AD Validation Guide Vol2 2016 EN

VALIDATION GUIDE

Advance Design

Validation Guide

Version: 2016

Tests passed on: 17 June 2015

Number of tests: 606

INTRODUCTION Before being officially released, each version of GRAITEC software, including Advance Design, undergoes a series of validation tests. This validation is performed in parallel and in addition to manual testing and beta testing, in order to obtain the "operational version" status. This document contains a description of the automatic tests, highlighting the theoretical background and the results we have obtained using the current software release.

Usually, a test is made of a reference (independent from the specific software version tested), a transformation (a calculation or a data processing scenario), a result (given by the specific software version tested) and a difference usually measured in percentage as a drift from a set of reference values. Depending on the cases, the used reference is either a theoretical calculation done manually, a sample taken from the technical literature, or the result of a previous version considered as good by experience.

Starting with version 2012, Graitec Advance has made significant steps forward in term of quality management by extending the scope and automating the testing process. While in previous versions, the tests were always about the calculation results which were compared to a reference set, starting with version 2012, tests have been extended to user interface behavior, import/export procedures, etc. The next major improvement is the capacity to pass the tests automatically. These current tests have obviously been passed on the “operational version”, but they are actually passed on a daily basis during the development process, which helps improve the daily quality by solving potential issues, immediately after they have been introduced in the code.

In the field of structural analysis and design, software users must keep in mind that the results highly depend on the modeling (especially when dealing with finite elements) and on the settings of the numerous assumptions and options available in the software. A software package cannot replace engineers experience and analysis. Despite all our efforts in term of quality management, we cannot guaranty the correct behavior and the validity of the results issued by Advance Design in any situation. With this validation guide, we are providing a set of concrete test cases showing the behavior of Advance Design in various areas and various conditions. The tests cover a wide field of expertise: modeling, climatic load generation according to Eurocode 1, combinations management, meshing, finite element calculation, reinforced concrete design according to Eurocode 2, steel member design according to Eurocode 3, steel connection design according to Eurocode 3, timber member design according to Eurocode 5, seismic analysis according to Eurocode 8, report generation, import / export procedures and user interface behavior.

We hope that this guide will highly contribute to the knowledge and the confidence you are placing in Advance Design.

Manuel LIEDOT

Chief Product Office

ADVANCE DESIGN VALIDATION GUIDE

6

Contents

13 STEEL DESIGN ....................................................................................................................... 9

13.1 Verifying the calculation results for steel cables (TTAD #11623) ................................................................ 10

13.2 Generating the shape sheet by system (TTAD #11471) ............................................................................. 10

13.3 Verifying shape sheet on S275 beam (TTAD #11731) ................................................................................ 10

13.4 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770) ......................... 10

13.5 Verifying the cross section optimization according to EC3 (TTAD #11516) ................................................ 10

13.6 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522) ................................. 11

13.7 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545) ........................................... 11

13.8 Verifying the shape sheet results for a column (TTAD #11550) .................................................................. 11

13.9 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method .................................. 11

13.10 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method ........................... 11

13.11 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method ............................. 12

13.12 Verifying the buckling coefficient Xy on a class 2 section ......................................................................... 12

13.13 EC3 Test 7: Class section classification and compression resistance for an IPE600 column ................... 13

13.14 Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873) . 19

13.15 EC3 Test 28: Verifying an user defined I section class 1, column fixed on base and without any other restraint ... 20

13.16 EC3 Test 22: Verifying the lateral torsional buckling of a IPE300 beam ................................................... 50

13.17 EC3 Test 23: Verifying a IPE400 column for compression, shear, bending moment, buckling, lateral torsional buckling and bending and axial compression ......................................................................................... 56

13.18 EC3 Test 20: Verifying the buckling resistance of a RC3020100 column ................................................. 92

13.19 EC3 Test 13: Verifying the resistance of a rectangular hollow section column subjected to bending and shear efforts ......................................................................................................................................................... 100

13.20 EC3 test 10: Verifying the classification and the bending resistance of a welded built-up beam ............ 107

13.21 EC3 test 11: Cross section classification and compression resistance verification of a rectangular hollow section column ..................................................................................................................................................... 113

13.22 EC3 Test 25: Verifying an user defined I section class 4 column fixed on the bottom and with a displacement restraint at 2.81m from the bottom ................................................................................................. 118

13.23 EC3 Test 19: Verifying the buckling resistance for a IPE300 column ...................................................... 152

13.24 EC3 Test 12: Verifying the design plastic shear resistance of a rectangular hollow section beam ......... 159

13.25 EC3 Test 27: Verifying an user defined I section class 3 beam simply supported with a displacement restraint .............................................................................................................................................. 162

13.26 EC3 Test 26: Verifying an user defined I section class 3 column fixed on the bottom ............................ 186

13.27 EC3 test 9: Verifying the classification and the compression resistance of a welded built-up column .... 209

13.28 EC3 Test 21: Verifying the buckling resistance of a CHS219.1x6.3H column ......................................... 216

13.29 EC3 Test 24: Verifying an user defined I section class 4 column fixed on the bottom and without any other restraint ............................................................................................................................... 222

13.30 EC3 Test 29: Verifying an user defined I section class 1, column hinged on base and restrained on top for the X, Y translation and Z rotation ........................................................................................................................ 248

13.31 EC3 Test 31: Verifying IPE450 column fixed on base subjected to axial compression and bending moment, both applied on top ................................................................................................................................ 289

13.32 EC3 Test 15: Verifying a rectangular hollow section column subjected to bending and axial efforts ....... 290

13.33 EC3 Test 33: Verifying UPN300 simple supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and punctual vertical force by Z axis ................................................................................. 297

13.34 EC3 Test 35: Verifying C310x30.8 class 4 cantilever, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load applied on the free end ......................................................... 297

13.35 EC3 Test 37: Verifying RHS350x150x8.5H class 3 column, loaded with centric compression, punctual lateral load and bending moment, all applied to the top of the column ................................................................. 297

13.36 EC3 Test 30: Verifying IPE300 beam, simply supported, loaded with centric compression and uniform linear efforts by Y and Z axis ................................................................................................................................ 297

13.37 EC3 Test 34: Verifying C310x30.8 class 3beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle ......................................................................... 298

13.38 EC3 Test 36: Verifying RHS300x150x9H class 1 simply supported beam, loaded with centric compression, uniform linear horizontal efforts and a vertical punctual load in the middle ..................................... 298

13.39 EC3 Test 14: Verifying the bending resistance of a rectangular hollow section column made of S235 steel ........................................................................................................................................................... 299

13.40 EC3 Test 32: Verifying IPE600 simple supported beam, loaded with centric compression and uniform linear efforts by Y and Z axis ................................................................................................................................ 305

13.41 EC3 Test 16: Verifying a simply supported rectangular hollow section beam subjected to biaxial bending ................................................................................................................................................................ 306

13.42 EC3 Test 18: Verifying a simply supported circular hollow section element subjected to torsional efforts 313

13.43 EC3 Test 39: Verifying CHS323.9x6.3H class 2 beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle ......................................................................... 313

13.44 EC3 Test 38: Verifying RHS350x150x5H class 4 column, loaded with centric compression, punctual horizontal force by Y and a bending moment, all applied to the top ..................................................................... 313

13.45 EC3 Test 45: Comparing the shear resistance of a welded built-up beam made from different steel materials ............................................................................................................................................................... 314

13.46 EC3 Test 43: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange .................................................................................................. 318

13.47 EC3 Test 41: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange .................................................................................................. 318

13.48 EC3 test 8: Verifying the classification and the resistance of a column subjected to bending and axial load ............................................................................................................................................................. 319

13.49 EC3 Test 17: Verifying a simply supported rectangular hollow section beam subjected to torsional efforts ........ 325

13.50 EC3 Test 44: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange ................................................................................................. 326

13.51 EC3 Test 40: Verifying CHS508x8H class 3, simply supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle ................................................... 332

13.52 EC3 Test 42: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange ................................................................................................. 332

13.53 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533) .......................... 332

13.54 Changing the steel design template for a linear element (TTAD #12491) ............................................... 332

13.55 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389) ..................................................................................................................................................... 332

13.56 EC3: Verifying the buckling length results (TTAD #11550)...................................................................... 333

13.57 EC3 fire verification: Verifying the work ratios after performing an optimization for steel profiles (TTAD #11975) 333


8

13.58 EC3 test 4: Class section classification and bending moment verification of an IPE300 column ............ 334

13.59 EC3 Test 2: Class section classification and shear verification of an IPE300 beam subjected to linear uniform loading ..................................................................................................................................................... 340

13.60 EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 beam 347

13.61 EC3 Test 1: Class section classification and compression verification of an IPE300 column ................. 354

13.62 EC3 Test 5: Class section classification and combined axial force with bending moment verification of an IPE300 column ..................................................................................................................................................... 360

13.63 EC3 Test 3: Class section classification, shear and bending moment verification of an IPE300 column 366

14 TIMBER DESIGN ................................................................................................................. 375

14.1 EC5: Verifying a timber purlin subjected to oblique bending ..................................................................... 376

14.2 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes ............................. 380

14.3 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending ..................................... 383

14.4 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression ......................................................................................................................................................... 388

14.5 EC5: Shear verification for a simply supported timber beam .................................................................... 393

14.6 Modifying the "Design experts" properties for timber linear elements (TTAD #12259) .............................. 393

14.7 Verifying the timber elements shape sheet (TTAD #12337) ...................................................................... 393

14.8 Verifying the units display in the timber shape sheet (TTAD #12445) ....................................................... 393

14.9 EC5: Verifying a timber beam subjected to simple bending ...................................................................... 394

14.10 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression ................................ 398

14.11 EC5: Verifying a timber column subjected to tensile forces .................................................................... 403

14.12 EC5: Verifying a timber column subjected to compression forces........................................................... 406

14.13 EC5: Verifying a timber beam subjected to combined bending and axial tension ................................... 410

14.14 EC5: Verifying a C24 timber beam subjected to shear force ................................................................... 415

13 Steel design


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13.1 Verifying the calculation results for steel cables (TTAD #11623)

Test ID: 3560

Test status: Passed

13.1.1 Description Performs the finite elements calculation and the steel calculation for a steel model with cables (D4) and a static nonlinear case. Generates the steel analysis report: data and results.

13.2 Generating the shape sheet by system (TTAD #11471)

Test ID: 3575

Test status: Passed

13.2.1 Description Generates shape sheets by system, on a model with 2 systems.

13.3 Verifying shape sheet on S275 beam (TTAD #11731)

Test ID: 3434

Test status: Passed

13.3.1 Description Performs the steel calculation for two horizontal bars and generates the shape sheets report.

The bars have cross sections from different catalogues (1016x305x487 UKB and UKB1016x305x487). They are made of the same material (S275); each is subjected to a -500.00 kN linear horizontal dead load and has two rigid supports at both ends.

13.4 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770)

Test ID: 3406

Test status: Passed

13.4.1 Description Performs the steel calculation for two vertical bars with different thicknesses and generates the shape sheets report.

The bars are made of the same material (S275 H - EN 10210-1), have rectangular hollow cross sections, but with different thicknesses (R80*40/4.1 and R80*40/3.9). Each is subjected to a -150.00 kN vertical live load and has a rigid support.

13.5 Verifying the cross section optimization according to EC3 (TTAD #11516)

Test ID: 3620

Test status: Passed

13.5.1 Description Verifies the cross section optimization of a steel element, according to Eurocodes 3.

Performs the finite elements calculation and the steel calculation. Generates the "Envelopes and shapes optimization" report.

The steel bar has a IPE360 cross section, a rigid hinge support at one end and a rigid support with translation restraints on X, Y and Z and rotation restraint on X. Two loads are applied: a punctual dead load of -1.00 kN on FZ and a punctual live load of -40.00 kN on FZ.


11

13.6 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522)

Test ID: 3612

Test status: Passed

13.6.1 Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet results report.

The structure is a simple vault consisting of three steel elements (S235 material, IPEA240 cross section) with two rigid fixed supports. The loads applied on the structure: self weight and a linear live load of -10.00kN on FZ.

13.7 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545)

Test ID: 3641

Test status: Passed

13.7.1 Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a horizontal steel element, verifies the cross section class.

The steel element (S235 material, IPE300 cross section) has a rigid fixed support at one end and a rigid support with translation restraints on Y and Z and rotation restraint on X at the other end. A linear live load of -10.00 kN on FX and a punctual live load of 1000 kN on FX are applied.

13.8 Verifying the shape sheet results for a column (TTAD #11550)

Test ID: 3640

Test status: Passed

13.8.1 Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a vertical steel element.

The steel element (S235 material, IPE300 cross section) has a rigid fixed support. A vertical live load of -200.00 kN is applied.

13.9 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method

Test ID: 3819

Test status: Passed

13.9.1 Description Verifies the buckling length for a steel portal frame, using the kA kB method, according to Eurocodes 3.

Generates the "Buckling and lateral-torsional buckling lengths" report.

13.10 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method

Test ID: 3814

Test status: Passed

13.10.1Description Verifies the buckling length for a steel portal frame with one level, using the roA roB method, according to CM66.



12

13.11 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method

Test ID: 3813

Test status: Passed

13.11.1Description Verifies the buckling length for a steel portal frame with one level, using the kA kB method, according to CM66.


13.12 Verifying the buckling coefficient Xy on a class 2 section

Test ID: 4443

Test status: Passed

13.12.1Description Performs the finite elements calculation and the steel elements calculation. Verifies the buckling coefficient Xy on a class 2 section and generates the shape sheets report.

The model consists of a vertical linear element (I26*0.71+15*1.07 cross section and S275 material) with a rigid hinge support at the base and a rigid support with translation restraints on X and Y and rotation restraint on Z, at the top. A punctual live load is applied.


13

13.13 EC3 Test 7: Class section classification and compression resistance for an IPE600 column

Test ID: 5620

Test status: Passed

13.13.1Description Verifies the classification and the compression resistance for an IPE 600 column made of S235 steel. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

13.13.2Background Classification and verification under compression efforts for an IPE 600 column made of S235 steel. The column is fixed at its base and free on the top. The column is subjected to a compression force (100 000 N) applied at its top. The dead load will be neglected.

This test was evaluated by the French control office SOCOTEC.

13.13.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:

■ Exploitation loadings (category A), Q: Fz = -100 000 N,

■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■ Cross section dimensions are in milimeters (mm).

Units

Metric System

Materials properties

S235 steel material is used. The following characteristics are used:

■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.

Boundary conditions

The boundary conditions are described below:

■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (x = 5.00).


14

■ Inner: None.

Loading

The column is subjected to the following loadings:

■ External: Point load at Z = 5.0: N = Fz = -100 000 N, ■ Internal: None.

13.13.2.2Reference results for calculating the cross section class The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.

In this case, the column is subjected to a punctual compression load, therefore the stresses distribution is like in the picture below:

Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the web class.


15

The section geometrical properties are described in the picture below:

The web class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1:

83.4212

224219600=

×−×−=

mmmmmmmm

tc

0.1235==

yfε


16

Therefore:

424283.42 =>= εtc

This means that the column web is Class 4.

Table 5.2 - sheet 2, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the flanges class.

The top flange class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 2:

21.419

2/)22412220(=

×−−=

mmmmmmmm

tc

0.1235==

yfε

Therefore:

9921.4 =≤= εtc

This means that the top column flange is Class 1. Having the same dimensions, the bottom column flange is also Class 1.

A cross-section is classified according to the least favorable classification of its compression elements (chapter 5.5.2(6) from EN 1993-1-1: 2001).

According to the calculation above, the column section have Class 4 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 4.


17

13.13.2.3Reference results for calculating the compression resistance of the cross-section The compression resistance for Class 4 cross-section is determined with the formula (6.11) from EN 1993-1-1:2001.

In order to verify the compression resistance for Class 4 cross-section, it is necessary to determine the effective area of the cross-section.

The effective area of the cross section takes into account the reduction factor, ρ, which is applying in this case only for the web of the IPE 600 cross-section.

The following parameters have to be determined in order to calculate the reduction factor: the buckling factor and the stress ratio, and the plate modified slenderness. They will be calculated considering only the cross-section web.

The buckling factor (kσ) and the stress ratio(Ψ)

Taking into account that the stress distribution on web is linear, the stress ratio becomes:

0.11

2 ==σσψ

0.4=σk

The plate modified slenderness (λp)

The formula used to determine the plate modified slenderness is:

( ) 754.00.40.14.28

12/2421926004.28

/=

×××−×−

==mmmmmmmm

ktb

pσε

λ

The reduction factor (ρ)

Because λp > 0.673, the reduction factor has the following formula:

( )0.1

3055.02 ≤

+×−=

p

p

λψλ

ρ

Effective area

The effective area is determined considering the following:

( ) ( ) ( ) 275.1522312242192600939.01156001 mmtbAA weff =××−×−×−−=××−−= ρ

Compression resistance of the cross section

For Class 4 cross-section, EN 1993-1-1: 2011 provides the following formula in order to calculate the compression resistance of the cross-section:

NMPammfAN

M

yeffRdc 3577581

0.123575.15223 2

0, =

×=

×=

γ

Work ratio

Work ratio = %79518.21003577581100000100

,

=×=×NN

NN

Rdc


18

Finite elements modeling

■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.

Finite elements results

Work ratio of the design resistance for uniform compression

Column subjected to bending and axial force Work ratio - Fx

13.13.2.4Reference results

Result name Result description Reference value Work ratio - Fx Compression resistance work ratio [%] 2.79518 %

13.13.3Calculated results Result name Result description Value Error Work ratio - Fx Compression resistance work ratio 2.79495 % -0.0082 %


19

13.14 Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873)

Test ID: 4289

Test status: Passed

13.14.1Description Verifies the steel shape optimization when using sections from Advance Steel Profiles database. Performs the finite elements calculation and the steel elements calculation and generates the steel shapes report.

The structure consists of columns with UKC152x152x23 cross section, beams with UKB254x102x22 cross section and roof beams with UKB127x76x13 cross section. Dead loads and live loads are applied on the structure.


20

13.15 EC3 Test 28: Verifying an user defined I section class 1, column fixed on base and without any other restraint

Test ID: 5720

Test status: Passed

13.15.1Description The test verifies a user defined cross section column.

The cross section has an “I symmetric” shape with: 260mm height; 150mm width; 7.1mm center thickness; 10.7mm flange thickness; 0mm fillet radius and 0mm rounding radius. The column is made of S275 steel.

The column is subjected to 328 kN compression axial force, 10 kNm bending moment over the X axis and 50 kNm bending moment over the Y axis. All the efforts are applied on the top of the column.

The calculations are made according to Eurocode 3 French Annex.

13.15.2Background An I260*7.1+150*10.7 shaped column subjected to compression and bending, made from S275 steel. The column has a 260x7.1mm web and 150x10.7mm flanges. The column is fixed at its base. The column is subjected to an axial compression load -328000 N, a 10000Nm bending moment after the X axis and a 50000Nm bending moment after the Y axis.



■ Exploitation loadings (category A): Fz=--328000N N; My=50000Nm; Mx=10000Nm; ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q ■ Cross section dimensions are in millimeters (mm).


21

Units

Metric System

Geometrical properties

■ Column length: L=5620mm

■ Cross section area: 206.4904 mmA =

■ Overall breadth: mmb 150=

■ Flange thickness: mmt f 70.10=

■ Root radius: mmr 0=

■ Web thickness: mmtw 10.7=

■ Depth of the web: mmhw 260=

■ Elastic modulus after the Y axis, 3, 63.445717 mmW yel =

■ Plastic modulus after the Y axis, 318.501177 mmWy =

■ Elastic modulus after the Z axis, 3, 89.80344 mmW zel =

■ Plastic modulus after the Z axis, 3, 96.123381 mmW zpl =

■ Flexion inertia moment around the Y axis: 464.57943291 mmI y =

■ Flexion inertia moment around the Z axis: 446.6025866 mmIz =

■ Torsional moment of inertia: 497.149294 mmIt =

■ Working inertial moment: 688.19351706542 mmIw =



■ Yield strength fy = 275 MPa, ■ Longitudinal elastic modulus: E = 210000 MPa. ■ Shear modulus of rigidity: G=80800MPa


22

Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at start point (x = 0) restrained in translation along X and Y axis, and restrained inrotation along Z

axis,

Loading

The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =-328000N; Mx=10000Nm and My=50000Nm


23

13.15.2.2Cross section Class According to Advance Design calculations:

Cross-class classification is made according to Table 5.2

-for beam web:

The web dimensions are 850x5mm.

1253.006.179

30.45

sup

inf −>−=−

==Mpa

Mpaσσψ


24

=×

=

=×

=⇒=

++

==73.190

36.22406.1796.238

175.4836.224

3.456.238

36.2246.238

06.17930.4506.17930.45 y

xyxyx

5.080.06.238

73.1906.238

>===xα


25

924.0275235235

===yf

ε

93.3818.013

924.0396113

3966.33924.0

61.331.7

7.102260=

−××

=−×

×≤=⇒

=

=×−

=α

ε

ε tc

mmmmmm

tc

therefore the beam

web is considered to be Class 1

-for beam flange:

316.8924.0968.6

924.0

68.67.10

21.7150

=×≤=⇒

=

=

−

=tc

tc

ε

therefore the haunch is considered to be Class1

In conclusion, the section is considered to be Class 1.

13.15.2.3Buckling verification a) over the strong axis of the section, y-y:

-the imperfection factor α will be selected according to the Table 6.1 and 6.2:


26

34.0=α

Coefficient corresponding to non-dimensional slenderness after Y-Y axis:

yχ coefficient corresponding to non-dimensional slenderness yλ will be determined from the relevant buckling

curve according to:

1122

≤−Φ+Φ

=yyy

yλ

χ (6.49)

yλ the non-dimensional slenderness corresponding to Class 1 cross-sections:

ycr

yy N

fA

,

*=λ

Cross section area: 206.4904 mmA =

Flexion inertia moment around the Y axis: 464.57943291 mmI y =

( ) kNNmm

mmmmNl

IEN

fy

yycr 33.380295.3802327

²562064.57943291/210000²

²² 42

, ==××

=××

=ππ

5956.095.3802327

/27506.4904 22

,

=×

=×

=N

mmNmmN

fA

ycr

yyλ

[ ] ( )[ ] 7446.05956.02.05956.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ

839.0

1

839.05956.07446.07446.0

112222 =⇒

≤

=−+

=−Φ+Φ

=y

y

yyy

yχ

χ

λχ

b) over the weak axis of the section, z-z:



27

49.0=α

Coefficient corresponding to non-dimensional slenderness after Z-Z axis:

zχ coefficient corresponding to non-dimensional slenderness zλ will be determined from the relevant buckling curve according to:

1122

≤−Φ+Φ

=zzz

zλ

χ (6.49)

zλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:

zcr

yz

NfA

,

*=λ

Flexion inertia moment around the Z axis: 446.6025866 mmI z =


( ) kNNmm

mmmmNl

IENfz

zzcr 43.39563.395426

²562046.6025866/210000²

²² 42

, ==××

=××

=ππ


28

847.163.395426

/27506.4904 22

,

=×

=×

=N

mmNmmN

fA

zcr

yzλ

[ ] ( )[ ] 609.2847.12.0847.149.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ

225.01

225.0847.1609.2609.2

112222 =⇒

≤

=−+

=−Φ+Φ

=z

z

zzz

zχ

χλ

χ

13.15.2.4Lateral torsional buckling verification The elastic moment for lateral-torsional buckling calculation, Mcr:

-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:

z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ

According to EN 1993-1-1-AN France; Chapter 2; …(3) -where:

C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure

²252.0423.0325.01

1 ψψ ++=C

According to EN 1993-1-1-AN France; Chapter 3; …(6)

77.10500

1,

, =⇒=== CM

M

topy

botomyψ


29



Torsional moment of inertia: 497.149294 mmIt =

Working inertial moment: 688.19351706542 mmI w =

Yield strength fy = 275 MPa,

Longitudinal elastic modulus: E = 210000 MPa.

Shear modulus of rigidity: G=80800MPa

Warping inertial moment (recalculated):

IW is the warping inertia (deformation inertia moment):

( )4

2fz

w

thII

−×=

h cross section height; mmh 260=

ft flange thickness; mmt f 7.10=

( ) 624

093627638294

7.1026046.6025866 mmmmmmmmI w =−×

=

-according to EN1993-1-1-AN France; Chapter 2 (…4)

Length of the column: L=5620mm

( )( )

kNmNmm

mmNmmmmN

mmmmNmmmmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

150181.150184702

58.21463.39542677.146.6025866/210000

97.149294/80800562046.6025866

09362763829

562046.6025866/21000077.1

²²

²²

422

422

4

6

2

422

1

==

=××=××

××+×

×××

×=××××

+×××

×=

π

ππ

π

958.01.150184702

/27518.501177 23, =

×==

NmmmmNmm

MfW

cr

yyplLTλ

Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:

1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=

The cross section buckling curve will be chose according to Table 6.4:

2733.1150260

≤==mmmm

bh


30

The imperfection factor α will be chose according to Table 6.3:

49.0=α

( )[ ] ( )[ ] 145.1²958.02.0958.049.015.0²2.015.0 =+−×+×=+−×+×= LTLTLTLT λλαφ

1564.0²958.0²145.1145.1

1²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ

13.15.2.5Internal factor, yyk , calculation

The internal factor yyk corresponding to a Class 4 section will be calculated according to Annex A, Table a.1, and will be

calculated separately for the two column parts separate by the middle torsional lateral restraint:

yy

ycr

Ed

ymLTmyyy C

NNCCk 1̀

1,

×−

××=µ


31

ycr

Edy

ycr

Ed

y

NN

NN

,

,

1

1

×−

−

=χ

µ

839.0=yχ (previously calculated)

kNNEd 328=

kNNl

IEN

fy

yycr 33.380295.3802327

²²

, ==××

=π

(previously calculated)

985.0

95.3802327328000839.01

95.38023273280001

1

1

,

, =×−

−=

×−

−

=

NN

NN

NN

NN

ycr

Edy

ycr

Ed

y

χµ

The myC will be calculated according to Table A.1:


32

Calculation of the 0λ term:

0

,0

cr

yypl

MfW ×

=λ

-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2 318.501177 mmWy =

The calculation the 0crM will be calculated using 11 =C and 02 =C , therefore:

( )( )

kNmNmm

mmNmmmmN

mmmmNmmmm

mm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

85.8427.84850646

58.21463.395426146.6025866/210000

97.149294/80800562046.6025866

88.19351706542

562046.6025866/2100001

²²

²²

422

422

4

6

2

422

1

==

=××=××

××+×

×××

×=××××

+×××

×=

π

ππ

π

274.127.84850646

/27518.501177 23,

0 =×

==Nmm

mmNmmM

fW

cr

yyplλ

Calculation of the 4

,,1 1120.0

−×

−××

TFcr

Ed

zcr

Ed

NN

NNC term:

Where:

-for a symmetrical section for the both axis, TcrTFcr NN ,, =

××+××=

²1

,

2

0,

Tcr

wtTcr L

IEIGI

N π

The mass moment of inertia 0I

44420 1.6396915846.602586664.57943291 mmmmmmIIzAIII zygzy =+=+=×++=


Working inertial moment: 688.19351706542 mmIw =

- the buckling length, TcrL , ,

mL Tcr 62.5, =

( )N

mmmmmmNmmmmN

mmmmN Tcr

607.1395246²5620

88.19351706542/21000097.149294/808001.63969158

06.4904 62242

4

2

,

=

=

××+××=

π

NNEd 328000=

NNN TcrTFcr 607.1395246,, ==


33

( ) kNNmm

mmmmNl

IENfz

zzcr 43.39563.395426

²562046.6025866/210000²

²² 42

, ==××

=××

=ππ


C1=1 for the top part of the column

For the top part of the column:

120.0

607.13952463280001

63.3954263280001120.01120.0 44

,,1

=

=

−×

−××=

−×

−××

NN

NN

NN

NNC

TFcr

Ed

zcr

Ed

Therefore:


( )

≥

−×

−

×=

=

×+

××−+=

⇒

=

−⋅

−⋅⋅>=

=

−×

−××

=

1

11

11

120.01120.0274.1

120.01120.0

274.1

,,

2

0,

0,0,

4

,,10

4

,,1

0

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

NN

NNC

ε

ε

λ

λ

The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.

The coefficient must be calculated considering the column over the entire height.

( )LTy

LTymymymy a

aCCC

×+

××−+=

ε

ε

11 0,0,

yeff

eff

Ed

Edyy W

AN

M

,

, ×=ξ

Elastic modulus after the Y axis, 3,, 63.445717 mmWW yeffyel ==

677.163.445717

06.4904328000

10503

26

,

, =××

=×=mm

mmN

NmmWA

NM

yeff

eff

Ed

Edyyξ

1997.064.57943291

97.14929411 4

4

≈=−=−=mm

mmIIa

y

tLT


34

The 0mC coefficient is defined according to the Table A.2:

The bending moment in null at one end of the column, therefore: 0=ψ

( )ycr

Ed

ycr

Edmy N

NNNC

,,0, 33.036.079.033.036.021.079.0 ××−=×−×+×+= ψψ

Where:

kNNl

IEN

fy

yycr 33.380295.3802327

²²

, ==××

=π


NNEd 328000=

780.095.3802327

32800033.036.079.00, =××−=N

NCmy

( ) ( ) 904.01677.11

1677.1780.01780.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ

Equivalent uniform moment factor, mLTC , calculation

256.2

1

256.2

607.13952463280001

63.3954263280001

997.0904.0

11

2

,,

2

=⇒

≥

=

−×

−

×=

=

−×

−

×=

mLT

mLT

Tcr

Ed

zcr

Ed

LTmymLT

C

CN

NN

N

NN

NN

aCC


35

The yyC coefficient is defined according to the Table A.1, Auxiliary terms:

ypl

yelLTplmy

ymy

yyyy W

WbnC

wC

wwC

,

,2

max2

max2 6.16.12)1(1 ≥

−×

××−××−×−+=

−−

λλ

1534.0/27596.123381

10000000/27518.501177564.0

50000000274.1997.05.0

5.05.0

23232

,

,

,

,2

0

,,

,

,,

,2

0

=

=×

×××

×××=

××

×××××=×

××××=

−−

mmNmmNmm

mmNmmNmm

fWM

fWM

aMM

MM

abyzpl

Edz

yyplLT

EdyLT

Rdzpl

Edz

RdyplLT

EdyLTLT χ

λχ

λ

5.1124.163.44571718.501177

3

3

,

, ≤===mmmm

WW

wyel

yply

243.0

1/27506.4904

32800022

1

=×

==mmNmm

NNNn

M

Rk

Edpl

γ

( ) 847.1847.1;5956.0max;maxmax ==

=

−−

zy λλλ

857.01534.0243.0²847.1²904.0124.16.1847.1²904.0

124.16.12)1124.1(1 =

−×

××−××−×−+=yyC


36

889.0889.018.50117763.445717

857.0

,

,

3

3

,

, =⇒

≥

==

=

yy

ypl

yelyy

ypl

yel

yy

C

WW

C

mmmm

WWC

kNNl

IEN

fy

yycr 33.380295.3802327

²²

, ==××

=π


Therefore the yyk term corresponding to the top part of the column will be:

47.2889.01

95.38023273280001

985.0256.2904.01̀

1,

=×−

××=×−

××=

NNC

NNCCk

yy

ycr

Ed

ymLTmyyy

µ

13.15.2.6Internal factor, yzk , calculation

y

z

yz

zcr

Ed

ymzyz w

wC

NNCk ×××

−×= 6.01

1,

µ

( ) ( ) 691.063.395426

32800033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz

985.0

95.3802327328000839.01

95.38023273280001

1

1

,

, =×−

−=

×−

−

=

NN

NN

NN

NN

ycr

Edy

ycr

Ed

y

χµ


kNNl

IENfz

zzcr 43.39563.395426

²²

, ==××

=π


−×

×

×−×−+=

−

LTplz

mzzyz cn

wCwC 5

2

max2

142)1(1 λ

Rdyplltmy

Edy

z

LTLT MCM

ac,,

,4

2

0

510

×××

+××=

−

−

χλ

λ

997.064.57943291

97.14929411 4

4

=−=−=mm

mmIIa

y

tLT (previously calculated)

274.127.84850646

/27518.501177 23,

0 =×

==Nmm

mmNmmM

fW

cr

yyplλ (previously calculated)

847.163.395426

/27506.4904 22

,

=×

=×

=N

mmNmmN

fA

zcr

yzλ (previously calculated)


37

NmM Edy 50000, =

( ) 904.01

1 0,0, =×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ (previously calculated)

1564.0²958.0²145.1145.1

1²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ (previously calculated)

NmmmmNmmfWM yyRdypl 5.137823724/27518.501177 23,, =×=×=

664.05.137823724564.0904.0

50000000847.15

²274.1997.010

510

4

,,

,4

2

0

=××

×+

××=

=××

×+

××=−

−

NmmNmm

MCM

acRdyplltmy

Edy

z

LTLT χλ

λ

−×

×

×−×−+=

−

LTplz

mzzyz cn

wCwC 5

2

max2

142)1(1 λ

5.15.1

5.1536.189.8034496.123381

3

3

,

,

=⇒

≤

≤===z

z

zel

zplz w

wmmmm

WW

w

kNNl

IENfz

zzcr 43.39563.395426

²²

, ==××

=π


( ) ( ) 691.063.395426

32800033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz

( ) 847.1847.1;5956.0max;maxmax ==

=

−−

zy λλλ

243.0

1

==

M

Rk

Edpl N

Nn

γ


532.0691.0243.05.1

847.1691.0142)15.1(1

142)1(1

5

22

5

2

max2

=

−×

××−×−+=

=

−×

×

×−×−+=−

LTplz

mzzyz cn

wCwC λ

20.5124.1

5.16.0532.01

63.3954263280001

985.0691.06.01

1,

=×××−

×=×××−

×=

NNw

wC

NNCk

y

z

yz

zcr

Ed

ymzyz

µ


38

13.15.2.7Internal factor, zyk , calculation

y

z

zy

ycr

Ed

zmLTmyzy w

wC

NNCCk ×××

−××= 6.01

1,

µ

210.0

63.395426328000225.01

63.3954263280001

1

1

,

, =×−

−=

×−

−=

NN

NN

NN

NN

zcr

Edz

zcr

Ed

z

χµ

225.0=zχ (previously calculated)

ypl

yel

z

yLTpl

y

myyzy W

Www

dnw

CwC

,

,5

2

max2

6.0142)1(1 ××≥

−×

×

×−×−+=

−

λ

NmmmmNmmfWM zzRdzpl 33930039/27596.123381 23,, =×=×=

771.033930039691.0

100000005.137823724564.0904.0

50000000847.11.0

274.1997.02

1.02

4

,,

,

,,

,4

0

=

=×

×××

×+

××=

=×

×××

×+

××=−

−

NmmNmm

NmmNmm

MCM

MCM

adRdzplmz

Edz

RdyplLTmy

Edy

z

LTLT χλ

λ

310.0771.0243.0124.1

847.1904.0142)1124.1(1

142)1(1

5

22

5

2

max2

=−×

××−×−+=

=

−×

×

×−×−+=

−

LTply

myyzy dn

wC

wCλ

462.018.50117763.445717

5.1124.16.06.0 3

3

,

, =××=××mmmm

WW

ww

ypl

yel

z

y

462.0

462.06.0

310.0

6.0

,

,

,

,

=⇒

=××

=

××≥

zy

ypl

yel

z

y

zy

ypl

yel

z

yzy

C

WW

ww

CWW

ww

C

256.2=mLTC (previously calculated)

529.05.1

124.16.0462.01

95.38023273280001

210.0256.2904.06.01

1,

=×××−

××=×××−

××=

NNw

wC

NNCCk

z

y

zy

ycr

Ed

zmLTmyzy

µ


39

13.15.2.8Internal factor, zzk , calculation

zz

zcr

Ed

zmzzz C

NNCk 1

1,

×−

×=µ

zpl

zelLTplmz

zmz

zzzz W

WenC

wC

wwC

,

,max

2max

2 6.16.12)1(1 ≥

−×

××−××−×−+= λλ

131.0/27518.501177564.0904.0

50000000847.11.0

274.1997.07.1

1.07.1

234

,,

,4

0

=×××

×+

××=

=××

×+

××=−

−

mmNmmNmm

MCM

aeRdyplltmy

Edy

z

LTLT χλ

λ

926.0131.0243.0847.1691.05.16.1847.1691.0

5.16.12)15.1(1

6.16.12)1(1

222

2max

2max

2

=

−×

××−××−×−+=

=

−×

××−××−×−+= LTplmz

zmz

zzzz enC

wC

wwC λλ

919.0926.01

63.3954263280001

210.0691.01

1,

=×−

×=×−

×=

NNC

NNCk

zz

zcr

Ed

zmzzz

µ

13.15.2.9Bending and axial compression verification

∆+×+

×

∆+×+

×

∆+×+

×

∆+×+

×

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rkz

RdzEdzzz

M

RkyLT

RdyEdyzy

M

Rkz

Ed

M

Rkz

RdzEdzyz

M

RkyLT

RdyEdyyy

M

Rky

Ed

MMM

kMMM

kNN

MMM

kMMM

kNN

γγχ

γχ

γγχ

γχ

iyRk AfN ×=


40

69.127.034.008.1

1/27596.123381

1010919.0

1/27518.501177564.0

1050529.0

1/27506.4904225.0

328000

41.353.159.129.0

1/27596.123381

101020.5

1/27518.501177564.0

1050479.2

1/27506.4904839.0

328000

23

6

23

6

22

23

6

23

6

22

=++=×

××+

+×

×

××+

××

=++=×

××+

+×

×

××+

××

mmNmmNmm

mmNmmNmm

mmNmmN

mmNmmNmm

mmNmmNmm

mmNmmN




41

yχ coefficient corresponding to non-dimensional slenderness yλ

Column subjected to axial and shear force to the top

yχ

zχ coefficient corresponding to non-dimensional slenderness zλ


zχ


42

Internal factor, yyk


yyk

Internal factor, yzk


yzk


43

Internal factor, zyk


zyk

Internal factor, zzk


zzk


44

Bending and axial compression verification term depending of the compression effort over the Y axis: SNy

Bending and axial compression verification term depending of the compression effort over the Y axis

SNy

Bending and axial compression verification term depending of the Y bending moment over the Y axis: SMyy

Bending and axial compression verification term depending of the Y bending moment over the Y axis

SMyy


45

Bending and axial compression verification term depending of the Z bending moment over the Y axis: SMyz

Bending and axial compression verification term depending of the Z bending moment over the Y axis

SMyz

Bending and axial compression verification term depending of the compression effort over the Z axis: SNz

Bending and axial compression verification term depending of the compression effort over the Z axis

SNz


46

Bending and axial compression verification term depending of the Y bending moment over the Z axis: SMzy

Bending and axial compression verification term depending of the Y bending moment over the Z axis

SMzy

Bending and axial compression verification term depending of the Z bending moment over the Z axis: SMzz

Bending and axial compression verification term depending of the Z bending moment over the Z axis

SMzz


47

Coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure


C1

The elastic moment for lateral-torsional buckling calculation

The elastic moment for lateral-torsional buckling calculation Mcr


48

The appropriate non-dimensional slenderness


LTχ


49


Result name Result description Reference value

yχ yχ coefficient corresponding to non-dimensional slenderness yλ

0.839

zχ zχ coefficient corresponding to non-dimensional slenderness zλ

0.225

yyk Internal factor, yyk

2.47

yzk Internal factor, yzk

5.20

zyk Internal factor, zyk

0.529

zzk Internal factor, zyk

0.919

SNy


0.29

SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis

1.59

SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis

1.53

SNz

Bending and axial compression verification term depending of the compression effort over the z axis

1.08

SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis

0.34

SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis

0.27 C1 Coefficient that depends of several parameters as: section properties;

support conditions; moment diagram allure 1.77

Mcr The elastic moment for lateral-torsional buckling calculation 150.18

LTχ The appropriate non-dimensional slenderness 0.564

Work ratio Stability work ratio (bending and axial compression verification) [%] 341 %

13.15.3Calculated results Result name Result description Value Error Xy Coefficient corresponding to non-dimensional slenderness

after Y-Y axis 0.839285 adim

0.0340 %

Xz Coefficient corresponding to non-dimensional slenderness after Z-Z axis

0.224656 adim

-0.1529 %

Kyy Internal factor kyy 2.47232 adim

0.0939 %

Kyz Internal factor kyz 5.20929 adim

0.1787 %

Kzy Internal factor kzy 0.525982 adim

-0.5705 %

Kzz Internal factor kzz 0.942941 adim

2.6051 %

Work ratio Stability work ratio 341.352 % 0.0000 %


50

13.16 EC3 Test 22: Verifying the lateral torsional buckling of a IPE300 beam

Test ID: 5702

Test status: Passed

13.16.1Description The test verifies the lateral torsional buckling of a IPE300 beam made of S235 steel.

