Examples To Eurocode 3 [PDF]

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ECCS CECM E KS

I!fflIIII 11JW"

EUROPEAN CONVENTION FOR CONSTRUCTIONAL STEELWORK CONVENTION EUROPEENNE DE LA CONSTRUCTION METALLIQUE

EUROPAISCHE

KONVENTION

FÜR

STAHLBAU

ECCS - Advisory Committee 5 Application of Eurocode 3

Examples to Eurocode 3

FIRST EDITION

1993

N°71 document r This contains

L

pages

Introduction

The European Convention for Structural Steeiwork (ECCS) has as one of its primary objectives the promotion of the safe and economical use of steel in structures. ECCS believes that the Introduction of the harmonized Eurocodeshas a greatvaluein achieving this objective; accordingly ECCS has set an up Advisory Committee, AC 5, charged with the task of promoting the Introduction and adoption of the Eurocodes.

The AC 5 Committee have considered how this process could be best achieved and concluded that a threestage approach was desirable.Thefirst stage was to produce a concise version of the Eurocodes whichcan be used for normal every day design; this part has already been issued as ECCS Publication No. 65.

Thesecond stage is the production of this document which gives design examplesto EC 3/1 and E-EC 3 and has been prepared as a design aid to facilitatethe use of EC 3/1 for thedesign of steel buildings during the ENV period. These examplesconcentrate on those aspects which are likely to be needed for daily practical design work.

Thethird and final stage wili be the production of a series of "DesignAids" which will enable the design process to be made morequickly by using tabulated or graphical values for the various design formula contained in EC 3/1.

The combination of these three documents will enable practising engineers to more easily to the useof the new Eurocodes and should have a beneficial help in their speedy Introduction. adopt

Scope These Design Examplesto EC 3/1 and E-EC3 have been prepared by the ECCS - Advisory Committee AC 5 as a design aid in supplement to the complete EC 3/1 to facilitate the use of EC 3/1 forthe design of steel buildings in the ENV-perlod. The Design Examplesonly contains examplesto EC 3/1 and E-EC3 that are likely to be needed for daily practical design work. The y values used in this document are the values recommended In EC 3 main document. These values may deviate fromthe values recommended in the National Application Documents (NAD) of the member states.

The ECCS - Advisory Committee 5 is at present composed of the following members: Aasen, B. Arda, T.S. Bock, H.

Danieli, S. Dowling, P.J. Falke, J. Gemperie, C. Gettins, H.J. (Chairman) Lequien, Ph. Lutteroth, A. Schleich, J.B. Sedlacek,G. Lundin, K

Norway Turkey United Kingdom Italy United Kingdom Germany Switzerland United Kingdom France Germany Luxembourg Germany Sweden

The Committee gratefully obtained contributions from: Braham, M. Gerardy, J.C. Grotmann, D. Taylor, J.C.

Luxembourg Luxembourg Germany United Kingdom

Also particular thanks are given to the ECCS TechnicalCommitteesTC 8 and TC 10 who have contributed to the work.

References (1) (2) (3) (4)

EC 3/1: ENV 1993-1-1 Eurocode 3: Part 1.1 E-EC 3: Essentials of Eurocode 3 - Design Manual for Steel Structures in Buildings, ECCSPublication No. 65 References to EC 3/1 are given in brackets [...] References to E-EC 3 are given without brackets

Ii'

Scope

Partl Load combination

1.1

. Contents

Introduction

.1

Example 1.1.1

60

T

3

Example 1.1.3

5

Example 1.1.4

page

Single storey frame

4

page

Crane girder

B6

('1

20.00

page

Purlin treated as a continuous beam

Braced portal frame

0

Example 1.1.2

page

2

f

16.0

6

LI

iI III

II

II

77 •i'-

1200

Example 1.1.5

page

Single storey frame including a crane girder 11

Methodsof analysis

1.2

14

Example 1.2.1 Continuous beam (plastic - plastic) I

-

I

I

60

.

I I

II

I

I

I

I

60

I I

I

I

I

I

Example 1.2.2

page

I I Lqd.1Z9kN/m

I

page

Continuous beam (elastic - plastic) with limited redistribution J.IIIIIIIIIIIIIIIIIIIIIIIJ.Q.9IW

60

60 .r

60

60

60

-r

Example 1.2.3 Continuous beam (elastic - plastic) withoutlimited redistribution I

I I

60

I

I

.

I

I

I

60

I

I

I

I

I I

60

I

I

r

I

I I

60

Example 1.2.4

page

I

1q.17.9kN/

page

Continuous beam (elastic - elastic) 17

-

60

.

18 60

.

60

-r

60

1.3

Frame analysis

19

Example 1.3.1 Calculation of a sway frame 200

I

Example 1.3.2

page

page

Determinationof frame imperfections

200

s0ioiJ.

0 20

.,,Ei2ooB2 100 1.4

23

.'

'

J.

Bracing systemanalysis Example 1.4.1

25

Example 1.4.2

page

A frame Is braced by a bracing

page

A frame is braced by a frame

system 1.(Th

26

4"

t—j I—_t'.__j trimsI

Example 1.4.3

page

A frame Is braced by a bracing

frim..2

Example 1.4.4

page

Bracing system imperfections for a wind bracing

system

sm

z:2;z::7!—.;i. 28

IS00

27

IQ0

29

i

wind bracrng

Part2 2.1

Members In compression

2.2

Members In bending

30 31

37

Example 2.2.1

page

Single span beam

// i1iii2'

7O

u1

•j;••

Example 2.2.3

38

page

7000

40

Example 2.2.4

page

Class 4 cross-section loaded in bending 42

I

lj

I I I I I I I I I I)LLLL 30*1

Single span beam with shear buckling verification

7000

page

Single span beam with lateral torsional buckling

D.thfl:

-:Ii

Example 2.2.2

44

Combined loading - Bending and compression

2.3

Example 2.3.1

47 Example 2.3.2

page

RHS column loaded in bending and axial compression

HEA profile loaded in bending and axial compression

IH

14

50

Example 2.3.3

Example 2.3.4

page

Column of a frame

page

Rafter of a frame

t

53

:' :. 2.4

page

56

cr•larr ra'..

'nit

?2

Local stresses

61

Example 2.4.1

Example 2.4.2

page

Design of transverse stiffeners

page

Design of intermediatetransverse stiffeners

,J

L

oo 7100

-H .

/-

62

ItG.N

64

't•O.•

7200

Example 2.4.3

Example 2.4.4

page

Axially loaded column supported by a beam

fl—Ti F,

Load introduction of wheel loads from cranes

j,

65

360

J IPC10(1 ) F,360

l'

Example 2.4.5

page

67

IM

.

ilrs

iz,.n

iliXit

f page

Beam supported by a beam (class 4 cross-section) 69

vI

rarti 3.1

71

Bolted

connections

72

Example 3.1.1 Bolted connection of a tension member to a gusset plate

l2CH

UtUtU.

ooI: 460

604. 4•_ii9._•4.

Example 3.1.3

page

Erection splice at mid span of a lattice girder

Is

0

Example 3.1.2

page

?6Ot2

Example 3.1.4

page

Angle connected to a gusset plate

1•'°

page

Fin plate connection to H section

column 77

Example 3.1.5

Example 3.1.7

Example 3.1.6

page

Fin plate connection to RHS column

"

nç

79

page

Flexible end plate connection

81

83 F.430

Example 3.1.8

page

Beam to beam connection with cleats

page

Splice of an unsymmetrical I-section

1 88

85

Example 3.1.9

page

Bolted end plate connection P0220 P.360

/ I

I

.' f 11 I°I:

91

Lri

VI'

3.2

Welded connections Example 3.2.1

96

page

Double angle welded to a gusset plate

Example 3.2.2 Bracket welded on a column

350*120

97

Example 3.2.3

page

page

10

3S0x110IO

Example 3.2.4

Welded beam to column connection

Welded beam to column connection

withoutstiffeners

with stiffeners 100

Example 3.2.5

98

page

103

page

Hollow section lattice girder joint

-4

3.3

105

pin connections Example 3.3.1

107

page

Pin connection NSd

j;;

108

Annex A

109

Tabled reduction factors for buckling curves a0, a, b, c, d

Viii

Part

1

Load combination Methods of analysis Frame analysis Bracing system analysis

1.1

Load combination

V, M,

etc) are These examples demonstrate how the design values of action effects (N, are not treated In this part. determinedfrom the load assumptions.The further stepsof design An actionis a force (load) applied to the structure or an imposed deformation (e.g. temperature effects or settlements).Characteristic (unfactored)values of these actions are specified in ENV 1991 Eurocode 1 or otherrelevant loadings codes. Thesevalues of actionsshall be multiplied by relevant safety factors and combination factors, see chapter "Combinations of actions" In ENV 1993 Eurocode 3 or Table 2.1 in the Essentialsof Eurocode 3 to determine the design values of the effects of actions. Thefollowing examplesshow the method of determination of the maximum effects of actions. Not all possible combinations of actions are presented nor are relevant combinations worked out. In practice, one will collect experienceto easily find out which load combination Is decisive for verification of the structures.

Thefollowing examplesare Included In this chapter: Example Example Example Example Example

1.1.1: 1.1.2: 1.1.3: 1.1.4: 1.1.5:

Braced portal frame Purlin treated as continuous beam Single storey frame withoutcrane girder Crane girder Single storey frame with a crane girder

2

3.3

3.2

3.1

3.

2.

6Q

j360

of internal forces duetoP

6O

frame System and load configuration

due to

w -05

P

S

due to imperfections equivalent horizontal unit force F

Load assumptions: G 480 kN P = 360 kN W= 6OkN Frame imperfections '1' = 1/200 Steel grade: Fe 360 Member (1) and (2): 2 I-profiles with welded flange plates, A = 104.102 mm' Member (3): IPE 400, A = 84,5.102 mm2 Member (4) and (5): IPE 200, A = 28,5.102 mm'

1.

= 2.(1,35.480 + 1,5.360)/200 = 11,88 kN Ne,, 1,35.480 + 1,5.360 + 11,88.1,0 = 1199,9 kN

N N,

=

480-1,5.60,0 = 390,OkN

for uplift (permanent - wind (1,00 - 1,5W))

uplift reaction force at support (A): —> no uplift effect

Relevant load combination

3

= 2.(1,35.480 + 1,35.360)/200 = 11,34 kN 1,35.67,082 + 11,34.1,118 = 103,2kN = 1,35.67,082 + 11,34.1,118 = 103,2kN = 1,35.(480 + 360 + 120) + 11,34.2,0 = 1318,7 kN

N, =

F,

for max compression/tensionIn the diagonals (permanent + imposed load + wind

F1

Imperfection: maximum compression force in diagonal (5): maximum tension force in diagonal (4): maximum reaction force at (B):

(1,35(G+P+W)))

Relevant load combination

Imperfection: maximum axial force in column (2):

Relevant load combinations Relevant load combination for max compression force In the column (permanent + Imposed load (1,35G + 1,5P))

duetoG

DistrIbution

Example 1.1.1: Braced portal

Problem, Sketch, Calculation

[2.3.3.1 .(1) and (5)]

Table 2.1

Table 2.1 [2.3.3.1 .(1) and (5))

[2.3.3.1 .(1) and (5)]

Table 2.1

Table 5.6

Reference

The effect of dead load counteracts the effect of wind loading in column (1).

The effect of loading is oriented in the same direction. The maximum reaction force is needed for base plate design. For foundation design characteristic values are needed.

The effect of loading is oriented in the same direction.

Internal forces and moments (effects of actions) are responsesof the structure to the loading (actions). Design values of the internal forces and moments are determined from the design (factored) values of loading using elastic or plastic global analysis. Hero: Using elastic global analysis, first order theory with unfactored loading by separating In single actions. This simplifies the decision of which load combination gives the worst effect.

Characteristic (unfactored) values of actions are specified In ENV 1991 Eurocode 1 or other relevant loading codes. The effects of Imperfections shall always be taken into account in both first and second order theory. Her.: Frame Imperfections are considered

Commentary

3.4

Imperfection: maximum compression force in beam (3):

F,

= 2.(1,35.480)/200 = 6,48 kN N,, 15.30,0 + 6,48.0,5 48,2 kN

4

Relevant load combination for max compressIon force In the beam (permanent + wind load (1,35G + 1,5W))

Problem, Sketch, Calculation Table 2.1 (2.3.3.1.(1)and (5)J

Reference rection.

The effect

of loading

l

oriented In the same dl-

Commentary

configuration

I

I

1

-

-

OO

OO

2: FIxing of purlin

__

oo•7'oo — r

rlIlIIIIIIlIItllI_lj,

DetaH

4

DetaIl 1: PurlIn

as a continuous beam

_______ 22,1 kNm)

3O,9•10 mm3

7,2 7,2 Relevant load combination for uplift (permanent + wind (1 ,OG 1,5W)) + = q 1,0 g 1,5w 1,0.1,0 1,5.2,0 = .2,0 kN/m

k0

\JY' \L'

-.—-.

