AD Validation Guide Vol II 2017 R2 En

VALIDATION GUIDE

2017 R2

Advance Design Validation Guide

Version: 2017 R2 Tests passed on: 29 August 2016 Number of tests: 610

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 Operating Officer

ADVANCE DESIGN VALIDATION GUIDE

Table of Contents

9

10

MESH ..................................................................................................................................... 15 9.1

Verifying the mesh for a model with generalized buckling (TTAD #11519) ...................................................16

9.2

Verifying mesh points (TTAD #11748) ..........................................................................................................16

9.3

Creating triangular mesh for planar elements (TTAD #11727) .....................................................................16

9.4

Verifying the mesh of a planar element influenced by peak smoothing. .......................................................16

9.5

Verifying the options to take into account loads in linear and planar elements mesh (TTAD #15251)..........16

REPORTS GENERATOR ...................................................................................................... 17 10.1

Generating a report with modal analysis results (TTAD #10849) ................................................................18

10.2 Reports - Global envelope for efforts in linear elements with Min/Max Values and coresponding absicsa position ...................................................................................................................................................................18 10.3

Verifying the steel shape sheet display (TTAD #12657) ..............................................................................18

10.4

Verifying the modal analysis report (TTAD #12718) ....................................................................................18

10.5

EC2 / NF EN 1992-1-1/NA - France: Verifying the EC2 calculation assumptions report (TTAD #11838) ...18

10.6

Verifying the shape sheet strings display (TTAD #12622) ...........................................................................19

10.7

Verifying the Max row on the user table report (TTAD #12512) ..................................................................19

10.8

Verifying the shape sheet report (TTAD #12353) ........................................................................................19

10.9

Verifying the shape sheet for a steel beam (TTAD #12455) ........................................................................19

10.10 Verifying the global envelope of linear elements forces result (on start and end of super element) (TTAD #12230) .....................................................................................................................................................19 10.11 Verifying the global envelope of linear elements stresses (on start and end of super element) (TTAD #12230) .....................................................................................................................................................20 10.12 Verifying the global envelope of linear elements displacements (on end points and middle of super element) (TTAD #12230) ........................................................................................................................................20 10.13 Verifying the global envelope of linear elements displacements (on start and end of super element) (TTAD #12230) .......................................................................................................................................................20 10.14 Verifying the global envelope of linear elements forces result (on end points and middle of super element) (TTAD #12230) .......................................................................................................................................................20 10.15 Verifying the global envelope of linear elements displacements (on all quarters of super element) (TTAD #12230) ......................................................................................................................................................21 10.16 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element) (TTAD #12230) ............................................................................................................................................ 21 10.17 Verifying the global envelope of linear elements displacements (on the start point of super element) (TTAD #12230) .......................................................................................................................................................21 10.18

Verifying the global envelope of linear elements stresses (on each 1/4 of mesh element) (TTAD #12230) ......21

10.19 Verifying the global envelope of linear elements stresses (on all quarters of super element) (TTAD #12230) ..................................................................................................................................................................22 10.20 Verifying the global envelope of linear elements stresses (on the end point of super element) (TTAD #12230, TTAD #12261) ..........................................................................................................................................22 10.21 Verifying the global envelope of linear elements stresses (on the start point of super element) (TTAD #12230) ..................................................................................................................................................................22 7

ADVANCE DESIGN VALIDATION GUIDE

10.22 Verifying the global envelope of linear elements displacements (on the end point of super element) (TTAD #12230, TTAD #12261) .............................................................................................................................. 22 10.23 Verifying the global envelope of linear elements stresses (on end points and middle of super element) (TTAD #12230) ...................................................................................................................................................... 23 10.24

Verifying the Min/Max values from the user reports (TTAD# 12231) ........................................................ 23

10.25

Modal analysis: eigen modes results for a structure with one level ........................................................... 23

10.26

Generating the critical magnification factors report (TTAD #11379) ......................................................... 23

10.27

System stability when the column releases interfere with support restraints (TTAD #10557) ................... 23

10.28

Verifying the model geometry report (TTAD #12201) ............................................................................... 24

10.29 Verifying the global envelope of linear elements forces result (on the end point of super element) (TTAD #12230, #12261) .................................................................................................................................................... 24 10.30 Verifying the global envelope of linear elements forces result (on the start point of super element) (TTAD #12230) ...................................................................................................................................................... 24 10.31 Verifying the global envelope of linear elements forces result (on all quarters of super element) (TTAD #12230)...................................................................................................................................................... 24 10.32 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element) (TTAD #12230) ....................................................................................................................................................... 25

11

10.33

Creating the steel materials description report (TTAD #11954) ................................................................ 25

10.34

Creating the rules table (TTAD #11802) ................................................................................................... 25

SEISMIC ANALYSIS ............................................................................................................. 27 11.1

EC8 / NF EN 1998-1-1 - France: Verifying the spectrum results for EC8 seism (TTAD #11478)................ 28

11.2 EC8 / CSN EN 1998-1 - Czech Republic: Verifying the displacements results of a linear element (DEV2012 #3.18).................................................................................................................................................... 28 11.3

Verifying signed concomitant linear elements envelopes on Fx report (TTAD #11517) .............................. 28

11.4

EC8 / SR EN 1998-1/NA - Romania: Verifying the spectrum results for EC8 seism (TTAD #12472) ......... 28

11.5

EC8 / NF EN 1998-1-1 - France: Verifying torsors on grouped walls from a multi-storey concrete structure ...... 29

11.6

EC8 / NF EN 1998-1-1 - France: Verifying torsors on walls ........................................................................ 29

11.7 EC8 / SR EN 1998-1-1 - Romania: Verifying action results and torsors per modes on point, linear and planar supports (TTAD #14840) ............................................................................................................................. 29 11.8

RPA99/2003 - Algeria: Verifying the displacements results of a linear element (DEV2013 #3.5) ............... 29

11.9 EC8 / EN 1998-1-1 - General: Verifying torsors on a 6 storey single concrete core subjected to horizontal forces and seismic action ....................................................................................................................................... 29 11.10 EC8 / NF EN 1998-1-1 - France: verifying torsors on walls, elastic linear supports and user-defined section cuts (TTAD #14460) ................................................................................................................................... 30 11.11 EC8 / EN 1998-1-1 - General: Verifying the displacements results of a linear element for spectrum with renewed building option (TTAD #14161) ................................................................................................................ 30 11.12 EC8 / NF EN 1998-1-1 - France: Verifying the sum of actions on supports and nodes restraints (TTAD #12706) ..................................................................................................................................... 30 11.13 EC8 / NF EN 1998-1-1 - France: Verifying seismic efforts on planar elements with Q4 and T3-Q4 mesh type (TTAD #14244) ............................................................................................................................................... 30 11.14 EC8 / EN 1998-1-1 - General: Verifying the displacements results of a linear element for an envelope spectrum (DEV2013 #8.2) ...................................................................................................................................... 31 11.15 PS92/2010 - France: Verifying the displacements results of a linear element for an envelope spectrum (DEV2013 #8.2)...................................................................................................................................................... 31

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ADVANCE DESIGN VALIDATION GUIDE

11.16

PS92 - France: Verifying efforts and torsors on planar elements (TTAD #12974).....................................31

11.17 EC8 / NF EN 1993-1-8/NA - France: Verifying the damping correction influence over the efforts in supports (TTAD #13011). .......................................................................................................................................31 11.18 EC8 / NF EN 1998-1-1 - France: Verifying seismic results when a design spectrum is used (TTAD #13778) ..................................................................................................................................................................32 11.19 EC8 / NF EN 1998-1-1 - France: Generating forces results per modes on linear and planar elements (TTAD #13797) .......................................................................................................................................................32 11.20 RPS 2011 - Morocco: Verifying the displacements results of a linear element for an envelope spectrum (DEV2013 #8.2) ......................................................................................................................................................32 11.21

RPS 2011 - Morocco: Verifying the displacements results of a linear element (DEV2013 #3.6) ...............32

11.22 EC8 / SR EN 1998-1-1 - Romania: Verifying the displacements results of a linear element for an envelope spectrum (DEV2013 #8.2) .......................................................................................................................33 11.23

EC8 / NF EN 1998-1-1 - France: Verifying torsors on walls with Seismic Loads (TTAD #16522) .............33

11.24 EC8 / NF EN 1998-1-1 - France: Verifying earthquake description report in analysis with Z axis down (TTAD #15095) .......................................................................................................................................................33

12

STEEL DESIGN ..................................................................................................................... 35 12.1 EC3 / EN 1993-1-1 - Romania: Verifying lateral-torsional buckling resistance of an I rolled beam laterally restrained at mid-span ............................................................................................................................................36 12.2

EC3 / CSN EN 1993-1-1 - Czech Republic: Compressed and bended profile ............................................36

12.3

AISC 360 2010 - United States of America: Compressive Force Resistance and Buckling Length (LRFD) 36

12.4 AISC 360-10 - United States of America: Verifying buckling lengths and lateral torsional buckling lengths values after imposing them before starting the steel verification (TFSAD #14487) ................................................36 12.5

EC3 / EN 1993-1-1 - Germany: Lateral-torsional buckling for a two field column with HEA 200 .................37

12.6 EC3 / SR EN 1993-1-1-2006 - Romania: Verifying the buckling resistance of a rectangular hollow section column (R50*100/1) ................................................................................................................................................37 12.7

EC3 / SR EN 1993-1-1-2006 - Romania: Stability check for an IPE300 single span beam, simply supported ....37

12.8

EC3 / NF EN 1993-1-6 - France: Buckling resistance of a class 4 circular hollow section ..........................37

12.9

EC3 / EN 1993-1-1 - United Kingdom: Verifying the compression checks for an I section column .............37

12.10 AISC 360 2010 - United States of America: Example H.1 W-shape subject to combined compression and bending (LRFD) ......................................................................................................................................................38 12.11

EC3 / EN 1993-1-1 - United Kingdom: Mcr & XLT stability values for laminated RHS beams ..................38

12.12

EC3 / EN 1993-1-1 - United Kingdom: Mcr & XLT stability values for welded RHS beams .......................38

12.13

EC3 / EN 1993-1-1 - United Kingdom: Fy shear checks for an I section column.......................................38

12.14

EC3 / EN 1993-1-1 - United Kingdom: k stability values for an I section column.......................................38

12.15

EC3 / EN 1993-1-1 - United Kingdom: Oblique bending checks for an I section column ..........................39

12.16

EC3 / EN 1993-1-1 - United Kingdom: Fz shear checks for an I section column.......................................39

12.17

EC3 / EN 1993-1-1 - United Kingdom: XLT stability value for an I section column....................................39

12.18

EC3 / EN 1993-1-1 - United Kingdom: Work Ratio stability value for an I section column ........................39

12.19

EC3 / EN 1993-1-1 - United Kingdom: X stability values for an I section column ......................................39

12.20

EC3 / EN 1993-1-1 - United Kingdom: C1, C2 & Mcr stability values for an I section column ...................40

12.21 EC3 / NF EN 1993-1-1/NA - France: Class section classification, shear and bending moment verification of an IPE300 column (evaluated by SOCOTEC France - ref. Test 3) .....................................................................41

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ADVANCE DESIGN VALIDATION GUIDE

