Design of steel beams LRFD-AISC

A Lecture edited

Assistant Prof Dr Ehab B Matar By

Design of steel beams LRFD-AISC

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

• 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, 5 ^th edition,

2009

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

with code ignorance

Result of

Provisions

Load and Resistance Factor Design (LRFD)

load service

relevant the

is

factor  overload

the is

capacity member

or strength

nominal the

is

factor  reduction

strength the

is

: where

. .

i i

n

i i n

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(L _r or S or R)

• 1.2D+1.6(L ^r or S or R)+(0.5L or 0.8W)

• 1.2D+1.3W+0.5L+0.5(L _r 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, L _r : 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 element frequently 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 bending can take the following modes:-

1. Occurrence of local buckling for the compression flange

2. Occurrence of lateral torsional buckling 3. Occurrence of warping

4. Occurrence of shear buckling

5. 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 PLATE ELASTIC CRITICAL BUCKLING OF STIFFENED PLATE

  _cr 

2

2 2

f   = k E

12(1- )(w / t)





FREQUENTLY USED k VALUES

FREQUENTLY 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 moment capacity 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

P

_crd

P

_y

local distortional

P

_crd

P

_y

local

distortional

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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  K _c =4/SQRT(d _w /t _w ) but shall not be taken less than 0.35 nor greater than 0.76

F _L =0.7F _y 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 S _xt /S _xc ^≥ 0.7 (where S _xt and S _xc are the elastic section modulus of tension and

compression flanges in symmetrical I section

  

 

  

 

   











 y  y

w

w

t   E     F  F 

d  /  3 . 76   /  1690 /

 NA

 y  y

 f  

 E     F  F 

t 

C  /  0 . 38   /  170 /

 y w

w

t     E  F 

d  /    5 . 7 /

 y w

w

t     E  F 

d  /    1 . 49 /

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

 y

 f  

   E  F 

t 

C  /    1 . 0 /

 L c

 f  

  k   E  F 

t 

C  /    0 . 95 /

7/24/2019

Width /thickness ratio 

Design of Steel Beams to AISC- LRFD

_r for non-compact sections for beams with different steel yield strength

Web h _w /t _w Welded sections

Rolled sections, un-

stiffened flange b _f  /2t _f  F _y (Mpa)

Flange stiffened box

section b _f  /2t _f  Flange un-

stiffened welded B.U.S.

b _f  /2t _f  Kc=4/SQR

T(h _w /t _w ) h/t _w

161.7 39.7

21.7 23.2 29.2 0.35

0.4 0.63 161.7

100 40 27.7

248

137.2 33.7

16.6 17.7 22.3 0.35

0.4 0.63 137.2

100 40 22.3

345

130.8 32.1

15.4 16.5 20.8 0.35

0.4 0.63 130.8

100 40 21.0

380

14.0 0.36

120.3

Lateral torsional buckling

Warping of beams

Stress distribution in beams at different stages of loading

• M

y

: yield moment =S

x

.F

y

• M _p : plastic moment= Z _x .F _y

• S _x , Z _x are the elastic and plastic section modulus,

• Shape factor  ^=Z _x ^/S _x 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

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Design of laterally supported steel beams i.e. L _uact =0 against Flexure

 A- Case of compact sections

• Governing equation is:

 ^.M _n ^≥M _u

 =0.9 M _n =Z.F _y

M _u = ultimate moment due to ultimate loads (N.mm)

Z= plastic section modulus mm ³

F _y = yield stress of steel (MPa i.e. N/mm ² )

C ti

7/24/2019

Continue

Design of Steel Beams to AISC- LRFD

B- Non-compact sections

For sections exactly satisfies the limitations for non-compact sections  _r for both of web and flanges, Governing equation is

 .M ⁿ ≥M ^u

 ^=0.9

M _n =M _r =S.(F _y -F _r )

M ^u = ultimate moment due to ultimate loads (N.mm)

S= elastic section modulus mm ³

F _y = yield stress of steel (MPa i.e. N/mm ² )

C- For partial compact sections

For partial compact sections with either flange or web slenderness ratios  lies in between compact   _p and non-compact  _r ^sections

 .M ⁿ ≥M ^u

 ^=0.9

sections compact

non or

compact

for or web

flanges for

limits relative

the are

or

c/t or /t

d either is

* ) (

r 

f  w

w

 

 

 

 

 

 

 

 p

 p r 

 p r 

 p  p

n  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

• Governing design equation:  .V _n ≥V _u

 =0.9

V _n =0.6.F _yw .A _w

F _yw = yield strength of web (Mpa)

A _w = web area (mm ² )

V _u = ultimate shear force (N)

• The above governing

equation in condition that there is no shear buckling in 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, an approximate value can be calculated as follows