The calculations are made according to Eurocode 3, French Annex.

13.16.2Background Lateral torsional buckling verification for an unrestrained IPE300 beam subjected to axis bending efforts, made of S235 steel. The beam is simply supported. The beam is subjected to a uniform vertical load (10 000 N) applied constantly on the entire length. The dead load will be neglected.



■ Exploitation loadings (category A): Q1 = -10 000 N, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■ Cross section dimensions are in millimeters (mm).

Units

Metric System


■ Beam length: 5m ■ Cross section area: A=5310mm2 ■ Flexion inertia moment around the Y axis: Iy=8356.00x104mm4 ■ Flexion inertia moment around the Z axis: Iz=603.80x104 mm4


51




Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at the end point (z = 3.00) restrained in translation along Y and Z axis and restrained rotation

along X axis. ■ Inner: None.

Loading


■ External: Linear load From X=0.00m to X=5.00m: FZ = N = -10 000 N, ■ Internal: None.

13.16.2.2Buckling in the strong inertia of the profile (along Y-Y) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling moment resistance of the bended element (Mb,Rd) from the designed value moment (MEd) produced by the linear force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.

%100100,

≤×Rdb

Ed

MM

(6.46)


■ for beam web:

7272014.351

014.351.76.248

=×≤=⇒

=

==ε

ε tc

mmmm

tc

therefore the beam web is considered to be

Class 1

■ for beam flange:

99276.51

276.57.1045.56

=×≤=⇒

=

==ε

ε tc

mmmm

tc



52

In conclusion, the section is considered to be Class 1

The buckling curve will be determined corresponding to Table 6.2:

22150300

≤==mmmm

bh

the buckling curve about Y-Y will be considered “a”

The design buckling resistance moment against lateral-torsional buckling is calculated according the next formula:

1,

M

yyLTRdb

fWM

γχ ××

= (6.55)

Where:

LTχ reduction factor for lateral-torsional buckling:

1122

≤−Φ+Φ

=LTLTLT

LTλ

χ (6.56)

Where:

[ ]2)2.0(15.0 LTLTLTLT λλαφ +−×+×=

LTα represents the imperfection factor; 21.0=LTα


53

LTλ the non-dimensional slenderness corresponding:

cr

yyLT

MfW ×

=λ

Mcr is the elastic critical moment for lateral-torsional buckling, is based on gross cross sectional properties and takes into account the loading conditions, the real moment distribution and the lateral restraints.

( )( ) ( )

×−×+

×××××

+×

×

×××

×= ggZ

tz

z

w

w

z

z

zcr zCzC

IEIGLk

II

kk

LkIECM 2

222

2

2

2

1 ππ

(1)

according to EN 1993-1-1-AN France; AN.3 Chapter 2

Where:

E is the Young’s module: E=210000N/mm2

G is the share modulus: G=80770N/mm2

Iz is the inertia of bending about the minor axis Z: Iz=603.8 x104mm4

It is the torsional inertia: It=20.12x104mm4

IW is the warping inertia (deformation inertia moment): Iw=12.59x1010mm6

L is the beam length: L=5000mm

kz and kw are buckling coefficients

zg is the distance between the point of load application and the share center (which coincide with the center of gravity)

C1 and C2 are coefficients depending on the load variation over the beam length

If the bending moment is linear along the bar, if there are no transversal loads or if the transverse load is applied to the center, then C2xxg=0 and the Mcr formula become:

z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ

The C1 coefficient is chosen from the Table2 of the EN 1993-1-1-AN France; AN.3 Chapter 3.3:


54

kNmNmmmmNmmmmN

mmmmNmmmmmm

mmmmmmNM cr

61.1308.13060942480.23044.50057813.11080.603/210000²

1012.20/80770)²5000(1080.603

1059.12

)²5000(1080.603/210000²13.1

442

442

44

610

442

==××=

=×××

×××+

××

×

××××

×=

π

π

therefore:

063.18.130609424

/235104.628 233

=××

=×

=Nmm

mmNmmM

fW

cr

yyLTλ

[ ] [ ] 156.1063.1)2.0063.1(21.015.0)2.0(15.0 22=+−×+×=+−×+×= LTLTLTLT λλαφ

1621.0²063.1²156.1156.1

1

²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ




The steel calculation results can be found in the Shape Sheet window. The “Class” tab shows the classification of the cross section and the effective characteristics (not applicable in this case, as the cross section is class 1).


55

Lateral torsional buckling coefficient

Simply supported beam subjected to bending efforts Lateral torsional buckling coefficient

Elastic critical moment for lateral-torsional buckling

Simply supported beam subjected to bending efforts Mcr



LTχ Lateral-torsional buckling coefficient [adim.] 0.621

Mcr Elastic critical moment for lateral-torsional buckling [kNm] 130.61

13.16.3Calculated results Result name Result description Value Error XLT Lateral-torsional buckling coefficient 0.621588

adim 0.0947 %

Mcr Elastic critical moment for lateral-torsional buckling 130.699 kN*m

0.0681 %


56

13.17 EC3 Test 23: Verifying a IPE400 column for compression, shear, bending moment, buckling, lateral torsional buckling and bending and axial compression

Test ID: 5703

Test status: Passed

13.17.1Description The test verfies a IPE400 column, made of S275 steel, subjected to compression, shear, bending moment, buckling, lateral torsional buckling and bending and axial compression

A lateral restraint is placed at 3m from the base.

The verifications are made according to Eurocode 3 French Annex.

13.17.2Background Unrestrained IPE400 column subjected to compression and bending, made from S275 steel. The column is fixed at its base and free on the top end. A lateral restraint is placed at 3m from the base. The column is subjected to an axial compression load (-125000 N) applied and to a lateral load after the X global axis (28330N). Both loads are applied on the top end of the column. The dead load will be neglected. The results will be compared with the ones obtained by the CTIM n4-2006.



■ Exploitation loadings (category A): Q1x= 28330 N, Q1z= -125000 N ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q


57


Units

Metric System



■ Cross section area: 28446mmA = ■ Overall breadth: mmb 180=





■ Elastic modulus after the Y axis, 33, 101156 mmW yel ×=

■ Plastic modulus after the Y axis, 33101307 mmWy ×=

■ Elastic modulus after the Z axis, 33, 1040.146 mmW zel ×=


58

■ Plastic modulus after the Z axis, 33, 10229 mmW zpl ×=

■ Flexion inertia moment around the Y axis: Iy=23130.00x104mm4 ■ Flexion inertia moment around the Z axis: Iz=1318.00x104 mm4 ■ Torsional moment of inertia: It=51.08x104 mm4 ■ Working inertial moment: Iw=490000x106mm6




Boundary conditions



along X axis. ■ Inner: lateral (xoz) restraint at z=3m

Loading

The column is subjected to the following loadings: ■ External: Point load From X=0.00m and Y=9.00m: FZ =-125000N and Fx=28330N ■ Internal: None.


59

CTICM model

The model is presented in the CTICM 2006-4-Resistance barre comprimee selon


60



-for beam web:

727249.381

49.386.8

331=×≤=⇒

=

==ε

ε tc

mmmm

tc

therefore the beam web is considered to be Class 1

-for beam flange:

9950.41

50.45.1347.67

=×≤=⇒

=

==ε

ε tc

mmmm

tc



According to CTICM document:

The cross section is considered to be Class 1. The column strength will be determined considering the plastic characteristics of the cross-section. Below can be seen the CTICM conclusion, extracted from CTICM 2006-4:


61

13.17.2.3Compression verification According to Advance Design calculations:

The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design axial compression resistance of the element (Nc,Rd) from the compression force applied to the element (NEd). The compressed resistance of the member, Nc,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.2.4.

%100100,

≤×Rdc

Ed

NN

(6.9)

The design resistance of the cross-section for uniform compression Nc,Rd is determined using the formula below:

0,

M

yRdc

fAN

γ×

= (6.10)

-where:

A is the section area: 28446mmA =

yf is the yielding strength: 2/275 mmNf y =

0Mγ is the partial safety factor: 10 =Mγ

kNNmmNmmfAN

M

yRdc 65.23222322650

1/2758446 22

0, ==

×=

×=

γ

NNEd 125000=

%100%38.51002322650125000

,

≤=×=N

NNN

Rdc

Ed


The compression resistance of the column is kNN Rdc 2324, = as it can be seen from conclusion extracted from

CTCIM 2006-4:


62

13.17.2.4Shear verification According to Advance Design calculations:

The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design shear resistance of the element (Vc,Rd) from the shear force applied to the element (VEd). The shear resistance of the member, Vc,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.2.6.

%100100,

≤×Rdc

Ed

VV

(6.17)

The design shear resistance of the element, Vc,Rd is determined using the formula below:

0,

3M

yV

Rdc

fA

Vγ

×

= (6.18)

-where:

AV is the shear area:

( ) wwfwfV thtrttbAA ××≥××++××−= η22

-where:

A is the cross section area: 28446mmA =

b is the overall breadth: mmb 180=

tf is the flange thickness: mmt f 5.13=

r is the root radius: mmr 21=

tw is the web thickness: mmtw 6.8=

hw is the depth of the web: mmhw 400=

1=η

( )( ) 22 1.42695.132126.85.1318028446

22

mmmmmmmmmmmmmm

trttbAA fwfV

=××++××−=

=××++××−=

234404006.81 mmmmmmth ww =××=××η

22 34401.4269 mmmmAV ≥=



N

mmNmmf

AV

M

yV

Rdc 66.6778101

3/2751.4269

3

22

0, =

×

=

×

=γ

NVEd 28330=

%100%180.410004179.010066.677810

28330100,

≤=×=×=×N

NVV

Rdc

Ed


63


The shear resistance of the column is kNV Rdzpl 8.677,, = as it can be seen from conclusion extracted from CTCIM

2006-4:

13.17.2.5Bending moment verification According to Advance Design calculations:

The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design bending moment resistance of the element (Mpl,Rd) from the bending moment effor applied to the element (MEd). The Bending moment resistance of the member, Mpl,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.2.5.

%100100,

≤×Rdpl

Ed

MM

(6.12)

-the shear force does not exceed 50% of the shear plastic resistance, therefore there is no influence of the shear on the composed bending;

-the axial compression force does not exceed 25% of the plastic resistance, therefore there is no influence of the compression on the composed bending

The design bending moment resistance of the element, Mpl,Rd is determined using the formula below:

0,

M

yplRdpl

fwM

γ×

= (6.13)

-where:

wpl is the plastic modulus: 31307000mmwpl =



NmmmmNmmfwM

M

yplRdpl 359425000

1/2751307000 23

0, =

×=

×=

γ

NmmM Ed 254970000=

%100%938.7010070938.0100359425000254970000100

,

≤=×=×=×NmmNmm

MM

Rdpl

Ed


The bending moment resistance of the column is kNmM Rdypl 7.359,, = as it can be seen from the conclusion

extracted from CTCIM 2006-4:


64

13.17.2.6Buckling verification According to Advance Design calculations:

a) over the strong axis of the section, y-y:

The cross section buckling curve will be chosen according to Table 6.2:

The imperfection factor α will be chosen according to Table 6.1:

21.0=α

Coefficient corresponding to non-dimensional slenderness after Y-Y axis


curve according to:

1122

≤−Φ+Φ

=yyy

yλ

χ (6.49)

yλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:

ycr

yy N

fA

,

*=λ

Where: A is the cross section area; A=8446mm2; fy is the yielding strength of the material; fy=275N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

( )N

mmmmMPa

LIE

Nfz

yycr 77.5918472

90001023130210000²

²²

2

44

, =×××

=××

=ππ

62645.077.5918472

/2758446 22

,

=×

=×

=N

mmNmmN

fA

ycr

yyλ

[ ] ( )[ ] 740997.062645.02.062645.021.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ

187968.062645.0740997.0740997.0

112222

≤=−+

=−Φ+Φ

=yyy

yλ

χ


65


The determined value for the coefficient corresponding to non-dimensional slenderness for the strong section, y-y

axis, yλ is: 8796.0=yχ as it can be observed from the conclusion extracted from CTCIM 2006-4:

b) over the strong axis of the section, z-z:



34.0=α


1122

≤−Φ+Φ

=zzz

zλ

χ (6.49)


66


zcr

yz

NfA

,

*=λ


Outside the frame, the calculation can be made with more than the safety of taking in account a buckling length equal to the grater length of the two beam sections, 6m. A more accurate calculation is to perform a modal analysis of the column buckling outside the frame. The first eigen mode of instability corresponds to an amplification factor equal to critical 15.9=crα . The normal critical force can be directly calculated:

( ) Nmm

mmmmNl

IENfy

zzcr 153.1142396

²4890101318/210000²

²² 442

, =×××

=××

=ππ

42588.1153.1142396

/2758446 22

,

=×

=×

=mmNmm

NfA

zcr

yzλ

[ ] ( )[ ] 72497.142588.12.042588.134.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ

137096.042588.172497.172497.1

112222

≤=−+

=−Φ+Φ

=zzz

zλ

χ


The determined value for the coefficient corresponding to non-dimensional slenderness for the strong section, z-z

axis, zλ is: 3711.0=zχ as it can be observed from the conclusion extracted from CTCIM 2006-4:


67

13.17.2.7Lateral torsional buckling verification According to Advance Design calculations:

a) for the 3m part of the column:

The elastic moment for lateral-torsional buckling calculation, Mcr:


z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ



²252.0423.0325.01

1 ψψ ++=C


ψ is the fraction of the bending moment from the column extremities: 66667.097.25498.169

==kNmkNmψ

17932.1²66667.0252.066667.0423.0325.0

1²252.0423.0325.0

11 =

×+×+=

++=

ψψC

■ Flexion inertia moment around the Y axis: Iy=23130.00x104mm4 ■ Flexion inertia moment around the Z axis: Iz=1318.00x104 mm4 ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa. ■ Torsional moment of inertia: It=51.08x104 mm4 ■ Working inertial moment: Iw=490000x106mm6 ■ Length of the column part: L=3000mm ■ Shear modulus of rigidity: G=80800MPa

( )( )

Nmm

mmNmmmmN

mmmmNmmmm

mm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

2.806585210

33396.22534.303523217932.1101318/210000²

1008.51/808003000101318

1049

3000101318/210000²17932.1

²²

²²

442

4422

44

610

2

442

1

=

=××=×××

×××+

××

×

××××

×=××××

+×××

×=

π

ππ

π

Calculation of the non-dimensional slenderness factor, LTλ :

cr

yyLT

MfW

=λ

Plastic modulus,33101307 mmWy ×=


68

66754.02.806585210

/275101307 233

=××

=×

=Nmm

mmNmmM

fW

cr

yyLTλ


1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×= The cross section buckling curve will be chosen according to Table 6.4:


34.0=α

( )[ ] 80228.0²66754.02.066754.034.015.0 =+−×+×=LTφ

180173.0²66754.0²80228.080228.0

1

²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ


The determined value for the coefficient corresponding to non-dimensional slenderness, LTλ is: 7877.0=LTχ as it can be observed from the conclusion extracted from CTCIM 2006-4:


69

b) for the 6m part of the column:



z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ



²252.0423.0325.01

1 ψψ ++=C


ψ is the fraction of the bending moment from the column extremities: 0=ψ

77.11 =C ■ Flexion inertia moment around the Y axis: Iy=23130.00x104mm4 ■ Flexion inertia moment around the Z axis: Iz=1318.00x104 mm4 ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa. ■ Torsional moment of inertia: It=51.08x104 mm4 ■ Working inertial moment: Iw=490000x106mm6 ■ Length of the column part: L=6000mm ■ Shear modulus of rigidity: G=80800MPa

Nmm

mmNmmmmN

mmmmNmmmm

mm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

406423987

604.302085.75880877.1101318/210000²

1008.51/80800²6000101318

1049

²6000101318/210000²77.1

²²

²²

442

4422

44

610

2

442

1

=

=××=×××

×××+

××

×

××××

×=××××

+×××

×=

π

ππ

π


70

Calculation of the non-dimensional slenderness factor, LTλ :

cr

yyLT

MfW

=λ

Plastic modulus,33101307 mmWy ×=

94040.0406423987

/275101307 233

=××

=×

=Nmm

mmNmmM

fW

cr

yyLTλ


1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×= The cross section buckling curve will be chosen according to Table 6.4:


34.0=α

( )[ ] 06804.1²94040.02.094040.034.015.0 =+−×+×=LTφ

163518.0²94040.0²06804.106804.1

1

²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ


71


The determined value for the coefficient corresponding to non-dimensional slenderness, LTλ is: 694.0=LTχ as it can be observed from the conclusion extracted from CTCIM 2006-4:

13.17.2.8Bending and axial compression verification According to Advance Design calculations:

1

1

,

,,

1

,

,,

1

≤∆+

×+×

∆+×+

×M

Rkz

EdzEdzyz

M

RdyLT

RdyEdyyy

M

Rky

Ed

MMM

kMMM

kNN

γγχ

γχ

(6.61)

1

1

,

,,

1

,

,,

1

≤∆+

×+×

∆+×+

×M

Rkz

EdzEdzzz

M

RdyLT

RdyEdyzy

M

Rkz

Ed

MMM

kMMM

kNN

γγχ

γχ

(6.62)

The formulae can be simplified because:

There is no bending on the small inertia axis: 0, =EdzM

The section is considered to be a Class1: 0, =∆ RdyM and 0, =∆ RdzM

Therefore the formulae are:

( )

( )62.600.1

61.600.1

1

,

,

1

1

,

,

1

≤×

×+×

≤×

×+×

M

RkyLT

Edyzy

M

Rkz

Ed

M

RkyLT

Edyyy

M

Rky

Ed

MM

kNN

MM

kNN

γχ

γχ

γχ

γχ


72

13.17.2.9Internal factor, yyk , calculation:

The internal factor yyk corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:

yy

ycr

Ed

ymLTmyyy C

NNCCk 1

1,

×−

××=µ

Auxiliary terms:

ycr

Edy

ycr

Ed

y

NN

NN

,

,

1

1

×−

−=

χµ

Where:


( ) Nmm

mmmmNl

IEN

fy

yycr 773.5918472

²900041023130/210000²

²² 42

, =×××

=××

=ππ

99741.0

773.591847212500087968.01

773.59184721250001

=×−

−= N

NN

yµ



73



LTC λλ ×= 10 -where:


²252.0423.0325.01

1 ψψ ++=C

According to EN 1993-1-1-AN France; Chapter 3; …(6) ψ is the fraction of the bending moment from the column extremities: 0=ψ

77.11 =C


74

66754.02.806585210

/275101307 233

=××

=×

=Nmm

mmNmmM

fW

cr

yyLTλ

88811.066754.077.1110 =×=×=×= LTLT CC λλλ


,,1 1120.0

−×

−××

TFcr

Ed

zcr

Ed

NN

NNC term:

Where:


××+××=

²,

2

0,

Tcr

wtTcr L

IEIGIAN π


20 gzy zAIII ⋅++=

Flexion inertia moment around the Y axis: Iy=23130.00x104mm4

Flexion inertia moment around the Z axis: Iz=1318.00x104 mm4

Cross section area: 28446mmA =

Distance between the section neutral axis and the section geometrical center: 0=gz

44444420 10244481013181023130 mmmmmmIIzAIII zygzy ×=×+×=+=⋅++=

-for simplification, it will be considered the same buckling length, TcrL , , for all the column parts:

mL Tcr 6, =

Torsional moment of inertia: It=51.08x104 mm4

Working inertial moment: Iw=490000x106mm6

Longitudinal elastic modulus: E = 210000 MPa


( )N

mmmmmmNmmmmN

mmmmN Tcr

788.2400423²6000

1049/2100001008.51/808001024448

8446 61022442

44

2

,

=

=

×××+×××

×=

π

NNN TcrTFcr 788.2400423,, ==

NN zcr 153.1142396, = (previously calculated)

25505.0

788.24004231250001

153.1142396125000177.120.01120.0 44

,,1

=

=

−×

−××=

−×

−××

NN

NN

NN

NNC

TFcr

Ed

zcr

Ed


75

Therefore:

( )

≥

−×

−

×=

=

×+

××−+=

⇒=

−×

−××>=

1

11

11

25505.01120.088811.0

,,

2

0,

0,0,

4

,,10

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

ε

ε

λ



( )LTy

LTymymymy a

aCCC

×+

××−+=

ε

ε

11 0,0,

yelEd

Edyy W

AN

M

,

, ×=ξ

Elastic modulus after the Y axis, 33, 101156 mmW yel ×=

90119.14101156

8446125000

1094.25433

26

,

, =×

××

=×=mm

mmNNmm

WA

NM

yelEd

Edyyξ

99779.01023130

1008.5111 44

44

=×⋅

−=−=mm

mmIyItaLT


The bending moment is null at the end of the column, therefore: 0=ψ

( )ycr

Edmy N

NC,

0, 33.036.021.079.0 ×−×+×+= ψψ

Where:

NN ycr 773.5918472, = (previously calculated)

( ) 78749.0773.5918472

12500033.0036.079.00, =×−×+=N

NCmy

( ) ( ) 95619.099779.090119.141

99779.090119.1478749.0178749.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ


76


The mLTC coefficient takes into account the laterally restrained parts of the column. The mLTC coefficient must be calculated individually for each of the column parts.

1

11,,

2 ≥

−×

−

×=

Tcr

Ed

zcr

Ed

LTmymLT

NN

NN

aCC

a) for the 3m part of the column:

( )LTy

LTymymymmy a

aCCC

ξ

ξ

+−+=

11 0,0,3.

90119.14=yξ (previously calculated)

99779.0=LTa (previously calculated)


ψ is the fraction of the bending moment from the column part extremities: 66667.097.25498.169

==kNmkNmψ


77

( )ycr

Edmy N

NC,

0, 33.036.021.079.0 ×−×+×+= ψψ

Where:


( ) 93256.0773.5918472

12500033.066667.036.066667.021.079.00, =×−×+×+=N

NCmy

( ) ( ) 98609.099779.090119.141

99779.090119.1493256.0193256.01

1 0,0,3, =×+

××−+=

×+

×−+=

LTy

LTymymymmy a

aCCC

ξ

ξ


NN Tcr 788.2400423, = (previously calculated)

99779.0=LTa (previously calculated)

05596.1

1

05596.1

788.24004231250001

153.11423961250001

99779.0²98609.0

11

3,

3,

,,

23,3,

=⇒

≥

=

−×

−

×=

=

−×

−

×=

mmLT

mmLT

Tcr

Ed

zcr

Ed

LTmmymmLT

C

CN

NN

N

NN

NN

aCC

b) for the 6m part of the column:

( )LTy

LTymymymmy a

aCCC

ξ

ξ

+−+=

11 0,0,6.

yelEd

Edyy W

AN

M

,

, ×=ξ

Elastic modulus after the Y axis, 33, 101156 mmW yel ×=

9353.9101156

8446125000

1098.16933

26

,

, =×

××

=×=mm

mmNNmm

WA

NM

yelEd

Edyyξ

99779.01023130

1008.5111 44

44

=×⋅

−=−=mm

mmIyItaLT



78

ψ is the fraction of the bending moment from the column part extremities: 0=ψ

( )ycr

Edmy N

NC,

0, 33.036.021.079.0 ×−×+×+= ψψ

Where:


( ) 78749.0773.5918472

12500033.0036.0021.079.00, =×−×+×+=N

NCmy

( ) ( ) 94873.099779.09353.91

99779.09353.978749.0178749.01

1 0,0,6, =×+

××−+=

×+

×−+=

LTy

LTymymymmy a

aCCC

ξ

ξ


NN Tcr 788.2400423, = (previously calculated)

1

1

97746.0

788.24004231250001

153.11423961250001

99779.0²94873.0

11

6,

6,

,,

26,6,

=⇒

≥

=

−×

−

×=

=

−×

−

×=

mmLT

mmLT

Tcr

Ed

zcr

Ed

LTmmymmLT

C

CN

NN

N

NN

NN

aCC

In conclusion:

=

==

105596.1

6,

3,

mmLT

mmLTmLT C

CC


79

The yyC coefficient is defined according to the Table A.1 on the “Auxiliary terms:” part:

ypl

yelLTplmy

ymy

yyyy W

WbnC

wC

wwC

,

,2max

2max

2 6.16.12)1(1 ≥

−×

××−××−×−+= λλ

Where:

■ 005.05.0,,

,20

,

,

,,

,20 =×

××××=×

××××=

RdyplLT

EdyLT

Rdpl

Edz

RdyplLT

EdyLTLT M

Ma

MM

MM

abχ

λχ

λ

■ 5.1,

, ≤=yel

yply W

Ww

■ Elastic modulus after the Y axis, 33, 101156 mmW yel ×=

■ Plastic modulus after the Y axis, 33, 101307 mmW ypl ×=

■ 5.113062.1101156101307

33

33

,

, ≤=××

==mmmm

WW

wyel

yply

■

1M

Rk

Edpl N

Nn

γ

=

■ NmmNmmfANN

M

yRdcRk 2322650

1/2758446 22

0, =

×=

×==

γ

■ 05382.0

12322650125000

1

=== NN

NNn

M

Rk

Edpl

γ


80

■ ( )zy λλλ ;maxmax =

■ 62645.0=yλ (previously calculated)

■ 42504.1=zλ (previously calculated)

■ ( ) ( ) 42504.142504.1;62645.0max;maxmax === zy λλλ

■ 95619.0=myC (previously calculated)

98511.0

005382.0²42504.1²95619.013062.1

6.142504.1²95619.013062.1

6.12)113062.1(1

=

=

−×

××−××−×−+=yyC

88447.010130710115698511.0 33

33

,

, =××

=≥=mmmm

WW

Cypl

yelyy

In conclusion:

=×−

××=×−

××=

=×−

××=×−

××=

=98902.0

98511.01

773.59184721250001

99741.0195619.01

1

04409.198511.0

1

773.59184721250001

99741.005569.195619.01

1

,

6,6,

,

3,3`,

NNC

NNCCk

NNC

NNCCk

k

yy

ycr

Ed

ymmLTmymyy

yy

ycr

Ed

ymmLTmymyy

yy µ

µ

According to CTICM 2006-4:


81

13.17.2.10Internal factor, zyk , calculation:

The internal factor zyk corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:

z

y

zy

ycr

Ed

zmLTmyzy w

wC

NNCCk ×××

−××= 6.01

1,

µ

Auxiliary terms:

zcr

Edz

zcr

Ed

z

NN

NN

,

,

1

1

×−

−=

χµ

Where:


( ) Nmm

mmmmNl

IENfy

zzcr 153.1142396

²4890101318/210000²

²² 442

, =×××

=××

=ππ

92826.0

153.114239612500037096.01

1533.11423961250001

=×−

−=

NN

NN

zµ



82


( )

≥

−×

−

×=

=

×+

××−+=

⇒=

−×

−××>=

1

11

11

208.01120.088811.0

,,

2

0,

0,0,

4

,,10

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

ε

ε

λ


90119.14101156

8446125000

1094.25433

26

,

, =×

××

=×=mm

mmNNmm

WA

NM

yelEd

Edyyξ (previously calculated)

99779.01023130

1008.5111 44

44

=×⋅

−=−=mm

mmIyItaLT (previously calculated)



83


( ) 78749.033.036.021.079.0,

0, =×−×+×+=ycr

Edmy N

NC ψψ


( ) 95619.01

1 0,0, =+

−+=LTy

LTymymymy a

aCCC

ξ

ξ


=

==

−×

−

×=1

05596.1

116,

3,

,,

2

mmLT

mmLT

Tcr

Ed

zcr

Ed

LTmymLT C

C

NN

NN

aCC


The zyC coefficient is defined according to the Table A.1 on the “Auxiliary terms:” part:

ypl

yel

z

yLTpl

y

myyzy W

Www

dnw

CwC

,

,5

2max

2

6.0142)1(1 ××≥

−×

××−×−+=

λ


84

Where:

001.0

21.0

2,,

,4

0

,,

,

,,

,4

0 =×××

×+

××=×

×××

×+

××=RdyplLTmy

Edy

zLT

Rdzplmz

Edz

RdyplLTmy

Edy

zLTLT MC

Ma

MCM

MCM

adχλ

λχλ

λ

5.113062.1101156101307

33

33

,

, ≤=××

==mmmm

WW

wyel

yply (previously calculated)

5.1,

, ≤=zel

zplz W

Ww

Elastic modulus after the Z axis, 33, 1040.146 mmW zel ×=

Plastic modulus after the Z axis, 33, 10229 mmW zpl ×=

5.15.1

564.11040.146

1022933

33

,

,

=⇒

≤

=×

×==

z

z

zel

zplz w

wmm

mmWW

w

1M

Rk

Edpl N

Nn

γ

=

05382.0

12322650125000

1

=== NN

NNn

M

Rk

Edpl

γ


( ) ( ) 42504.142504.1;62645.0max;maxmax === zy λλλ (previously calculated)

90887.0005382.013062.1

42504.198609.0142)113062.1(1 5

22

=

−×

××−×−+=zyC

46073.0101307101156

5.113062.16.06.0 33

33

,

, =××

××=××mmmm

WW

ww

ypl

yel

z

y

46073.06.090887.0142)1(1,

,5

2max

2

=××≥=

−×

××−×−+=

ypl

yel

z

yLTpl

y

myyzy W

Www

dnw

CwC

λ

In conclusion:

56307.0

5.113062.16.0

90887.01

773.59184721250001

923830.005569.198609.06.01

1,

=

=×××−

××=×××−

××=

NNw

wC

NNCCk

z

y

zy

ycr

Ed

zmLTmyzy

µ


85

=×××−

××=

=×××−

××=

=×××−

××=

=×××−

××=

=

50752.05.1

13062.16.098511.0

1

773.59184721250001

99741.0195619.0

6.01

1

52306.05.1

13062.16.090887.0

1

773.59184721250001

99741.005569.195619.0

6.01

1

,

6,6,

,

3,3`,

NN

ww

CNNCCk

NN

ww

CNNCCk

k

z

y

zy

zcr

Ed

ymmLTmymzy

z

y

zy

zcr

Ed

ymmLTmymzy

zyµ

µ


The bending and axial compression verifications are:

-for the 3m column part:

160789.046281.014508.01

35942500080173.0

25497000052306.0

1232265037096.0

125000

198501.092383.006119.01

35942500080173.0

25497000004409.1

1232265087968.0

125000

1

,

,

1

1

,

,

1

≤=+=

=×

×+×

=

=×

×+×

≤=+=

=×

×+×

=

=×

×+×

NmmNmm

NN

MM

kNN

NmmNmm

NN

MM

kNN

M

RkyLT

Edyzy

M

Rkz

Ed

M

RkyLT

Edyyy

M

Rky

Ed

γχ

γχ

γχ

γχ


86

-for the 6m column part:

152073.037565.014508.01

35942500063518.0

16898000050752.0

1232265037096.0

125000

179322.073204.006118.01

35942500063518.0

16898000098902.0

1232265087968.0

125000

1

,

,

1

1

,

,

1

≤=+=

=×

×+×

=

=×

×+×

≤=+=

=×

×+×

=

=×

×+×

NmmNmm

NN

MM

kNN

NmmNmm

NN

MM

kNN

M

RkyLT

Edyzy

M

Rkz

Ed

M

RkyLT

Edyyy

M

Rky

Ed

γχ

γχ

γχ

γχ





87

Ratio of the design normal force to design compresion resistance

Column subjected to axial and shear force to the top Work ratio - Fx

Ratio of the design share force to design share resistance

Column subjected to axial and shear force to the top Work ratio - Fz


88

Ratio of the design share force to design share resistance

Column subjected to axial and shear force to the top Work ratio - oblique



yχ


89



zχ



yyk


90



yzk


Result name Result description Reference value Work ratio - Fx Ratio of the design normal force to design compression resistance

5.38

Work ratio - Fz Ratio of the design share force to design share resistance

4.18 Work ratio - Oblique Ratio of the design moment resistance to design bending resistance

one the principal axis

70.94


0.88


0.37

yyk Internal factor, yyk for the 3m segment

1.04

yyk Internal factor, yyk for the 6m segment

0.99

zyk Internal factor, zyk for the 3m segment

0.52

zyk Internal factor, zyk for the 6m segment

0.51


91

13.17.3Calculated results Result name Result description Value Error Work ratio - Fx Ratio of the design normal force to design compression

resistance 5.38178 % 0.0331 %

Work ratio - Fz Ratio of the design share force to design share resistance 4.17973 % -0.0065 % Work ratio - Oblique

Ratio of the design moment resistance to design bending resistance one the principal axis

70.9383 % -0.0024 %

Xy Coefficient corresponding to non-dimensional slenderness 0.879684 adim

-0.0359 %

Xz Coefficient corresponding to non-dimensional slenderness 0.370957 adim

0.2586 %

Kyy Internal factor,kyy for the 3m segment 1.03159 adim

-0.8087 %

Kyy Internal factor,kyy for the 6m segment 0.983324 adim

-0.6743 %

Kzy Internal factor,kzy for the 3m segment 0.537037 adim

3.2763 %

Kzy Internal factor,kzy for the 6m segment 0.511305 adim

0.2559 %


92

13.18 EC3 Test 20: Verifying the buckling resistance of a RC3020100 column

Test ID: 5700

Test status: Passed

13.18.1Description The test verifies the buckling of a RC3020100 column made of S355 steel.