=

3,6•5,O-1,;10

4. Relevant load combinations for serviceability limit state 4.1 Serviceability limit state requirement (example only) Requirement: 67 < 1/200 (for snow only) q = 1,0 s = 1,5kN/m 6 = 18,0 mm = L/280 < L/200

3.2

W1,1

chosen: IPE 140 (W,,, = 88,3.10 mm3; Moment distribution due to q 3,6 kN/m

Plastic global analysis:

g=

— snowoa. Sk— wind load (uplift): w, =

permanent load:

Purlin (3-span-beam): Steel grade: Each span length:

3. Relevant load combinations for ultimate limit state verificationof purlins 3.1 Relevant load combination for moment distribution (permanent + snow (1,35G+ 1,5S)) q = 1,35g + 1,5s 1,35.1,0 + 1,5.1,5 = 3,6kN/m

-

I

TLfllTh______ _______ nII:TiFrrrrIIII:Io

2. Load

Example 1.1.2: Purfin treated 1. System

Problem, Sketch, Calculation

5

2,0

kN/m

1,0 kN/m

I

Fe 430 = 5,00 m

IPE 140

Table 4.2; (2.3.4]

4.2.1.(1);

Table 2.3;

Table 2.1 (2.3.3.1.(1)and (5)]

Table 2.1 (2.3.3.1.(1)and (5)]

Reference

ng

For roofs in general less stringent values may be used [4.2.1.(2)j.

R

This load combination leads to the following effects: - M,, (lateral torsional buckling) (reaction)

In view of erection an IPE 140 is chosen.

:

No Imperfection effects This load combination leads to the following effects: - max M, (strength, lateral torsional buckling)

U

Purlins are laterally restrained by

cladding —i sufficient resistance to prevent minor axis

Commentary

System

I

I

l.i1

I

from left side

11,1

wind

I

snow:

ILI_,LS

It

1260

.1,16..

i-2:51200J

storey frame

2. Load configuration and distribution of moments permanent:

1.

Example 1.1.3: SIngle

,9.6

t72.O

252

9660 12ZO

= 3,0

kN/m

HE 180A IPE 400 Fe 430

IRE 450 IRE 400

w4 w5

6

= 1,4kN/m = 1,2kN/m

= 1,0 kN/m or 1,4 kN/m (whichever gives a larger effect) w3 = 1,0 kN/m or 1,4 kN/m (whichever gives a larger effect)

w,

w, = 1,9kN/m

s = 4,5 kN/m

g

column (1) and (4): beam (2) and (3): column (6): beam (5): Steel grade:

Problem, Sketch, Calculation

Reference

y

the relevant load combination

Using first order theory for determining the internal = 1,0 for finding out forces and moments with

Commentary

3.1.1

3.1

3.

•

EV EV

Equivalent horizontal forces F6 = 1,35.[60,6 + 96,6) due to g: 212,2 kN = due to s: 1,50.[45,2 + 72,2) = 176,1 kM F = 212,2 + 176,1 = 388,3 kN F = •.F = 388,3/200 = 1,9 kN

n

*0 67.7

for ultimate limit state verification

DetermInation of the initial sway imperfections Column (3) does not contribute to the resistance = 2 ; n, = 1 = (see table 5.6 in E-EC3) = 1/200

Relevant load combInations permanent + snow i,35g + 1,5s

169

'L

Imperfections (equivalent horizontal unit force)

W2

wind from right side

"p

J,25.2

25.2

I.'

Problem, Sketch, Calculation = 1,9kN/m

7

w2 1,0kN/m or 1,4 kN/m (whichever gives a larger effect) 1,0 kN/m or 1,4 kN/m w3 (whichever gives a larger effect) w4 = 1,4 kN/m w6 = 1,2kM/rn

w,

Table 5.5; [Figure 5.2.4)

5.2.3.1.2; Table 5.6 (5.2.4.3)

5.2.1.2; (5.2.1.21

Table 2.1; [2.3.3.1.(5))

Reference

N

RR

This load combination leads to following effects: - max at column (4) and (6) • max M,, at 2, 3, 4 and 6 - max at 5 and 9 is needed for base plate design. For foundation design characteristic values are needed.

Commentary

1.0

kN/m

8

Not all possible load combinations are given in this example. It could be that for different elements different load combinations are relevant. E.g. for a beam-column two checks are necessary: max M and associated N (1) max N and associated M (2) Therefore the load combination permanent + wind could also be relevant.

Furthercommentary

3.3

w2 and w3

Maximum effects due to load combination permanent + snow + wind = 1,35.(14,1 + 21,0 + 35) + 0,5.1,6 52,9 kN

.F

EquIvalent horIzontal forces F1 due to g: EV,4 = 1,35.(60,6 + 96,6) = 212,2 kN due to s: !V,4 1,35.(45,2 + 72,2 158,5 kN due tow: EV,4 = 1,35.(.11,2 .25,2) = -49,1 kN F = 212,2 + 158,5 - 49,1 = 321,6 kN = = 321,6/200 = 1,6kN F1

'

DeterminatIon of the initial sway Imperfections Column (3) does not contribute to the resistance 1/200 (as demonstrated in 3.1.1)

permanent+ snow + wind 1,35 (g + s + w) wind is considered from the right side with

MaxImum effects due to the load combination permanent + snow = max N at column (4): N,4 135.48,2 + 1,5.720 + 0,3.1,9 173,6 kN max N at column (6): N,4 1,35.18,0 + 1,5.27,0 173,6kN max M at 2 and 4: M,4 1,35.84,9 + 1,5.126,0 + 3,0.1,9 309,3 kNm max M at 3: M,4 1,35.52,2 + 1,5.77,9 = 187,3kNm max Mat 6: M,4 m 1,35.54,0 + 1,5.81,0 194,4kNm

3.2.2

3.2.1

3.2

3.1.2

Problem, Sketch, Calculation

Table 5.5; [Figure 5.2.4]

[52.4.3]

5.2.3.1.2; Table 5.6

5.2.1.2; [5.2.1.2]

Table 2.1 [2.3.3.1 .(5)]

Reference

ThIs load combination leads to the following effect: max R1,4 (reaction horizontal). R1,4Is needed for base plate design.

max R,,4 at 5 and øare equal N,4 at column (4) and (6).

Commentary

'I, I

II

I

I I I I

System and load configuration

Detail: Crane girder (two span-beam) wheel base = 3,00 m

3.1

3.

(lOg

,

g: 1,0.0,375.2,0.6,0 = 4,5 kN P: . 1,50.0,14.1,2.130 = -32,8 kN 4,5 - 32,8 = - 28,3 kN B,

+ 1,5P)

Relevant load combination for maximum uplift at the end support

Relevant load combination for maximum moment at the inner support (1,35g + 1,5(P + H)) - 1,35.0,125.2,0.6,02 - 12,2 kNm P: - 1,50.0,14.1,2.130.6,0 = - 196,6 kNm M,,, = - 12,2- 196,6 = - 208,8 kNm H: M,,, -1,50.0,14.13.6,0 = 16,4 kNm

.i, z ,4'H

9

Relevant load combinations for ultimate limit state verification Load position at mid-span see sketch Relevant load combination for max mid-span moment and max reaction force at support (3) (1,35g + 1,5(P + H)) g: 1,35.0,0703.2,0.6,02 = 6,8 kNm .1 P: 1,50.0,23.1,2.130.6,0 = 322,9 kNm _________________ M,,, = 6,8 + 322,9 = 329,7 kNm (2) 13) (1) = 1,50.0,23.13.6,0 = 26,9 kNm H:

2. Load assumptions permanent: g = 2,0 kN/m wheel load: P = 130kW wheel loads must be increased by a dynamic coefficient 1,2 (purely for calculation) Lateral force: H = 13 kN (lateral forces due to crabbing, surging, etc) L = 15 kN Braking force:

1.

Example 1.1.4: Crane girder

Problem, Sketch, Calculation Reference

No imperfection effects For the determination of the internal forces and moments Influence lines are used

Lateral forces and wheel loads can act together Braking forces are independent from lateral forces

No fatigue Is Included here

Commentary

3.4

—

—

+

+ H))

V

g: 1,35.0.625.2,0.6,0 10,1 kN P: 1,50.1,60.1,2.130 — 374,4 kN —. = 10,1 + 374,4 = 384,5 kN = 1,50.1,6.13 = 31,2 kN H:

1,5(P

perpendicular 1,359 + 1,5H

to crane girder

10

Relevant load combination for determining the horizontal forces which must be applied at the bracing system parallel to crane girder 1,35(g + P + H + L)

F'

(1,35g

Load position at inner support see sketch Relevant load combination for max shear force at support (2)

RR

3.3

4 1V L

Load position at inner support see sketch Relevant load combination for max reaction force at support (2) (1,35g + 1,5(P + H)) g: 1,35.1,25.2,0.6,0 = 20,3 kN F' P: 1,50.1,82.1,2.130 — 425,9 kN = 20,3 + 425.9 446,2 kN —. = 1,50.1,82.13 = 35,5 kN H:

3.2

'

Problem, Sketch, Calculation

Reference Commentary

n

System

o

Wi

Support load from crane girder (vertical)

wind from left side

snow:

814

Including a crane girder

2. Load configuration and distribution of moments permanent:

1.

Example 1.1.5: sIngle storey frame

Ii...'

11

60 kN 446 kN

1,2 kN/m

1,OkN/m

1,0 kN/m

R2 =

w2 w3 w4

=

= 4,5 kN/m

w1 = 1,9 kN/m

S

g = 3,0 kN/m

column (1) and (4): beam (2) and (3): Steel grade:

Problem, Sketch, Calculation HE 400 A IPE 600 Fe 430

Reference

Loads of crane girder are already multiplied with partial safety factors.

Using first order theory for determining the internal forces and moments with 1,0 for finding out the relevant load combination for each element.

Commentary

3.1.1

3.1

3.

35

so

132,'

(see table 5.6 In E.EC3)

.F

Equivalent horizontal forces F1 due to g: EV 1,35.30,4.2 = 82,1 kN due to s: EV 1,50.45,2.2 = 135,6 kN F = 82,1 + 135,6 = 217,7 kN = = 217,7/200 = 1,1 kN F,

.----

of the initial sway imperfections

2; n = 1 I k0k,0 '1' = 1/200

DeterminatIon

Relevant load combinations for ultimate limit state verification permanent + snow including crane loads 1,35 (g + s) + R0 +

Imperfections (equivalent horizontal unit force)

Support load from crane girder (horizontal)

Problem, Sketch, Calculation

R2

12

35,5 kN

Table 5.5; [Figure 5.2.4]

5.2.3.1.2; Table 5.6 [5.2.4.3]

5.2.1.2; [5.2.1.2]

Table 2.1; [2.3.3.1 .(5))

Reference

R

N

shall be considered.

If this criterion is fulfilled member imperfections

1> 0,5

with the criterion:

ber imperfections may be neglected [5.2.4.2.(4))

For the columns it must be checked whether mem-

For the determination of the initial sway Imperfections only self-weight and snow is considered.

This load combination leads to the following effects: - max at column (4) • max M, at 3, 4 and 5 - max at 7

Commentary

• •

Furthercomentary

3.3

13

Not all possible load combinations are given In this example. It could be that for different elements different load combinations are relevant, e.g. "permanent and snow without crane loads" (1,35 g + 1.5 s).

Effectsof load combinationpermanent and wind max M, at 2: M, 1,35.(42,2 + 53,8) + 78,0 + 155,7 + 0,4.3,5 = 364,7 kNm max M,, at 6: M,, = 1,35.(42,2 .87,3) - 145,1 - 92,8 - 0,4.3,5 = -300,2 kNm

Equivalent horizontalforces F1 due to g: EV = 1,35.30,4.2 = 82,1 kN F1 •.EV,, = 82,1/200 = 0,4 kN

n

DetermInation of the Initial away Imperfections = 2; n, 1 = k6k,cc (see table 5.6 in E-EC3) = 1/200

+ R,,,

3.2.2

3.2.1

1,35(g + w)

+

permanent + wind

M

3.2

N

Effects of load combinationpermanent and snow = max in column: Ne,, 1,35.(30,4 + 45,2) + 436,4 + 12,4 + 1,10,5 = 551.4 kN maxM,,,at4 M,, 1,35.(85,4 + 127,2) + 4,0 + 8,3 = 299,3kNm = max M,, at 3 and 5 1,35.(60,3 + 89,9) + 81,4 + 116,0 + 1,1.5,0 = 405,7

3.1.2

Problem, Sketch, Calculation

Table 5.5; [Figure 5.2.4)

[5.2.4.3)

5.2.3.1.2; Table 5.6

5.2.1.2; [5.2.1.21

Table 2.1 [2.3.3.1.(5fl

Reference

This load combination leads to the following effect:

N in column.

-max Mat2and6

Is equal to

Commentary

1.2

Methodsof analysis These examplesdemonstrate on continuous beams how the design values of action effects are determined using either plastic global analysis or elastic global analysis and plastic or elastic stress distribution. AU of the methods of analysis presented mayalso be applied on frames. Thefurther steps of design are not treated in this part. Note:

For plastic global analysisspecial requirementsspecified in [5.2.7, 5.3.3 and 3.2.2.2] shall be satisfied.

Thefollowing examples are included in this chapter: Example Example Example Example

1.2.1: 1.2.2: 1.2.3: 1.2.4:

Continuous beam (plastic - plastic) Continuous beam (elastic- plastic) with limited redistribution Continuous beam (elastic - elastic) withoutredistribution Continuous beam (elastic - elastic)

14

.

L!

I

6.0

I

I

I

I

.

