Steel Chap5 Ultimate Limit State

8/3/2019 Steel Chap5 Ultimate Limit State

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CENG 6507-Steel Structures

CENG 6507 - Steel Structures

Course Instructor: Dr.-techn. Medhanye B.

STRUCTURAL ENGINEERING CHAIR

ETHIOPIAN INSTITUTE OF TECHNOLOGY

MEKELLE UNIVERSITY

5. Ultimate limit state of

buckling: beam-columns,

built-up members

Introduction

Limit state design requires the structure to satisfytwo principal criteria: the ultimate limit state (ULS)and the serviceability limit state (SLS).

A limit state is a set of performance criteria (e.g.

vibration levels, deflection, strength, stability,buckling, twisting, collapse) that must be met whenthe structure is subject to loads.

All engineering design criteria have a common goal:that of ensuring a safe structure and ensuring thefunctionality of the structure.


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Ultimate Limit State

To satisfy the ultimate limit state, the structuremust not collapse when subjected to the peakdesign load for which it was designed.

A structure is deemed to satisfy the ultimate limitstate criteria if all factored bending, shear andtensile or compressive stresses are below thefactored resistance calculated for the sectionunder consideration.

Whereas Magnification Factor is used for theloads, and Reduction Factor for the resistance ofmembers.

Serviceability Limit State

To satisfy the serviceability limit state criteria, astructure must remain functional for its intendeduse subject to routine (everyday) loading, and assuch the structure must not cause occupantdiscomfort under routine conditions.

Examples: Deflection, Vibration, Crack widths inconcrete


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ULS: TYPES OF STEEL FAILURE

Failure can happen because ofa) yielding of members

b) failure of connections- most common cause of steel buildingscollapse

c) buckling of members

- buckling of columns rarely has resultedin structural failure

Tension members are linear members inwhich axial forces act so as to elongate

(stretch) the member.

Tension members carry loads mostefficiently, since the entire cross section issubjected to uniform stress.

Unlike compression members, they do notfail by buckling

Tension MembersDesign consideration of tension members

Tension members are efficient structural elements but its

efficiency may be affected by:- End connection (e.g. Bolt holes reduce member section)

Reversal of load (buckling)

Bending moments

Design of Axially loaded Tension members (EBCS-3, 1995)

The design value of the axial tension force isWhere Nt,Rd is the design tension resistance capacity

of the x-section taken as the smaller of:

1. The design plastic resistance of the gross section is

2. The design ultimate resistance of the net section

at the bolt hole is

RdRdRdRdt,t,t,t,SdSdSdSdt,t,t,t,NNNNNNNN

1

,

M

y

Rdpl

AfN

=

2

,

9.0

M

ueff

Rdu

fAN

=


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BUCKLING OF STEEL MEMBERS

Buckling of steel may happen in

a) members under compression

b) members under bending

c) members under combined loading:compression + bending

Local Buckling and Section Classification

Internal or outstand elementsinternal

are the webs of open beams or the flanges of boxes

outstand

are the flanges of open sections and the legs of angles

Outstand

Internal

Web

Flange

Web

Internal

Flange

(a) Rolled I-section (b) Hollow section

Flange

(c) Welded box section

InternalOutstand

InternalWeb

Local buckling

As the plate elements in structuralsections are relatively thin compared withtheir width, when loaded in compression

they may buckle locally

Local buckling within the cross-sectionmay limit the load carrying capacity of thesection by preventing the attainment ofyield strength.


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Classes of cross-sections

Premature failure (arising from the effects oflocal buckling) may be avoided by limiting thewidth to thickness ratio - or slenderness- ofindividual elements within the cross section.

This is the basis of the section classificationapproach.

EBCS-3, 1995 defines four classes of cross-

sections. The class into which a particular cross-section

falls depends upon the slenderness of eachelement and the compressive stress distribution.

Four classes of cross-section are:-1. Class 1 or plastic cross-sections:-

Design of Steel Structures for plastic elements.

