CV3012 EC3 Design of Bolted Welded Connections Xx1

8/17/2019 CV3012 EC3 Design of Bolted Welded Connections Xx1

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CV3012: Steel Design – Eurocode 3

Loadings –

one way slabs for steel structures Members subjected to axial loads only

Fully restrained beams

Completely unrestrained beams

Columns subjected to axial load and nominal moments

Columns subjected to axial load and moments

Angle, channel and tee sections under tension and compression

Typical structural connections

CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013 1


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Pinned joints

A pinned joint should be capable of transmitting the internalforces, without developing significant moments which mightadversely affect the members or the structure as a whole.

A nominally pinned joint should be capable of accepting the

resulting rotations under the design loads.

Rigid joints

Joints classified as rigid may be assumed to have sufficient

rotational stiffness to justify analysis based on full continuity.

Types of Joints or Connections

2CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013


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Typical Simple (Pinned) Connections

(a) Web Cleats (b) End Plate (c) Fin Plates

NEd

3

NEd NEd

CV3012 – STEEL DESIGN – EUROCODE 3 – LIE SENG TJHEN – AY2012/2013 – JANUARY 2013


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Typical Rigid (Moment) Connections

(a) Haunch Connection

(b) Extended End Plate

4

MEd

NEd

NEd

MEd



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A Typical Pinned Connection

5

NEd



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Bolt Subjected to Shear and Tension

6

VEd

N t,Ed



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A Typical Rigid Connection

7

NEd

MEd F 1

F 1

N t,Ed



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Failure of Plate by Block Tearing and Plain Shear

NEd

8

NEd



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A Typical Connection – Bracket Plate

Components check – bolt group under resultant shear, plain shear and blocktearing of bracket plate.

NEd = 75 kN

203 203 UC 60(T f = 14.2)

40

40

50

50 50

M20 class 8.8 bolt

50

40

12 mm



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A Typical Connection – Welded End Plate

Components check – bolt group under direct shear, end plate under plain shear &block tearing, strength of fillet welds, beam web under local shear across partialdepth D .

D

NEd = 65 kN



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A Typical Connection – Double Web Cleats

Components check – bolt group A under resultant shear, bolt group B under directshear, web cleat under plain shear and block tearing, block tearing of cross beam,and reduced moment capacity at notched cross beam end

Bolt group Athrough crossbeam

Bolt group Bthrough mainbeam

Cross beam

Web cleat

Main beam

NEd = 165 kN



12/93

A Typical Connection – Extended End Plate

Components check – compression zone, tension zone, horizontal shear zone andvertical shear zone

Tension zone checks:bolts in tension, flange to end plate weld,end plate in bending, column unstiffenedflange, column web tension.

Vertical shear zone checks:Bolts in direct shear, beam web toend plate weld.

Compression zone checks:column unstiffened flange, column webcrushing, column web bearing.

Horizontal shear zone check:column web panel shear.

MEdVEd

N t,Ed

NEd

NEd



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Nc,Ed

t

w1

w2

f ck = 30 N/mm 2

A Typical Connection – Base Plate Components check – base plate thickness and holding down bolts



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Eurocode 3: Part 1-8: Design of Joints

Clause 3.1 – Bolts, nuts and washers

Clause 3.4 – Categories of bolted connections

Clause 3.5 – Positioning of holes for bolts and rivets

Clause 3.5(1) – Minimum spacing, end and edge distances

Clause 3.5(2) – Maximum spacing, end and edge distances

Clause 3.6 – Design resistance of individual fasteners

Clause 3.8 – Long joints

Clause 3.9 – Slip-resistant connections using 8.8 or 10.9 bolts

Clause 3.10 – Deductions for fastener holes

Clause 3.10.2 – Design for block tearing

Clause 3.11(1) – Prying forces



15/93

15

Eurocode 3: Part 1-8: Design of Joints

Clause 4.5.3 – Design resistance of fillet welds

Clause 4.5.3.1 – General method

Clause 4.5.3.2 – Directional method

Clause 4.5.3.3 – Simplified method for design resistance of fillet weld

Clause 4.7 – Design resistance of butt welds

Clause 4.7.1 – Full penetration butt welds

Clause 6.2.6.12 – Anchor bolt in tension

Clause 6.2.7 – Design moment resistance of beam-to-column joints and spices

Clause 6.2.8 – Design resistance of column bases with base plates

Clause 6.2.8.2 – Column base subjected only to axial forces

Clause 6.2.8.3 – Column base subjected to axial forces and bending moments



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Base Plates - Effective Edge DistanceEC3:1-8 – Clause 6.2.5(4) and Clause 6.2.8