12.22 EC3 / NF EN 1993-1-1/NA - France: Class section classification and compression resistance for an IPE600 column (evaluated by SOCOTEC France - ref. Test 7) ............................................................................. 49 12.23 EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling resistance of a CHS219.1x6.3H column (evaluated by SOCOTEC France - ref. Test 21) .................................................................................................... 55 12.24 EC3 / NF EN 1993-1-1/NA - France: Verifying the lateral torsional buckling of a IPE300 beam (evaluated by SOCOTEC France - ref. Test 22) ...................................................................................................................... 60 12.25 EC3 / NF EN 1993-1-1/NA - France: Verifying the design plastic shear resistance of a rectangular hollow section beam (evaluated by SOCOTEC France - ref. Test 12) .............................................................................. 66 12.26 EC3 / NF EN 1993-1-1/NA - France: Verifying the resistance of a rectangular hollow section column subjected to bending and shear efforts (evaluated by SOCOTEC France - ref. Test 13) ....................................... 69 12.27 EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling resistance of a RC3020100 column (evaluated by SOCOTEC France - ref. Test 20) ...................................................................................................................... 76 12.28 EC3 / NF EN 1993-1-1/NA - France: Cross section classification and compression resistance verification of a rectangular hollow section column (evaluated by SOCOTEC France - ref. Test 11) ....................................... 84 12.29 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 4 column fixed on the bottom and with a displacement restraint at 2.81m from the bottom (evaluated by SOCOTEC France - ref. Test 25) ..................... 89 12.30 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 3 column fixed on the bottom (evaluated by SOCOTEC France - ref. Test 26)....................................................................................... 121 12.31 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 3 beam simply supported with a displacement restraint (evaluated by SOCOTEC France - ref. Test 27) .................................................... 144 12.32 EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling resistance for a IPE300 column (evaluated by SOCOTEC France - ref. Test 19) ......................................................................................................................... 168 12.33 EC3 / NF EN 1993-1-1/NA - France: Verifying a IPE400 column for compression, shear, bending moment, buckling, lateral torsional buckling and bending and axial compression (evaluated by SOCOTEC France - ref. Test 23) ........................................................................................................................................................ 175 12.34 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 4 column fixed on the bottom and without any other restraint (evaluated by SOCOTEC France - ref. Test 24)...................................... 211 12.35 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 1, column hinged on base and restrained on top for the X, Y translation and Z rotation (evaluated by SOCOTEC France - ref. Test 29) .... 235 12.36 EC3 / NF EN 1993-1-1/NA - France: Verifying a rectangular hollow section column subjected to bending and axial efforts (evaluated by SOCOTEC France - ref. Test 15) ........................................................................ 275 12.37 EC3 / NF EN 1993-1-1/NA - France: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange (evaluated by SOCOTEC France - ref. Test 44) ....... 282 12.38 EC3 / NF EN 1993-1-1/NA - France: Class section classification and compression verification of an IPE300 column (evaluated by SOCOTEC France - ref. Test 1) ........................................................................... 288 12.39 EC3 / NF EN 1993-1-1/NA - France: Class section classification and shear verification of an IPE300 beam subjected to linear uniform loading (evaluated by SOCOTEC France - ref. Test 2) ................................... 294 12.40 EC3 / NF EN 1993-1-1/NA - France: Class section classification and combined axial force with bending moment verification of an IPE300 column (evaluated by SOCOTEC France - ref. Test 5) .................................. 301 12.41 EC3 / NF EN 1993-1-1/NA - France: Class section classification and combined biaxial bending verification of an IPE300 beam (evaluated by SOCOTEC France - ref. Test 6) ................................................... 307 12.42 EC3 / NF EN 1993-1-1/NA - France: Verifying the bending resistance of a rectangular hollow section column made of S235 steel (evaluated by SOCOTEC France - ref. Test 14) ...................................................... 314 12.43 EC3 / NF EN 1993-1-1/NA - France: Comparing the shear resistance of a welded built-up beam made from different steel materials (evaluated by SOCOTEC France - ref. Test 45) .................................................... 320 12.44 EC3 / NF EN 1993-1-1/NA - France: Class section classification and bending moment verification of an IPE300 column (evaluated by SOCOTEC France - ref. Test 4) ........................................................................... 324 12.45 EC3 / NF EN 1993-1-1/NA - France: Verifying the classification and the resistance of a column subjected to bending and axial load (evaluated by SOCOTEC France - ref. Test 8) ............................................................ 330

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ADVANCE DESIGN VALIDATION GUIDE

12.46 EC3 / NF EN 1993-1-1/NA - France: Verifying the classification and the compression resistance of a welded built-up column (evaluated by SOCOTEC France - ref. Test 9) ...............................................................336 12.47 EC3 / NF EN 1993-1-1/NA - France: Verifying the classification and the bending resistance of a welded built-up beam (evaluated by SOCOTEC France - ref. Test 10) ............................................................................343 12.48

NTC 2008 - Italy: Stability check for steel column hinged base ...............................................................349

12.49

EC3 / NF EN 1993-1-1 - France: Buckling resistance of a steel column .................................................353

12.50

EC3 / NF EN 1993-1-2 - France: Verifying the bending resistance of a purlin for a 15min duration ........357

12.51 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 1, column fixed on base and without any other restraint (evaluated by SOCOTEC France - ref. Test 28) ..................................................360 12.52 EC3 / NF EN 1993-1-1/NA - France: Verifying a simply supported rectangular hollow section beam subjected to biaxial bending (evaluated by SOCOTEC France - ref. Test 16) ......................................................391 12.53

EC3 / CSN EN 1993-1-1 - Czech Republic: Compressed IPE300 ..........................................................398

12.54

NTC 2008 - Italy: Deflection check on simply supported beam ...............................................................404

12.55 AISC 360 2010 - United States of America: Deflection and Strength for a W-Shape Continuously Braced Flexural Member in Strong-Axis Bending (LRFD) .................................................................................................406 12.56

EC3 / NF EN 1993-1-1/NA - France: Verifying critical moment Mcr on a beam with intermediate restraints .. 409

12.57

EC3 / CSN EN1993-1-1 - Czech Republic: Bended beam without stability failure ..................................413

12.58

AISC 360 2010 - United States of America: Example G.1 W-shape in strong axis shear (LRFD) ...........418

12.59 EC3 / EN 1993-1-1 - United Kingdom: Simply supported laterally restrained (from P374 Hollow Sections Example 3) ...........................................................................................................................................................421 12.60

EC3 / EN 1993-1-1 - United Kingdom: Pin-ended column (from P374 Hollow Sections Example 2).......426

12.61

EC3 / EN 1993-1-1 - United Kingdom: Laterally unrestrained beam (from P374 Hollow Sections Example 4) ...430

12.62

NTC 2008 - Italy: Verifying the lateral torsional buckling of a IPE300 beam ..........................................433

12.63

EC3 / CSN EN 1993-1-1 - Czech Republic: Compressed U_profile ........................................................440

12.64 EC3 / NF EN 1993-1-1/NA - France: Verifying critical bending moment on a IPE200 beam with complex loading .................................................................................................................................. 444 12.65 EC3 / NF EN 1993-1-1/NA - France: Verifying critical moment Mcr of a beam with intermediate restraints subject to upwards loading ...................................................................................................................................444 12.66

EC3 / CSN EN1993-1-1 - Czech Republic: Tensioned diagonal .............................................................452

12.67 EC3 / EN 1993-1-1 - United Kingdom: Simply supported laterally restrained (from P364 Open Sections Example 2) ...........................................................................................................................................................454 12.68

NTC 2008 - Italy: Strenght verification of a steel linear hollow section ....................................................456

12.69

NTC 2008 - Italy: Lateral torsional buckling verification of a steel column ...............................................462

12.70 EC3 / NF EN 1993-1-1/NA France: Verifying the stability of an asymmetric I-shape subjected to axial force and bending moment ...................................................................................................................................491 12.71 EC3 / NF EN 1993-1-1/NA - France: Verifying IPE450 column fixed on base subjected to axial compression and bending moment, both applied on top (ref. Test 31) .................................................................505 12.72 EC3 / NF EN 1993-1-1/NA - France: Verifying IPE600 simple supported beam, loaded with centric compression and uniform linear efforts by Y and Z axis (ref. Test 32) ..................................................................505 12.73 EC3 / NF EN 1993-1-1/NA - France: Verifying RHS300x150x9H class 1 simply supported beam, loaded with centric compression, uniform linear horizontal efforts and a vertical punctual load in the middle (ref. Test 36) ............505 12.74 EC3 / NF EN 1993-1-1/NA - France: 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 (ref. Test 35) .....................505 12.75 EC3 / NF EN 1993-1-1/NA - France: Verifying IPE300 beam, simply supported, loaded with centric compression and uniform linear efforts by Y and Z axis (ref. Test 30) .................................................................506 11

ADVANCE DESIGN VALIDATION GUIDE

12.76 EC3 / NF EN 1993-1-1/NA - France: Verifying a simply supported rectangular hollow section beam subjected to torsional efforts (ref. Test 17) ........................................................................................................... 506 12.77 EC3 / NF EN 1993-1-1/NA - France: 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 (ref. Test 39) ....... 506 12.78 EC3 / NF EN 1993-1-1/NA - France: Verifying C310x30.8 class 3beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle (ref. Test 34) ............................. 506 12.79 EC3 / NF EN 1993-1-1/NA - France: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange (ref. Test 42) ........................................ 507 12.80 EC3 / NF EN 1993-1-1/NA - France: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange (ref. Test 43) .................................................. 507 12.81 EC3 / NF EN 1993-1-1/NA - France: Verifying RHS350x150x5H class 4 column, loaded with centric compression, punctual horizontal force by Y and a bending moment, all applied to the top (ref. Test 38) ........... 507 12.82 EC3 / NF EN 1993-1-1/NA - France: 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 (ref. Test 40) .... 507 12.83 EC3 / NF EN 1993-1-1/NA - France: Verifying UPN300 simple supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and punctual vertical force by Z axis (ref. Test 33) .............. 508 12.84 EC3 / NF EN 1993-1-1/NA - France: Verifying RHS350x150x8.5H class 3 column, loaded with centric compression, punctual lateral load and bending moment, all applied to the top of the column (ref. Test 37)....... 508 12.85 EC3 / NF EN 1993-1-1/NA - France: Verifying a simply supported circular hollow section element subjected to torsional efforts (ref. Test 18) ........................................................................................................... 508 12.86 EC3 / NF EN 1993-1-1/NA - France: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange (ref. Test 41) .................................................. 508 12.87 EC3 / NF EN 1993-1-1/NA - France: Changing the steel design template for a linear element (TTAD #12491) ................................................................................................................................................... 509 12.88

Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533) ......................... 509

12.89 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389) .................................................................................................................................................... 509 12.90

Generating the shape sheet by system (TTAD #11471) ......................................................................... 509

12.91

Verifying the shape sheet results for the elements of a simple vault (TTAD #11522) ............................. 509

12.92 EC3 / NF EN 1993-1-1/NA - France: Verifying the cross section optimization of a steel element (TTAD #11516).................................................................................................................................................... 510 12.93 EC3 / NF EN 1993-1-1/NA - France: Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545).................................................................................................................................................... 510 12.94

Verifying the shape sheet results for a column (TTAD #11550) .............................................................. 510

12.95

Verifying shape sheet on S275 beam (TTAD #11731) ............................................................................ 510

12.96

Verifying results on square hollowed beam 275H according to thickness (TTAD #11770) ..................... 511

12.97

Verifying the calculation results for steel cables (TTAD #11623) ............................................................ 511

12.98 EC3 / NF EN 1993-1-1/NA: Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873) ............................................................................................................... 511 12.99 CM66 (steel design) - France: Verifying the buckling length for a steel portal frame, using the roA roB method ................................................................................................................................................ 511 12.100 EC3 / EN 1993-1-1 - General: Verifying the buckling length for a steel portal frame, using the kA kB method ........................................................................................................................................................... 512 12.101 CM66 (steel design) - France: Verifying the buckling length for a steel portal frame, using the kA kB method ........................................................................................................................................................... 512 12.102

12

EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling coefficient Xy on a class 2 section ........... 512

ADVANCE DESIGN VALIDATION GUIDE

12.103 EC3 / SR EN 1993-1-1-2006 - Romania: Fire verification: verifying the work ratios after performing an optimization for steel profiles (TTAD #11975) .......................................................................................................512 12.104

13

EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling length results (TTAD #11550) ..................513

TIMBER DESIGN ................................................................................................................. 515 13.1

Modifying the "Design experts" properties for timber linear elements (TTAD #12259) ..............................516

13.2

EC5 / NF EN 1995-1-1 - France: Verifying the units display in the timber shape sheet (TTAD #12445) ...516

13.3

EC5 / NF EN 1995-1-1 - France: Verifying the timber elements shape sheet (TTAD #12337) ..................516

13.4

EC5 / NF EN 1995-1-1 - France: Shear verification for a simply supported timber beam .........................516

13.5 EC5 / NF EN 1995-1-1 - France: Verifying a timber purlin subjected to biaxial bending and axial compression .........................................................................................................................................................517 13.6

EC5 / NF EN 1995-1-1 - France: Verifying a timber purlin subjected to oblique bending ..........................522

13.7

EC5 / NF EN 1995-1-1 - France: Verifying a timber column subjected to tensile forces ...........................526

13.8 EC5 / NF EN 1995-1-2 - France: Verifying the residual section of a timber column exposed to fire for 60 minutes .................................................................................................................................................................529 13.9

EC5 / NF EN 1995-1-2 - France: Verifying the fire resistance of a timber purlin subjected to simple bending... 532

13.10

EC5 / NF EN 1995-1-1 - France: Verifying a timber column subjected to compression forces ...............537

13.11 EC5 / NF EN 1995-1-1 - France: Verifying a timber beam subjected to combined bending and axial tension 541 13.12 EC5 / NF EN 1995-1-1 - France: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression ............................................................................................................547 13.13

EC5 / NF EN 1995-1-1 - France: Verifying a C24 timber beam subjected to shear force ........................552

13.14

EC5 / SR EN-1995-1-1-2004 - Romania: Timber column subjected to shear stress and torsion ............556

13.15

EC5 / SR EN-1995-1-1-2004 - Romania: Timber column subjected to compression ..............................556

13.16

EC5 / SR EN-1995-1-1-2004 - Romania: Timber beam subjected to simple bending .............................556

13.17 EC5 / SR EN 1995-1-1 - Romania: Verifying compression strength for C14 circular column with fixed base ............................................................................................................................................................556 13.18

EC5 / NF EN 1995-1-1 - France: Verifying a timber beam subjected to simple bending .........................557

13

9 Mesh

ADVANCE DESIGN VALIDATION GUIDE

9.1

Verifying the mesh for a model with generalized buckling (TTAD #11519) Test ID: 3649 Test status: Passed

9.1.1 Description Performs the finite elements calculation and verifies the mesh for a model with generalized buckling.

9.2

Verifying mesh points (TTAD #11748) Test ID: 3458 Test status: Passed

9.2.1 Description Performs the finite elements calculation and verifies the mesh nodes of a concrete structure. The structure consists of concrete linear elements (R20*20 cross section) and rigid supports; the loads applied on the structure: dead loads, live loads, wind loads and snow loads, according to Eurocodes.

