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Structural Steelwork Eurocodes – Development of a Trans-National ApproachWorked examples Design of a 3-storey unbraced frame

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Structural Steelwork Eurocodes Development of

A Trans-national ApproachCourse: Eurocode 3Module 7 : Worked Examples

Lecture 22 : Design of an unbraced sway frame with rigid jointsSummary:

NOTE – This example is a draft version••••

Pre-requisites:•

Notes for Tutors:This material comprises one 60 minute lecture.

Objectives:• To explain the main principles of EC3 by practical worked example.

References:• Eurocode 3: Design of steel structures Part 1.1 General rules and rules for

buildings••

Contents:


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WORKED EXAMPLE 3Design of a Sway Frame

The frame consists of three storeys and three bays. The frames are at 10 m spacing. The beam span is 6,5 m and the total height is 10,5 m, each storey being 3,5 m high. It is assumed that the column foot is pinned at the foundation.

Internalcolumns

Internalcolumns

Externalcolumns

Externalcolumns

Roof beams

Floor beams

6,5 m 6,5 m 6,5 m

3,5 m

3,5 m

3,5 m

Figure 1 Typical Cross Section of FrameThe structure is assumed to be braced out of its plane and to be unbraced in its plane. In the longitudinal direction of the building, i.e. in the direction perpendicular to the frame plane, a bracing does exist so that the tops of


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the columns are held in place. The lateral support for the floor beams is provided by the floor slabs.All the beam-to-column joints are assumed to be perfectly rigid. The connections must be capable of transmitting the forces and moments calculated in design.With these assumptions, the frame is classified as ‘continuous’, and the internal forces and moments are determined using a global elastic analysis which assumes the members to be effectively held in position.

6.4.2.2(3)

5.2.2.3

The steel grade selected for beams, columns and joints is Fe360 (fy = 235 N/mm2).

Table 3.1

Characteristic LoadsFloor: Variable actions, Qk = 1,8 kN/m2, Permanent actions, Gk = 3,0 kN/m2

Roof: Variable actions, Qk = 0,6 kN/m2, Permanent actions, Gk = 2,0 kN/m2

The wind loads are applied as point loads of 10,5 kN at roof level and 21 kN at the first and second storey levels.The basic loading cases, shown in Figure 2, have been considered in appropriate combinations.

Permanent Loading (G) Wind Loading (W)

loading case 1 (N1) loading case 2 (N2) loading case 3 (N3)

Imposed Loading Cases

20 kN/m

30 kN/m

30 kN/m

10,5 kN

21 kN

21 kN

6 kN/m 6 kN/m 6 kN/m 6 kN/m

18 kN/m

18 kN/m

18 kN/m 18 kN/m

18 kN/m

18 kN/m

18 kN/m 18 kN/m

Figure 2 Loading Cases


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Frame ImperfectionsFrame imperfections are considered by means of equivalent horizontal loads. The initial sway imperfection is given as:φ = kcksφ0

where k = 0,5+1nc

c

but kc ≤ 1,0,

kns

s

= +0 21

, but ks ≤ 1,0, and

φ 01

200= .

In this case, the number of full height columns per floor, nc, is 4 and the number of storeys in the frame, ns, is 3.

Therefore k c = + =0 514

0 866, , , and

k s = + =0 213

0 73, , .

Substituting into the above equation:

φ = 0,866 x 0,73 x 1200

=1

315

5.2.4.3(1)

The equivalent horizontal load, H, at each storey of the frame is derived from the initial sway, φ, and the total design vertical load, N, in any storey for a given load case. Therefore H = φN.The relevant values are listed in Table 1 for all the basic loading cases.

Basic loading case

(see Figure 2)

Storey N(kN)

φN(kN)

G Roof 390 1,242nd Floor 585 1,861st Floor 585 1,86

N1 Roof 117 0,372nd Floor 351 1,111st Floor 351 1,11



Table 1 Equivalent Horizontal ForcesThe equivalent horizontal forces must also be multiplied be the appropriate partial safety factors for actions.


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Load CombinationsIt was decided to use the simplified combinations for the ultimate limit state and the serviceability limit state.The basic load cases are combined at the ULS as summarised in Table 2.

2.3.3.1(5) and 2.3.4(5)

Load Case 1 1,35G + 1,5WLoad Case 2 1,35G + 1,5N1Load Case 3 1,35G + 1,5N2Load Case 4 1,35G + 1,5N3Load Case 5 1,35G + 1,35W + 1,35N1Load Case 6 1,35G + 1,35W + 1,35N2Load Case 7 1,35G + 1,35W + 1,35N3

Table 2 Load Combination Cases at the Ultimate Limit State

The basic load cases are combined at the SLS as summarised in Table 3.Load Case 1 1,0G + 1,0WLoad Case 2 1,0G + 1,0N1Load Case 3 1,0G + 1,0N2Load Case 4 1,0G + 1,0N3Load Case 5 1,0G + 0,9W + 0,9N1Load Case 6 1,0G + 0,9W + 0,9N2Load Case 7 1,0G + 0,9W + 0,9N3

Table 3 Load Combination Cases at the Serviceability Limit StatePartial Safety Factors for Strength

The following partial safety factors for strength have been adopted during the design:• Resistance of Class 1,2 or 3 cross-section, γM0 = 1,1• Resistance of member to buckling, γM1 = 1,1• Resistance of bolted connections, γMb = 1,25

2.3.3.2(1)

5.1.1(2)5.1.1(2)6.1.1(2)

Trial SectionsIn order for a global elastic analysis of the structure to be carried out, initial section sizes must be assumed and allocated to the structural members. The analysis must then be carried out and the members checked for the relevant failure modes. The sections will then need to be modified and the structure re-analysed. This can be a long winded iterative process.The engineer may have his own method of selecting initial section sizes.As a guideline, for this example, columns were selected by assuming an average stress of approximately 100 N/mm2 under axial forces. Axial forces can be estimated by approximating the floor area supported by that column.Generally, the bending moments withing the beams are critical. Simple bending moment diagrams can be constructed, assuming fixed end moments, and the maximum bending moment can then be estimated. An initial section size can then be determined.The trial member sizes for this example are:Inner columns: HEB 260Outer columns: HEB 220


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Floor beams: IPE 450Roof beams: IPE 360

InternalcolumnsHEB 260

InternalcolumnsHEB 260

ExternalcolumnsHEB 220

ExternalcolumnsHEB 220

Roof beamsIPE 360

Floor beamsIPE 450

6,5 m 6,5 m 6,5 m

3,5 m

3,5 m

3,5 m

Figure 3 Trial Member SizesDetermination of Design Moments and Forces

This worked example uses the amplified sway moments method of analysis.

5.2.6.2(1)

An alternative method is to calculate the sway-mode buckling lengths of members, and then carry out a first order linear elastic analysis.

5.2.6.2(1)

However, it may be possible to directly use a second order elastic analysis.

5.2.6.2(1)

A global linear elastic analysis is carried out on the sway frame in order to determine the moments, axial forces and shear forces in each member.Props are then applied to the structure in order to prevent any horizontal movement (i.e. as if the structure was now braced), and the analysis carried out again.The moments induced in the members from the propped case are then subtracted from the moments determined from the sway case. The resultant moments are those induced by “pure sway”.The “pure sway” moments are then amplified to take account of second order effects ignored by the linear elastic analysis. The amplified “pure sway” moments are then added to the moments obtained from analysis of the propped structure. These are the design moments which each member must be able to resist.The members must also be checked for the axial and shear forces determined from the initial analysis of the sway frame, and in certain cases the interaction of the design moments with the axial forces and/or shear forces must also be checked.This needs to be carried out for every load case so that the critical conditions for each individual member can be identified.


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Original Sway Case MomentsMinus

Propped Case MomentsEquals

"Pure Sway" Moments(To Be Amplified)


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Figure 4 Determination of “Pure Sway” MomentsCalculation of Amplification Factors

The sway moments should be increased by multiplying them by the ratio:1

1− V VSd cr/where VSd is the design value of the total vertical load, andVcr is its elastic critical value for failure in a sway mode.

5.2.6.2(3)

Instead of determining VSd / Vcr directly, the following approximation may be used:VV h

VH

Sd

cr

=

δ

where δ is the horizontal displacement at the top of the storey, relative to the bottom of the storey,h is the storey height,V is the total vertical reaction at the bottom of the storey, andH is the total horizontal reaction at the bottom of the storey.This approximation may not be used if VSd / Vcr is greater than 0,25.

5.2.6.2(6)

5.2.5.2(4)

5.2.6.2(4)

If VSd / Vcr is greater than 0,25, then the frame may be more susceptible to buckling. It is therefore necessary to carry out analysis using a direct second order analysis. It may also be necessary to stiffen the frame - for example increasing the column sizes.Alternatively, if the value of VSd / Vcr is less than 0,1 then the structure is classified as non-sway.

5.2.5.2(3)

The amplification factor will be different for each storey of the structure. The maximum factor should be used to multiply the moments at all levels of the structure. This is essentially a conservative method as it corresponds to the critical elastic load of the whole structure.The amplification factor for load case 5 of this example was determined as follows:


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δ

h

V

H

δ

δδδ = - = 15,6 mm

h = 3500 mm

V = 3296 kN

H = 81 kN

2

2

1

1

Figure 5 Determination of Amplification Factor

Therefore VV h

VH

15,6 x 32963500 x 81

Sd

cr

=

= =δ 0 18,

The amplification factor =−

=−

=1

11

1 0 181 22

V VSd cr/ ,,

All the “pure sway” moments for load case 5 were amplified by a factor of 1,22.

