Structural Engineering Imperfections

7/27/2019 Structural Engineering Imperfections

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Im erfections

Introduction into analysis

global sway imperfections

bracing system imperfections Example Portal frame

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Geometrical Im erfections

Variance of dimensions of a structure or a memberb

flatness of a member

e0

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Geometrical Im erfections

N

L eo,d

Frame imperfection Member Imperfection

Frame imperfection always to be allowed for

Member imperfection: only for slender members (rare)

,

covered in the relevant buckling

curves

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Material & Structural Im erfections

Residual stresses: fy

Distribution in a I-section rr

Apply load eccentricities in joints of the structure

e0

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In order to cover the effects of imperfections, engineers

need to make appropriate allowances in the structural

analysis.

The following imperfections should be taken into account:

Cover lack of verticality for frames or straightness

of structure restrained b bracin s

Local imperfections for individual members

Cover lack of straightness or flatness of a memberand residual stresses of the member

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rame mper ec on can

be replaced by an

e uivalent closed s stem

F1

of horizontal forcesapplied at the floor levels

F2nc u ng e oun a on

level).F2

h

For building frames sway

imperfections, ,

may be disregardedwhere:

(F1+F2)/2 (F1+F2)/2

Equivalent forcesEdEd V15.0

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The factor can be calculated from equation (5.5) of EN 1993:

mh

0=

Where, 0 is the basic value:h is the reduction factor for height h applicable to columns:

13

2but

2=

hh

h

is the height of the structure in meters

m is the reduction factor for the number of columns in a row:

+= mm 15.0

s e num er o co umns n a row

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equivalent closed system of horizontal forces, introduced

for each column.

EdEd

L

eNEd 04

2

08

L

eNEd0e

eNEd 04

NEd NEd

L

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of members for flexural buckling e0/Lare given in Table

5.1 in EN 1993.

ere, s e mem er eng

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Usuall local im erfections are i nored in lobal

analysis and covered by reduction factors and LT in

member checks.

for the frame is sensitive to 2nd order effects, local bow

im erfections should be incor orated into structural

analysis when:

A member has at least one moment resistant end joint;

The slenderness

Ed

y

N

f5.0>

W ere, NEd s t e es gn va ue o t e compress on orce

is the slenderness for the member considered as .

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Bracing systems may provide lateral stability of a strut in

compression or a beam in bending

For members to be restrained, the strut/beam should beconsidered with a geometrical imperfection (initial bow)

an initial bow imperfection.

=

Where, Lis the span of the bracing system

0 m

+= mm 115.0

In which m is the number of members to be restrained

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The bow with amplitudee0 may be replaced by transverse

uniform loadingqd

he loading qd corresponds to impact of sum of theamplitude e0 and the in-plane deflection of the bracing

q.

e0NEd NEdMember in

q08e

Nq

compression

(or compression

flan e force

of a beam)

L

Bracing

system

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-

he members of the ortal frame are ro osed as made of

rolled rods (beam: 533210 UB109, column: 305305

UC158).

hr=2000

=

L=24000

Figure 1

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The combinations of loads are anal zed refer to Fi ure 2:

self-weight & imposed load and wind load

w = 10kN/m

40kN40kNim 1

Figure 2

+3kN/m -2kN/m

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For formulas see EN 1993-1-1

( ) kNVkNHEdEd

4880241015.015.04085 =+=


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Global imperfections:

kN96.0)802410(003.01imp =+== V

Internal forces (loading + imperfections):

NEd (kN) VEd (kN) MEd (kNm)

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Local imperfections for global analysis only if simultaneously

(column concerned):

One end of the member features a rigid connection: OK

Slenderness for systemic length is greater than:

8.21024.168

5,05,03

Ed

y =

=>N

In this case,

8000crh

(this condition is not met)8.256.094.09.939.931


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Structural behaviour

Types of structural analysis

2nd order analysis

Elastic global analysis

as c g o a ana ys s

Simple modelling of structural analysis

Example - 2nd order effects in portal frame

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Structural Modelin for Anal sis

BeamsJoint

-

Joints

Beam-column

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Framing and joints

Continuous framing: rigid joint

Semi-continuous framing: semi-rigid joint

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Load

Load parameter

Peak load

Frame Elastic limit

Displacement parameter

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c ua response o e rame s non near

Linear behaviour limited

Non-linear behaviour due to:

(second order effects)

Material yielding

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Sway

Load

D sp acement

HEdHEd

NEd NEd

xx

h

FrameM(x) = HEdxM(h) = HEdh

M(x) = HEdx + p + P (x/h)M(x) = HEdh + P

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P- effect :

due to floor swa

1st order frame stiffness modified

dominant effect

P- effect : due to beam-column deflection

1st order member stiffness modified

significant only for relatively slender members which is

rare

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Aims of global frame analysis

Determine the corresponding deformations

Means Adequate models incorporating assumptions about the

behaviour of the structure and its com onent:

members and joints

Basic principles to be satisfied:

Equilibrium throughout the structure

ompa y o e orma on e ween e rame componen s

Constitutive lawsfor the frame components

Frame model - element model

must satisfy the basic principles

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T f St t l A l i


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Type of Structural Analysis

The internal forces & moments may be determined usinge er:

First order analysis:

,

Second-order analysis:

deformation of the structure, e.g. both the sway

effect (P- effect) and the P- effect (memberdeflection effect).

he second-order effects should be considered if they

increase the action effects significantly or modify.

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Elastic global analysis should be based on the

assumption that the stress-strain behaviour is linear,

.

,

if the resistance of the section is based on its plastic

resistance.

Elastic global analysis may also be used for cross

sec ons e res s ance o w c are m e y oca

buckling.

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Elastic Global Analysis


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Elastic Global Analysis

General types of Elastic analysis:

LA: Linear elastic analysis;

LBA: Linear bifurcation anal sis

GNA: Geometrically non-linear analysis;GNIA: Geometrically non-linear with imperfections analysis.

Simplified scheme of elastic analyses:

LBAFcr

e

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Method of Structural Analysis


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1st order elastic analysis Indefinite linear elastic response of member sections and of

joints

configuration

MElastic

ElasticMj

Mj

MM

Moment rotation characteristic of the section Moment rotation characteristic of the joint

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Load parameter 2n order elastic analysis

cr

-

elastic response ofmember sections ando n s

Equilibrium

deformedstructure

-

n or er e as c ana ys s

and, if necessary, forP- effect

Displacement parameter

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Plastic lobal anal sis ma be used where:

the members are capable of sufficient rotation capacity to

enable the re uired redistributions of bendin moments to

develop.

the stability of members at plastic hinges can be assured.

Rigid plastic analysis may be applied if no effects of the

deformed geometry have to be considered.

Frigid-plastic analysis

non-linear plasticelastic-plastic analysisplastic hinge

Simplified scheme of

plastic analyses

fibre plasticity

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Method of Structural Analysis Rigid-plastic global analysis

Rigid-plastic member section behaviour

Rigid plastic Rigid plastic

Rigid-plastic joint behaviour when plastic hinges are allowedthere

M

Mpl.Rd Mpl.Rd

Mj,Rd p

M l.Rd

p j

Plastic hingeM

j,RdPlastic hinge

Moment rotation characteristics of the member Moment rotation characteristics of the joint

p p

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Method of Structural AnalysisLoad parameter

1-

Plastic mechanism

3

2

Critical collapse load

LRP3

analysis

W

H

B D B D

h

H

W

w

Easy application for

sim le frames e. .

Beam mechanism

1

Swa mechanism

2

A

C

E A E

industrial portal frames

Serviceabilit deflection

W

H

BD

h

w

check

3

plastic hinge location

A

C

E

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Elastic-perfectly plastic global analysis

nd

Deterioration of frame stability as plastic hinges form

M Mj.Rdj

M

M Mpl.Rd

Plastic hinge.

Plastic hingepl.Rd

Elastic-perfectly plastic response of

member sections

Elastic-perfectly plastic response of

joints

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Simple Modeling of Structures


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First-order analysis may be used, if the following criterion issatisfied:

For elastic analysis: 10=Ed

crcr

F

F

.

For plastic analysis: 15=Ed

crcr

F

F

Where,

or using Eq. 5.2 in EN 1993-1-1.

cr

is the design loading on the structure

is the elastic critical buckling load for global instability

Ed

crF

mode based on initial elastic stiffnesses.