17

Effective edges offset by a distance c

2c+t w2c+t f 2c+t w



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Effective area, EC3:1-8 – Clause 6.2.5(4)

area)(sectionperimeter)(section c4c)t2tht(2bt)c2t4b(2h4c)t2tc2tht2bt2bc(hb

)4c2hc2bc(hb)t)(b2t2c(h2c)2c)(b(hAArea,Effective

2wf wf w2wf wwf

2wf

eff

(b + 2c) – (tw + 2c)

= (b – tw)

= 2a( b+2c ) b

h

(tw+2c )

(h+2c )

(tf +2c )

twtf

(h-2c-2t f )

c

a

a



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The required area A req = N Ed / f jd where N ed = applied force and f jd = design bearingstrength

With reference to EC2:1.1 – Clause 3.1.6(1), the compressive strength of concrete f cd is defined as f cd =( cc f ck )/ c where c=partial safety factor of concrete(1.5 for persistent and transient design), cc =0.85 for compression,f ck =characteristic cylinder strength

It is shown that f cd = f jd Eurocode 3:1-8 Clause 6.2.5(7)

Using the effective area method, equate effective area A eff with required area A reqto obtain the value of c

Substitute c into the following equation to get the base plates thickness t

EC3:1-8 – Clause 6.2.5(4)

where M0 = 1.0 (UK National Annex) and f y = yield strength of the base plate

Base plates thickness, EC3:1-8 – Clause 6.2.5(4)

0.5

y

M 0 jd

f

3f

ct



20/93

Base plate thickness t

Assume unit width cantilever strip,

and, bending moment at column face assumingsingle action under the baseplate. For a cantilever subjected under UDL, M y,Ed = wl 2/2 where l = c,

Therefore,

6

t

6

d1.0

d

2

12

bd

y

IW

223

maxyel,

y

2

y

22

2

y

2yel,y

f 3wc

f 2wc6t

6tf 2

wc

Wf MEdy,

0.5

y

M 0 jd0.5

y f 3f

c f 3w ct

w

t

c

c

x

x

NEd

1.0

c

1.0 t = d



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t

305 x 305 UC 137

600

600

4300 kN

f ck,cube = 30 N/mm 2

Example 1



22/93

Example 1

22

mm500.255

)0.1)(17.14(34.122f

3f ct

mm122.4cgivesAASolving mm1003.5317.14

104300

f

NA

17243.261850.2c4c)t2tht(2bt)c2t4b(2h4cA

N/mm14.17f f N/mm14.175.1)25(85.0

f

0.851.5

N/mm52f

N/mm30f

f f

0.50.5

y

M 0 jd

eff req

233

jd

Edreq

2wf wf w

2eff

2cd jd 2cd

cc

c

2ck

2cubeck,

c

ck cccd

Therefore,

Now,

NA.1leAnnex TabNationalUKon)(compressi2.1N Table2.4.2.4(1)Clause1.1:EC2 design)transientandt(persisten