The global linear elastic analysis and amplification of sway moments was carried out for all seven load cases. Tables 4a and 4b shows the maximum forces in each member.

A B C D

E F G H

J K L M

N P Q R

1 2 3 4

5 6 78 9 10 11

12 13 1415 16 17 18

19 20 21

9G

= Element= Node

Figure 6 Labelling of Members within the Structure


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Member

Node Tension(kN)

Compression

(kN)

Shear(kN)

Bending Moment(kNm)

Maximum forces

1 A - 517,6 C2 -13,1 C4

0

E - -514,3 C2 13,1 C4 44,7 C42 B - 1196,4 C2 35,9 C7 0 Axial ld + BM

F - -1192,1 C2 -35,9 C7

-142,1 C7 internal col

3 C - 1196,3 C2 30,8 C6 0G - -1192,0 C2 -30,8

C6-124,0 C6

4 D - 529,0 C5 27,6 C7 0 Axial ld + BM

H - -525,7 C5 -27,6 C7

-105,5 C7 external col

8 E - 317,9 C2 -43,9 C2

79,8 C2

J - -314,6 C2 43,9 C2 73,4 C29 F - 722,8 C2 25,8 C5 58,6 C7

K - -718,5 C2 -25,8 C5

-69,2 C6

10 G - 722,8 C2 11,5 C1 31,2 C4L - -718,5 C2 -11,5

C1-45,9 C7

11 H - 320,0 C2 51,3 C5 -91,7 C5M - -316,7 C2 -51,3

C5-93,7 C5

15 J - 113,0 C4 -43,0 C2

73,5 C3

N - -109,7 C4 43,0 C2 77,3 C216 K - 253,9 C2 15,2 C6 -41,9 C6

P - -249,6 C2 -15,2 C6

-36,7 C6

17 L - 253,9 C2 -9,8 C3 34,6 C3Q - -249,6 C2 9,8 C3 24,5 C4

18 M - 113,5 C4 44,8 C5 -76,0 C5R - -110,1 C4 -44,8

C5-82,4 C5

Table 4a Maximum Axial Forces, Shear Force and Bending Moment in Each Column

Shaded boxes indicate the critical case that each section type will be designed to.For example, internal columns are the same section for every storey, therefore the worst case will obviously be one of the base columns.


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Member

Node Tension(kN)

Compression

(kN)

Shear(kN)

Bending Moment(kNm)

Maximum forces

5 E -28,9 C2

8,5 C1 201,8 C4

-118,5 C4 Internal BM

F 28,9 C2 -8,5 C1 255,2 C5

341,6 C5 on floor beam

6 F -23,4 C2

1,4 C1 220,8 C3

-241,2 C2

G 23,4 C2 -1,4 C1 231,6 C5

310,7 C5

7 G -31,5 C2

- 244,2 C2

-271,7 C2 External BM

H 31,5 C2 - 213,3 C7

194,4 C5 on floor beam

12 J -0,6 G 28,1 C5 207,1 C3

-146,8 C2

K 0,6 G -28,1 C5 242,7 C2

288,6 C5

13 K -1,3 C4 15,6 C5 222,0 C4

-242,6 C2

L 1,3 C4 -15,6 C5 223,6 C2

261,3 C5

14 L -1,9 G 6,5 C5 241,0 C2

-271,5 C2

M 1,9 G -6,5 C5 208,5 C3

169,7 C5

19 N - 55,8 C5 109,7 C4

-77,3 C2 Internal BM

P - -55,8 C5 130,4 C2

150,2 C5 on roof beam

20 P - 42,4 C5 119,3 C3

-131,7 C2

Q - -42,4 C5 119,7 C2

135,8 C5

21 Q - 44,8 C5 129,9 C2

-146,7 C2 External BM

R - -44,8 C5 110,1 C4

82,4 C5 on roof beam

Table 4b Maximum Axial Force, Shear Force and Bending Moment in Each Beam


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Shaded boxes indicate the critical case that each section type will be designed to.Generally, the beams will be designed to the maximum bending moment, and the columns will be designed for maximum axial load. However, it may also be necessary to check the interaction of the bending moments and axial forces for certain load cases.

Floor Beam - Fully RestrainedThe maximum moment any floor beam has to resist is 341,6 kNm at node F, under load case 5. The beam is in tension for this load case, therefore the interaction of the bending moment and axial force will not need to be checked.The maximum compression force any floor beam has to resist is 28,1 kN (member 12) under load case 5. The beam also has to resist a bending moment of 288,6 kNm but as the axial force is low it is unlikely that the interaction will be critical.The maximum shear force any floor beam has to resist is 255,2 kN at node F, under load case 5.The maximum tensile force any floor beam has to resist is 31,5 kN (member 7) under load case 2. This will not be the critical condition for the floor beam design.IPE 450Section properties:Depth, h = 450 mm, Width, b = 190 mmWeb thickness, tw = 9,4 mm Flange thickness, tf = 14,6 mmPlastic modulus, Wpl = 1702 cm3

5.4.5.1

This notation conforms with Figure 1.1 in Eurocode 3: Part 1.1.Classification of Cross-section

Figure 7 shows a typical cross-section for an IPE.


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tw

t

d

c

f

Figure 7 A Typical Cross-Section

Flange BucklingClass 1 limiting value of c/tf for an outstand of a rolled section is 10ε.ε = 235 / fy and fy = 235 N/mm2, therefore ε =1.

Calculate the ratio ct f

, where c is half

the width of the flange = 95 mm, and tfis the flange thickness = 14,6 mm (if the flange is tapered, tf should be taken as the average thickness).ctf

= =95

14 66 5

,,

Table 5.3.1 (Sheet 3)

Web BucklingClass 1 limiting value of d/tw for a web of a rolled section under bending is 72ε.ε = 235 / fy and fy = 235 N/mm2, therefore ε =1.

Calculate the ratio dtw

, where d is the depth between root radii = 378,8 mm

and tw is the web thickness = 9,4 mm.dtw

= =378 89 4

40 3,,

,

ct

10f

< ε and dt

72w

< ε

∴ Section is Class 1.


Table 5.3.1 (Sheets 1 and 3)

The beam shown in Figure 8 is fixed at both ends and is fully restrained along its length.For the critical load cases given above, check the beam, assuming it is grade Fe360, and that it is carrying plaster, or a similar brittle finish.

6,5 m

Load varies dependingon appropriate load case

Figure 8 Loading on Fully Restrained Floor BeamCheck Moment Capacity

Design moment, MSd = 341,6 kNm (load case 5 - member 5)The design bending moment, MSd, must be less than or equal to the design moment resistance of the cross section, Mc.Rd:MSd ≤ Mc.Rd

Mc.Rd = Mpl.y.Rd =W fpl y

M0γ

5.4.5.1(1)


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where Wpl is the plastic section modulus = 1702 cm3,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

Table 3.15.1.1(2)

Therefore, Mpl.y.Rd =W fpl y

M0γ= 1702x10 x 235

1,1364 kNm

3

=

MSd = 341,6 kNm ≤ Mpl.y.Rd = 364 kNmTherefore the section is satisfactory.

Check Interaction of Maximum Axial Force and Bending Moment

Interaction case: Design moment, MSd = 341,6 kNm and design axial force, NSd = 28,1 kN (load case 5 - member 12)Lateral support to the floor beams is provided by the floor slabs, therefore there is no need to check for failure due to flexural or lateral torsional buckling.A class 1 member subject to moment about the major axis only should satisfy the following:

MM

NN

Sd

pl.Rd

Sd

pl.Rd

+

≤2

1

Applied bending moment, MSd, = 341,6 kNm,design bending moment resistance of the section, Mpl.Rd = 364 kNm,applied axial load, NSd, = 28,1 kN, and

design compression resistance of the cross-section, Npl.Rd =Afy

M0γwhere A is the cross-sectional area of the section = 9880 mm2,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

Therefore, Npl.Rd = =9880 x 235 kN

112111

,Substituting into the above equation:341 6 28 1 0 94

2, , ,x10

364x10x10

2111x10

6

6

3

3+

=

Therefore the section is satisfactory under combined axial load and bending moment

5.4.8.1(2)

Table 3.15.1.1(2)

Lateral Torsional BucklingAlthough lateral restraint is provided by the floor slabs, it is necessary to check for lateral torsional buckling over the length of the beam which is in hogging (approximately 1,7 m). This is because the floor beams are restrained only by the upper flange, while the lower flange which is under compression, has no lateral restraint.As this is over such a short length of the beam, it is unlikely that this will be a critical condition. For completeness, however, the check is included in this worked example.A class 1 section should satisfy the following: 5.5.4(2)


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NAf

k MW f

Sd

z y M1

LT y.Sd

LT pl.y y M1χ γ χ γ/ /,+ ≤ 1 0

Applied axial force, NSd = 28,1 kNDetermination of χz

Slenderness, λz = l/izwhere l is taken as the length of beam in hogging, and i is the radius of gyration about the appropriate axis.Slenderness, λz = 1700 / 41,2 = 41,3

Non-dimensional slenderness, λλλ

βzz

1A

0,5=

where λ1 = 93,9ε = 93,9 x 1 = 93,9, andβA = 1 for class 1 members

Non-dimensional slenderness, λ z = =41 393 9

0 44,,

,

From Table 5.5.2, using buckling curve b, reduction factor, χz = 0,91

5.5.1.45.5.1.5(2)

5.5.1.2(1)5.5.1.1(1)

Table 5.5.3 Table5.5.2

Cross-sectional area, A = 9880 mm2,Yield strength of the steel, fy = 235 N/ mm2, andPartial material safety factor for buckling resistance, γM1 = 1,1.