9.3

Creating triangular mesh for planar elements (TTAD #11727) Test ID: 3423 Test status: Passed

9.3.1 Description Creates a triangular mesh on a planar element with rigid supports and self weight.

9.4

Verifying the mesh of a planar element influenced by peak smoothing. Test ID: 6190 Test status: Passed

9.4.1 Description The model consists in a c25/30 concrete planar element supported by three concrete columns (2 x R20/30, 1 x D40) and one steel column (IPE400). Self-weight of elements is taken into account and 2 live loads of -20KN/m2, respectively -100 KN/m2, are applied on the planar element.

9.5

Verifying the options to take into account loads in linear and planar elements mesh (TTAD #15251) Test ID: 6215 Test status: Passed

9.5.1 Description Verifies the options to take into account loads in linear and planar elements mesh.

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10 Reports Generator

ADVANCE DESIGN VALIDATION GUIDE

10.1 Generating a report with modal analysis results (TTAD #10849) Test ID: 3734 Test status: Passed

10.1.1Description Generates a report with modal results for a model with seismic actions.

10.2 Reports - Global envelope for efforts in linear elements with Min/Max Values and coresponding absicsa position Test ID: 6363 Test status: Passed

10.2.1Description The purpose of this test is to check the content of a report which returns a global envelope for efforts on linear elements, with the Max/Min Valus. The model consisits in a concrete simple frame, 3 LoadCases and several combinations. The report returns the Max and Min values of the efforts, per linear element, and the position/abiscisa of these Max/Min values.

10.3 Verifying the steel shape sheet display (TTAD #12657) Test ID: 4562 Test status: Passed

10.3.1Description Verifies the steel shape sheet display when the fire calculation is disabled.

10.4 Verifying the modal analysis report (TTAD #12718) Test ID: 4576 Test status: Passed

10.4.1Description Generates and verifies the modal analysis report.

10.5 EC2 / NF EN 1992-1-1/NA - France: Verifying the EC2 calculation assumptions report (TTAD #11838) Test ID: 4544 Test status: Passed

10.5.1Description Verifies the EC2 calculation assumptions report.

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ADVANCE DESIGN VALIDATION GUIDE

10.6 Verifying the shape sheet strings display (TTAD #12622) Test ID: 4559 Test status: Passed

10.6.1Description Verifies the shape sheet strings display for a steel beam with circular hollow cross-section.

10.7 Verifying the Max row on the user table report (TTAD #12512) Test ID: 4558 Test status: Passed

10.7.1Description Verifies the Max row on the user table report.

10.8 Verifying the shape sheet report (TTAD #12353) Test ID: 4545 Test status: Passed

10.8.1Description Generates and verifies the shape sheet report.

10.9 Verifying the shape sheet for a steel beam (TTAD #12455) Test ID: 4535 Test status: Passed

10.9.1Description Verifies the shape sheet for a steel beam.

10.10 Verifying the global envelope of linear elements forces result (on start and end of super element) (TTAD #12230) Test ID: 4489 Test status: Passed

10.10.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the start and end of super element by generating the "Global envelope of linear elements forces result report". The model consists of a concrete portal frame with rigid fixed supports.

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ADVANCE DESIGN VALIDATION GUIDE

10.11 Verifying the global envelope of linear elements stresses (on start and end of super element) (TTAD #12230) Test ID: 4501 Test status: Passed

10.11.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on start and end of super element by generating the "Global envelope of linear elements stresses report". The model consists of a concrete portal frame with rigid fixed supports.

10.12 Verifying the global envelope of linear elements displacements (on end points and middle of super element) (TTAD #12230) Test ID: 4494 Test status: Passed

10.12.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on end points and middle of super element by generating the "Global envelope of linear elements displacements report". The model consists of a concrete portal frame with rigid fixed supports.

10.13 Verifying the global envelope of linear elements displacements (on start and end of super element) (TTAD #12230) Test ID: 4495 Test status: Passed

10.13.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on start and end of super element by generating the "Global envelope of linear elements displacements report". The model consists of a concrete portal frame with rigid fixed supports.

10.14 Verifying the global envelope of linear elements forces result (on end points and middle of super element) (TTAD #12230) Test ID: 4488 Test status: Passed

10.14.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on end points and middle of super element by generating the "Global envelope of linear elements forces result report". The model consists of a concrete portal frame with rigid fixed supports.

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ADVANCE DESIGN VALIDATION GUIDE

10.15 Verifying the global envelope of linear elements displacements (on all quarters of super element) (TTAD #12230) Test ID: 4493 Test status: Passed

10.15.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on all quarters of super element by generating the "Global envelope of linear elements displacements report". The model consists of a concrete portal frame with rigid fixed supports.

10.16 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element) (TTAD #12230) Test ID: 4492 Test status: Passed

10.16.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on each 1/4 of mesh element by generating the "Global envelope of linear elements displacements report". The model consists of a concrete portal frame with rigid fixed supports.

10.17 Verifying the global envelope of linear elements displacements (on the start point of super element) (TTAD #12230) Test ID: 4496 Test status: Passed

10.17.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on the start point of super element by generating the "Global envelope of linear elements displacements report". The model consists of a concrete portal frame with rigid fixed supports.

10.18 Verifying the global envelope of linear elements stresses (on each 1/4 of mesh element) (TTAD #12230) Test ID: 4498 Test status: Passed

10.18.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on each 1/4 of mesh element by generating the "Global envelope of linear elements stresses report". The model consists of a concrete portal frame with rigid fixed supports.

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ADVANCE DESIGN VALIDATION GUIDE

10.19 Verifying the global envelope of linear elements stresses (on all quarters of super element) (TTAD #12230) Test ID: 4499 Test status: Passed

10.19.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on all quarters of super element by generating the "Global envelope of linear elements stresses report". The model consists of a concrete portal frame with rigid fixed supports.

10.20 Verifying the global envelope of linear elements stresses (on the end point of super element) (TTAD #12230, TTAD #12261) Test ID: 4503 Test status: Passed

10.20.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on the end point of super element by generating the "Global envelope of linear elements stresses report". The model consists of a concrete portal frame with rigid fixed supports.

10.21 Verifying the global envelope of linear elements stresses (on the start point of super element) (TTAD #12230) Test ID: 4502 Test status: Passed

10.21.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on the start point of super element by generating the "Global envelope of linear elements stresses report". The model consists of a concrete portal frame with rigid fixed supports.

10.22 Verifying the global envelope of linear elements displacements (on the end point of super element) (TTAD #12230, TTAD #12261) Test ID: 4497 Test status: Passed

10.22.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on the end point of super element by generating the "Global envelope of linear elements displacements report". The model consists of a concrete portal frame with rigid fixed supports.

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ADVANCE DESIGN VALIDATION GUIDE

10.23 Verifying the global envelope of linear elements stresses (on end points and middle of super element) (TTAD #12230) Test ID: 4500 Test status: Passed

10.23.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on end points and middle of super element by generating the "Global envelope of linear elements stresses report". The model consists of a concrete portal frame with rigid fixed supports.

10.24 Verifying the Min/Max values from the user reports (TTAD# 12231) Test ID: 4505 Test status: Passed

10.24.1Description Performs the finite elements calculation and generates a user report containing the results of Min/Max values.

10.25 Modal analysis: eigen modes results for a structure with one level Test ID: 3668 Test status: Passed

10.25.1Description Performs the finite elements calculation and generates the "Characteristic values of eigen modes" report. The one-level structure consists of linear and planar concrete elements with rigid supports. A modal analysis is defined.

10.26 Generating the critical magnification factors report (TTAD #11379) Test ID: 3647 Test status: Passed

10.26.1Description Performs the generalised buckling calculation for a steel structure hall, and generates a critical magnification factors report.

10.27 System stability when the column releases interfere with support restraints (TTAD #10557) Test ID: 3717 Test status: Passed

10.27.1Description Performs the finite elements calculation and generates the systems description report for a structure which has column releases that interfere with the supports restraints. The structure consists of steel beams and steel columns (S235 material, HEA550 cross section) with rigid fixed supports.

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ADVANCE DESIGN VALIDATION GUIDE

10.28 Verifying the model geometry report (TTAD #12201) Test ID: 4467 Test status: Passed

10.28.1Description Generates the "Model geometry" report to verify the model properties: total weight, largest structure dimensions, center of gravity.

10.29 Verifying the global envelope of linear elements forces result (on the end point of super element) (TTAD #12230, #12261) Test ID: 4491 Test status: Passed

10.29.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the end point of super element by generating the "Global envelope of linear elements forces result report". The model consists of a concrete portal frame with rigid fixed supports.

10.30 Verifying the global envelope of linear elements forces result (on the start point of super element) (TTAD #12230) Test ID: 4490 Test status: Passed

10.30.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the start point of super element by generating the "Global envelope of linear elements forces result report". The model consists of a concrete portal frame with rigid fixed supports.

10.31 Verifying the global envelope of linear elements forces result (on all quarters of super element) (TTAD #12230) Test ID: 4487 Test status: Passed

10.31.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on all quarters of super element by generating the "Global envelope of linear elements forces result report". The model consists of a concrete portal frame with rigid fixed supports.

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ADVANCE DESIGN VALIDATION GUIDE

10.32 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element) (TTAD #12230) Test ID: 4486 Test status: Passed

10.32.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on each 1/4 of mesh element by generating the "Global envelope of linear elements forces result report". The model consists of a concrete portal frame with rigid fixed supports.

10.33 Creating the steel materials description report (TTAD #11954) Test ID: 4100 Test status: Passed

10.33.1Description Generates the "Steel materials" report as a .txt file. The model consists of a steel structure with supports and a base plate connection.

10.34 Creating the rules table (TTAD #11802) Test ID: 4099 Test status: Passed

10.34.1Description Generates the "Rules description" report as .rtf and .txt file. The model consists of a steel structure with supports and a base plate connection. Two rules were defined for the steel calculation.

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11 Seismic analysis

ADVANCE DESIGN VALIDATION GUIDE

11.1 EC8 / NF EN 1998-1-1 - France: Verifying the spectrum results for EC8 seism (TTAD #11478) Test ID: 3703 Test status: Passed

11.1.1Description Performs the finite elements calculation and generates the "Displacements of linear elements by load case" report for a concrete beam with rectangular cross section R20*30 with fixed rigid punctual support. Model loads: self weight and seismic loads according to Eurocodes 8 French standards.

11.2 EC8 / CSN EN 1998-1 - Czech Republic: Verifying the displacements results of a linear element (DEV2012 #3.18) Test ID: 3626 Test status: Passed

11.2.1Description Verifies the displacements results of an inclined linear element according to Eurocodes 8 Czech standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports. The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads (CSN EN 1998-1).

11.3 Verifying signed concomitant linear elements envelopes on Fx report (TTAD #11517) Test ID: 3576 Test status: Passed

11.3.1Description Performs the finite elements calculation on a concrete structure. Generates the "Signed concomitant linear elements envelopes on Fx report". The structure has concrete beams and columns, two concrete walls and a windwall. Loads applied on the structure: self weight and a planar live load of -40.00 kN.

11.4 EC8 / SR EN 1998-1/NA - Romania: Verifying the spectrum results for EC8 seism (TTAD #12472) Test ID: 4537 Test status: Passed

11.4.1Description Performs the finite elements calculation and generates the "Displacements of linear elements by load case" report for a concrete beam with rectangular cross section R20*30 with fixed rigid punctual support. Model loads: self weight and seismic loads according to Eurocodes 8 Romanian standards (SR EN 1998-1/NA).

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ADVANCE DESIGN VALIDATION GUIDE

11.5 EC8 / NF EN 1998-1-1 - France: Verifying torsors on grouped walls from a multi-storey concrete structure Test ID: 4810 Test status: Passed

11.5.1Description Verifies torsors on grouped walls from a multi-storey concrete structure. EC8 with French Annex is used.

11.6 EC8 / NF EN 1998-1-1 - France: Verifying torsors on walls Test ID: 4803 Test status: Passed

11.6.1Description Verifies the torsors on walls. Eurocode 8 with French Annex is used.

11.7 EC8 / SR EN 1998-1-1 - Romania: Verifying action results and torsors per modes on point, linear and planar supports (TTAD #14840) Test ID: 5860 Test status: Passed

11.7.1Description Verifies action results and torsors per modes on point, linear and planar supports of a simple concrete structure. The seismic action is generated according to Eurocode 8 norm (Romanian annex).