3.1 Table3.1.2(3)Clause1.1:EC2

3.16(1)Clause1.1:EC2



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EdEdEd

EdEdEd

FT Ne)(hFMe)2h( N

:A pointaboutmomentTaking

Ned=1380 kN

Med=185 kNm

TEd FEd

b + 2c

tf + 2c

Assume uniform pressureand symmetrical around thecompression flange

Effective portion of thecompression zone, Aeff

w

e h/2 h/2

A

Extension to eccentric base plate

23

mm70e

z



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Example 2

24

kN1324.233=F

1079.653F=185+1070)2

320.5(1380 ,

mm.657932

21.7 700.5322

t ehz ,

N/mm14.17f f N/mm14.175.1)25(85.0

f

0.851.5

N/mm52f N/mm30f

f f

Edr,c,

3- Edr,c,

3-

f

2cd jd

2cd

cc

c

2ck

2cubeck,

c

ck cccd

Aaboutmoment Taking

6.2.8.3(1)Clause1.8:EC3armLever

NA.1leAnnex TabNationalUKon)(compressi2.1N Table2.4.2.4(1)Clause1.1:EC2 design)transientandt(persisten

3.1 Table3.1.2(3)Clause1.1:EC2

3.16(1)Clause1.1:EC2



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Example 2

25

(O.K.)

(O.K.)and

Now

6.2.8.2(1)Clause1.8:3Eurocode

mm40mm517.43265.0.0)3(14.17)(1181.68

f 3f

c tTherefore,

mm600 < 2c+ b 2c+h

mm86.181cgives AASolving mm10453.3914.17

10233.3241

f

NA ,

6709.64+661.8c+4c=

bt+2b)c+(2t+4c=2c)+2c)(b+(t=A

0.50.5

y

M 0 jd

eff req23

3

jd

Edreq

2

f f 2

f eff

γ



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Anchor bolts in tension – EC3: Part 1-8 – Clause 6.2.6.12

Holding down bolts should be designed for the effects of factored loading. Where they are required to resist tension they should be properly anchored into the foundation by a washer plate or other load distributing member embedded in the concrete.

F t,Ed F t,Rd

where F t,Rd is design tensionresistance of the bolt

26

Shear key

Base plate

Concrete foundation

Grout



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Rigid beam-to-column connections – EC3: Part 1-8 –Clause 6.2.7

27

The applied design moment M j,Ed should satisfy:

where M j,Rd is the design moment resistance of the joint.

1.0 M

M

j,Rd

j,Ed

F Rd is total resistance of the fillet welds.

Mj d



28/93

Rigid beam-to-column connections – EC3: Part 1-8 –Clause 6.2.7

28

F Rd is the total tensionresistance of the bolts.

F Rd is the total resistance

of the fillet welds or totaltension resistance of thebolts, whichever is lower.

F Rd is the total tensionresistance of the bolts.



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Plain Shear – EC3: Clause 6.2.3

29

The usual approach is to use the plastic shear resistance V pl,Rd inpractice.

The plastic shear resistance is essentially defined as the yieldstrength in shear multiplied by a shear area A v, i.e.

where A v = effective shear area, i.e. A v = A –

Ah where A = grossarea and A h = total area of holes

M0

yV

Rd pl,

3f A

V



30/93

Block Tearing – EC3: Clause 3.10.2

Block tearing consists of failure in shear at the row of bolts alongthe shear face of the hole group accompanied by tensile rupturealong the line of bolt holes on the tension face of the bolt group

Block Tearing

Ned = design tension force1 small tension force2 large shear force

3 small shear force4 large tension force



31/93

Block Shear – EC3: Clause 3.10.2

For a symmetric bolt group subject to concentric loading thedesign block tearing resistance, V

eff,1,Rd is given by

Eqn. (3.9)

For a bolt group subject to eccentric loading the design blockshear tearing resistance Veff,2,Rd is given by

Eqn. (3.10)

where Ant = net area subjected to tension;Anv = net area subjected to shear

M0

nvy

M2

ntuRdeff,1,

Af

31Af V

M0

nvy

M2

ntuRdeff,2,

Af

31Af 5.0V



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End PlateLt L t

LvLv

Veff,1,Rd

Block Shear – EC3: Clause 3.10.2(2)

32

M0nvyM2ntuRdeff,1, /Af )3/1(/Af V :capacityshearBlock

Veff,1,Rd



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33

Block Shear – EC3: Clause 3.10.2(3)

