ec3 - steel-prof walker

7/30/2019 EC3 - Steel-Prof Walker

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Department of Architecture & Civil Engineering


EUROCODE 3


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References

Trahair NS, Bradford MA, Nethercot DA andGardner L (2009). The behaviour and design

of steel structures to EC3. 4th Ed. Taylor &Francis

IStructE Design Manual to EC3

Arya C (2009) Design of structural elements.3rd edition. Spon.


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BS EN 1993: Eurocode 3:Part 1.1-Design of steel structures, general rulesand rules for buildings

1.1 Design of steel structures,general rules and rules for

buildings

1.2 Fire Resistance

1.3 Cold formed thin gauge members

1.4 Stainless steel

1.5 Plated structural elements

1.6 Strength and stability of shell

structures

1.7 Strength and stability of planar

plated structures transversely

loaded

Parts 1.8 Design of joints1.9 Fatigue strength of steel

structures1.10 Selection of steel for fracture

toughness and through thickness

properties

1.11 Design of structures with tension

components made of steel

1.12 Supplementary rules for highstrength steel


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Introduction

Basis of design

Materials

Serviceability limit states

Ultimate limit states

Connections

Fabrication

Deign assisted by testing

Fatigue

Volume 1

Volume 2

Lots of Annexes

Symbols, Wel

Grades? E, G

Deflections

Calculations

Chapters based on design

criteria not element type as

in BS 5950

1.1 Design of steel structures,

general rules and rules for

buildings

Important data and National

Application Documents

(NAD)

EC 3 Layout


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Member axes A matter of consistency

with the other Eurocodes

Member definition


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Reading Symbols Schematic but cumbersome!

Rd.y.plMabout the y-y axis (thats the

major one)

Design Plastic Moment of Resistance

New Symbols!


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Limit States Two principallimit states,ULS and SLS with the samemeanings as BS5950.

Clause 2.3.2Clause 2.3.2

Static equilibrium

Rupture or excessivedeformation

Transformation intomechanism

Instability due to 2nd

ordereffects( sway)

Fatigue

Accidental damage

Deformations which affectappearance

Vibration/sway which causediscomfort

Damage to finishes

PLUS a 3rd limit

state of Durability

Roughly the same as Table 4.1 BS 5950


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ACTIONSnot LOADS

EC1: Actions on structures

Fd = gf FkDesign action

Partial safetyfactor

Characteristicaction

BS 5950

Dead load (gg=1.4)

Live load (gq=1.6)

EC 3

Permanent actions (gg=1.35)

Variable actions (gq=1.5)


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Cross section Resistance

not

Material Design Strength

EC3: Clause 2.2.3.3

Rd = Rk/ gMDesignresistance

Partial safety factorfor the resistance

Characteristicresistance

gM is not the same as gm(BS5950, generally 1.0)

gM0 Resistance of section = 1.00

gM1 Buckling resistance = 1.00

gM2 Net section resistance = 1.10/1.25


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Classification of sections

Depends upon the

proportionsof each of its

compression elementsClass 1 Cross sections plastic

Class 2 Cross section compact

Class 3 Cross section semi-compact

Class 4 Cross section slender

EC3 BS 5950

c/tffor flanges

d/twfor webs

These are defined

differently from BS5950


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Department of Architecture & Civil EngineeringSection classification

Maximum width to thickness ratios for compression elements

For rolled sections EC3 is

More onerous for webs

Less onerous for outstand flanges

Type of elementClass of element

Class 1 Class 2 Class 3

Outstand flange

Web (neutral axis

at mid depth)Web subject to

compression

fy

e

c/tf10ec/tf 11e c/tf 15e

d/tw 72e d/tw 83e d/tw 124e

d/tw 33e d/tw 38e d/tw 42e

235 275 355

1 0.92 0.81

e = (235/ fy) 0.5


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Material CoefficientsSteel got stiffer (+2.4%)

210GPa versus 205GPa

Characteristic

strengths

Materials


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Design of fully

restrained beams

Resistance of cross section to bendingand shear

Shear buckling resistance Flange induced buckling

Resistance of web to transverse forces(web buckling)

Deflection (SLS)


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Design of laterally restrained beam (EC3)

Check the suitability of 356 x 171 x 51 UB

section to span (simply supported) 8.0 metres

in S275 steel loaded by UDLs gk= 8 kN/mand qk= 6 kN/m.