Where

M s = moment at mid span M _a , M _b = moments at

interior supports

)) (

1 . 0 48 (

5 ²

b a

 s    M  M 

 EI   M 

 L  



 

Example 1

• Given: the shown beam, fully laterally supported

• Required: select the lightest section to

sustain the shown loads using steel A36

• Solution:

• Loads and straining actions

W _u =1.2W _d.L +1.6W _L.L =1.2*2 .9+1.6*11.67=22.16KN/

m’

M ^u =W ^u *L

2

/8=99.698KN.m

7/24/2019

Continue

Design of Steel Beams to AISC- LRFD

• Selecting a trial section For steel A36, F ^y =250MPa

 M _n ≥M _u

Assuming compact section, then

 *Z _x *F _y  ≥M _u Z _x  ≥M _u /(  *F _y )

Z ^x  ≥99.698E6/(0.9*250) ≥443102mm ³

Try HEA 220, b _f  =220mm, t _f  =11mm, d _w =152mm, t _w =7mm, Z _x =W _pl.y =568.5cm ³ , own weight= 50.5 kg/m’

• Checking cross section class

b _f  /2t _f  =220/(2*11)=10<10.8 flange is compact d _w /t _w =152/7=21.7<107 web is compact

Whole section is compact

Continue

• Checking section capacity

M ⁿ = plastic section moment capacity as the section is compact= Z _x *F _y

 ^*Z _x ^*F _y =0.9*568.5E3*250=127.91E6N.mm=127.91KN.

m

Considering the own weight of steel beam, then

M _u =(1.2*(2.9+0.5)+1.6*11.67)*6 ² /8=102.384KN.m

Then ,  ^M _n ^>M _u section is safe for flexure

Example 2

• Given: shown simply supported beam, full laterally supported

• Required: select the lightest section, considering

deflection using A572 Gr 50

• Solution:

• Loads and straining actions:

a- service loads

W=7.3+14.59=21.89KN/m’

b- ultimate loads

W _u =1.2*7.3+1.6*14.59=32.104 KN/m’

M _u =32.104*12.6 ² /8=637.1KN.

Continue

• Selecting a trial section

For steel A572 Gr 50, F _y =350MPa

 M ⁿ ≥M ^u

Assuming compact section, then

 *Z _x *F _y  ≥M _u Z _x  ≥M _u /(  *F _y )

Z _x  ≥637.1E6/(0.9*350) ≥2022552mm ³

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

 _max =5/384*W*L ⁴ /(EI)≤L/360 5/384*21.89*12600 ⁴ /(2E5*I)

≤12600/360 I≥1,026,286,655mm ⁴

Try HEA 550, b _f  =300mm, t _f  =24mm, d _w =438mm, t _w =12.5mm, Z _x =W _pl.y =4622cm ³ , I=111900cm ⁴ , own weight= 166 kg/m’

• Checking cross section class

b ^f  /2t ^f  =300/(2*24)=6.25<9.2 flange is compact

d _w /t _w =438/12.5=35.04<90.5 web is compact

Whole section is compact

7/24/2019

Continue

Design of Steel Beams to AISC- LRFD

• Checking section capacity

M ⁿ compact= Z = plastic section moment capacity as the section is _x *F _y

 ^*Z _x ^*F _y =0.9*4622E3*350=1455.9E6N.mm=1455.9KN.

m

Considering the own weight of steel beam, then

M _u =(1.2*(7.3+1.66)+1.6*14.59)*12.6 ² /8=676.635KN.m Then ,  ^M _n ^>M _u section is safe for flexure

it should be noted that the cross section is constrained

by serviceability requirements not by ultimate limit

states

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 M _cr is multiplied by the moment

gradient factor C _b

























   

  

 

  

) . .(

3 constant 1

torsional J

 bending axis

minor about

inertia of 

moment I

/4 .h I /2 .h I constant torsional

warping C

75800MPa ))

E/(2(1 modulus

shear G

modulus s

Young' E

. . . .

.

3 y

2 y 2

f  w

2

i i  y  y

w cr 

t  b

 J  G  I   E   I 

 L C   E   M   L

 

 

 

7/24/2019

Zone 1: when L _u <L _pd then plastic moment capacity is

Design of Steel Beams to AISC- LRFD

reached with large plastic rotation capacity- plastic analysis is permitted- section should be compact

flange n

compressio of 

gyration of 

radius

Z.F capacity

moment  plastic

M

curvature) reverse

when (

segment unbraced

 

laterally the

of  ends at the moment

smaller

/ 15200 24800

y  p

1

1











 

ry  M 

 F  r 

 M 

 L  M 

 _y

 y

 p  pd 

9 . 0

. .







 

 

 y n

u n

 F   Z   M 

 M 

 M 

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little rotation capacity- section should be compact

 y  y

 p   r 

 L 790  F 



9 . 0

. .