M0nvyM2ntuRdeff,2, /Af )3/1(/A0.5f V :capacityshearBlock

Cut-off topflange of beam

Lv

Lt

Veff,2,Rd



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Example 3

34

M20 Grade 8.86 mm thick S275 fin plate

40

70

70

4070

120

Lv

Lt

Veff,1,Rd

Determine bolt shear resistance, bolt bearing resistance, plain shearresistance and block tear resistance of the connection



35/93

Example 3

kN354.22/1.0)1230)(275)(3/1(/1.25)462(304V Rdeff,1,

mm12302)](6)2.5(2070)70[(120A

mm4622)](6)1.5(2040)[(70A

.01 ; N/mm275f .25;1 ; N/mm430f

2nv

2nt

M 02

yM 22

u:resistancetearBlock

kN 222.911.0

)31404(275/)3/(f AV

mm14042)(6)3(20300(6)A .0;1 ; N/mm275f

M 0

yvRd pl,

2vM0

2y

:resistanceshearPlain

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Bearing on Plate and Bolt

Failure Modes of Bolt and Plate

Bolt under Shear

NEd NEd

Top Plate Bottom Plate

Bearing

Shear

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Shear and tension resistance of fasterners – EC3:1-8 –Clause 3.6.1

37

Shear resistance per shear plane, F v,Rd is given by

where f ub = ultimate tensile strength (Table 3.1); A = A s if the shear planepasses through the threaded portion of the bolt; A = A if the shear planepasses through the unthreaded portion of the bolt; v = 0.6 for classes4.6, 5.6 and 8.8 and v = 0.5 for classes 4.8, 5.8, 6.8 and 10.9.

Tension resistance per tensile stress area, F t,Rd is given by

where f ub = ultimate tensile strength (Table 3.1); A s is the tensile stressarea of the bolt; k 2 = 0.63 for countersunk bolt; and k 2 = 0.9 forhexagonal bolt.

M 2

ubvRdv,

Af F

M 2

sub2Rdt,

Af k F



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38

Nominal values of ultimate tensile strength f ub for bolts



39/93

Shear resistance of bolts

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40

Bearing and tension resistances of bolts

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41

Bolt resistance – Non Preloaded, Class 4.6

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When the distance L j between centers of two end bolts in a joint exceedsmore than 15d (d=nominal bolt diameter), the design shear resistance

F v,Rd of all fasteners should be reduced by multiplying it by a reductionfactor Lf given by:

0.75 and 1.0 but 200d

15dL 1 Lf Lf

jLf

Long joints – EC3: Part 1-8 – Clause 3.8

L j

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45

Packing plates – EC3: Part 1-8 – Clause 3.6.1

Where the fasteners transmitting load in shear and bearing pass throughpacking of total thickness t p greater than one-third of the nominal

diameter d, the design shear resistance F v,Rd of all fasteners should bemultiplied by a reduction factor p given by:

1.0 but t3d8

d9p

pp

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Positioning of holes for bolts – EC3:1-8 – Clause 3.5

46

Symbols for end and edge distances and spacing of fasteners

End, edge distances and spacing End and edge distances for slotted holes

e4

e3

d0

0.5d 0

p1e1

e2

p2Ft,Ed

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Bearing resistance of bolts – EC3:1-8 – Clause 3.6.1

47

Bearing resistance per shear plane, F b,Rd is given by

where b is the smallest of d ; f ub /f u or 1.0;

parallel to the direction of load transfer:

- for end bolts: ; for inner bolts:

perpendicular to the direction of load transfer:

- for edge bolts: k 1 is the smallest of

- for inner bolts: k 1 is the smallest of

M 2

u b1Rd b,

dtf k F

0

1d

3de

4

3d p 1

0

1d

2.5 1.7d

e2.8

0

2 or

2.5 1.7d

p4.1

0

2 or

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E l 3


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Example 3

48

kN564.4894.086 F bolts,sixFor

94.08kN101.25

45)0.6(800)(2 F bolt,oneFor

mm245AA

N/mm800f 25.1

8.8Classfor0.6

Af F

Rdv,

3Rdv,

2s

2

ub

M2

v

M2

ubvRdv,

TableBoltEC3.1 Table3.1.1(3)Clause-8-1:3EC

2.2(2)Clause-8-1:3EC

3.4 Table3.6.1(1)Clause-8-1:EC3

:boltstheof resistanceShear

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Example 3

49

boltsinnerforororof smallestboltsendforororof smallest

boltsinnerfor

boltsendfor

:transferloadof directionthetoParallel

3.1 Table3.2.1(1)Clause-1-1:3EC

3.4 Table3.6.1(1)Clause-8-1:EC3

:boltsof resistanceBearing

0.8111.01.860.811 1.01.01.861.818

0.81141

3(22)70

41

3d p

1.8183(22)120

3de

86.1430800

f f

N/mm430f

dtf

F

d

d

0

1d

0

1d

u

ub

2u

M2

u b1Rd b,

k

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E l 3


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50

kN541.184(83.695)2(103.2)

kN.695381.25

6)(430)(20)(2.5(0.811)dtf F

kN2.0311.25

30)(20)(6)2.5(1.0)(4dtf F

2.5 2.5 2.75 3.39 k

227022

40

M2

u b1Rd b,

M2

u b1Rd b,

1

k

k

2.75 1.71.4 1.7d

p4.1

3.39 1.72.8 1.7d

e2.8

0

2

0

2

bolts6of resistanceBearing

boltsinnerfor

boltsendfor

ororof smallest

boltsinnerfor

boltsedgefor

:transferloadof directionthetolarPerpendicu

Example 3

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Minimum and maximum spacing, end and edge distances

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l


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Bolts Components

Bolt

Washer

Nut

22mm

< 22 mm

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Categories of Bolt Connections – EC3:1-8 – Clause 3.4

Tension connections – EC3:1-8 – Clause 3.4.2Bolted connection loaded in tension should be designed as one of the

following:

a) Category D: non-preloadedIn this category bolts from class 4.6 up to class 10.9 should be used. Nopreloading is required.

b) Category E: preloadedIn this category preloaded 8.8 and 10.9 bolts with controlled tightening inconformity with 1.2.7 Reference Standards: Group 7 should be used. Thedesign checks for these connections are summarized in Table 3.2.

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B lt d Si l d D bl Sh


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Single Shear

Double Shear

Bolts under Single and Double Shear

56

NEd

Shear

NEd

Shear

NEd NEd/2

NEd/2

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B l d C i


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Bolted Connections

Non-preloaded bolts of Class 4.6, 8.8 and 10.9 in S275

Diameters are 12 mm, 16 mm, 20 mm, 24 mm and 30 mm

Direct shear or direct tension connections

57

Bolts in single and double shear joints

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The shear stress in the bolts is given by

Single shear:

Double shear:

sbEd

s

Edb

Af nN or

)(AAreaRoot(n)boltsof No.)(NLoad

f stress,Shear

Single Shear and Double Shear

58

sbEd

s

Edb

Af n2N or

)(AAreaRoot(n)boltsof No.2)(NLoad

f stress,Shear

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Eccentric Bolted Connections


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Eccentric Bolted Connections

There are two types of eccentrically loaded connections:

bolt group in direct shear and torsion bolt group in direct shear and tension

M = P × eT = P × e

P

e

P

e

P

G

P

59

(a) Bolt group indirect shear and torsion

(b) Bolt group indirect shear and tension

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B lt G i Di t Sh d T i


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Bolt Group in Direct Shear and Torsion

The moment applied in the plane of the connection tends to rotate the side plateabout the bolt group centre of gravity.

A linear variation of loading due to moment is assumed with the both farthest fromthe centre of gravity of the group carrying the greatest load.

The direct shear is divided equally between the bolts.

cos = r 2/r 1

Therefore,

F T cos = FT r 2/r 1

G

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Let the force due to the moment on bolt “A” be F T . Then, the force on any boltr 2 from the centre of rotation is F T r 2 /r 1 , and so on for all the other bolts in thegroup.