Table 3.15.1.1(2)

( )i I I

Wx10 x 791x10

1702x10mmLt

z w

ply

6 9

3=

=

=2

0 25

2

0 25

16 76 46 3, ,

, , F.2.2(3)

a II

x1066,9x10

mmLtw

t

9

4=

=

=

0 5 0 5791 1087

, , F.2.2(1)


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To Calculate kLT

k NAfLT

LT Sd

z y

= −1 µχ

but kLT ≤ 1,0.

where µ λ β µLT z M.LT LTbut = − ≤0 15 0 15 0 9, , ,λ z = 0,44βM.LT = 1,8 (ψ = 0)∴µLT = (0,15 x 0,44 x 1,8) - 0,15 = -0,03

∴ = −−

=kx 28,1x10

x 9880 x 235LT

3

10 03

0 911 0

( , ),

,

Applied moment, My.Sd = 341,6 kNm

5.5.4(2)

5.5.4(7) and Figure 5.5.3

To Calculate Reduction Factor, χLTThe value of χLT can be determined from Table 5.5.2 for the appropriate value of the non-dimensional slenderness, λ LT .

λλλ

βLTLT

1w

0,5=

where λ1 = 93,9 ε = 93,9 x 1 = 93,9,βw = 1 for class 1 sections, and

( )λ LT

LT

1LT

L / i

CL / a

=

+

0 52 0 25

125 66

,

,

,

where L is the length = 1700 mm,iLT = 46,3 mm (from section properties),ψ = 0, k = 1,0, therefore C1 = 1,879, andaLT = 1087 mm (from section properties).

5.5.2(4)

5.5.2(5)

5.5.2(1)

Equation F.15


( )λ LT =

+

=1700 46 3

1 879 11700 1087

25 66

26 20 5

2 0 25/ ,

,/,

,,

,

∴ = = =λλλ

βLTLT

1w

0 5 0 526 293 9

1 0 0 28, ,,,

, ,5.5.2(5)

Where the non-dimensional slenderness λ LT ≤ 0 4, , no allowance for lateral torsional buckling is necessary.

5.5.2(7)

Therefore, this section is satisfactory for lateral torsional buckling.Shear on Web

Design shear force, VSd = 255,2 kN (load case 5 - member 5)The shear resistance of the web must be checked. The design shear force, VSd, must be less than or equal to the design plastic shear resistance, Vpl.Rd:VSd ≤ Vpl.Rd

where Vpl.Rd is given by Af / 3

vy

M0γ

5.4.6


15/02/07 48

For rolled I and H sections loaded parallel to the web,shear area, Av = 1,04 h tw,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

5.4.6(4)

Table 3.15.1.1(2)

∴ =V1,04ht f

3xpl.Rdw y

M0γ= 1 04 543, x 450 x 9,4 x 235

3 x 1,1kN=

This is greater than the shear on the section (255,2 kN).The section is satisfactory under shear.A further check is sometimes required, especially when there are significant point loads, cantilevers or continuity, to ensure that the shear will not have a significant effect on the moment resistance. This check is carried out for the moment and shear at the same point. The moment resistance of the web is reduced if the shear is more than 50% of the shear resistance of the section. With a uniform load, the maximum moment and shear are not coincident and this check is not required for beams without web openings.

5.4.7(3)

Check Tension Resistance of MemberMaximum applied tensile force, NSd = 31,5 kN (load case 2 - member 7)For members in axial tension, the design value of the tensile force, NSdmust be less than the design tension resistance of the cross-section, Nt.Rd.

Nt.Rd = Npl.Rd = A fy

M0γwhere A is the cross-sectional area of the section = 9880 mm2,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

Therefore N 9880 x 235 2110 kNpl.Rd = =11,

5.4.3(1)

Table 3.15.1.1(2)

The design tension resistance of the cross-section, Nt.Rd = 2110 kN, is greater than the applied tensile force, NSd = 31,5 kN, therefore the section is OK.

Deflection CheckFor a plaster or similar brittle finish, the deflection limit is L/250 for δmax. Deflection checks are based on the serviceability loading.

Table 4.1 Figure 4.1

The maximum deflection for any floor beam is 6,2 mm which occurs under serviceability load case 4 at member 5.

Deflection limit for δ maxL

250mm= = =

6500250

26

The actual deflection is less than the allowable deflection: 6,2 mm < 26mm, therefore the section is OK.

Table 4.1

The calculated deflections are less than the limits, so no pre-camber is required.

4.3.2(2)


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SummaryThe trial section IPE 450 is satisfactory.

Roof BeamThe maximum moment any roof beam has to resist is 150,2 kNm at node P (member 19), under load case 5.The maximum compression force any roof beam has to resist is 55,8 kN (member 19) also under load case 5. The interaction of the axial force and bending moment will therefore be critical.The maximum shear force any roof beam has to resist is 130,4 kN at node P, under load case 2.The roof beams experience no tensile force.IPE 360Section properties:Depth, h = 360 mm, Width, b = 170 mmWeb thickness, tw = 8,0 mm Flange thickness, tf = 12,7 mmPlastic modulus, Wpl = 1019 cm3

5.4.5.1

This notation conforms with Figure 1.1 in Eurocode 3: Part1.1.Classification of Cross-Section

Flange bucklingClass 1 limiting value of c/tf for an outstand of a rolled section is 10ε.ε = 235 / fy and fy = 235 N/mm2, therefore ε =1.

Calculate the ratio ct f

, where c is half the width of the flange = 85 mm,

and tf is the flange thickness = 12,7 mm (if the flange is tapered, tf should be taken as the average thickness).ct f

= =85

12 76 7

,,


Web bucklingClass 1 limiting value of d/tw for a web of a rolled section under bending is 72ε.ε = 235 / fy and fy = 235 N/mm2, therefore ε =1.

Calculate the ratio dtw

, where d is the depth between root radii = 298,6 mm

and tw is the web thickness = 8,0 mm.dtw

= =298 68 0

37 3,,

,

ct

10f

< ε and dt

72w

< ε

∴ Section is Class 1.


Table 5.3.1 (Sheets 1 and 3)

For the critical load cases given above, check the beam, assuming it is grade Fe360, and that it is for a general roof.

Check Interaction of Maximum Axial Force and Bending Moment

Interaction case: Design moment, MSd = 150,2 kNm and design axial


15/02/07 50

force, NSd = 55,8 kN (load case 5 - member 19).There is no lateral support provided for the roof beams, therefore they must be checked for failure due to flexural and lateral torsional buckling.

Flexural BucklingA class 1 member subject to moment about the major axis only should satisfy the following:

NAf /

k MW f /

1,0sd

min y M1

y y.sd

pl.y y M1χ γ γ+ ≤

Applied axial force, NSd, = 55,8 kN

5.5.4(1)

To Calculate Reduction Factor, χminThe reduction factor χmin is the lesser of χy and χz, where χy and χz are the reduction factors from clause 5.5.1 for the y-y and z-z axes respectively.

Determination of χyThe reduction factor χy depends on the slenderness about the y-y axis.Assuming that the connections at the beam ends are rigid, then the slenderness about the y-y axis is:λy = l/iywhere l is taken as the system length, and i is the radius of gyration about the appropriate axis.Slenderness, λy = 6500 / 150 = 43,3λ1 = 93,9ε = 93,9 x 1,0 = 93,9

Non-dimensional slenderness, λλ

λβy

y

1A

0,5=

where βA = 1 for class 1 members

Non-dimensional slenderness, λ y = =43 393 9

0 46,,

,

From Table 5.5.2, using buckling curve a, the reduction factor, χy = 0,935

5.5.1.5

5.5.1.45.5.1.5(2)

5.5.1.2(1)

5.5.1.1(1)

Table 5.5.3 Table 5.5.2

Determination of χzSlenderness, λz = l/izwhere l is taken as the system length, and i is the radius of gyration about the appropriate axis.Slenderness, λz = 6500 / 37,9 = 171,5


βzz

1A

0,5=



1 8,,

,

From Table 5.5.2, using buckling curve b, reduction factor, χz = 0,2521

5.5.1.45.5.1.5(2)

5.5.1.2(1)5.5.1.1(1)


∴ Reduction factor, χmin = χz = 0,2521A is the cross-sectional area = 7270 mm2,fy is the yield strength = 235 N/mm2, andγM1 is the partial material safety factor for buckling resistance = 1,1.

Table 3.15.1.1(2)


15/02/07 51

To Calculate ky

k 1NAfy

y sd

y y

= −µ

χbut ky ≤ 1,5

where

µ λ βy y MYpl.y el.y

el.y

(2 4)W W

W= − +

−but µy ≤ 0,90

βMy is an equivalent uniform moment factor for flexural buckling

βMy = βM,ψ + M

MQ

MQ∆( ),β β ψ− M

where βM,ψ = 1,8 - 0,7ψ, and ψ = 0,47Therefore βM,ψ = 1,8 - (0,7 x 0,47) = 1,47,MQ = 187 kNm,∆M = 231 kNm, andβM,Q = 1,3.Therefore βMy = 1,47 + (187/231) x (1,3 - 1,47) = 1,33

µ y x 1,33 - 4) + 1019 - 904= = −0 46 2

9040 49, ( ,

k(-0,49) x 55,8x10

x 7270 x 235y

3

= − =10 935

1 02,

,

My.Sd is the design applied moment = 150,2 kNm, andWpl.y is the plastic section modulus = 1019 x 103 mm3.