I

0

I

I

I

I

I

—

6.0

tf

,rcx

A

I I

d/t =

—

55.3

6.0

I I

1

1

—# A. = A; W,1 = W,,

— class — class

IjMPtRd

lIMpLRd

-*

nfl-

I 4,q=17.9kN/m

\.J_'

A

55,3

mm3

,

I

no shear buckling verification is required

28,4 < 69€ = 63,5

I

nfl-

I

Fe 430 —.— 0,92 Flange: c/t 50/8,5 = 5,9 < 10€ = 9,2 Web: d/t = 159/5,6 = 28,4 < 72€ = 66,2

2.3 WIdth-to-thickness ratios

55.3

A

55,3

I

- 220,9-1O

221.10 mm3)

17,9.6,02.1,1 •10 11.666275

I

2.2 PlastIc bending momentdistribution for IPE 200 max q, = 17,91 kN/m

— chosen: 1PE 200

qd'-yu0 11,666f

2. Determination of internal forces and moments 2.1 Plastic global analysis

1. System

Problem, Sketch, Calculation Example 1.2.1: Continuousbeam (plastic - plastic)

15

I I IPE Fe 430

5.3.3; [5.4.6)

[5.33(4); Table 5.3.1)

Reference

275.221.102.10/1,1

Plastic global analysis may be used only where the cross-sectionssatisfy the requirements specified in [5.3.3).

= f.W,,,/'y,,,, = 55,3 kNm

The plastic hinges occur at the first inner support and in the outer span at 2,568 m from the outer support.

The beam Is laterally restrainedto prevent lateral torsional buckling. Further, lateral restraints have to be provided at all plastic hinge locations,see [5.2.t4 and 5.3.3]. Fe 430 satisfies the requirements specified in [3.2.2.2]

Commentary

1

1A'

d/t =

d/t

Web:

cit

234,61O mm3

—. e

58,7

69,0

= 0,92

54.1

30,0 < 72 = 66,2 --—. A.,, = A; We,, = Wp, 30,0 < 69€ = 63,5 --— no shear buckling verification is required

177/5,9

55/9,2 = 6,0 < 10€ = 9,2

Flange:

Fe 430

2.2 Width-to-thIckness ratios

tt,

58,65•1.1 .10' 275

28,7

460

46,0

chosen: IPE 220 (W,, = 285.10 mm3)

w M

\Thy/

587

2.1.2 Moment after redistribution

69.0

2. DeterminatIon of internal forces and moments 2.1 Elastic global analysis 2.1.1 Moment before redistribution

1. System see example

16

0,15.69,0 = 10,4 kNm 49,7 + 0,42.10,4 = 54,1 kNm 18,5 + 1,14.22,0 = 43,6 kN ---—. no interaction bending - shear is required

Vu0

v4 - _____

0,5VRa

= 431,4 kN

21

For comparison: Moment distribution (calculation by computer)

= 376,2 kN > 200 + 1,14.27,5 = 231,4 kM no Interaction bending - shear is required

V,..,,,

5.4 AxIal compression

Column:

Beam:

5.3 Shear resistance of the cross-section

- W,,, —-

5.2 Moment resistance

N°4,, = 400,0 + 1,14.27,5 = 431,4 kN = 200,0 + 1,14.27,5 = 231,4 kN 18,5 + 1,14.22 = 43,6kN

—

av • 1-0.125 -114

M'4, = 92,6 + 1,14.110

1

-

5. VerificatIon using 1' order theorywith amplified sway moments 5.1 internal forces and moments

Problem, Sketch, Calculation

5.3.1 [5.4.4]

5.3.3 (5.4.6]

53.2; [5.4.5]

5.2.31.3; (5.2.6.2]

Reference

1

hEH

6Ev

I —

i.M! H

I

In the amplified sway moment method the sway internal forces and moments found by first order elastic analysis should be Increased by multiplying them by the following factor:

Commentary

=

1,11

{1

-

N)

-

1.11

I

E

1

).

112

93,9

=

o,

S9).v4i

3,9810 .J_.

-

1

+

wyrA

252,7

ktlm> 92,6

= 0,183 +

1283

.io3.j

1,005 218,0

10' —

0,98 < 1

1283.10/1150.1t? - 1 = 0,116 + a = 0,36 (2.1,8-4) + 0.116 = -0,028 < 0,9 + 0,028.0,183/1,1 1,005< 1,5—.k = 1,005

!L.

K, M2

1

W1,1/W, 1 A2(28M -4)

—

• 11O•1,14-218,0 kNm

22

this check is fulfilled

- 0,9839.118.102.235 10 = 2354,3 kN > 431,0 kN

NSJN- = 431,0/2354.3 = 0,183

=

-pn/Ml =

n = a = p =

N,,,

using buckling curve b — Xb = 0,9339

-

5.6 Member resistanceof the column Buckling length of column: = 3,98 m From the moment distribution —i = 1.8

5.5 InteractIon

bending - axial compression

Problem, Sketch, Calculation

5.4.2; (5.5.1; 5.2.6.2.(2)]

5.3.4; [5.4.91

Reference

When second order elastic global analysis is used, in-plane buckling lengths for the non-sway mode may be used for member design.

Commentary

I

I II

I

I

I

OW

II I I I

kJLI

I

II I I_I_I_liP

LLL4

I1I1lIIII'9,

-

1* 1IIIIIIIIt_I

111111111

I I II

j

15.00

J.

15.00

1,V

lS.O0

I

s

p

2.1.2

1

+

1,35.[6,0 + 4,5 + 2.(10,0 + 5)1.15,0.3,0 = 2460,4 kN

s)

6,0 kN/m = 10,0 kN/m — 5,0 kN/m = 4,5 kN/m —

= 1,35.[6,0

+ 4,5

+ 10,0

+ 5].15,0.3,0 = 1549,1 kN

n0

23

The axial force of each column at this floor is higher than 193,6 kN, therefore all columns are included in n0

Half of the "mean reaction force' per column: 0,5.V = 0,5.EV,/4 = 0,5.1549,1/4 = 193,6 kN

EV

floor

The axial force of each column at this floor is higher than 307,6 kN, therefore all columns are included in

Half of the mean reaction force per column: O,5.V 0,5.EV,,/4 = 0,5.2460,4/4 = 307,6 kN

EV =

2. DetermInation of the Imperfections for the load case 1,35 (g + p 2.1 Total vertical reaction per storey 2.1.1 Ground floor

'S

j 11IIII1lI Fillilill J.IllllIII I I fI I I I I I I_I',, ,,uIII I Ill_LII Lii u

"I

I I I II 4IlI1IIlI

F111I11111 11II1I11I

TI

1. System and loading

Example 1.3.2: DetermInation of frame imperfection.

Problem, Sketch, Calculation

5.2.3.1.2; (5.2.4.3]

5.2.3.1.2; [5.2.4.3]

Reference

Commentary

l,35.[6,0

+ 451.15,0.3,0

= 637,9 kN

!!.

2.!,

*

3

• • = 1/315

4; n

=

=

= 2,0 kN = 2,9kN

n

F0, = 911,3/315 = 2,9 kN

= 637,9/315 F, F,, 911,3/315

(see table 5.6 in E.EC3)

'iii

=

2.2 DetermInation of the Initial sway imperfectIon All columns extend through all storeys and therefore are included In

24

The axial force of each column at this floor is higher than 79,7 kN, therefore all columns are included in n,

Half of the "mean reaction force" per column: = 0,5.EV/4 = 0,5.637,9/4 = 79,7 kN

EV

2' floor

2.1.3

Problem, Sketch, Calculation

[5.2.4.3]

5.2.3.1.2;

5.2.3.1.2; [5.2.4.3)

Reference Commentary

1.4

Bracing systemanalysis The first three examples demonstrate the application of the criterion "braced - unbraced". The fourth example demonstrates how bracing Imperfectionsare determined. The further steps of design are not treated in this part.

Thefollowing examples are included in this chapter: Example Example Example Example

1.4.1: 1.4.2: 1.4.3: 1.4.4:

A frame is braced by a bracing system A frame is braced by a frame A frame is braced by a bracing system Bracing system imperfectionsfor a wind bracing

25

of the frame

1500

pE400

7

=

2,55,0

1.0

= 15336 kN/m

21000010,2102oos226,6

dx

1177,6 kN/m

2.lO

co

10,0 cos26,6

•10

- 0,000065 rikN

- 0,00085

column: beam: column 1: column 2: beam: diagonals:

2,52.7,5 •21D 3•2100O0•23130•10

Assumption: EA

Bracing system:

Frame:

m/kN

HE12OA IPE 400 4' 36

IPE 500 IPE 400 HE 120 A

——.

4t1

JJ

26

= 15336/1177,6 = 13,02 > 5 the continued system can be calculated as a unique system or separated into sub-systems where the frame may be considered as braced and the bracing system is subjected with the horizontal forces

4. CriterIon

=

M'

321000048200•10

S--

6

s-f!

—

or

system

46' • _________

j• M

S,,

JOE 500

3. Stiffness of the bracing system

2. StIffness

500

1. System

Example 1.4.1: A frame Is braced by a bracing

Problem, Sketch, Calculation

5.2.3; Table 5.3 [5.2.5.3]

Reference

A frame may be classified as braced if the stiffness of the bracing system Is at least 5 times greater than the stiffness of the frame.

Only the bracing diagonal loaded in tension Is used to calculate the system stiffness (since the compression diagonal has insignificant resistance.

laterally.

A braced frame may be treated as fully supported

rizontal loads are transferred (i.e. any horizontal loads plus the Initial sway Imperfections applied on a braced frame are treated as affecting only the bracing system). However It Is possible that the bracing system itself could be classified as sway.

The criterion braced - unbraoedgives an indication

to which part of a structure or sub-structure the ho-

Commentary

of the frame.2

3. StIffness

-1

—1

o

frame.2

6

=

6=

= 454,0

4Ji

.— frame.1 may be considered as

H

kN/m

32100005920010

5,02•10,0

210+

210° +

5,02.6,0

3210000•9208010

5,0210,0

00

3•2100O011770•10

Assumption: EA

3•210000•1045010

• 85,9 kN/m

Frame 2:

Frame 1:

Assumption:EAoo

,ll

5,02.10,0

.f-!dx

j

::

S,.2/Sh.l = 454/85,9 = 529 > braced

4. CriterIon

of the frame.1

2. StIffness

frame.1

jHE2soA

1. System

Example 1.4.2: A frame Is braced by a frame

Problem, Sketch, Calculation

=

0,0116 m/kN

27

210 = 0,0022 m/kN

210'

column: HE 260 A beam: 1PE 330 column: HE 300 M beam: IPE 600

5.2.3; Table 5.3 [5.2.5.3]

Reference

Frame.1 may be classified as braced if the stiff. ness of lrame.2 Is 5 times greater than the stiffness of the frame.1. The continued system may be separated into subframes where the frame.1 may be considered as braced and the frame.2 which Is designed to resist the horizontal forces.

when It Is sufficiently stiff to resist all horizontal loads.

A frame may be assumed as a bracing system

Commentary

Sfr,mSI

dx

= 4058,3/293,8

Sfr52

-

1,332.5,0.2.109

+

kN/m

3210000•23130•10

13,8 > 5 —. the frame could be considered as braced

4. Criterion

dx4f.ME

293,8 kNjm

+

co

-

IPE 400 IPE 400 IPE 120

1PE 400

IPE 400

•2•10' 0,0034 mfkN

column: beam: column: beam: Diagonals:

3•2100002313010

4,027,5

Assumption: EA

•2•10'

8,00

Bracing system:

Frame:

+

1,332.2,5.2.103 3•210000•23130•10

28

= 0,00024641 m/kN

5,O•10 6— 0,272.8,0.2.103 + 0,942.9,434 •2•10 + + 0,52.5,0.103 + 210000.84,5.103 210000.13,2.103 21000084,5•102 210000.84,5,102

6=1

4,03.8,0

321000023130•10

-+

=

H6 =1 --dx +f .MM dx

is braced by a bracing system

3. Stiffness of the bracing system

2. Stiffness of the frame

1. System

Example 1.4.3: A frame

Problem, Sketch, Calculation

15.2.5.31

5.2.3; Table 5.3

Reference

The frame may be classified as braced if the stiffness of the bracing system is at least 5 times greater than the stiffness of the frame. The frame may be considered as braced and the bracing system is designed to resist the horizontal forces.

For the calculation of the system - stiffness both diagonals are considered (The connections of the diagonals must be formed such that the compression forces can be transferred).

Commentary

JN2sa

,LN2Sd

J?Sd

IN2sd

tN2sd

2Sd

1-

Jjl2Sd

—

,tRSd

,!.$Sd

J35d

JSd

,j$Sd

,LllllllIIlllllll_,tw

N

Frames: column: beams: Bracing: Diagonals: Chord 1: Chord 2: IPE 400

L 60x40x5 HE 120A

IPE 500 IPE 400

w = 1,5.2,0 3,0 kN/m Mid - span moment of beam: M,, = 310 kNm Axial force applied on the bracing system: = M,,,/h = 310/0,4 = 775 kN

2. Loading

for a wind bracing

ll

imperfections

67,9

3100,0 20,0 2,28 kN/m

--—-.

6=9,5mm 600 kN

gCk2

go

A

3. DetermInation of the resIstance 3.1 Diameter-to-thIckness ratio Fe360— = 1,0 d/t = 219,1/4,5 = 48,7
2900 kN

10 = 3009,6 kN > 2900 kN

035

Range: cit = 150/19 = 7,9< 15€ = 15 Web: d/t = 208/11 = 18,9 < 42€ 42

= 2900 kN

— 4,20 m 0,88.3,0 m = 2,64 m

= 0,7.6,0 m

Fe 360 L = 6,0 m

of the resistance

Wdth.to-thlcknessratio Fe360—.e = 1,0

)z15

3.1

3. DetermInation

2. Loading Compression force:

Column: HE300B Steel grade: System length: Buckling length:

using buckling curve b

=

3.2 Member resistance 3.2.1 Buckling about the strong axis = 4,20 m Buckling length:

1. System

Example 2.1.2: HEB profile as column

Problem, Sketch, Calculation

33

Table 5.5.3]

(5.5.1;

5.4.2; Table 5.21

Table 5.5.3)

(5.5.1;

5.4.2; Table 5.21

Table 5.3.1]

(5.3;

5.3.1; Table 5.9;

Reference

Determination of

,

x

see tables in Appendix A

The stability of compression members needs to be checked according to the two principal axes of the section with appropriate effective lengths. Determination of see tables in Appendix A

The whole cross.section is effectiveto resist the loadings.