A plastic hinge can be developed with sufficient rotationcapacity to allow redistribution of moments within the structure.Only Class 1 section may be used for plastic design.

2. Class 2 or compact cross-sections:

The full plastic moments capacity can be developed but local

buckling may prevent development of a plastic hinge withsufficient rotation capacity to permit plastic design. Class 2sections can be used without restriction except for plasticdesign.

3. Class 3 or semi-compact sections:-

The stress at the extreme fibers can reach the design strengthbut local buckling may prevent the development of the fullplastic moment. Class 3 sections are subjected to limitationson their capacity.

4. Class 4 or thin-walled sections:-

Local buckling may prevent the stress in a thin-walled sectionfrom reaching the design strength. Design of Class 4 sectionsrequires special attention.


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(b) Class 4 cross-sections - bending moment

(a) Class 4 cross-sections - axial force

Gross cross-section

Gross cross-section

Centroidal axis ofgross cross-section

Centroidal axis of

gross cross-section

Centroidal axis ofeffective cross-section

Non-effective zones

Non-effective zone

Non-effective zone

Centroidal axis

Centroidal axisCentroidal axis ofeffective section

e N

eM

e M

Centroidal axis ofeffective section

Effective cross-

sections for class4 in compressionand bending(class 4 x-section )

Stress distribution(compression positive)

Effective width beff

c

beff

1

2

1 0> :

beff= c

1

2

bt bc

beff

< 0 :

b b ceff c= = / ( )1

=2 1/

Buckling factor k

1

0,43

0

0,57

-1

0,85

1 1

0 5 7 0 2 1 0 07 2, , , +

c

beff

1

2

1 0> :

beff= c

beff

1

2

btbc

< 0 :

b b ceff c= = / ( )1

=2 1/

Buckling factor k

1

0,43

1 0> >

0578

0 34

,

, +

0

1,70

0 1> >

1 7 5 17 1 2, , +

-1

23,8

Effective widths ofoutstandcompressionelements

(EBCS-3, Tab. 4.4)

Stress distribution

(compression positive)Effective width b eff

be1 be2

1

2

b

= 1:

b = b - 3t

beff=b

be1 = 0,5 beff

be2 = 0,5 beff

be1 be2

1

2

b

1 0> > :

b = b - 3t

beff=b

be1 =2

5

beff

-

be2 = beff- be1

be1 be2

1

2

b

bc bt < 0:

b = b - 3t

b b beff c= = - / ( )1

be1 = 0,4b eff

be2 = 0,6b eff

=2 1/

Buckling

factor k

1

4,0

1 0> >

8 2

1 05

,

, +

0

7,81

0 1> > -

7 81 6 92 9 78 2, , ,- +

-1

23,9

- > > -1 2

598 1 2, ( )-

Alternatively, for 1 1-> > : k =

+ + - + +

16

1 0112 1 12 2 0 5[( ) , ( ) ] ( ),

Illustrated as rhs.

For other sections b = d for webs

b= b for internal flange elements (except rhs)

_

_ _

Effective widths of

internalcompressionelements

(EBCS-3, Tab. 4.3)

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Compression Members The strength of steel compression members is

usually limited by their tendency to buckle. If a 6mm diameter steel rod 1m long is placed in

a testing machine subjected to a pull, it will befound to carry a load of about 7KNbefore failureoccurs. If on the other hand this same rod hadbeen subjected to compression, then themaximum load, which would have been carried,would be about 0.035KN, a very big difference

Failure in the first test occurs by the fracture ofthe member, in the second it is due to bendingout of the line of action of the load.

The load at which a compression memberbecomes unstable is the buckling load.

The buckling load depends on the length, cross-section, and end conditions of the column andthe stiffness of the material.

Buckling Load

Pcr is the load at which the compressionmember becomes unstable

E is modulus of elasticity of steel

I is moment of inertia of the cross section

L is the length of the compression member

K is the effective length factor

2

2

2

2

)(KL

EI

L

EIP

e

cr

==

EFFECTIVE LENGTH OF COLUMNS

The effective length, Le, of a

member hinged at its ends is thedistance between the axes of thehinges.