The moment of resistance of the bolt group is given by

The load F T due to the moment on the maximum loaded bolt “A” is given by

The load F S due to direct shear is given by

Pe)yx(r

Fr

r

F

...)r (r r

F ...r

r

r Fr

r

r FM

22

1

T2

1

T

22

21

1

T2

1

2T1

1

1TR

221

T yx

Per F

Boltsof No.P FS

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Resolving the load F T vertically and horizontally produces

Vertical load acting on bolt A = F S + F T cos Horizontal load acting on bolt A = F T sin

Resultant load on bolt A is equal to

The total shear stress f b = F R/A s should not exceed the permissible shear stressof the bolt as defined in EC3: Part 1-8 – Clause 3.1.1.

1/2TS

2S

2T

1/222TTS

2S

22T

1/22TS

2Tv,EdR

)cosF2FFF(

)cosFcosF2FFsinF(])cosFF()sinF([FF

62CV3012–

STEEL DESIGN–

EUROCODE 3–

LIE SENG TJHEN–

AY2012/2013–

JANUARY 2013

Bolt Group in Direct Shear and Tension


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Bolt Group in Direct Shear and Tension

Pey

y2F .)]..yy[(y

y

2F .)..

y

y F

y

y F

y

y(F2M

1

2

T232221

1

T

1

2

3T

1

2

2T

1

2

1TR

The centre of rotation is assumed to be at the bottom bolt of the group, and theloads vary linearly.

A bracket subjected to a load P at an eccentricity e .

The moment of resistance M R of the bolt group is given by

A

Centre of rotation

63CV3012–

STEEL DESIGN–

EUROCODE 3–

LIE SENG TJHEN–

AY2012/2013–

JANUARY 2013


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The load F T due to the moment on the maximum loaded bolt “A” is given by

The load F S due to direct shear is given by

Tensile stress is given by

Shear stress is given by

The combined tensile and shear stresses should also satisfy the conditions givenin EC3: Part 1-8 – Clause 3.6.1.

21

T y2

Pey F

Boltsof No.P FS

stEdt, T s

T t Af FF A

F f

ssEdv,Ss

S

sAf FF

A

F f

64

1.01.4FF

FF

Rdt,

Edt,

Rdv,

Edv,

CV3012–

STEEL DESIGN–

EUROCODE 3–

LIE SENG TJHEN–

AY2012/2013–

JANUARY 2013

i i i


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Tension – No Prying Action

65

Mode – 1

Flange is fully rigid.

There is no prying

force. Bolt failure.

2F t,Rd

F t,Rd F t,Rd


Tension – Prying Action


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Tension Prying Action

66

2F t,Rd

Ft,Rd+Q

F t,Rd+Q

Q Q

Prying action due to flexible end plate

Mode - 2

Flange is flexible.

There is prying force.

Total bolt force

= Ft,Rd + Q Tension in bolts is 90% of

resistance value.

Bolt failure with flange

yielding.



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P l d d Hi h St th F i ti G i (HSFG)


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Preloaded or High Strength Friction Grip (HSFG)Bolted Connections

Preloaded Bolts Connections


Diff b t N l d d d P l d d B lt


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Difference between Non-preloaded and Preloaded Bolts

Non-preloaded Bolted Connection in Shear

Shear Bearing

NEd

Friction

Preloaded Bolted Connection in Shear

69

NEd



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Preloaded Bolts Tension and Tightening


P l d d Hi h S h F i i G i (HSFG) B l


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HSFG bolts consist of high-strength steel bolts, nuts and hardened steelwashers.

The bolts are tightened to a predetermined shank tension so that theclamping force will transmit the force in the connected members by friction.

The bolts do not act in shear or bearing as in non-preloaded bolted

connections.

There is no slip or movement between the connected parts, hence this typeof joint is useful where rigid connections are required.

If there is a slip the bolts are then in tension and shear, and theconditions given in EC3: Part 1-8 – Clause 3.6.1 has to be checkedaccordingly.