'

The determination of the buckling length Is based on a simpleassumption (Annex E.2.1J:

Commentary

Detail

.—. A,, = A

l— =

N

210000 280,9

859 '

=

>

1.1

io

641

04885.6,41 .10228,09

using buckling curve c .—. x. = 0,4685

A1 =

k n t2

'

1 23,1

85,9

2,25.1021

76.7 kN > 70 kN

I A1

- 360 N/mm2

34

- 1,13

3OrJk, = 56,5

aid = 7200/900

—'

= 3,0 kN/m

Beam:

Dead load: g Reaction force: P 1585,0 kN Relevant load combination of this load case: 1,35.g + 1,5.P

4. DeterminatIon of the resistance 4.1 Width-to-thIckness ratios Flange: tf = 50 mm > 40 mm —i Fe 510 —. e = 1(235/335) = 0,84 c = 300- 12/2 - 12.7 = 284,1 mm c/t = 284,1/50 = 5,7 < 14 11,8 Web: Fe 510 e 0,81 d/t = 900/12 75 < 124o 100,4

72279

L

liii II II III.

4'____________________H I___

I4

1

3. Internal forces and moments

2. Loading

1. System

Example 2.2.3: Single span girder with

Problem, Sketch, Calculation

[5.4.6.(7)J

Table 5.13;

[5.3; Table 5.3.1)

[2.3.3.1 .(5)J

2.3.2.2; Table 2.1

Reference

Note that where the plate thickness is greater than 40 mm the yield strength has to be reduced. In the classification of welded cross-sectionsthe welds are taken into account.

Dead load includes the weight of floor, etc.

The beam is continuously laterally restrained. Cross-section propertIes A = 708.102 mm2 I = 1427900.10kmm4 throatthickness between flange and web: a = 7 mm

Commentary

- J4 - 0,55.210

—

37,4 •0,81

- _____________ - 1 '

206,9>75

=

=

1,1

900'12171 .10-' = 1678,9 kN> 1217,9 kN

f/13 (1,5- 0,625 A) = 35,5/13 (1.5- 0,625.1,065).10 = 171 N/mm2

s vc

4.7 interaction bending - shear buckling verification ——. not necessary because of M,, < Mf Rd

4.6 Determination of the design plastic moment 9002.12/4.355.loa/l,l = 784,2 kNm MWRd = d2tf/4/1,1 = = + M, Rd (M Rd MRd) 8679,5 + 784,2 9463,7 kNm

4.5 Bending moment resistance W, = 28558.10 mm3 = 28558103•33510R/1,1 = 8697,2 kN > 8664,OkNm M,Rd = WdfV/IMO

4.4 DeterminatIon of the design moment resistanceconsistingthe flanges only = kNm M, Rd = A (h - t) f,/'M0 = 600.50.(1000 50).335.10'/l,l 8679,5 kNm > 8664,0

1188,0 > O,SVtM = 839,5 kN —+ Interaction bending - shear is required

v,,4

Tb.

374

d!t

4.3 Shear buckling verificationof web

- 73,4 1,0—e k =5,34 (a/d)2

—_±-—

=5,34

mm;

1,0812

- 0.22

= 203,9.102 mm2;

e

= 20,55 mm

W = 8542.1O

1

138,8 e

mm3

= 177 mm = 265mm

t

.!.

= 0,737

dlt

28,4

0,4.p.d/2 = 0,4.0,737.1200/2 0,6.p.d/2 = 0,6.0,743.1200/2

p

'

1,081

1,0

- 0.22 -

1200

i/o =

= 547148,5.10 mm4

d,,,7

df1

p

• I

b=d=

Fe 360 —.

138.8

- 150,0._.i!._ -

> 0,673

44

1,081

+ —--— 1,752

=6646

Welded section Dimensionssee sketch Steel grade: Fe 360

d/t = 150,0 > 30€/k, = 77,3 —+ shear buckling verification is required

=

1200/8 = 150,0 > 124e = 124,0 Determination of W0

a/d = 2100/1200

d/t =

Web:

V

Part of a Beam:

Bending moment distribution: see sketch Shear force: 700,0 kN

- 8/2 - 12.5 138,9 mm 138,9/20 = 6,9 < 14€ = 14,0 Flange Is fully effective

150

cit =

=

Range:

C

of W, of the web

'1177

3.2 Determination

1200 KNm

loaded In bending

Fe360—'€ = 1,0

3. DetermInation of the resistance 3.1 Width-to-thickness ratios

2. LoadIng

4L(

1. System

Example 2.2.4: Class 4 cross-section

Problem, Sketch, Calculation

5.3.5; [5.3.51

5.3.3; [5.4.6)

5.4.5.2; [5.6.31

Table 5.3.1)

(5.3;

Reference

The effective width of the web is calculated on the basis of f, and 'P = -1.

Flange is fully effective = (265+177+593÷212.5).8 + 300.20.2 — 203,9.102 mm2

welds are taken Into account.

By the classification of welded cross-sections the

Cross-sectionpropertIes 216.102 mm2 I — 561760.10kmm4 throatthickness between flange and web: a 5 mm

A

Commentary

d/

=

-

150

374

= 1

556> 1,2

1,1

—

f=/131/A2 = 235/13.1/1,5562

=

=

0,25btf,

[

-

(btffYMO)I

0,25btf - (bt,,j2]

=

/f - 3,

+

-

=

235 -

356,02 + 53,52

-

-

167,1

(30

- 53,5 =

- 0,25.300.202.235

0,25.300.202.235

3.4.3 Determinationof the strength of the tension field 8 = arctan d/a = arctan 1200/2100 = 29,7 conservative assumption: 8/1,5 = 19,8 = 1,5r sin(24) = 1,5.56,0.sin(2.19,8) = 53,5 N/mm2

M,

MrJtflk

5/1,1)]

Nmm2

3.4.2 Determination of the reduced plastic moment resistance M1,: On the right: N,,, = M5J(h -- t,) = 1200/(1240.- 20).10== 983,6 kN left: On the N1f, = M/(h tf) 620/(1240 20).10 508,2 kN

3.4.1 lnitiaj shear buckling strength 1,556 > 1,25 —. Thb =

3.4 Shear buckling verificationusing tension field method Requirement: 1 700 kN

= 644,2 mm

sin19,8

19,8

I594•10'6mm 8167,1

8•167,1

I 2,90 •10'

considering the flanges only 300.20.(1200 + 20).235.10/1,1 1563,8 kNm > 1200,0 kNm

of the design resistancemoment

A, (d + t,)

2

sln4'

= 1200•8•56,0+0,9•644,2•8•167,1 1,1

3.4.8 Interactionbending - shear buckling —' not necessary because of < M,,,4

M=

3.4.7 DetermInation

=

4' = 1200.cos 19,8

3Lt

= 359,8 kN 700 > —.- interaction bending - shear is required

0,5V

-

3.4.6 Shear buckling resistance

3.4.5 Width

of the tension field g = d.cos 4' - (a- s,, - s,).sin

3.4.4Anchorage lengths of the tension field

Problem, Sketch, Calculation

.; (5.6.7.3.(1)J

.; [5.6.7.3.(1)]

-; [5.6.4.1.(1)J

-; [5.6.4.1.(3)1

-; [5.6.4.1.(3)J

Reference

Commentary

2.3

Combined loading - Bending and compression These examplesdemonstrate the verification of members loaded by the combination of bending and compression assuming design values of action effects (N. etc) which have been calculated by an analysisof the structure and these valuesalready Include 4v, YF etc.Thesecond order effects are consideredby usingfirst order elastic analysiswith sway-mode buckling lengths. The results presentedin the examplesare rounded values. For the purpose of easyre-calculation each formula and each check is calculated with the rounded values.

V, M,

The following examplesare included in this chapter: Example 2.3.1: Example 2.3.2: Example 2.3.3: Example 2.3.4:

RHS column loaded In bending and axial compression HEA profile loaded In bending and axial compression Column of a frame Rafter of a frame

47

0 0 0

800

400

36

18

Detail:

and axial compression

W11

=

W

= 76,4

_i_

-

96,7

86,8

.

10,8510

= 0,4760' 91.7.102.275 1,1

using buckling curve a —. x. = 0,4760

I

I-!

= 10,85 m

io-

—i no shear buckling verification is required

d/t = 22 < 69e

A 63,5A;

= 22< 83c d/t= 220/10 —4 =

Web:

d/t=220/10=22 200 kNm

- 1874,9 kNm

= 0,9501

+ 0,039.1,52.101.26,5.104

> 0,4 using buckling curve a —. Xir..

4410

1,52.1012

. 1,1992210000

!I!_!i - 0,9501 . 907,010-355

1

= XLTA

+ 0,039

=133/200—.C1=1,199

k=k_=1,0;

1,5 m

—

Buckling length

3.2.3 Interaction Bending - axial compression without lateral torsional buckling phenomena = N/NYR,I = 100,0/719,2 = 0,14

M

3.2.2.2 SectIon 2

Problem, Sketch, Calculation

5.4.4; Table 5.26 [5.5.4]

5.4.4; Table 5.26 [5.5.4]

5.4.4; Table 5.26 [5.5.4]

5.4.3; [5.5.2)

Reference

Commentary

Z?SO

(LU

t

Detail of lateral restraints

mm;

0.78

0.78 - 0,22

28,4

cit

0,92.94 = 86,5 mm

- 0,22

= 94

C• = p.c

-

b=c

Flange:

- 092

t

2

18,6 e

1

f

11,8 >

14 11,3

94 mm

d/t

56

Determinationof A,,, and W•, 150 > 69 = 55,9 —. shearbuckling verification is requi. red

d/t=600/4—150>124e=100,4

Web:

c/t = 94/8 =

3. DeterminatIon of the resistance 3.1 Width-to-thIckness ratios Fe 510 —. e — 0,81 Flange: c = 100 -2- J2.3

see sketch

2. LoadIng

Welded section Steel grade Fe 510 Dimensions: see sketch Buckling length: 14,2875 m Rafter laterally restraint at A, B, C, E, F

- 11,8.1 - 0.78 > 0,673 18,6

3.2 DeterminatIon of effective propertiesof the cross-section 3.2.1 A,,,, for compressIon Fe 510—' 1/e = 1,23

_________H20®

25O

Z2O

L1 I

1. System

Example 2.3.4: Rafter of a frame

Problem, Sketch, Calculation

5.3.5; Table 5.18 [5.3.5]

5.3.3 [5.4.6J

Table 5.3.1]

[5.3;

5.3.1; Table 5.10

Reference

l

-

The verification of shear buckling Is not contained in this example.

The classification with the exact stress distribution leads also to a class 4 section.

l

Weld between flange and web: a = 3 mm A 56.102 mm2 = 36775.10' mm4 256.3 mm I, = 1067.10kmm4 43,6 mm I, The buckling length is calculated separately.

Commentary

-

- 0.22 - 029

0,29.592 = 171,7 mm

3,202

3,20

284 ' efV

t

- 0.22

-

-

28,4

dit

1,31 0,22 1,312

—

0,64

t

-4

-

138,8 e

1 —

138,8

148.._1!_

W•

= 33698.10k mm4 = 33698.10/331,7

lA

i

A

[

N

256,3

NYb

=

.

0,8420.36,8102355 .Ø-3

0.59

= 1000,0 kN> 120,OkN

36,8.102 76,4 N 56.102

14,2875101•

= 14,2875 m

using buckling curve b ----- Xb = 0,8420

'=

f

1016.10 mm3

— .

cr::

9Z5

tJ=

"

:::CR:::

23.7

113.7

t179.s 1106.7

'I'

57

+ 327,22.179.8 + 285,82.80.4 + 413,9.4.69,452

50,5.102 mm2

- 1,31 > 0,673

56,8 - 3,20 > 0,873

dff1 = 0,4.p.d, = 0,4.0,64.05.592 75,8 mm = 0,6.p.d, = 0,6.0,64.0,5.592 113,7 mm dm112 = Af, (2.86,5 + 12 + 200).8 + (75,8 + 113,7 + 0,5.592 + 8).4) = e = (608.185.8 + 80.4.564 + 413,9.4.211)/505 = 280,3 mm eM = 304 . 280,3 = 23,70 mm = 4.(80 + 413,9)/12 + 2.8.(179 + 200)/12 + 280,32.200.8

p

b = d = 592 mm;

—

56,8 e

36,8.102 mm2

Flange: The compressed flange is reduced as above = p.c = 0,92.94 = 86,5 mm Web:

W for bending

3.3 Member resistance verification 3.3.1 Axial compressionresistance Strong axis: Buckling length:

3.2.2

= p.d

- 0,22

-

A = 2.(2.86,5 + 12).8 + (171,7 + 8).4

d•f1

= A0

b = d = 592 mm;

Web:

Problem, Sketch, Calculation

5.4.2; [5.5.1)

5.3.5; Table 5.18 (5.3.5)

Reference

) a class 4 section the radius of gyration about the relevant axis from the gross cross-section should always be used. For determining the slenderness of

Commentary

A

1A

A1 N

Yin

-

Yin

using buckling curve

A

1

A1 N

lI

E

=

c,

n2E

l

M€

w.