K, is the ratio of the length (Le) ofthe equivalent column to the actuallength (L); and the length of theequivalent column is the distancebetween two consecutive points ofcontraflexure (points of zeromoment) in the actual column.

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Effective Length FactorsBuckling of a column in a non sway frame

Buckling of a column in a sway frame COLUMNS OF NON-SWAY FRAMES

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Design steps for Axially Compressed Members1. Determine the axial load, Nsd.2. Determine the buckling length,

3. Select a trial section (take into consideration economy, i.e.least weight per unit length).4. Determine the Class of the section according to Section

4.3.2 and Table 4.1. If the cross-section is classified asClass 4, determine Aeff according to Section 4.3.4 andTable 4.4 (Sec 4.5.4.3)

5. Determine the non-dimensional slenderness ratio(Section 4.5.4.3)

6. Determine the appropriate buckling curve, Table 4.11

7. Determine the value of . Interpolation must be used to

determine more exact values, Table.4.98. Calculate the design buckling resistance Nb,Rd of themember. Buckling about both principal axes must bechecked.

9. Check the computed buckling resistance against theapplied load. If the calculated value is inadequate or is toohigh, select another section and go back to Step 4.

y

zP

z

x

P

x

y

Example-1:The column B E on the Figure shown below is under theaction of NSd = 2800 kN. Both sides are pinned. Check theresistance of the column. Steel grade Fe 430 is used.

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Solution:

Step 1: Axial load NSd = 2800 kN.

Step 2:Buckling length L = 4000 mm (pinned endboth sides. Frame non-sway mode).

Step 3: The section is given.

Step 4:Determine the class of the cross-sectionand check for a local buckling. The sectionis subjected to uniform compression.

For the section to be classified as at least class 3, in order

to avoid any modification to the full cross sectional areadue to local buckling, the limiting width to thickness ratio for

class 3 section are (See Table 4.1 EBCS-3).

Outstand element of compression flange: c/tf 15 .

Web subject to compression only: d / tw 42 .

For Fe 430 steel grade fy = 275 N / mm2.Thus

This gives the following limiting values:

Outstand element of compression flange:

c / tf = (254/2) / 16.3 = 7.78 < 15 x 0.92 = 13.8 OK.

Web subject to compression only:

d/tw = (310-2(33))/9.1 = 26.8 < 42 x 0.92 = 38.64 OK.

Therefore, the section belongs to at least Class 3.Thus, A = 1.0

92.0275

235 ==

Step-5: Determine the non-dimensionalslenderness ratio.

For Fe 430 steel grade,

1 = 93.9 = 93.9 x 0.92 = 86.39

Slenderness ratio about y-axis:

y = L / iy = 4000/135 = 29.63

Slenderness ratio about z-axis:

z = L / iz = 4000/63.6 = 62.89

Hence, the non-dimensional slenderness ratio isdetermined as:

( )

( ) 73.0139.86

89.62

34.0139.86

63.29

1

1

==

=

==

=

Az

z

Ay

y

Step-6: Determine the appropriate column curves(Table 4.11 EBCS-3).

Use curve a for buckling about y-axis and curve bfor buckling about z-axis.

Step 7: Determine value of . Using Table 4.9 andinterpolating:

For y-axis: curve a for

For z-axis: curve b for

Therefore, buckling about the z-axis becomes critical.

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310


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Step-8: Calculate the design buckling resistance.

Step 9: Because 2800 kN > 2117.5 kN, the column

do not resist.

N.B. Please note that even the plastic capacity ofthe section (2750 kN) is less than the applied load(2800 kN)

kNNxxxAf

NM

yA

Rdb 5.211721175001.1

27511000177.0

1

, ====

(Only to show the method!!) Add an additionalhinged support at mid-height to increase the

resistance about the minor axis.Go to Step 5.