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Simple Modeling of structures


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Simple Modeling of structures

A approximative formula for cr given by Clause 5.2.1(4)

Portals with shallow roof slopes;

Beam and column frames

(5.2)EdH,

VEd

Ed

cr=

,

HEd is the horizontal reaction at bottom of the storey

VEd is the total vertical load at bottom of the storey

H,Ed is the storey sway when frame loaded with

horizontal loads

h is the storey height

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Limits on portal frames given in NOTE 1B & 2B of Clause5.2.1:

Note 1B: the portal rafter slope is not steeper than 1:2

(26) Note 2B: the axial load in the rafters or beams is not

significant. when

EdN. or rearrange o .cr

NEd

y

N3.0

, cr

2EI=

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10cr

= FF

cr ifStrength design using 1st order elastic analysis

Axial loading is so low that no instability of membersor the frame would occur.

Neglect equivalent imperfections or fictitious forces.

From EC 3 cl. 6.3.1.2 simple compression (1) is

appropriate for a member with Ncr/NEd 25.

Need to check EC 3 equation (5.3) for in-plane beamAf

EdN,

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p g

If 103


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p g

If 3


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p g

,

such as

Strength design using 1st order plastic collapse analysis.15

cr

be ignored in the analysis.

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nd Order Effects in Portal Frame


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Order Effects in Portal Frame

he framing spacing is 10m c/c, and 1/200 equivalent

horizontal load was applied.

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Method 1: Buckling Analysis


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g y

theoretical buckling strength of the portal frame.

an eigenvalue buckling analysis of a column will matchthe classical Euler solution.

The design load and constraints were used to calculate

the eigenvalue, because each load has an associatedbuckled mode.

he software ABAQUS was used to analysis the

buckling model

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Eigenvalue from buckling analysis

crFloadBuckling

Eigenvalue ===

Ed

oaes gn

he boundary conditions:

column base is pinned; All the beam-column joints are rigid;

The torsion & out-of-plan rotation of rafter is constrained;

the out-plan displacement of crane beam is limited.

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Buckling Modes


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3 first buckling modes are shown for demonstrationIncluding horizontal loads to simulate global imperfections

.

1,

=

cr

.

,2

=

cr

.,3=

cr

Beam, Column, Column,

in-plane buckling Out-plane buckling Out-plane buckling

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Method 2: Eq. 5.2 in EC3


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Limits on portal frames using formula 5.2

Slope of rafter: (15.12-13):21.5=1:10 (5.5) is less than

26, the formula is available.

The largest axial load of rafter is 50kN, and the elastic

critical load of thus rafter is 1843kN.

So,09.0027.0

1843

50


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Taking Equivalent Horizontal Loads into account to simulate

global imperfection:

kN3.11200

2256

200

loadverticalEHF ==

=

2.2kNF1detailedFor

==

.

F2

F1

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Method 2: Eq. 5.2 in EC3


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H,Ed is calculated by the software, and

mm49EdH, =

So,33.1

130001===

hHEd

cr

EdH,Ed

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Method 3: BS 5950-1


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This method is based on the plastic analysis, and wasused in BS 5950. (see reference)

For simplified, the axial load is calculated by software.

2 9

2342

37

492

316

D E

250

558B

Axial load (kN)A

F

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-


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For the left hand bay (ABC):

Ic = 14300cm4 Pc =250kN

= 4 =r r

Lr= 16.078m

276.1422.8

078.16

21400

14300===

h

L

I

IR r

r

c

2.1

3

= r

cr

EI

0.2102142103

.

9

==

rrcr

R

16078200003.0842210250276.1

.116078 3

+

+

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-


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For the right hand bay (DEF):

Ic = 126000cm4 Pc =505kN (conservatively)

= 4 =r r

Lr= 20.36m

8.413

36.20

41100

126000===

hIR r

r

c

3.02.1

1

3

+

+

=

rrcr

r

cr

LphpL

I

51.11041121039

==20360340003.01013505

8.4

.120360 6

+

+

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, ,


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Three method were used to calculate the factor cr

Method 1:29.1=

cr

Method 2:

Method 3:33.1=

cr51.1==

crcr

e ac or cr s ess an , an e secon or er e ec smust be considered in design.

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Shear resistance

Ben ng moment res stance

Compression resistance.

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The design shear force is denoted by VEd (shear force design

effect).

The design shear resistance of a cross-section is denoted by

c, p ,

elastic distribution of shear stress.