The verifications are made according to Eurocode3 French Annex.

13.18.2Background Verification of buckling under compression efforts for a rectangular hollow, RC3020100 column made of S235 steel. The column is fixed at its base and free on the top. The column is subjected to a compression force (200 000 N) applied at its top. The dead load will be neglected.





Units

Metric System


■ Cross section area: A=9490mm2 ■ Flexion inertia moment around the Y axis: Iy=11819x104mm4 ■ Flexion inertia moment around the Z axis: Iz=6278x104 mm4


93




Boundary conditions



■ Inner: None. ■ Buckling lengths Lfy and Lfz are both imposed (10m)

Loading


■ External: Point load at Z = 5.0: FZ = N = -200 000 N, ■ Internal: None.


94

13.18.2.2Buckling in the strong inertia of the profile (along Y-Y) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.

%100100,

≤×Rdb

Ed

NN

(6.46)

The design buckling resistance of the compressed element is calculated using the next formula:

1,

M

yRdb

fAN

γχ ××

= (6.47)

Where:

Coefficient corresponding to non-dimensional slenderness after the Y-Y axis

χ coefficient corresponding to non-dimensional slenderness λ will be determined from the relevant buckling curve according to:

1122

≤−Φ+Φ

=λ

χ (6.49)

λ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:

cr

y

NfA*

=λ


( )N

mmmmMPa

LIEN

fy

zcr 943.2449625

100001011819210000²

²²

2

44

=×××

=××

=ππ

954.0943.2449625

/2359490 22

=×

=×

=N

mmNmmN

fA

cr

yλ

[ ]²)2.0(15.0 λλαφ +−+=

It will be used the following buckling curve:

The imperfection factor α corresponding to the appropriate buckling curve will be 0.21:

[ ] ( )[ ] 034.1954.02.0954.021.015.0²)2.0(15.0 2 =+−×+×=+−×+×= λλαφ


95

Therefore:

1698.0954.0034.1034.1

112222

≤=−+

=−+

=λφφ

χ

1Mγ is a safety coefficient, 11 =Mγ

NmmNmmN Rdb 7.15566441

/2359490698.0 22

, =××

=

NNEd 200000=

%848.121007.1556644

200000100,

=×=×N

NNN

Rdb

Ed

13.18.2.3Buckling in the weak inertia of the profile (along Z-Z) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.

%100100,

≤×Rdb

Ed

NN

(6.46)


1,

M

yRdb

fAN

γχ ××

= (6.47)

Where:

Coefficient corresponding to non-dimensional slenderness after the Z-Z axis


1122

≤−Φ+Φ

=λ

χ (6.49)


cr

y

NfA*

=λ


( )N

mmmmMPa

LIEN

fy

zcr 905.1301188

10000106278210000²

²²

2

44

=×××

=××

=ππ

309.1905.1301188

/2359490 22

=×

=×

=N

mmNmmN

fA

cr

yλ

[ ]²)2.0(15.0 λλαφ +−+=



96


[ ] ( )[ ] 473.1309.12.0309.121.015.0²)2.0(15.0 2 =+−×+×=+−×+×= λλαφ

Therefore:

1465.0309.1473.1473.1

112222

≤=−+

=−+

=λφφ

χ



/2359490465.0 22

, =××

=

NNEd 200000=

%286.1910075.1037019

200000100,

=×=×N

NNN

Rdb

Ed




97



Buckling of a column subjected to compression force Non-dimensional slenderness after Y-Y axis

Coefficient corresponding to non-dimensional slenderness after Z-Z axis

Buckling of a column subjected to compression force Non-dimensional slenderness after Z-Z axis


98

Ratio of the design normal force to design buckling resistance (strong inertia)

Buckling of a column subjected to compression force Work ratio (y-y)

Ratio of the design normal force to design buckling resistance (weak inertia)

Buckling of a column subjected to compression force Work ratio (z-z)


99



yχ coefficient corresponding to non-dimensional slenderness after Y-Y axis

0.698

zχ coefficient corresponding to non-dimensional slenderness after Z-Z axis

0.465

Work ratio (y-y) Ratio of the design normal force to design buckling resistance (strong inertia) [%]

12.85%

Work ratio (z-z) Ratio of the design normal force to design buckling resistance (weak inertia) [%]

19.29%

13.18.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness


0.0000 %

Xz coefficient corresponding to non-dimensional slenderness after Z-Z axis

0.465226 adim

0.0000 %

SNy Ratio of the design normal force to design buckling resistance in the strong inertia of the profile

0.128586 adim

0.0000 %

SNz Ratio of the design normal force to design buckling resistance in the weak inertia of the profile

0.192767 adim

0.0000 %


100

13.19 EC3 Test 13: Verifying the resistance of a rectangular hollow section column subjected to bending and shear efforts

Test ID: 5707

Test status: Passed

13.19.1Description Verifies the resistance of a rectangular hollow section column (made of S235 steel) subjected to bending and shear efforts.

The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

13.19.2Background Verifies the adequacy of a rectangular hollow section column made of S235 steel to resist shear and bending efforts. Verification of the shear resistance at ultimate limit state, as well as the design resistance for bending, is realised. The name of the cross-section is RC3020100 and can be found in the Advance Design OTUA library. The column is fixed at its base and it is subjected to a punctual horizontal load applied to the middle height (200 000 N). The dead load will be neglected.



■ Exploitation loadings (category A), Q: Fx = 200 000 N,




101

Units

Metric System

Geometry

Below are described the column cross section characteristics:

■ Height: h = 300 mm, ■ Width: b = 200 mm, ■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm, ■ Column height: L = 5000 mm, ■ Section area: A = 9490 mm2 , ■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,

■ Partial factor for resistance of cross sections: 0.10 =Mγ .




Boundary conditions


■ Outer: Support at start point (z = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (z = 5.00).

■ Inner: None.

Loading


■ External: Point load at z = 2.5: V= Fx = 200 000 N,

■ Internal: None.


102

13.19.2.2Reference results for calculating the design plastic shear resistance of the cross-section The design plastic shear resistance of the cross-section is determined with formula (6.18) from EN 1993-1-1:2001. Before using it, the shear area (Av) has to be determined.

Shear area of the cross section

For a rectangular hollow section of uniform thickness the shear area is determined according to chapter 6.2.6 (3) from EN 1993-1-1. As the load is parallel to depth, the shear area is:

22

56942003003009490 mm

mmmmmmmm

hbhAAv =

+×

=+×

=

Design plastic shear resistance of the cross section

EN 1993-1-1: 2011 provides the following formula to calculate the design plastic shear resistance of the cross-section:

N

MPammf

AV

M

yv

Rdpl 6.7725460.1

32355694

32

0, =

×=

×=

γ

Work ratio

The verification of the design plastic shear resistance is done with relationship (6.17) from EN 1993-1-1. The corresponding work ratio is:

Work ratio = %89.251006.772546

200000100,

=×=×RdplV

V

13.19.2.3Reference results for calculating the design resistance for bending Before calculating the design resistance for bending, the cross section class has to be determined.

Cross section class

The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.

In this case, the column is subjected to bending efforts, therefore the stresses distribution is like in the picture below:


103

Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class for compressed parts. The picture below shows an extract from this table.


104

Taking into account that the top wing part is subjected to compression stresses, its class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to compression).

1510

10215220022=

×−×−=

×−×−=

mmmmmmmm

ttrb

tc

0.1235==

yfε

Therefore:

333315 =≤= εtc

This means that the top wing is Class 1. Because the bottom wing is tensioned, it will be classified as Class 1.

The left/right web is subjected to bending stresses. Their class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to bending):

2510

10215230022=

×−×−=

×−×−=

mmmmmmmm

ttrh

tc

0.1235==

yfε

Therefore:

727225 =≤= εtc

This means that the left/right web is Class 1.

Because a cross-section is classified according to the least favorable classification of its compression elements (chapter 5.5.2(6) from EN 1993-1-1: 2001), this means that the cross-section is Class 1.

Design resistance for bending

The design resistance for bending, for Class 1 cross-section, is determined with formula (6.13) from EN 1993-1-1:2001.

NmmMPammfWM

M

yyplRdc 224660000

0.1235956000 3

0

,, =

×=

×=

γ

Work ratio

The verification of the design resistance for bending is done with relationship (6.12) from EN 1993-1-1. The corresponding work ratio is:

Work ratio = %56.222100224660000

25000200000

1002100,,

=××

=××

=×Nmm

mmN

M

LV

MM

RdcRdc

Ed




105


Work ratio of the design shear resistance

Column subjected to a punctual horizontal load applied to the middle height Work ratio - Fz

Work ratio of the design resistance for bending

Column subjected to a punctual horizontal load applied to the middle height Work ratio – Oblique


106


Result name Result description Reference value Work ratio - Fz Work ratio of the design plastic shear resistance [%] 25.89 %

Work ratio - Oblique Work ratio of the design resistance for bending [%] 222.56 %

13.19.3Calculated results Result name Result description Value Error Work ratio - Fz Work ratio of the design plastic shear resistance 25.8793 % -0.0413 % Work ratio - Oblique

Work ratio of the design resistance for bending 222.559 % -0.0004 %


107

13.20 EC3 test 10: Verifying the classification and the bending resistance of a welded built-up beam

Test ID: 5692

Test status: Passed

13.20.1Description Verifies the classification and the bending resistance of a welded built-up beam made of S355 steel. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

13.20.2Background Classification and bending resistance verification of a welded built-up beam made of S355 steel. The beam is simply supported and it is loaded by a uniformly distributed load (15 000 N/ml). The dead load will be neglected.



■ Exploitation loadings (category A), Q: Fz = -15 000 N/ml,



Units

Metric System

Geometry

Below are described the beam cross section characteristics:

■ Height: h = 630 mm, ■ Flange width: b = 500 mm, ■ Flange thickness: tf = 18 mm, ■ Web thickness: tw = 8 mm, ■ Beam length: L = 5000 mm, ■ Section area: A = 22752 mm2,


108





Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation along Y, Z axis and restrained in rotation along

X axis. ■ Inner: None.

Loading


■ External: Uniformly distributed load: q = Fz = -15 000 N/ml,

■ Internal: None.


109

13.20.2.2Cross-section classification Before calculating the design resistance for bending, the cross-section class has to be determined.


In this case, the beam is subjected to a uniformly distributed load; therefore the stresses distribution is like in the picture below:

Table 5.2 - sheet 2, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class of the compressed flange (top flange). The picture below shows an extract from this table.

The top flange class can be determined by considering the cross-section geometrical properties and the conditions described in Table 5.2 - sheet 2 (above extract):

67.1318

2/)8500(=

−=

mmmmmm

tc

8136.0235==

yfε

Therefore:


110

39.111467.13 =>= εtc

This means that the top flange is Class 4. Because the bottom flange is tensioned, it will be classified as Class 1.

Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the web class. The picture below shows an extract from this table. The web part is subjected to bending stresses.

The web class can be determined by considering the cross-section geometrical properties and the conditions described in Table 5.2 - sheet 1 (above extract):

25.748

218630=

×−=

mmmmmm

tc

8136.0235==

yfε

Therefore:

89.10012425.74 =≤= εtc

This means that the beam web is Class 3.


According to the calculation above, the beam section have Class 4 for top flange, Class 3 for web and Class 1 for bottom flange; therefore the class section for the entire beam section will be considered Class 4.

13.20.2.3Reference results for calculating the design resistance for bending The design resistance for bending for Class 4 cross-section is determined with the formula (6.15) from EN 1993-1-1:2001.

Before verifying this formula, it is necessary to determine the effective section modulus of the cross-section.

The effective section modulus of the cross section takes into account the reduction factor, ρ, which is applying only to parts in compression (top flange in this case).

The following parameters have to be determined in order to calculate the reduction factor: the buckling factor, the stress ratio and the plate modified slenderness.

The buckling factor (kσ) and the stress ratio(Ψ) - for flanges


111

Table 4.2 from EN 1993-1-5 offers detailed information about determining the buckling factor and the stress ratio for flange. The below picture presents an extract from this table.

Taking into account that the stress distribution on the top flange is linear, the stress ratio becomes:

43.00.11

2 =→== σσσψ k

The plate modified slenderness (λp)

The formula used to determine the plate modified slenderness for top flange is:

( )( ) 902.043.08136.04.2818/2/8500

4.28/

=××

−==

mmmmmmk

tcp

σελ

The reduction factor (ρ)

The reduction factor for top flange is determined with relationship (4.3) from EN 1993-1-5. Because λp > 0.748, the reduction factor has the following formula:

0.1188.0

2 ≤−

=p

p

λλ

ρ

The effective width of the flange part can now be calculated:

( ) mmmmmmcb feff 89.2152

85008776.0, =−

×=×= ρ

Effective section modulus

The effective section modulus is determined considering the following cross-section:

■ Top flange width: beff,t = beff,f + tw + beff,f = 439.78 mm; ■ Top flange thickness: tf = 18 mm; ■ Web and bottom flange have the same dimensions as the original section.


112

3,sup, 91.5204392 mmW yeff =

3,inf, 4.5736064 mmW yeff =

( ) 3,inf,,sup,min,, 91.5204392,min mmWWW yeffyeffyeff ==


For Class 4 cross-section, EN 1993-1-1: 2001 provides (6.15) formula in order to calculate the design resistance for bending:

NmmMPammfWM

M

yeffRdc 1847559483

0.135591.5204392 3

0

min,, =

×=

×=

γ

Work ratio

Work ratio = ( ) %54.2100

18475594838/5000/151008/100

2

,

2

,

=××

=××

=×NmmmmmmN

MLq

MM

RdcRdc




Beam subjected to uniformly distributed load Work ratio - Oblique


Result name Result description Reference value Work ratio - Oblique Design resistance for bending work ratio [%] 2.54 %

13.20.3Calculated results Result name Result description Value Error Work ratio - Oblique

Design resistance for bending work ratio 2.53713 % -0.1130 %


113

13.21 EC3 test 11: Cross section classification and compression resistance verification of a rectangular hollow section column

Test ID: 5705

Test status: Passed

13.21.1Description Verifies the cross section classification and the compression resistance of a rectangular hollow section column.


13.21.2Background Classification and verification under compression efforts of a hot rolled rectangular hollow section column made of S235 steel. The name of the cross-section is RC3020100 and can be found in the Advance Design OTUA library. The column is fixed at its base and free on the top. It is subjected to a compression force (100 000 N) applied at its top. The dead load will be neglected.






Units

Metric System

Geometry


■ Height: h = 300 mm, ■ Width: b = 200 mm,


114

■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm, ■ Column length: L = 5000 mm, ■ Section area: A = 9490 mm2 ,

■ Partial factor for resistance of cross sections: .




Boundary conditions



■ Inner: None.

Loading


■ External: Point load at Z = 5.0: N = Fz = -100 000 N,

■ Internal: None.


115

13.21.2.2Reference results for calculating the cross section class The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.


Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class for compressed parts. The picture below shows an extract from this table. The entire cross-section is subjected to compression stresses.

The cross-section class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1, and it is calculated for the most defavourable compressed part:

2510

10215230022=

×−×−=

×−×−=

mmmmmmmm

ttrh

tc


116

0.1235==

yfε

Therefore:

333325 =≤= εtc


13.21.2.3Reference results for calculating the compression resistance of the cross-section The compression resistance for Class 1 cross-section is determined with formula (6.10) from EN 1993-1-1:2001.


For Class 1 cross-section, EN 1993-1-1: 2011 provides the following formula in order to calculate the compression resistance of the cross-section:

NMPammfAN

M

yRdc 2230150

0.12359490 2

0, =

×=

×=

γ

Work ratio

Work ratio = %48.41002230150100000100

,

=×=×N

NN

N

Rdc




117






13.21.3Calculated results Result name Result description Value Error Work ratio - Fx Compression resistance work ratio 4.484 % 0.0893 %


118

13.22 EC3 Test 25: Verifying an user defined I section class 4 column fixed on the bottom and with a displacement restraint at 2.81m from the bottom

Test ID: 5712

Test status: Passed


The cross section has an “I symmetric” shape with: 880mm height; 220mm width; 5mm center thickness; 15mm flange thickness; 0mm fillet radius and 0mm rounding radius.

The column is subjected to a -328kN axial compression force; 1274 kNm bending moment after the Y axis and 127.4 kNm bending moment after the Z axis. All the efforts are applied on the top of the column. The column height is 5.62m and has a restraint of displacement at 2.81m from the bottom over the weak axis.


13.22.2Background An I880*5+220*15 shaped column subjected to compression and bending, made from S275 steel. The column has a 880x5mm web and 220x15mm flanges. The column is hinged at its base and at his top the end is translation is permitted only on vertical direction and the rotation is blocked for the long axis of the column. The column is subjected to an axial compression load -328000 N, a 127400Nm bending moment after the X axis and a1274000Nm bending moment after the Y axis.

The column has lateral restraints against torsional buckling placed in at 2.81m from the column end (in the middle).



119


■ Exploitation loadings (category A): Fz=-328000N N, Mx= 127400Nm; My=1274000Nm ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q ■ Cross section dimensions are in millimeters (mm).

Units

Metric System



■ Cross section area: 210850mmA =


■ Flange thickness: mmt f 15=


■ Web thickness: mmtw 5=


■ Elastic modulus after the Y axis, 33, 1066.3387 mmW yel ×=

■ Plastic modulus after the Y axis, 331062.3757 mmWy ×=


■ Plastic modulus after the Z axis, 33, 1031.368 mmW zpl ×=

■ Flexion inertia moment around the Y axis: Iy=149058.04x104mm4 ■ Flexion inertia moment around the Z axis: Iz=2662.89x104 mm4 ■ Torsional moment of inertia: It=51.46x104 mm4 ■ Working inertial moment: Iw=4979437.37x106mm6





120

Boundary conditions



along X axis. ■ Inner:

Lateral buckling restraint in the middle of the column (z=2.81).

Loading

The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =--328000N; Mx=127400Nm and My=1274000Nm


121



-for beam web:


178.012750109.0

328.0212 −>−=−×

⋅=−⋅

⋅=y

Ed

fANψ

5.064.0005.085.0275

328.01211

21

>=

××+⋅=

××+⋅=

dtfN

y

Edα


122

924.0275235235

===yf

ε

06.94)78.0(33.067.0

924.04233.067.0

42170924

1705

152880=

−×+×

=×+

×>=⇒

=

=×−

=ψ

ε

ε tc

mmmmmm

tc


- for beam flange:

316.8924.0961.7924

61.715

5.107=×≤=⇒

=

==tc

tc

ε therefore the haunch is considered to be Class1.



123

13.22.2.3Effective cross-sections of Class4 cross-sections - the section is composed from Class 4 web and Class 1 flanges, therefore will start the web calculation:

- in order to simplify the calculations the web will be considered compressed only

1705

152880=

×−=

mmmmmm

tc

:

41 =⇒= σψ k

According to EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1

σελ

kt

bw

p ××=

4.28

wb is the width of the web; mmbw 850=

t is the web thickness; t=5mm

9244.0275235235

===yf

ε

261.349244.04.28

5850

=××

= mmmm

pλ

According to EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4

-the web is considered to be an internal compression element, therefore:

( )286.0

261.34055.0261.33055.0

043

673.0261.322 =

×−=

+×−=⇒

≥=+

>=

p

pp

λ

ψλρ

ψ

λ

=×==×=

=×=

⇒

×=

×=

×=

mmmmbmmmmbmmmmb

bbbb

bb

e

e

eff

effe

effe

weff

55.1211.2435.055.1211.2435.0

1.243850286.0

5.0

5.0

2

1

2

1

ρ

221, 5.121555.121555.1215 mmmmmmmmmmbtbtA ewewwebeff =×+×=×+×=

2, 330022015 mmmmmmbtA ffflangeeff =×=×=

222,, 2.7815330025.12152 mmmmmmAAA flangeeffwebeffeff =×+=×+=


124

13.22.2.4Effective elastic section modulus of Class4 cross-sections - In order to simplify the calculation the section will be considered in pure bending

mmmmbb tc 4252

8501sup

inf ===⇒−==σσψ

9.231 =⇒−= σψ k

According to EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1

1705

152880=

×−=

mmmmmm

tc

σελ

kt

bw

p ××=

4.28



9244.0275235235

===yf

ε

325.19.239244.04.28

5850

=××

= mmmm

pλ

According to EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4

( )692.0

325.12055.0325.13055.0

023

673.0325.122 =

×−=

+×−=⇒

≥=+

>=

p

pp

λ

ψλρ

ψ

λ

( )

=×==×=

=−−

×=

⇒

×=

×=−

×=×=

mmmmbmmmmb

mmmmb

bbbb

bbb

e

e

eff

effe

effe

wceff

46.1761.2946.064.1171.2944.0

1.29411

850692.0

6.0

4.01

2

1

2

1

ψρρ


125

-the weight center coordinate is:

( ) ( ) ( ) ( )

mm

yG

53.155.10195095.158330

15220546.601564.1171522027.124546.6015.4321522018.366564.1175.43215220

−=−

=

=×+×+×+×

××−××−××+××=

-the inertial moment along the strong axis is:

4

23

23

23

23

14482344295.6993800726.748854356

97.4161522012

2201574.107546.60112

546.601

71.381564.11712

564.11703.4481522012

22015

mm

I y

=+=

=××+×

+××+×

+

+××+×

+××+×

=

423

23

23

23

63.2662749001522012

152200546.60112

46.6015

0564.11712

64.117501522012

15220

mm

I z

=××+×

+××+×

+

+××+×

+××+×

=

34

max, 533.3179229

53.4551448234429 mm

mmmm

zI

W yyel ===

34

max, 10.242068

11063.26627490 mm

mmmm

yIW z

zel ===


126


- the imperfection factor α will be selected according to Tables 6.1 and 6.2:

34.0=α


127



curve according to:

1122

≤−Φ+Φ

=yyy

yλ

χ (6.49)


ycr

yeffy N

fA

,

*=λ

Where: A is the effective cross section area; 22.7815 mmAeff = ; fy is the yielding strength of the material;

fy=275N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

( ) kNNmm

mmmmNl

IEN

fy

yycr 37.9503544.95035371

²56201448234429/210000²

²² 42

, ==××

=××

=ππ

15.044.95035371

/2752.7815 22

,

=×

=×

=N

mmNmmN

fA

ycr

yeffyλ

[ ] ( )[ ] 503.015.02.015.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ

1

1

017.115.0503.0503.0

112222 =⇒

≥

=−+

=−Φ+Φ

=y

y

yyy

yχ

χ

λχ


128



49.0=α



1122

≤−Φ+Φ

=zzz

zλ

χ (6.49)


zcr

yeffz

NfA

,

*=λ



ml fz 81.2= because of the torsional buckling restraint from the middle of the column


129

( ) kNNmm

mmmmNl

IENfz

zzcr 35.698962.6989347

²281063.26627490/210000²

²² 42

, ==××

=××

=ππ

555.062.6989347

/2752.7815 22

,

=×

=×

=N

mmNmmN

fA

zcr

yeffzλ

[ ] ( )[ ] 741.0555.02.0555.049.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ

812.0

1

812.0555.0741.0741.0

112222 =⇒

≤

=−+

=−Φ+Φ

=z

z

zzz

zχ

χλ

χ

13.22.2.6Lateral-torsional buckling verification a) for the top part of the column:


- the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:

z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ


- where:


²252.0423.0325.01

1 ψψ ++=C


ψ is the fraction of the bending moment from the column extremities: 50.01274637

==kNmkNmψ


130

31.15.01274637

1 =⇒== CkNmkNmψ

According to EN 1993-1-1-AN France; Chapter 3.2; Table 1

Flexion inertia moment around the Y axis: 41448234429mmI y =


Longitudinal elastic modulus: E = 210000 N/mm2.

Torsional moment of inertia: It=514614.75mm4

Warping inertial moment:


( )4

2fz

w

thII

−×=

h cross section height; h=880mm

ft flange thickness; mmt f 15=

( ) 61124

10808.494

15880mm 326627490.6 mmmmmmIw ×=−×

=

According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column part: L=2810mm

Shear modulus of rigidity: G=80800N/mm2

( )( )

kNmNmm

mmNmmmmNmmmmNmm

mmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

43.40224022433856

32.439626.698934731.163.26627490/210000

514614/80800281063.26627490

10808.49

281063.26627490/21000031.1

²²

²²

422

422

4

611

2

422

1

==

=××=××

××+

××

×××

×=××××

+×××

×=

π

ππ

π

The elastic modulus : 34

max, 533.3179229

53.4551448234429 mm

mmmm

zI

W yyel ===

466.04022433856

/275533.3179229 23, =

×==

NmmmmNmm

MfW

cr

yyeffLTλ


1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=


24220880

>==mmmm

bh


131


76.0=α

( )[ ] ( )[ ] 710.0²466.02.0466.076.015.0²2.015.0 =+−×+=+−×+= LTLTLTLT λλαφ

1803.0²466.0²710.0710.0

1

²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ

b) for the bottom part of the column:



z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ


- where:

C1 is a coefficient that depends on several parameters, such as: section properties; support conditions; moment diagram allure

²252.0423.0325.01

1 ψψ ++=C



132

ψ is the fraction of the bending moment from the column extremities: 0637

0==

kNmψ

77.10 1 =⇒= Cψ








( )4

2fz

w

thII

−×=



( ) 61124

10808.494

15880mm 326627490.6 mmmmmmIw ×=−×

=

According to EN1993-1-1-AN France; Chapter 2 (…4)

Length of the column part: L=2810mm



133

( )( )

kNmNmm

mmNmmmmNmmmmNmm

mmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

89.54345434891269

32.439626.698934777.163.26627490/210000

514614/80800281063.26627490

10808.49

281063.26627490/21000077.1

²²

²²

422

422

4

611

2

422

1

==

=××=××

××+

××

×××

×=××××

+×××

×=

π

ππ

π


max, 533.3179229

53.4551448234429 mm

mmmm

zI

W yyel ===

401.05434891269

/275533.3179229 23, =

×==

NmmmmNmm

MfW

cr

yyeffLTλ

Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with the formula:

1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=


24220880

>==mmmm

bh


76.0=α

( )[ ] ( )[ ] 657.0²401.02.0401.076.015.0²2.015.0 =+−×+=+−×+= LTLTLTLT λλαφ

1849.0²401.0²657.0657.0

1

²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ





134

a) for the top part of the column:

ycr

Ed

ymLTmyyy

NNCCk

,

1−××=

µ

ycr

Edy

ycr

Ed

y

NN

NN

,

,

1

1

×−

−

=χ

µ

1=yχ (previously calculated)

kNNEd 328=

kNNl

IEN

fy

yycr 37.9503544.95035371

²²

, ==××

=π


1

44.9503537132800011

44.950353713280001

1

1

,

, =×−

−=

×−

−

=

NN

NN

NN

NN

ycr

Edy

ycr

Ed

y

χµ


135



0

,0

cr

yyeff

MfW ×

=λ

According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2

34

max, 533.3179229

53.4551448234429 mm

mmmm

zI

W yyel ===


( )( )

kNmNmm

mmNmmmmNmmmmNmm

mmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

56.30703070560199

32.439626.6989347163.26627490/210000

514614/80800281063.26627490

10808.49

281063.26627490/2100001

²²

²²

422

422

4

611

2

422

10

==

=××=××

××+

××

×××

×=××××

+×××

×=

π

ππ

π

534.03070560199

/275533.3179229 23

0

,0 =

×=

×=

NmmmmNmm

MfW

cr

yyeffλ


136


,,1 1120.0

−×

−××

TFcr

Ed

zcr

Ed

NN

NNC term:

Where:

- for a symmetrical section for the both axis, TcrTFcr NN ,, =

××+××=

²,

2

0,

Tcr

wtTcr L

IEIGIAN π


20 gzy zAIII ⋅++=





44420 151720927017.2662885467.1490580416 mmmmmmIIzAIII zygzy =+=+=⋅++=


mL Tcr 81.2, =


Working inertial moment: 61110808.49 mmIw ×= (previously calculated)



( )N

mmmmmmNmmmmN

mmmmN Tcr

24.9646886²2810

10808.49/21000075.514614/808001517209270

10850 6112242

4

2

,

=

=

×××+××=

π

NNEd 328000=

NNN TcrTFcr 24.9646886,, ==

kNNl

IENfz

zzcr 35.698962.6989347

²²

, ==××

=π


C1=1.31 for the top part of the column

C1=1.77 for the bottom part of the column


224.0

24.96468863280001

62.6989347328000131.120.01120.0 44

,,1

=

=

−×

−××=

−×

−××

NN

NN

NN

NNC

TFcr

Ed

zcr

Ed


137

Therefore: For the top part of the column:

( )

≥

−×

−

×=

=

×+

××−+=

⇒

=

−⋅

−⋅⋅>=

=

−×

−××

=

1

11

11

224.01120.0534.0

224.01120.0

534.0

,,

2

0,

0,0,

4

,,10

4

,,1

0

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

NN

NNC

ε

ε

λ

λ



( )LTy

LTymymymy a

aCCC

×+

××−+=

ε

ε

11 0,0,

yeff

eff

Ed

Edyy W

AN

M

,

, ×=ξ


55.9533.3179229

2.7815328000

1012743

26

,

, =××

=×=mm

mmNNmm

WA

NM

yeff

eff

Ed

Edyyξ

19996.01448234429

75.51461411 4

4

≈=−=−=mm

mmIIa

y

tLT



( )ycr

Edmy N

NC,

0, 33.036.021.079.0 ×−×+×+= ψψ


138

Where:

kNNl

IEN

fy

yycr 37.9503544.95035371

²²

, ==××

=π


( ) 79.044.95035371

32800033.0036.079.00, =×−×+=N

NCmy

( ) ( ) 949.0155.91

155.979.0179.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ


- mLTC must be calculated separately for each column part, separated by the lateral buckling restraint

1

11,,

2 ≥

−×

−

×=

Tcr

Ed

zcr

Ed

LTmymLT

NN

NN

aCC

-the myC term used for mLTC calculation, must be recalculated for the corresponding column part (in this case the top column part)

-this being the case, the myC will be calculated using 5.0=ψ :

( )LTy

LTymymymy a

aCCC

ξ

ξ

+−+=

11 0,0,

( ) ( ) 895.044.95035371

32800033.05.036.05.021.079.033.036.021.079.0,

0, =×−×+×+=×−×+×+=N

NNNC

ycr

Edmy ψψ

55.9533.3179229

2.7815328000

1012743

26

,

, =××

=×=mm

mmNNmm

WA

NM

yeff

eff

Ed


( ) ( ) 974.0155.91

155.9895.01895.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ

1

1

989.0

644.9328.01

9897.6328.01

1974.0

11

2

,,

2

=⇒

≤

=

−×

−

×=

−×

−

×=

mLT

mLT

Tcr

Ed

zcr

Ed

LTmymLT

C

C

NN

NN

aCC


952.0

44.950353713280001

11949.01

,

=−

××=−

××=

NN

NN

CCk

ycr

Ed

ymLTmyyy

µ


139


ycr

Ed

ymLTmyyy

NNCCk

,

1−××=

µ

534.03070560199

/275533.3179229 23

0

,0 =

×=

×=

NmmmmNmm

MfW

cr

yyeffλ

For the bottom part of the column:

260.0

24.96468863280001

62.6989347328000177.120.01120.0 44

,,1

=

=

−×

−××=

−×

−××

NN

NN

NN

NNC

TFcr

Ed

zcr

Ed


140

Therefore:

For the bottom part of the column:

( )

≥

−×

−

×=

=

×+

××−+=

⇒

=

−⋅

−⋅⋅>=

=

−×

−××

=

1

11

11

260.01120.0534.0

260.01120.0

534.0

,,

2

0,

0,0,

4

,,10

4

,,1

0

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

NN

NNC

ε

ε

λ

λ



( )LTy

LTymymymy a

aCCC

×+

××−+=

ε

ε

11 0,0,

yeff

eff

Ed

Edyy W

AN

M

,

, ×=ξ


55.9533.3179229

2.7815328000

1012743

26

,

, =××

=×=mm

mmNNmm

WA

NM

yeff

eff

Ed

Edyyξ

19996.01448234429

75.51461411 4

4

≈=−=−=mm

mmIIa

y

tLT


The bending moment is null at the end of the column, therefore: 01274

0

,sup

,inf ===kNmM

M

Ed

Edψ


141

( )ycr

Edmy N

NC,

0, 33.036.021.079.0 ×−×+×+= ψψ

Where:

kNNl

IEN

fy

yycr 37.9503544.95035371

²²

, ==××

=π


( ) 79.044.95035371

32800033.0036.079.00, =×−×+=N

NCmy

( ) ( ) 949.0155.91

155.979.0179.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ



1

11,,

2 ≥

−×

−

×=

Tcr

Ed

zcr

Ed

LTmymLT

NN

NN

aCC

-the myC term used for mLTC calculation, must be recalculated for the corresponding column part (in this case the top column part)

-this being the case, the myC will be calculated using 5.0=ψ :

( )LTy

LTymymymy a

aCCC

ξ

ξ

+−+=

11 0,0,

( ) ( ) 895.044.95035371

32800033.05.036.05.021.079.033.036.021.079.0,

0, =×−×+×+=×−×+×+=N

NNNC

ycr

Edmy ψψ

77.4533.3179229

2.7815328000

106373

26

,

, =××

=×=mm

mmN

NmmWA

NM

yeff

eff

Ed

Edyyξ

( ) ( ) 967.0177.41

177.4895.01895.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ

1

1

974.0

24.96468863280001

62.69893473280001

1967.0

11

2

,,

2

=⇒

≤

=

−×

−

×=

−×

−

×=

mLT

mLT

Tcr

Ed

zcr

Ed

LTmymLT

C

C

NN

NN

NN

NN

aCC

952.0

44.950353713280001

11949.01

,

=−

××=−

××=

NN

NN

CCk

ycr

Ed

ymLTmyyy

µ


142

Note:

The software does not give the results of the lower section because it is not the most solicited segment.



zcr

Ed

ymzyz

NN

Ck

,

1−×=

µ

The

mzC term must be calculated for the hole column length

=== 0

12740

,sup

,inf

kNmMM

Ed

Edψ

( ) ( ) 784.062.6989347

32800033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz

823.0

62.69893473280001

1784.01

,

=−

×=−

×=

NN

NN

Ck

zcr

Ed

ymzyz

µ


823.0

62.69893473280001

1784.01

,

=−

×=−

×=

NN

NN

Ck

zcr

Ed

ymzyz

µ

Note:




ycr

Ed

zmLTmyzy

NN

CCk

,

1−××=

µ

991.0

62.6989347328000812.01

62.69893473280001

1

1

,

, =×−

−=

×−

−=

NN

NN

NN

NN

zcr

Edz

zcr

Ed

z

χµ

944.0

44.950353713280001

991.01949.01

,

=−

××=−

××=

NN

NN

CCk

ycr

Ed

zmLTmyzy

µ


143


944.0

44.950353713280001

991.01949.01

,

=−

××=−

××=

NN

NN

CCk

ycr

Ed

zmLTmyzy

µ

Note:




815.0

62.69893473280001

991.0784.01

,

=−

×=−

×=

NN

NN

Ck

zcr

Ed

zmzzz

µ


815.0

62.69893473280001

991.0784.01

,

=−

×=−

×=

NN

NN

Ck

zcr

Ed

zmzzz

µ

Note:



∆+×+

×

∆+×+

×

∆+×+

×

∆+×+

×

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rkz

RdzEdzzz

M

RkyLT

RdyEdyzy

M

Rkz

Ed

M

Rkz

RdzEdzyz

M

RkyLT

RdyEdyyy

M

Rky

Ed

MMM

kMMM

kNN

MMM

kMMM

kNN

γγχ

γχ

γγχ

γχ

iyRk AfN ×=


144


56.356.171.119.0

1/27510.242068

104.127815.0

1/275533.3179229803.0

101274944.0

1/2752.7815812.0

328000

46.358.173.115.0

1/27510.242068

104.127823.0

1/275533.3179229803.0

101274952.0

1/2752.78151

328000

23

6

23

6

22

23

6

23

6

22

=++=×

××+

+×

×

××+

××

=++=×

××+

+×

×

××+

××

mmNmmNmm

mmNmmNmm

mmNmmN

mmNmmNmm

mmNmmNmm

mmNmmN


56.356.171.119.0

1/27510.242068

104.127815.0

1/275533.3179229803.0

101274944.0

1/2752.7815812.0

328000

46.358.173.119.0

1/27510.242068

104.127823.0

1/275533.3179229803.0

101274952.0

1/2752.7815812.0

328000

23

6

23

6

22

23

6

23

6

22

=++=×

××+

+×

×

××+

××

=++=×

××+

+×

×

××+

××

mmNmmNmm

mmNmmNmm

mmNmmN

mmNmmNmm

mmNmmNmm

mmNmmN

Note:





145



yχ



zχ


146



yyk



yzk


147



zyk



zzk


148



SNy



SMyy


149



SMyz



SNz


150



SMzy



SMzz


151




1


0.81


0.95


0.82


0.94


0.82

SNy


0.15


1.72


1.58

SNz


0.19


1.71


1.56

13.22.3Calculated results Result name Result description Value Error Xy Coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 % Xz Coefficient corresponding to non-dimensional slenderness 0.811841

adim 0.2273 %

Kyy Internal factor, kyy 0.950358 adim

0.0377 %

Kyz Internal factor, kyz 0.823048 adim

0.3717 %

Kzy Internal factor, kzy 0.941635 adim

0.1739 %

Kzz Internal factor, kzz 0.815493 adim

-0.5496 %

SNy Bending and axial compression verification term depending of the compression effort over the Y axis

0.152455 adim

1.6367 %


1.72357 adim

0.2076 %


1.57508 adim

-0.3114 %

SNz Bending and axial compression verification term depending of the compression effort over the z axis

0.187789 adim

-1.1637 %


1.70775 adim

-0.1316 %

SMzy Bending and axial compression verification term depending of the Z bending moment over the Z axis

1.70775 adim

-0.1316 %


152

13.23 EC3 Test 19: Verifying the buckling resistance for a IPE300 column

Test ID: 5699

Test status: Passed

13.23.1Description The test verifies the buckling resistance for a IPE300 column made of S235 steel.