11.8 RPA99/2003 - Algeria: Verifying the displacements results of a linear element (DEV2013 #3.5) Test ID: 5559 Test status: Passed

11.8.1Description Verifies the displacements results of a vertical linear element according to Algerian seismic standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports. The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads (RPA 99/2003).

11.9 EC8 / EN 1998-1-1 - General: Verifying torsors on a 6 storey single concrete core subjected to horizontal forces and seismic action Test ID: 6088 Test status: Passed

11.9.1Description Verifies torsors on a 6 storey single concrete (C25/30) core subjected to horizontal forces and seismic action. The calculation spectrum is generated considering Eurocode 8 General Annex rules. The walls describing the core are grouped at each level.

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ADVANCE DESIGN VALIDATION GUIDE

11.10 EC8 / NF EN 1998-1-1 - France: verifying torsors on walls, elastic linear supports and userdefined section cuts (TTAD #14460) Test ID: 5861 Test status: Passed

11.10.1Description Verifies torsors on walls, elastic linear supports and user-defined section cuts for a concrete structure which is subjected to seismic action defined according Eurocode 8 norm (French annex).

11.11 EC8 / EN 1998-1-1 - General: Verifying the displacements results of a linear element for spectrum with renewed building option (TTAD #14161) Test ID: 6192 Test status: Passed

11.11.1Description Verifies the displacements results of a linear element for spectrum with renewed building option, according to Eurocode EC8 standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports. The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads for an envelope spectrum.

11.12 EC8 / NF EN 1998-1-1 - France: Verifying the sum of actions on supports and nodes restraints (TTAD #12706) Test ID: 4859 Test status: Passed

11.12.1Description Verifies the sum of actions on supports and nodes restraints for a simple structure subjected to seismic action according to EC8 French annex.

11.13 EC8 / NF EN 1998-1-1 - France: Verifying seismic efforts on planar elements with Q4 and T3-Q4 mesh type (TTAD #14244) Test ID: 5492 Test status: Passed

11.13.1Description Generates seismic results on planar element meshed with T3-Q4 mesh type and only Q4 mesh type.

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ADVANCE DESIGN VALIDATION GUIDE

11.14 EC8 / EN 1998-1-1 - General: Verifying the displacements results of a linear element for an envelope spectrum (DEV2013 #8.2) Test ID: 5600 Test status: Passed

11.14.1Description Verifies the displacements results of a vertical linear element according to Eurocode EC8 standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports. The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads for an envelope spectrum.

11.15 PS92/2010 - France: Verifying the displacements results of a linear element for an envelope spectrum (DEV2013 #8.2) Test ID: 5599 Test status: Passed

11.15.1Description Verifies the displacements results of a vertical linear element according to French PS92/2010 standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports. The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads for an envelope spectrum.

11.16 PS92 - France: Verifying efforts and torsors on planar elements (TTAD #12974) Test ID: 4858 Test status: Passed

11.16.1Description Verifies efforts and torsors on several planar elements of a concrete structure subjected to horizontal seismic action (according to PS92 norm).

11.17 EC8 / NF EN 1993-1-8/NA - France: Verifying the damping correction influence over the efforts in supports (TTAD #13011). Test ID: 4853 Test status: Passed

11.17.1Description Verifies the damping correction influence over the efforts in supports. The model has 2 seismic cases. Only one case uses the damping correction. The seismic spectrum is generated according to the Eurocodes 8 - French standard (NF EN 1993-1-8/NA).

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ADVANCE DESIGN VALIDATION GUIDE

11.18 EC8 / NF EN 1998-1-1 - France: Verifying seismic results when a design spectrum is used (TTAD #13778) Test ID: 5425 Test status: Passed

11.18.1Description Verifies the seismic results according to EC8 French Annex for a single bay single story structure made of concrete.

11.19 EC8 / NF EN 1998-1-1 - France: Generating forces results per modes on linear and planar elements (TTAD #13797) Test ID: 5455 Test status: Passed

11.19.1Description Generates reports with forces results per modes on a selection of elements (linear and planar elements) from a concrete structure subjected to seismic action (EC8 French Annex).

11.20 RPS 2011 - Morocco: Verifying the displacements results of a linear element for an envelope spectrum (DEV2013 #8.2) Test ID: 5598 Test status: Passed

11.20.1Description Verifies the displacements results of a vertical linear element according to Marocco seismic standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports. The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads for an envelope spectrum (RPS 2011).

11.21 RPS 2011 - Morocco: Verifying the displacements results of a linear element (DEV2013 #3.6) Test ID: 5597 Test status: Passed

11.21.1Description Verifies the displacements results of a vertical linear element according to Marocco seismic standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports. The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads (RPS 2011).

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ADVANCE DESIGN VALIDATION GUIDE

11.22 EC8 / SR EN 1998-1-1 - Romania: Verifying the displacements results of a linear element for an envelope spectrum (DEV2013 #8.2) Test ID: 5601 Test status: Passed

11.22.1Description Verifies the displacements results of a vertical linear element according to Romanian EC8 appendix. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports. The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads for an envelope spectrum.

11.23 EC8 / NF EN 1998-1-1 - France: Verifying torsors on walls with Seismic Loads (TTAD #16522) Test ID: 6382 Test status: Passed

11.23.1Description Concrete walls subjected to EQ loads with different q factors on the 2 directions X, Y; Verification is made according to EC8 - FR NA.

11.24 EC8 / NF EN 1998-1-1 - France: Verifying earthquake description report in analysis with Z axis down (TTAD #15095) Test ID: 6207 Test status: Passed

11.24.1Description Verifying earthquake description report in analysis with Z axis down, according to the Eurocodes EC8 standard.

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12 Steel design

ADVANCE DESIGN VALIDATION GUIDE

12.1 EC3 / EN 1993-1-1 - Romania: Verifying lateral-torsional buckling resistance of an I rolled beam laterally restrained at mid-span Test ID: 6231 Test status: Passed

12.1.1Description Verifies lateral torsional-buckling resistance of a simply supported I rolled beam (HEA 400) made of S355 steel. The beam, laterally restrained on supports and at its middle span, is subjected to bending efforts about its strong axis. The verification is made according to Romanian Annex of EN 1993-1-1.

12.2 EC3 / CSN EN 1993-1-1 - Czech Republic: Compressed and bended profile Test ID: 6280 Test status: Passed

12.2.1Description Class section classification and combined axial force with bending moment verification of an IPE300 column

12.3 AISC 360 2010 - United States of America: Compressive Force Resistance and Buckling Length (LRFD) Test ID: 6268 Test status: Passed

12.3.1Description Member AISC W14x132 ASTM A992 (Fy = 50 ksi) W-shape column to carry an axial dead load of 140 kips and live load of 420 kips. The column is 30 ft long and is pinned top and bottom in both axes. (LRFD)

12.4 AISC 360-10 - United States of America: Verifying buckling lengths and lateral torsional buckling lengths values after imposing them before starting the steel verification (TFSAD #14487) Test ID: 6233 Test status: Passed

12.4.1Description Verifies buckling lengths and lateral torsional buckling lengths values on a W18x60 column made of ASTM A992 steel. Buckling lengths and lateral torsional buckling lengths values were manually imposed before starting AISC steel verification.

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ADVANCE DESIGN VALIDATION GUIDE

12.5 EC3 / EN 1993-1-1 - Germany: Lateral-torsional buckling for a two field column with HEA 200 Test ID: 6286 Test status: Passed

12.5.1Description Lateral-torsional buckling for a two field column with HEA 200. The system concerned a one field system (8m) in one, and a 2 field system (4m + 4m) in the other direction. The assumptions for buckling and especially for Lateraltorsional buckling are set manually. The value XLT is chosen to be the reference value.

12.6 EC3 / SR EN 1993-1-1-2006 - Romania: Verifying the buckling resistance of a rectangular hollow section column (R50*100/1) Test ID: 6241 Test status: Passed

12.6.1Description Description: The test verifies the buckling resistance for a R50*100/1 column made of S235 steel. The verifications are made according to Eurocode3 Romanian Annex.

12.7 EC3 / SR EN 1993-1-1-2006 - Romania: Stability check for an IPE300 single span beam, simply supported Test ID: 6284 Test status: Passed

12.7.1Description Verification of an IPE300 steel beam with a distributed load applied along the linear element; the beam is made of S335 material and it is simply supported. The beam is subjected to a -30 kN/m uniform linear effort applied vertically.

12.8 EC3 / NF EN 1993-1-6 - France: Buckling resistance of a class 4 circular hollow section Test ID: 6307 Test status: Passed

12.8.1Description Calculates the buckling resistance of a class 4 circular hollow section in pure compression and compares the value with the results from the document from CTICM ("Revue Construction Métallique n°4 - 2011").

12.9 EC3 / EN 1993-1-1 - United Kingdom: Verifying the compression checks for an I section column Test ID: 6315 Test status: Passed

12.9.1Description Verifying the compression steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

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ADVANCE DESIGN VALIDATION GUIDE

12.10 AISC 360 2010 - United States of America: Example H.1 W-shape subject to combined compression and bending (LRFD) Test ID: 6292 Test status: Passed

12.10.1Description Determines the available compressive and flexural strengths. Checks for the combined effects of an ASTM A992 W14x99 and whether it has sufficient available strength to support the axial forces and moments together, obtained from a second-order analysis that includes P-delta effects. The unbraced length is 14 ft and the member has pinned ends. KLx = KLy = Lb= 14.0 ft.

12.11 EC3 / EN 1993-1-1 - United Kingdom: Mcr & XLT stability values for laminated RHS beams Test ID: 6324 Test status: Passed

12.11.1Description Verifying the Mcr & XLT stability values for 3 laminated RHS beams, R40*75/3, R40*80/3 and R40*150/3.

12.12 EC3 / EN 1993-1-1 - United Kingdom: Mcr & XLT stability values for welded RHS beams Test ID: 6325 Test status: Passed

12.12.1Description Verifying the Mcr & XLT stability values for 3 welded RHS beams, R40*75/3, R40*80/3 and R40*150/3.

12.13 EC3 / EN 1993-1-1 - United Kingdom: Fy shear checks for an I section column Test ID: 6316 Test status: Passed

12.13.1Description Verifying the Fy shear steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

12.14 EC3 / EN 1993-1-1 - United Kingdom: k stability values for an I section column Test ID: 6320 Test status: Passed

12.14.1Description Verifying the kyy, kyz, kzy & kzz stability values from steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

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ADVANCE DESIGN VALIDATION GUIDE

12.15 EC3 / EN 1993-1-1 - United Kingdom: Oblique bending checks for an I section column Test ID: 6318 Test status: Passed

12.15.1Description Verifying the oblique bending steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

12.16 EC3 / EN 1993-1-1 - United Kingdom: Fz shear checks for an I section column Test ID: 6317 Test status: Passed

12.16.1Description Verifying the Fz shear steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

12.17 EC3 / EN 1993-1-1 - United Kingdom: XLT stability value for an I section column Test ID: 6319 Test status: Passed

12.17.1Description Verifying the XLT stability value from steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

12.18 EC3 / EN 1993-1-1 - United Kingdom: Work Ratio stability value for an I section column Test ID: 6323 Test status: Passed

12.18.1Description Verifying the Work Ratio stability value from steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

12.19 EC3 / EN 1993-1-1 - United Kingdom: X stability values for an I section column Test ID: 6322 Test status: Passed

12.19.1Description Verifying the Xyy and Xzz stability values from steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

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ADVANCE DESIGN VALIDATION GUIDE

12.20 EC3 / EN 1993-1-1 - United Kingdom: C1, C2 & Mcr stability values for an I section column Test ID: 6321 Test status: Passed

12.20.1Description Verifying the C1, C2 and Mcr stability values from steel checks for a vertical UKB356x171x67 S355 column subject to axial force and bending moment

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ADVANCE DESIGN VALIDATION GUIDE

12.21 EC3 / NF EN 1993-1-1/NA - France: Class section classification, shear and bending moment verification of an IPE300 column (evaluated by SOCOTEC France - ref. Test 3) Test ID: 5411 Test status: Passed

12.21.1Description 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 200 kN force applied on the web direction, defined as a live load. The dead load will be neglected.

12.21.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.21.2.1 Model 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 = -200kN, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units Metric System Materials properties

41

ADVANCE DESIGN VALIDATION GUIDE

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, Support at end point (z = 5.00) free. Inner: None.

► ►

■

12.21.2.2 Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 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:

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.

42

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

43

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

This means that the column web is Class 1. 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.

The section geometrical properties are described in the picture below:

44

ADVANCE DESIGN VALIDATION GUIDE

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

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).