The design shear force, VEd, should satisfy:

1.0V

Ed

,

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p ,


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p ,

e usua approac s o use e p as c s ear res s ance

Vpl,Rd in practice

The plastic shear resistance is essentially defined as theyield strength in shear multiplied by a shear area A

The yield strength in shear is related the yield strength in

yf

A

M0

V

Rdpl,

3V

=

.0=M

For BS 5950, Pv = 0.6pyAv

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v


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The shear area Av is in effect the area of the cross-

section that can be mobilized to resist the design

redistribution.

For sections where the action is applied parallel tothe web, this is essentially the area of the web (with

some allowance for the root radii in rolled sections).

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v


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EC 3 BS 5950Rolled I and H sections,load parallel to web

Rolled I, H and Channelsections, load parallel to web

wwfwfv r ++=

SS)toNto(refer1.0= Av = tD

Rolled channel sections,

trtbtAA 2 ++=

oa para e o we

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v


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EC 3 BS 5950

e e , ox sect ons,load parallel to web

= th SStoNtorefer1.0=

Welded I, & box sections,load parallel to web

Av = td

Welded I H & box sections

=

wwv

thAA load parallel to flange

Av =2 td

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v


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EC 3 BS 5950

Rectangular hollow sections,load parallel to depth

Rectangular hollow sections,load parallel to depth

)( hbAhAv += / Av =AD(D+B)

Rectangular hollow sections,

)( hbAbAv += /

oa para e to w t

Circular hollow sections

/2=

=

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v


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- -bending) and member buckling resistance must bechecked, i.e.

Cross-section check (In-plane bending)Rdc,Ed MM

Member-buckling checkb,RdEd MM

MEd = design bending moment

Mc,Rd = design bending resistance about one principal axis

of a cross-sectionM = desi n bucklin resistance moment,

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EC 3 BS 5950

Class 1 & 2 cross-sections: Class 1 & 2 cross-sections:

=fW

c y . y

0M

Rd,plRd,c MM

==

C ass 3 cross-sect ons: C ass 3 cross-sect ons:

Mc = pyZ

ymin,le

Rd,ElRd,c

fWMM ==

Class 4 cross-sections:

or

Mc =pySeff1.2pyZ

0M

Class 4 cross-sections:

M = Zymin,eff

Rd,cfWM =

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Subscripts are used to differentiate between the plastic,

elastic or effective section modulus.

Plastic modulus Wpl (S in BS 5950)

Elastic modulus Wel (Z in BS 5950)

Effective modulus Weff (Zeff in BS 5950)

The partial factor M0 is applied to all cross-section bending

resistances , and equal 1.0.

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When the design value of the shear force is less than 50% of

the design plastic shear resistance, i.e. VEd0.5 Vpl,Rd, its

.

When the design value of the shear force exceeds 50% of the

design plastic shear resistance i.e. VEd> 0.5 Vpl,Rd, the yield

strengthfyshould be reduced by (1 r) in the determination

, c,Rd.

yyr ff )1( =

2

12 = EdVwhereRdpl,

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EC 3 BS 5950When, VEd > 0,5Vpl,Rd

When, Fv > 0.6 Pv

available to determine thereduced design plastic

or ass or cross sect ons

Mc =py (S Sv)

res stance moment or c ass 1

and 2 I-sections. For Class 3 semi-comp

sections

Mc =py (S Sv/1.5)

0

)0.25(

M

ywwRdypl

RdyV

ftAWM

/,,,,

=

or ass s en er sec ons

Mc =py (Zeff Sv/1.5)RdycRdyV MM ,,,, But

Where, = [2Fv/Pv 1.0]2

, www2

12

= EdV

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Rdpl,


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follows:

RdtEdt NN ,, (Tension check)

Nt,Ed is the tension design effect

N is the design tension resistance,

Design tension resistance Nt Rd is limited either by:

Yielding of the gross cross-section Npl,Rd

- u,R

Whichever is the lesser.