Slenderness ratio about y-axis = 29.63 (doesnt vary)

Slenderness ratio about z-axis = 2000/63.6 = 31.45

Non dimensional slenderness ratio

(doesnt vary)

Values of :

y-axis: y = 0.97 doesnt vary

z-axis: Curve b for

Hence buckling about the z-axis becomes critical

34.0=y

( ) 36.01

39.86

45.31==z

94.036.0 == zz

resisttdoesnkNkNxx

N Rdb '.280025851.1

2751100094.0,

==

Example-2:Determine the design buckling resistance ofa 457 x 152 x 52 UB used as a pin-ended

column. The column is 3m long and its steelgrade is Fe 360.

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Solution:Step 1: buckling strength is required!

Step 2: Buckling length = 3000 mm.Step 3: The section is given.

Step 4: Determine the class of the cross-sectionand check for local buckling.

For Fe-360 steel grade fy = 235 N / mm2.

Thus,

These the limiting values for an I-section are:

Outstand element of compression flange: c/tf15=15

Web subject to compression only: d / tw 42 = 42

1235 ==yf

For the 457x152x52UB profile, the actual values are:

Outstand element of compression flange:

c / tf = (152.4 / 2) / 10.9 = 7 < 15 OK.Web subject to compression only:

d / tw =(449.8 2x10.9 2x10.2) / 7.6 = 53.60 > 42

Therefore, the flange satisfies the Class 3requirement, but the web is Class 4 section.Consequently, there must be a reduction in the

strength of the section to allow for the local bucklingwhich will take place in the web. Therefore, theeffective area, Aeffmust be determined for the web!

The effective width is beff = reduction factor x b = x b.

The method to calculate the effective area (Aeff) isexplained in section 4.3.4 of EBCS-3.

The reduction factor is given by,

a) = 1; if

b)

Where

In which: t is the relevant thickness.

k

is the buckling factor corresponding tothe stress ratio from Table 4.3 or 4.4as appropriate.

for webs.

673.0p

( )673.0

22.02 >

= p

p

p if

( ) kt

bp 4.28

=

db =

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In this example, since the column is axially loadedthe stress distribution is uniform, i.e. 1 = 2.

Table 4.3 is used to calculate the effective width.Thus, 1/2 = 1 & k = 4.0 (see lower part of Table 4.3)

6.536.7

6.407

6.407

==

==

wtb

mmdb

( )( ) ( )

mmxbbAnd

xx

eff

p

p

p

2.3316.407812.0

812.0944.0

22.0944.022.0

673.0944.0

414.28

6.53

22

===

=

=

=

>==

Therefore the area that should be ignored at thecenter of the web is:

And then

Step 5: Determine the non-dimensional slendernessratio (axis-z govern).

Hence the non dimensional slenderness ratiobecomes

( )2

4.5816.72.3317.407 mmxA ==

( )913.0

66504.5816650

=

==A

AeffA

9.939.93

5.961.31

3000

1 ==

==

z

( )98.0913.0

9.935.96

1

==

= Azz

Step 6: Appropriate column curve:

For h/b = 449.8/152.4 = 2.95 > 1.2; & tf = 10.9 < 40 mm;

use curve b for buckling about z-axis.

Step 7: Determine the value of .

Using Table 4.9 and interpolating, z-axis: curve b for

Step 8: Design buckling resistance:

Answer:The design buckling resistance will then be

6034.098.0 == zz

NxxxAf

NM

yA

Rdb 7826601.1

2356650913.06034.0

1

, ===

.66.782, kNN Rdb =

Strength of flexural members are limited by: Local buckling of a cross section

Lateraltorsional buckling of the entire member

Development of a plastic hinge at a particularcross section

General Stability (Lateral torsional buckling)Stability Due to shear on the web.