-

j

58.102

I, = 4,50 m

055

—' C,

1016102.355 2515,3•10e

- 0,38 200 kN

15.7.4.(2)J

5.6.4(2);

5.6.4(1); [5.7.4.(1)]

5.6.3; (5.7.3]

-; 5.7.3.(4)

Reference

the member Is subject to bending moments, the interaction criterion has to be fulfilled Where

Where the details are such that there Is doubt over which mode (crushing, crippling or buckling) governs, all three modes should be considered. kR is a constant taken as 3,25 for crane rails mounted directly on the flange

I. — 45296.1O mm4 Throat thickness of the weld connecting the flange and the web: a — 5 mm The crane rail is not welded on the flange.

Commentary

I 11

1_L

3,5

93,9

0,36•101

A

Viii

1.1

-- - 0,484243,2-10•- •iO

= 0,4842

110

2O7

104,2 N/mm2

_(i)2+(.i2)

= N,j(s•tJ = 200.10/(16O.12) =

(+()2_

Compression:

s•=2.80=l6Omm

,t128

I

o = 235/1,1 =

68

=0,23 +0,24-O,48•0,49 =0.23 200 kN

3.4 Cross-section resistance verification = M,,.z/i = 366.10'.128/45296.1O = Compression:

-

using buckling curve c .—.

i

3.3.2 BucklIng resistanceverification Assumption: buckling length = 0,360 m A — 360.12 = 43,2.102 mm2; I = 360.12/12 5,184.10k mm4 I = 15.184.104/43,2.102 = 3,5mm

3.3 BucklIng resistance 3.3.1 DeterminatIon of the effective breadth of the web buff • h = 360 mm

Problem, Sketch, Calculation

-; [5.4.10]

5.4.2; [5.5.1]

5.6.5; [5.7.5]

5.6.5; [5.7.5]

Reference

The effect of the transverse force on the bending moment resistance of the member should also be considered.

When the detail Is such that there Is a doubt as to whether buckling mode governs, it should be considered.

Where forces are applied through one flange and resisted by shear forces in the web, the buckling resistance of the web to transverse forces need not be considered.

Commentary

(ij

IPE 160

KEII0A

jj

(j

supported

by a beam

(s,

+

RYRd

=

3.1.2.3 Verification

(S,

3.1.2 Beam '2" (1PE 160)

=

3.1.1.3 Verification

____________

YMI

=

s) t, .-'Ml

=

M/W.

of; =

=

IL -

twf1

.?.

>60 kN

29,5

)]

______ "'°)1 - 27,4 f,i

= 147 N/mm2

43,5).5•. •10 = 92,0 kN > 60 kN

2t0

25t1

16.10°/109.10

+

103,3 kN

1

187 N/mm2

30,3 mm

4.15.(1 - 12/2) = 42,6mm

= 82 mm < 185mm =

(42,6 +

b1

=

3.1.2.2 Determination

+ 2.9,5

YMo

f,

01 Ed

•iO -

f[ -(

-- [

50,3).6.

2t1 I

s, = 6,0

= (30,3 +

•

3.1.2.1 BearIng stiff length

s)t

b, =

of

4.9.(1 - 12/2)

180 —

-

21

)

69

r]

55 kNm 16 kNm

43,5 mm

1871.1 235 1150,3mm

14711 ( 235

-(

2. Loading Reaction force: R,d = 60 kN Bending moment: Beam '1": Beam 2":

55.10/294.10 = 180 mm < 237,5 mm = 25t1

=

3.1.1.2 Determination

3. Determinationof the resistance 3.1 CrushIng resistance 3.1.1 Beam "1' (HE 180 A) 3.1.1.1 BearIng stiff length 5, = 5,0 + 2.7,4 +

1. System

Example 2.4.5: Beam

Problem, Sketch, Calculation

5.6.3; [5.7.3(1)]

5.6.3; [5.7.3(1)]

5.6.2; [5.7.2.(3)]

5.6.3; (5.7.3.(1)]

5.6.3; (5.7.3.(1)]

[5.7.2.(3)J

5.6.2;

Reference

For rolled sections: t_ + 2t0 + 4r(1

s,

For rolled sections: = t_ + 2t6 + 4r(1

- 12/2)

- 12/2)

Commentary

M

+_!! ._12_ 188,2

89,4

—111

55 kNm

3.2.1.1 interaction

=

R =O.5t

3.2 Crippling resistance 3.2.1 Beam (HE 180 A) sjd = 30,3/122 = 0,25 > 0,2 —i sjd = 0,2

Problem, Sketch, Calculation

..

[541oJ

-; [5.4.10]

(5.7.4.(2)]

5.6.4.(2)

[5.7.4.(1)J

5.6.4.(1)

5.6.4.(2) [5.7.4.(2)J

[5.7.4.(1)]

5.6.4.(1)

Reference

Her.: Local buckling does not need to be checked

Commentary

Part 3 Botted connections Welded connections Pin connections

71

3.1

Bolted connections These examples demonstrate the verification of bolted connections assuming design values of action effects etc) which have been calculated by an analysis of the structure and these values already include p, YF etc. Theresults presented in the examplesare roundedvalues. Forthe purpose of easy re-calculation each formula and each check is calculated with the rounded values.

(N, V, M,

Thefollowing examplesare included in this chapter: Example Example Example Example Example Example Example Example Example

3.1.1: 3.1.2: 3.1.3: 3.1.4: 3.1.5: 3.1.6: 3.1.7: 3.1.8: 3.1.9:

Bolted connection of a tension member to a gusset plate Erection splice at mid span of a lattice girder Angle connected to a gusset plate Fin plate connection to H section column Fin plate connection to RHS column Flexible end plate connection Beam to beam connection with cleats Splice of an unsymmetrical I-section Bolted end plate connection

72

4.60.1.

120

a5.

aS

Yuo

15,17.102.235 1,1

.L& -

1,1

120•12-235

•iO

N -0,9

A,,.

Shear resistance of bolts:

0,6

f,

169,4

-

kN>

1,25

130,0 kN

o,68o03,53i02 •io-3

135,5

A3

i0 =292,9kN>260kN

--- .

1,25

307,6 kN > 260 kN

.0,9 11,3.102360

4. Determination of the connection resistance 4.1 Bolted connection

11

Tension force:

2. Loading

135,6 kN>

.

N = 260 kN

CHS 101,6x5,0 120x 12 120x 12 M 24, 8.8 —i d0 = 26 mm weld length: • = 120mm throat thickness:a = 5 mm

Member: Plate: Gusset plate: Bolts:

a gusset plate 1. System

•10 - 324,1 kN > 260 kN

3.3 Net section (Plate and gusset plate) A,, = (120.26).12 = 11,3.102 mm2

I

+

3.2 Gross section (Plate and gusset plate)

N

3. Determinationof the resistance of the members 3.1 Gross section (Circular hollow section)

12o]1

°

80

101.6'S

aS b.

Example 3.1.1: Bolted connection of a tension bar on

Problem, Sketch, Calculation

73

- 130,0 kN

Steel grade: Steel grade:

Steel grade:

Fe 360 Fe 360 Fe 360

Table 6.7

6.2.3.2;

[6.5.5; Table 6.5.3)

5.5.1; [5.4.3)

5.5.1; [5.4.3)

5.5.1; (5.4.3)

Table 6.7; [7.5.2)

Reference

e

Shear in thread

is covered by the design resiThe eccentricity stances for bolts and need not to be taken into account

Assumption:

Commentary

4

1,25100

aL

- 4 tL

YUo

—

80/(3.26) - 1/4 —

0,77; —

-

235

10

- 159,6 kN

•o -

130,0 kN

296,0 kN > 260,0 kN

> 260 kN

1,25

iO- - 498,8 kN

498,8 kN > 260 kN

1,25

1,25

YMb

74

- 2,5ctdt 2,50,fl3802412 •iO - 159,7 kN >

0,77; p1/3db. 1/4

. ° .!Li

Fb,w

60/(3.26)

- 45120

e1/3d.

Local shear resistance for CHS

FWpd

-

4.2 Welded connection (Plate - CHS) Resistance of the fillet weld

Bearing resistance:

Problem, Sketch, Calculation —

- 130,0 kN

0,77

5.3.3; [5.4.6]

Table 6.15

6.3.4.3;

[6.6.5.3)

Table 6.6

6.2.3.1;

Table 6.5.3)

(6.5.5;

Reference

(F < Fb,j.

CHS is critical for local shear failure. The verification for the plate may be neglected.

high deformation capacity is needed, the shear resistance must be increased.

Th. shear resistance is governing

Commentary

+

+

+

+

+

+

+

..L.I.

.1.

.1.

.1.

.L Sk

.1.

.1.

.1.

_T

N,,.,...,

H

+

+

ir:1K

=

. 0,9

E

U2

(A;

A1 -

A - max

•

- o,g

4p(

- 202

88,32.102 . 510 1.25

75

i0- = 3243.1 kN > 3000 kN

mm2

N = 3000 kN

Welded section Steel grade: Fe 510 see sketch 880x340x 16 mm thick Steel grade: Fe 510 M 24, 8.8 —' = 26 mm

- 2048.102

2. Loading Tension force:

Bolts:

Splice plates:

Member:

1. System

48016).2 (26.16.4 108,8.102 - 20,48.102 = 88,32.102 mm7

-

16,64.102 mm2

- 3511,3 kN > 3000 kN

= 4.26.16

108,8.102.355 .W-3

+ +

Yuo

OaOIBo.tBOsO

•i

':. H-

3.2 Net section

84..

34 Ox 16

260x12

Welded section: A = 340.16.2 + 260.12 = 140,0.102 mm2 Splice plates: A = 2.340.16 108,8.102 mm2

3. DeterminatIon

3.1

+

-i-

of the resistance of the members Gross section I -

I

+

O8O8O.8O48O4.6OO.IRO18O BO.I.BOOl.

+

+

/7\/\/\//\\//\ +f :::::::t:j :i

Example 3.1.2: ErectIonsplice at mid span of a lattice girder

Problem, Sketch, Calculation

5.5.1; [5.4.3]

[5.4.2.2. (4)]

5.5.1; [5.4.3]

Table 6.7; [7.5.2]

Reference

A,.

s 1,11

u Yuo verification is necessary.

If

A then only net section

Commentary

Bearing resistance:

Shear resistance of bolts:

bAd

Fb.Rd

—

10

235,4,16

Yin,

2,5atdt

301,3 kN> 150,0 kN

1,25

- 1/4 =

= 150,0 kN

1,8; .—. a = 0,77

210

76

•io = 301,6 kN> 150,0 kN

160/(3.26)

•iO = 173,6 kN >

2,50,77-51024'16

0,77; ps/3d, - 1/4

1,25

0,6 t1 A = O.600.4,52.102

e1/3d, = 60/ (3.26)

—

4. DeterminatIon of the connection resistance 4.1 Bolted connection

Problem, Sketch, Calculation

6.2.3.1; Table 6.6

[6.5.5; Table 6.5.3]

6.2.3.2 [6.5.5; Table 6.5.3]

Reference

—A

—

rd2/4

—

Assumptlon:Shear In shank 4,52.102 mm2

Commentary

I

L70x7

I

I

j

I

r ________________

Yo

1,1

100 15275

•io-

2 9,4.102.275

+

P2

- 0,4

N

P2

=

(80

20,536

5•22 - 2,522

0,3

1,25

7,86102 .

° iO

- 2.5'22) - 0,536

77

Steel grade: Steel grade:

N = 285 kN

289,9 kN > 285,0 kN

375,0 kN > 285 kN

> 285 kN

2. LoadIng Tension force:

M 20, 8.8

Bolts:

d, = 22 mm

lOOxl5mm

Plate:

Member: 2 L 70 x 7

1. System

. 470,0 kN

= 94.102 -22.7 = 7,86.102 mm2 Determination of 62 by linear interpolation:

3.3 Net section (Angle)

3.2 Gross section (plate)

N

3. DetermInation of the resistance 3.1 Gross section (angIe)

1oo

ttm

Example 3.1.3: Angle connected on a gusset plate

Problem, Sketch, Calculation

Fe 430 Fe 430

5.5.1; Table 5.33 [5.4.3; 6.5.2.3)

5.5.1; [5.4.3]

5.5.1; [5.4.3)

Table 6.7; [7.5.2)

Reference

Angles connected by a single row of bolts in one leg may be treated as concentrically loaded if the design ultimate resistance Is determined as 5.5.1.(2); [6.5.2.31, I.e. the effect of eccentricities is included in the joint design.

In the case of joints with angles connected by at least two bolts, the setting out lines for the bolts In the angles may be substituted for the centroldal axes for the purpose of Intersection at the joint.

Commentary

or

= 30mm =

1,36db

resistance required

< 1,5d,

-

1,25 10

- 91,2 kN

3

91,5 , (2 1,36 1,50

188,2 kN

=

196,1

-

.

>

2

.? 22

78

= 71,3 kN

0,96; —' a

kN>

91,5 kN

- 1,20 1 \ = - 1,20 -') 77,3 kN

Reduktion of the bearing resistance:

Fbpd

1,25

2,5afdt . 2,5'0,764-3020•7 .10"

TUb

—

= 0,76; p1/3d, - 1/4 = 80/3.22.1/4

1,25

2,5•0,76•430•2015 •iO -

Bearing resistance without reduction:

Fb

- 142,5 kN

io •o

0,76; p1/3d,- 1/4 = 80/3.22 - 1/4 = 0,96; —. a

>

1.25

2 0.6 800 2,45

—. Reduction of the bearing

a2

e1/3d0 = 50/3.22

Angles:

TUb

2,5afdt

—

= 188,2 kN

A1

50/3.22 =

1.25

,

1

e,/3d

=

-2

0.8

Plate:

Bearing resistance:

or

Shear resistance of bolts:

-2

4. DetermInation of the connection resistance 4.1 Bolted connection

Problem, Sketch, Calculation

0,76

142,5 kN

= 0,76

= 142.5 kN

[6.5.5(10)]

Table 6.6

6,2.3.1;

[6.5.5; Table 6.5.3]

Table 6.6

6.2.3.1;

[6.5.5; Table 6.5.3]

Table 6.7

6.2.3.2;

Table 6.5.3]

(6.5.5;

Reference

The bearing resIstance Is calculated per member (here: per angle).

calculation Is necessary.