Preloaded or High Strength Friction Grip (HSFG) Bolts


Slip resistant connections using class 8 8 or 10 9 bolts


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The design slip resistance at serviceability limit state (SLS) and

ultimate limit state (ULS) of a preloaded class 8.8 or 10.9 boltshould be taken as

wherek s = 1.0 for bolts in normal holes; n = number of friction surfaces; = 0.5 slip factor (coefficient of friction) and F p,C = 0.7f ub Aspreloading force to be used in the above equation. f ub is obtained

from Table 3.1 and A s is the bolts root area. M3, service is 1.1 atserviceability limit state and M3 is 1.25 at ultimate limit state.

Slip-resistant connections using class 8.8 or 10.9 boltsEC3: Part 1-8 – Clause 3.9.1

72

C p,

M 3

s

Rds,C p,

serviceM 3,

s

Rds,F

nk FF

nk F or


Slip resistant connections using class 8 8 or 10 9 bolts


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Combined tension and shear

If a slip-resistant connection is subjected to an applied tensile force, F t,Edor F t,Ed,ser , in addition to the shear force, F v,Ed or F v,Ed,ser , tending to produceslip, the design slip resistance per bolt should be taken as follows:

It applies for both slip-resistant at serviceability and slip-resistant

at ultimate limit state.

Slip-resistant connections using class 8.8 or 10.9 boltsEC3: Part 1-8 – Clause 3.9.2

73

M 3

Edt,C p ,

sRds,

)0.8FF(nk F


Example 4


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p

A bracket shown below is constructed from a cut 300 x 200 x 77.3 kg/m UBwith a 15 mm thick plate welded on top if it. The bracket is subjected with afactored vertical design load of 380 kN acting at an eccentricity of 160 mmfrom the face of the column.

A total of 8 numbers of M20 class 8.8 preloaded higher grade high strengthfriction grip (HSFG) bolts in S275 and designed to be non-slip in service, and

assuming the slip factor = 0.5 and there is no prying force.

Assuming the centre of rotation is at point A, the bottom of the bolts group,and the loads vary linearly, show that the proposed 8-bolt group is adequateunder combined shear and tension action.


Example 4


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70

70

70

380 kN

160Two bolts onein each side of flange of column

8 No. M20class 8.8bolts

A60

50

75

p

Maximum load dueto moment F t

y1y2

y3

Ft × y1/y1

Ft × y2/y1

Ft × y3/y1


Example 4


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The moment of resistance MR of the bolt group is given by

Applied moment is equal to the moment of resistance MR of the bolt group,

Therefore, the tensile load per bolt F t due to the moment on the maximum loadedbolt is given by



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Schematic Representation of Welding Process


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Electrode

Electrode, Core Wire, Flux, Slag, Heat-affected Zone, Completed Weld

Welding is a process of permanently joining metal parts, where heat is applied to thework pieces to melt and fuse to form a permanent bond. Because of its strength, weldingis used to construct and repair parts of many on-shore and offshore steel structures.


Manual Metal Arc Welding (MMAW) – This is the simplest form of welding process. The electrode ishand held and fed into the weld pool The wire is covered with an extruded flux coating It requires a


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The MMAW Welding ProcessDifferent Type of Electrodes

hand held and fed into the weld pool. The wire is covered with an extruded flux coating. It requires askilled welder.

It is flexible as it can be manipulated in many situations. Quality of welding is closely related tooperator skill. Typical electrode diameters vary from 2.5 to 6.0 mm and length from 350 to 450 mm.Open-circuit voltages are from 50 to 90 V, and as soon as the arc is struck, the voltage falls from 20to 35 V. Typical currents range from 50 to 400 amps and the deposition rates is from 20 to 100g/min.


Flux Core Arc Welding (FCAW) – This is a more automated method of welding as compared to


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FCAW Machines Mechanism and Spool of Electrode Wire ofFCAW Machine

Flux Core Arc Welding (FCAW) This is a more automated method of welding as compared toMMAW. This method consists of the emission of wire and gas (CO 2 , argon or helium). The wiresare supplied on a spool which is placed in a FCAW welding compartment.