The verifications are made according to Eurocode3 French Annex.

13.23.2Background Classification and verification under compression efforts for an IPE 300 column made of S235 steel. The column is fixed at its base and free on the top. The column is subjected to a compression force (200 000 N) applied at its top. The dead load will be neglected.





Units

Metric System


■ Cross section area: A=5380mm2 ■ Flexion inertia moment around the Y axis: Iy=603.80x104mm4 ■ Flexion inertia moment around the Z axis: Iz=8356x104 mm4





153

Boundary conditions



■ Inner: None. ■ Buckling lengths Lfy and Lfz are doth imposed with 10m value

Loading



13.23.2.2Buckling in the strong inertia of the profile (along Y-Y) The calculations are made in order to obtain the buckling resistance work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.

%100100,

≤×Rdb

Ed

NN

(6.46)


1,

M

yRdb

fAN

γχ ××

= (6.47)

Where:

Coefficient corresponding to non-dimensional slenderness for Y-Y axis


1122

≤−Φ+Φ

=λ

χ (6.49)


cr

y

NfA*

=λ


( )N

mmmmMPa

LIEN

fy

zcr 70.1731878

10000108356210000²

²²

2

44

=×××

=××

=ππ

854.070.1731878

/2355380 22

=×

=×

=N

mmNmmN

fA

cr

yλ

[ ]²)2.0(15.0 λλαφ +−+=


154

The following buckling curve will be used:


[ ] ( )[ ] 933.0854.02.0854.021.015.0²)2.0(15.0 2 =+−×+×=+−×+×= λλαφ

Therefore:

1764.0854.0933.0933.0

112222

≤=−+

=−+

=λφφ

χ



/2355380764.0 22

, =××

=

NNEd 200000=

%703.20100089.966051

200000100,

=×=×N

NNN

Rdb

Ed


155

13.23.2.3Buckling in the weak inertia of the profile (along Z-Z) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.

%100100,

≤×Rdb

Ed

NN

(6.46)


1,

M

yRdb

fAN

γχ ××

= (6.47)

Where:

Coefficient corresponding to non-dimensional slenderness for Z-Z axis


1122

≤−Φ+Φ

=λ

χ (6.49)


cr

y

NfA*

=λ


( )N

mmmmMPa

LIEN

fy

zcr 610.125144

100001080.603210000²

²²

2

44

=×××

=××

=ππ

178.3610.125144

/2355380 22

=×

=×

=N

mmNmmN

fA

cr

yλ

[ ]²)2.0(15.0 λλαφ +−+=



156


[ ] ( )[ ] 056.6178.32.0178.334.015.0²)2.0(15.0 2 =+−×+×=+−×+×= λλαφ

Therefore:

1089.0178.3056.6056.6

112222

≤=−+

=−+

=λφφ

χ



/2355380089.0 22

, =××

=

NNEd 200000=

%349.17710078.112771

200000100,

=×=×N

NNN

Rdb

Ed




157



Buckling of a column subjected to compression force Non-dimensional slenderness after Y-Y axis

Ratio of the design normal force to design buckling resistance (strong inertia)

Buckling of a column subjected to compression force Work ratio (y-y)


158



yχ coefficient corresponding to non-dimensional slenderness after Y-Y axis

0.764

zχ coefficient corresponding to non-dimensional slenderness after Z-Z axis

0.089

Work ratio (y-y) Ratio of the design normal force to design buckling resistance (strong inertia) [%]

0.21 %

Work ratio (z-z) Ratio of the design normal force to design buckling resistance (weak inertia) [%]

1.77 %

13.23.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness


-0.1140 %

Xz coefficient corresponding to non-dimensional slenderness after Z-Z axis

0.0891543 adim

0.1734 %


0.207253 adim

-1.3081 %

SNz Ratio of the design normal force to design buckling resistance in the weak inertia of the profile

1.77401 adim

0.2266 %


159

13.24 EC3 Test 12: Verifying the design plastic shear resistance of a rectangular hollow section beam

Test ID: 5706

Test status: Passed

13.24.1Description Verifies the design plastic shear resistance of a rectangular hollow section beam made of S275 steel.


13.24.2Background Verifies the adequacy of a rectangular hollow section beam made of S275 steel to resist shear. Verification of the shear resistance at ultimate limit state is realised. The name of the cross-section is RC3020100 and can be found in the Advance Design OTUA library. The beam is simply supported and it is subjected to an uniformly distributed load (50 000 N/ml) applied at its top. The dead load will be neglected.



■ Exploitation loadings (category A), Q: Fz = -50 000 N/ml,



Units

Metric System

Geometry


■ Height: h = 300 mm, ■ Width: b = 200 mm, ■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm, ■ Beam length: L = 5000 mm, ■ Section area: A = 9490 mm2 ,



160




Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation along Y, Z axis and restrained in rotation along

X axis. ■ Inner: None.

Loading

The beam is subjected to the following loadings:

■ External: Uniformly distributed load: q = Fz = -50 000 N/ml,

■ Internal: None.

13.24.2.2Reference results for calculating the design plastic shear resistance of the cross-section The design plastic shear resistance of the cross-section is determined with formula (6.18) from EN 1993-1-1:2001. Before using it, the shear area (Av) has to be determined.

Shear area of the cross section

For a rectangular hollow section of uniform thickness the shear area is determined according to chapter 6.2.6 (3) from EN 1993-1-1. As the load is parallel to depth, the shear area is:

22

56942003003009490 mm

mmmmmmmm

hbhAAv =

+×

=+×

=

Design plastic shear resistance of the cross section

EN 1993-1-1: 2011 provides the following formula to calculate the design plastic shear resistance of the cross-section:

N

MPammf

AV

M

yv

Rdpl 9040440.1

32755694

32

0, =

×=

×=

γ

Work ratio

The verification of the design plastic shear resistance is done with relationship (6.17) from EN 1993-1-1. The corresponding work ratio is:

Work ratio = %83.13100904044125000100

9040442

5000/50

1002100,,

=×=×

×

=×

×

=×NN

N

mmmmN

V

Lq

VV

RdplRdpl

Ed




161


Work ratio of the design shear resistance

Beam subjected to uniformly distributed load Work ratio - Fz


Result name Result description Reference value Work ratio - Fz Work ratio of the design shear resistance [%] 13.83 %

13.24.3Calculated results Result name Result description Value Error Work ratio - Fz Shear resistance work ratio 13.8219 % 0.1587 %


162

13.25 EC3 Test 27: Verifying an user defined I section class 3 beam simply supported with a displacement restraint

Test ID: 5717

Test status: Passed

13.25.1Description The test verifies a user defined cross section beam. The beam is hinged at one end and the translations over the Y and Z axis and rotation after the X axis are blocked.


The beam is subjected to 10 kN/m linear force applied vertically, 5 kN/m linear force applied horizontally and 3700 kN punctual force applied on the end of the beam.


13.25.2Background An I53*1.2+22*1.9 beam column subjected to axial compression, uniform distributed vertical force and uniform distributed horizontal force, made from S235 steel. The beam has a 53x12mm web and 220x19mm flanges. The beam is simply supported. The beam is subjected to an axial compression load 3700000 N, 10000 N/m uniform distributed load over the Z axis and 5000 N/m horizontal uniform distributed force after the Y axis.



■ Exploitation loadings (category A): Fx=-3700000N; Fy=-5000N/m; Fz=-10000N/m ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q ■ Cross section dimensions are in millimeters (mm).


163

Units

Metric System














■ Flexion inertia moment around the Z axis: 467.33789514 mmI z =


■ Working inertial moment: 667.6662201162989 mmI w =




Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis,


164

Support at the end point (x = 5) restrained in translation along Y and Z axis and restrained rotation along X axis.

■ Inner: Lateral buckling restraint in the middle of the column (x=2.50).

Loading

The beam is subjected to the following loadings: ■ External: Point load from X=f.00m and z=.00m: Fx =-3700000N; ■ External: vertical uniform distributed linear load from X=0.00 to X=5.00: Fz=-10000N/m ■ External: horizontal uniform distributed linear load from X=0.00 to X=5.00: Fy=-5000N/m


165



-for beam web:


194.24694.246

sup

inf ===MpaMpa

σσψ


166

1235235235

===yf

ε


167

42424138381

4112

192530=×≤=<=×⇒

=

=×−

=εε

ε tc

mmmmmm

tc

therefore the beam web is

considered to be Class 3

-for beam flange:

91957.4

1

47.5192

12220

=×≤=⇒

=

=

−

=tc

tc

ε





34.0=α


168



curve according to:

1122

≤−Φ+Φ

=yyy

yλ

χ (6.49)


ycr

yy N

fA

,

*=λ



( ) kNNmm

mmmmNl

IEN

fy

yycr 06.5513921.55139061

²500067665089874./210000²

²² 42

, ==××

=××

=ππ

247.021.55139061

/23514264 22

,

=×

=×

=N

mmNmmN

fA

ycr

yyλ

[ ] ( )[ ] 538.0247.02.0247.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ

984.0

1

984.0247.0538.0538.0

112222 =⇒

≤

=−+

=−Φ+Φ

=y

y

yyy

yχ

χ

λχ




169

49.0=α



1122

≤−Φ+Φ

=zzz

zλ

χ (6.49)


zcr

yz

NfA

,

*=λ

ml fz 50.2=




170

( ) kNNmm

mmmmNl

IEN

fz

zzcr 235.1120519.11205235

²250067.33789514/210000²

²² 42

, ==××

=××

=ππ

547.019.11205235

/23514264 22

,

=×

=×

=N

mmNmmN

fA

zcr

yzλ

[ ] ( )[ ] 735.0547.02.0547.049.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ

816.0

1

819.0547.0735.0735.0

112222 =⇒

≤

=−+

=−Φ+Φ

=z

z

zzz

zχ

χλ

χ


-it must be studied separately for each beam segment

-however, the two sections are symmetrical, the same result will be obtained

µ is the isotactic moment report (for simply supported bar) due to Q load ant the maxim moment value

( ) 25.01025.318

²2500/100008

²3 =

×××

=××

=Nm

mmmNMLqµ


z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ





171

0=ψ therefore:


172

31.11 =C



Longitudinal elastic modulus: 2/210000 mmNE =



Shear modulus of rigidity: 2/80800 mmNG =

Buckling length of the beam mmL 2500=

Elastic modulus after the Y axis, 3, 11.2509773 mmW yel =

( )( )

Nmm

mmNmmmmN

mmmmNmmmmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

4001163141

58.27219.1120523531.167.33789514/210000

73.1269555/80800250067.337895141001163.22

250067.33789514/21000031.1

²²

²²

422

422

4

611

2

422

1

=

=××=××

××+

××

×××

×=××××

+×××

×=

π

ππ

π

384.04001163141

/23511.2509773 23, =

×==

NmmmmNmm

MfW

cr

yyeffLTλ

Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined below:

Note:

36

1081.74001163141

1025.31 −×=×

=Nmm

MM

cr

Ed

0156.000781.0

0156.0125.05302203.03.0

04.01081.7

20.0384.0

20,

20,0,

20,3

=≤=⇒

=⇒=×=×=

≤⇒

≤×=

>=

−

LT

cr

Ed

LTLT

LT

cr

Ed

cr

Ed

LT

MM

hb

MM

MM

λ

λλ

λλ

According to EN 1993-1-1-AN France; AN.3; Chapter 6.3.2.2(4)

For slendernesses 2

0,LT

cr

Ed

MM λ≤ (see 6.3.2.3) lateral torsional buckling effects may be ignored and only cross sectional checks

apply.

-therefore:

1=LTχ

According to EN 1993-1-1-AN (2005); Chapter 6.3.2.2(4)


173




ycr

Ed

ymLTmyyy

NNCCk

,

1−××=

µ

ycr

Edy

ycr

Ed

y

NN

NN

,

,

1

1

×−

−

=χ

µ


kNN Ed 3700=

( ) Nmm

mmmmNl

IEN

fy

yycr 21.55139061

²500067665089874./210000²

²² 42

, =××

=××

=ππ

(previously

calculated)

999.0

21.551309613700000984.01

21.5513096137000001

1

1

,

, =×−

−=

×−

−

=

NN

NN

NN

NN

ycr

Edy

ycr

Ed

y

χµ

- Cmy coefficient takes into account the behavior in the plane of bending (buckling in the plan and distribution of the bending moment).

- Must be calculated considering the beam along its length.


174



0

,0

cr

yyeff

MfW ×

=λ

-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2


175

34

max, 11.2509773

26567.665089874 mm

mmmm

zI

W yyel ===


( )( )

Nmm

mmNmmmmN

mmmmNmmmmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

3054323008

58.27219.11205235167.33789514/210000

73.1269555/80800250067.337895141001163.22

250067.33789514/2100001

²²

²²

422

422

4

611

2

422

10,

=

=××=××

××+

××

×××

×=××××

+×××

×=

π

ππ

π

439.03054323008

/23511.2509773 23

0,

,0 =

×==

NmmmmNmm

MfW

cr

yyeffλ


,,1 1120.0

−×

−××

TFcr

Ed

zcr

Ed

NN

NNC term:

Where:


××+××=

²1

,

2

20

,Tcr

wtTcr L

IEIGi

N π

0491.020

20

2220 =+++= zyiii zy

Torsional moment of inertia: 473.1269555 mmI t =



mL Tcr 50.2, =

( )N

mmmmmmNmmmmN

mmmmN Tcr

13

62242

4

2

,

10696.1

²250067.6662201162989/21000073.1269555/80800

0491.01

×=

=

××+××=

π

NN Ed 3700000=

NNN TcrTFcr13

,, 10696.1 ×==

( ) Nmm

mmmmNl

IENfz

zzcr 19.11205235

²250067.33789514/210000²

²² 42

, =××

=××

=ππ


C1=1

181.0

10696.137000001

19.1120523537000001120.01120.0 4

134

,,1

=

=

×−×

−××=

−×

−××

NN

NN

NN

NN

CTFcr

Ed

zcr

Ed


176

Therefore:

( )

≥

−×

−

×=

=

×+

××−+=

⇒

=

−⋅

−⋅⋅>=

=

−×

−××

=

1

11

11

181.01120.0439.0

181.01120.0

439.0

,,

2

0,

0,0,

4

,,10

4

,,1

0

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

NN

NNC

ε

ε

λ

λ



( )LTy

LTymymymy a

aCCC

×+

××−+=

ε

ε

11 0,0,

yeff

eff

Ed

Edyy W

AN

M

,

, ×=ξ


048.011.2509773

142643700000

1025.313

26

,

, =××

=×=mm

mmNNmm

WA

NM

yeff

eff

Ed

Edyyξ

998.067.665089874

73.126955511 4

4

=−=−=mm

mmIIa

y

tLT


ycr

Edmy N

NC,

0, 03.01 ×+=


177

Where:

( ) Nmm

mmmmNl

IEN

fy

yycr 21.55139061

²500067665089874./210000²

²² 42

, =××

=××

=ππ


NN Ed 3700000=

002.121.55139061

370000003.010, =×+=N

NCmy

( ) ( ) 002.1100481

1048.0002.11002.11

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ


- It must be calculated for each of the two sections.

−×

−

×=

Tcr

Ed

zcr

Ed

LTmymLT

NN

NN

aCC

,,

2

11

It must again calculate the coefficient Cmy, but only for the left section.

mz 0001396.0−=δ


178

( )999.0

21.5513906137000001

1025.3125000001396.067665089874./2100001

11

62

422

,,2

2

0,

=×

−

×××××

+=

=×

−

×

×××+=

NN

NmmmmmmmmN

NN

MLIE

Cycr

Ed

Edy

ymy

π

δπ

( ) ( ) 999.0998.0219.01

998.0219.0999.01999.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ε

ε

377.1

10696.137000001

19.1120523537000001

998.0²999.0

11 13,,

2 =

×−×

−

×=

−×

−

×=

NN

NN

NN

NN

aCC

Tcr

Ed

zcr

Ed

LTmymLT


476.1

21.5513096137000001

1377.1002.11

,

=−

××=−

××=

NN

NN

CCk

ycr

Ed

ymLTmyyy

µ


zcr

Ed

ymzyz

NN

Ck

,

1−×=

µ

Cmz coefficient must be calculated considering the beam along its length.

010.119.11205235

370000003.0103.01,

0, =×+=×+==N

NNN

CCzcr

Edmzmz

506.1

19.1120523537000001

1776.01

,

=−

×=−

×=

NN

NN

Ck

zcr

Ed

ymzyz

µ


179


ycr

Ed

zmLTmyzy

NN

CCk

,

1−××=

µ

917.0

19.112052353700000816.01

19.1120523537000001

1

1

,

, =×−

−=

×−

−=

NN

NN

NN

NN

zcr

Edz

zcr

Ed

z

χµ


355.1

21.5513096137000001

917.0377.1002.11

,

=−

××=−

=

NN

NN

CCk

ycr

Ed

zmLTmyzy

µ


383.1

19.1120523537000001

917.0776.01

,

=−

×=−

×=

NN

NNCk

zcr

Ed

zmzzz

µ


∆+×+

×

∆+×+

×

∆+×+

×

∆+×+

×

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rkz

RdzEdzzz

M

RkyLT

RdyEdyzy

M

Rkz

Ed

M

Rkz

RdzEdzyz

M

RkyLT

RdyEdyyy

M

Rky

Ed

MMM

kMMM

kNN

MMM

kMMM

kNN

γγχ

γχ

γγχ

γχ

iyRk AfN ×=


180

72.130.007.035.1

1/23541.307177

1062.15383.1

1/23511.25097731

1025.31355.1

1/23514264816.0

3700000

53.033.008.012.0

1/23541.307177

1062.15506.1

1/23511.25097731

1025.31476.1

1/23514264984.0

3700000

23

6

23

6

22

23

6

23

6

22

=++=×

××+

+×

×

××+

××

=++=×

××+

+×

×

××+

××

mmNmmNmm

mmNmmNmm

mmNmmN

mmNmmNmm

mmNmmNmm

mmNmmN





yχ



zχ


181



yyk



yzk



zyk


182



zzk



SNy



SMyy


183



SMyz



SNz



SMzy


184



SMzz




0.98


0.82


1.47


1.51


1.35

zzk Internal factor, zzk

1.38

SNy


1.12


0.08


0.33

SNz


1.35


0.07


0.30


185

13.25.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness 0.983441

adim 0.3511 %

Xz coefficient corresponding to non-dimensional slenderness 0.816369 adim

-0.4428 %

Kyy Internal factor, kyy 1.47276 adim

0.1878 %

Kyz Internal factor, kyz 1.50599 adim

-0.2656 %

Kzy Internal factor, kzy 1.35211 adim

0.1563 %

Kzz Internal factor, kzz 1.38261 adim

0.1891 %

SNy Bending and axial compression verification term depending of the compression effort over the Y

1.12239 adim

0.2134 %


0.078033 adim

-2.4587 %


0.325974 adim

-1.2200 %


1.35209 adim

0.1548 %


0.0716405 adim

2.3436 %


0.29927 adim

-0.2433 %


186

13.26 EC3 Test 26: Verifying an user defined I section class 3 column fixed on the bottom

Test ID: 5714

Test status: Passed


The cross section has an “I symmetric” shape with: 408mm height; 190mm width; 9.4mm center thickness; 14.6mm flange thickness; 0mm fillet radius and 0mm rounding radius.

The column is subjected to 1000kN axial compression force and a 200kNm bending moment after the Y axis. All the efforts are applied on the top of the column.


13.26.2Background An I40.8*0.94+19*1.46 shaped column subjected to compression and bending, made from S275 steel. The column has a 40.8x9.4mm web and 190x14.6mm flanges. The column is fixed at it’s base The column is subjected to an axial compression load -1000000 N, a 200000Nm bending moment after the Y axis and a 5000N lateral force after the Y axis.



■ Exploitation loadings (category A): Fz=-1000000N N; My=200000Nm


187

■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q ■ Cross section dimensions are in millimeters (mm).

Units

Metric System












■ Plastic modulus after the Z axis, 69.271897, =zplW

■ Flexion inertia moment around the Y axis: Iy=257332751mm4 ■ Flexion inertia moment around the Z axis: Iz=16716452.10 mm4 ■ Torsional moment of inertia: It=492581.13 mm4 ■ Working inertial moment: Iw=645759981974.33mm6




Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,


188

Loading

The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=2.00m: FZ =-1000000N; Mx=200000Nm and Fy=5000N



- for beam web:

The web dimensions are 378.8x9.4mm.

120.012750091.0

000.1212 −>−=−×

⋅=−⋅

⋅=y

Ed

fANψ

5.001.10094.03788.0275

11211

21

>=

××+⋅=

××+⋅=

dtfN

y

Edα


189

924.0275235235

===yf

ε

25.64)20.0(33.067.0

924.04233.067.0

42170924.0

30.404.9

6.142408=

−×+×

=×+

×>=⇒

=

=×−

=ψ

ε

ε tc

mmmmmm

tc

therefore the beam web is considered to be Class 3.

- for beam flange:

316.8924.0918.6924.0

18.66.1430.90

=×≤=⇒

=

==tc

tc

ε therefore the haunch is considered to be Class1



190



34.0=α



curve according to:

1122

≤−Φ+Φ

=yyy

yλ

χ (6.49)


ycr

yy N

fA

,

*=λ


Flexion inertia moment around the Y axis: Iy=257332751mm4


191

( ) kNNmm

mmmmNl

IEN

fy

yycr 05.1333387.133338053

²2000257332751/210000²

²² 42

, ==××

=××

=ππ

137.07.133338053

/27572.9108 22

,

=×

=×

=N

mmNmmN

fA

ycr

yyλ

[ ] ( )[ ] 499.0137.02.0137.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ

1

1

022.1137.0499.0499.0

112222 =⇒

≤

=−+

=−Φ+Φ

=y

y

yyy

yχ

χ

λχ



49.0=α




192

1122

≤−Φ+Φ

=zzz

zλ

χ (6.49)


zcr

yz

NfA

,

*=λ

ml fz 00.2=

Flexion inertia moment around the Z axis: Iz=16716452.10 mm4


( ) kNNmm

mmmmNl

IENfz

zzcr 70.866138.8661700

²200010.16716452/210000²

²² 42

, ==××

=××

=ππ

538.038.8661700

/27572.9108 22

,

=×

=×

=N

mmNmmN

fA

zcr

yzλ

[ ] ( )[ ] 728.0538.02.0538.049.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ

821.0

1

821.0538.0728.0728.0

112222 =⇒

≤

=−+

=−Φ+Φ

=z

z

zzz

zχ

χλ

χ



z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ


-where:


²252.0423.0325.01

1 ψψ ++=C



193

- in order to simplify the calculation, we will consider 11 =C

Flexion inertia moment around the Y axis: Iy=257332751mm4

Flexion inertia moment around the Z axis: Iz=16716452.10 mm4


Torsional moment of inertia: It=492581.13 mm4



( )4

2fz

w

thII

−×=



( ) 61124

1046774.64

6.144080mm16716452.1 mmmmmmIw ×=−×

=

According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column: L=2000mm


( )( )

kNmNmm

mmNmmmmN

mmmmNmmmmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

089.18021802088994

052.208384.8661700110.16716452/210000

13.492581/80800200010.16716452

1046774.6

200010.16716452/2100001

²²

²²

422

422

4

611

2

422

1

==

=××=××

××+

××

×××

×=××××

+×××

×=

π

ππ

π


194


max, 172.1261434

204257332751 mm

mmmm

zI

W yyel ===

439.01802088994

/275172.1261434 23, =

×==

NmmmmNmm

MfW

cr

yyeffLTλ


1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=


2147.2190408

>==mmmm

bh


( )[ ] ( )[ ] 687.0²439.02.0439.076.015.0²2.015.0 =+−×+×=+−×+×= LTLTLTLT λλαφ

1813.0²439.0²687.0687.0

1

²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ

The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.


195




ycr

Ed

ymLTmyyy

NNCCk

,

1−××=

µ

ycr

Edy

ycr

Ed

y

NN

NN

,

,

1

1

×−

−

=χ

µ


kNNEd 1000=

kNNl

IEN

fy

yycr 05.1333387.133338053

²²

, ==××

=π


1

7.133338053100000011

7.13333805310000001

1

1

,

, =×−

−=

×−

−

=

NNN

N

NN

NN

ycr

Edy

ycr

Ed

y

χµ


196



0

,0

cr

yyeff

MfW ×

=λ

According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2

34

max, 172.1261434

204257332751 mm

mmmm

zI

W yyel ===


( )( )

kNmNmm

mmNmmmmN

mmmmNmmmmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

089.18021802088994

052.208384.8661700110.16716452/210000

13.492581/80800200010.16716452

1046774.6

200010.16716452/2100001

²²

²²

422

422

4

611

2

422

1

==

=××=××

××+

××

×××

×=××××

+×××

×=

π

ππ

π

439.01802088994

/275172.1261434 23,

0 =×

==Nmm

mmNmmM

fW

cr

yyeffλ


197


,,1 1120.0

−×

−××

TFcr

Ed

zcr

Ed

NN

NNC term:

Where:


××+××=

²1

,

2

20

,Tcr

wtTcr L

IEIGi

N π


0301.020

20

2220 =+++= zyiii zy

Torsional moment of inertia: It=492581.13 mm4

Working inertial moment: Iw=645759981974.33mm6


mL Tcr 00.2, =

( )N

mmmmmmNmmmmN

mmmmN Tcr

13

62242

4

2

,

10244.1

²200033.746457599819/21000013.492581/80800

0301.01

×=

=

××+××=

π

NNEd 1000000=

NNN TcrTFcr13

,, 10244.1 ×==

( ) Nmm

mmmmNl

IENfz

zzcr 38.8661700

²200010.16716452/210000²

²² 42

, =××

=××

=ππ




172.0

10244.110000001

38.866170010000001120.01120.0 4

134

,,1

=

=

×−×

−××=

−×

−××

NN

NN

NN

NNC

TFcr

Ed

zcr

Ed


198

Therefore:


( )

≥

−×

−

×=

=

×+

××−+=

⇒

=

−⋅

−⋅⋅>=

=

−×

−××

=

1

11

11

172.01120.0469.0

172.01120.0

469.0

,,

2

0,

0,0,

4

,,10

4

,,1

0

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

NN

NNC

ε

ε

λ

λ



( )LTy

LTymymymy a

aCCC

×+

××−+=

ε

ε

11 0,0,

yeff

eff

Ed

Edyy W

AN

M

,

, ×=ξ


444.106.1261435

72.91081000000

102003

26

,

, =××

=×=mm

mmN

NmmWA

NM

yeff

eff

Ed

Edyyξ

1998.0257332751

492581.1311 4

4

≈=−=−=mmmm

IIa

y

tLT



199

The bending moment has the same value on both ends of the column: 1=ψ

( )ycr

Edmy N

NC,

0, 33.036.021.079.0 ×−×+×+= ψψ

Where:

( ) Nmm

mmmmNl

IEN

fy

yycr 7.133338053

²2000257332751/210000²

²² 42

, =××

=××

=ππ


NNEd 1000000=

( ) 002.17.133338053

100000033.0136.0121.079.00, =×−×+×+=N

NCmy

( ) ( ) 001.11444.11

1444.1002.11002.11

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ


065.1

1

065.1

10244.110000001

38.866170010000001

1001.1

11 13

2

,,

2

=⇒

≥

=

×−×

−

×=

−×

−

×=

mLT

mLT

Tcr

Ed

zcr

Ed

LTmymLT

C

C

NN

NN

NN

NN

aCC


074.1

7.13333805310000001

1065.1001.11

,

=−

××=−

××=

NN

NN

CCk

ycr

Ed

ymLTmyyy

µ


zcr

Ed

ymzyz

NN

Ck

,

1−×=

µ

mzC term must be calculated for the whole column length

=== 1

200200

,sup

,inf

kNmkNm

MM

Ed

Edψ

( ) kNNmm

mmmmNl

IENfz

zzcr 70.866138.8661700

²200010.16716452/210000²

²² 42

, ==××

=××

=ππ

( ) ( ) 776.038.8661700

100000033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz

878.0

38.866170010000001

1776.01

,

=−

×=−

×=

NN

NNCk

zcr

Ed

ymzyz

µ


200


ycr

Ed

zmLTmyzy

NN

CCk

,

1−××=

µ

799.0

38.86617001000000821.01

38.866170010000001

1

1

,

, =×−

−=

×−

−

=

NN

NN

NN

NN

zcr

Edz

zcr

Ed

z

χµ


050.1

7.13333805310000001

977.0065.1001.11

,

=−

××=−

××=

NN

NN

CCk

ycr

Ed

zmLTmyzy

µ


857.0

38.866170010000001

977.0776.01

,

=−

×=−

×=

NN

NN

Ck

zcr

Ed

zmzzz

µ


∆+×+

×

∆+×+

×

∆+×+

×

∆+×+

×

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rkz

RdzEdzzz

M

RkyLT

RdyEdyzy

M

Rkz

Ed

M

Rkz

RdzEdzyz

M

RkyLT

RdyEdyyy

M

Rky

Ed

MMM

kMMM

kNN

MMM

kMMM

kNN

γγχ

γχ

γγχ

γχ

iyRk AfN ×=


201

41.118.074.049.0

1/2755.175962

1010857.0

1/27506.1261435813.0

10200050.1

1/27572.9108821.0

1000000

34.118.076.040.0

1/2755.175962

1010878.0

1/27506.1261435813.0

10200074.1

1/27572.91081

1000000

23

6

23

6

222

23

6

23

6

22

=++=×

××+

+×

×

××+

××

=++=×

××+

+×

×

××+

××

mmNmmNmm

mmNmmNmm

mmNmmmmN

mmNmmNmm

mmNmmNmm

mmNmmN




202



yχ



zχ


203



yyk



yzk


204



zyk



zzk


205



SNy



SMyy


206



SMyz



SNz


207



SMzy



SMzz


208




1


0.821


1.074


0.878


1.050


0.857

SNy


0.40


0.76


0.18

SNz


0.49


0.74


0.18

13.26.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 % Xz coefficient corresponding to non-dimensional slenderness 0.821634

adim 0.0772 %


4.0661 %

Kyz Internal factor kyz 0.877605 adim

-0.0450 %

Kzy Internal factor kzy 1.09224 adim

4.0229 %


0.0746 %

#SNy Bending and axial compression verification term depending of the compression effort over the Y axis

0.399218 adim

-0.1955 %


0.783306 adim

3.0666 %


0.181362 adim

0.7567 %


0.485883 adim

-0.8402 %


0.765485 adim

3.4439 %


0.177236 adim

-1.5356 %


209

13.27 EC3 test 9: Verifying the classification and the compression resistance of a welded built-up column

Test ID: 5674

Test status: Passed

13.27.1Description Verifies the cross-section classification and the compression resistance of a welded built-up column made of S355 steel. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

13.27.2Background Classification and verification of a welded built-up column made of S355 steel. The column is fixed at its base and free on the top. It is loaded by a compression force (100 000 N), applied at its top. The dead load will be neglected.






Units

Metric System

Geometry


■ Height: h = 630 mm, ■ Flange width: b = 500 mm, ■ Flange thickness: tf = 18 mm,


210

■ Web thickness: tw = 8 mm, ■ Column length: L = 5000 mm, ■ Section area: A = 22752 mm2 ,





Boundary conditions



■ Inner: None.

Loading


■ External: Point load at Z = 5.0: N = FZ = -100 000 N,

■ Internal: None.


211

13.27.2.2Cross-section classification Before calculating the compression resistance, the cross-section class has to be determined.



Table 5.2 - sheet 2, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the flanges class. The picture below shows an extract from this table.

The top flange class can be determined by considering the cross-section geometrical properties and the conditions described in Table 5.2 - sheet 2:

67.1318

2/)8500(=

−=

mmmmmm

tc

81.0235==

yfε

Therefore:

34.111467.13 =>= εtc


212

This means that the top column flange is Class 4. Having the same dimensions, the bottom column flange is also Class 4.

Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the web class. The picture below shows an extract from this table.