NAf /

k MW f /

1,0sd

min y M1

y y.sd

pl.y y M1χ γ γ+ ≤

55,8x100,2521 x 7270 x 235 / 1,1

x 150,2x101019x10 x 235 / 1,1

3 6

3+ =1 02

0 85,

,

Therefore this section is satisfactory for flexural buckling

5.5.4(1)

5.5.4(7) and Figure 5.5.3

Lateral Torsional BucklingA class 1 section should satisfy the following:

NAf

k MW f

Sd

z y M1

LT y.Sd

LT pl.y y M1χ γ χ γ/ /,+ ≤ 1 0

Applied axial force, NSd = 55,8 kN,Reduction factor, χz = 0,2521,Cross-sectional area, A = 7270 mm2,Yield strength of the steel, fy = 235 N/ mm2, andPartial material safety factor for buckling resistance, γM1 = 1,1.

5.5.4(2)

Table 3.15.1.1(2)

( )i I I

Wx10 x 314x10

1019x10mmLt

z w

ply

6 9

3=

=

=2

0 25

2

0 25

10 43 42 1, ,

, , F.2.2(3)

a II

x1037,3x10

mmLtw

t

9

4=

=

=

0 5 0 5314 917 5

, ,

,F.2.2(1)


15/02/07 52

To Calculate kLT

k NAfLT

LT Sd

z y

= −1 µχ

but kLT ≤ 1,0.

where µ λ β µLT z M.LT LTbut = − ≤0 15 0 15 0 9, , ,λ z = 1,8βM.LT = βM,ψ + (MQ/∆M)(βM,Q - βM,ψ),where βM,ψ = 1,8 - 0,7ψ, and ψ = 0,47, therefore βM,ψ = 1,8 - (0,7 x 0,47) = 1,47,MQ = 187 kNm,∆M = 231 kNm, andβM,Q = 1,3.Therefore βM.LT = 1,47 + (187/231) (1,3 - 1,47) = 1,33.∴µLT = (0,15 x 1,8 x 1,33) - 0,15 = 0,21.

∴ = − =kx 55,8x10x 7270 x 235LT

3

10 21

0 25210 97

,,

,


5.5.4(2)

5.5.4(7) and Figure 5.5.3


λλλ

βLTLT

1w

0,5=


( )λ LT

LT

1LT

L / i

CL / a

=

+

0 52 0 25

125 66

,

,

,

where L is the length = 6500 mm,iLT = 42,1 mm (from section properties),ψ = 0,47, k = 1,0, therefore C1 = 1,107, andaLT = 917,5 mm (from section properties).

5.5.2(4)

5.5.2(5)

5.5.2(1)

Equation F.15

F.1.2(6)


( )λ LT =

+

=6500 42 1

1107 16500 917 5

25 66

111 90 5

2 0 25/ ,

,/ ,,

,,

,

∴ = = =λλλ

βLTLT

1w

0 5 0 5111 993 9

1 0 1 2, ,,,

, ,5.5.2(5)

Therefore, from Table 5.5.2, using buckling curve a (for rolled sections), the reduction factor, χLT = 0,53The plastic section modulus, Wpl.y = 1019 x 103 mm3.

5.5.3(4), Table 5.5.3 Table 5.5.2


15/02/07 53

NAf

k MW f

Sd

z y M1

LT y.Sd

LT pl.y y M1χ γ χ γ/ /,+ ≤ 1 0

55 8 0 971 41 1

, ,,

x100,2521 x 7270 x 235 / 1,1

x 150,2x100,53 x 1019x10 x 235 / 1,1

3 6

3+ = ≥

5.5.4(2)

Therefore, this section is not satisfactory for lateral torsional buckling.It is highly unusual to have, in practice, unrestrained roof beams. It is, therefore, asssumed that beams/purlins are present at quarter points along the beam, reducing the unrestrained length to 1,625 m. The maximum change in moment occurs between the 3rd restraint and node P on member 19. When the check is carried out again, λ LT ≤ 0 4, , therefore no allowance for lateral torsional buckling is necessary.

5.5.2(7)

Shear on WebDesign shear force, VSd = 130,4 kN (load case 2 - member 19)The shear resistance of the web must be checked. The design shear force, VSd, must be less than or equal to the design plastic shear resistance, Vpl.Rd:VSd ≤ Vpl.Rd


vy

M0γ

5.4.6

For rolled I and H sections loaded parallel to the web,shear area, Av = 1,04 h tw,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

5.4.6(4)

Table 3.15.1.1(2)

∴ =V1,04ht f

3xpl.Rdw y

M0γ= 1 04 369, x 360 x 8,0 x 235

3 x 1,1kN=

This is greater than the shear on the section (130,4 kN), therefore this section is satisfactory under shear.A further check is sometimes required, especially when there are significant point loads, cantilevers or continuity, to ensure that the shear will not have a significant effect on the moment resistance. This check is carried out for the moment and shear at the same point. The moment resistance of the web is reduced if the shear is more than 50% of the shear resistance of the section. With a uniform load, the maximummoment and shear are not coincident and this check is not required for beams without web openings.

5.4.7(3)

Deflection CheckFor a roof generally the deflection limit is L/200 for δmax. Deflection checks are based on the serviceability loading.

Table 4.1 Figure 4.1

The maximum deflection for any roof beam is 6,5 mm which occurs under serviceability load case 4 at member 19.

Deflection limit for δ maxL

200mm= = =

6500200

32 5,

The actual deflection is less than the allowable deflection: 6,5 mm < 32,5 mm therefore the section is OK.

Table 4.1

The calculated deflections are less than the limits, so no pre-camber is 4.3.2(2)


15/02/07 54

required.Summary

The trial section IPE 360 is satisfactory.

ColumnsExternal Column

The maximum compression force any external column has to resist is 529 kN (member 4) under load case 5.The maximum moment any external column has to resist is 106 kNm at node H, under load case 7. The column is also subject to an axial load of 458,4 kN under this load case. However, the critical interaction case is likely to be under load case 5, with a maximum axial force of 529 kN and a bending moment of 103 kNm. It is this case that is likely to be critical.The maximum shear force any external column has to resist is 51,3 kN at member 11 under load case 5.

Section PropertiesAll external columns are HE 220 B grade Fe360h = 220 mm b = 220 mmtw = 9,5 mm tf = 16 mmd/tw = 16,0 c/tf = 6,9A = 9100 mm2 Iy = 80,91 x 106 mm4

Iw = 295 x 109 mm6 Iz = 28,43 x 106 mm4

It = 76,6 x 104 mm4 Wpl.y = 827 x 103 mm3

Wel.y = 736 x 103 mm3 iy = 94,3 mmiz = 55,9 mm

( )i I I

Wx10 x 754x10

1283x10mmLt

z w

ply

6 9

3=

=

=2

0 25

2

0 25

51 4 69 7, ,

, , F.2.2(3)

a II

x10124x10

mmLtw

t

9

4=

=

=

0 5 0 5754 779 8

, ,

,F.2.2(1)

All the above properties can be obtained from section property tables.Classification of Cross Section

This section is designed to withstand moments in addition to axial force. (Note that the section is always in compression.)

5.3

Flange (subject to compression)Class 1 limiting value of c/tf for an outstand of a rolled section is 10ε.ε = 235 / fy where fy = 235 N/mm2, ∴ ε = 1.10ε = 10 x 1 = 10From section properties, c/tf = 6,9


Web (subject to compression)Class 1 limiting value of d/tw for a web subject to compression only is 33ε.33ε = 33 x 1 = 33From section properties, d/tw = 16

Table 5.3.1(Sheet 1)

c/tf ≤ 10ε and d/tw ≤ 33εTherefore the section is Class 1.


15/02/07 55

Resistance of Cross-Section 5.4.4It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example.For members in axial compression, the design value of the compressive force, NSd, at each cross-section shall satisfy NSd ≤ Nc.Rd

5.4.4(1)

For a class 1 cross-section, the design compression resistance of the cross-section, Nc.Rd, may be determined as:

NAf

c.Rdy

M0

=γ

where A is cross-sectional area = 9100 mm2,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

N 9100 x 235x10

kNc.Rd 3= =11

1944,

NSd = 529 kN, therefore Nsd < Nc.Rd.The section can resist the applied axial load.

5.4.4(2)

Table 3.15.1.1(2)


15/02/07 56

Buckling Resistance of MemberA class 1 member subject to combined bending and axial compression should be checked for the following modes of failure:• Flexural buckling (Clause 5.5.4(1)), and • Lateral Torsional Buckling (Clause 5.5.4(2)).