Assume beam is fully laterally restrained and

beam sits on 100 mm bearings.

(Arya, 3rd Edition, Example 9.1)


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Design moment

FEd = (1.35 x 8 + 1.50 x 6) x 8.0

= 158.4 kN

MEd= FEd.L/8

= 158.4 x 8/8

= 158.4 kNm


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Section classification

Flange thickness tf= 11.5 mm, hence fy(S275) = 275 N/mm2.

e = (235/fy)0.5 = (235/275)0.5 = 0.92

Flange outstand, c = (b

tw-2r)/2 = 71.9 mm

c/tf= 71.9/11.5 = 6.25 < 9e (= 8.28)

Web (d= 312.3 mm)

d/tw= 312.3/7.3 = 42.8 < 72e (= 66.2)

Therefore section CLASS 1.


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Moment of resistance

Plastic and Compact sections

BS5950 EC 3

Class 1&2 sections

Mc=pyS

Reduction for shear if

Fv


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Shear resistanceBS5950 EC 3

Pv=0.6 PyAv

Shear Buckling if

Av =A-2btf+(tw+2r)tf

d/t > 70 e

Vpl,Rd= Av (fy/3)/g M0

Av =tD

Check shear

Buckling if

hw/tw > 72 e/h(h = 1 mostly)


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Cross-section resistance

Bending moment:

Mpl,Rd = (Wpl,yfy)/gM0

= (895 x 103

x 275)/1.0= 246.1 kNm

> MEd(=158.4 kNm) OK


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Shear resistance:VEd= FEd/2 = 79.2 kN

For class 1 section use plastic shear resistance:

Vpl,Rd= Av(fy/3)/gM0.Av = A 2btf+ (tw+ 2r)tfhhwtw(=2428 mm2)

= 64.6 x 102 2 x 171.5 x 11.5 + (7.3 + 2 x 10.2) x 11.5

= 2834 mm2

Hence, Vpl,Rd = 2834 (275/3)/1.00= 540.4 kN > VEd


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Flange induced buckling

hw/tw k (E/fyf)[Aw/Afc] 0.5

where:

Aw= web areaAfc= compression flange area

fyf= compression flange yield strengthk= 0.3 (class 1); 0.4 (class 2); 0.55 (class 3 and 4)


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Resistance of web to transverse forces

1 Crushing of webs + plastic deformation of flange

2 Crippling and crushing web + plastic deformation of flange

3 Buckling of web over most of depth

Force resisted by web shear

Force transmitted

through web

Use smaller of 1 or 2 Use smaller of 1 or 3


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Deflections

BS5950 EC 3

Imposed load only

Span/360 Brittle finishes

Span/200 General

Permanent action, d1

Variable action, d2

Pre-camber, d0

dmax

=(d1 +d2d0)


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Compression

members

Resistance of cross sectionto compression

Buckling resistance ofmembers

Nc.Rd= A fy/gM0Nc.Rd= Aefffy/gM0

BS5950 EC 3

Pc.= Ag pc

A function of theslenderness l

Quite a few tables !

Nb.Rd=cA fy/g M1A reduction factor

pronounced (Hee)

c depends upon thenon-dimensional

slenderness l


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Compression members-Buckling Resistance

Nb.Rd=cA fy/g M1The process to get c

Calculate the slenderness

Calculate the non

dimensional slenderness

Calculate phi

l (Afy/Ncr)0.5

f 0.5(1+ a(l-0.2)+l2

Or Table

5.5.2!