The web class can be determined by considering the cross-section geometrical properties and the conditions described in Table 5.2 - sheet 1:

25.748

218630=

×−=

mmmmmm

tc

81.0235==

yfε

Therefore:

02.344225.74 =>= εtc



According to the calculation above, the column section have Class 4 web and Class 4 flanges; therefore the class section for the entire column section will be considered Class 4.

13.27.2.3Reference results for calculating the compression resistance of the cross-section The compression resistance for Class 4 cross-section is determined with the formula (6.11) from EN 1993-1-1:2001.

In order to verify the compression resistance for Class 4 cross-section, it is necessary to determine the effective area of the cross-section.

The effective area of the cross section takes into account the reduction factor, ρ, which is applying to both parts in compression (flanges and web).

The following parameters have to be determined, for each part in compression, in order to calculate the reduction factor: the buckling factor, the stress ratio and the plate modified slenderness.

The buckling factor (kσ) and the stress ratio(Ψ) - for flanges

Table 4.2 from EN 1993-1-5 offers detailed information about determining the buckling factor and the stress ratio for flanges. The below picture presents an extract from this table.


213

Taking into account that the stress distribution on flanges is linear, the stress ratio becomes:

43.00.11

2 =→== σσσψ k

The buckling factor (kσ) and the stress ratio(Ψ) - for web

Table 4.1 from EN 1993-1-5 offers detailed information about determining the buckling factor and the stress ratio for web. The below picture presents an extract from this table.

Taking into account that the stress distribution on web is linear, the stress ratio becomes:

0.40.11

2 =→== σσσψ k

The plate modified slenderness (λp) – for flanges

The formula used to determine the plate modified slenderness for flanges is:

( )( ) 906.043.081.04.2818/2/8500

4.28/

=××

−==

mmmmmmk

tcp

σελ

The plate modified slenderness (λp) – for web

The formula used to determine the plate modified slenderness for web is:

( ) 614.10.481.04.288/182630

4.28/

=××

×−==

mmmmmmk

tbp

σελ

The reduction factor (ρ) – for flanges

The reduction factor for flanges is determined with relationship (4.3) from EN 1993-1-5. Because λp > 0.748, the reduction factor has the following formula:

0.1188.0

2 ≤−

=p

p

λλ

ρ

The effective width of the flange part can now be calculated:

( ) mmmmmmcb feff 25.2152

8500875.0, =−

×=×= ρ

The reduction factor (ρ) – for web

The reduction factor for web is determined with relationship (4.2) from EN 1993-1-5. Because λp > 0.673, the reduction factor has the following formula:


214

( )0.1

3055.02 ≤

+×−=

p

p

λψλ

ρ

The effective width of the web can now be calculated:

( ) mmmmmmbb weff 8.317182630535.0, =×−×=×= ρ

Effective area

The effective area is determined considering the following:

weffwwfefffefffeff bttbbtA ,,, )(2 ×+++××=


For Class 4 cross-section, EN 1993-1-1: 2001 provides (6.11) formula in order to calculate the compression resistance of the cross-section:

NMPammfAN

M

yeffRdc 6506582

0.13554.18328 2

0, =

×=

×=

γ

Work ratio

Work ratio = %54.11006506582100000100

,

=×=×N

NN

NRdc




215


Column subjected to compression axial force Work ratio - Fx



13.27.3Calculated results Result name Result description Value Error Work ratio - Fx Compression resistance work ratio 1.5322 % 0.1438 %


216

13.28 EC3 Test 21: Verifying the buckling resistance of a CHS219.1x6.3H column

Test ID: 5701

Test status: Passed

13.28.1Description The test verifies the buckling resistance of a CHS219.1x6.3H made of S355.

The tests are made according to Eurocode 3 French Annex.

13.28.2Background Buckling verification under compression efforts for an CHS219.1x6.3H column made of S355 steel. The column is fixed at its base and free on the top. The column is subjected to a compression force (100 000 N) applied at its top. The dead load will be neglected.





Units

Metric System


■ Tube wall thickness: t=6.3mm ■ Tube diameter: d=219.1mm ■ Cross section area: A=4210mm2 ■ Radius of gyration about the relevant axis: i=75.283mm ■ Flexion inertia moment around the Y axis: Iy=2366x104mm4 ■ Flexion inertia moment around the Z axis: Iz=2366x104 mm4


217




Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (z = 3.00).

■ Inner: None. ■ Buckling lengths Lfy and Lfz are both imposed with 6m value

Loading




218

13.28.2.2Buckling in the strong inertia of the profile (along Y-Y) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.

%100100,

≤×Rdb

Ed

NN

(6.46)


778.343.6

1.219==

mmmm

td

814.0355235235

===yf

ε

381.46814.05080778.34 2 =×=×≤= εtd

therefore the section is considered to be Class 2

It will be used the following buckling curve corresponding to Table 6.2:


219



1,

M

yRdb

fAN

γχ ××

= (6.47)

Where:

Coefficient corresponding to non-dimensional slenderness after the Y-Y axis


1122

≤−Φ+Φ

=yyy

yλ

χ (6.49)


1

*λ

λ×

==iL

NfA cr

cr

yy (6.50)

41.76355

2100001 === ππλ

yfE

mmmm

mmAI

i y 283.754210

238600002

4

===

043.141.76283.75

6000

1

=×

=×

=mm

mmiLcr

y λλ

[ ] [ ] 132.1²043.1)2.0043.1(21.015.0)2.0(15.0 2 =+−×+×=+−+×= yyy λλαφ

1636.0²043.1²132.1132.1

1≤=

−+=yχ

A is the cross section area; A=4210mm2; fy is the yielding strength of the material; fy=355N/mm2 and 1Mγ is a safety

coefficient, 11 =Mγ

The design buckling resistance of the compression member will be:

NmmNmmfAN

M

yyRdb 8.950533

1/3554210636.0 22

1, =

××=

××=

γχ

NNEd 100000=

%520.101008.950533

100000100,

=×=×N

NNN

Rdb

Ed


220






LTχ


221

Ratio of the design normal force to design buckling resistance in the strong inertia of the profile

Column subjected to axial force Adimensional - SNy




0.636

ySN Ratio of the design normal force to design buckling resistance in the strong inertia of the profile

0.1052

13.28.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness 0.635463

adim 0.0000 %


0.105293 adim

0.0000 %


222

13.29 EC3 Test 24: Verifying an user defined I section class 4 column fixed on the bottom and without any other restraint

Test ID: 5709

Test status: Passed

13.29.1Description The test verifies an user defined cross section column.


The column is subjected to 328 kN axial compression force; 1274 kNm bending moment after the Y axis and 127.4 kNm bending moment after the Z axis. All the efforts are applied on the top of the column. The column height is 5.62m and has no restraints over its length.



223

13.29.2Background An I880*5+220*15 shaped column subjected to compression and bending, made from S275 steel. The column has a 880x5mm web and 220x15mm flanges. The column is hinged at its base and, at his top end, translation is permitted only on vertical direction and the rotation is blocked for the long axis of the column. The column is subjected to an axial compression load 328000 N, 127400Nm bending moment after the X axis and 1274000Nm bending moment after the Y axis.



■ Exploitation loadings (category A): Fz=-328000N N, Mx= 127400Nm; My=1274000Nm ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q ■ Cross section dimensions are in millimeters (mm).

Units

Metric System


224









■ Elastic modulus after the Y axis, 33, 1066.3387 mmW yel ×=

■ Plastic modulus after the Y axis, 331062.3757 mmWy ×=


■ Plastic modulus after the Z axis, 33, 1031.368 mmW zpl ×=

■ Flexion inertia moment around the Y axis: Iy=149058.04x104mm4 ■ Flexion inertia moment around the Z axis: Iz=2662.89x104 mm4 ■ Torsional moment of inertia: It=51.46x104 mm4 ■ Working inertial moment: Iw=4979437.37x106mm6




Boundary conditions



along X axis.


225

Loading

The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =--328000N; Mx=127400Nm and My=1274000Nm



-for beam web:


178.012750109.0

328.0212 −>−=−×

⋅=−⋅

⋅=y

Ed

fANψ

5.064.0005.0850.0275

328.01211

21

>=

××+⋅=

××+⋅=

dtfN

y

Edα


226

924.0275235235

===yf

ε

057.94)78.0(33.067.0

924.04233.067.0

421701

1705

152880=

−×+×

=×+

×>=⇒

=

=×−

=ψ

ε

ε tc

mmmmmm

tc


-for beam flange:

9961.71

61.715

5.107=×≤=⇒

=

==ε

ε tc

tc



13.29.2.3Effective cross-sections of Class4 cross-sections -the section is composed from Class 4 web and Class 1 flanges, therefore will start the web calculation:

-in order to simplify the calculations the web will be considered compressed only

1705

152880=

×−=

mmmmmm

tc

:

41 =⇒= σψ k

-according EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1


227

σελ

kt

bw

p ××=

4.28



9244.0275235235

===yf

ε

261.349244.04.28

5850

=××

= mmmm

pλ

-according EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4

-the web is considered to be an internal compression element, therefore:

( )286.0

261.34055.0261.33055.0

043

673.0261.322 =

×−=

+×−=⇒

≥=+

>=

p

pp

λ

ψλρ

ψ

λ

=×==×=

=×=

⇒

×=

×=

×=

mmmmbmmmmbmmmmb

bbbb

bb

e

e

eff

effe

effe

weff

55.1211.2435.055.1211.2435.0

1.243850286.0

5.0

5.0

2

1

2

1

ρ

221, 5.121555.121555.1215 mmmmmmmmmmbtbtA ewewwebeff =×+×=×+×=

2, 330022015 mmmmmmbtA ffflangeeff =×=×=

222,, 2.7815330025.12152 mmmmmmAAA flangeeffwebeffeff =×+=×+=

13.29.2.4Effective elastic section modulus of Class4 cross-sections -In order to simplify the calculation the section will be considered in pure bending

mmmmbb tc 4252

8501sup

inf ===⇒−==σσψ

9.231 =⇒−= σψ k

-according EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1

1705

152880=

×−=

mmmmmm

tc

σελ

kt

bw

p ××=

4.28




228

9244.0275235235

===yf

ε

325.19.239244.04.28

5850

=××

= mmmm

pλ

-according EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4

( )692.0

325.12055.0325.13055.0

023

673.0325.122 =

×−=

+×−=⇒

≥=+

>=

p

pp

λ

ψλρ

ψ

λ

( )

=×==×=

=−−

×=

⇒

×=

×=−

×=×=

mmmmbmmmmb

mmmmb

bbbb

bbb

e

e

eff

effe

effe

wceff

46.1761.2946.064.1171.2944.0

1.29411

850692.0

6.0

4.01

2

1

2

1

ψρρ

-the weight center coordinate is:

( ) ( ) ( ) ( )

mm

yG

53.155.10195095.158330

15220546.601564.1171522027.124546.6015.4321522018.366564.1175.43215220

−=−

=

=×+×+×+×

××−××−××+××=

-the inertial moment along the strong axis is:


229

4

23

23

23

23

14482344295.6993800726.748854356

97.4161522012

2201574.107546.60112

546.601

71.381564.11712

564.11703.4481522012

22015

mm

I y

=+=

=××+×

+××+×

+

+××+×

+××+×

=

423

23

23

23

63.2662749001522012

152200546.60112

46.6015

0564.11712

64.117501522012

15220

mm

I z

=××+×

+××+×

+

+××+×

+××+×

=

34

max, 533.3179229

53.4551448234429 mm

mmmm

zI

W yyel ===

34

max, 10.242068

11063.26627490 mm

mmmm

yIW z

zel ===




230

34.0=α



curve according to:

1122

≤−Φ+Φ

=yyy

yλ

χ (6.49)


ycr

yeffy N

fA

,

*=λ



( ) kNNmm

mmmmNl

IEN

fy

yycr 37.9503544.95035371

²56201448234429/210000²

²² 42

, ==××

=××

=ππ

15.044.95035371

/2752.7815 22

,

=×

=×

=N

mmNmmN

fA

ycr

yeffyλ

[ ] ( )[ ] 503.015.02.015.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ

1

1

017.115.0503.0503.0

112222 =⇒

≥

=−+

=−Φ+Φ

=y

y

yyy

yχ

χ

λχ




231

49.0=α



1122

≤−Φ+Φ

=zzz

zλ

χ (6.49)


zcr

yeffz

NfA

,

*=λ



( ) kNNmm

mmmmNl

IENfz

zzcr 34.1747905.1747336

²562063.26627490/210000²

²² 42

, ==××

=××

=ππ


232

109.1905.1747336

/2752.7815 22

,

=×

=×

=N

mmNmmN

fA

zcr

yeffzλ

[ ] ( )[ ] 338.1109.12.0109.149.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ

479.0

1

479.0109.1338.1338.1

112222 =⇒

≤

=−+

=−Φ+Φ

=z

z

zzz

zχ

χλ

χ

13.29.2.6Lateral torsional buckling verification -the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:

z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ



²252.0423.0325.01

1 ψψ ++=C



0==

kNmkNmψ

77.101274

01 =⇒== C

kNmkNmψ



233







( )4

2fz

w

thII

−×=



( ) 61124

10808.494

15880mm 326627490.6 mmmmmmIw ×=−×

=

-according to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column part: L=5620mm


( )( )

kNmNmm

mmNmmmmNmmmmNmm

mmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

163.14201420163158

185.459905.174733677.163.26627490/210000

514614/80800562063.26627490

10808.49

562063.26627490/21000077.1

²²

²²

422

422

4

611

2

422

1

==

=××=××

××+

××

×××

×=××××

+×××

×=

π

ππ

π


max, 533.3179229

53.4551448234429 mm

mmmm

zI

W yyel ===

785.01420163158

/275533.3179229 23, =

×==

NmmmmNmm

MfW

cr

yyeffLTλ


1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=


234


24220880

>==mmmm

bh


76.0=α

( )[ ] ( )[ ] 030.1²785.02.0785.076.015.0²2.015.0 =+−×+=+−×+= LTLTLTLT λλαφ

1589.0²785.0²030.1030.1

1

²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ


The internal factor yyk corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:


235

ycr

Ed

ymLTmyyy

NNCCk

,

1−××=

µ

ycr

Edy

ycr

Ed

y

NN

NN

,

,

1

1

×−

−

=χ

µ

1=yχ (previously calculated)

kNNEd 328=

kNNl

IEN

fy

yycr 37.9503544.95035371

²²

, ==××

=π


1

44.9503537132800011

44.950353713280001

1

1

,

, =×−

−=

×−

−

=

NN

NN

NN

NN

ycr

Edy

ycr

Ed

y

χµ



236


0

,0

cr

yyeff

MfW ×

=λ

-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2 33

, 1066.3387 mmW yel ×=

kNmNmmM cr 163.14201420163158 == (previously calculated)

810.01420163158

/2753387660 23

0

,0 =

×=

×=

NmmmmNmm

MfW

cr

yyeffλ


,,1 1120.0

−×

−××

TFcr

Ed

zcr

Ed

NN

NNC term:

Where:


××+××=

²,

2

0,

Tcr

wtTcr L

IEIGIAN π


20 gzy zAIII ⋅++=





44420 151720927017.2662885467.1490580416 mmmmmmIIzAIII zygzy =+=+=⋅++=


mL Tcr 62.5, =


Working inertial moment: 61110808.49 mmIw ×= (previously calculated)



( )N

mmmmmmNmmmmN

mmmmN Tcr

14.2634739²5620

10808.49/21000075.514614/808001517209270

10850 6112242

4

2

,

=

=

×××+××=

π

NNEd 328000=

NNN TcrTFcr 14.2634739,, ==


237

Nl

IENfz

zzcr 905.1747336

²²

, =××

=π


244.0

14.26347393280001

905.1747336328000177.120.01120.0 44

,,1

=

=

−×

−××=

−×

−××

NN

NN

NN

NNC

TFcr

Ed

zcr

Ed

Therefore:

( )

≥

−×

−

×=

=

×+

××−+=

⇒

=

−⋅

−⋅⋅>=

=

−×

−××

=

1

11

11

244.01120.0810.0

244.01120.0

810.0

,,

2

0,

0,0,

4

,,10

4

,,1

0

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

a

aCCC

NN

NNC

NN

NNC

ε

ε

λ

λ



( )LTy

LTymymymy a

aCCC

×+

××−+=

ε

ε

11 0,0,

yeff

eff

Ed

Edyy W

AN

M

,

, ×=ξ


55.9533.3179229

2.7815328000

1012743

26

,

, =××

=×=mm

mmNNmm

WA

NM

yeff

eff

Ed

Edyyξ

19996.01448234429

75.51461411 4

4

≈=−=−=mm

mmIIa

y

tLT


238



( )ycr

Edmy N

NC,

0, 33.036.021.079.0 ×−×+×+= ψψ

Where:

kNNl

IEN

fy

yycr 37.9503544.95035371

²²

, ==××

=π


( ) 79.044.95035371

32800033.0036.079.00, =×−×+=N

NCmy

( ) ( ) 949.0155.91

155.979.0179.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ

1

11,,

2 ≥

−×

−

×=

Tcr

Ed

zcr

Ed

LTmymLT

NN

NN

aCC

068.1

14.26347393280001

905.17473363280001

1949.0 2 =

−×

−

×=

NN

NN

CmLT

0161.1

44.950353713280001

1068.1949.01

,

=−

××=−

××=

NN

NNCCk

ycr

Ed

ymLTmyyy

µ


239


zcr

Ed

ymzyz

NN

Ck

,

1−×=

µ

( ) ( ) 7677.0905.1747336

32800033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz

945.0

905.17473363280001

17677.01

,

=−

×=−

×=

NN

NNCk

zcr

Ed

ymzyz

µ


ycr

Ed

zmLTmyzy

NN

CCk

,

1−××=

µ

893.0

905.1747336328000479.01

905.17473363280001

1

1

,

, =×−

−=

×−

−=

NN

NN

NN

NN

zcr

Edz

zcr

Ed

z

χµ

908.0

44.950353713280001

893.0068.1949.01

,

=−

××=−

××=

NN

NNCCk

ycr

Ed

zmLTmyzy

µ


844.0

905.17473363280001

893.07677.01

,

=−

×=−

×=

NN

NNCk

zcr

Ed

zmzzz

µ


∆+×+

×

∆+×+

×

∆+×+

×

∆+×+

×

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rkz

RdzEdzzz

M

RkyLT

RdyEdyzy

M

Rkz

Ed

M

Rkz

RdzEdzyz

M

RkyLT

RdyEdyyy

M

Rky

Ed

MMM

kMMM

kNN

MMM

kMMM

kNN

γγχ

γχ

γγχ

γχ

iyRk AfN ×=


240

14.461.124.232.0

1/27510.242068

104.127844.0

1/275533.3179229589.0

101274908.0

1/2752.7815479.0

328000

47.4808.151.215.0

1/27510.242068

104.127945.0

1/275533.3179229589.0

1012740161.1

1/2752.78151

328000

23

6

23

6

22

23

6

23

6

22

=++=×

××+

+×

×

××+

××

=++=×

××+

+×

×

××+

××

mmNmmNmm

mmNmmNmm

mmNmmN

mmNmmNmm

mmNmmNmm

mmNmmN




241



yχ



zχ


242



yyk



yzk


243



zyk



zzk


244



SNy



SMyy


245



SMyz



SNz


246



SMzy



SMzz


247




1


0.48


1.011


0.95


0.902


0.84

SNy


0.15


2.50


1.81


2.23


1.61

13.29.3Calculated results Result name Result description Value Error Xy Coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 % Xz Coefficient corresponding to non-dimensional slenderness 0.479165

adim -0.1740 %


-0.0336 %

Kyz Internal factor kyz 0.9451 adim -0.5158 % Kzy Internal factor kzy 0.902094

adim 0.0104 %


0.4254 %

#SNy Bending and axial compression verification term depending of the compression effort over the Y axis

0.152455 adim

1.6367 %

#SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis

2.49874 adim

-0.0504 %

#SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis

1.80865 adim

-0.0746 %

#SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis

2.23031 adim

0.0139 %

#SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis

1.61436 adim

0.2708 %


248

13.30 EC3 Test 29: Verifying an user defined I section class 1, column hinged on base and restrained on top for the X, Y translation and Z rotation

Test ID: 5729

Test status: Passed


The column is an “I symmetric” shape with: 260mm height; 150mm width; 7.1mm web thickness; 10.7mm flange thickness; 0mm fillet radius and 0mm rounding radius. The column is made of S275 steel.

The column is subjected to 328 kN axial compression force and 50kNm bending moment after the Y axis and 10kNm bending moment after the Z axis. All the efforts are applied to the top of the column.


13.30.2Background An I260*7.1+150*10.7 shaped column subjected to compression and bending, made from S275 steel. The column has a 260x7.1mm web and 150x10.7mm flanges. The column is fixed for all translations and free for all rotations, at its base, and on the top end, the translations over the X and Y axes and the rotation over the Z axis are not permitted. In the middle of the column there is a restraint over the Y axis, therefore the bucking length for the XY plane is equal to half of the column length. The column is subjected to an axial compression load -328000 N, a 10000Nm bending moment after the X axis and a 50000Nm bending moment after the Y axis



■ Exploitation loadings (category A): Fz=-328000N N; My=50000Nm; Mx=10000Nm; ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q ■ Cross section dimensions are in millimeters (mm).


249

Units

Metric System














■ Flexion inertia moment around the Z axis: 446.6025866 mmIz =


■ Working inertial moment: 688.19351706542 mmIw =





250

Boundary conditions


■ Outer: ► Support at start point (x = 0) restrained in translation along X, Y and Z axis, ► Restraint of translation over the Y axis at half (z=2.81) ► Support at start point (z = 5.62) restrained in translation along X and Y axis, and restrained in rotation

along Z axis,

Loading

The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =-328000N; Mx=10000Nm and My=50000Nm


251



- for beam web:

The web dimensions are 238.6x7.1mm.

178.012750109.0

328.0212 −>−=−×

⋅=−⋅

⋅=y

Ed

fANψ

5.085.00071.02386.0275

328.01211

21

>=

××+⋅=

××+⋅=

dtfN

y

Edα


252

924.0275235235

===yf

ε

41.36185.013

924.0396113

3966.33924.0

61.331.7

7.102260=

−××

=−×

×≤=⇒

=

=×−

=α

ε

ε tc

mmmmmm

tc

therefore the beam

web is considered to be Class 1

-for beam flange:

316.8924.0968.6

924.0

68.67.10

21.7150

=×≤=⇒

=

=

−

=tc

tc

ε




253



34.0=α



curve according to:

1122

≤−Φ+Φ

=yyy

yλ

χ (6.49)


ycr

yy N

fA

,

*=λ




254

( ) kNNmm

mmmmNl

IEN

fy

yycr 33.380295.3802327

²562064.57943291/210000²

²² 42

, ==××

=××

=ππ

5956.095.3802327

/27506.4904 22

,

=×

=×

=N

mmNmmN

fA

ycr

yyλ

[ ] ( )[ ] 7446.05956.02.05956.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ

839.0

1

839.05956.07446.07446.0

112222 =⇒

≤

=−+

=−Φ+Φ

=y

y

yyy

yχ

χ

λχ



49.0=α



255


1122

≤−Φ+Φ

=zzz

zλ

χ (6.49)


zcr

yz

NfA

,

*=λ



( ) kNNmm

mmmmNl

IENfz

zzcr 71.3158151.1581706

²281046.6025866/210000²

²² 42

, ==××

=××

=ππ

923.051.1581706

/27506.4904 22

,

=×

=×

=N

mmNmmN

fA

zcr

yzλ

[ ] ( )[ ] 103.1923.02.0923.049.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ

586.01

586.0923.0103.1103.1

112222 =⇒

≤

=−+

=−Φ+Φ

=z

z

zzz

zχ

χλ

χ


256

13.30.2.4Lateral torsional buckling verification a) for the top part of the column:



z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ


- where:


²252.0423.0325.01

1 ψψ ++=C


31.15.02525

1,

, =⇒=== CkNkN

MM

topy

botomyψ


257










( )4

2fz

w

thII

−×=



( ) 624

093627638294

7.1026046.6025866 mmmmmmmmI w =−×

=


( )( )

kNmNmm

mmNmmmmN

mmmmNmmmmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

36.31574.315363380

20.15251.158170631.146.6025866/210000

97.149294/80800281046.6025866

09362763829

281046.6025866/21000031.1

²²

²²

422

422

4

6

2

422

1

==

=××=××

××+×

×××

×=××××

+×××

×=

π

ππ

π

661.074.315363380

/27518.501177 23, =

×==

NmmmmNmm

MfW

cr

yyplLTλ


1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=


258


2733.1150260

≤==mmmm

bh


49.0=LTα

( )[ ] ( )[ ] 831.0²661.02.0661.049.015.0²2.015.0 =+−×+×=+−×+×= LTLTLTLT λλαφ

1749.0²661.0²831.0831.0

1²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ




z

t

z

wzcr IE

IGLII

LIECM

××××

+×××

×=²²

²²

1 ππ


- where:

C1 is a coefficient that depends of several parameters, such as: section properties; support conditions; moment diagram allure

²252.0423.0325.01

1 ψψ ++=C



259


0==

kNmψ

77.10 1 =⇒= Cψ According to EN 1993-1-1-AN France; Chapter 3.2; Table 1










( )4

2fz

w

thII

−×=



( ) 624

093627638294

7.1026046.6025866 mmmmmmmmI w =−×

=



260

( )( )

kNmNmm

mmNmmmmN

mmmmNmmmmmm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

10.4266.426102243

20.15251.158170677.146.6025866/210000

97.149294/80800281046.6025866

09362763829

281046.6025866/21000077.1

²²

²²

422

422

4

6

2

422

1

==

=××=××

××+×

×××

×=××××

+×××

×=

π

ππ

π

569.06.426102243

/27518.501177 23, =

×==

NmmmmNmm

MfW

cr

yyplLTλ


1²²

1≤

−+=

LTLTLT

LTλφφ

χ (6.56)

( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=


2733.1150260

≤==mmmm

bh


49.0=LTα

( )[ ] ( )[ ] 752.0²569.02.0569.049.015.0²2.015.0 =+−×+×=+−×+×= LTLTLTLT λλαφ

1804.0²569.0²752.0752.0

1²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ


261


The internal factor yyk corresponding to a Class 1 section will be calculated according to Annex A, Table a.1, and will

be calculated separately for the two column parts separate by the middle torsional lateral restraint:


yy

ycr

Ed

ymLTmyyy C

NNCCk 1̀

1,

×−

××=µ

ycr

Edy

ycr

Ed

y

NN

NN

,

,

1

1

×−

−

=χ

µ


kNNEd 328=

kNNl

IEN

fy

yycr 33.380295.3802327

²²

, ==××

=π


985.0

95.3802327328000839.01

95.38023273280001

1

1

,

, =×−

−=

×−

−=

NN

NN

NN

NN

ycr

Edy

ycr

Ed

y

χµ


262



0

,0

cr

yypl

MfW ×

=λ

According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2 318.501177 mmWy =


( )( )

kNmNmm

mmNmmmmN

mmmmNmmmm

mm

mmmmmmN

IEIGL

II

LIECM

z

t

z

wzcr

73.2408.240735730

20.15251.1581706146.6025866/210000

97.149294/80800281046.6025866

88.19351706542

281046.6025866/2100001

²²

²²

422

422

4

6

2

422

1

==

=××=××

××+×

×××

×=××××

+×××

×=

π

ππ

π

757.08.240735730

/27518.501177 23,

0 =×

==Nmm

mmNmmM

fW

cr

yyplλ


263


,,1 1120.0

−×

−××

TFcr

Ed

zcr

Ed

NN

NNC term:

Where:


××+××=

²1

,

2

0,

Tcr

wtTcr L

IEIGI

N π


44420 1.6396915846.602586664.57943291 mmmmmmIIzAIII zygzy =+=+=×++=


Working inertial moment: 688.19351706542 mmIw =


mL Tcr 81.2, =

( )kNN

mmmmmmNmmmmN

mmmmN Tcr

63.280668.2806625²2810

88.19351706542/21000097.149294/808001.63969158

06.4904 62242

4

2

,

==

=

××+××=

π

NNEd 328000=

NNN TcrTFcr 68.2806625,, ==

kNNl

IENfz

zzcr 71.3158151.1581706

²²

, ==××

=π




183.0

68.28066253280001

51.15817063280001120.01120.0 44

,,1

=

=

−×

−××=

−×

−××

NN

NN

NN

NNC

TFcr

Ed

zcr

Ed


264

Therefore:


( )

≥

−×

−

×=

=

×+

××−+=

⇒

=

−⋅

−⋅⋅>=

=

−×

−××

=

1

11

11

183.01120.0757.0

183.01120.0

757.0

,,

2

0,

0,0,

4

,,10

4

,,1

0

Tcr

Ed

zcr

Ed

LTmymLT

mzmz

LTy

LTymymymy

TFcr

Ed

zcr

Ed

TFcr

Ed

zcr

Ed

NN

NN

aCC

CC

aa

CCC

NN

NNC

NN

NNC

ε

ε

λ

λ



( )LTy

LTymymymy a

aCCC

×+

××−+=

ε

ε

11 0,0,

yeff

eff

Ed

Edyy W

AN

M

,

, ×=ξ


677.163.445717

06.4904328000

10503

26

,

, =××

=×=mm

mmN

NmmWA

NM

yeff

eff

Ed

Edyyξ

1997.064.57943291

97.14929411 4

4

≈=−=−=mm

mmIIa

y

tLT



265

The bending moment in null at one end of the column, therefore: 0=ψ

( )ycr

Ed

ycr

Edmy N

NNNC

,,0, 33.036.079.033.036.021.079.0 ××−=×−×+×+= ψψ

Where:

kNNl

IEN

fy

yycr 33.380295.3802327

²²

, ==××

=π


NNEd 328000=

780.095.3802327

32800033.036.079.00, =××−=N

NCmy

( ) ( ) 904.01677.11

1677.1780.01780.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ




1

11,,

2 ≥

−×

−

×=

Tcr

Ed

zcr

Ed

LTmymLT

NN

NN

aCC

- the myC term used for mLTC calculation, must be recalculated for the corresponding column part (in this case the top column

part)

- this being the case, myC will be calculated using 5.0=ψ :

( )LTy

LTymymymy a

aCCC

ξ

ξ

+−+=

11 0,0,

( ) ( ) 900.095.3802327

23800033.05.036.05.021.079.033.036.021.079.0,

0, =×−×+×+=×−×+×+=N

NNNC

ycr

Edmy ψψ

677.163.445717

06.4904328000

10503

26

,

, =××

=×=mm

mmN

NmmWA

NM

yeff

eff

Ed


( ) ( ) 956.01677.11

1677.1900.01900.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ


266

089.1

1

089.1

68.28066253280001

51.15817063280001

997.0956.0

11

2

,,

2

=⇒

≥

=

−×

−

×=

=

−×

−

×=

mLT

mLT

Tcr

Ed

zcr

Ed

LTmymLT

C

CN

NN

N

NN

NN

aCC

The yyC coefficient is defined according to the Table A.1, Auxiliary terms:

ypl

yelLTplmy

ymy

yyyy W

WbnC

wC

wwC

,

,2

max2

max2 6.16.12)1(1 ≥

−×

××−××−×−+=

−−

λλ

Rdzpl

Edz

RdyplLT

EdyLTLT M

MM

Mab

,,

,

,,

,2

05.0 ××

×××=−

χλ

- LTb must be calculated separately for each of the two column parts, depending on 0

−

λ and LTχ :

757.0661.031.110 =×=×=−−

LTC λλ (for the top part of the column)


267

041.0/27596.123381

10000000/27518.501177749.0

50000000757.0997.05.0

5.05.0

23232

,

,

,

,2

0

,,

,

,,

,2

0

=

=×

×××

×××=

××

×××××=×

××××=

−−

mmNmmNmm

mmNmmNmm

fWM

fWM

aMM

MM

abyzpl

Edz

yyplLT

EdyLT

Rdzpl

Edz

RdyplLT

EdyLTLT χ

λχ

λ

5.1124.163.44571718.501177

3

3

,

, ≤===mmmm

WW

wyel

yply

243.0

1/27506.4904

32800022

1

=×

==mmNmm

NNNn

M

Rk

Edpl

γ

( ) 923.0923.0;5956.0max;maxmax ==

=

−−

zy λλλ

993.0041.0243.0²923.0²904.0124.16.1923.0²904.0

124.16.12)1124.1(1 =

−×

××−××−×−+=yyC

993.0889.018.50117763.445717

993.0

,

,

3

3

,

, =⇒

≥

==

=

yy

ypl

yelyy

ypl

yel

yy

C

WW

C

mmmm

WW

C

kNNl

IEN

fy

yycr 33.380295.3802327

²²

, ==××

=π


Therefore, the yyk term corresponding to the top part of the column will be:

069.1993.01

95.38023273280001

985.0089.1904.01̀

1,

=×−

××=×−

××=

NNC

NNCCk

yy

ycr

Ed

ymLTmyyy

µ


268


yy

ycr

Ed

ymLTmyyy C

NNCCk 1̀

1,

×−

××=µ

- the terms: mLTC ; LTb and yyC must be recalculated:

1

11,,

2 ≥

−×

−

×=

Tcr

Ed

zcr

Ed

LTmymLT

NN

NN

aCC

- the myC term must be calculated corresponding to the bottom part of the column (with )0=ψ :

( ) ( ) 780.095.3802327

23800033.036.079.033.036.021.079.0,

0, =×−×+=×−×+×+=N

NNNC

ycr

Edmy ψψ

839.063.445717

06.4904328000

10253

26

,

, =××

=×=mm

mmN

NmmWA

NM

yeff

eff

Ed

Edyyξ

( ) ( ) 885.01839.01

1839.0780.01780.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ

ξ

1

1

933.0

68.28066253280001

51.15817063280001

997.0885.0

11

2

,,

2

=⇒

≥

=

−×

−

×=

=

−×

−

×=

mLT

mLT

Tcr

Ed

zcr

Ed

LTmymLT

C

CN

NN

N

NN

NN

aCC


269

ypl

yelLTplmy

ymy

yyyy W

WbnC

wC

wwC

,

,2

max2

max2 6.16.12)1(1 ≥

−×

××−××−×−+=

−−

λλ

Rdzpl

Edz

RdyplLT

EdyLTLT M

MM

Mab

,,

,

,,

,2

05.0 ××

×××=−

χλ

-the LTb must be calculated separately for each of the two column parts, depending of 0

−

λ and LTχ :

757.08.240735730

/27518.501177 23,

0 =×

==Nmm

mmNmmM

fW

cr

yyplλ (for the bottom part of the column)

1804.0²569.0²752.0752.0

1²²

1≤=

−+=

−+=

LTLTLT

LTλφφ

χ (for the bottom part of the column)

0095.0/27596.123381

5000000/27518.501177804.0

25000000757.0997.05.0

5.05.0

23232

,

,

,

,2

0

,,

,

,,

,2

0

=

=×

×××

×××=

××

×××××=×

××××=

−−

mmNmmNmm

mmNmmNmm

fWM

fWM

aMM

MM

abyzpl

Edz

yyplLT

EdyLT

Rdzpl

Edz

RdyplLT

EdyLTLT χ

λχ

λ

5.1124.163.44571718.501177

3

3

,

, ≤===mmmm

WW

wyel

yply

243.0

1/27506.4904

32800022

1

=×

==mmNmm

NNNn

M

Rk

Edpl

γ

( ) 923.0923.0;5956.0max;maxmax ==

=

−−

zy λλλ

997.00095.0243.0²923.0²904.0124.16.1923.0²904.0

124.16.12)1124.1(1 =

−×

××−××−×−+=yyC

997.0889.018.50117763.445717

997.0

,

,

3

3

,

, =⇒

≥

==

=

yy

ypl

yelyy

ypl

yel

yy

C

WW

C

mmmm

WW

C

kNNl

IEN

fy

yycr 33.380295.3802327

²²

, ==××

=π



270

Therefore the yyk term corresponding to the bottom part of the column will be:

977.0997.01

95.38023273280001

985.01904.01̀

1,

=×−

××=×−

××=

NNC

NNCCk

yy

ycr

Ed

ymLTmyyy

µ

Note: The software does not give the results of the lower section because it is not the most requested segment.