If the end distance Is equal then only one

Two shear planes Shear In thread Assumption:

Her.:

Commentary

•

0,6

f, A

=

A

Vu0

=

11,210

355

- 45 kN

Steel grade: Steel grade: Steel grade:

Fe 430 Fe 510 Fe 510

2,5•0,57•510•20•10 1,25

1,25

.g-5 = 208,7 kN > 90.0 kN

79

2.5afdt - ________________.g4 - 101.7 kN > 45 kN

a = 0,89

10 = 116,3 kN > 45 kN

0,57; p1/3d, - 1/4 = 75/3.22- 1/4 = 0,89 ; —i a = 0,57

p1/3d - 1/4 = 75/3.22- 1/4 = 0,89; —.

2,5afdt

e1/3d = 37,5/3.22

Fb

= 90 kN

0,6.800.2,45.102 .ft)-3 = 94,1 kN > 2 1.25

F_ii!6_94,1kN>._45kN

—

V

—. d, = 22 mm

M 20, 8.8

150x ilOx 10

HE 200 B IPE 200

3.2 Shear resistanceof the beam 11,2.102 - 2.22.5,6 = 8,7.102 mm2 A.f/f, = 11,2.102.355/510 = 7,8.102 mm2 —---. fastener holes need not be allowed for in shear verification

L_________ 1

°

Beam web:

Bearing resistance: Fin plate:

or

Shear resistance of bolts:

Column member: Beam member Fin plate: Bolts:

I-section

2. LoadIng Shear force:

3. DeterminatIon of the resistance of the connection: 3.1 DesIgn resistanceof the connection (fin plate - beam)

OkN

I

1. System:

Example 3.1.4: Fin plate connection with supporting member as an

Problem, Sketch, Calculation

5.3.3; (5.4.6]

Table 6.6 [6.5.5; Table 6.5.3]

6.2.3.1;

Table 6.7

6.2.3.2;

[6.5.5; Table 6.5.3]

Reference

Shear

in thread

= h.t_ = 11,2.102 mm2 ft Is allowed to consider larger values by using the formulae given In EC 3 (5.4.6]

The edge distance e1 should be taken as the minimum value of both sides when the acting force effects not parallel to one of both edge distances

Assumption:

Fin plate connections ate assumed as hinged connections. The location of the hinge Is controlled by the stiffness of the supporting member. Here: The column web Is a rigid support of the fin plate. Therefore it may be assumed that the hinge occurs at the bolts.

Commentary

of fin plate

M,.. =

V,,
90,0kN

jv

_______

> 5,4 kNm

3.5 Resistance

M

- i..,

L,8 = L +

'YMO

-

= 90.103/12.102 = 75 N/mm2 MS,J/WW = 5,4.108/30.103 = 180 N/mm2

=

1fr2 +

v y,

.

=

v0,851,25

430

< L3

224,1 mm

= 200mm —.

80

iO = 201,9 kN > 90 kN

= 193,5mm

= 234 N/mm2 > 195 N/mm2

,72) = J(752 + 1802) = 195 N/mm2

V/A =

=

+ 56

193.55.6

= 75 + 62,5

L3= L÷a1 +a3=75+62,5+62,5=200mm < (L + a, + a3- nd,)(fJf) = (200 - 2.22)510/355

Geometrical data of the weld: a 4 mm 2.150.4 = 12.102 mm2 W,, = A.,.h/6 = 12.102.150/6 = 30.10 mm3

Weld stresses:

10.4.102 mm2

= a1 62,5 mm 5,4 kNm

82

•10 - 208.7 kN >90 kN

234 N/mm2> 195 N/mm2

= 7,5 kNm

0,851,25

to RHS . chord

P.

________

Action effects at fillet weld: = 90 kN Shear force: = 90.60.i0 5,4 kNm Bending moment: Weld stresses: = V/A, = 90.103/12.102 = 75 N/mm2 = MJW,. = 5,4.1O/3Q.1O 180 N/mm2 Ifr,2 + c,j2) = 1(752 + 1802) = 195 N/mm2

of the filletweld

—

—

= a, = 62,5 mm < Sd = 5.20 = 100 —. L = 62,5 mm = (a2. 1.2 kd•,)(fjf,) (55. 0,5.22)510/355 = 63,2 mm = L. + a, + a3 = 75 + 62,5 + 62,5 = 200mm 1.3 (L + a1 + a3 - nd.)(fJf) = (200. 2.22)510/355 = 224,1 mm = L + L, +L2 = 200 mm —. 75 + 62,5 + 63,2 = 200,7 mm >

L1

3.6 Joint resistanceof fin plate welded

3.5 Resistance

y1.1

•10 - 279.5 kN > 90.0 kN

—. no interaction bending - shear is required

Yuo

- 15.0.102.

1502.10/6 =37,5.10 mm3 = 37,5.103.355.106/1,1 WJV/'YM,

-A

3.4 Block shear resistance

M,,,

W

V 37,5 kN

•io - 81,5 kN > 37,5 kN

2,5•0,74430•16•8

3.4 Bearing resistanceof the column flange not relevant because of t, >

or

FbP4

- 189.3 kN>

60,3 kN > 37,5 kN

- 60,3 kN > 37,5 kN

1,25

3135 •10

Shear force: per bolt:

2. Loading V,,

mm

= 150 kN = 37,5 kN

d, = 18

M 16, 8.8

HE 200 B IPE 400 135x1 15x8

150,0 kN

1. System Column member: Beam member: End plate: Bolts:

3.3 Bearing resistanceof the endpiate e1/3d = 40/3.18 = 0,74 p1/3d, - 1/4 = 55/3.18. 1/4 = 0,77—. a = 0,74

or

Ya)

- 2•

0.6157800 •i0

aL

- 0.6 A1 t.e -

3.2 Shear resistanceof bolts

=

3. DetermInation of the resistance of the connection: 3.1 Weld resistance throat thickness chosen to be a = 3

Fe 430

t = 8 mm

Example 3.1.6: flexible end plate connection

Problem, Sketch, Calculation

83

Steel grade:

Steel grade: Steel grade: Fe 430 Fe 430 Fe 430

6.2.3.1; Table 6.6 [6.5.5; Table 6.5.3]

6.2.3.2 Table 6.7 [6.5.5; Table 6.5.3]

6.3.4.3; Table6.15 [6.6.5.3]

Table6.7 [7.5.2]

Reference

e

Shear in thread

The choice of = 40 mm is conservative In view of the presence of the transverseweld between flange and flexible endpiate. f,,/f, is normally greater than 1,0; however a IS equal or lesser than 1,0.

Assumption:

From practical evidence flexible end plate connec-

tionsare assumed as hinged connections.

Commentary

-

- 21.8.102

275

Vp1.

- A. .—!Y_—

- ii.i -102-

275

3.6 Local shear resistanceof the beam web = 8,6.135 11,61.102 mm2

Vp1.

3.5 Shear resistance of the endpiate 2.(135 - 2.18).8 15,8.102 mm2

•iO 167,6 kN> 150,0 kN

Af/f = (135.8.2).275/430 10 - 311.8 kN > 150.0 kN

>

13,8.102 mm2

Problem, Sketch, Calculation

84

5.3.3 [5.4.6]

Reference

Commentary

beam connection with cleats

1,25

- 116,3 kN > .2L

1,25

0,6 ub A, = 0,8800•303

= 49.4 kN

.Ws - 116,4 kN >

VSd

- 49,4 kN

= 890 kN

24 mm

-

—

grade:

Fe 430

Steel grade: Steel grade:

YMb

2.5afdt 1,25

= 2,5-0,58-430-22-9

—

2,5-0.58-430-22-18,5 1,25

o = 203,0 kN

)

=

2160.10' mm2

Horizontal shear force: Vertical shear force per bolt: Resultant shear force:

I,

M

F

1(98,92

58,62)

115,0 kN

52,7.10°.240/2160.10 = 58,6 kN

V = 890/9 = +98,9 kN =

M.z/i =

85

9

• 98,9 kN

Shear in thread

V

The bearing resistance Is calculated per member.

Assumption:

The distribution of internal forces between fasteners needs to be proportional to the distance from the centre of rotation in slip-resistant connections polar moment of inertia of the bolt configuration: = Ey2 + Ez' = 2.(602+1202+1802+2401 I,, = 2160.10 mm2

YMb

2,5afdt

Table 6.6 [6.5.5; Table 6.5.3]

6.2.3.1;

Table 6.7

6.2.3.2;

[6.5.5; Table 6.5.3]

Table 6.7 [7.5.2]

33 Design resistanceof the connection (angle. side girder)

—

29

= 49,4 kN

Fe 430 Fe 430

hinged connections. Torsional effects may be neglected because of the large length of the cleats.

cleated connections on web only are assumed as

Commentary

The reaction force is applied to the connection at the notched main girder - angle cleat position

Fb

- 95,4 kN

a = 0,58

•iO-

p1/3d,. 1/4 = 601(3.24)- 1/4 = 0,58; —.

=

e1/3d0 = 40/(3.24) = 0,56; p1/3d, - 1/4 = 601(3.24).1/4 = 0,58; — a = 0,56

-

—

d0

M 22, 8.8

2 L90x9 Steel

HE 900 B HE 800 A

2. Loading Shear force:

Main girder: Side girder: Angles: Bolts:

1. System

Reference

3.2 DetermInation of the moment due to eccentricity of the bolt group (angle - side girder) = 59,25.10-.890 52,7 kNm

Beam web:

Bearing resistance: Angle:

or

Shear resistance of bolts:

3. DeterminatIon of the resistance of the connection 3.1 DesIgn resistanceof the connection (angle - main girder)

Example 3.1.7: Beam to

Problem, Sketch, Calculation

m,

0,6 i,,, A, =

1,25

2 0,6 800 303

•1o — 232,7 kN> 115,0 kN

1,25

Fb

2,50,5043022-15 1,25

2,5afdt YMb

A __!L._ . 48,5.102.

275

I-

- 700,0 kN> 2

300

j:Fo

- 4.45,0 kN

=

86

mm1

• 141,9 kN > 115,0 kN

712.15/12 + 300.28.207,12 + 712.15.162,92 = 109486,8.10 W,,, = 109486,8.10/518,9 = 2110.10 mm3 A,, = 712.15 = 106,8.102 mm2

= 712-15 712/2

14 + 14 = 221,1 mm 190,8.102

3.5 Design resistanceof the notched beam at cut I = 890 kN --— M1 = 890.155.10 = 138 kNm Vs,, Cross-sectional dimensions at cut I - I A, = 300.28 + 712.15 = 190,8.102 mm2

V,Rd =

fJf

- 57,5

—+a

- 2,5sfdt - 2,50,56430'•9 'iO' - 95,4 kN> Ji.!0

;

e1/3d, = 36/(3.24) = 0,5; p1/3d, - 1/4 = 60/3.24 - 1/4 = 0,58; —i a =

Fb,.

= = e1/3d, = 40/(3.24) = 0,56; p1/3d,, - 1/4 601(3.24) 1/4 0,58

Fp22232,6kN>115,0kN 1,25

—

3.4 Shear resistanceof the angle = 560.9 - 9.24.9 = 31,0.102 mm2 A,.f/f,, = 560.9.275/430 = 32,2.102 mm2 —. fastener holes have to be allowed for in shear verification = 31,0.102 .430/275 = 48,5.102 mm2 < 50,4.102 mm2 = A,,,, = A,,,,,

Beam web:

Bearing resistance: Angle:

Shear resistance of bolts:

Problem, Sketch, Calculation

0,56

5.3.3; [5.4.6]

6.2.2; [6.5.4)

Table 6.6 [6.5.5; Table 6.5.3)

6.2.3.1;

Table 6.7 [6.5.5; Table 6.5.3)

6.2.3.2;

Reference

.

Two angle cleats connect the side girder to the main girder. The shear resistance is calculated per member.

The edge distance e, should be taken as the minimum value of both sides when the acting force effects not parallel to one of both edge distances.

The bearing resistance is calculated per member.

Assumptions: m, number of members; here: two shear planes - Shear In thread - elastic distribution of internal forces

Commentary

(1

- p)

V-

A, —Z——

/•i,i

106,8.102.275

_1)2.o.024

•1O = 770,8 kN 138,0kNm

V,,.4

(L

=

L, t

a1

. 106,8.102

275

10 - 1541,5 kN> 890.0 kN

>

87

mm'

890 kN

= 68,3.10'

- l,6IcN

- 28).15.275/430

275

(740

- 627,5•15•

÷a3=480÷130+130=740mm + a3 - nd,)(fJf) = (740 - 9.24)430/275 = 819,3 mm +1.2=480+ 110+37,5=627,5mm 5d = 5.22 = 110 —' = 110mm = (a2. kd01) (f0/f) = (36 - 0,5.24)430/275 = 37,5 mm

527,5 (1 - 0,024) = 514,8 kNm > 138 kNm

P2__1r-(2890 t V ) 1,2770,8

=

L3=+a, +

L2

L1

Yo

- 05

3.7 DesIgn shear resIstance of side girder = (740 - 28).15 - 9.24.15 = 74,4.102 mm' A,.f/f,, = A,,,., —e fastener holes need not be allowed for in shear verification

3.6 Block shear resIstance

V - 05

1,1

211010275

Interaction Bending . shear is required

0,5

M

Problem, Sketch, Calculation

5.3.3; [5.4.6)

5.5.2; [6.5.2.2)

Table 5.16

5.3.3; [5.4.6]

5.3.2; [5.4.5)

Reference

Though It looks like a long joint, it is not because the rules for long joints refer to load Introduction problems only. The progressive Introduction of the load rather than simultaneous introduction as is the case here.