NA f /

k MW f /

Sd

min y M1

y y.Sd

pl.y y M1χ γ γ+ ≤ 1 0,

Applied axial force, NSd, = 529 kN

5.5.4(1)


Determination of χyThe reduction factor χy depends on the slenderness about the y-y axis.The connections at the column base are effectively pinned, then the slenderness about the y-y axis is:λy = l/iywhere l is equal to the system length and i is the radius of gyration about the appropriate axis.(Use the system length as using the amplified sway moment method. If using the effective length method, this is where the effective lengths would be greater than 1)Slenderness, λy = 3500 / 94,3 = 37,1λ1 = 93,9ε = 93,9 x 1,0 = 93,9


λβy

y

1A

0,5=



0 4,,

,

From Table 5.5.2, using buckling curve b, the reduction factor, χy = 0,9261

5.5.1.5

5.5.1.45.5.1.5(1)

5.5.1.2(1)

5.5.1.1(1)


Determination of χzSlenderness, λz = l/izwhere l is equal to the system length and i is the radius of gyration about the appropriate axis.(Use the system length as using the amplified sway moment method. If using the effective length method, this is where the effective lengths would be greater than 1)Slenderness, λz = 3500 / 55,9 = 62,6


βzz

1A

0,5=


5.5.1.45.5.1.5(2)

5.5.1.2(1)

5.5.1.1(1)

Table


15/02/07 57


0 67,,

,

From Table 5.5.2, using buckling curve c, reduction factor, χz = 0,73

5.5.3 Table 5.5.2


Table 3.15.1.1(2)


15/02/07 58

To Calculate ky

k 1NAfy

y sd

y y

= −µ

χbut ky ≤ 1,5

where

µ λ βy y Mypl.y el.y

el.y

W WW

= − +−

( )2 4 but µy ≤ 0,90

βMy is an equivalent uniform moment factor for flexural buckling (ψ = 0)= 1,8 - 0 = 1,8.

µ y =−

= −0 4 2736

0 04, ( ,x 1,8 - 4) +827x10 x10

736x10

3 3

3

( )∴ = −

−=k

x 529x10x9100 x 235y

3

10 04

0 92611 01

,,

,

My.Sd is the design applied moment = 103 kNm, andWpl.y is the plastic section modulus = 827 x 103 mm3.

NA f /

k MW f /

Sd

min y M1

y y.Sd

pl.y y M1χ γ γ+ ≤ 1 0,

529 1 010 96

x100,73 x 9100 x 235 / 1,1

x 103x10827x10 x 235 / 1,1

3 6

3+ =,

,

Therefore the section is satisfactory for flexural buckling

5.5.4(1)

5.5.4(7) and Figure 5.5.3

Lateral Torsional BucklingA class 1 section should satisfy the following:

NAf

k MW f

Sd

z y M1

LT y.Sd

LT pl.y y M1χ γ χ γ/ /,+ ≤ 1 0

Applied axial force, NSd = 529 kN,Reduction factor, χz = 0,73,Cross-sectional area, A = 9100 mm2,Yield strength of the steel, fy = 235 N/ mm2, andPartial material safety factor for buckling resistance, γM1 = 1,1.

5.5.4(2)

Table 3.15.1.1(2)

To Calculate kLT

k NAfLT

LT Sd

z y

= −1 µχ

but kLT ≤ 1,0.

where µ λ β µLT z M.LT LTbut = − ≤0 15 0 15 0 9, , ,λ z = 0,67βM.LT = 1,8 (ψ = 0)∴µLT = (0,15 x 0,67 x 1,8) - 0,15 = 0,03

∴ = − =kx 529x10

x 9100 x 235LT

3

10 03

0 730 99

,,

,


5.5.4(2)

5.5.4(7) and Figure 5.5.3


5.5.2(4)


15/02/07 59

λλλ

βLTLT

1w

0,5=


( )λ LT

LT

1LT

L / i

CL / a

=

+

0 52 0 25

125 66

,

,

,

where L is the length = 3500 mm,iLT = 69,7 mm (from section properties),ψ = 0, k = 1,0, therefore C1 = 1,879, andaLT = 779,8 mm (from section properties).

5.5.2(5)

5.5.2(1)

Equation F.15


( )λ LT =

+

=3500 69 7

1879 13500 779 8

25 66

40 10 5

2 0 25/ ,

,/ ,,

,,

,

∴ = = =λλλ

βLTLT

1w

0 5 0 540 193 9

1 0 0 43, ,,,

, ,5.5.2(5)

Therefore, from Table 5.5.2, using buckling curve a (for rolled sections), the reduction factor, χLT = 0,94The plastic section modulus, Wpl.y = 827 x 103 mm3.

5.5.3(4), Table 5.5.3 Table 5.5.2

NAf

k MW f

Sd

z y M1

LT y.Sd

LT pl.y y M1χ γ χ γ/ /,+ ≤ 1 0

529 0 990 98

x100,73 x 9100 x 235 / 1,1

x 102,8x100,94 x 827x10 x 235 / 1,1

3 6

3+ =,

,

5.5.4(2)

Therefore, this section is satisfactory for lateral torsional buckling.This section is satisfactory for both flexural buckling and lateral torsional buckling.

Shear on WebGenerally, a shear check on a column web would only be carried out if the structure is to be subjected to considerable seismic loading. For completeness, the check will be included in this worked example.Design shear force, VSd = 51,3 kN (load case 5 - member 11)The shear resistance of the web must be checked. The design shear force, VSd, must be less than or equal to the design plastic shear resistance, Vpl.Rd:VSd ≤ Vpl.Rd


vy

M0γ

5.4.6

For rolled I and H sections loaded parallel to the web,shear area, Av = 1,04 h tw,

5.4.6(4)


15/02/07 60

fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

Table 3.15.1.1(2)

∴ =V1,04ht f

3xpl.Rdw y

M0γ= 1 04 268 1, ,x 220 x 9,5 x 235

3 x 1,1kN=

This is greater than the shear on the section (51,3 kN).The section is satisfactory under shear.The column is satisfactory under all loading conditions.

Internal ColumnThe maximum compression force any internal column has to resist is 1196 kN (member 2) under load case 2.The maximum moment any internal column has to resist is 142 kNm at node F, under load case 7. The column is also subject to an axial load of 961 kN under this load case. However, the critical interaction case is likely to be under load case 5, with an axial force of 1149,7 kN and a bending moment of 123,9 kNm. It is this case that is likely to be critical.The maximum shear force any internal column has to resist is 35,9 kN at member 2 under load case 7.

Section PropertiesAll internal columns are HE 260 B grade Fe360h = 260 mm b = 260 mmtw = 10 mm tf = 17,5 mmd/tw = 17,7 c/tf = 7,4A = 1180 mm2 Iy = 149,2 x 106 mm4

Iw = 754 x 109 mm6 Iz = 51,4 x 106 mm4

It = 124 x 104 mm4 Wpl.y = 1283 x 103 mm3

Wel.y = 1148 x 103 mm3 iy = 112 mmiz = 65,8 mm

( )i I I

Wx10 x 754x10

1283x10mmLt

z w

ply

6 9

3=

=

=2

0 25

2

0 25

51 4 69 7, ,

, , F.2.2(3)

a II

x10124x10

mmLtw

t

9

4=

=

=

0 5 0 5754 779 8

, ,

,F.2.2(1)

All the above properties can be obtained from section property tables.


15/02/07 61

Classification of Cross Section 5.3Flange (subject to compression)

Class 1 limiting value of c/tf for an outstand of a rolled section is 10ε.ε = 235 / fy where fy = 235 N/mm2, therefore ε = 1.From section properties, c/tf = 7,4


Web (subject to compression)Class 1 limiting value of d/tw for a web subject to compression only is 33ε.From section properties, d/tw = 17,7c/tf ≤ 10ε and d/tw ≤ 33εTherefore the section is Class 1.

Table 5.3.1(Sheet 1)Table 3.5.1 (Sheets 1 and 3)

Resistance of Cross-Section 5.4.4It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example.For members in axial compression, the design value of the compressive force, NSd, at each cross-section shall satisfy NSd ≤ Nc.Rd

5.4.4(1)

For a class 1 cross-section, the design compression resistance of the cross-section, Nc.Rd, may be determined as:

NAf

c.Rdy

M0

=γ

where A is cross-sectional area = 11800 mm2,fy is the yield strength = 235 N/mm2, andγM0 is the partial material safety factor = 1,1.

N 11800 x 235x10

kNc.Rd 3= =11

2521,

NSd = 1196 kN, therefore Nsd < Nc.Rd.The section can resist the applied axial load.

5.4.4(2)

Table 3.15.1.1(2)


15/02/07 62

Buckling Resistance of MemberA class 1 member subject to combined bending and axial compression should be checked for the following modes of failure:• Flexural buckling (Clause 5.5.4(1)), and • Lateral Torsional Buckling (Clause 5.5.4(2)).