y

z

yz

zcr

Ed

ymzyz w

wC

NNCk ×××

−×= 6.01

1,

µ

-the mzC ter will be considered for the entire column length (with )0=ψ :

( ) ( ) 765.051.1581706

32800033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz

985.0

95.3802327328000839.01

95.38023273280001

1

1

,

, =×−

−=

×−

−

=

NN

NN

NN

NN

ycr

Edy

ycr

Ed

y

χµ


Nl

IENfz

zzcr 51.1581706

²²

, =××

=π


−×

×

×−×−+=−

LTplz

mzzyz cn

wCwC 5

2

max2

142)1(1 λ

Rdyplltmy

Edy

z

LTLT MCM

ac,,

,4

2

0

510

×××

+××=

−

−

χλ

λ

997.064.57943291

97.14929411 4

4

=−=−=mm

mmIIa

y

tLT (previously calculated)

757.08.240735730

/27518.501177 23,

0 =×

==Nmm

mmNmmM

fW

cr

yyplλ (previously calculated)

923.051.1581706

/27506.4904 22

,

=×

=×

=N

mmNmmN

fA

zcr

yzλ (previously calculated)

NmM Edy 50000, =

5.15.1

5.1536.189.8034496.123381

3

3

,

,

=⇒

≤

≤===z

z

zel

zplz w

wmmmm

WW

w


271

5.1124.163.44571718.501177

3

3

,

, ≤===mmmm

WW

wyel

yply

- the myC term will be considered separately for each column part:


( ) ( ) 956.01677.11

1677.1900.01900.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ


1749.0²661.0²831.0831.0

1²²

1≤=

−+=

−+=

LTLTLT

LTλφφ



506.05.137823724749.0956.0

50000000923.05

²757.0997.010

510

4

,,

,4

2

0

=××

×+

××=

=××

×+

××=−

−

NmmNmm

MCM

acRdyplltmy

Edy

z

LTLT χλ

λ

Nl

IENfz

zzcr 51.1581706

²²

, =××

=π

(previously calculated

( ) ( ) 765.051.1581706

32800033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz

( ) 923.0923.0;5956.0max;maxmax ==

=

−−

zy λλλ

243.0

1

==

M

Rk

Edpl N

Nn

γ


- Therefore:

( ) 878.0506.0243.05.1

923.0765.014215.11

142)1(1

5

22

5

2

max2

=

−×

××−×−+=

=

−×

×

×−×−+=

−

LTplz

mzzyz cn

wCwC λ

750.0124.1

5.16.0878.01

51.15817063280001

985.0765.06.01

1,

=×××−

×=×××−

×=

NNw

wC

NNCk

y

z

yz

zcr

Ed

ymzyz

µ


272


( ) ( ) 885.01839.01

1839.0780.01780.01

1 0,0, =×+

××−+=

×+

××−+=

LTy

LTymymymy a

aCCC

ξ


1804.0²569.0²752.0752.0

1²²

1≤=

−+=

−+=

LTLTLT

LTλφφ



254.05.137823724804.0885.0

25000000923.05

²757.0997.010

510

4

,,

,4

2

0

=××

×+

××=

=××

×+

××=−

−

NmmNmm

MCM

acRdyplltmy

Edy

z

LTLT χλ

λ

Nl

IENfz

zzcr 51.1581706

²²

, =××

=π

(previously calculated

( ) ( ) 765.051.1581706

32800033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz

( ) 923.0923.0;5956.0max;maxmax ==

=

−−

zy λλλ

243.0

1

==

M

Rk

Edpl N

Nn

γ


-Therefore:

( ) 0043.1254.0243.05.1

923.0765.014215.11

142)1(1

5

22

5

2

max2

=

−×

××−×−+=

=

−×

×

×−×−+=

−

LTplz

mzzyz cn

wCwC λ

656.0124.1

5.16.00043.11

51.15817063280001

985.0765.06.01

1,

=×××−

×=×××−

×=

NNw

wC

NNCk

y

z

yz

zcr

Ed

ymzyz

µ



273



y

z

zy

ycr

Ed

zmLTmyzy w

wC

NNCCk ×××

−××= 6.01

1,

µ

902.0

51.1581706328000586.01

51.15817063280001

1

1

,

, =×−

−=

×−

−=

NN

NN

NN

NN

zcr

Edz

zcr

Ed

z

χµ


ypl

yel

z

yLTpl

y

myyzy W

Www

dnw

CwC

,

,5

2

max2

6.0142)1(1 ××≥

−×

×

×−×−+=

−

λ



-in order to calculate the LTd term, the terms myC and mzC must be recalculated for each column part;

-the term mzC must be recalculated for the top column part only, using 5.0=ψ :

( )

( ) 908.051.1581706

32800033.05.036.05.021.079.0

33.05.036.05.021.079.0,

0,

=×−×+×+=

=×−×+×+==

NN

NNCC

zcr

Edmzmz

301.033930039908.0

100000005.137823724749.0956.0

50000000923.01.0

757.0997.02

1.02

4

,,

,

,,

,4

0

=

=×

×××

×+

××=

=×

×××

×+

××=−

−

NmmNmm

NmmNmm

MCM

MCM

adRdzplmz

Edz

RdyplLTmy

Edy

z

LTLT χλ

λ


274

859.0

6.0

616.018.50117763.445717

124.15.16.06.0

859.0301.0243.0124.1

847.1904.0142)1124.1(1

142)1(1

,

,

3

3

,

,

5

22

5

2

max2

=⇒

××≥

=××=××

=

−×

××−×−+=

=

−×

×

×−×−+=

−

zy

ypl

yel

z

yzy

ypl

yel

z

y

LTply

myyzy

C

WW

ww

C

mmmm

WW

ww

dnw

CwC

λ

- for the calculation of the zyC term, myC will be used for the entire column and LTd will be used for the top column part:

588.05.1

124.16.0859.01

95.38023273280001

902.0089.1904.06.01

1,

=×××−

××=×××−

××=

NNw

wC

NNCCk

z

y

zy

ycr

Ed

zmLTmyzy

µ


y

z

zy

ycr

Ed

zmLTmyzy w

wC

NNCCk ×××

−××= 6.01

1,

µ

902.0

51.1581706328000586.01

51.15817063280001

1

1

,

, =×−

−=

×−

−=

NN

NN

NN

NN

zcr

Edz

zcr

Ed

z

χµ


ypl

yel

z

yLTpl

y

myyzy W

Www

dnw

CwC

,

,5

2

max2

6.0142)1(1 ××≥

−×

×

×−×−+=

−

λ



- in order to calculate the LTd term, the terms myC and mzC must be recalculated for each column part;

- the term mzC must be recalculated for the top column part only, using 0=ψ :

( ) ( ) 765.051.1581706

32800033.036.079.033.036.079.0,

0, =×−×+=×−×+==N

NNNCC

zcr

Edmzmz


275

090.033930039765.0

50000005.137823724804.0885.0

25000000923.01.0

757.0997.02

1.02

4

,,

,

,,

,4

0

=

=×

×××

×+

××=

=×

×××

×+

××=−

−

NmmNmm

NmmNmm

MCM

MCM

adRdzplmz

Edz

RdyplLTmy

Edy

z

LTLT χλ

λ

( ) 923.0923.0;5956.0max;maxmax ==

=

−−

zy λλλ

892.0

6.0

462.018.50117763.445717

5.1124.16.06.0

892.0090.0243.0124.1

923.0885.0142)1124.1(1

142)1(1

,

,

3

3

,

,

5

22

5

2

max2

=⇒

××≥

=××=××

=

−×

××−×−+=

=

−×

×

×−×−+=

−

zy

ypl

yel

z

yzy

ypl

yel

z

y

LTply

myyzy

C

WW

ww

C

mmmm

WW

ww

dnw

CwC

λ

- for the calculation of the

zyC term, myC will be used for the entire column and LTd will be used for the top column part:

566.05.1

124.16.0892.01

95.38023273280001

902.0089.1904.06.01

1,

=×××−

××=×××−

××=

NNw

wC

NNCCk

z

y

zy

ycr

Ed

zmLTmyzy

µ




zz

zcr

Ed

zmzzz C

NNCk 1

1,

×−

×=µ

zpl

zelLTplmz

zmz

zzzz W

WenC

wC

wwC

,

,max

2max

2 6.16.12)1(1 ≥

−×

××−××−×−+= λλ


- in calculating the LTe , the myC term must be used accordingly with the corresponding column part

787.0/27518.501177749.0956.0

50000000923.01.0

757.0997.07.1

1.07.1

234

,,

,4

0

=×××

×+

××=

=××

×+

××=−

−

mmNmmNmm

MCM

aeRdyplltmy

Edy

z

LTLT χλ

λ


276


zzC term, mzC will be used for the entire column and LTe will be used for the top column part:

013.1651.0

96.12338189.80344

013.1243.0787.0847.1765.05.16.1847.1765.0

5.16.12)15.1(1

6.16.12)1(1

,

,

3

3

,

,

222

2max

2max

2

=⇒

≥

==

=×

−

××−××−×−+=

=×

−

××−××−×−+=

zz

zpl

zelzz

zpl

zel

plLTmzz

mzz

zzz

C

WW

C

mmmm

WW

neCw

Cw

wC λλ

860.0013.11

51.15817063280001

902.0765.01

1,

=×−

×=×−

×=

NNC

NNCk

zz

zcr

Ed

zmzzz

µ


zz

zcr

Ed

zmzzz C

NNCk 1

1,

×−

×=µ

zpl

zelLTplmz

zmz

zzzz W

WenC

wC

wwC

,

,max

2max

2 6.16.12)1(1 ≥

−×

××−××−×−+= λλ


- in calculating the LTe , the myC term must be used according to the corresponding column part

396.0/27518.501177804.0885.0

25000000923.01.0

757.0997.07.1

1.07.1

234

,,

,4

0

=×××

×+

××=

=××

×+

××=−

−

mmNmmNmm

MCM

aeRdyplltmy

Edy

z

LTLT χλ

λ


zzC term, mzC will be used for the entire column and LTe will be used for the top column part:

( ) 923.0923.0;5956.0max;maxmax ==

=

−−

zy λλλ

060.1651.0

96.12338189.80344

060.1243.0396.0923.0765.05.16.1923.0765.0

5.16.12)15.1(1

6.16.12)1(1

,

,

3

3

,

,

222

2max

2max

2

=⇒

≥

==

=×

−

××−××−×−+=

=×

−

××−××−×−+=

zz

zpl

zelzz

zpl

zel

plLTmzz

mzz

zzz

C

WW

C

mmmm

WW

neCw

Cw

wC λλ


277

821.0060.11

51.15817063280001

902.0765.01

1,

=×−

×=×−

×=

NNC

NNCk

zz

zcr

Ed

zmzzz

µ



∆+×+

×

∆+×+

×

∆+×+

×

∆+×+

×

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rkz

RdzEdzzz

M

RkyLT

RdyEdyzy

M

Rkz

Ed

M

Rkz

RdzEdzyz

M

RkyLT

RdyEdyyy

M

Rky

Ed

MMM

kMMM

kNN

MMM

kMMM

kNN

γγχ

γχ

γγχ

γχ

iyRk AfN ×=


93.025.027.041.0

1/27596.123381

1010860.0

1/27518.501177804.0

1050588.0

1/27506.4904586.0

328000

02.121.052.029.0

1/27596.123381

1010750.0

1/27518.501177749.0

1050069.1

1/27506.4904839.0

328000

23

6

23

6

22

23

6

23

6

22

=++=×

××+

+×

×

××+

××

=++=×

××+

+×

×

××+

××

mmNmmNmm

mmNmmNmm

mmNmmN

mmNmmNmm

mmNmmNmm

mmNmmN


278


67.012.013.042.0

1/27596.123381

105821.0

1/27518.501177804.0

1025566.0

1/27506.4904586.0

328000

30.197.022.029.0

1/27596.123381

105656.0

1/27518.501177804.0

1025977.0

1/27506.4904839.0

328000

23

6

23

6

22

23

6

23

6

22

=++=×

××+

+×

×

××+

××

=++=×

××+

+×

×

××+

××

mmNmmNmm

mmNmmNmm

mmNmmN

mmNmmNmm

mmNmmNmm

mmNmmN




279



yχ



zχ


280



yyk



yzk


281



zyk



zzk


282



SNy



SMyy


283



SMyz



SNz


284



SMzy



SMzz


285



C1

The elastic moment for lateral-torsional buckling calculation

The elastic moment for lateral-torsional buckling calculation Mcr


286



LTχ


287

13.30.2.10Reference results a) for the top part of the column:



0.839


0.586


1.069


0.750


0.588


0.860

SNy


0.29


0.52


0.21

SNz


0.41


0.27


0.25 C1 Coefficient that depends of several parameters as: section properties; support

conditions; moment diagram allure 1.77




288




0.839


0.586


0.977


0.656


0.566


0.821

SNy


0.29


0.22


0.97

SNz


0.42


0.13


0.12 C1 Coefficient that depends of several parameters as: section properties;

support conditions; moment diagram allure 1.77




13.30.3Calculated results Result name Result description Value Error Xy Coefficient corresponding to non-dimensional slenderness

by Y axis 0.839285 adim

0.0340 %

Xz Coefficient corresponding to non-dimensional slenderness by the Z axis

0.585533 adim

-0.0797 %

Kyy Internal coefficient kyy 1.07027 adim

0.1188 %

Kyy Internal coefficient kyy (bottom) 0.954475 adim

-2.3055 %

Kyz Internal coefficient kyz 0.750217 adim

0.0289 %

Kyz Internal coefficient kyz (bottom) 0.656481 adim

0.0733 %

Kzy Internal coefficient kzy 0.593445 adim

0.9260 %

Kzy Internal coefficient kzy (bottom) 0.508819 adim

-0.0356 %

Kzz Internal coefficient kzz 0.860237 adim

0.0276 %

Kzz Internal coefficient kzz (bottom) 0.821717 adim

0.0873 %


289

13.31 EC3 Test 31: Verifying IPE450 column fixed on base subjected to axial compression and bending moment, both applied on top

Test ID: 5731

Test status: Passed

13.31.1Description The test verifies an IPE450 column made of S275 steel.

The column is subjected to a -1000kN compression effort and a 200kNm bending moment by the Y axis.



290

13.32 EC3 Test 15: Verifying a rectangular hollow section column subjected to bending and axial efforts

Test ID: 5735

Test status: Passed

13.32.1Description Verifies a rectangular hollow section column made of S235 steel subjected to bending and axial efforts.


13.32.2Background Verifies the adequacy of a rectangular hollow section column made of S235 steel to resist bending and axial efforts. The name of the cross-section is RC3020100 and it can be found in the Advance Design OTUA library. The column is fixed at its base and it is subjected to a uniformly distributed load over its height and a punctual axial load applied on the top. The dead load will be neglected.



■ Exploitation loadings (category A), Q: Fz = - 500 000 N, Fx = 5 000 N/ml,




291

Units

Metric System

Geometry


■ Height: h = 300 mm, ■ Width: b = 200 mm, ■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm, ■ Column height: L = 5000 mm, ■ Section area:A = 9490 mm2, ■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,





Boundary conditions



■ Inner: None.

Loading


■ External: Point load at z = 5.0: Fz = - 500 000 N, Uniformly distributed load: q = Fx = 5 000 N/ml

■ Internal: None.


292

13.32.2.2Reference results for calculating the column subjected to bending and axial force In order to verify the steel column subjected to bending and axial force, the design resistance for uniform compression (Nc,Rd) and also the design plastic moment resistance (Mpl,Rd) have to be compared with the design values of the corresponding efforts.

The design resistance for uniform compression is verified considering the relationship (6.9) from chapter 6.2.4 (EN 1993-1-1), while the design plastic moment resistance is verified considering the criterion (6.12) from chapter 6.2.5 (EN 1993-1-1).

Before starting the above verifications, the cross-section class has to be determined.

Cross section class


In this case, the stresses distribution is like in the picture below:



293


1510

10215220022=

×−×−=

×−×−=

mmmmmmmm

ttrb

tc

0.1235==

yfε

Therefore:

333315 =≤= εtc


The left/right web is subjected to bending stresses. Their class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to bending and compression). It is also necessary to determine which portion of the web is compressed (α). α is determined considering the stresses distribution on the web.

5.0832.063.26

1321

>=→=−

αα

αMPa

MPa

2510

10215230022=

×−×−=

×−×−=

mmmmmmmm

ttrh

tc

0.1235==

yfε

Therefore:

34.40113

39625 =−

≤=α

εtc




294

Verifying the design resistance for uniform compression

The design resistance for uniform compression, for Class 1 cross-section, is determined with formula (6.10) from EN 1993-1-1:2001.

NMPammfAN

M

yRdc 2230150

0.12359490 2

0, =

×=

×=

γ

The verification of the design resistance for uniform compression is done with relationship (6.9) from EN 1993-1-1. The corresponding work ratio is:

Work ratio = %42.221002230150500000100100

,,

=×=×=×N

NN

FNN

Rdc

z

Rdc

Ed

Verifying the design plastic moment resistance

The design plastic moment resistance, for Class 1 cross-section, is determined with formula (6.13) from EN 1993-1-1:2001.

NmmMPammfWMM

M

yyplRdplRdc 224660000

0.1235956000 3

0

,,, =

×=

×==

γ


Work ratio = %82.27100224660000

250005000/5

1002100,,

=×××

=×××

=×Nmm

mmmmmmN

M

LLq

MM

RdcRdc

Ed




295



Column subjected to bending and axial efforts Work ratio – Fx


Column subjected to bending and axial efforts Work ratio – Oblique


296


Result name Result description Reference value Work ratio – Fx Compression resistance work ratio [%] 22.42 %

Work ratio – Oblique Work ratio of the design resistance for bending 27.82 %

13.32.3Calculated results Result name Result description Value Error Work ratio - Fx Compression resistance work ratio 22.42 % 0.0000 % Work ratio - Oblique



297

13.33 EC3 Test 33: Verifying UPN300 simple supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and punctual vertical force by Z axis

Test ID: 5733

Test status: Passed

13.33.1Description The test verifies an upn300 beam made of S235 steel.

The beam is subjected to 20 kN compression force, 50 kN PUNCTUAL vertical load applied to the middle of the beam and 5kN/m linear uniform horizontal load.


13.34 EC3 Test 35: Verifying C310x30.8 class 4 cantilever, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load applied on the free end

Test ID: 5737

Test status: Passed

13.34.1Description The test verifies a C310x30.8 beam made of S355 steel.

The beam is subjected to 3.00 kN compression force, 1.80 kN punctual vertical load applied on the free end of the beam and 1.2.kN/m linear uniform horizontal load.


13.35 EC3 Test 37: Verifying RHS350x150x8.5H class 3 column, loaded with centric compression, punctual lateral load and bending moment, all applied to the top of the column

Test ID: 5739

Test status: Passed

13.35.1Description The test verifies a RHS350x150x8.5H column made of S275 steel.

The column is subjected to 680 kN compression force, 5 kN horizontal load applied on Y axis direction and 200 kNm bending moment after the Y axis. All loads are applied on the top of the column.


13.36 EC3 Test 30: Verifying IPE300 beam, simply supported, loaded with centric compression and uniform linear efforts by Y and Z axis

Test ID: 5730

Test status: Passed

13.36.1Description The test verifies an IPE 300 beam made of

The beam is subjected to a 20kN compression effort, a -10kN/m uniform linear effort applied vertically and a -5kN/m linear uniform load applied horizontal.



298

13.37 EC3 Test 34: Verifying C310x30.8 class 3beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle

Test ID: 5734

Test status: Passed

13.37.1Description The test verifies an C310x30.8 beam made of S235 steel.

The beam is subjected to a 12 kN compression force, 8 kN PUNCTUAL vertical load applied to the middle of the beam and 1.5 kN/m linear uniform horizontal load.


13.38 EC3 Test 36: Verifying RHS300x150x9H class 1 simply supported beam, loaded with centric compression, uniform linear horizontal efforts and a vertical punctual load in the middle

Test ID: 5738

Test status: Passed

13.38.1Description The test verifies an RHS300x150x9H beam made of S275 steel.

The beam is subjected to 12 kN axial compression force, 7 kN punctual vertical load applied to the middle of the beam and 3 kN/m linear uniform horizontal load.



299

13.39 EC3 Test 14: Verifying the bending resistance of a rectangular hollow section column made of S235 steel

Test ID: 5728

Test status: Passed

13.39.1Description Verifies the design resistance for bending of a rectangular hollow section column made of S235 steel.

The verification is made according to Eurocode 3 (EN 1993-1-1) French annex.

13.39.2Background Verifies the adequacy of a rectangular hollow section column made of S235 steel to resist bending efforts. Verification of the design resistance for bending at ultimate limit state is realised. The name of the cross-section is RC3020100 and can be found in the Advance Design OTUA library. The column is fixed at its base and it is subjected to a punctual horizontal load applied to the middle height (50 000 N). The dead load will be neglected.



■ Exploitation loadings (category A), Q: Fx = 50 000 N,




300

Units

Metric System

Geometry


■ Height: h = 300 mm, ■ Width: b = 200 mm, ■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm, ■ Column height: L = 5000 mm, ■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,





Boundary conditions



■ Inner: None.

Loading


■ External: Point load at z = 2.5: V= Fx = 50 000 N,

■ Internal: None.


301

13.39.2.2Reference results for calculating the design resistance for bending Before calculating the design resistance for bending, the cross section class has to be determined.

Cross section class


In this case, the column is subjected to bending efforts, therefore the stresses distribution is like in the picture below:



302


1510

10215220022=

×−×−=

×−×−=

mmmmmmmm

ttrb

tc

0.1235==

yfε

Therefore:

333315 =≤= εtc


The left/right web is subjected to bending stresses. Their class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to bending):

2510

10215230022=

×−×−=

×−×−=

mmmmmmmm

ttrh

tc

0.1235==

yfε

Therefore:

727225 =≤= εtc




303


The design resistance for bending, for Class 1 cross-section, is determined with formula (6.13) from EN 1993-1-1:2001.

NmmMPammfWM

M

yyplRdc 224660000

0.1235956000 3

0

,, =

×=

×=

γ

Work ratio


Work ratio = %64.55100224660000

2500050000

1002100,,

=××

=××

=×Nmm

mmN

M

LV

MM

RdcRdc

Ed




304



Column subjected to a punctual horizontal load applied to the middle height Work ratio – Oblique


Result name Result description Reference value Work ratio - Oblique Work ratio of the design resistance for bending [%] 55.64 %




305

13.40 EC3 Test 32: Verifying IPE600 simple supported beam, loaded with centric compression and uniform linear efforts by Y and Z axis

Test ID: 5732

Test status: Passed

13.40.1Description The test verifies an IPE600 beam made of S275 steel.

The beam is subjected to a -3700kN compression force, a -10kN/m linear uniform vertical load and a -5kN/m linear uniform horizontal load.



306

13.41 EC3 Test 16: Verifying a simply supported rectangular hollow section beam subjected to biaxial bending

Test ID: 5736

Test status: Passed

13.41.1Description Verifies a simply supported rectangular hollow section beam made of S235 steel subjected to biaxial bending.


13.41.2Background Verifies the adequacy of a rectangular hollow section beam made of S235 steel to resist bi-axial bending efforts. The name of the cross-section is RC3020100 and it can be found in the Advance Design OTUA library. The beam is simply supported and it is subjected to uniformly distributed loads over its length. The dead load will be neglected.


13.41.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear; ■ Element type: linear. The following load case and load combination are used:

■ Exploitation loadings (category A), Q: Fz = - 10 000 N/ml, Fy = 10 000 N/ml,



Units

Metric System

Geometry


■ Height: h = 300 mm, ■ Width: b = 200 mm, ■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm,


307

■ Beam length: L = 5000 mm, ■ Section area: A = 9490 mm2, ■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3, ■ Plastic section modulus about z-z axis: Wpl,z = 721000 mm3,





Boundary conditions


■ Outer: Support at start point (z = 0) restrained in translation along X, Y and Z axis, Support at end point (z = 5.00) restrained in translation along Y and Z axis, and restrained in rotation

about the X axis. ■ Inner: None.

Loading


■ External: Uniformly distributed load: q1 = Fz = -10 000 N/ml Uniformly distributed load: q2 = Fy = 10 000 N/ml

■ Internal: None.


308

13.41.2.2Reference results for calculating the beam subjected to bi-axial bending In order to verify the steel beam subjected to bi-axial bending, the criterion (6.41) from chapter 6.2.9.1 (EN 1993-1-1) has to be used.

Before verifying this criterion, the cross-section class has to be determined.

Cross section class


In this case, the stresses distribution is like in the picture below:



309

Taking into account that the entire cross-section is subjected to bending stresses, its class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to bending).

2510

10215230022=

×−×−=

×−×−=

mmmmmmmm

ttrb

tc

0.1235==

yfε

Therefore:

727225 =≤= εtc

This means that the left/right web is Class 1. As the dimensions for top/bottom wing are smaller than the left/right web, they will be also classified as Class 1.


Determining the design plastic moment resistance

Before verifying for bi-axial bending a rectangular structural hollow section of uniform thickness, the design plastic moment resistance reduced due to the axial force (MN,Rd) needs to be determined. Its determination has to be made about 2 axes (according to the bending efforts) and it will be done with formulae (6.39) and (6.40) from EN 1993-1-1. Other terms involved in calculation have to be determined: aw, af, n.

■ Ratio of design normal force to design plastic resistance to normal forces, n:

→=Rdpl

Ed

NNn

,

as the beam is not subjected to axial efforts n = 0.

■ Determination of aw for hollow section:

5.05.02=→≤

××−= ww a

AtbAa


310

■ Determination of af for hollow section:

3678.05.02=→≤

××−= ff a

AthAa

■ Determination of aw for hollow section:

5.05.02=→≤

××−= ww a

AtbAa

■ Determination of design plastic moment resistance (about y-y axis) reduced due to the axial force, MN,y,Rd:

( )( ) Rdypl

w

RdyplRdyN M

anM

M ,,,,

,, 5.011

≤×−

−×=

In order to fulfill the above relationship MN,y,Rd must be equal to Mpl,y,Rd.

NmmMPammfWMM

M

yyplRdyplRdyN

43

0

,,,,, 1022466

0.1235956000

×=×

=×

==γ

■ Determination of design plastic moment resistance (about z-z axis) reduced due to the axial force, MN,z,Rd:

( )( ) Rdzpl

f

RdzplRdzN M

anM

M ,,,,

,, 5.011

≤×−

−×=

In order to fulfill the above relationship MN,z,Rd must be equal to Mpl,z,Rd.

NmmMPammfWMM

M

yzplRdzplRdzN

43

0

,,,,, 105.16943

0.1235721000

×=×

=×

==γ

Verifying for bi-axial bending

Criterion (6.41) from EN 1993-1-1 has to be fulfilled:

0.1,,

,

,,

, ≤

+

βα

RdzN

Edz

RdyN

Edy

MM

MM

■ Determination of constants α and β for rectangular hollow section:

66.1613.1166.1

2 ==→≤×−

== βαβαn

■ Determination of design bending moments (My,Ed and Mz,Ed) at the middle of the beam:

NmmmmmmNLqM Edy

4222

, 10312585000/10

81

×=×

=×

=

NmmmmmmNLqM Edz

4222

, 10312585000/10

82

×=×

=×

=

■ Verifying criterion (6.41) from EN 1993-1-1:

0.1098.00604.00378.0105.16943

1031251022466

10312566.1

4

466.1

4

4

≤=+=

×

×+

×

×Nmm

NmmNmm

Nmm

Work ratio = %8.9100

0.1098.0

=×


311

Verifying for simple bending about y-y axis

For class 1 cross-section, design resistance for simple bending about y-y axis is verified using the criterion (6.12) from EN 1993-1-1:

0.1139.01022466

1031254

4

,,

, ≤=×

×=

NmmNmm

MM

Rdypl

Edy


0.1139.0

=×

Verifying for simple bending about z-z axis

For class 1 cross-section, design resistance for simple bending about z-z axis is verified using the criterion (6.12) from EN 1993-1-1:

0.11844.0105.16943

1031254

4

,,

, ≤=×

×=

NmmNmm

MM

Rdzpl

Edz


0.11844.0

=×

As this work ratio is bigger than the others, we can consider it as reference.





Beam subjected to bending efforts Work ratio – Oblique


312


Result name Result description Reference value Work ratio - Oblique Work ratio of the design resistance for biaxial bending 18.44 %


Work ratio of the design resistance for biaxial bending 18.4437 % 0.2375 %


313

13.42 EC3 Test 18: Verifying a simply supported circular hollow section element subjected to torsional efforts

Test ID: 5743

Test status: Passed

13.42.1Description Verifies a simply supported circular hollow section element made of S235 steel subjected to torsional efforts.

The verification is made according to Eurocode3 (EN 1993-1-1) French Annex.

13.43 EC3 Test 39: Verifying CHS323.9x6.3H class 2 beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle

Test ID: 5741

Test status: Passed

13.43.1Description The test verifies a CHS323.9x6.3H beam made of S275 steel.



13.44 EC3 Test 38: Verifying RHS350x150x5H class 4 column, loaded with centric compression, punctual horizontal force by Y and a bending moment, all applied to the top

Test ID: 5740

Test status: Passed

13.44.1Description The test verifies a RHS350x150x5H column made of S355 steel.

The column is subjected to 680 kN compression force, 5 kN punctual horizontal load and 200 kNm bending moment, all applied to the top.



314

13.45 EC3 Test 45: Comparing the shear resistance of a welded built-up beam made from different steel materials

Test ID: 5745

Test status: Passed

13.45.1Description The shear resistance of a welded built-up beam made of S275 steel is compared with the shear resistance of the same built-up beam made of a user-defined steel material.


13.45.2Background Verifies the shear resistance of a welded built-up beam made of 500 MPa yield strength user-defined steel. The beam is simply supported and it is subjected to a uniformly distributed load (20 000 N/ml) applied over its length. The dead load will be neglected.

Also verifies the shear resistance of the same welded built-up beam made of S275 steel. The loading and support conditions are the same.



■ Exploitation loadings (category A), Q: Fz = - 20 000 N/ml,



Units

Metric System


315

Geometry


■ Height: h = 300 mm, ■ Flange width: b = 150 mm, ■ Flange thickness: tf = 10.7 mm, ■ Web thickness: tw = 7.1 mm, ■ Beam length: L = 5000 mm, ■ Section area: A = 5188 mm2,



500 MPa yield strength user-defined material and S275 steel are used. The following characteristics are used:

■ Yield strength fy = 500 MPa, ■ Yield strength (for S275 steel) fy = 275 MPa.

Boundary conditions


■ Outer: Support at start point (z = 0) restrained in translation along X, Y and Z axis, Support at end point (z = 5.00) restrained in translation along X, Y and Z axis.

► Inner: None.

Loading


■ External: Uniformly distributed load: q = Fz = -20 000 N/ml

■ Internal: None.

13.45.2.2Reference results for calculating the design plastic shear resistance of the cross section In order to verify the steel beam subjected to shear, the criterion (6.18) from chapter 6.2.6 (EN 1993-1-1) has to be used:

0.1,

≤Rdpl

Ed

VV

■ VEd represents the design value of the shear force:

NmmmlNLqVEd 500002

5000/200002

=×

=×

=

■ Vpl,Rd represents the design plastic shear resistance. The design plastic shear resistance of the cross-section is determined with formula (6.18) from EN 1993-1-1:2001. Before using it, the shear area (Av) has to be determined.


316

Shear area of the cross section made of 500 MPa yield strength user-defined material

According to chapter 5.1 from EN 1993-1-5, because the steel grade used for beam is higher than S460, the factor for shear area (η) may be conservatively taken equal 1.0.

For a welded I sections, the shear area is determined according to chapter 6.2.6 (3) from EN 1993-1-1. As the load is parallel to web, the shear area is:

( )∑ =××=××= 206.1978)1.76.278(0.1 mmmmmmthA wwv η

Shear area of the cross section made of S275 steel

According to chapter 5.1 from EN 1993-1-5, because the steel grade used for beam is up to S460, the factor for shear area (η) is 1.2.

As the load is parallel to web, the shear area becomes:

( )∑ =××=××= 267.2373)1.76.278(2.1 mmmmmmthA wwv η

Design plastic shear resistance of the cross section made of 500 MPa yield strength user-defined material

EN 1993-1-1 provides the following formula to calculate the design plastic shear resistance of the cross-section:

N

MPammf

AV

M

yv

Rdpl 7.5710160.1

350006.1978

32

0, =

×=

×=

γ

The verification of the design plastic shear resistance of the cross section is done with criterion (6.18) from EN 1993-1-1:

0.10876.07.571016

50000

,

≤==N

NVV

Rdpl

Ed

The corresponding work ratio is:

Work ratio = %76.81007.571016

50000100,

=×=×N

NVV

Rdpl

Ed

Design plastic shear resistance of the cross section made of S275 steel

EN 1993-1-1 provides the following formula to calculate the design plastic shear resistance of the cross-section:

N

MPammf

AV

M

yv

Rdpl 7.3768700.1

327567.2373

32

0, =

×=

×=

γ

The verification of the design plastic shear resistance of the cross section is done with criterion (6.18) from EN 1993-1-1:

0.1133.07.376870

50000

,

≤==N

NVV

Rdpl

Ed

The corresponding work ratio is:

Work ratio = %27.131007.376870

50000100,

=×=×N

NVV

Rdpl

Ed




317


Work ratio of the design shear resistance (beam made of 500 MPa yield strength user-defined material)

Beam subjected to uniformly distributed load applied over its length Work ratio – Fz

Work ratio of the design shear resistance (beam made of S275 steel)

Beam subjected to uniformly distributed load applied over its length Work ratio – Fz


Result name Result description Reference value Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 275 MPa) 13.27 %

Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 500 MPa) 8.76 %

13.45.3Calculated results Result name Result description Value Error Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 275

MPa) 13.2671 % -0.0219 %

Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 500 MPa)

8.75631 % -0.0421 %


318

13.46 EC3 Test 43: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange

Test ID: 5750

Test status: Passed

13.46.1Description Determines the lateral torsional buckling parameters for a I-shaped laminated beam made of S235 steel, considering the load applied on the lower flange.

The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.

13.47 EC3 Test 41: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange

Test ID: 5753

Test status: Passed

13.47.1Description Determines the lateral torsional buckling parameters for a I-shaped laminated beam made of S235 steel, considering the load applied on the lower flange. The loadings applied on the beam are: a uniformly distributed load and 2 punctual bending moments, acting opposite to each other, applied at beam extremities.



319

13.48 EC3 test 8: Verifying the classification and the resistance of a column subjected to bending and axial load

Test ID: 5632

Test status: Passed

13.48.1Description Verifies the classification and the resistance for an IPE 600 column made of S235 steel subjected to bending and axial force. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

13.48.2Background Classification and verification of an IPE 600 column, made of S235 steel, subjected to bending and axial force. The column is fixed at its base and free on the top. The column is loaded by a compression force (1 000 000 N), applied at its top, and a uniformly distributed load (50 000 N/ml). The dead load will be neglected.


13.48.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case (Q1) and load combination are used:

■ Exploitation loadings (category A), Q1: Fz = -1 000 000 N, Fx = 50 000 N/ml,

■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q1


Units

Metric System

Geometry


■ Height: h = 600 mm, ■ Flange width: b = 220 mm, ■ Flange thickness: tf = 19 mm, ■ Column length: L = 5000 mm, ■ Section area: A = 15600 mm2 ,

■ Plastic section modulus about the strong y-y axis: 3, 3512000mmW ypl = ,



320




Boundary conditions



■ Inner: None.

Loading


■ External: Point load at Z = 5.0: N = FZ = -1 000 000 N, Uniformly distributed load: q = Fx = 50 000 N/ml

■ Internal: None.


321

13.48.2.2Reference results for calculating the column subjected to bending and axial force In order to verify the steel column subjected to bending and axial force, the design resistance for uniform compression (Nc,Rd) and also the design plastic moment resistance reduced due to the axial force (MN,Rd) have to be compared with the design values of the corresponding efforts.