Look up imperfection factora

f + [f2

-l2

]

0.5

1c

l (Lcr/ icr)(1/93.9e)


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Example: Column subject to axial load

Check suitability of 203 x 203 x 60 UCin S275 steel to resist design axialcompression force of 1400 kN.

Assume column is pinned at both endsand is 6.0 m metres high.

(Arya, 3rd Edition, Example 9.7).


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Section properties

From steel section tables:

Area of section, A = 7580 mm2

Flange thickness, tf= 14.2 mm Radius of gyration about major axis, iy= 89.6 mm

Radius of gyration abut minor axis, iz= 51.9 mm


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Section classification

Strength classification

tf= 14.2 mm, steel grade S275

Hence, fy= 275 N/mm2Section classification

e = (235/275)0.5 = 0.92c/tf= 87.75/14.2 = 6.18 < 9e (9 x 0.92 = 8.28)

d/tw= 160.9/9.3 = 17.3 < 33e (33 x 0.92 = 30.36)

Hence section class 1.


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Compression resistance of section

Nc,Rd = Afy/gM0= (7580 x 275)/1.00

= 2084 kN > NEd= 1400 kN


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Buckling resistance

Effective length of column about both axes:

Lcr = Lcry= Lcrz= 1.0L= 6000 mm

Minor axis therefore more critical

Relative slenderness l1 given by:l1 = 93.9e = 93.9 x 0.92

= 86.4

Non-dimensional slenderness about minor axis:= ((Afy)/Ncr) = (Lcr/iz).(1/l1)= (6000/51.9) . (1/86.4)

= 1.33

l


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Determine buckling curve:h/b= 209.6/205.2 = 1.02 < 1.2

tf = 14.2 mm < 100 mm

Use buckling about z-z axis buckling

curve c. Use a = 0.49


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_ _ = 0.5 . [1 +a(l- 0.2) + l2]

= 0.5 . [1 + 0.49 x (1.33 0.2) + 1.332]

= 1.66 _

c = 1/[+ (2 -l2)]= 1/[1.66 + (1.662 1.332)]= 0.377


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Hence design buckling resistance:

Nb,Rd =cAfy/gM1

= (0.377 x 7580 x 275)/1.00

= 785.8 kN < 1400 kN

Therefore section is unsuitable.


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Lateral Torsional Buckling of beams

Designer must

ensure

Mb.RdMSd

Similar process as for buckling

identical buckling curves

Geometric slenderness ratio lLT

Slenderness ratio lLT Buckling FactorcLT Buckling Resistance Mb.Rd

Mb.Rd

=cLT

Wpl.y

fy

/gM1

Partial safety factorfor Buckling

A reduction factor for Lateral

torsional buckling


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How to obtain cLT

Calculate the non

dimensional slenderness

Calculate phi

Look up imperfection factora LT

Lateral Torsional Buckling Resistance

fLT 0.5(1+ a LT (l LT -0.2)+lLT2

a LT =0.21

fLT+ [fLT2-blLT2]0.51cLT

lLT (Wpl.yfy/ Mcr)0.5


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Approximately

Lateral Torsional Buckling ResistanceHow to obtain lLT

For Hot rolledFully restrained flangesNo destabilising loadslLT L / 96 iz

By Calculating Mcr

Mcr=p2EIzC1

Lcr2

g kw

kIz

Iw (C2zg)2

p2EIz

LcrGIT++ -C2zg

2

0.5

Accounts of ratio of

end moments

Accounts for load position

from shear centre

Vertical distance of load

from shear centre

Z axis fixity

warping fixity

Increase factor for

buckling accounting for

flexural curvature


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Connections

Connection design covered in BS EN1993-1-8.

Guidance more comprehensive than BS 5950.

Results for bolted and welded connections moreconservative than BS 5950 (M = 1.10).

Slip factor for friction grip fasteners taken as 0.2rather than 0.45.