

• Mp=Zx.Fy=3216E3*250=804E6N.mm

• Mr=Sx(Fy-Fr)=2896E3(250-68.95)=524E6N.mm

• Mn=1.3[804-(804-524){(7.5-3.64)/(11.286-3.64)}]=861.4KN.m804E6N.mm

• Mn=0.9*804=723.6KN.m>Mu=671.625KN.

m o.k. safe for flexural limit states

Documents

Design of Steel Beams to AISC- LRFD