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-


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or mem ers connec e y we ng

Desi n tension resistance N is e ual to the ield,

strength of the cross-section, Npl,Rd

,

y

Rdpl

AfN =

A is the gross area of the cross-section

M0 = 1.0 from Singapore NA

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-


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For members connected b boltin , N is reduced due to,holes and is the lesser of:

1. Yield resistance of gross section

0

Rdpl,y

AfN

=

2. Ultimate strength of net section

M2

unet

Rdu, 9.0

fAN =0.9 is a reduction

factor for

ne the cross-section

eccentricity, stress

concentration etc

u

tension strength

M2 is the partial factor for fracture, and

M2 = 1.10 in Singapore NA

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The Net Area of the cross-section is the

gross areadeduction for bolt holes

For each fastener hole, deduction is the

ross cross sectional area of the hole

Angles connected through one leg;

T sections and channels connected through outstands

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-


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When the fastener holes are not sta ered, the total area tobe deducted for fastener holes should be the maximum sum

of the sectional areas of the holes in any cross-section

.p1

e2=p2

net 0B

n = number of holes0 ae c ness t

t

net

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Wh f t h l t d th t t l t b d d t d i th


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When fastener holes are staggered, the total area to be deducted is the

1. The area for holes crossing a perpendicular cross section

2. the sum of the areas of all holes in an dia onal or zi -za line acrossthe member less s2t/4p for each gauge space in the chain of holes

s

s

d0 b a

- , 0

On section a-b, Net Area = A 2d0t s2

t/4p

Net cross-sectional area is the lesser of:tnnet 0=

=

s2

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ne

4

BS 5950 EC 3


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BS 5950 EC 3

Tension resistance

RduRdplRdt NNN ,,, ;min=

Tension capacity

AP =

0

Rdpl,

M

y

AfN

= Ae is the sum of the effective net area ae

M2

unet

Rdu, 9.0

fAN =

e n g

For grade S275: Ke = 1.2

tndAAnet 0=For grade S355: Ke = 1.1

For grade S460: Ke = 1.0

tndAAnet 0=

2

Staggered

fastenersMIN

e s . y

a is the ross area of the element

= p

ndtnet4

0an is the net area of the element

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The desi n value of the com ression force N ateach cross-section shall satisfy:

RdcEd ,

For Class 1, 2 & 3 cross-sections

Unaffected by local buckling

Design compression resistance Nc,Rd equals theplastic resistance Npl,Rd:

0M

y

Rdc

N

=,

M1 = 1.00, in NA to SS EN 1993-1

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For Class 4 sections: Local buckling of Class 4 prevents the attainment of

the squash load

es gn compress on res s ance m e o oca

buckling resistance,

0M

yeff

RdcN

=,

Aeffis the area of the effective cross-section

If Class 4 section is unsymmetrical, it has to be designed

as beam-column due to the additional moment arising.

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Web bearing and buckling

Resistance to transverse forces

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Web Bearing and Web Buckling are modes of failure that

arise from concentrated forces being transversely applied

.

Web bearing (a) and buckling (b) failure modes

a b

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Plate Stiffener


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79CV3202 CSP 08 - 2012 79DOUBLE STRUT WITH SPLAY

Strut-Waler Connection with Plate Stiffener

Plate Stiffener


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Waler Flange

StrutWaler

80CV3202 CSP 08 - 2012Waler Web


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most vulnerable location, close to the root radius at the

junction of the web and flange where the force is

concentrated.

Buckling happens when the web behaves like a slender

. ,is assumed that the web is adequately restrained by the

flange to which it is connected in the lateral direction, and

therefore, the web can neither rotate nor move laterally.

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Design Treatment is very different from

current Approach


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- requ res a wo n epen en c ec s are carr e

for web bearing and buckling.

EC3 resents a sin le check to deal with these two failure

modes. This single check accounts for the bearing and

uc ng o t e we w en t e mem er s su ecte to a

transverse force.

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Resistance to Transverse Force


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EC1993-1-5 distinguishes between two types of forces applied

through a flange on to the web:

i) forces resisted by shear in the web [load types (a) and (c)]

flange [load type (b)].

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Load T es & k

Buckling coefficients for different types of load application


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If there are no web stiffeners, a=, hw/a=0

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Web Failure

For load type (a) and (c) the web is likely to fail as a result of:


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i. Crushing of the web close to the flange accompanied by

yielding of the flange.

ii. Localised buckling and crushing of the web beneath the

flange, the combined effect sometimes referred to as web.

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Web Failure For loading type (b) the web is likely to fail as a result of:


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. e rus ng;

ii. Buckling of the web over most of the depth of the member.

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Web Bucklin Failure


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Resistance to Transverse Force

For unstiffened or stiffened webs the design resistance to


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tf

1

RdF

=

Where, tw is the thickness of the web;

yw

M1 is the partial factor for resistance of members

(M1 = 1.0 in SG National Annex)

Leff is the effective length of web.