12.21.2.3 Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd , is determined as follows:

Av *

V pl , Rd =

fy

γM0

3 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(2)

Where: Av section shear area for rolled profiles A

cross-section area A=53.81cm2

b

overall breadth b=150mm

h

overall depth h=300mm

Av = A − 2 * b * t f + (t w + 2 * r ) * t f

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

Av = A − 2 * b * t f + (t w + 2 * r ) * t f = 53.81 − 2 * 15 * 17 + (0.71 + 2 * 1.5) * 1.07 = 25.68cm 2 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3) fy

nominal yielding strength for S235 fy=235MPa

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

V pl , Rd =

Av *

fy

γM0

3 =

25.68 * 104 * 1

235 3 = 0.3484MN = 348.42kN

12.21.2.4 Reference results in calculating the bending moment resistance

VEd 200 = = 57.4% > 50% V pl , Rd 348.42

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ADVANCE DESIGN VALIDATION GUIDE

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:

2

 2V   2 * 0.200  2 − 1 = 0.0223 ρ =  Ed − 1 =  V 0 . 348   , pl Rd   According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(3)

Av = A − 2 * b * t f + (t w + 2 * r ) * t f 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

Av = A − 2 * b * t f + (t w + 2 * r ) * t f = 53.81 − 2 * 15 * 17 + (0.71 + 2 * 1.5) * 1.07 = 25.68cm 2 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3) fy

nominal yielding strength for S235 fy=235MPa

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

M y ,V , Rd

  ρ * Av ²  0.0223 * (25.68 * 10−4 )²   * f y  628.40 * 10− 6 −  w pl −  * 235 4t w  4 * 0.0071    = 0.146 MNm = = 1 γM0

M Ed = 2.5m * 200kN = 500kNm = 0.5MNm M Ed 0.500 = = 342% M y ,V , Rd 0.146 Finite elements modeling ■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

Finite elements results Shear z direction work ratio

46

ADVANCE DESIGN VALIDATION GUIDE

Shear z direction work ratio Work ratio - Fz

Combined oblique bending Combined oblique bending Work ratio - Oblique

47

ADVANCE DESIGN VALIDATION GUIDE

12.21.2.5 Reference results Result name

Result description

Reference value

Shear z work ratio

direction

Work ratio - Fz

57 %

Combined bending

oblique

Work ratio - Oblique

341 %

12.21.3Calculated results Result name Work ratio - Fz Work ratio Oblique

48

-

Result description

Value

Error

Work ratio Fz

57.4021 %

0.7054 %

Work ratio - Oblique

341.348 %

0.1021 %

ADVANCE DESIGN VALIDATION GUIDE

12.22 EC3 / NF EN 1993-1-1/NA - France: Class section classification and compression resistance for an IPE600 column (evaluated by SOCOTEC France - ref. Test 7) Test ID: 5620 Test status: Passed

12.22.1Description 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.

12.22.2Background 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.

12.22.2.1 Model 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, 5 Longitudinal elastic modulus: E = 2.1 x 10 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, 49

ADVANCE DESIGN VALIDATION GUIDE

Free at end point (x = 5.00). 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.

12.22.2.2 Reference 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.

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ADVANCE DESIGN VALIDATION GUIDE

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:

c 600mm − 19mm × 2 − 24mm × 2 = = 42.83 t 12mm

ε=

235 = 1.0 fy

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ADVANCE DESIGN VALIDATION GUIDE

Therefore:

c = 42.83 > 42ε = 42 t 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:

c (220mm − 12mm − 24mm × 2) / 2 = 4.21 = 19mm t

ε=

235 = 1.0 fy

Therefore:

c = 4.21 ≤ 9ε = 9 t 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.

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ADVANCE DESIGN VALIDATION GUIDE

12.22.2.3Reference 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:

ψ=

σ2 = 1.0 σ1

kσ = 4.0 The plate modified slenderness (λp) The formula used to determine the plate modified slenderness is:

λp =

(600mm − 2 ×19mm − 2 × 24mm ) / 12mm = 0.754 b/t = 28.4 × 1.0 × 4.0 28.4ε kσ

The reduction factor (ρ) Because λp > 0.673, the reduction factor has the following formula:

ρ=

λ p − 0.055 × (3 +ψ ) ≤ 1.0 λ2p

Effective area The effective area is determined considering the following:

Aeff = A − (1 − ρ )× b × t w = 15600 − (1 − 0.939 )× (600 − 2 ×19 − 2 × 24 )×12 = 15223.75mm 2 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:

N c , Rd =

Aeff × f

γM0

y

=

15223.75mm 2 × 235MPa = 3577581N 1.0

Work ratio Work ratio =

100000 N N × 100 = 2.79518% × 100 = 3577581N N c , Rd

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ADVANCE DESIGN VALIDATION GUIDE

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

12.22.2.4 Reference results Result name

Result description

Reference value

Work ratio - Fx

Compression resistance work ratio [%]

2.79518 %

12.22.3Calculated results

54

Result name

Result description

Value

Error

Work ratio - Fx

Compression resistance work ratio

2.79495 %

-0.0082 %

ADVANCE DESIGN VALIDATION GUIDE

12.23 EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling resistance of a CHS219.1x6.3H column (evaluated by SOCOTEC France - ref. Test 21) Test ID: 5701 Test status: Passed

12.23.1Description The test verifies the buckling resistance of a CHS219.1x6.3H made of S355. The tests are made according to Eurocode 3 French Annex.

12.23.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.23.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■

Exploitation loadings (category A): Q1 = -100 000 N, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in millimeters (mm).

Units Metric System Geometrical properties ■ ■ ■ ■ ■ ■

Tube wall thickness: t=6.3mm Tube diameter: d=219.1mm 2 Cross section area: A=4210mm Radius of gyration about the relevant axis: i=75.283mm 4 4 Flexion inertia moment around the Y axis: Iy=2366x10 mm 4 4 Flexion inertia moment around the Z axis: Iz=2366x10 mm

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ADVANCE DESIGN VALIDATION GUIDE

Materials properties S235 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 355 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 (z = 3.00). Inner: None. Buckling lengths Lfy and Lfz are both imposed with 6m value ► ►

Loading The column is subjected to the following loadings: ■ ■

External: Point load at Z = 3.0: FZ = N = -100 000 N, Internal: None.

12.23.2.2 Buckling 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.

N Ed × 100 ≤ 100% N b , Rd

(6.46)

Cross-class classification is made according to Table 5.2

d 219.1mm = = 34.778 t 6.3mm

ε=

235 235 = = 0.814 fy 355

d = 34.778 ≤ 80 × ε 2 = 50 × 0.814 = 46.381 therefore the section is considered to be Class 2 t

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ADVANCE DESIGN VALIDATION GUIDE

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

The imperfection factor

α

corresponding to the appropriate buckling curve will be 0.21:

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

N b , Rd =

χ × A× fy γ M1

(6.47)

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

χ coefficient corresponding to non-dimensional slenderness according to:

χy =

1 2

Φ y + Φ y − λy

2

λ

will be determined from the relevant buckling curve

≤1

(6.49)

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

λy =

N cr

λ1 = π i=

λy =

Iy A

=

Lcr i × λ1

(6.50)

210000 E =π = 76.41 355 fy

=

23860000mm 4 = 75.283mm 4210mm 2

Lcr 6000mm = 1.043 = i × λ1 75.283mm × 76.41

φ y = 0.5 × [1 + α (λ y − 0.2) + λ2y ] = 0.5 × [1 + 0.21 × (1.043 − 0.2) + 1.043² ] = 1.132

χy =

1 = 0.636 ≤ 1 1.132 + 1.132² − 1.043²

2 2 A is the cross section area; A=4210mm ; fy is the yielding strength of the material; fy=355N/mm and

coefficient,

γ M1 = 1

γ M1

is a safety

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ADVANCE DESIGN VALIDATION GUIDE

The design buckling resistance of the compression member will be:

N b , Rd

χ y × A × f y 0.636 × 4210mm 2 × 355 N / mm 2 = = = 950533.8 N γ M1 1

N Ed = 100000 N N Ed 100000 N × 100 = × 100 = 10.520% N b , Rd 950533.8 N Finite elements modeling ■ ■ ■

Linear element: S beam, 4 nodes, 1 linear element.

Finite elements results The appropriate non-dimensional slenderness The appropriate non-dimensional slenderness

χ LT

58

ADVANCE DESIGN VALIDATION GUIDE

Ratio of the design normal force to design buckling resistance in the strong inertia of the profile Column subjected to axial force Adimensional - SNy

12.23.2.3 Reference results Result name

Result description

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

SN y

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

Reference value 0.636 0.1052

12.23.3Calculated results Result name

Result description

Value

Error

Xy

coefficient corresponding to non-dimensional slenderness

0.635463 adim

0.0000 %

SNy

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

0.105293 adim

0.0000 %

59

ADVANCE DESIGN VALIDATION GUIDE

12.24 EC3 / NF EN 1993-1-1/NA - France: Verifying the lateral torsional buckling of a IPE300 beam (evaluated by SOCOTEC France - ref. Test 22) Test ID: 5702 Test status: Passed

12.24.1Description 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.

12.24.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.24.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■

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 Geometrical properties ■ ■ ■ ■

Beam length: 5m Cross section area: A=5310mm2 4 4 Flexion inertia moment around the Y axis: Iy=8356.00x10 mm 4 4 Flexion inertia moment around the Z axis: Iz=603.80x10 mm

Materials properties 60

ADVANCE DESIGN VALIDATION GUIDE

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

Yield strength fy = 235 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa.

Boundary conditions The boundary conditions are described below: ■

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 The column is subjected to the following loadings: ■ ■

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

12.24.2.2 Buckling 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.

M Ed × 100 ≤ 100% M b , Rd

(6.46)

Cross-class classification is made according to Table 5.2 ■

for beam web:

c 248.6mm  = = 35.014 c 7.1mm t  ⇒ = 35.014 ≤ 72 × ε = 72 t  ε =1

therefore the beam web is considered to be

Class 1 ■

for beam flange:

c 56.45mm  = = 5.276 c t 10.7 mm  ⇒ = 5.276 ≤ 9 × ε = 9 t ε =1 

therefore the haunch is considered to be Class1

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ADVANCE DESIGN VALIDATION GUIDE

In conclusion, the section is considered to be Class 1 The buckling curve will be determined corresponding to Table 6.2:

h 300mm = = 2 ≤ 2 the buckling curve about Y-Y will be considered “a” b 150mm

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

M b , Rd =

χ LT × W y × f y γ M1

(6.55)

Where:

χ LT reduction factor for lateral-torsional buckling:

χ LT = Where:

1 2

Φ LT + Φ LT − λ LT

2

≤1

[

(6.56)

2

φ LT = 0.5 × 1 + α LT × (λ LT − 0.2) + λ LT

α LT

62

represents the imperfection factor;

]

α LT = 0.21

ADVANCE DESIGN VALIDATION GUIDE

λ LT the non-dimensional slenderness corresponding:

λ LT =

Wy × f y M cr

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.

M cr = C1 ×

π 2 × E × I z   k z  I w

(k z × L )2

×

 ×   k w  I z 

2  ( k z × L) × G × I t 2  ( ) + C z C z + × − × 2 2 g g 2 

π × E × IZ

(1)



according to EN 1993-1-1-AN France; AN.3 Chapter 2 Where: E is the Young’s module: E=210000N/mm

2

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:

M cr = C1 ×

I L² × G × I t π ²× E × Iz × w+ L² Iz π ²× E × Iz

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

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ADVANCE DESIGN VALIDATION GUIDE

M cr = 1.13 ×

π ² × 210000 N / mm 2 × 603.80 × 10 4 mm 4 (5000mm)²

×

12.59 × 1010 mm 6 (5000mm)² × 80770 N / mm 2 × 20.12 × 10 4 mm 4 = + 603.80 × 10 4 mm 4 π ² × 210000 N / mm 2 × 603.80 × 10 4 mm 4 = 1.13 × 500578.44 N × 230.80mm = 130609424.8 Nmm = 130.61kNm

×

therefore:

λ LT =

Wy × f y M cr

=

[

628.4 × 10 3 mm 3 × 235 N / mm 2 = 1.063 130609424.8 Nmm

]

φ LT = 0.5 × 1 + α LT × (λ LT − 0.2) + λ LT = 0.5 × [1 + 0.21× (1.063 − 0.2) + 1.0632 ] = 1.156

χ LT =

2

1

φ LT + φ LT ² − λ LT ²

=

1 = 0.621 ≤ 1 1.156 + 1.156² − 1.063²

Finite elements modeling ■ ■ ■

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

Finite elements results 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).

Lateral torsional buckling coefficient Simply supported beam subjected to bending efforts Lateral torsional buckling coefficient

64

ADVANCE DESIGN VALIDATION GUIDE

Elastic critical moment for lateral-torsional buckling Simply supported beam subjected to bending efforts Mcr

12.24.2.3 Reference results Result name

Result description

Reference value

χ LT

Lateral-torsional buckling coefficient [adim.]