The design resistance for uniform compression is verified considering the relationship (6.9) from chapter 6.2.4 (EN 1993-1-1), while for bi-axial bending, the criterion (6.41) from chapter 6.2.9.1 (EN 1993-1-1) has to be satisfied.

Before starting the above verifications, the cross-section class has to be determined.

Cross section class

Considering that the column is subjected to combined bending and axial compression, and also that its axial effort is bigger than 835 kN, the following classification is made according to the CTICM journal no. 4 – 2005 (extracted of journal):

So, according to this table, the column cross-section is Class 2.

Verifying the design resistance for uniform compression

Expression (6.10) from EN 1993-1-1 is used to determine the design compression resistance, Nc,Rd:

NMPammfAN

M

yRdc 3666000

0.123515600 2

0, =

×=

×=

γ

In order to verify the design resistance for uniform compression, the criterion (6.9) from chapter 6.2.4 (EN 1993-1-1) has to be satisfied:

%100%3.270.1273.036660001000000

,,

≤→≤===NN

NN

NN

RdcRdc

Ed

Verifying the column subjected to bending and axial force

According to paragraph 6.2.9.1 (4) from EN 1993-1-1, allowance will not be made for the effect of the axial force on the plastic resistance moment about the y-y axis if relationship (6.33) is fulfilled.

NNNNN RdplEd 916500366600025.0100000025.0 , =×>→×≤

In this case, because the above verification is not fulfilled, the axial force has an impact on the plastic resistance moment about the y-y axis.

In order to verify the column subjected to bending and axial force, the criterion (6.41) from EN 1993-1-1 has to be used. Supplementary terms need to be determined: design resistance for bending (Mpl,Rd), ratio of design normal force to design plastic resistance to normal force of the gross cross-section (n), ratio of web area to gross area (a), design plastic moment resistance reduced due to the axial force (MN,Rd).


322

■ Design plastic moment resistance:

NmmMPammfWM

M

yyplRdypl 825320000

0.12353512000 3

0

,,, =

×=

×=

γ

■ Ratio of design normal force to design plastic resistance to normal force of the gross cross-section:

273.036660001000000

,

===NN

NNn

Rdpl

■ Ratio of web area to gross area:

464.015600

1922021560022

2

=××−

=××−

=mm

mmmmmmA

tbAa f

■ Design plastic moment resistance reduced due to the axial force is determined according to expression (6.36) from EN 1993-1-1:

anMM RdyplRdyN ×−

−×=

5.011

,,,, but RdyplRdyN MM ,,,, ≤

NmmNmmNmmM RdyN 825320000781259948464.05.01

273.01825320000,, ≤=×−

−×=

■ The column subjected to bending and axial force is verified with criterion (6.41) from EN 1993-1-1:

0.1,,

,

,,

, ≤

+

βα

RdzN

Edz

RdyN

Edy

MM

MM

Because the column doesn’t have bending moment about z axis, the second term from criterion (6.41) is

neglected. The verification becomes: 0.12/

,,

2

,,

, ≤

×=

RdyNRdyN

Edy

MLq

MM

( ) %100%9.790.1799.0781259948

2/5000/502/ 2

,,

2

<→≤=

×=

×Nmm

mmmmNM

Lq

RdyN




323



Work ratio of the design resistance for oblique bending

Column subjected to bending and axial force Work ratio - Oblique


324


Result name Result description Reference value Work ratio - Fx Work ratio of the design resistance for uniform compression [%] 27.3 % Work ratio - Oblique Work ratio of the design resistance for oblique bending [%] 79.9 %

13.48.3Calculated results

Result name Result description Value Error Work ratio - Fx Work ratio of the design resistance for uniform

compression 27.2777 % -0.0084 %

Work ratio - Oblique

Work ratio of the design resistance for oblique bending 79.9691 % -0.0386 %


325

13.49 EC3 Test 17: Verifying a simply supported rectangular hollow section beam subjected to torsional efforts

Test ID: 5742

Test status: Passed

13.49.1Description Verifies a simply supported rectangular hollow section beam made of S235 steel subjected to torsional efforts.



326

13.50 EC3 Test 44: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange

Test ID: 5749

Test status: Passed

13.50.1Description Determines the lateral torsional buckling parameters for a I-shaped welded built-up beam made of S235 steel, considering the load applied on the upper flange. The loadings applied on the beam are: a uniformly distributed load and 2 punctual bending moments, acting opposite to each other, applied at beam extremities.


13.50.2Background Determines the elastic critical moment (Mcr) and factors (C1, C2, χLT) involved in the torsional buckling verification for a simply supported steel beam. The beam is made of S235 steel and it is subjected to a uniformly distributed load (50 000 N/ml) applied over its length and concentrated bending moments applied at its extremities (loads are applied to the upper fibre). The dead load will be neglected.



■ Exploitation loadings (category A), Q: Fz = - 50 000 N/ml, My,1 = 142 x 106 Nmm, My,2 = - 113.6 x 106 Nmm,




327

Units

Metric System

Geometry


■ Height: h = 260 mm, ■ Flange width: b = 150 mm, ■ Flange thickness: tf = 10.7 mm, ■ Web thickness: tw = 7.1 mm, ■ Beam length: L = 5000 mm, ■ Section area: A = 5188 mm2, ■ Flexion inertia moment about the z axis: Iz = 6025866.46 mm4, ■ Torsional moment of inertia: It = 149294.97 mm4, ■ Warping constant: Iw = 93517065421.88 mm6, ■ Plastic modulus about the y axis: Wy = 501177.18 mm3




■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa; ■ Shear modulus of rigidity: G=80800MPa.

Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation along Y and Z axis and restrained rotation

along X axis. ■ Inner: None.

Loading


■ External: Uniformly distributed load over its length: q = Fz = -50 000 N/ml Bending moment at x=0: My,1 = 142 x 106 Nmm Bending moment at x=5: My,2 = - 113.6 x 106 Nmm,

■ Internal: None.


328

13.50.2.2Reference results for calculating the elastic critical moment of the cross section In order to determine the elastic critical moment of the cross section (Mcr), factors C1 and C2 have to be calculated. They are determined considering the method provided at chapter 3.5 from French Annex of EN 1993-1-1. C1 and C2 coefficients are depending on the bending moment diagram along the member segment between lateral restraints.

The simply supported beam has the following bending moment diagram (the values are in “Newton x meter”):

For a beam subjected to uniformly distributed load and concentrated bending moments applied at its extremities, the moments distribution is defined considering two parameters:

■ Ratio between the moments at extremities:

8.0142000113600

=−−

=NmNmψ

■ Ratio between the moment given by uniformly distributed load and the biggest bending moment from extremity:

( ) 1.11014285000/50

8 6

22

=×××

=××

=Nmm

mmmmNMLqµ

Its value is positive as both loadings are deforming the beam about the same fibre (chapter 3.4 from French Annex of EN 1993-1-1).

In order to determine C1 and C2 parameters, factors β, γ, a, b, c, A, B, d1, e1, r1, ξ, m, C10, d2, e2, r2 need to be calculated considering the analytical relationships provided in chapter 3.5 from French Annex of EN 1993-1-1:

■ 2.411.148.014 =−×+=−×+= µψβ

■ 84.81.182.48 22 =×−=×−= µβγ

■ ( ) 738.19126223.06960364.01413364.015.0 2 =+−++×= µβµγβa

■ ( ) 4765.14281556.19240091.01603341.015.0 2 =+−++×= µβµγβb

■ 0602.09352904.05940757.00900633.01801266.0 2 =−+−−= µβµγβc

■ 5625.22 =−×= cbaA

■ 2143.42

2 =+×=baB

■ ( ) 036.2152.01 =+×+= ψµd

■ 3.01 =e

■ As d1 > e1, the factor r1 is equal to 1.0

■ 0.14773.08

15.0 ≤=−

−=µψξ

■ ( ) ( ) 0.1002.21411 ≥=−×××+−×−= ξξµψξm

■ 5363.02

42

110 =×

×−−×=

AABBrC

■ 065.2675.0425.02 =×++= ψµd

■ 37.035.065.02 =×−= ψe


329

■ As d2 > e2, the factor r2 is equal to 1.0 Having the above factors, C1 and C2 coefficients become:

■ 074.1101 =×= CmC

■ 235.0398.0 1022 =×××= CrC µ

The load being applied at the top fibre it tends to accentuate the lateral torsional buckling, so it will reduce the value of elastic critical moment. In this case, the distance from the shear centre to the point of load application (zg) will be positive:

■ mmzg 130+=

The French Annex of EN 1993-1-1 provides the analytical relationship used to determine the value of the elastic critical moment:

■ ( ) NmmzCzCIEIGL

II

LIECM gg

z

t

z

wzcr

62

222

2

2

2

1 1071772.91 ×=

×−×+××××

+×××

×=π

π

13.50.2.3Reference results for calculating the reduction factor for lateral torsional buckling The calculation of the reduction factor for lateral torsional buckling (χLT) is done using the formula (6.56) from chapter 6.3.2.2 (EN 1993-1-1).

Before determining the reduction factor for lateral torsional buckling (χLT), the following terms should be determined:

LTλ , imperfection factor αLT, φLT.

■ Non dimensional slenderness for lateral torsional buckling, LTλ :

133.11071772.91

/23518.5011776

23, =

××

=×

=Nmm

mmNmmM

fW

cr

yyplLTλ

■ In order to determine the imperfection factor αLT, the buckling curve must be chosen. According to table 6.4

from EN 1993-1-1, for welded I-sections which have the ratio h / b ≤ 2, the recommended lateral torsional buckling curve is “c”. In this case, table 6.3 from EN 1993-1-1 recommends the value for imperfection factor αLT:

49.0=LTα

■ The value used to determine the reduction factor χLT, φLT, becomes:

( )[ ] ( )[ ] 370.1133.12.0133.149.015.02.015.0 22 =+−×+×=+−×+×= LTLTLTLT λλαφ

■ The reduction factor for lateral torsional buckling is calculated using the formula (6.56) from EN 1993-1-1:

0.1467.0133.137.137.1

112222

≤=−+

=−+

=LTLTLT

LTλφφ

χ




330


C1 parameter

Simply supported beam C1

C2 parameter

Simply supported beam C2

Elastic critical moment

Simply supported beam Mcr

Reduction factor for lateral torsional buckling

Simply supported beam XLT


331


Result name Result description Reference value C1 C1 parameter [adim.] 1.074

C2 C2 parameter [adim.] 0.235

Mcr Elastic critical moment [kNm] 91.72 kNm

XLT Reduction factor for lateral torsional buckling [adim.] 0.467

13.50.3Calculated results Result name Result description Value Error C1 C1 parameter 1.07375

adim -0.0233 %

C2 C2 parameter 0.234812 adim

-0.0800 %

Mcr Elastic critical moment 91.72 kN*m 0.0000 % XLT Reduction factor for lateral torsional buckling 0.466895

adim -0.0225 %


332

13.51 EC3 Test 40: Verifying CHS508x8H class 3, simply supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle

Test ID: 5744

Test status: Passed

13.51.1Description The test verifies a CHS508x8H beam made of S235 steel.



13.52 EC3 Test 42: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange

Test ID: 5752

Test status: Passed

13.52.1Description Determines the lateral torsional buckling parameters for a I-shaped welded built-up beam made of S235 steel, considering the load applied on the upper flange. The loadings applied on the beam are: a uniformly distributed load and 2 punctual negative bending moments applied at beam extremities.


13.53 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533)

Test ID: 4549

Test status: Passed

13.53.1Description Verifies the shape sheet for a steel beam with circular cross-section when the lateral torsional buckling is computed and when it is not.

13.54 Changing the steel design template for a linear element (TTAD #12491)

Test ID: 4540

Test status: Passed

13.54.1Description Selects a different design template for steel linear elements.

13.55 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389)

Test ID: 4529

Test status: Passed

13.55.1Description Verifies the program behavior when the "Shape sheet" command is used for elements which were excluded from the specialized calculation (chords).


333

13.56 EC3: Verifying the buckling length results (TTAD #11550)

Test ID: 4481

Test status: Passed

13.56.1Description Performs the steel calculation and verifies the buckling length results according to Eurocodes 3 - French standards. The shape sheet report is generated.

The model consists of a vertical linear element (IPE300 cross section, S275 material) with a rigid fixed support at the base. A punctual live load of 200.00 kN is applied.

13.57 EC3 fire verification: Verifying the work ratios after performing an optimization for steel profiles (TTAD #11975)

Test ID: 4484

Test status: Passed

13.57.1Description Runs the Steel elements verification and generates the "Envelopes and optimizing profiles" report in order to verify the work ratios. The verification is performed using the EC3 norm with Romanian annex.


334

13.58 EC3 test 4: Class section classification and bending moment verification of an IPE300 column

Test ID: 5412

Test status: Passed

13.58.1Description Classification and verification of an IPE 300 column made of S235 steel.

The column is connected to the ground by a fixed connection and is free on the top part.

In the middle, the column is subjected to a 50 kN force applied on the web direction, defined as a live load.

The dead load will be neglected.

13.58.2Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 50kN force applied on the web direction, defined as a live load. The dead load will be neglected.


13.58.2.1Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

■ Exploitation loadings (category A): Q = 50kN, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q


335

Units

Metric System


Boundary conditions


■ Outer: Support at start point (x=0) restrained in translation and rotation along X, Y and Z axis, Support at end point (z = 5.00) free.

■ Inner: None.

13.58.2.2Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2

In this case, the column is subjected to a lateral load, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below.

The Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.


336


According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:


337

Therefore:


Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.


According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:


338

Therefore:

This means that the column flanges are Class 1.

A cross-section is classified by quoting the heist (least favorable) class of its compression elements.

According to the calculation above, the column section have a Class 1 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 1.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)

13.58.2.3Reference results in calculating the bending moment resistance

1,

≤Rdc

Ed

MM


0,

*

M

yplRdc

fwM

γ= for Class1 cross sections


Where:

340.628 cmwpl =

fy nominal yielding strength for S235 fy=235MPa

0Mγ partial safety coefficient 10 =Mγ

Therefore:

MNmfw

MM

yplRdVy 147674.0

1235*10*40.628* 6

0,, ===

−

γ

MNmkNmkNmM Ed 125.012550*5.2 ===

%84148.0125.0

,,

==RdVy

Ed

MM




339


Combined oblique bending

Combined oblique bending Work ratio - Oblique


Result name Result description Reference value Combined oblique bending

Work ratio - Oblique 85 %


Work ratio - Oblique 84.6459 % -0.4166 %


340

13.59 EC3 Test 2: Class section classification and shear verification of an IPE300 beam subjected to linear uniform loading

Test ID: 5410

Test status: Passed

13.59.1Description Classification and verification of an IPE 300 beam made of S235 steel.

The beam is subjected to a 50 kN/m linear uniform load applied gravitationally.

The force is considered to be a live load and the dead load is neglected.

13.59.2Background Classification and verification of sections for an IPE 300 beam made from S235 steel. The beam is subjected to a 50 kN/m linear uniform load applied gravitationally. The force is considered to be a live load and the dead load is neglected.



■ Exploitation loadings (category A): Q = -50kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System



341

Boundary conditions


■ Outer: Support at start point (x=0) restrained in translation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation along Y and Z axis and rotation restrained on X

axis ■ Inner: None.


In this case the stresses distribution along the section is like in the picture below:

■ compression for the top flange ■ compression and tension for the web ■ tension for the bottom flange

To determine the web class it will be used the Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.


342


According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:


343

Therefore:


To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2


According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:

276.57.1045.56

==mmmm

tc

92.0=ε


344

Therefore:

28.892.0*9*9276.57.1045.56

==≤== εmmmm

tc

this means that the column flanges are Class 1.


According to the calculation above, the beam section have a Class 1 web and Class 1 flanges; therefore the class section for the entire beam section will be considered Class 1.


13.59.2.3Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd shall be determined as follows:

0,

3*

M

yv

Rdpl

fA

Vγ

=


Where:

Av: section shear area for rolled profiles fwfv trttbAA *)*2(**2 ++−=

A: cross-section area A=53.81cm2

b: overall breadth b=150mm

h: overall depth h=300mm

hw: depth of the web hw=248.6mm

r: root radius r=15mm

tf: flange thickness tf=10.7mm

tw: web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=


fy: nominal yielding strength for S275 fy=275MPa

0Mγ : partial safety coefficient 10 =Mγ

Therefore:

kNMN

fA

VM

yv

Rdpl 7.4074077.01

3275*10*68.25

3* 4

0, ====

γ


345

For more:

Verification of the shear buckling resistance for webs without stiffeners:

εη*72≤

w

w

th


20.111*20.1*20.1

0

1 ===M

M

γγη

92.0275235235

===yf

ε

91.9392.020.1*72*7201.35

1.76.248

==≤==εη

w

w

th

There is no need for shear buckling resistance verification

According to: EC3 Part 1,5 EN 1993-1-5-2004 Chapter 5.1(2)




Shear resistance work ratio Work ratio - Fz


346


Result name Result description Reference value Fz Shear force 125 kN Work ratio Work ratio - Fz 31 %

13.59.3Calculated results Result name Result description Value Error Fz Fz -125 kN 0.0000 % Work ratio - Fz Work ratio Fz 30.6579 % -1.1035 %


347

13.60 EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 beam

Test ID: 5424

Test status: Passed

13.60.1Description Classification and verification on combined bending of an IPE 300 beam made of S235 steel.

The beam is connected to its ends by a connection with all translation blocked and on the other end by a connection with translation blocked on the Y and Z axis and rotation blocked along X axis.

The beam is subjected to a -10 kN/m uniform linear force applied along the beam gravitational along the Z local axis, and a 10kN/m uniform linear force applied along the beam on the Y axis.

Both forces are considered as live loads.


13.60.2Background Classification and verification on combined bending of sections for an IPE 300 beam made from S235 steel. The beam is connected to its ends by a connection with all translation blocked and on the other end by a connection with translation blocked on the Y and Z axis and rotation blocked along X axis. The beam is subjected to a -10kN/m uniform linear force applied along the beam gravitational along the Z local axis, and a 10kN/m uniform linear force applied along the beam on the Y axis. Both forces are considered live loads. The dead load will be neglected.



■ Exploitation loadings (category A): Q1 = -10kN/m, Q2 = 10kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System


348


Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation and rotation along Y, Z axis and rotation

blocked along X axis. ■ Inner: None.


In this case, the beam is subjected to linear uniform equal loads, one vertical and one horizontal, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below:

Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.


349




350

Therefore:

This means that the beam web is Class 1.



According to Table 5.2 and the beam section geometrical properties, the next conclusions can be found:


351

Therefore:

This means that the beam left top flanges are Class 1.

Overall the beam top flange cross-section class is Class 1.

In the same way will be determined that the beam bottom flange cross-section class is also Class 1




13.60.2.3Reference results for calculating the combined biaxial bending

1,

,

,

, ≤

+

βα

EdNz

Edz

RdNy

EdY

MM

MM

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(5)

In which α and β are constants, which may conservatively be taken as unity, otherwise as follows:

For I and H sections:

2=α

)1;max(n=β

00,,

===RdplRdpl

Ed

NNNn therefore 1=β

Bending around Y:

For cross-sections without bolts holes, the following approximations may be used for standard rolled I or H sections and for welded I or H sections with equal flanges:

anM

M RdyplRdNy *5.01

)1(*,,, −

−= but RdplyRdNy MM ,, ≤


00,,

===RdplRdpl

Ed

NNNn

5.0403.010*81.53

)0107.0*15.0*210*81.53()**2(4

4

≤=−

=−

= −

−

AtbA

a f

8.0)403.0*5.01(,,,,

,,RdyplRdypl

RdyN

MMM =

−=

RdyplRdyNRdyplRdyN MMMM ,,,,,,,,*8.0 >⇒= but RdplyRdNy MM ,, ≤


352

Therefore, it will be considered:

Bending around Y:

For cross-sections without bolts holes, the following approximations may be used for standard rolled I or H sections and for welded I or H sections with equal flanges:

anM

M RdzplRdNz *5.01

)1(*,,, −

−= but RdplzRdNz MM ,, ≤


RdzplRdzNRdzplRdzN MMMM ,,,,,,,,*8.0 >⇒= but RdplzRdNz MM ,, ≤ therefore it will be considered:

MNmfw

MMM

yzplRdzplRdzN 030.0

1235*10*20.125* 6

0

,,,,, ====

−

γ In conclusion:

11086.1029375.003125.0

148.003125.0 12

,

,

,

, >=

+

=

+

βα

EdNz

Edz

RdNy

EdY

MM

MM

The coresponding work ratio is:

WR = 1.1086 x 100 = 110.86 %




Work ratio – oblique bending

Beam subjected to combined bending

Work ratio – Oblique [%]


353


Result name Result description Reference value Combined oblique bending

Combined oblique bending [%] 110.86 %


Work ratio-Oblique 110.691 % -0.2784 %


354

13.61 EC3 Test 1: Class section classification and compression verification of an IPE300 column

Test ID: 5383

Test status: Passed



On top, the column is subjected to a 100 kN force applied gravitationally, defined as a live load.


13.61.2Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. On top, the column is subjected to a 100kN force applied gravitationally, defined as a live load. The dead load will be neglected.



■ Exploitation loadings (category A): Q = -100kN, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System


355


Boundary conditions



■ Inner: None.

13.61.2.2Reference results for calculating the cross section class In this case, the column is subjected only to compression, therefore the distribution of stresses along the section is like in the picture below:

To determine the web class, we use Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2


356



1=ε Therefore:


To determine the flanges class, we will use Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2


357



276.57.1045.56

==mmmm

tc

1=ε

Therefore:

9*9276.57.1045.56

=≤== εmmmm

tc

this means that the column flanges are Class 1.





358

13.61.2.3Reference results in calculating the compressive resistance Nc,Rd The design resistance of the cross-section force Nc,Rd shall be determined as follows:

For Class 1, 2 or 3 cross-section 0

,

*

M

yRdc

fAN

γ=

Where:

A section area A=53.81cm2

Fy nominal yielding strength for S235 fy=235MPa


Therefore:

kNMNfA

NM

yRdc 54.1264264535.1

1235*10*81.53* 4

0, ====

−

γ





359


Compressive resistance work ratio

Column subjected to compressive load Work ratio [%]


Result name Result description Reference value Work ratio Compressive resistance work ratio [%] 8 %

13.61.3Calculated results Result name Result description Value Error Work ratio - Fx Work ratio Fx 7.90805 % -1.1494 %


360

13.62 EC3 Test 5: Class section classification and combined axial force with bending moment verification of an IPE300 column

Test ID: 5421

Test status: Passed



The column is subjected to a 500 kN compressive force applied on top and a 5 kN/m uniform linear load applied on all the length of the column, on the web direction, both defined as live loads.


13.62.2Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. The column is subjected to a 500kN compressive force applied on top and a 5kN/m uniform linear load applied for all the length of the column, on the web direction, both defined as live loads. The dead load will be neglected.



■ Exploitation loadings (category A): Q1 = 500kN, Q2 = 5kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System


361


Boundary conditions



■ Inner: None.


In this case the column is subjected to compression and lateral load, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below.

Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.


362



121.01235005381.0

500.0212 −>−=−×

⋅=−⋅

⋅=y

Ed

fANψ

5.010.10071.02486.0235

5.01211

21

>=

××+⋅=

××+⋅=

dtfN

y

Edα


363

Therefore:

92.69)21.0(*33.067.0

1*4233.067.0

42=

−+=

+ ψε

Therefore:

92.6933.067.0

42014.35 =+

≤=ψ

εtc




364



Therefore:





Cross sections for class 3, the maximum longitudinal stress should check:


In this case:

In absence of shear force, for Class 3 cross-sections the maximum longitudinal stress shall satisfy the criterion:


This means:

The corresponding work ratio is: WR = 0.8728 x 100 = 87.28 %


365




Work ratio – bending and axial compression

Column subjected to bending and axial compression

Work ratio – Oblique [%]


Result name Result description Reference value Bending and axial compression

Work ratio – oblique [%] 87.28 %


Work ratio- Oblique 87.2799 % 0.3217 %


366

13.63 EC3 Test 3: Class section classification, shear and bending moment verification of an IPE300 column

Test ID: 5411

Test status: Passed



In the middle, the column is subjected to a 200 kN force applied on the web direction, defined as a live load.


13.63.2Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 200kN force applied on the web direction, defined as a live load. The dead load will be neglected.



■ Exploitation loadings (category A): Q = -200kN, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System


367


Boundary conditions


■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, Support at end point (z = 5.00) free.

■ Inner: None.


In case the column is subjected to a lateral load, the stresses distribution on the most stressed point (the column base) is like in the picture below:


368

To determine the web class, we will use Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.




369

Therefore:


To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.



370


Therefore:




According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6).

13.63.2.3Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd , is determined as follows:

0,

3*

M

yv

Rdpl

fA

Vγ

=


Where:

Av section shear area for rolled profiles fwfv trttbAA *)*2(**2 ++−=

A cross-section area A=53.81cm2

b overall breadth b=150mm

h overall depth h=300mm

hw depth of the web hw=248.6mm

r root radius r=15mm

tf flange thickness tf=10.7mm

tw web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=




Therefore:

kNMN

fA

VM

yv

Rdpl 42.3483484.01

3235*10*68.25

3* 4

0, ====

γ


371

13.63.2.4Reference results in calculating the bending moment resistance

%50%4.5742.348

200,

>==Rdpl

Ed

VV

The shear force is greater than half of the plastic shear resistance. Its effect on the moment resistance must be taken into account.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(1)(2)

Where:

0223.01348.0

200.0*212 22

,

=

−=

−=

Rdpl

Ed

VVρ


fwfv trttbAA *)*2(**2 ++−=

Av section shear area for rolled profiles

A cross-section area A=53.81cm2

b overall breadth b=150mm

h overall depth h=300mm

hw depth of the web hw=248.6mm

r root radius r=15mm

tf flange thickness tf=10.7mm

tw web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=




Therefore:

MNmf

tAw

MM

yw

vpl

RdVy 146.01

235*0071.0*4

)²10*68.25(*0223.010*40.628*4

²* 46

0,, =

−

=

−

=

−−

γ

ρ

MNmkNmkNmM Ed 5.0500200*5.2 ===

%342146.0500.0

,,

==RdVy

Ed

MM


372




Shear z direction work ratio

Shear z direction work ratio Work ratio - Fz


373


Combined oblique bending Work ratio - Oblique


Result name Result description Reference value Shear z direction work ratio

Work ratio - Fz 57 %


Work ratio - Oblique 341 %

13.63.3Calculated results Result name Result description Value Error Work ratio - Fz Work ratio Fz 57.4021 % 0.7054 % Work ratio - Oblique

Work ratio - Oblique 341.348 % 0.1021 %

14 Timber design


376

14.1 EC5: Verifying a timber purlin subjected to oblique bending

Test ID: 4878

Test status: Passed

14.1.1 Description Verifies a rectangular timber purlin made from solid timber C24 to resist oblique bending. The verification is made following the rules from Eurocode 5 French annex.

14.1.2 Background Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to oblique bending. The verification of the bending stresses at ultimate limit state is performed.

14.1.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test E.3; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent purlins (span): d = 1.8 m. The following load cases and load combination are used:

■ Loadings from the structure: G = 550 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2; All loads will be projected on the purlin direction since the roof slope is 17°.

Simply supported purlin subjected to loadings

Units

Metric System

Geometry

Purlin cross section characteristics:

■ Height: h = 0.20 m, ■ Width: b = 0.10 m, ■ Length: L = 3.5 m, ■ Section area: A = 0.02 m2 ,

■ Elastic section modulus about the strong axis, y: 322

000666.06

20.01.06

mhbWy =⋅

=×

= ,

■ Elastic section modulus about the strong axis, z: 322

000333.06

20.01.06

mhbWz =⋅

=×

= .


377


Rectangular solid timber C24 is used. The following characteristics are used in relation to this material:

■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Service class 2.

Boundary conditions


■ Outer: Support at start point (z=0) restrained in translation along X, Y, Z; Support at end point (z = 3.5) restrained in translation along X, Y, Z and restrained in rotation along X.

■ Inner: None.

Loading

The purlin is subjected to the following projected loadings (at ultimate limit state):

■ External: Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x sin17° =

1101.22 N/m, Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x cos17° =

3601.92 N/m, ■ Internal: None.

14.1.2.2 Reference results in calculating the timber purlin subjected to oblique bending In order to verify the timber purlin subjected to oblique bending at ultimate limit state, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, have to be determined. After this, the reference solution (which includes the design bending stress about y axis, the design bending stress about z axis and the maximum work ratio for strength verification) is calculated.

Reference solution for ultimate limit state verification

Before calculating the reference solution (design bending stress about y axis, design bending stress about z axis and maximum work ratio for strength verification) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).

■ Modification factor for duration of load (short term) and moisture content: kmod = 0.9 (according to table 3.1 from EN 1995-1-1)

■ Partial factor for material properties: γM = 1.3

■ Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0

■ System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)

■ Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections): km = 0.7

■ Design bending stress about y axis (induced by uniformly distributed load, qz):

σm,y,d = Pam

mmN

WLq

WM

y

z

y

y 63

222

102814.8000666.08

5.392.3601

8×=

×

×=

××

=

■ Design bending stress about z axis (induced by uniformly distributed load, qy):

σm,z,d = Pam

mmN

WLq

WM

z

y

z

z 63

222

100638.5000333.08

5.322.1101

8×=

×

×=

×

×=


378

■ Design bending strength:

fm,y,d = fm,z,d = Pakkkf hsysM

km66mod

, 10615.160.10.13.19.01024 ×=××××=×××

γ

■ Maximum work ratio for strength verification; it represents the maximum value between the work ratios obtained with formulae 6.17 and 6.18 from EN 1995-1-1 norm:

1max

,,

,,

,,

,,

,,

,,

,,

,,

≤

+

+

dzm

dzm

dzm

dzmm

dym

dymm

dym

dym

f

fk

fk

fσ

σ

σ

σ



Stress SMy diagram

Simply supported purlin subjected to uniformly distributed load, qz Stress SMy [Pa]

Stress SMz diagram

Simply supported purlin subjected to uniformly distributed load, qz Stress SMz [Pa]

Maximum work ratio for strength verification

Strength of a simply supported purlin subjected to oblique bending Work ratio [%]


379

14.1.2.3 Reference results

Result name Result description Reference value Smy Design bending stress about y axis [Pa] 8281441 Pa SMz Design bending stress about z axis [Pa] 5063793 Pa

Work ratio Maximum work ratio for strength verification [%] 71.2 %

14.1.3 Calculated results Result name Result description Value Error Stress SMy Design bending stress about y axis 8.47672e+006 Pa 0.0000 % Stress SMz Design bending stress about z axis 5.18319e+006 Pa 0.0000 % Work ratio Maximum work ratio for strength verification 71.153 % 0.0000 %


380

14.2 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes

Test ID: 4896

Test status: Passed

14.2.1 Description Verifies the residual cross section of a column exposed to fire for 60 minutes. The column is made from glued laminated timber GL24 and it has only 3 faces exposed to fire. The verification is made according to chapter 4.2.2 (Reduced cross section method) from EN 1995-1-2 norm.

14.2.2 Background Verifies the adequacy of the cross sectional resistance for a rectangular cross section, which is made from glued laminated timber GL24, exposed to fire for 60 minutes on 3 faces. The verification is made according to chapter 4.2.2 from EN 1995-1-2 norm.

14.2.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test F.1; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

Timber column with fixed base

Units

Metric System

Geometry


■ Depth: h = 0.60 m, ■ Width: b = 0.20 m, ■ Section area: A = 0.12 m2 ■ Height: H = 5.00 m


381


Glued laminated timber GL24 is used. The following characteristics are used in relation to this material:

■ Density: ρ = 380 kg/m3, ■ Design charring rate: βn = 0.7 x 10-3 m/min,

Boundary conditions


■ Outer: Fixed at base (Z = 0), Support at top (Z = 5.00) restrained in translation along X and Y,

■ Inner: None.

Loading


■ External: Point load at Z = 5.00: Fz = N = - 100000 N, ■ Internal: None.

14.2.2.2 Reference results in calculating the cross sectional resistance of a timber column exposed to fire

Reference solution

The reference solution (residual cross section) is determined by reducing the initial cross section dimensions by the effective charring depth according to chapter 4.2.2 from EN 1995-1-2. Before calculating the effective charring depth we need to determine some parameters involved in calculations (dchar,n, k0, d0).

■ Depth of layer with assumed zero strength and stiffness: d0 = 7 x 10-3 m; ■ Coefficient depending of fire resistance time and also depending if the members are protected or not:

k0 = 1.0 (according to table 4.1 from EN 1995-1-2)

■ Notional design charring depth:

dchar,n = m.minminm.tn 0420601070β 3 =⋅⋅=⋅ − (according to relation 3.2 from EN 1995-1-2)

■ Effective charring depth: def = 00 dkd n,char ⋅+

■ Residual cross section:

Afi = )db()dh( efef ⋅−×− 2




382

Residual cross section

Column with fixed base exposed to fire for 60 minutes Afi


Result name Result description Reference value Afi Residual cross section [m2] 0.056202 m2

14.2.3 Calculated results Result name Result description Value Error Afi Residual cross section 0.056202 m² 0.0000 %


383

14.3 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending

Test ID: 4901

Test status: Passed

14.3.1 Description Verifies the fire resistance of a rectangular cross section purlin made from solid timber C24 to resist simple bending. The purlin is exposed to fire on 3 faces for 30 minutes. The verification is made according to chapter 4.2.2 (Reduced cross section method) from EN 1995-1-2 norm.

14.3.2 Background Verifies the adequacy of the fire resistance for a rectangular cross section purlin made from solid timber C24 to resist simple bending. The purlin is exposed to fire on 3 faces for 30 minutes (the top of the purlin is not exposed to fire). Verification of the bending stresses corresponding to frequent combination of actions is realized.

Chapter 1.1.1.3 presents the results obtained with the theoretical background explained at chapter 1.1.1.2.

14.3.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test F.2; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

Loadings from the structure: G = 500 N/m2, Snow load: S = 700 N/m2, Frequent combination of actions: CFQ = 1.0 x G + 0.5 x S = 850 N/m2

Purlin with 3 supports

Units

Metric System

Geometry


■ Height: h = 0.20 m, ■ Width: b = 0.075 m, ■ Length: L = 3.30 m, ■ Distance between adjacent purlins (span): d = 1.5 m, ■ Section area: A = 15.0 x 10-3 m2 ,


384


Rectangular solid timber C24 is used.The following characteristics are used in relation to this material:

■ Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Density: ρ = 350 kg/m3, ■ Design charring rate (softwood): βn = 0.8 x 10-3 m/min, ■ Service class 1.

Boundary conditions


■ Outer: Support at start point (X = 0) restrained in translation along X, Y and Z, Support at middle point (X = 3.30) restrained in translation along X, Y and Z, Support at end point (X = 6.60) restrained in translation along X, Y, Z and restrained in rotation along

X. ■ Inner: None.

Loading


■ External: Uniformly distributed load: q = CFQ x d = 850 N/m2 x 1.5 m = 1275 N/m,

■ Internal: None.

14.3.2.2 Reference results in calculating the fire resistance of a timber purlin subjected to simple bending

In order to verify the fire resistance for a timber purlin subjected to simple bending it is necessary to determine the residual cross section. After this, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod,fi, γM,fi, kfi, km, have to be determined.

Residual cross section

The residual cross section is determined by reducing the initial cross section dimensions by the effective charring depth according to chapter 4.2.2 from EN 1995-1-2. Before calculating the effective charring depth we need to determine some parameters involved in calculations (dchar,n, k0, d0).

■ Depth of layer with assumed zero strength and stiffness: d0 = 7 x 10-3 m; ■ Coefficient depending of fire resistance time and also depending if the members are protected or not:

k0 = 1.0 (according to table 4.1 from EN 1995-1-2)

■ Notional design charring depth:

dchar,n = mmtn 024.0min30min

108.0 3 =⋅⋅=⋅ −β (according to relation 3.2 from EN 1995-1-2)

■ Effective charring depth:

def = 00, dkd nchar ⋅+

■ Residual cross section:

Afi = )2()( efefefef dbdhbh ⋅−×−=×


385

Reference solution for frequent combination of actions

Before calculating the reference solution (maximum work ratio for fire verification based on formulae (6.17) and (6.18) from EN 1995-1-1 norm) it is necessary to determine the design bending stress (taking into account the residual cross section), the design bending strength and some parameters involved in calculations (kmod,fi, γM,fi, kfi).