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Effective Len th for Resistance

h ff ti l th f b h ld b d t i d f


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he effective len th of web should be determined from:

eff= f y

where, f is the reduction factor due to local buckling

y s t e e ect ve oa e engt , appropr ate to t elength of the stiff bearing ss.

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Stiff Bearin Len th

The stiff bearing length ss

on the flange should be taken as the


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sdistance over which the applied load is effectively distributed

at a slope of 1:1

ss is less than hw

90 CSP 08 - 2012

Stiff Bearin Len th

The stiff bearing length is not given in EC3; it can be calculated


91/106

r s

Ss Ss

ss=t+1.6r+2T ss=t+1.6s+2T

91 CSP 08 - 2012

Stiff Bearin Len th

If the bearin surface of the a lied load rests at an an le on


92/106

If the bearin surface of the a lied load rests at an an le onthe flange surface, as shown below, ss should be taken as

zero.

92 CSP 08 - 2012

Effective Loaded Len th


93/106

y

calculated as follow:

For type (a) and (b)

but l distance between adjacent transverse

21 mms fsy =

stiffeners.

93 CSP 08 - 2012

Effective Loaded Len th

y


94/106

y

value of the following equations.

2

2

1 mlmtll efey +++=

mmtll ++=

2

Where, cshf

l s

wyw

wFe +=

2

94 CSP 08 - 2012

Non-Dimensional Factors m & m


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wyw

fyf

tfm =1

where,fyf is the yield strength of the flange

fyw is the yield strength of the web

2

5.0if02.02 >

= Ff

w

tm

5.0if02


96/106

F

w bh 2

sFaa

= .-.

, 1

distance between the loaded flange and the stiffener

This equation is valid for 3.0and,3.005.0 11 bb

And loading according to Type (a)w

96 CSP 08 - 2012

Lon itudinal Stiffener k


97/106

+

= baIsl

s1

3

3

1,-3.0210139.10

awww

Where, Isl,1 is the second moment of area of the

stiffener closet to the loaded flan e includin

contributing parts of the web according.

97 CSP 08 - 2012

Reduction Factor

The reduction factor should be obtained from:


98/106

The reduction factorF should be obtained from:

0.15.0

=F

F

ywwy ftl

crF F=,

w

w

Fcr

h

tEkF 9.0=

98 CSP 08 - 2012

Verification under Transverse Force

The verification should be performed as follow:


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The verification should be performed as follow:

0.12 =weffyw

d

tLf

1M

, e Leff is the effective length for resistance to transverse

forces;

tw is the thickness of the plate.

99 CSP 08 - 2012

Verification under Interaction

If the girder is subjected to a concentrated transverse


100/106

If the girder is subjected to a concentrated transverse

force and bending , the combined effect should satisfy

the following:

4.18.0 12 +

Where, 0.11 =ply

Ed

Wf

M

0M

100 CSP 08 - 2012

Example - Resistance to Transverse

Force

Determine the resistance to transverse force of a 406 140 UB39


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Determine the resistance to transverse force of a 406 140 UB39

Grade S355 steel section under the type of load application shown

below:

ss = 50 mm

c = 2725 mm

5500 mm

101 CSP 08 - 2012

b


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b

z h = 398.0mm

r

f

b = 141.8mm

=

h dhw

w .

tf= 8.6mm

tw

s

r = 10.2mm

z

tf= 8.6mm


103/106

The load application is type (a)

,

a=, hw/a=0

2

=

= hw

aF

103 CSP 08 - 2012

-

tw3

=

hwFcr .


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hw

4.6 3

780649

8.380.

=

2.228.141355

1 =

== fyfbf

m

.wyw

5.0Assuming >F

39.2

8.380

02.002.0

22

2 =

=

= wh

m .

104 CSP 08 - 2012

y

( )


105/106

( )mmtsl fsy 12 21+++=mm2022.392.2216.8250 =+++=

ywwy ftl

3554.6202

crFok

..780649

==

105 CSP 08 - 2012

5.0


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5.0=

F

F

0.165.077.0

5.0

== ok

mmlL yFeff 13120265.0 ===

tLfF

weffyw=

kN

M

298104.6131355 3-

1

==

.

106 CSP 08 - 2012

Structural Engineering Imperfections

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