0.621

Mcr

Elastic critical moment for lateral-torsional buckling [kNm]

130.61

12.24.3Calculated 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 %

65

ADVANCE DESIGN VALIDATION GUIDE

12.25 EC3 / NF EN 1993-1-1/NA - France: Verifying the design plastic shear resistance of a rectangular hollow section beam (evaluated by SOCOTEC France - ref. Test 12) Test ID: 5706 Test status: Passed

12.25.1Description Verifies the design plastic shear resistance of a rectangular hollow section beam made of S275 steel. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

12.25.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.25.2.1 Model 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 = -50 000 N/ml, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

►

Units Metric System Geometry Below are described the beam cross section characteristics:

66

■ ■ ■ ■ ■ ■

Height: h = 300 mm, Width: b = 200 mm, Thickness: t = 10 mm, Outer radius: r = 15 mm, Beam length: L = 5000 mm, 2 Section area: A = 9490 mm ,

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

ADVANCE DESIGN VALIDATION GUIDE

Materials properties S275 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 275 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 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.

►

■

12.25.2.2 Reference 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:

A × h 9490mm 2 × 300mm Av = = = 5694mm 2 b+h 300mm + 200mm 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 crosssection:

V pl , Rd =

Av ×

f

γ M0

y

3 =

5694mm 2 × 1.0

275MPa 3 = 904044 N

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 =

q× L 50 N / mm × 5000mm VEd 125000 N 2 × 100 = 2 × 100 = × 100 = × 100 = 13.83% 904044 N V pl , Rd V pl , Rd 904044 N

Finite elements modeling ■ ■ ■

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

67

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Work ratio of the design shear resistance Beam subjected to uniformly distributed load Work ratio - Fz

12.25.2.3 Reference results Result name

Result description

Reference value

Work ratio - Fz

Work ratio of the design shear resistance [%]

13.83 %

12.25.3Calculated results

68

Result name

Result description

Value

Error

Work ratio - Fz

Shear resistance work ratio

13.8219 %

0.1587 %

ADVANCE DESIGN VALIDATION GUIDE

12.26 EC3 / NF EN 1993-1-1/NA - France: Verifying the resistance of a rectangular hollow section column subjected to bending and shear efforts (evaluated by SOCOTEC France - ref. Test 13) Test ID: 5707 Test status: Passed

12.26.1Description 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.

12.26.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.26.2.1 Model 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: Fx = 200 000 N, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

►

69

ADVANCE DESIGN VALIDATION GUIDE

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, 2 Section area: A = 9490 mm , 3 Plastic section modulus about y-y axis: Wpl,y = 956000 mm ,

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

Materials properties S235 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 235 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa.

Boundary conditions The boundary conditions are described below: ■

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 The column is subjected to the following loadings: ■

External: Point load at z = 2.5: V= Fx = 200 000 N, Internal: None. ►

■

70

ADVANCE DESIGN VALIDATION GUIDE

12.26.2.2 Reference 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:

Av =

A × h 9490mm 2 × 300mm = 5694mm 2 = 300mm + 200mm b+h

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 crosssection:

V pl , Rd =

Av ×

f

γ M0

y

3 =

5694mm 2 × 1 .0

235MPa 3 = 772546.6 N

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 =

V V pl , Rd

× 100 =

200000 × 100 = 25.89% 772546.6

12.26.2.3 Reference 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:

71

ADVANCE DESIGN VALIDATION GUIDE

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.

72

ADVANCE DESIGN VALIDATION GUIDE

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).

c b − 2 × r − 2 × t 200mm − 2 × 15mm − 2 × 10mm = = = 15 t t 10mm

ε=

235 = 1.0 fy

Therefore:

c = 15 ≤ 33ε = 33 t 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):

c h − 2 × r − 2 × t 300mm − 2 × 15mm − 2 × 10mm = = = 25 t t 10mm

ε=

235 = 1.0 fy

Therefore:

c = 25 ≤ 72ε = 72 t 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 1993-11:2001.

M c , Rd =

W pl , y × f

γ M0

y

=

956000mm 3 × 235MPa = 224660000 Nmm 1.0

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 =

M Ed M c , Rd

L 5000mm 200000 N × 2 × 100 = 2 × 100 = × 100 = 222.56% M c , Rd 224660000 Nmm V×

Finite elements modeling ■ ■ ■

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

73

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results 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

74

ADVANCE DESIGN VALIDATION GUIDE

12.26.2.4 Reference results 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 %

12.26.3Calculated results Result name Work ratio - Fz Work ratio Oblique

-

Result description

Value

Error

Work ratio of the design plastic shear resistance

25.8793 %

-0.0413 %

Work ratio of the design resistance for bending

222.559 %

-0.0004 %

75

ADVANCE DESIGN VALIDATION GUIDE

12.27 EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling resistance of a RC3020100 column (evaluated by SOCOTEC France - ref. Test 20) Test ID: 5700 Test status: Passed

12.27.1Description The test verifies the buckling of a RC3020100 column made of S355 steel. The verifications are made according to Eurocode3 French Annex.

12.27.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.27.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■

Exploitation loadings (category A): Q1 = -200 000 N, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

Units Metric System Geometrical properties ■ ■ ■

76

Cross section area: A=9490mm2 Flexion inertia moment around the Y axis: Iy=11819x104mm4 4 4 Flexion inertia moment around the Z axis: Iz=6278x10 mm

ADVANCE DESIGN VALIDATION GUIDE

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

Loading The column is subjected to the following loadings: ■ ■

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

77

ADVANCE DESIGN VALIDATION GUIDE

12.27.2.2 Buckling 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.

N Ed × 100 ≤ 100% N b , Rd

(6.46)

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

χ × A× fy γ M1

N b , Rd =

(6.47)

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

χ coefficient corresponding to non-dimensional slenderness according to:

χ=

1 Φ+ Φ −λ 2

2

λ

will be determined from the relevant buckling curve

≤1

(6.49)

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

λ=

A* f y N cr

Where: A is the cross section area; A=5380mm2; fy is the yielding strength of the material; fy=235N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

N cr =

λ=

π ² × E × Iz L fy ²

A× fy N cr

=

=

π ² × 210000MPa × 11819 × 104 mm 4

(10000mm )2

= 2449625.943 N

9490mm2 × 235 N / mm2 = 0.954 2449625.943 N

[

φ = 0.5 1 + α (λ − 0.2) + λ ²

]

It will be used the following buckling curve:

The imperfection factor

[

α

corresponding to the appropriate buckling curve will be 0.21:

]

φ = 0.5 × 1 + α × (λ − 0.2) + λ ² = 0.5 × [1 + 0.21× (0.954 − 0.2) + 0.9542 ] = 1.034

78

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

χ= γ M1

1

φ + φ2 − λ

=

2

1 1.034 + 1.0342 − 0.9542

is a safety coefficient,

= 0.698 ≤ 1

γ M1 = 1

0.698 × 9490mm2 × 235 N / mm2 = 1556644.7 N 1

N b , Rd =

N Ed = 200000 N N Ed 200000 N × 100 = × 100 = 12.848% N b , Rd 1556644.7 N 12.27.2.3 Buckling 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.

N Ed × 100 ≤ 100% N b , Rd

(6.46)

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

χ × A× fy γ M1

N b , Rd =

(6.47)

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

χ coefficient corresponding to non-dimensional slenderness λ

will be determined from the relevant buckling curve

according to:

χ=

1 Φ+ Φ −λ 2

2

≤1

(6.49)

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

λ=

A* f y N cr

Where: A is the cross section area; A=5380mm2; fy is the yielding strength of the material; fy=235N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

N cr =

λ=

π ² × E × Iz L fy ²

A× fy N cr

[

=

=

π ² × 210000MPa × 6278 × 104 mm 4

(10000mm )2

= 1301188.905 N

9490mm 2 × 235 N / mm 2 = 1.309 1301188.905 N

φ = 0.5 1 + α (λ − 0.2) + λ ²

]

It will be used the following buckling curve:

79

ADVANCE DESIGN VALIDATION GUIDE

The imperfection factor

α

corresponding to the appropriate buckling curve will be 0.21:

[

]

φ = 0.5 × 1 + α × (λ − 0.2) + λ ² = 0.5 × [1 + 0.21 × (1.309 − 0.2) + 1.3092 ] = 1.473 Therefore:

χ= γ M1

1

φ + φ −λ 2

2

=

1 1.473 + 1.4732 − 1.3092

is a safety coefficient,

N b , Rd =

γ M1 = 1

0.465 × 9490mm2 × 235 N / mm2 = 1037019.75 N 1

N Ed = 200000 N N Ed 200000 N × 100 = 19.286% × 100 = 1037019.75 N N b , Rd Finite elements modeling ■ ■ ■

80

= 0.465 ≤ 1

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

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Coefficient corresponding to non-dimensional slenderness after Y-Y axis 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

81

ADVANCE DESIGN VALIDATION GUIDE

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)

82

ADVANCE DESIGN VALIDATION GUIDE

12.27.2.4 Reference results Result name

Result description

Reference value

χy

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

0.698

χz

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%

12.27.3Calculated results Result name

Result description

Value

Error

Xy

coefficient corresponding slenderness after Y-Y axis

to

non-dimensional

0.697433 adim

0.0000 %

Xz

coefficient corresponding slenderness after Z-Z axis

to

non-dimensional

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 %

83

ADVANCE DESIGN VALIDATION GUIDE

12.28 EC3 / NF EN 1993-1-1/NA - France: Cross section classification and compression resistance verification of a rectangular hollow section column (evaluated by SOCOTEC France - ref. Test 11) Test ID: 5705 Test status: Passed

12.28.1Description Verifies the cross section classification and the compression resistance of a rectangular hollow section column. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

12.28.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.28.2.1 Model 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 Geometry Below are described the column cross section characteristics: ■

84

Height: h = 300 mm,

ADVANCE DESIGN VALIDATION GUIDE

■ ■ ■ ■ ■

Width: b = 200 mm, 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:

.

Materials properties S235 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 235 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa.

Boundary conditions The boundary conditions are described below: ■

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 The column is subjected to the following loadings: ■ ■

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

85

ADVANCE DESIGN VALIDATION GUIDE

12.28.2.2 Reference 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 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:

c h − 2 × r − 2 × t 300mm − 2 × 15mm − 2 × 10mm = = = 25 t t 10mm

86

ADVANCE DESIGN VALIDATION GUIDE

ε=

235 = 1.0 fy

Therefore:

c = 25 ≤ 33ε = 33 t 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.

12.28.2.3 Reference 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. Compression resistance of the cross section 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:

N c , Rd =

A× f

γ M0

y

9490mm 2 × 235MPa = 2230150 N = 1.0

Work ratio Work ratio =

100000 N N × 100 = × 100 = 4.48% 2230150 N N c , Rd

Finite elements modeling ■ ■ ■

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

87

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Work ratio of the design resistance for uniform compression Column subjected to bending and axial force Work ratio - Fx

12.28.2.4 Reference results Result name

Result description

Reference value

Work ratio - Fx

Compression resistance work ratio [%]

4.48 %

12.28.3Calculated results

88

Result name

Result description

Value

Error

Work ratio - Fx

Compression resistance work ratio

4.484 %

0.0893 %

ADVANCE DESIGN VALIDATION GUIDE

12.29 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 4 column fixed on the bottom and with a displacement restraint at 2.81m from the bottom (evaluated by SOCOTEC France - ref. Test 25) Test ID: 5712 Test status: Passed

12.29.1Description The test verifies a user defined cross section column. 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. The calculations are made according to Eurocode 3 French Annex.

12.29.2Background 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 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). This test was evaluated by the French control office SOCOTEC.

89

ADVANCE DESIGN VALIDATION GUIDE

12.29.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■ ■

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 Geometrical properties ■

Column length: L=5620mm

■ ■

A = 10850mm 2 Overall breadth: b = 220mm Flange thickness: t f = 15mm

■

Root radius:

■

Web thickness: t w

= 5mm

■

Depth of the web:

hw = 880mm

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, W y

■

Elastic modulus after the Z axis, Wel , z

= 242.08 × 103 mm 3

■

Plastic modulus after the Z axis, W pl , z

= 368.31 × 103 mm 3

■ ■ ■ ■

4 4 Flexion inertia moment around the Y axis: Iy=149058.04x10 mm 4 4 Flexion inertia moment around the Z axis: Iz=2662.89x10 mm Torsional moment of inertia: It=51.46x104 mm4 6 6 Working inertial moment: Iw=4979437.37x10 mm

■

Cross section area:

r = 0mm

= 3387.66 × 103 mm 3

= 3757.62 × 103 mm 3

Materials properties S275 steel material is used. The following characteristics are used: ■ ■ ■

90

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

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions The boundary conditions are described below: ■

Outer: ► ►

■

Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at the end point (z = 5.62) restrained in translation along Y and Z axis and restrained rotation 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

12.29.2.2 Cross section Class According to Advance Design calculations: Cross-class classification is made according to Table 5.2

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ADVANCE DESIGN VALIDATION GUIDE

- for beam web: The web dimensions are 850x5mm.