 

 

 y n

u n  Z   F   M 

 M   M 

Case 3: when L < L ≤L - lateral torsional buckling

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

Case 3: when L _p < L _u ≤L _r lateral torsional buckling of compact sections may occur in the inelastic range (M _p >M _n ≥M _r )

68.95MPa stress

residual F

flange n

compressio of 

strength yield

F

axis minor about

inertia of 

moment I

flange n

compressio of 

gyration of 

radius r 

modulus shear

G

200000MPa modulus

s Young' E

section the

of  area sectional cross

A

constant torsional

J

constant warping

4

2

) (

1 ) 1

( .

r  yf  y y

2 2

1

2 2

1























    

  

 









 



w

 x  y

w  x

r   yf   r 

 yf    y r 

C 

GJ  S  C   I 

 X 

 EGJA  X  S 

 F   F 

 F   X   F 

 X   L r 

 

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Continue case 3

length unbraced

of  length 3/4

and half  quarter,

at moment  bending

M , M , M

length unbraced

hin moment wit  bending

max.

3 4

3 5

.

2 12 . 5

)

C B

A max

max

max









 

 







 





     

 

 

 

     

 

 

 



 







 M 

 M   M 

 M 

 M   M 

C 

 L  M   L

 L  M   L

 M   M 

C   M 

 M   M 

C   B

 A b

 p  p

r 

 p u

r   p

 p b

n

u

  n

7/24/2019

Case 4: L _p <L _u ≤L _r General limit state

Design of Steel Beams to AISC- LRFD

for any section where nominal moment strength M _n occurs in the inelastic range

• When L _p <L _u ≤L _r or  _p <  <

 _r whether for flange or web, then

 

 

 





 





     

 

 

 



 





     

 

 

 



 







 p r 

 p u

r   p

 p b

n

 p r 

 p r 

 p  p

n

 L  L

 L  M   L

 M   M 

C   M 

 M   M 

 M   M 

 buckling torsional

lateral for

state Limit

 buckling local

of  states limit .M ⁿ M ^u

 

 

 

 

 

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

moment strength M _n equals the elastic buckling strength M _cr

 J  G  I   E   I 

 L C   E  C   L

 M  Mn

 M   M 

 y  y

w u

u b cr 

u n

. . . .

.

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

• W _u =14.6(1.2*0.2+1.6*0.8)=22.2 KN/m’

•

M ^u =22.2*15

^2

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

- Z _x ^≥M _u  /(  *F  _y  )  ^≥ 624.4E6/(0.9*250)=2775E3mm ³

• Try HEA 450, Z _x =3216cm ³ , S _x =2896cm ³ ,

I _y =9465cm ⁴ , r _y =7.29cm, b _f  =300mm, t _f  =21mm, d _w =344mm, t _w =11.5mm, A=178cm ² , G=140 Kg/m’, J=I _t =243.8cm ⁴ , C _w =I _w =4.148cm ⁶

• Checking sec. class: flange is compact flange; b ^f  /2t ^f  =7.14<10.8 ^

• d _w /t _w =29.91<107 _ web is compact, _ whole

sec. is compact

7/24/2019

continue

Design of Steel Beams to AISC- LRFD

• 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 

 X  C 

 E   E 

 E   E 

 EGJA  X  S 

 F   F   X 

 X   L r 

mm  L

mm  F  r 

 L















 





   

  

 

     

  

 

  

 





















11286 )

95 . 68 250

(

* 11 305

. 4 1 ) 1

95 . 68 250

(

19820

* 9 . 72

11 305

. 4 4

8 . 243

* 75800

3 2896 4

9465

6 148 . 4

* 4 4

19820 2

* ) 3 . 0 1 (

2 * 243 . 8 4 * 178 2 )

5 2 ( 3 2896 2

. 1

. 1

7500 3642

9 . 72 250 *

790 .

790

2 2 2

2

2 1

2 2

1

 

 

• Calculating moment gradient factor 

3 . 1

. 65

. 629

. 72

. 503

. 84

. 293

. 625

. 671 8

/ 15

* ) 4 . 1

* 2 . 1 ( 4 . 624

3 4

3 5

.

2 12 . 5

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 

7/24/2019

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Design of Steel Beams to AISC- LRFD

• Moment capacity for the section:-

• M ⁿ =C ^b [M ^p -(M ^p -M ^r ){(L ^uact -L ^p )/(L ^r -L ^p )}]

• M _p =Z _x .F _y =3216E3*250=804E6N.mm

• M _r =S _x (F _y -F _r )=2896E3(250-68.95)=524E6N.mm

• M ⁿ =1.3[804-(804-524){(7.5-3.64)/(11.286- 3.64)}]=861.4KN.m _ 804E6N.mm

•    ^M _n =0.9*804=723.6KN.m>M _u =671.625KN.

m _ o.k. safe for flexural limit states