■ Modification factor in case of a verification done with residual section: kmod,fi = 1.0 (according to paragraph 5 from chapter 4.2.2 from EN 1995-1-2)

■ Partial safety factor for timber in fire situations: γM,fi = 1.0

■ Factor kfi, taken from table 2.1 (EN 1995-1-2): kfi = 1.25 (for solid timber)

■ Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections) – according to paragraph 2 from chapter 6.1.6 (EN 1995-1-1): km = 0.7

■ Design bending stress (taking into account the residual cross section):

σm,d = 2

6

efef

y

y

y

hbM

WM

×

×=

The picture below shows the bending moment diagram (kNm). My from the above formula represents the maximum bending moment achieved from frequent combination of actions.

■ Design bending strength (for fire situation):

fm,d,fi = Pak

fkfiM

fikmfi

66

,

mod,, 1030

0.10.1102425.1 ×=×××=××

γ

■ Work ratio according to formulae 6.17 from EN 1995-1-1 norm (considering that the axial effort, as well as the bending moment about z axis, are null):

0.1,

, ≤dm

dm

fσ

■ Work ratio according to formulae 6.18 from EN 1995-1-1 norm (considering that the axial effort, as well as the bending moment about z axis, are null):

0.1,

, ≤×dm

dmm f

kσ

■ Maximum work ratio for bending verification for fire situation:

100100;max,

,

,

,

,

, ×=×

×=

dm

dm

dm

dmm

dm

dm

ffk

fWR

σσσ




386

Residual cross-section area

Simply supported beam subjected to bending (fire situation) Residual area [m2]

Design bending stress taking into account the residual cross-section

Simply supported beam subjected to bending (fire situation) Design bending stress for residual cross-section [Pa]

Maximum work ratio for bending verification (fire situation)

Simply supported beam subjected to bending (fire situation) Work ratio [%]


387


Result name Result description Reference value Afi Residual area [m2] 0.002197 m2

Stress Design bending stress for residual cross-section [Pa] 27568524 Pa Work ratio Maximum work ratio for fire verification [%] 91.9 %

14.3.3 Calculated results Result name Result description Value Error Afi Residual area 0.002197 m² 0.0000 % Stress Design bending stress for residual cross-section 2.75626e+007 Pa 0.0000 % Work ratio Maximum work ratio for fire verification 91.8752 % 0.0000 %


388

14.4 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression

Test ID: 4877

Test status: Passed

14.4.1 Description Verifies the lateral torsional stability for a rectangular timber beam subjected to combined bending and axial compression. The verification is made following the rules from Eurocode 5 French annex.

14.4.2 Background Verifies the lateral torsional stability of a rectangular cross section made from solid timber C24 subjected to simple bending (about the strong axis) and axial compression.

14.4.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test E.2; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent rafters (span): d = 0.5 m. The following load cases and load combination are used:

■ Loadings from the structure: G = 450 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m2; All loads will be projected on the rafter direction, since its slope is 50% (26.6°).

Simply supported rafter subjected to projected loadings

Units

Metric System

Geometry


■ Height: h = 0.20 m, ■ Width: b = 0.05 m, ■ Length: L = 5.00 m, ■ Section area: A = 10 x 10-3 m2 ,


000333.06

20.005.06

mhbWy =⋅

=×

= .


389



■ Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Longitudinal elastic modulus: E = 1.1 x 1010 Pa, ■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 1010 Pa, ■ Service class 2.

Boundary conditions


■ Outer: Support at start point (z = 0) restrained in translation along Y, Z and restrained in rotation along X. Support at end point (z = 5.00) restrained in translation along X, Y, Z.

■ Inner: None.

Loading

The rafter is subjected to the following projected loadings (at ultimate limit state):

■ External: Uniformly distributed load: q = Cmax x d x cos26.6° = 1957.5 N/m2 x 0.5 m x cos26.6° = 875.15 N/m, Compressive load component: N = Cmax x d x sin26.6° x L = 1957.5 N/m2 x 0.5m x sin26.6° x 5.00m =

= 2191.22 N ■ Internal: None.

14.4.2.2 Reference results in calculating the timber beam subjected to combined stresses In order to verify the lateral torsional stability of a timber beam subjected to combined stresses at ultimate limit state, the formula (6.35) from EN 1995-1-1 norm is used. Before applying this formula we need to determine some parameters involved in calculations like: slenderness ratios, relative slenderness ratios, instability factors. After this, we calculate the reference solution which includes: the design compressive stress, the design bending stress and the work ratio based on formula (6.35) from EN 1995-1-1.

Slenderness ratios

In professional practice the slenderness ratio is limited to 120. The slenderness ratios corresponding to bending about y and z axes are determined as follows:

■ Slenderness ratio corresponding to bending about the z axis:

4.34605.051121212 =

×=

×==

mm

blm

bl gc

zλ

It is necessary to reduce the buckling length about the z axis, because λz exceeded the value 120. A restraint is placed in each tierce of the rafter, so that the slenderness ratio corresponding to bending about the z axis become:

1205.11505.0667.11121212 <=

×=

×==

mm

blm

bl gc

zλ

■ Slenderness ratio corresponding to bending about the y axis:

6.862.051121212 =

×=

×==

mm

hlm

hl gc

yλ


390

Relative slenderness ratios

The relative slenderness ratios are:

■ Relative slenderness ratio corresponding to bending about the z axis:

958.11074.0

10215.115

,10

6

050

,0,, =

××

==Pa

PaEf kcz

zrel ππλλ

■ Relative slenderness ratio corresponding to bending about the y axis:

468.11074.0

10216.86

,10

6

050

,0,, =

××

==Pa

PaEf kcy

yrel ππλ

λ

■ Maximum relative slenderness ratio:

( ) 958.1,max ,,max, == yrelzrelrel λλλ

So there is a risk of buckling because λrel,max ≥ 0.3.

Relative slenderness for bending

The effective length of the beam and the critical bending stress must be determined before calculating the relative slenderness for bending.

■ Effective length of the beam; its calculation is made according to table 6.1 from EN 1995-1-1 and it is based on the loading type and support conditions. The effective length is increased by “2⋅h” because the load is applied at the compressed fiber of the beam:

mmmhlleff 9.42.020.59.029.0 =⋅+⋅=⋅+⋅=

■ Critical bending stress (determined according to formula 6.32 from EN 1995-1-1):

σm,y,crit = Pamm

Pamlh

Ebef

61022

05.02

10724.149.42.0

1074.005.078.078.0×=

⋅⋅⋅⋅

=⋅

⋅

■ Relative slenderness for bending (determined according to formula 6.30 from EN 1995-1-1):

λrel,m = 277.110724.14

10246

6

,

, =⋅

⋅=

PaPaf

critm

km

σ

Instability factors

In order to determine the instability factors we need to determine the βc factor. It is a factor for solid timber members within the straightness limits defined in Section 10 from EN 1995-1-1:

βc = 0.2 (according to relation 6.29 from EN 1995-1-1)

The instability factors are: kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z

2] (according to relation 6.28 from EN 1995-1-1)

ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y2] (according to relation 6.27 from EN 1995-1-1)

2,

2,1

zrelzz

zckk

kλ−+

= (according to relation 6.26 from EN 1995-1-1)

2,

2,1

yrelyy

yckk

kλ−+



391


Before calculating the reference solution (the design compressive stress, the design bending stress and the work ratio based on formula (6.35) from EN 1995-1-1) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km, kcrit).





■ Factor which takes into account the reduced bending strength due to lateral buckling: Kcrit = 1.56 – 0.75λrel,m (because 0.75 < λrel,m < 1.4)

■ Design compressive stress (induced by the compressive component, N):

σc,d = PamN

AN 219122

101022.2191

23 =×

= −

■ Design compressive strength:

fc,0,d = PakfM

kc66mod

,0, 10538.143.19.01021 ×=××=×

γ

■ Design bending stress (induced by uniformly distributed load, q):

σm,d = Pam

mmN

WLq

WM

yy

y 63

222

10213.8000333.08

00.515.875

8×=

×

×=

××

=


fm,y,d = Pakkkf hsysM

km66mod

, 10615.160.10.13.19.01024 ×=××××=×××

γ

■ Work ratio according to formula 6.35 from EN 1995-1-1 norm:

1,0,,

,

2

,,

, ≤⋅

+

⋅ dczc

dc

dymcrit

dm

fkfkσσ




392

Stress SFx diagram

Simply supported rafter subjected to compressive component of the applied forces Stress SFx [Pa]

Stress SMy diagram

Simply supported rafter subjected to uniformly distributed loads Stress SMy [Pa]

Lateral-torsional stability work ratio

Stability of a simply supported rafter subjected to combined stresses Work ratio [%]


Result name Result description Reference value SFx Design compressive stress [Pa] 219122 Pa SMy Design bending stress [Pa] 8212744 Pa kcrit Kcrit factor 0.602

Work ratio Lateral-torsional stability work ratio [%] 76 %

14.4.3 Calculated results Result name Result description Value Error Stress SFx Design compressive stress 219124 Pa 0.0000 % Stress SMy Design bending stress 8.20519e+006 Pa 0.0000 % Kcrit kcrit factor 0.592598 adim 0.0000 % Work ratio Work ratio according to formula 6.35 0.759538 adim 0.0000 %


393

14.5 EC5: Shear verification for a simply supported timber beam

Test ID: 4822

Test status: Passed

14.5.1 Description Verifies a rectangular cross section beam made from solid timber C24 to shear efforts. The verification of the shear stresses at ultimate limit state is performed.

14.6 Modifying the "Design experts" properties for timber linear elements (TTAD #12259)

Test ID: 4509

Test status: Passed

14.6.1 Description Defines the "Design experts" properties for a timber linear element, in a model created with a previous version of the program.

14.7 Verifying the timber elements shape sheet (TTAD #12337)

Test ID: 4538

Test status: Passed

14.7.1 Description Verifies the timber elements shape sheet.

14.8 Verifying the units display in the timber shape sheet (TTAD #12445)

Test ID: 4539

Test status: Passed

14.8.1 Description Verifies the Afi units display in the timber shape sheet.


394

14.9 EC5: Verifying a timber beam subjected to simple bending

Test ID: 4682

Test status: Passed

14.9.1 Description Verifies a rectangular cross section beam made from solid timber C24 to resist simple bending. Verifies the bending stresses at ultimate limit state, as well as the deflections at serviceability limit state.

14.9.2 Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending. Verification of the bending stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit state are performed.

14.9.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test C; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

■ Loadings from the structure: G = 0.5 kN/m2, ■ Exploitation loadings (category A): Q = 1.5 kN/m2, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2 ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q

Simply supported beam

Units

Metric System

Geometry


■ Height: h = 0.20 m, ■ Width: b = 0.075 m, ■ Length: L = 4.50 m, ■ Distance between adjacent beams (span): d = 0.5 m, ■ Section area: A = 15.0 x 10-3 m2 ,

■ Elastic section modulus about the strong axis y: 322

0005.06

20.0075.06

mhbWy =⋅

=×

=


395



■ Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Service class 1.

Boundary conditions


■ Outer: Support at start point (z=0) restrained in translation along X, Y and Z, Support at end point (z = 4.5) restrained in translation along X, Y, Z and restrained in rotation along X.

■ Inner: None.

Loading


■ External: Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,

■ Internal: None.

14.9.2.2 Reference results in calculating the timber beam subjected to uniformly distributed loads In order to verify the timber beam bending stresses at ultimate limit state, the formulae (6.11) and (6.12) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, must be calculated. After this, the reference solution, which includes the design bending stress about the principal y axis, the design bending strength and the corresponding work ratios, is calculated.

A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the effective values with the limiting values for deflections specified in EN 1995-1-1 norm.


Before calculating the reference solution (design bending stress, design bending strength and work ratios) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).

■ Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)





■ Design bending stress (induced by the applied forces):

σm,d = PahbLq

WM

y

y 62

2

2

2

104039.72.0075.085.44625.16

86

×=××××

=××××

=


fm,d = Pakkkf hsysM

km66mod

, 10769.140.10.13.18.01024 ×=××××=×××

γ


396

■ Work ratio according to formulae 6.11 from EN 1995-1-1 norm:

0.1,

, ≤dm

dm

fσ


0.1,

, ≤×dm

dmm f

kσ

Reference solution for serviceability limit state verification

The following limiting values for instantaneous deflection (for a base variable action), final deflection and net deflection are considered:

300)( LQwinst ≤

125Lwfin ≤

200,Lw finnet ≤

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are the followings:

■ Instantaneous deflection (for a base variable action):

8.600)(00749.0)( LQwmQw instinst =⇒=

■ Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

45.45000999.0 Lwmdw instCQinst =⇒==

■ In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the

deformation factor (kdef) has to be chosen:

6.0=defk (calculated value for service class 1, according to table 3.2 from EN 1995-1-1)

95.157800285.000475.06.06.0 Lwmmdw creepQPcreep =⇒=×=×=

■ Final deflection:

47.35001284.000285.000999.0 Lwmmmwww fincreepinstfin =⇒=+=+=

■ Net deflection:

47.35001284.0001284.0 ,,

Lwmmmwww finnetcfinfinnet =⇒=+=+=




397

Work ratio diagram

Simply supported beam subjected to bending Strength work ratio


Result name Result description Reference value σm,d Design bending stress [Pa] 7403906.25 Pa Strength work ratio Work ratio (6.11) [%] 50 % winst (Q) Deflection for a base variable action [m] 0.00749 m

dCQ Deflection for a characteristic combination of actions [m] 0.00999 m

winst Instantaneous deflection [m] 0.00999 m kdef Deformation coefficient 0.6 dQP Deflection for a quasi-permanent combination of actions [m] 0.00475 m wfin Final deflection [m] 0.01284 m wnet,fin Net deflection [m] 0.01284 m

14.9.3 Calculated results Result name Result description Value Error Stress Design bending stress 7.40391e+006

Pa 0.0000 %

Work ratio Work ratio (6.11) 50.1306 % 0.0000 % D Deflection for a base variable action 0.00749224 m 0.0000 % D Deflection for a characteristic combination of actions 0.00998966 m 0.0000 % Winst Instantaneous deflection 0.00998966 m 0.0000 % Kdef Deformation coefficient 0.6 adim 0.0000 % D Deflection for a quasi-permanent combination of actions 0.00474509 m 0.0000 % Wfin Final deflection 0.0128367 m 0.0000 % Wnet,fin Net deflection 0.0128367 m 0.0000 %


398

14.10 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression

Test ID: 4879

Test status: Passed

14.10.1Description Verifies the stability of a rectangular timber purlin made from solid timber C24 subjected to biaxial bending and axial compression. The verification is made following the rules from Eurocode 5 French annex.

14.10.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to biaxial bending and axial compression. The verification is made according to formulae (6.23) and (6.24) from EN 1995-1-1 norm.

14.10.2.1Model description ■ Reference: Guide de validation Eurocode 5, test E.4; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent purlins (span): d = 1.8 m. The following load cases and load combination are used:

■ The ultimate limit state (ULS) combination is: CULS = 1.35 x G + 1.5 x S + 0.9 x W; ■ Loadings from the structure: G = 550 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ Axial compression force due to wind effect on the supporting elements: W = 15000 N; ■ Uniformly distributed load corresponding to the ultimate limit state combination:

Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2.

All loads will be projected on the purlin direction since its slope is 30% (17°).

Simply supported purlin subjected to loadings

Units

Metric System

Geometry


■ Height: h = 0.20 m, ■ Width: b = 0.10 m, ■ Length: L = 3.50 m, ■ Section area: A = 0.02 m2 ,


000666.06

20.01.06

mhbWy =⋅

=×

= ,


399

■ Elastic section modulus about the strong axis, z: 322

000333.06

20.01.06

mhbWz =⋅

=×

= .


Rectangular solid timber C24 is used. The followings characteristics are used in relation to this material:

■ Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Longitudinal elastic modulus: E = 1.1 x 1010 Pa, ■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 1010 Pa, ■ Service class 2.

Boundary conditions


■ Outer: Support at start point (z = 0) restrained in translation along X, Y, Z; Support at end point (z = 3.50) restrained in translation along Y, Z and restrained in rotation along X.

■ Inner: None.

Loading

The purlin is subjected to the following projected loadings (corresponding to the ultimate limit state combination):

■ External: Axial compressive load: N =0.9 x W = 13500 N; Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x sin17° =

1101.22 N/m, Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x cos17° =

3601.92 N/m, ■ Internal: None.

14.10.2.2Reference results in calculating the timber purlin subjected to combined stresses In order to verify the stability of a timber purlin subjected to biaxial bending and axial compression at ultimate limit state, the formulae (6.23) and (6.24) from EN 1995-1-1 norm are used. Before applying these formulae we need to determine some parameters involved in calculations like: slenderness ratios, relative slenderness ratios, instability factors. After this, we’ll calculate the maximum work ratio for stability verification, which represents in fact the reference solution.

Slenderness ratios

The slenderness ratios corresponding to bending about y and z axes are determined as follows:


24.1211.0

5.31121212 =×

=×

==m

mb

lmbl gc

zλ

■ Slenderness ratio corresponding to bending about the y axis:

62.602.0

5.31121212 =×

=×

==m

mh

lmhl gc

yλ


400




056.21074.0

102124.12110

6

05,0

,0,, =

××

==Pa

PaEf kcz

zrel ππλλ

■ Relative slenderness ratio corresponding to bending about the y axis:

028.11074.0

102162.60

,10

6

050

,0,, =

××

==Pa

PaEf kcy

yrel ππλ

λ

■ Maximum relative slenderness ratio:

( ) 056.2,max ,,max, == yrelzrelrel λλλ

So there is a risk of buckling because λrel,max ≥ 0.3.

Instability factors



The instability factors are: kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z

2] (according to relation 6.28 from EN 1995-1-1)

ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y2] (according to relation 6.27 from EN 1995-1-1)

2,

2,1

zrelzz

zckk

kλ−+


2,

2,1

yrelyy

yckk

kλ−+



Before calculating the reference solution (the maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1) it is necessary to determine the design compressive stress, the design compressive strength, the design bending stress, the design bending strength and some parameters involved in calculations (kmod, γM, kh, ksys, km).

■ Design compressive stress (induced by the axial compressive load from the corresponding ULS combination, N):

σc,0,d = Pam

NAN 675000

02.013500

2 ==

■ Design bending stress about the y axis (induced by uniformly distributed load, qz):

σm,y,d = Pam

mmN

WLq

WM

y

z

y

y 63

222

102814.8000666.08

50.392.3601

8×=

×

×=

××

=

■ Design bending stress about the z axis (induced by uniformly distributed load, qy):

σm,z,d = Pam

mmN

WLq

WM

z

y

z

z 63

222

100638.5000333.08

50.322.1101

8×=

×

×=

×

×=


401

■ Modification factor for duration of load (instantaneous action) and moisture content (service class 2):

kmod = 1.1 (according to table 3.1 from EN 1995-1-1)





fc,0,d = PakfM

kc66mod

,0, 10769.173.11.11021 ×=××=×

γ


fm,y,d = fm,z,d = Pakkkf hsysM

km66mod

, 10308.200.10.13.11.11024 ×=××××=×××

γ

■ Maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1:

1max

,,

,,

,,

,,

,0,,

,0,

,,

,,

,,

,,

,0,,

,0,

≤

+⋅+⋅

⋅++⋅

dzm

dzm

dym

dymm

dczc

dc

dzm

dzmm

dym

dym

dcyc

dc

ffk

fk

fk

ffk

σσσ

σσσ




402

Instability factor, kcy

Simply supported purlin subjected to biaxial bending and axial compression Kc,y

Instability factor, kcz

Simply supported purlin subjected to biaxial bending and axial compression Kc,z

Maximum work ratio for stability verification

Simply supported purlin subjected to biaxial bending and axial compression Work ratio [%]


Result name Result description Reference value Kc,y Instability factor, kc,y 0.67 Kc,z Instability factor, kc,z 0.21

Work ratio Maximum work ratio for stability verification [%] 71.1 %

14.10.3Calculated results Result name Result description Value Error Kc,y Instability factor, kc,y 0.665025 adim 0.0000 % Kc,z Instability factor, kc,z 0.212166 adim 0.0000 % Work ratio Maximum work ratio for stability verification 70.6586 % -0.6208 %


403

14.11 EC5: Verifying a timber column subjected to tensile forces

Test ID: 4693

Test status: Passed

14.11.1Description Verifies the tensile resistance of a rectangular cross section column (fixed at base) made from solid timber C24.

14.11.2Background Verifies the adequacy of the tension resistance for a rectangular cross section made from solid timber C24. The verification is made according to formula (6.1) from EN 1995-1-1 norm.

14.11.2.1Model description ■ Reference: Guide de validation Eurocode 5, test A; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

Column with fixed base

Units

Metric System

Geometry

Cross section characteristics:

■ Height: h = 0.122 m, ■ Width: b = 0.036m, ■ Section area: A = 43.92 x 10-4 m2 ■ I = 5.4475 x 10-6 m4.


404



■ Longitudinal elastic modulus: E = 1.1 x 1010 Pa, ■ Characteristic tensile strength along the grain: ft,0,k = 14 x 106 Pa, ■ Service class 2.

Boundary conditions

The boundary conditions:

■ Outer: Fixed at base (z = 0),

Free at top (z = 5),

■ Inner: None.

Loading


■ External: Point load at z = 5: Fz = N = 10000 N, ■ Internal: None.

14.11.2.2Reference results in calculating the timber column subjected to tension force

Reference solution

The reference solution is determined by formula (6.1) from EN 1995-1-1. Before applying this formula we need to determine some parameters involved in calculations (kmod, γM, kh). After this, the design tensile stress, the design tensile strength and the corresponding work ratio are calculated.

■ Modification factor for duration of load and moisture content: kmod = 0.9 ■ Partial factor for material properties: γM = 1.3 ■ Depth factor (“h” represents the width, because the element is tensioned):

kh = min

3.1

150 2.0

h

■ Design tensile stress (induced by the ultimate limit state force, N):

σt,0,d = AN

■ Design tensile strength:

ft,0,d = hM

kt kkf ××γ

mod,0,

■ Work ratio:

SFx = 0.1,0,

,0, ≤dt

dt

fσ

(according to relation 6.1 from EN 1995-1-1)




405

Work ratio SFx diagram

Column with fixed base, subjected to tension force Work ratio SFx


Result name Result description Reference value σt,0,d Design tensile stress [Pa] 2276867.03 Pa SFx Work ratio [%] 18 %

14.11.3Calculated results Result name Result description Value Error Stress SFx Design tensile stress 2.27687e+006 Pa 0.0000 % Work ratio Work ratio 18.0704 % 0.3911 %


406

14.12 EC5: Verifying a timber column subjected to compression forces

Test ID: 4823

Test status: Passed

14.12.1Description Verifies the compressive resistance of a rectangular cross section column (hinged at base) made from solid timber C18.

14.12.2Background Verifies the adequacy of the compressive resistance for a rectangular cross section made from solid timber C18. The verification is made according to formula (6.35) from EN 1995-1-1 norm.

14.12.2.1Model description ■ Reference: Guide de validation Eurocode 5, test B; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

Simply supported column

Units

Metric System

Geometry

Column cross section characteristics:

■ Height: h = 0.15 m, ■ Width: b = 0.10 m, ■ Section area: A = 15.0 x 10-3 m2



■ Longitudinal elastic modulus: E = 0.9 x 1010 Pa,


407

■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.6 x 1010 Pa, ■ Characteristic compressive strength along the grain: fc,0,k = 18 x 106 Pa, ■ Service class 1.

Boundary conditions

The boundary conditions:

■ Outer: Support at base (z=0) restrained in translation along X, Y and Z, Support at top (z = 3.2) restrained in translation along X, Y and restrained in rotation along Z.

■ Inner: None.

Loading


■ External: Point load at z = 3.2: Fz = N = -20000 N, ■ Internal: None.

14.12.2.2Reference results in calculating the timber column subjected to compression force The formula (6.35) from EN 1995-1-1 is used in order to verify a timber column subjected to compression force. Before applying this formula we need to determine some parameters involved in calculations, such as: slenderness ratios, relative slenderness ratios, instability factors. After this, the reference solution is calculated. This includes: the design compressive stress, the design compressive strength and the corresponding work ratio.

Slenderness ratios

The most important slenderness is calculated relative to the z axis, as it will be the axis of rotation if the column buckles.


85.1101.0

2.31121212 =×

=×

==m

mb

lmbl gc

zλ

■ Slenderness ratio corresponding to bending about the y axis (informative):

9.7315.0

2.31121212 =×

=×

==mm

hlm

hl gc

yλ




933.1106.01018

1.0122.31

,

12

,10

6

050

,0,

050

,0,, =

××

×××

=×

××==

PaPa

mm

Ef

blm

Ef kcgkcz

zrel πππλλ

■ Relative slenderness ratio corresponding to bending about the y axis (informative):

288.1106.01018

15.0122.31

,

12

,10

6

050

,0,

050

,0,, =

××

×××

=×

××==

PaPa

mm

Ef

hlm

Ef kcgkcy

yrel πππλ

λ

So there is a risk of buckling, because λrel,max ≥ 0.3.

Instability factors




408

The instability factors are: ■ kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z

2] (according to relation 6.28 from EN 1995-1-1) ■ ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y

2] (informative)

■ 2

,2,1

zrelzz

zckk

kλ−+


■ 2

,2,1

yrelyy

yckk

kλ−+

= (informative)

Reference solution

Before calculating the reference solution (design compressive stress, design compressive strength and work ratio) we need to determine some parameters involved in calculations (kmod, γM).


■ Partial factor for material properties: γM = 1.3 ■ Design compressive stress (induced by the applied forces):

σc,0,d = AN


fc,0,d = M

kckfγ

mod,0,

■ Work ratio:

Work ratio = 0.1,0,,

,0, ≤× dczc

dc

fkσ





409

Work ratio diagram

Simply supported column subjected to compression force Work ratio


Result name Result description Reference value kc,z Instability factor 0.2400246 kc,y Instability factor (informative) 0.488869 σc,0,d Design compressive stress [Pa] 1333333 Pa Work ratio Work ratio [%] 50 %

14.12.3Calculated results Result name Result description Value Error Kc,z Instability factor 0.240107 adim 0.0000 % Kc,y Instability factor 0.488612 adim 0.0000 % Stress SFx Design compressive stress 1.33333e+006 Pa 0.0000 % Work ratio Work ratio 50.1319 % 0.2638 %


410

14.13 EC5: Verifying a timber beam subjected to combined bending and axial tension

Test ID: 4872

Test status: Passed

14.13.1Description Verifies a rectangular cross section rafter made from solid timber C24 to resist combined bending and axial tension. The verification of the cross-section subjected to combined stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit state are performed.

14.13.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending and axial tension. The verification of the deflections at serviceability limit state is also performed.

14.13.2.1Model description ■ Reference: Guide de validation Eurocode 5, test E; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent rafters (span): d = 0.5 m. The following load cases and load combination are used:

■ Loadings from the structure: G = 450 N/m2; ■ Snow load (structure is located at an altitude > 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m2; ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x S; ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.2 x S. All loads will be projected on the rafter direction since its slope is 50% (26.6°).

Simply supported rafter subjected to projected loadings

Units

Metric System

Geometry


■ Height: h = 0.20 m, ■ Width: b = 0.05 m, ■ Length: L = 5.00 m, ■ Section area: A = 10 x 10-3 m2 ,

■ Elastic section modulus about the strong axis y: 322

000333.06

20.005.06

mhbWy =⋅

=×

= .


411



■ Characteristic tensile strength along the grain: ft,0,k = 14 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Service class 2.

Boundary conditions


■ Outer: Support at start point (z=0) restrained in translation along Y, Z and restrained in rotation along X. Support at end point (z = 5.00) restrained in translation along X, Y, Z.

■ Inner: None.

Loading

The rafter is subjected to the following projected loadings (at ultimate limit state):

■ External: Uniformly distributed load: q = Cmax x d x cos26.6° = 1957.5 N/m2 x 0.5 m x cos26.6° = 875.15 N/m, Tensile load component: N = Cmax x d x sin26.6° x L = 1957.5 N/m2 x 0.5m x sin26.6° x 5.00m =

= 2191.22 N ■ Internal: None.

14.13.2.2Reference results in calculating the timber beam subjected to combined stresses In order to verify the timber beam subjected to combined stresses at ultimate limit state, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, must be determined. After this the reference solution, which consists of the design tensile stress, the design tensile strength and the corresponding work ratio and also the work ratios of the combined stresses, is calculated.

A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the effective values with the limiting values for deflections specified in EN 1995-1-1 norm.


Before calculating the reference solution (the design tensile stress, the design tensile strength and the corresponding work ratio, and also the work ratios of the combined stresses) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).



■ Depth factor (“h” represents the width in millimeters because the element is tensioned):

kh = min 25.13.125.1

min3.150

150min

3.1

150 2.02.0

=

=

=

h



■ Design tensile stress (induced by the ultimate limit state force, N):

σt,0,d = PamN

AN 219122

101022.2191

23 =×

= −


412

■ Design tensile strength:

ft,0,d = Pakkf hM

kt66mod

,0, 10115.1225.13.19.01014 ×=×××=××

γ

■ Work ratio:

SFx = 0.1,0,

,0, ≤dt

dt

fσ


■ Design bending stress (induced by the applied forces):

σm,y,d = Pamm

mmN

hbLq

WM

y

y 622

22

2

2

102045.82.005.08

00.515.8756

86

×=××

××=

××××

=


fm,y,d = Pakkkf hsysM

km66mod

, 10615.160.10.13.19.01024 ×=××××=×××

γ


1,,

,,

,,

,,

,0,

,0, ≤++dzm

dzmm

dym

dym

dt

dt

fk

ffσσσ


1,,

,,

,,

,,

,0,

,0, ≤++dzm

dzm

dym

dymm

dt

dt

ffk

fσσσ

Reference solution for serviceability limit state verification

The following limiting values for instantaneous deflection (for a base variable action), final deflection and net deflection are considered:

300)( LQwinst ≤

125Lwfin ≤

200,Lw finnet ≤

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are:

■ Instantaneous deflection (for a base variable action):

05.547)(00914.0)( LQwmQw instinst =⇒=

■ Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

7.36401371.0 Lwmdw instCQinst =⇒==

■ In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the

deformation factor (kdef) has to be chosen:

8.0=defk (value determined for service class 2, according to table 3.2 from EN 1995-1-1)

6.97600512.00064.08.08.0 Lwmmdw creepQPcreep =⇒=×=×=


413

■ Final deflection:

5.26501883.000512.001371.0 Lwmmmwww fincreepinstfin =⇒=+=+=

■ Net deflection:

5.26501883.0001883.0 ,,

Lwmmmwww finnetcfinfinnet =⇒=+=+=



SFx work ratio diagram

Simply supported beam subjected to tensile forces Work ratio SFx

Strength work ratio diagram

Simply supported beam subjected to combined stresses Strength work ratio

Instantaneous deflection winst(Q)

Simply supported beam subjected to snow loads Instantaneous deflection winst(Q) [m]

Instantaneous deflection winst(CQ)

Simply supported beam subjected to a characteristic load combination of actions Instantaneous deflection winst(CQ) [m]


414

Final deflection wfin

Simply supported beam subjected to characteristic load combination of actions Instantaneous deflection winst(CQ) [m]

Net deflection wnet,fin

Simply supported beam subjected to characteristic load combination of actions Instantaneous deflection winst(CQ) [m]


Result name Result description Reference value SFx SFx work ratio [%] 1.808 % Strength work ratio Work ratio (6.17) [%] 51.19 % winst (Q) Deflection for a base variable action [m] 0.00914 m

dCQ Deflection for a characteristic combination of actions [m] 0.01371 m

winst Instantaneous deflection [m] 0.01371 m kdef Deformation coefficient 0.8 dQP Deflection for a quasi-permanent combination of actions [m] 0.0064 m wfin Final deflection [m] 0.01883 m wnet,fin Net deflection [m] 0.01883 m

14.13.3Calculated results Result name Result description Value Error Work ratio SFx SFx work ratio 1.81483 % 0.3778 % Work ratio Strength work ratio 51.1941 % 0.0000 % D w_inst(Q) 0.00914038 m 0.0000 % D deflection for a characteristic combination 0.0137106 m 0.0000 % Winst instantaneous deflection 0.0137105 m 0.0000 % Kdef deformation coefficient 0.8 adim 0.0000 % D deformation for a quasi-permanent combination 0.00639828 m 0.0000 % Wfin final deflection 0.0188291 m 0.0000 % Wnet,fin net final deflection 0.0188291 m 0.0000 %


415

14.14 EC5: Verifying a C24 timber beam subjected to shear force

Test ID: 5036

Test status: Passed

14.14.1Description Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification of the shear stresses at ultimate limit state is performed.

14.14.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification of the shear stresses at ultimate limit state is performed.

14.14.2.1Model description ■ Reference: Guide de validation Eurocode 5, test D; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

■ Loadings from the structure: G = 0.5 kN/m2, ■ Exploitation loadings (category A): Q = 1.5 kN/m2, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2

Simply supported beam

Units

Metric System

Geometry

Beam cross section characteristics:

■ Height: h = 0.225 m, ■ Width: b = 0.075 m, ■ Length: L = 5.00 m, ■ Distance between adjacent beams (span): d = 0.5 m, ■ Section area: A = 16.875 x 10-3 m2 ,



■ Characteristic shear strength: fv,k = 2.5 x 106 Pa, ■ Service class 1.


416

Boundary conditions


■ Outer: Support at start point (z=0) restrained in translation along X, Y and Z, Support at end point (z = 5.00) restrained in translation along X, Y, Z and restrained in rotation along X.

■ Inner: None.

Loading


■ External: Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,

■ Internal: None.

14.14.2.2Reference results in calculating the timber beam subjected to uniformly distributed loads In order to verify the timber beam shear stresses at ultimate limit state, the formula (6.13) from EN 1995-1-1 norm is used. Before using it, some parameters involved in calculations, like kmod, kcr, γM, kf, beff, heff, have to be determined. After this the reference solution, which includes the design shear stress about the principal y axis, the design shear strength and the corresponding work ratios, is calculated.


Before calculating the reference solution (design shear stress, design shear strength and work ratio) it is necessary to determine some parameters involved in calculations (kmod, γM, kcr, kf, beff, heff).



■ Cracking factor, kcr : kcr = 0.67 (for solid timber)

■ Factor depending on the shape of the cross section, kf: kf = 3/2 (for a rectangular cross section)

■ Effective width, beff: beff = kcr x b = 0.67 x 0.075m = 0.05025m

■ Effective height, heff: heff = h = 0.225m

■ Design shear stress (induced by the applied forces):

τd = Pamm

N

hbFk

effeff

dvf 6, 10485075.0225.005025.0

25.365623

×=×

×=

×

×

■ Design shear strength:

fv,d = PaPakfM

kv66mod

, 10538.13.18.0105.2 ×=××=×

γ


0.1,

≤dv

d

fτ


417



Shear force, Fz, diagram

Simply supported beam subjected to bending Shear force diagram [N]

Design shear stress diagram

Simply supported beam subjected to bending Design shear stress [Pa]

Shear strength work ratio diagram

Simply supported beam subjected to bending Work ratio S_d [%]


418


Result name Result description Reference value Fz Shear force [kN] 3.65625 kN Stress S_d Design shear stress [Pa] 485074.63 Pa

Work ratio S_d Shear work ratio (6.13) [%] 32 %

14.14.3Calculated results Result name Result description Value Error Fz Shear force -3.65625 kN 0.0000 % Stress S_d Design shear stress 485075 Pa 0.0001 % Working ratio S_d Shear strength work ratio 31.5299 % -1.4691 %

031516-0409-0622

AD Validation Guide Vol2 2016 EN

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