ψ = 2⋅

N Ed 0.328 −1 = 2 ⋅ − 1 = −0.78 > −1 0.0109 × 275 A⋅ fy

1 

N



1 

0.328



Ed  = ⋅ 1 + α = ⋅ 1 +  = 0.64 > 0.5 f y × t × d  2  275 × 0.85 × 0.005  2 

ε=

235 = fy

235 = 0.924 275

c 880mm − 2 ×15mm  = = 170 c 42 × ε 42 × 0.924 = = 94.06 t 5mm  ⇒ = 170 > t ψ 0 . 67 + 0 . 33 × 0 . 67 + 0 . 33 × ( − 0 . 78 )  ε = 924 therefore the beam web is considered to be Class 4 - for beam flange:

c 107.5  = = 7.61 c 15 t  ⇒ = 7.61 ≤ 9 × 0.924 = 8.316 t  ε = 924 In conclusion, the section is considered to be Class 4.

92

therefore the haunch is considered to be Class1.

ADVANCE DESIGN VALIDATION GUIDE

12.29.2.3 Effective 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

c 880mm − 2 × 15mm = = 170 : t 5mm

ψ = 1 ⇒ kσ = 4 According to EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1

bw t λp = 28.4 × ε × kσ

bw

is the width of the web;

bw = 850mm

t is the web thickness; t=5mm

ε=

235 235 = = 0.9244 fy 275

850mm 5mm = 3.261 λp = 28.4 × 0.9244 × 4 According to EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4 -the web is considered to be an internal compression element, therefore:

λ p = 3.261 > 0.673 3 +ψ = 4 ≥ 0

⇒ ρ = 

λ p − 0.055 × (3 +ψ ) 3.261 − 0.055 × 4 λp

2

=

3.2612

= 0.286

beff = ρ × bw  beff = 0.286 × 850mm = 243.1mm   be1 = 0.5 × beff  ⇒ be1 = 0.5 × 243.1mm = 121.55mm   be 2 = 0.5 × beff  be 2 = 0.5 × 243.1mm = 121.55mm Aeff ,web = t w × be1 + t w × be 2 = 5mm ×121.55mm + 5mm ×121.55mm = 1215.5mm 2 Aeff , flange = t f × b f = 15mm × 220mm = 3300mm 2

Aeff = Aeff , web + 2 × Aeff , flange = 1215.5mm 2 + 2 × 3300mm 2 = 7815.2mm 2

93

ADVANCE DESIGN VALIDATION GUIDE

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

ψ =

σ inf 850mm = −1 ⇒ bc = bt = = 425mm 2 σ sup

ψ = −1 ⇒ kσ = 23.9 According to EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1

c 880mm − 2 × 15mm = = 170 t 5mm

bw t λp = 28.4 × ε × kσ

bw

is the width of the web;

bw = 850mm

t is the web thickness; t=5mm

ε=

235 235 = = 0.9244 275 fy

850mm 5mm = 1.325 λp = 28.4 × 0.9244 × 23.9 According to EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4

λ p = 1.325 > 0.673 3 +ψ = 2 ≥ 0 beff = ρ × bc = ρ × be1 = 0.4 × beff be 2 = 0.6 × beff

94

⇒ ρ = 

λ p − 0.055 × (3 +ψ ) 1.325 − 0.055 × 2 λp

2

=

1.325 2

bw   850mm beff = 0.692 × = 294.1mm   1 − (− 1) 1 −ψ    ⇒ be1 = 0.4 × 294.1mm = 117.64mm  b = 0.6 × 294.1mm = 176.46mm   e2  

= 0.692

ADVANCE DESIGN VALIDATION GUIDE

-the weight center coordinate is:

yG = =

(220 ×15)× 432.5 + (117.64 × 5)× 366.18 − (220 ×15)× 432.5 − (601.46 × 5)×124.27 = 220 × 15 + 117.64 × 5 + 601.46 × 5 + 220 × 15

− 158330.095 = −15.53mm 10195.5

-the inertial moment along the strong axis is:

15 3 × 220 117.64 3 × 5 + 220 × 15 × 448.03 2 + + 117.64 × 5 × 381.712 + 12 12 3 3 601.46 × 5 15 × 220 + + 601.46 × 5 × 107.74 2 + + 220 × 15 × 416.97 2 = 12 12 = 748854356.6 + 699380072.5 = 1448234429mm 4 Iy =

220 3 ×15 53 ×117.64 + 220 ×15 × 0 2 + + 117.64 × 5 × 0 2 + 12 12 3 5 × 601.46 220 3 ×15 2 + + 601.46 × 5 × 0 + + 220 ×15 × 0 2 = 26627490.63mm 4 12 12 Iz =

Wel , y = Wel , z

Iy z max

1448234429mm 4 = = 3179229.533mm 3 455.53mm

Iz 26627490.63mm 4 = = = 242068.10mm 3 y max 110mm

95

ADVANCE DESIGN VALIDATION GUIDE

12.29.2.5 Buckling verification a) over the strong axis of the section, y-y: - the imperfection factor α will be selected according to Tables 6.1 and 6.2:

α = 0.34 Coefficient corresponding to non-dimensional slenderness after Y-Y axis:

χ y coefficient

corresponding to non-dimensional slenderness

λy

will be determined from the relevant buckling

curve according to:

χy =

1 2

Φ y + Φ y − λy

2

≤1

λ y the non-dimensional slenderness corresponding to Class 4 cross-sections:

λy =

96

Aeff * f y N cr , y

(6.49)

ADVANCE DESIGN VALIDATION GUIDE

Where:

A is the effective cross section area; 2

Aeff = 7815.2mm 2 ;

fy is the yielding strength of the material;

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

N cr , y =

π²× E× Iy l fy ²

Aeff × f y

λy =

N cr , y

=

=

π ² × 210000 N / mm 2 × 1448234429mm 4

(5620mm)²

= 95035371.44 N = 95035.37kN

7815.2mm 2 × 275 N / mm 2 = 0.15 95035371.44 N

[

]

φ y = 0.5 × 1 + α × (λ y − 0.2) + λ y ² = 0.5 × [1 + 0.34 × (0.15 − 0.2 ) + 0.15 2 ] = 0.503

χy =

1 2

Φy + Φy −λy

χy ≥1

2

=

1

 = 1.017  0.503 + 0.5032 − 0.15 2  ⇒ χy =1  

b) over the weak axis of the section, z-z: - the imperfection factor α will be selected according to Tables 6.1 and 6.2:

α = 0.49

97

ADVANCE DESIGN VALIDATION GUIDE

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

χ z coefficient corresponding to non-dimensional slenderness according to:

χz =

1 2

Φz + Φz − λ z

2

λz

will be determined from the relevant buckling curve

≤1

(6.49)

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

Aeff * f y

λz =

N cr , z

Where:

Aeff = 7815.2mm 2 ;

A is the effective cross section area; 2

fy is the yielding strength of the material;

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

l fz = 2.81m

N cr , z =

because of the torsional buckling restraint from the middle of the column

π ²× E × Iz l fz ²

Aeff × f y

λz =

=

π ² × 210000 N / mm 2 × 26627490.63mm 4

(2810mm )²

= 6989347.62 N = 6989.35kN

7815.2mm 2 × 275 N / mm 2 = = 0.555 6989347.62 N

N cr , z

[

]

φ z = 0.5 × 1 + α × (λ z − 0.2) + λ z ² = 0.5 × [1 + 0.49 × (0.555 − 0.2) + 0.555 2 ] = 0.741

χz =

1 2

Φz + Φz − λ z

χz ≤ 1

2

=

1

 = 0.812 0.741 + 0.741 − 0.555  ⇒ χ z = 0.812   2

2

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

M cr = C1 ×

I L² × G × It π ² × E × Iz × w+ L² Iz π ² × E × Iz 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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1-AN France; Chapter 3; …(6)

ψ

98

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

=

637 kNm = 0.50 1274kNm

ADVANCE DESIGN VALIDATION GUIDE

ψ=

637 kNm = 0.5 ⇒ C1 = 1.31 1274kNm According to EN 1993-1-1-AN France; Chapter 3.2; Table 1

Flexion inertia moment around the Y axis:

I y = 1448234429mm 4

Flexion inertia moment around the Z axis:

I z = 26627490.63mm 4

Longitudinal elastic modulus: E = 210000 N/mm2. Torsional moment of inertia: It=514614.75mm4 Warping inertial moment: IW is the warping inertia (deformation inertia moment):

Iw =

I z × (h − t f

)

2

4

h cross section height; h=880mm

tf

flange thickness;

t f = 15mm

26627490.63 mm 4 × (880mm − 15mm ) Iw = = 49.808 × 1011 mm 6 4 2

According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column part: L=2810mm Shear modulus of rigidity: G=80800N/mm2

99

ADVANCE DESIGN VALIDATION GUIDE

M cr = C1 ×

π ²× E × Iz

I w L² × G × I t π 2 × 210000 N / mm 2 × 26627490.63mm 4 × + = 1.31× × Iz π ²× E × Iz (2810mm )2

L²

49.808 ×1011 mm 6 (2810mm ) × 80800 N / mm 2 × 514614mm 4 × + = 1.31× 6989347.626 N × 439.32mm = 26627490.63mm 4 π 2 × 210000 N / mm 2 × 26627490.63mm 4 = 4022433856 Nmm = 4022.43kNm Iy 1448234429mm 4 = = 3179229.533mm 3 The elastic modulus : Wel , y = z max 455.53mm 2

λ LT =

Weff , y f y M cr

Calculation of the

χ LT =

3179229.533mm 3 ×275 N / mm 2 = 0.466 4022433856 Nmm

=

χ LT for appropriate non-dimensional slenderness λ LT

1

φLT + φLT ² − λ LT ²

≤1

will be determined with formula:

(6.56)

[

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

]

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

h 880mm = =4>2 b 220mm

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

α = 0.76

[

(

)

]

φ LT = 0.5 1 + α LT × λ LT − 0.2 + λ LT ² = 0.5[1 + 0.76 × (0.466 − 0.2) + 0.466² ] = 0.710

χ LT =

1

φ LT + φ LT ² − λ LT ²

=

1 = 0.803 ≤ 1 0.710 + 0.710² − 0.466²

b) for the bottom part of the column: 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:

100

ADVANCE DESIGN VALIDATION GUIDE

M cr = C1 ×

π ² × E × Iz I L² × G × It × w+ L² Iz π ² × E × Iz According to EN 1993-1-1-AN France; Chapter 2; …(3)

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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1-AN France; Chapter 3; …(6)

ψ

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

=

0 =0 637 kNm

ψ = 0 ⇒ C1 = 1.77 According to EN 1993-1-1-AN France; Chapter 3.2; Table 1 Flexion inertia moment around the Y axis:

I y = 1448234429mm 4

Flexion inertia moment around the Z axis:

I z = 26627490.63mm 4

Longitudinal elastic modulus: E = 210000 N/mm2. Torsional moment of inertia: It=514614.75mm4 Warping inertial moment: IW is the warping inertia (deformation inertia moment):

Iw =

I z × (h − t f

)

2

4

h cross section height; h=880mm

tf

flange thickness;

t f = 15mm 101

ADVANCE DESIGN VALIDATION GUIDE

26627490.63 mm 4 × (880mm − 15mm ) = 49.808 × 1011 mm 6 Iw = 4 2

According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column part: L=2810mm Shear modulus of rigidity: G=80800N/mm2

M cr = C1 ×

π ²× E × Iz L²

I w L² × G × I t π 2 × 210000 N / mm 2 × 26627490.63mm 4 + = 1.77 × × Iz π ²× E × Iz (2810mm )2

×

49.808 ×1011 mm 6 (2810mm ) × 80800 N / mm 2 × 514614mm 4 × + = 1.77 × 6989347.626 N × 439.32mm = 26627490.63mm 4 π 2 × 210000 N / mm 2 × 26627490.63mm 4 = 5434891269 Nmm = 5434.89kNm 2

The elastic modulus : Wel , y

λ LT =

Weff , y f y M cr

Calculation of the

χ LT =

=

Iy z max

=

1448234429mm 4 = 3179229.533mm 3 455.53mm

3179229.533mm3 ×275 N / mm 2 = 0.401 5434891269 Nmm

=

χ LT for appropriate non-dimensional slenderness λ LT

1

φLT + φLT ² − λ LT ²

≤1

will be determined with the formula:

(6.56)

[

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

]

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

h 880mm = =4>2 b 220mm

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

α = 0.76

[

(

)

]

φ LT = 0.5 1 + α LT × λ LT − 0.2 + λ LT ² = 0.5[1 + 0.76 × (0.401 − 0.2 ) + 0.401² ] = 0.657 102