The wire is emitted from a nozzle with a trigger to eject it. Compared to the MMAW method of


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FCAW Welding Machine Welder Operating theFCAW Welding Machine

welding, welds can be continuous and lengthy, without the need to change electrodes. Thus thismethod is more convenient. The FCAW machines can be connected to a special travelingmachine which facilitate the welding of lengthy welds such as a T-beam as shown below.


Submerged Arc Welding (SMAW) – SMAW is a form of welding that utilizes the use of flux. It ismainly used for long straight welds usually on flat surfaces The machine is made up of a flux


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Flux Used in the SAWProcess

SAW Machine

mainly used for long straight welds usually on flat surfaces. The machine is made up of a fluxhopper, which holds the flux, a compartment to store the spool of electrode wire and a control panelwhich allows the operator to control the speed of travel, the height of welding nozzle and voltage of the SAW machine. Advantages of the SMAW method is that its easily automated, minimum cleaningrequired unlike normal welding processes and there is no visible arc light since the welding action issubmerged under the flux that is used.



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SMAW Control PanelSpool of Electrode Wire for SMAW Operating the SMAW with Ease

Hardened Slag

SMAW Producing an ExcellentQuality Welds

Close up of Hardened Slag


Weld Types – Fillet and Butt Welds


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84

Types of fillet and butt welds


Classification of Fillet Welds


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Total weld strength Total weld strength == 2 x 2 x 50mm x 0.942 kN/mm 2 x 60mm x 1.155 kN/mm = 138.6 kN= 184.8 kN = 138.6 kN

F

F

F

F F

F

F /2

F /2

F /2

F /2

60 mm

50 mm

We ld leg length = 6 mm

(a) Side Shear (b) End Tension


Design resistances of fillet welds


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g


Design resistance of fillet weldsEC3: Part 1-8 – Clause 4 5 3 3


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EC3: Part 1-8 Clause 4.5.3.3

Simplified method

The design resistance of a fillet weld may be assumed to be adequate if, at every point along its length, the resultant of all the forces, not necessarily along theweld axis, per unit length transmitted by the weld should satisfy the followingcriteria:

where F w,Ed is the design value of the weld force per unit length; F w,Rd is thedesign weld resistance per unit length.

The design resistance per unit length F w,Rd should be determined from:

where f vw,d is the design strength of the weld; a = throat thickness of the filletwelds.

87

Rdw,Edw, F F

af F dvw,Rdw,




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EC3: Part 1 8 Clause 4.5.3.3

Simplified method

The design shear strength f vw,d of the weld should be determined from:

where f u = 410 N/mm 2 for S275; w = 0.85 from Table 4.1; M2 = 1.25

88

M2w

udvw,

3f f




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Directional method

In this method, the forces transmitted by a unit length of weld are resolved

into components parallel and transverse to the longitudinal axis of the weldand stresses normal and transverse to the plane of its throat.

Transverse welds are stronger than longitudinal welds up to 25% more

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Longitudinal resistance,F w,L,Rd = f vw,d a × 1.0

Transverse resistance,

F w,T,Rd = K f vw,d a ×

1.0

F T

0when1.083 K and 45when1.25 K ;cos1.0

1.51.25K 2

90



Design resistance of fillet weldsEC3: Part 1-8 – Clause 4.5.3.2


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6mm

6mmwt = 6 x cos 45 °

45 °

Longitudinal capacity, F w,L,Rd = f vw,d a

As an example, a 6-mm fillet weld strength is = 222.80 x 6 x cos 45= 222.80 x 6 x 0.707 = 0.945 kN/mm run

91




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Design resistance of butt weldsEC3 P 1 8 Cl 4 7


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The strength of butt welds should be taken as equal to that of the parent metalprovided matching electrodes are used.

Matching electrodes should have specified minimum tensile, yield strengths,elongation and impact values each equivalent or better than those specified for theparent materials.

Backing plate

Throatthickness

h

Root opening

Bevel angle

EC3: Part 1-8 – Clause 4.7

CV3012 EC3 Design of Bolted Welded Connections Xx1

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