Local Stability

Due to compressive stress onthe flange

Beams can be

Laterally restrained (supported)

Laterally non-restrained (unsupported)

Bending/Flexural Members

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Restrained Beams: Beams which are unable to move laterally are

termed restrained. and are unaffected by out-of-plane buckling (lateral-torsional instability).

Beams may often be designed on basis ofbending moment resistance.

Moment resistance is dependent on sectionclassification.

Co-existent shear forces, affect momentresistance.

Restrained &unrestrained Beams Restrained Beams (Contd)

Beams may be considered restrained if

Full lateral restraint is provided by for example positiveattachment of a floor system to the top flange of a simplysupported beam

Adequate torsional restraint of the compression flange isprovided

Closely spaced bracing elements are provided such thatthe minor axis slenderness is low

Moment resistance

In a simple single span beam, failure occurs when thedesign value of the bending moment (Msd) exceeds thedesign moment resistance of the cross-section (MC,Rd) ,

The magnitude of which depends on the section shape,material strength and section classification

Resistance to bending moment

According to EBCS-3, for bending about oneaxis in the absence of shearing force, the designvalue of bending moment .

Resistance to shear

The design value of the shear force VSd at eachcross-section shall satisfy

RdcSd MM ,

RdplSd VV ,

Restrained Beams

The design moment resistance (Mc,Rd) may be taken as:For class 1 & 2 cross-sections, the design plasticresistance moment of the gross section

Mc.Rd= Mpl.Rd=

For a class 3 cross-section, the design elastic resistancemoment of the gross section

Mc.Rd= Mel.Rd=

For a class 4 cross-section, the design local bucklingresistance

Mc.Rd= Mo.Rd=Shear resistance

the plastic shear resistance, Vpl.Rd of a shear area (Av)

W fpl y

M

.

0

0

.

M

yelfW

1

.

M

yefffW

MO

y

vRdpl

fAV

)3/(. =

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Resistance for bending and shear

The theoretical plastic resistance moment of a cross-section is reduced by the presence of the shear.

For small values of the shear force this reduction is notsignificant and may be neglected. However, when theshear force exceeds half of the plastic shear resistance,

allowance shall be made for its effect on plasticresistance moment.

That is, if the value of the shear force VSd does notexceed 50% of the design plastic shear resistance noreduction need be made in the resistance moments

When VSd exceeds 50% of Vpl the design resistancemoment of the cross-section should be reduced to Mv,Rdobtained as stated in EBCS Article 4.6.1.3 (3).

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Local Stability

During bending, part of the web and at least one flange isunder compressive stress, therefore can be subjected to loss

of stability.

1) Shear buckling resistance: Near the support, where thereis a considerable acting shear force, the web of the beamcan lose its stability through the formation of web folds of

buckles.

This problem is prevented by providing transverse

stiffeners as shown in the figure below.

2. Flange induced buckling

As we can see in the figure below, the upperflange is under the action of the compressive

stress and may lose it local stability

Laterally Unrestrained Beams

Beams bent about the major axis may fail by buckling in amore flexible plane

This form of buckling involves both lateral deflection andtwisting - Lateral-torsional buckling

When the beam has a higher bending stiffness in the verticalplane compared with the horizontal plane, the beam can twistsideways under the action of the load as shown in the Figure

below. The applied moment at which a beam buckles by deflectinglaterally and twisting reached is the elastic criticalmoment

Dead weight

load applied

vertically

Buckled

position

Unloaded

position

Clamp at

root

slender cantilever beamloaded by a vertical end load.