+

19,312.1,5.10,23/3).(1- 0,024) + + 2,8.30.(19,31+ 2,8/2). .(11,72 + 10,23)/2]/(100.1,1) = 519,0 kNm

[(51892.15.27,5/3 +

—

More accurate method

Commentary

I.

OO)L0

.

4_

r.iI

•

Npi

Ii

_i11

N

t,—_3Qo)t

.32OX12

4_65OX2O

gross section:

Compression resistance

net section:

gross section:

' io .

!Z

y.O

btit—

- 0,9AIs.

450

Vie

15 1.1

•i0'

817,3 kN > 309,6 kN

> 309,6 kN

88

10 - 1687,5 kN > 392,9 kN

09 .26,4.102

1,25

itt-

= 750,0 kN

158kN

= 8,2kNm M V=0,= NWSd

237 kN

= 52,2 kN

M, =

VWSd

kN N, == 78,3 12,3 kNm

158 kN = 87,3 kNm

V=

M0,1 = 131 kNm Serviceability limit state:

= 300.10 - 2.18.10 = 264.102 mm2

= bt._!Z. = 300 NpAd

Serviceability limit state: N,(sd = 261,9 kN = 206,4 kN

see sketch Steel grade: Fe 430 Fe 430 Steel grade: M 16, 8.8 preloaded d, = 18 mm

V,, = 237 kN

2. Loading Ultimate limit state:

Surface: class A

All plates: Bolts:

Member:

1. System

Distribution of action effects on web and flanges Ultimate limit state: = 392,9 kN = 309,6 kN

of the resistanceof connection

0;e0;5;

—

I r —

4.125,!.2OO,12S,!

3.1 Flange cover plate 3.1.1 Member resistance of the flange cover plate Tension resistance

6.63

3. DeterminatIon

°

490

unsymmetrical I-section

e .o

}.QJ.55ISSI55I

I.

Example 3.1.8: Splice of an

Problem, Sketch, Calculation

5.4.2 [5.5.1]

[5.4.3]

5.5.1

Table 6.7 [7.5.2]

Reference —

Requirement: No slip at serviceabilitylimit state Assumptions: (1) Shear force is transmitted by the web alone (2) Due to unsymetrical section bending moment and normal force exists in web

flange cover plate (tot): 490 x 450 x 15 flange cover plate (boO: 490 x 300 x 10 web cover plates: 280 x 260 x 10

2760.10' mm3 tot - top flange 1483.10' mm3 bof - bottom flange

164,4.10' mm' 33953.iO mm4

W, W,

I

A

Commentary

f

A 0,6•800•201 1.2

,-s = 77,2 kN>

. 38.7 kN

F•

F$Ad

Fb,. 1,25

1,10

17,62,5

—

-

iO

= 39,5 kN 237,0.70.10 + 12,3

= 237 kN = 28,9 kNm

1(3952

81,92)

90,9 kN

F = 1(26,32 + 54,72)

19,3.90.103/378.102 + 52,2/6 60,7 kN

M == 158.70.10' + 8,2 = 19,3 kNm =

158/6 = 26,3 kN

kN V = 158 =

F

.

275 •1cr3

- 750,6 kN > 237,0 kN

237,0 kN < 0,5.750,6 = 375,3 kN no interaction bending - shear is required

V1YMo

- A .-_.!_ - 2 260.10

89

3.2.1 Member resistanceof the web cover plate = 2.(260 - 3.18).10 = 41,2.102 mm' > 2.260.10.275/430 = 33,3.102 mm2 Shear resistance: —. no reduction of A Is required

A,,

= 25,8 kN

0,74

54,7 kN

+ 78,3/6 = 81,9 kN H == 28,9.90.103/378.102 +

M

V/n= = 237,0/6

V.1

8

101,8 kN > 38.7 kN

= 40,0 kN

t,

206,4 = 25,8 kN —i--—

1,10

0,7•0,5800157

40,0 kN

________

Verification not needed because of t1 >

YMb

2,5efdt - 2,50,7443016-10 •io

e1/3d0 = 40/(3.18) = 0,74; p1/3d0 - 1/4 = 55/(3.18) - 1/4 = 0,77 ; —i a

0.6

3.2 Web cover plate Ultimate limit state: Vertical shear force: Vertical shear force per bolt: Moment due to vertical shear: Horizontal shear force per bolt: Resultant shear force at the outer bolt: Serviceability limit state: Vertical shear force Vertical shear force per bolt: Torsional moment Horizontal shear force per bolt: Resultant shear force at the outer bolt:

Slip resistance:

Serviceability limit state:

Beam flange:

Range cover plate:

Bearing resistance:

Shear resistance of bolts:

3.1.2 ResIstance of the connection Ultimate limit state:

Problem, Sketch, Calculation

5.3.3 [5.4.6]

6.2.2 [6.5.4]

4.3; 6.2.4; Table 4.5 [6.5.8]

Table 6.7 [6.5.5; Table 6.5.3]

6.2.3.1

Table 6.6; [6.5.5; Table 6.5.3]

6.2.3.2;

Reference

Shear In shank

A = ir.162/4 = 2,01.10'mm2

Two web cover plates are used.

The distribution of internal forces between fasteners need to be proportional to the distance from the centre of rotation in slip-resistantconnections polar moment of inertia of the bolt configuration: Ey' + Ez' = 6.302 + 4.9Q2 = 378.102 mm'

Forces acting on the bolt group on each side of

the splice.

No slip shall occur at the serviceability limit state

The edge distance e, should be taken as the minimum value of both sides when the acting force effects are not parallel to one of the edge distances

Assumption:

Commentary

Slip resistance:

limit state

or

FbRd

e,/3d,

FbRd

=

F,pd

=

1,25

2 1,10

17,62,5

=

122,2

- 80,0 kN > 60,7 kN

Lu,,,


60,7 kN

= 434,7 mm 350mm

L3

i0 =

kN

90

iO - 595,8kN > 237 kN

344mm

f11

180 + 80 + 84,4

L+a,+a3= 180+85+85=350mm + a, + a3- ndj(fJf) = (350 - 4.18)430/275 = = =
90,9

2 0,70,5800157 1,10

•iO

>

kN

- 1/4 = 0,86; -----.

= 101,8 kN

= 60/(3.18)

2,5•0,74•430•16•12 1,25

0,7) f,A3

> 90,9

= 601(3.18) - 1/4 = 0,86 ;

= 154.4 kN

2,50,744301610 •1O

= 0,74; p,/3d,,. 1/4

=

= 0,74 ; p,/3d, - 1/4

FIRd = m,•

TUb

2,5afdt

= 40/(3.18)

YMb

. 2,5cttdt

e,/3d0 = 40/(3.18)

2 0,6800201 1,25

3.3 Block shear resistance of the beam member L, = a, = 90mm > 5d = 5.16 = 80 ——. L, = 80mm = 84 mm L2 = (a2 - kd,,,)(f,/f) = (99 - 2,5.18)430/275

Serviceability

Beam web:

Bearing resistance: Web cover plate:

Shear resistance of bolts:

0,6 f,,, A

-io - 76,4 kNm > 28,9 kNm

Fv.Rd = m a

- 305,6-10•

3.2.3 Resistance of the connection Ultimate limit state:

- w1,

3.2.2 BendIng resistanceof the web cover plate 2.[2602.10/4 - 18.10.90] = 305,6.10 mm3 W,,,

Problem, Sketch, Calculation

5.5.2 [6.5.2.2]

[6.5.8]

4.3; 6.2.4

6.2.3.1 Table 6.7 [6.5.5; Table 6.5.3]

Table 6.6;

6.2.3.2;

5.3.2 [5.4.5)

Reference

No slip shall occur at the serviceability limit state rn = number of friction interfaces Here: Two friction interfaces are available

Shear in shank. Assumption: m• - number of shear planes Here: two shear planes) The edge distance e, should be taken as the minimum value of both sides when the acting force effects are not parallel to one of the edge distances The bearing resistance is calculated per member.

Commentary

to ol to ot

Ok

t=

l,,,t_&_k 'ru0

;

BIRd

=

12 mm

F,, /-..,

> 8 mm

113,0/1,25 = 90,4kN

= 0,5d

1,1

1,65 kNm

assumption: kr

- 0.25•215,1•12••10 -

Mpp -0,25

215,1 mm

- 1,0

=

2"' bolt row (end bolt)

1.' bolt row (end bolt)

= = = =

91

0,5p + 2m + O,625e 0,5.70 + 2.26,9 + 0,625.30 = 107,55 mm 0,5p + 2m + 0,625e 05.70 + 2.26,9 + 0,625.30 107,55 mm

Dimensions of a T-stub n = e_, = e = 30 mm m = (80 - 7 - 2.0,8.12)/2 = 26,9 mm

V

IRE 220, Fe 360 HE 140 B, Fe 360 305 x 140 x 12, Fe 360 M 16, 8.8

2. Loading Bending moment: M,,, = 20,0 kNm = 120 kN Shear force:

Beam: Column: End plate: Bolts:

1. System

of the 1-stub with 2 bolt rows

55

140

70

Calculation of the plate bending momentof the 1-stub

Relevant bolt pattern

(1) Determination of the total effective length

I—

3. DetermInation of the moment resIstance 3.1 TensIon zone 3.1.1 Unstlffened column flange

j"

2Ad

Example 3.1.9: Bolted end plate connection

Problem, Sketch, Calculation

Table 6.8 [6.5.5.(4))

6.2.3.3(2)

[J.3•31

[J.3.4.1; Figure J.3.4J

[Figure J.3.1J

Reference

bolt plate assembly.

B,,.4 is the design tension resistance of a single

o

Assumption: The compressive normal stress In the column flange due to axial force and bending moment does not exceed 180 N/mm2 at the location of the tension zone. — The reduction factor k, — 1,0.

The effective length for each row of bolts is based on the yield line patterns and should be taken as the smallest value of the different failure modes. Here: The relevant failure mode is the combined bolt group pattern

SimplIfication: Calculation of an equivalent 1-stub for bolt rows I and 2.

This example Illustrates, how a bolted beam-tocolumn connection Is designed by a simplified hand calculation using the 1-stub model. When the moment resistance of complex bolted beam-tocolumn connections should be calculated based on the full procedure given In Annex J it is recommended to develop a software programme.

Commentary

Relevant bolt pattern

0

oo

o

140

•.

T-tub

(2) Determination of the total effective length of the T-stub with

3.1.2 End plate (1) Bolt outside tension flange

Design tension resistance of the T-stub

.2).

2).

Determination of the failure mode of the 1-stub

41.65 26,910'4-90.4 0,679

- 0,690

1,115

=

—

io - 245,4 kN

e,

1

where n

= 50 mm

b0

= 140mm

f•ff =

92

Dimensions of a 1-stub = e = 40 mm n = m = (70-9,2 - 2.0,8.8.12)12 = 21,3 mm

4•1.65 26,9

m

4M

0,679 < 0690 —+ failure mode

2 bolt rows

F

B =

-

1

30 2•1,115 + 2•1.115

4M

—

mEB

+ 2A

6-

I

2

n

Problem, Sketch,_Calculation

[J,3.4.4; Figure J.3.8]

[J.3.3; Figure J.3.2]

Figure J.3.3)

(J.3.3;

Reference

The extended end plate is considered as a simple T-stub, see sketch, neglecting the stiffening effect of the beam web.

The design tension resistance of the T-stub Is dependent on the failure mode. The effects of prying forces are taken Into account.

Commentary

It

—s'-

F/

1.08 kNm

0,5d = 113,0/1,25 = 90,4 kN

-

1,1

- 7330- 10= 493,5

YMo

= 330 mm

tb

= 220.1,5

3.1.3 ResIstance of column web

Design tension resistance of the T-stub

kN

______________

Determination of the failure mode of the T-stub

B,

12mm >8mm

1.1

- 0,25.140.122..10

- 0.25

Calcutatlon of the plate bending moment

B

1

2t3

-

21,878 +2•1.878

-

21.3104•90,4

41 .08 0.561

- 0,79

1.878

21 3

io -

4108

m

4M11, 202,8 kN

0,561 120 kN

0,81,25

1,25

•io-

- 1/4

2

138,2 kN

1401(3.18)

60,3 kN >

220,7 kN > 120 kN

2,51,03601612

p1/3d - 1/4 =

YMb

31772 .10-' -

—

60,0 kN

10

2.5afdt

1,01 ;

e1/3d = 551(3.18) =

1,25

0.6600157

kN>

A,

60,3

= 220,6

aL -

0,6

- .Zd

F,, -

of the web weld resistance

Bearing resistance:

or

Shear resistance of bolts:

3.5 Shear resistance of the bolted connection

= 120,9.(220 - 9,2).1O 25,5 kNm M,,,4 W1,1f,/-0 = 285.10.235.10/1,1 = 60,9 kNm — The beam-to-column connection is classified as partial strength [6.4.3.3; 6.9.6.3)

M ==max Fh

> 2

--

95

- 60,0 kN

= 2,34 ; —p a

- 60,0 kN

Distribution of the bolt forces according to the tension resistance of the unstiffened column flange and the end plate The moment resistance of the connection is determined to:

max

Maximum potential resistance of the tension zone, shear zone and compression zone:

3.4 Moment resistanceof the connection

Problem, Sketch, Calculation

1,0

6.3.4.3 Table 6.15

[6.6.5; 6.6.9]

-

[6.5.5; Table 6.5.3] 6.2.3.1; Table 6.6

Table 6.7

6.2.3.2;

[6.5.5; Table 6.5.3]

[J.3.4.5J

-

(J.3.4.5)

Reference

e

The eccentricity is covered by the design resistances for bolts and need not to be taken into account

Shear in thread Assumption: Only the lower two bolts are used to transfer the shear force from the beam into the column.