NA f /

k MW f /

Sd

min y M1

y y.Sd

pl.y y M1χ γ γ+ ≤ 1 0,

Applied axial force, NSd, = 1149,7 kN

5.5.4(1)


Determination of χyThe reduction factor χy depends on the slenderness about the y-y axis.The connections at the column base are effectively pinned, then the slenderness about the y-y axis is:λy = l/iywhere l is equal to the system length and i is the radius of gyration about the appropriate axis.(Use the system length as using the amplified sway moment method. If using the effective length method, this is where the effective lengths would be greater than 1)Slenderness, λy = 3500 / 112 = 31,3λ1 = 93,9ε = 93,9 x 1,0 = 93,9


λβy

y

1A

0,5=



0 33,,

,

From Table 5.5.2, using buckling curve b, the reduction factor, χy = 0,97

5.5.1.5

5.5.1.45.5.1.5(1)

5.5.1.2(1)

5.5.1.1(1)


Determination of χzSlenderness, λz = l/izwhere l is equal to the system length and i is the radius of gyration about the appropriate axis.(Use the system length as using the amplified sway moment method. If using the effective length method, this is where the effective lengths would be greater than 1)Slenderness, λz = 3500 / 65,8 = 53,2


βzz

1A

0,5=


5.5.1.45.5.1.5(2)

5.5.1.2(1)

5.5.1.1(1)

Table


15/02/07 63


0 57,,

,

From Table 5.5.2, using buckling curve c, reduction factor, χz = 0,803

5.5.3 Table 5.5.2


Table 3.15.1.1(2)

To Calculate ky

kNA fy

y Sd

y y

= −1µ

χbut ky ≤ 1,5

where

µ λ βy y Mypl.y el.y

el.y

W WW

= − +−

( )2 4 but µy ≤ 0,90

βMy is an equivalent uniform moment factor for flexural buckling (ψ = 0)= 1,8 - 0 = 1,8.

µ y =−

= −0 33 21148

0 01, ( ,x 1,8 - 4) +1283x10 x10

1148x10

3 3

3

( )∴ = −

−=k

x 1154x10x 11800 x 235y

3

10 01

0 971 0

,,

,

My.Sd is the design applied moment = 123,9 kNm, andWpl.y is the plastic section modulus = 1283 x 103 mm3.

NA f /

k MW f /

Sd

min y M1

y y.Sd

pl.y y M1χ γ γ+ ≤ 1 0,

1149 7 1 01 02

, ,,

x100,803 x 11800 x 235 / 1,1

x 123,9x101283x10 x 235 / 1,1

3 6

3+ =

Therefore the section fails. The internal column section needs to be increased and the analysis and checks carried out on the modified structure.

5.5.4(1)

5.5.4(7) and Figure 5.5.3


15/02/07 64

1. Joint characterisation1.1. General

An important step when designing a frame consists of the characterisation of the rotational response of the joints, i.e. the evaluation of the mechanical properties in terms of stiffness, strength and ductility.

Three main approaches may be followed :

• experimental• numerical• analytical.The only practical option for the designer is the analytical approach. Analytical procedures have been developed which enable a prediction of the joint response based on the knowledge of the mechanical and geometrical properties of the joint components.In this section a general analytical procedure, termed component method, is introduced. It applies to any type of steel or composite joints, whatever the geometrical configuration, the type of loading (axial force and/or bending moment, ...) and the type of member sections.The method is used in the Lecture on “Practical procedures for the characterisation of the response of moment resisting joints” where the mechanical properties of joints subjected to bending moment and shear force are computed.1.2. Introduction to the component methodThe component method considers any joint as a set of individual basic components. For the particular joint shown in Figure 2.b. (joint with an extended end-plate connection subject to bending), the relevant components are the following :

Compression in zone :

• column web in compression;• beam flange and web in compression;Tension zone :• column web in tension;• column flange in bending;• bolts in tension;• end-plate in bending;• beam web in tension;Shear zone :• column web panel in shear.

Each of these basic components possesses its own strength and stiffness either in tension or in compression or in shear. The column web is subject to coincident compression, tension and shear. This coexistence of several components within the same joint element can obviously lead to stress interactions that are likely to decrease the

J.1.5


15/02/07 65

resistance of the individual basic components.The application of the component method requires the following steps :a) identification of the active components in the joint being considered;b) evaluation of the stiffness and/or resistance characteristics for each individual basic

component (specific characteristics - initial stiffness, design resistance, ... - or the whole deformability curve) ;

c) assembly of all the constituent components and evaluation of the stiffness and/or resistance characteristics of the whole joint (specific characteristics - initial stiffness, design resistance, ... - or the whole deformability curve).

In Figure 1, the principles of the component method are illustrated in the specific case of a beam-to-column joint with a welded connection.

COMPONENT METHOD

Three steps

First step:Identification of thecomponents

Second step:Response of thecomponents

Third step:Assembly of thecomponents

Column web Column web Column webin shear in compression in tension

Stiffness coefficient ki of each componentResistance FRdi of each component

Stiffness of the joint Sj,ini = Ez²/ΣkiResistance of the joint MRd = min(FRdi).z

F

F

M=Fz

F

∆1

FRd1Ek1

F

∆1

FRd2Ek2

F

∆1

FRd3Ek3

M

MRd

φSj,ini

Figure 1 Application of the component method to a welded joint

The assembly procedure consists in deriving the mechanical properties of the whole joint from those of all the individual constituent


15/02/07 66

components. This requires a preliminary distribution of the forces acting on the joint into internal forces acting on the components in a way that satisfies equilibrium.In Eurocode 3 Annex J, the analytical assembly procedures are described for the evaluation of the initial stiffness and the design moment resistance of the joint. These two properties enable the designer to determine the design joint moment-rotation characteristic whatever the type of analysis (Figures 4 to Figure 6). In Annex A of the present lecture, information is provided on how the stiffness and strength assembly is carried out.

The application of the component method requires a sufficient knowledge of the behaviour of the basic components. Those covered by Eurocode 3 are listed in Table 1. The combination of these components allows one to cover a wide range of joint configurations, which should be sufficient to satisfy the needs of practitioners as far as beam-to-column joints and beam splices in bending are concerned. Examples of such joints are given in Figure 2.Some fields of application can also be contemplated :• Joints subject to bending moment (and shear) and axial force;• Column bases subject to coincident bending moment, shear force and axial force where the

components such as :- concrete foundation in compression;- end-plates with specific geometries;- anchorages in tension;- contact between soil and foundation,

will be activated.

These situations are however not yet covered, or only partially covered, by Eurocode 3.


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N° Component

1 Column web panel in shear VSd

VSd

2 Column web in compression

Fc.Sd

3 Beam flange and web in compression

Fc.Sd

4 Column flange in bendingFt.Sd

5 Column web in tensionFt.Sd

6 End-plate in bending Ft.Sd

7 Beam web in tension Ft.Sd

8 Flange cleat in bending Ft.Sd

Table J.1


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9 Bolts in tension Ft.Sd

10 Bolts in shear Fv.Sd

11 Bolts in bearing (on beam flange,column flange, end-plate or cleat)

Fb.Sd

12 Plate in tension or compression Ft.Sd

Fc.Sd

Table 1 List of components covered by Eurocode 3


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(a) Welded joint (b) Bolted joint with extended end-plate

(c) Two joints with flush end-plates (d) Joint with flush end-plate(Double-sided configuration)

(e) End-plate type beam splice (f) Cover-joint type beam splice

(g) Bolted joint with angle flange cleats (h) Two beam-to-beam joints(Double-sided configuration)

Figure 2 Examples of joints covered by Eurocode 3

2. Joint idealisation

Figure J.4


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The non-linear behaviour of the isolated flexural spring which characterises the actual joint response does not lend itself towards everyday design practice. However the moment-rotation characteristic curve may be idealised without significant loss of accuracy. One of the most simple idealisations possible is the elastic-perfectly plastic relationship (Figure 3.a). This modelling has the advantage of being quite similar to that used traditionally for the modelling of member cross-sections subject to bending (Figure 3.b).The moment Mj,Rd that corresponds to the yield plateau is termed the design moment resistance in Eurocode 3. It may be considered as the pseudo-plastic moment resistance of the joint. Strain-hardening effects and possible membrane effects are henceforth neglected, which explains the difference in Figure 3 between the actual M-φ characteristic and the yield plateau of the idealisation.

Sj,ini/η

φ

Mj

Mj,Rd

EI/L

φ

Mb , Mc

Mpl,Rd

Actual M-φ characteristicIdealised M-φ characteristic

(a) Joint (b) Member

Figure 3 Bi-linearisation of moment-rotation curves

The value of the constant stiffness Sj.ini/η is discussed below.In fact there are different possible ways to idealise a joint M-φ characteristic. The choice of one of them is dependent upon the type of frame analysis which is contemplated:

- Elastic idealisation for an elastic analysis (Figure 4) :

The principal joint characteristic is the constant rotational stiffness.


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Sj,ini/η

φ

Mj

Mj,Rd

φ

Mj

Mj,Rd

2/3 Mj,Rd

Sj,ini

(a) For elastic verification (b) For plastic verification

Actual M-φ curve

Idealised representation

Figure 4 Linear representation of a M-φ curve

Two possibilities are offered in Eurocode 3-(revised) Annex J :• Elastic verification of the joint resistance (Figure 4.a) : the constant

stiffness is taken equal to the initial stiffness Sj.ini; at the end of the frame analysis, a check that the design moment MSd experienced by the joint is less than the maximum elastic joint moment resistance defined as 2/3 Mj,Rd.

• Plastic verification of the joint resistance (Figure 4.b) : the constant stiffness is taken equal to a fictitious stiffness, the value of which is intermediate between the initial stiffness and the secant stiffness relative to Mj,Rd; it is defined as Sj.ini/η. This idealisation is valid for MSd

values less than or equal to Mj,Rd . The values of η are as follows (Table 2) :

Table 2 Values of η

- Rigid-plastic idealisation for a rigid-plastic analysis (Figure 5).

Only the design resistance Mj,Rd is needed. In order to allow the possible plastic hinges to form and rotate at the joint locations, it is necessary to check that the joint has a sufficient rotation capacity.