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A reduction factorfor lateral torsional buckling, LT

is

used to determine the capacity

The design buckling resistance moment(Mb.Rd) of alaterally unrestrained beam is thus taken as:

5,0

2

2

2

2

+=

z

t

z

wzcr

EI

GIL

I

I

L

EIM

( )

( ) ( )

+

+

= g2

5.0

2g2

z2

t2

z

w

2

w2

z2

1cr zCzCEI

GIkL

I

I

k

k

kL

EICM

1,, / MyyplwLTRdb fWM =

The elastic critical moment(Mcr) for LTB is

EBCS-3,Section 4.6.3P(78-84)

The values of the imperfection factor LT for lateraltorsional buckling should be taken as:LT = 0,21 for rolled sectionsLT = 0,49 for welded sections

Refer to the same table like those of membersunder compression and with: for rolled sections:

curve a ( = 0,21) for welded sections:

curve c ( = 0,49)

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The resistance of an unstiffened web to transverse forcesapplied through a flange, is governed by one of the followingmodes of failure:

1. Crushing of the web close to the flange, accompanied by plasticdeformation of the flange. See (a).

2. Crippling of the webin the form of localized buckling and crushingof the web close to the flange, accompanied by deformation of the

flange. See (b).3. Buckling of the webover most of the member. See (c).

Resistance of web to transverse forces

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Example:Bending members

A simply supported beam 7.0 m span is laterallysupported at the third points and carriesunfactored uniform loads of 18.5 kN/m imposed

load and 9.4 kN/m permanent load. In addition, thebeam carries unfactored concentrated loads of 50kN permanent load and 50 kN imposed load at midspan . Find a universal beam of grade Fe 430.

Solution:Geometry, materials and loads

Factored loads:

Imposed loads: q = 1.6 x 18.50 = 29.60 kN/m

Q = 1.6 x 50.00 = 80.00 kN.

Permanent loads: g = 1.3 x 9.40 = 12.20 kN/m

G = 1.3 x 50.00 = 65.00 kN.

Fe 430; fy = 275 N/mm2

(assume t 40 mm)

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Step 1: Maximum bending moment and shear

force:

Step 2: Required plastic modulus:

Try 533 x 210 x 92 UB.

( ) ( )

( ) ( )kN

xV

mkNxx

M

8.2182

6580

2

72.126.29max

8.5094

76580

8

72.126.29max

2

=+

++

=

=+

++

=

3

2

2

1

20391.1)/(5.27

)(108.509cm

cmkN

cmkNx

f

MW

My

el =

==

Step 3: Selection of the profile:

The relevant section properties are:

Class of section

The section is at least Class 2.

The relevant section properties are:

h = 533.1 mm w = 0.92 kN/m It = 76.2 cm4

d = 476.5 mm A = 118 cm2 Iw = 1.6 x 106 cm6

tf= 15.6 mm Iy = 55400 cm4

tw = 10.2 mm Iz = 2390 cm4

b = 209.3 mm Wel,y = 2080 cm3

Wpl,y = 2370 cm3

92.0275

235==

OKx

OKx

92.0837.462.10

5.476

92.01170.66.15

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Step 7: Check for lateral torsional buckling.

The section is Class 1 and

Determination of Mcr.

Lateral support to the beam is provided at the ends at thethird points. Therefore the effective buckling length is:

L = span/3 = 7000/3 = 2333 mm.

The critical moment for lateral-torsional buckling is:

Take C1 = 1.132; G = 80 Gpa

and

2.992.01070.66.15

3.209.1

,

, ====== LTandx

9250.0=LT

mkNmkNxxxxx

M Rdb >== 51754810

1.1

27510237019250.0 63

,

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Beam-columns are defined as members subject tocombined bending and compression. In principle,all members in frame structures are actually beam-columns

Beam-Column members

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Buckling Resistance of compressionmembers with moments.

The total stress due by the combined action of axial forceand bending moment is:

, then we can write

and finally:

Now, taking into account the problem of the loss of stability,

the design according with EBCS-3 is as follows:

1

,,

M

y

zMyMN

f

++

111

,

1

++My

Mz

My

yM

My

N

fff

1111

+

+ Myz

z

Myy

y

My fWM

fWM

fAN

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Example:A 4.00 m pin-ended column supports a beamwith a reaction of 100 kN permanent load and 150 imposedload. Assuming the beam reaction to be applied 75 mmfrom the face of the flange. Check the adequacy of a 203 x203 x 46 UC grade 430 steel profile.