The maximum potential design resistance of the column flange Is generally not the same as the maximum potential design resistance of the beam end plate. In order to determine the actual design resistance of the tension zone a compatible distribution of bolt-row forces should be obtained.

Commentary

3.2

Welded connections These examplesdemonstrate the verification of welded connections assuming design values of action effects (N, etc) which have been calculated by an analysis of the structure and include these values already ip, Y etc. In the Theresults presented examplesare rounded values. For the purpose of easy re-calculation each formula and each check Is calculated with the rounded values.

V, M,

Thefollowing examplesare included in this chapter: Example 3.2.1: Example 3.2.2: Example 3.2.3: Example 3.2.4: Example 3.2.5:

Double angle welded to a gusset plate Bracket welded on a column Welded beam to column connection withoutstiffeners Welded beam to column connection with stiffeners Hollow section lattice girderjoint

96

=

2

=

0,8 1,25

AT

1•YkIO

+

A

=

=

A

YMO

—-

(255.tan3O

50)

iO

).io-s

= 242,5

199.5 kN> 168 kN

188 kN

+

Steel grade: Steelgrade:

2.50 = 320 mm

kN> 188 kN

N=

a =

3 mm • = 4 x 55

= 10 mm

2L50x5

1,1

•io = 242.5 kN > 188 kN

97

of the gusset plate may be done by assuming a load dispersion with an

235 +

332O

(2.88.10.

As an alternative the check of Tension resistance see sketch.

=

.-

a•L

77,9 320 = 199,4 kN > 188 kN

-i--——

3.2 Check of Tear-out

or

2. LoadIng Tension force:

Gusset plate: Angle: throat thickness: weld length:

1. System

max{ 6.a=6.3=i8mm} 0,7•120 - 84 mm —.. the joint need not be stiffened

3.1 Effective breadth In tension zone:

3. DetermInation of the weld resistance:

Fe 430

IPE 300

______________

,[,20kN

Example 3.2.2: Bracket welded on a column

Problem, Sketch, Calculation

Fe 430 Fe 430 Fe 430

5.3.3 [5.4.6]

5.3.2 [5.4.5]

6.3.6 [6.6.8]

Reference

V

No Interation bending . shear is required because — 15 kN < of 0,5V,,.4 79,4 kN (5.3.4 [5.4.7])

Elastic stress distribution in the cross-section is assumed.

welds.

A reduced effective breadth shall be taken Into account both for the parent material and for the

Load factors are included

Elastic stress distribution in cross-section and in the weld Is assumed.

Commentary

W

1

A

-

r

110

S

l2a

- V5,S

172,3 N/mm2

iO iO

310,710'•23

15 -32.1

32,1.10 mm3

- 233,7 N/mm2>

= 112.10.28,7

0,85 -1.25

430

Tuw Pw

- _________

3.6 Weld of the section

tw.Rd

3.5 Verification

- 10)5110 + 1105255 • 84,7 mm

10,7.102 + 11,0.102

+ (112

25.8 N/mm2 < 233,7 N/mm2

= 221,6.10'

99

= 2.110.5/12 + 2.29,72.110.5 + 25,32.(112. 10).5 + 35,32.112.5 = 221 ,6.10'/84,7 = 26,2.10 mm3

1125-120

= 5.112 + 5.(112 - 10) = 10,7.102 mm2 = 2.1 10.5 = 11.0.102 mm2

of the weld

3.4 DeterminatIon of the stresses in the weld = 15.103/11,0.102 = 13,6 N/mm2 Cs,, = 4,5.108/26,2.103 = 171,8 N/mm2 CwSd = I(13,6 + 171,82) = 172,3 N/mm2

-r

112 __________

3.3 Area and section modulus

Problem, Sketch, Calculation

mm'

6.3.4.3 [6.6.5.3)

Reference

Elastic stress distribution is assumed.

No deformation capacity Is necessary for a bracket. —i flange weld need not to develop the full design resistance of the beam flange.

Assumption: Elastic stress distribution In weld

As elastic stress distribution Is assumed in the cross-section, It is also assumed for the weld.

Commentary

F,Rd

F,fld

(t

+

= 291,5kM = 291,5 kN

t

ft,.'b.Il/1M

212,5 kN

= 181,6 kN

F, = 291,5 kN

= + 2/2 ab + 5 (t1+r) b1, = 8,5 + 212. 6 + 5 (14 + 27) = 230,5mm = 235.8,5.230,5.10/1,1= 418,6 kN

=

-10"

6170.4360 •10' 0,81,25

= 100 •8,5-235

?'. Ypw

Yo

--—. Resistance of the tension zone:

F,Rd

F,Ad

,

_,,,L

3.1.2 Unstiffened column web:

FWRd =

b

100

2. LoadIng Bending moment beam: Bending moment column: Shear force: Axial force:

1. System column: HE300A beam: IPE 200 throat thickness: tension flange: web: compression flange:

0,7.235.8,5.100.10-'/l,l =

no stiffening required

> 127,1 kN =

2, + 7tfC)/'YM. 235.8,5(8,5 + 2.27 + 7.14).10/1,1

fYbt, =

Verification of the weld in tension zone: L = 100 + 100. 5,6 - 2.12 = 170,4 mm

or

t

3.1 Tension zone: 3.1.1 Unstlffenedcolumn flange: = (fb (t_,, + 2,,,) + 7 (fyn tfC))/1M F,Rd = (235.8,5.(8,5 + 2.27) + 7.(235.142)).l0'/1,1 = 406,6 kN

3.Determlnatlonof the moment resistance

3SkN in

Example 3.2.3: Welded beam-to-column connection without stiffeners

Problem, Sketch, Calculation

30 kN

45 kNm 35 kNm

N= llOkN

=

= 6 mm = 3 mm 3 mm

Vbs,I

a a a

Steel grade: Fe 360 Steel grade: Fe 360

[J.2.3.2]

6.3.6.(5) [J.2.3.1 (3)]

[J.2.3.2. (2)]

6.3.6 (J.2.3.1.(1))

Reference

The welds connecting the beam flange to the column should be designed to develop the full design resistance of the beam flange equal to

Commentary

-

= 131 mm

1,17>

t,.

2{2.3 + 5(14

+ 27) =

222cm

= f,, 235 • 8,5. 222.i0/i,i = 403,1 kN

=

8,5+

=t,+2I2a.+5(t+r) =

125-0,5.1,1.35/235

1,0

.z = NJA + MJl 110.103/113.102 + 35.10.131/l8260.l0 = 35 N/mm2 = =

!

bff

'1'

- 1,484104 mm4

1,484.10'

2,454 93,9

076

N 24,65102

- 2,454 mm

—

Resistance of the compression zone:

Using buckling curve c —. x = 0,6873 = 0,6873.8,5.29.235.1O/1,1 = 361,9 kN

I

i. NA

A = 290.8,5 = 24,65.102 mm'

12

2908,5

h

= 361,9 kN

101

3.2.2 DesIgn resistanceof the column web to buckling: Non-sway mode: bfl = = 290 mm 290 .1 K The buckling length of the virtual compression member should be determined from the conditions lateral and rotational restraint at the flanges at the point of load application: here: = 0,6.h = 0,6.290 = 174 mm

F

=

_L

0.5} b, Tuo

of unstiffenedcolumn web

z = ½(h - 2t) = ½(290 -2.14)

F-

3.2.1 DesIgn crushing resistance

3.2 CompressIon zone

Problem, Sketch, Calculation

of at

5.6.5 [J.2.4.1.(3); 5.7.51

[J.2.4.1 J

Reference

The buckling resistance R,,, should be obtained theweb as a compression member by considering with an effectivebreadth be,, obtained from Table 5.39 [Figure 5.7.3].

The sway mode should normally be prevented by constructional restraint.

the maximumcompressive normal stress in the web of the column due to axial force and bending. The compression force Is introduced by contact.Therefore In the compression zone the minimum throatthickness may be chosen as a 3 mm.

-

Commentary

208/8.5 = 24,5 < 69

of the shear zone:

—

69

Resistance of the shear zone:

tj

.fld = 291,5 kN

or

=

F6 =

FW

r,

-

aL

3.6 Verification of the "web weld'

F

303,4 kN

YMO

- Af

Yo

-

198,2 kN> 110 kN

47,2 kNm

>

45 kNm

= 198,3 kN> 110 kN

—

no interaction bending - shear Is needed

iO

—

•io--

= V,,,

31592 .10

221

0,81.25

- W,

> 30 kN

V1.1

•iO- = 138,1

> 45,OkNm

11,2•10235

= 55,8kNm

= 69,1 kN 0,5VRd

v

Moment resistance: M,.,,

Shear resistance:

F,4

Tension zone F,,,.d = 291,5 kN Compression zone FORd = 361,9 kN Shear zone 303,4 kN

3.5 Cross-sectIon resistance of the beam

M = 291,5.(200-8,5) i0

=

Potential failure modes:

M,. = F,,,, (hb

3.4 Bending moment of the connection:

-.—-.

kN > 30 kN

—. no shear buckling verification required

A = 235.24,6102 •io - 303,4 kN v, - Mo

d/t

3.3 Resistance

Problem, Sketch, Calculation

102

6.3.4.3 Table 6.15

[6,6.5.3]

5.3.2 (5.4.5]

5.3.4 (5.4.7]

5.3.3 [5.4.6)

5.3.3 (J.2.5]

Reference

Commentary

HCA

300

'\5Øiç,

2. Loading Bending moment beam: Bending moment column: Shear force: Axial force:

column: HE300A beam: IPE 200 throat thickness: tension flange: web: compression flange:

1. System

N,,,,= llOkN

Vb,, = 30 kN

= 6 mm = 3 mm 3 mm

= 45 kNm Mbsd = 35 kNm

a a a

Steel grade: Fe 360 Steel grade: Fe 360

190

=

b'

0.81.25

181.6 kN

6170.4360 •10 = 212.5 kN

1,1

1008,5235 •iO-

t6,

360

b/0,8•1.25 W,,,

- 207.8 N/mm2

103

A,,, = 2.160.3 = 9,6.102 mm2 2.1602.3/6 = 25,6.10 mm3

= 235.8,5.18,75.10/1,1 = 34,0 kN; b/-y = 34,0.36,375.10 = 1,24 kNm =

=

34,0.103/9,6.102 = 35,4 N/mm2 1,24.106/25,6.103 48,4 N/mm2 = /(3542 + 48,42) = 60,0 N/mm2 < 207,8 N/mm2 =

f

___

F,,, =

-

= max

M,=

max

w.Rd

of the weld between the stlffeners and the column web

1875 62.5 18J5

Verification

Yi,w

u

=

aL

F

YM0

N . b,

Verification of the weld between the stiffeners and the column flanges in tension zone: L = 100 + 100-5,6-2.12 = 170,4mm

3.Determinatlon of the moment resistance 3.1 Tension and compression zone: The design resistance of a stiffened column subject to a transverse tensile or compressiveforce is at least equal to the design resistance of the beam fiange, provided the thickness of the stiffener is not smaller than the flange thickness of the beam. The welds need sufficient deformation capacity.

350N,,

JIO (N

______

33

tpooo

=33_____

Example 3.2.4: Welded beam-to-column connection with stiffeners

Problem, Sketch, Calculation

[J.2.3.3)

(J.2.3.1.(3)J

6.3.6.(5)

(J.2.3.2.(2)J

(J.2.3.1.(1)J

6.3.6

Reference

St

- stiffener

The welds connecting the beam flange to the column should be designed to develop the full design resistance of the beam flange equal to

Commentary

V'•11

235.24,6102

Resistance of the shear zone:

-

.

F,,,,,, (hb t%)

i03 =

v

=

=

W,,,

._!L.

- 221 io-

VM

3•1,1

69,1 kN > 30 kN

VYuo

138,1 kN

> 30 kN

•io8 = 47,2 kNm

> 45 kNm

—. no interaction bending - shear is needed

i0 -

—. the beam-to-column connection is classified as full strength [6.4.3.2; 6.9.6.31

M

O,5VRd

Moment resistance:

Shear resistance:

Af -

11,2.102235

58,1 kNm > 45,0 kNm

Shear zone FRd = 303,4 kN

3.5 Cross-sectIon resIstance of the beam

MM = 303,4.(200 8,5)

—. F,, = Rd = 303,4 kN

Potential failure modes:

—'

FM = 303,4 kN

•io - 303,4 kN

69€ — no shear buckling verification required

3.4 Bending momentof the connection:

-.-—.

'YMO

= tvc Av

3.3 Resistance of the shear zone: d/t 208/8,5 — 24,5 < 69

Problem, Sketch, Calculation

104

5.3.2 [5.4.5]

5.3.4 [5.4.7]

5.3.3 [5.4.61

[J.2.5J

5.3.3

Reference

Commentary

O7kN

87kN

'1

5.00

"

5.00

"

\I'Dotailf

87kN

5.00

87kN

2

b1

b1 + b2

—

1,3