J.2.1.2

Type of connection Beam-to-column joint Other types of jointWelded 2 3Bolted end-plate 2 3Bolted cleat 2 3,5


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φ

Mj

Mj,Rd

Figure 5 Rigid-plastic representation of a M-φ curve

- Non-linear idealisation for an elastic-plastic analysis (Figure 6).

The stiffness and resistance properties are of equal importance in this case. The possible idealisations range from bi-linear or tri-linear representations to the fully non-linear curve. Again rotation capacity is required in joints where plastic hinges are likely to form and rotate.

J.2.1.3


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(a) Bi-linear (b) Tri-linear

Sj,ini/η

φ

Mj

Mj,Rd

φ

Mj

Mj,Rd

φ

Mj

Mj,Rd

(c) Non linear

Figure 6 Non-linear representations of a M-φ curve

Annex A Evaluation of the stiffness and resistance properties of the joints according to Eurocode 3-(revised) Annex J

The component method presented in this lecture is a three step procedure for evaluating the stiffness and resistance properties of structural joints. In the first step, the list of constitutive components is established. The stiffness and/or resistance properties of the components are then derived.

The assembly of the components constitutes the third and last step of the method. As its name indicates, it consists in assembling the individual components so as to derive the mechanical properties of the whole joint. The relationship between component properties and joint properties is based on what is commonly called the "distribution of internal forces in the joint". The latter consists of determining, for a given set of external forces acting on the joint, the manner in which these forces are distributed between all the constitutive components. The force to which any given component is subjected is termed the "internal

J.2.1.4


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force".

This notion applies not only to structural joints but also to any cross-section in a member and a parallel between these situations is now drawn. In any cross-section of beams and columns, the distribution of internal forces in it is required so as to permit computation of its flexural rigidity or its resistance in bending, shear, torsion and/or axial compression or tension. That is why, in the following paragraphs, the word "cross-section" covers beam and column sections as much as joint sections.

The distribution of internal forces has to be carried out in a rational way, and has, from a theoretical point of view, to fulfil the following requirements :• the internal forces given by the distribution have to be in equilibrium

with the external forces acting on the cross-section;• the compatibility of the displacements between the constitutive parts

of the cross-section - namely components in the case of a joint – has to be satisfied;

• each part of the cross-section has to be able to transfer the internal force to which it is subjected;

• the maximum deformation capacity of each part of the cross-section must not be exceeded.

In the case of a beam or column H or I cross-section subjected to a bending moment, the distribution of internal forces - read stresses in this particular case - in the elastic range is generally assumed to follow the Navier rule (Figure A.1).

Hy

M

σ =M y

I.

(a) Beam (b) Internal stresses (c) External force

Figure A.1 Elastic distribution of internal forces (here stresses) in a beam profile in bending (I = second moment of area)


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fy

mγ

Figure A.2 Internal forces – here stresses – corresponding to the maximum elastic design moment resistance of the cross-section

fy

mγ

Figure A.3 Internal forces - stresses here – corresponding to the plastic moment resistance of the cross-section

When the maximum internal stresses (at y = ±H/2) reach the value of the yield stress fy of the constitutive steel divided by a partial safety factor γM, the maximum elastic moment resistance of the cross-section is obtained (Figure A.2). The design moment resistance is expressed as :

MyM

yRd /Wf

f2/H

IM γγ

== (A.1)

where W is the elastic modulus of the cross-section in bending.

To profit from the extra resistance provided by the plastification of the cross-section, reference has therefore to be made to the distribution represented at Figure A.3. The design resistance then becomes :

MyRd /ZfM γ= (A.2)where Z denotes the plastic modulus of the cross-section in bending.

In these distributions, Bernoulli’s assumption has been considered so as to ensure a compatibility between the elongation - or shortening - of all the constitutive fibers of the cross-section. Both elastic and plastic distributions illustrated in Figure A.2 and Figure A.3 are in equilibrium with the applied external forces and respect the plasticity criterion (σ ≤fy/γM). Therefore, they satisfy the first three of the four above-mentioned

6.1.4


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criteria which any distribution of internal forces should fulfil.

To reach the elastic or plastic moment resistance, the constitutive fibers of the section have to possess a sufficient deformation capacity so as to reach the yield stress in the case of the elastic distribution, or, in the case of the plastic distribution, to reach the yield stress and also to allow a plastic redistribution of the stresses between the adjacent fibers. This means that no premature local buckling of one of the section walls in compression and that no rupture of the material in tension must occur which would limit the moment capacity of the section. Specific criteria are provided in the codes so as to prevent the user from overestimating the resistance when such limitations apply. In these conditions, the fourth criterion on ductility is also fulfilled.

In a H or I profile, it is also common practice to replace the bending moment applied to the cross-section by a couple of forces located at the level of the centre lines of the beam flanges. The intensity of the forces is limited to the design resistance of the flanges in tension and compression, due attention being paid to the possible local buckling of the flange in compression. The web, the bending resistance of which is neglected, is usually devoted to the transfer of shear forces. Three of the four identified criteria are therefore satisfied. However the fourth condition (compatibility) is neglected. This so-called static approach is known to lead to a safe estimation of the design resistance of the section and is usually followed for sake of simplicity.

The procedure to distribute internal forces within structural joints is quite similar to that described in the foregoing paragraphs for beam and column cross-sections. In the next paragraphs, the procedure followed by Eurocode 3-(revised) Annex J is described; it applies to steel beam-to-column joints and beam splices where the beam(s) is (are) subject to bending moments and shear forces. For sake of simplicity and to allow for a hand calculation, two separate distribution procedures are detailed, one for the evaluation of the elastic initial stiffness and the other for the assessment of the design resistance of the joint.

The initial elastic stiffness and the design resistance are considered by Eurocode 3 as the two main parameters characterizing the response of a joint in bending. Based on these two values, a full M-φ curve can then be derived as shown in Figure A.4.

Provided that the non-linear M-φ curve of (revised)-Annex J is not limited by the rotational capacity (φCd), this curve consists of three parts. Up to a level of 2/3 of the design moment resistance MRd , the curve is assumed to be linear elastic and the corresponding stiffness is the so-called initial stiffness Sj,ini. Between 2/3 MRd and MRd, the curve is non-linear. Once the moment in the joint reaches MRd, a yield plateau develops under further imposed rotations of the joint.


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Figure A.4 Non-linear M-φ curve according to Annex J

The model assumes a fixed ratio between the initial stiffness Sj,ini and the secant stiffness at the intersection between the non-linear part and the yield plateau (Sj at level MRd). For end-plate and welded joints, this ratio is equal to 3. For flange cleat joints, this ratio is 3,5.The shape of the non-linear part between 2/3 MRd and MRd is given by the following interpolation formula :

ψ

=

Rd

Sd

ini,jj

MM5,1

SS (A.3)

where ψ = 2,7 for end-plate and welded joints and 3,1 for flange cleat joints.In this interpolation formula, the value of Sj is dependent on MSd.

Stiffness assemblyAs (revised)-Annex J refers to the so-called component method, the rotational response of a joint is based on the mechanical properties of its different constitutive components. The advantage is that an engineer is able to calculate the mechanical properties of any joint by decomposing the joint into relevant components. Annex J gives direct guidance for end-plate, welded and flange cleat joints for this decomposition. Table A.1 shows an overview of components to be taken into account when calculating the initial stiffness for these types of joints.


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Component Number End-plated Welded Flange cleatedColumn web panel in shear 1 x x xColumn web in compression 2 x x xColumn flange in bending 3 x xColumn web in tension 4 x x xEnd-plate in bending 5 xFlange cleat in bending 6 xBolts in tension 7 x xBolts in shear 8 xBolts in bearing 9 x

Table A-1 Overview of components for different joints

In the model, it is assumed that the deformations of the following components : a) beam flange and web in compression, b) beam web in tension and c) plate in tension or compression are included in the deformations of the beam in bending. Consequently they are not assumed to contribute to the flexibility of the joint. Haunches, when present, also fall in this category.

The initial stiffness Sj,ini is derived from the elastic stiffnesses of the components. The elastic behaviour of each component is represented by an extensional spring. The force-deformation relationship of this spring is given by :

Fi = ki .E . ∆ i (A.4)

Where: Fi is the force in the spring i;ki is the stiffness coefficient of the component i;E is the Young modulus;∆i is the deformation of the spring i.

The spring components in a joint are combined into a spring model. Figure A.5 shows for example the spring model for an unstiffened welded beam-to-column joint.

J.1.4

J.4.1

J.4.2


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Figure A.5 Spring model for an unstiffened welded joint

The force in each spring is equal to F. The moment M acting on the spring model is equal to F·z, where z is distance between the centre of tension (for welded joints, located in the centre of the upper beam flange) and the centre of compression (for welded joints, located in the centre of the lower beam flange). The rotation φ in the joint is equal to (∆1+∆2+∆4) / z. In other words:

i

2

i

2

i

zini,j

k1

Ez

k1

EF

Fz

z

FMSΣΣ

Σ∆φ==== (A.5)

The same formula applies for an end-plate joint with a single bolt-row in tension and for a flange cleat joint. However, the components to be taken into account are different, see Table A.1.