Table 4.1

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Sec 4.5.4.3

Table 4.11 Tab.4.9

BM-Coln

Fig 4.13

BenMem

bding

Table4.12

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Built-up compressionmembers

Build-up members, made

out by coupling two or moresimple profiles for obtainingstronger and stiffer sectionare very common in steelstructures, usually forrealizing members whichare usually undercompression.

Two of the most commonarrangements for built-upmembers are representedin the figure to the right.

Although the discrete nature of the connections betweenthe two members connected by lacings and/or battenings,the models that will be described in the following foranalyzing and checking built-up members are based on theassumption that the member is regular and smearedmechanical properties (such as, flexural stiffness) can be

assumed throughout the member axis and utilized incalculations.

Consequently, some regularity requirements are usuallyimposed in designing these members and can be listedbelow as a matter of principles:

- the lacings or battenings consist of equal modules with parallelchords;

- the minimum numbers of modules in a member is three.

The key models which can be utilized for the stability

check of this kind of members will be discussed in thissection.

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Shear Flexibility of members and critical load:

While shear flexibility can usually be neglected in memberswith solid sections, built-up members are hugely affectedby these parameters as a result of the axial deformation oflacings and out-of-plane flexural flexibility of the chordmembers.

Consequently, the critical load of built-up members have tobe evaluated taking into account the role of shear stiffnessSv. By definition, the shear stiffness, Sv, is the force

resulting in a unit value of the shear deformation .

Since imperfections play an even important role in bothstrength and stability checks of built-up members, aconventional eccentricity e0 is usually introduced forsimulating their effect in amplifying the axial action N

Sd. For

instance, EBCS-3, 1995 provides the following value ofeccentricity as a function of the member span length L:

Nevertheless, it is worth emphasizing the role of shearflexibility on the value of eccentricity to be adopted inverifications; indeed, since second order effects are usuallyof concern, the following magnified value of eccentricity hasto be considered to take into account its total value:

The expression of the magnification factor considered inthe above equation is based on consideration of the role ofshear flexibility.

As a matter of principle, the above eccentricity of the axialforce results in an external moment M which can bedefined as follows:

Then the compression in one of the two connected

members can be estimated as follows:

where h0 is the distance between the centroids of the two

chord members.

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Laced compression members:

Laced members are made out of two chords connected bya bracing system with inclined lacings, in which each

segment of longitudinal profile between two braced nodescan be regarded as an isolated beam-column, whoselateral slenderness is considerably reduced at leastthroughout the plane of lacings.

The stability check, along with all the structural verificationdealing with members and connections, have to be carriedout by assuming an accidental eccentricity due to

imperfections. Consequently the design action in the singlechord of a laced member with total axial force is NSd can be

derived according to the following equation:

The value of the critical load Ncr has to be determined byneglecting the shear flexibility influence. Consequently, theusual expression can be considered:

where the effective moment of inertia Ieff is defined for oneof the two axes which does not cross all the connectedchord sections:

Af being the area of the single chord section and h0 the distance

between their centroids.

A virtual shear force VS, for the strength check of the

connections can be determined as a function of the aboveeccentricity e

0

:

Moreover, the diagonal members have to be checkedconsidering the following value for axial force Nd:

v

sd

cr

sd

sdss

S

N

N

N

eN

LL

MV

=

=

10

Laced

compression

member

Example:

Let us consider the laced member shown below stressed incompression under an axial force whose design value isNSd=3500 kN. The member is 10 m high and simply hingedat its ends. Chord members

are IPE 450 profiles while

diagonals consists of

steel plates with 60x12 mm2

rectangular section both made

out of grade Fe-360 steel.

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Cross-Sectional Properties:

Solution:

Stability check according to EBCS-3 provisions:

Step #1: classifying the transverse section:Since the thicknesses of the elements are less than 40mm, theyield strength of the steel used is fy=235 Mpa and the value =1.

The following values of the length-to-thickness ratios can beevaluated for flange and web:

Finally, the profile IPE 450 made out of steel FE-360 is in class3 if loaded in compression.

Step #2: evaluating the design actions:

The eccentricity e0 which has to be considered for taking intoaccount the possible imperfection effects is:

The effective value of the moment of inertia of the built-up

section Ieff can be also calculated as:

The shear stiffness for the current configuration will be given by

The elastic critical load Ncr can be then easily evaluated as:

in which the overall effective length L=10000 mm has been

considered since the calculation is aimed at deriving the totaleffect of the eccentricity e0 on the beam-column as a whole.Indeed, the moment MS induced by the eccentricity e0 can beevaluated as follows:

taking into account the magnification effect due to second orderdisplacements.

Finally, the actions on the various members can be easilyderived as follows:

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Step #3: performing the stability check of the chord members:

The reduction factor due to the slenderness

of the member has to be calculated by looking

after the possible instability in both the principal

direction y and z.

Step #3.1: calculation of y for instability in the z direction:

As far as the possible instability in the plane orthogonal to the y-axis (namely, buckling in z direction) is considered, the value ofthe effective length coincides with the overall span length of themember, since no lacings restraints buckling in the considereddirection, lying the diagonal members in a plane parallel to they-axis. Consequently Lo,y=10000 mm and the moment of inertiaof the single longitudinal chord member is Iy:

The non-dimensional slenderness can be derived as a functionof the elastic critical load Ncr,y as follows:

The profile follows the curve a and, consequently, the followingvalue of the reduction factor y can be evaluated:

Step #3.2: calculation of z for instability in the y direction:

On the contrary, as far as the possible instability in the plane

orthogonal to the z-axis (namely, buckling in y direction) isconsidered, the value of the effective length coincides with thediagonal spacing, since buckling in y direction is forced bylacings to develop only between two adjacent nodes.

Consequently L0,z=1000 mm and the moment of inertia of the

single longitudinal chord member is Iz, have to be determinedaccording to:

The non-dimensional slenderness can be derived as a function

of the elastic critical load Ncr,z as follows:

The profile follows the curve b and, consequently, the followingvalue of the reduction factor y can be evaluated as:

Step #3.3: calculating the axial bearing capacity :

Since y


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Battened members:

Battened members are widely utilized as a technologicalsolution for realizing beam-columns in industrial buildings.

Due to the significant flexural stiffness of battenings and therelated connections with the chord members, the nodes

between the longitudinal profiles and the horizontal battenings isusually assumed as completely fixed, rather than hinges asusually considered for laced members.

Consequently, the lateral shear flexibility of this kind of members

can be determined considering the following three contributions:- bending in longitudinal members;

- bending in battening;

- shear strains in battenings.

Since battenings are usually assumed infinitely stiff with respectto the chord sections, EBCS-3 provided a lower bound for theirmoment of inertia Ib with respect to one of the chord member

and other geometrical parameters:

..abeing the battening spacing.

The compressive force Nf,Sd

acting on the single chord membercan be determined as follows for taking into account the effectof eccentricity e0 due to imperfections:

where the moment Ms is defined as follows:

and the effective moment of inertia can be estimated as follows:

The parameter is basically defined as a function ofslenderness :

where the mentioned slenderness is defined as:

and

with I1 equal to Ieff assuming =1

The elastic critical load Ncr can be evaluated according to the

following expression:

In the general case, the shear stiffness Sv has to be evaluatedas follows:

The shear force to be considered in local verifications according

to the equilibrium conditions can be evaluated as alreadydescribed for laced members.

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Finally, EBCS-3 classifies the battened members on the basis ofthe distance between the longitudinal chord members. Inparticular, for closely spaced members, the above provision

does not apply and the general procedure given for the usualmembers described in the previous chapters can be applied.

Examples of closely spaced battened members are representedin the figure below and considering different types oflongitudinal profiles.

CHAPTER

END!

Steel Chap5 Ultimate Limit State

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