Figure A.6.a shows the spring model adopted for end-plate joints with two or more bolt-rows in tension. It is assumed that the bolt-row deformations for all rows are proportional to the distance to the point of compression, but that the elastic forces in each row are dependent on the stiffness of the components. Figure A.6.b shows how the deformations ki,r of components 3, 4, 5 and 7 are added to an effective spring per bolt-row, with an effective stiffness coefficient keff,r (r is the index of the row number). Figure A.6.c indicates how these effective springs per bolt-row are replaced by an equivalent spring acting at a lever arm z. The stiffness coefficient of this effective spring is keq. The effective stiffness coefficient keq can directly be applied in Formula (A.5). The formulae to determine keff,r, z and keq are as follows :

∑=

i r,i

r,eff

k1

1k (A.6)

∑∑

=

rrr,eff

r

2rr,eff

hk

hkz (A.7)


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z

hhk r

rr,eff

equ

∑= (A.8)

They can be derived from the sketches of Figure A.6. The basis for these formulae is that the moment-rotation behaviour of each of the systems in Figure A.6.a to A.6.c are equal. An additional condition is that the compressive force in the lower rigid bar is equal in each of these systems.

Figure A.6 Spring model for a beam-to-column end-plated joint with more than one bolt-row in tension

In this stiffness model :• the internal forces are in equilibrium with the bending moment;• the compatibility of the displacements is ensured through the

assumption of an infinitely rigid transverse stiffness of the beam cross-section;

• the plasticity criterion is fulfilled as long as the elastic resistance of the springs is not reached;

• no ductility requirement is likely to limit the deformation capacity of the springs in the elastic range of behaviour as far as Eurocode 3-(revised) Annex J components are concerned.

The solution provided by Eurocode 3-(revised) Annex J for the initial stiffness prediction fulfils the four main requirements which any distribution of internal forces should satisfy, from a theoretical point of view, and can therefore be considered as an “ exact ” one.

Strength assembly

Formula (J.34)

J.4.3


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The procedure for strength assembly as suggested in Eurocode 3-(revised) Annex J is aimed at deriving the value of the so-called design resistance of the joint. For the sake of clarity, it is not presented here in a general way but is illustrated in the particular case of beam splices with flush end-plates.

For the connection represented in Figure A.7, the distribution of internal forces is quite easy to obtain : the compressive force is transferred at the centroïd of the beam flange, while the tension force is at the level of the upper bolt-row. The resistance possibly associated with the lower bolt-row is usually neglected as it contributes in a very modest way to the transfer of bending moment in the joint (small level arm).

The design resistance of the joint MRd is associated with the design resistance FRd of the weakest joint component which can be one of the following : the beam and web in compression, the beam web in tension, the plate in bending or the bolts in tension. For the two last components (plate and bolts), reference is made to the concept of "idealized T-stub" introduced in Eurocode 3-(revised) Annex J. The bending resistance becomes :

z.FM RdRd = (A.9)

where z = h is the lever arm.

MFRd h

Figure A.7 Joint with one bolt-row in tension

When more than one bolt-row has to be considered in the tension zone (Figure A.8), the distribution of internal forces is more complex.

M

Figure A.8 Joint with more than one bolt-row in tensionJ.3.6.1


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Assume, initially, that the design of the joint leads to the adoption of a particularly thick end-plate in comparison to the bolt diameter (Figure A.9). In this case, the distribution of internal forces between the different bolt-rows is linear according to the distance from the centre of compression. The compression force Fc which equilibrates the tension forces acts at the level of the centroïd of the lower beam flange. For the sake of clarity, it is shown only in Figure A.9, but not in the subsequent figures.

h1

h2

hi

FRd

Fc

Figure A.9 Joint with a thick end-plate

The design resistance MRd of the joint is reached as soon as the bolt-row subjected to the highest stresses - in reality that which is located the farthest from the centre of compression - reaches its design resistance in tension 2Bt.Rd.

Because of a limited deformation capacity of the bolts in tension no redistribution of forces is allowed to take place between bolt-rows.

It is assumed here that the design resistance of the beam flange and web in compression is sufficient to transfer the compression Fc force. The tensile resistance of the beam web is also assumed not to limit the design resistance of the joint. MRd is so expressed as (Figure A.9) :

∑= 2i

1

RdRd h

hF

M (A.10)

For thinner end-plates, the distribution of internal forces requires much more attention. When initial moment is applied to the joint, the forces distribute between the bolt-rows according to the relative stiffnesses of the latter. This stiffness is namely associated to that of the part of the end-plate adjacent to the considered bolt-row. In the particular case of Figure A.10, the upper bolt-row is characterized by a higher stiffness because of the presence of the beam flange and the web welded to the end-plate.

Because of the higher stiffness, the upper bolt-row is capable of transferring a higher load than the lower bolt-rows (Figure A.10.b)

J.3.2

J.3.6.2


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M

(a) Configuration

h1

h2

hi

(b) Distribution of the internal forces at the beginning of the loading

Figure A.10 Joint with a thin end-plate

In Eurocode 3-(revised) Annex J, it is assumed that the upper bolt-row will reach its design resistance first. This assumption is quite justified in the particular case being considered but is probably less justified for other connection types such as for end-plate connections with an extended part in the tension zone where it is usual that the second bolt-row - that just below the flange in tension - reaches its design resistance first.

The design resistance of the upper bolt-row may be associated with one of the following components : the bolts only, the end-plate only, the bolts-plate assembly or the beam web in tension. If its failure mode is ductile, a redistribution of forces between the bolt-rows can take place : as soon as the upper bolt-row reaches its design resistance, any additional bending moment applied to the joint will be carried by the lower bolt-rows (the second, then the third, etc.), each of which in their turn may reach its own design resistance.

The failure may occur in three different ways :i. The plastic redistribution of the internal forces extends to all bolt-rows

when they have sufficient deformation capacity. The redistribution is said to be "complete" and the resulting distribution of internal forces is called "plastic".The design moment resistance MRd is expressed as (Figure A.11) :

∑=i

ii,RdRd hFM (A.11)

The plastic forces FRd,i vary from one bolt-row to another according to the failure modes (bolts, plate, bolt-plate assembly, beam web, ...).


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the failure modes (bolts, plate, bolt-plate assembly, beam web, ...).

Eurocode 3 considers that a bolt-row possesses a sufficient deformation capacity to allow a plastic redistribution of internal force to take place when :

• Frd,i is associated to the failure of the beam web in tension or;• FRd,i is associated to the failure of the bolt-plate assembly

(including failure of the bolts alone or of the plate alone) and :

Rd.ti,Rd B9,1F ≤h1

hi

FRd

FRd,i

Figure A.11 Plastic distribution of forces

ii. The plastic redistribution of forces is interrupted because of the lack of deformation capacity in the last bolt-row which has reached its design resistance ( Rd.tk,Rd B9,1F > and linked to the failure of the bolts or of the bolt-plate assembly).

In the bolt-rows located lower than bolt-row k, the forces are then linearly distributed according to their distance to the point of compression (Figure A.12).

The design moment resistance equals :

∑ ∑= +=

+=k,1i n,1kj

2j

k

k,Rdii,RdRd h

hF

hFM (A.12)

where : n is the total number of bolt-rows;k is the number of the bolt-row, the deformation

capacity of which is not sufficient.

In this case, the distribution is said "elasto-plastic".h1

hk

hj

FRd,1

FRd,k


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Figure A.12 Elasto-plastic distribution of internal forces

iii. The plastic or elasto-plastic distribution of internal forces is interrupted because the compression force cF attains the design resistance of the beam flange and web in compression. The moment resistance MRd is evaluated with formulae similar to (A.11) and (A.12) in which, obviously, only a limited number of bolt-rows are taken into consideration. These bolts rows are such that :

∑=

=m,1

Rd.cFFl

l

where : m is the number of the last bolt-row transferring a tensile force;

lF is the tensile force in bolt-row number l ;Fc.Rd is the design resistance of the beam flange and web

in compression

The application of the above-described principles to beam-to-column joints is quite similar. The design moment resistance MRd is, as for the beam splices, likely to be limited by the resistance of :• the end-plate in bending,• the bolts in tension,• the beam web in tension,• the beam flange and web in compression,but also by that of :• the column web in tension,• the column flange in bending,• the column web in compression,• the column web panel in shear.

In Eurocode 3-(revised) Annex J, evaluation formulae are provided for each of these components. Annex J also presents a full example showing how to distribute the internal forces in a beam-to-column joint with a "multi-bolt-rows" end-plate connection.

This example also highlights the concept of individual and group yield mechanisms. When adjacent bolt-rows are subjected to tension forces, various yield mechanisms are likely to form in the connected plates (end-plate or column flange).• individual mechanisms (see Figure 3-A.13.a) which develop when the

distance between the bolt-rows are sufficiently large;• group mechanisms (see Figure 3-A.13.b) including more than one

adjacent bolt rows.

To each of these mechanisms are associated specific design resistances.

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When distributing the internal forces, Eurocode 3 recommends that, in a given bolt-row, the model never transfers :• a higher load than that which can be carried when it is assumed that

the considered bolt-row is the only one able to transfer tensile forces (individual resistance);

• a load such that the resistance of the whole group to which the bolt-row belongs is exceeded.

The whole procedure is rather long to apply but in the Lecture 18 on "Practical procedures for the characterisation of the response of moment resisting joints", calculation tools compatible with designer's expectations will be presented.

(a) Individual (b) Group mechanism

Figure A.13 Plastic mechanisms

J.3.5

J.3.5.4 andJ.3.5.5

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