ADVANCE DESIGN VALIDATION GUIDE

χ LT =

1

φ LT + φ LT ² − λ LT ²

=

1 = 0.849 ≤ 1 0.657 + 0.657² − 0.401²

12.29.2.7 Internal factor, k yy , calculation The internal factor k yy 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: a) for the top part of the column:

k yy = C my × C mLT ×

µy 1−

N Ed N cr , y

N Ed N cr , y µy = N 1 − χ y × Ed N cr , y 1−

χy =1

(previously calculated)

N Ed = 328kN N cr , y =

π ²× E× Iy l fy ²

= 95035371.44 N = 95035.37 kN (previously calculated)

N Ed 328000 N 1− N cr , y 95035371.44 N = 1 = µy = N Ed 328000 N 1 − 1× 1− χ y × 95035371.44 N N cr , y 1−

103

ADVANCE DESIGN VALIDATION GUIDE

The

Cmy will be calculated according to Table A.1:

Calculation of the λ 0 term:

λ0 =

Weff , y × f y M cr 0 According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2

Wel , y =

Iy z max

=

1448234429mm = 3179229.533mm 3 455.53mm

The calculation the

M cr 0 = C1 ×

4

M cr 0

will be calculated using

π ²× E × Iz L²

×

C1 = 1 and C2 = 0 , therefore:

I w L² × G × I t π 2 × 210000 N / mm 2 × 26627490.63mm 4 + = 1× × Iz π ²× E × Iz (2810mm )2

49.808 ×1011 mm 6 (2810mm ) × 80800 N / mm 2 × 514614mm 4 × + = 1× 6989347.626 N × 439.32mm = 26627490.63mm 4 π 2 × 210000 N / mm 2 × 26627490.63mm 4 = 3070560199 Nmm = 3070.56kNm 2

λ0 =

104

Weff , y × f y M cr 0

=

3179229.533mm 3 × 275 N / mm 2 = 0.534 3070560199 Nmm

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed N cr , z   N cr ,TF 

   

term:

Where: - for a symmetrical section for the both axis,

N cr ,T =

N cr ,TF = N cr ,T

A  π 2 × E × I w  ×  G × It + I0  Lcr ,T ² 

The mass moment of inertia I 0

I 0 = I y + I z + A ⋅ z g2 Flexion inertia moment around the Y axis:

I y = 1490580416.67 mm 4

Flexion inertia moment around the Z axis:

I z = 26628854.17 mm 4

Cross section area:

A = 10850mm 2

Distance between the section neutral axis and the section geometrical center: z g

=0

I 0 = I y + I z + A ⋅ z g2 = I y + I z = 1490580416.67 mm 4 + 26628854.17 mm 4 = 1517209270mm 4 - the buckling length,

Lcr ,T , Lcr ,T = 2.81m

Torsional moment of inertia: Working inertial moment:

I t = 514614.75mm 4

I w = 49.808 × 1011 mm 6

(previously calculated)

Longitudinal elastic modulus: E = 210000 MPa Shear modulus of rigidity: G=80800MPa

N cr ,T =

10850mm 2 1517209270mm 4

 π 2 × 210000 N / mm 2 × 49.808 ×1011 mm 6   = ×  80800 N / mm 2 × 514614.75mm 4 + (2810mm )²  

= 9646886.24 N N Ed = 328000 N N cr ,TF = N cr ,T = 9646886.24 N

N cr , z =

π ²× E × Iz l fz ²

= 6989347.62 N = 6989.35kN

(previously calculated)

C1=1.31 for the top part of the column C1=1.77 for the bottom part of the column For the top part of the column:

 N 0.20 × C1 × 4 1 − Ed N cr , z  = 0.224

   × 1 − N Ed   N cr ,TF  

  = 0.20 × 1.31 × 4 1 − 328000 N  × 1 − 328000 N  =   6989347.62 N   9646886.24 N  

105

ADVANCE DESIGN VALIDATION GUIDE

Therefore: For the top part of the column:

    ε y × a LT  C my = C my , 0 + (1 − C my , 0 )× λ 0 = 0.534   1 + ε y × a LT       N N   0.20 × C1 × 4 1 − Ed  × 1 − Ed  = 0.224  ⇒ C mz = C mz , 0  N cr , z   N cr ,TF    a LT 2  C mLT = C my × ≥1       N N     λ0 = 0.534 > 0.20 ⋅ C1 ⋅ 4 1 − Ed  ⋅ 1 − Ed  = 0.224  1 − N Ed  × 1 − N Ed  N N       cr ,TF  cr , z     N cr , z   N cr ,T   The

Cmy 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.

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξy =

M y , Ed N Ed

×

ε y × aLT 1 + ε y × aLT

Aeff Weff , y

Elastic modulus after the Y axis,

ξy =

M y , Ed N Ed

×

Aeff Weff , y

=

Wel , y = Weff , y = 3179229.533mm 3

1274 ×10 6 Nmm 7815.2mm 2 × = 9.55 328000 N 3179229.533mm 3

a LT

It 514614.75mm 4 = 1− = 1− = 0.9996 ≈ 1 Iy 1448234429mm 4

The

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

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

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) × Where:

106

N Ed N cr , y

=0

ADVANCE DESIGN VALIDATION GUIDE

N cr , y =

π ²× E× Iy l fy ²

= 95035371.44 N = 95035.37 kN (previously calculated)

C my , 0 = 0.79 + 0.36 × (0 − 0.33)×

C my = C my , 0 + (1 − C my , 0 )×

ξ y × a LT

= 0.79 + (1 − 0.79 )×

1 + ξ y × a LT

Equivalent uniform moment factor, -

328000 N = 0.79 95035371.44 N

9.55 ×1 = 0.949 1 + 9.55 ×1

CmLT , calculation

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

2 CmLT = Cmy ×

-the

C my

a LT  1 − N Ed  N cr , z 

   × 1 − N Ed   N cr ,T  

   

≥1

CmLT calculation, must be recalculated for the corresponding column part (in this case the top column part)

term used for

-this being the case, the

C my

Cmy = Cmy , 0 + (1 − Cmy , 0 )

will be calculated using ψ

ξ y a LT 1 + ξ y a LT

C my , 0 = 0.79 + 0.21×ψ + 0.36 × (ψ − 0.33)×

ξy =

M y , Ed N Ed

×

Aeff Weff , y

=

ξ y × a LT 1 + ξ y × a LT

a LT  1 − N Ed  N cr , z 

   × 1 − N Ed   N cr ,T  

   

= 0.895 + (1 − 0.895) ×

= 0.974 2 ×

C mLT ≤ 1 Therefore the

N Ed 328000 N = 0.79 + 0.21× 0.5 + 0.36 × (0.5 − 0.33)× = 0.895 95035371.44 N N cr , y

1274 ×10 6 Nmm 7815.2mm 2 × = 9.55 328000 N 3179229.533mm 3

Cmy = Cmy , 0 + (1 − Cmy , 0 ) × 2 C mLT = C my ×

= 0.5 :

(previously calculated)

9.55 × 1 = 0.974 1 + 9.55 × 1

1

 = 0.989 0.328   0.328    1 −  × 1 −   ⇒ C mLT = 1  6.9897   9.644    

k yy term corresponding to the top part of the column will be:

k yy = C my × C mLT ×

µy 1−

N Ed N cr , y

= 0.949 × 1 ×

1 = 0.952 328000 N 1− 95035371.44 N

b) for the bottom part of the column:

k yy = C my × C mLT ×

µy 1−

N Ed N cr , y 107

ADVANCE DESIGN VALIDATION GUIDE

λ0 =

Weff , y × f y M cr 0

=

3179229.533mm 3 × 275 N / mm 2 = 0.534 3070560199 Nmm

For the bottom part of the column:

 N 0.20 × C1 × 4 1 − Ed  N cr , z = 0.260

   × 1 − N Ed   N cr ,TF  

  = 0.20 × 1.77 × 4 1 − 328000 N  × 1 − 328000 N  =   6989347.62 N   9646886.24 N  

Therefore: For the bottom part of the column:

    ε y × a LT  C my = C my , 0 + (1 − C my , 0 )× λ 0 = 0.534   1 + ε y × a LT       N N   0.20 × C1 × 4 1 − Ed  × 1 − Ed  = 0.260  ⇒ C mz = C mz , 0    N cr , z   N cr ,TF  a LT 2  C mLT = C my ≥1 ×       N N     λ0 = 0.534 > 0.20 ⋅ C1 ⋅ 4 1 − Ed  ⋅ 1 − Ed  = 0.260  1 − N Ed  × 1 − N Ed  N N  N   N  cr , z   cr ,TF     cr , z   cr ,T  

108

ADVANCE DESIGN VALIDATION GUIDE

The

Cmy 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.

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξy =

M y , Ed N Ed

×

ε y × aLT 1 + ε y × aLT

Aeff Weff , y

Elastic modulus after the Y axis,

ξy =

M y , Ed N Ed

a LT = 1 −

The

×

Aeff Weff , y

=

Wel , y = Weff , y = 3179229.533mm 3

1274 ×10 6 Nmm 7815.2mm 2 × = 9.55 328000 N 3179229.533mm 3

It 514614.75mm 4 = 0.9996 ≈ 1 = 1− Iy 1448234429mm 4

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

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

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) ×

=

M Ed ,inf M Ed ,sup

=

0 =0 1274kNm

N Ed N cr , y

Where:

N cr , y =

π ²× E× Iy l fy ²

= 95035371.44 N = 95035.37 kN (previously calculated)

C my , 0 = 0.79 + 0.36 × (0 − 0.33)×

C my = C my , 0 + (1 − C my , 0 )×

328000 N = 0.79 95035371.44 N

ξ y × a LT 1 + ξ y × a LT

= 0.79 + (1 − 0.79 )×

9.55 ×1 = 0.949 1 + 9.55 ×1

109

ADVANCE DESIGN VALIDATION GUIDE

Equivalent uniform moment factor, -

CmLT , calculation

CmLT must be calculated separately for each column part, separated by the lateral buckling restraint a LT

2 × CmLT = Cmy

-the

C my

 1 − N Ed  N cr , z 

   × 1 − N Ed   N cr ,T  

   

≥1

CmLT calculation, must be recalculated for the corresponding column part (in this case the top column part)

term used for

-this being the case, the

will be calculated using ψ

C my

Cmy = Cmy , 0 + (1 − Cmy , 0 )

ξ y a LT 1 + ξ y a LT

C my , 0 = 0.79 + 0.21×ψ + 0.36 × (ψ − 0.33)×

ξy =

M y , Ed N Ed

×

Aeff Weff , y

=

N Ed 328000 N = 0.79 + 0.21× 0.5 + 0.36 × (0.5 − 0.33)× = 0.895 N cr , y 95035371.44 N

637 ×10 6 Nmm 7815.2mm 2 × = 4.77 328000 N 3179229.533mm 3

ξ y × a LT

C my = C my , 0 + (1 − C my , 0 )× 2 × C mLT = C my

1 + ξ y × a LT

a LT  1 − N Ed  N cr , z 

   × 1 − N Ed   N cr ,T  

   

= 0.895 + (1 − 0.895)×

= 0.967 2 ×

C mLT ≤ 1 k yy = C my × C mLT ×

= 0 .5 :

µy

N 1 − Ed N cr , y

= 0.949 × 1 ×

4.77 ×1 = 0.967 1 + 4.77 ×1

1

 = 0.974 328000 N   328000 N      × 1 − 1 −  ⇒ C mLT = 1  6989347.62 N   9646886.24 N    

1 = 0.952 328000 N 1− 95035371.44 N

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

12.29.2.8 Internal factor, k yz , calculation a) for the top part of the column:

k yz = C mz ×

The

110

Cmz

µy 1−

N Ed N cr , z

term must be calculated for the hole column length

  M 0 ψ = Ed ,inf = = 0   M Ed ,sup 1274kNm  

ADVANCE DESIGN VALIDATION GUIDE

C mz = C mz , 0 = 0.79 + 0.36 × (− 0.33)× k yz = C mz ×

µy 1−

N Ed N cr , z

= 0.784 ×

N Ed 328000 N = 0.79 + 0.36 × (− 0.33)× = 0.784 N cr , z 6989347.62 N

1 = 0.823 328000 N 1− 6989347.62 N

b) for the bottom part of the column:

k yz = C mz ×

µy 1−

N Ed N cr , z

= 0.784 ×

1 = 0.823 328000 N 1− 6989347.62 N

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

12.29.2.9 Internal factor, k zy , calculation a) for the top part of the column:

k zy = C my × C mLT ×

µz 1−

N Ed N cr , y

N Ed 328000 N 1− N cr , z 6989347.62 N = µz = = 0.991 N Ed 328000 N 1 − 0.812 × 1− χz × 6989347.62 N N cr , z 1−

k zy = C my × C mLT ×

µz 1−

N Ed N cr , y

= 0.949 × 1 ×

0.991 = 0.944 328000 N 1− 95035371.44 N

b) for the bottom part of the column:

k zy = C my × C mLT ×

µz 1−

N Ed N cr , y

= 0.949 × 1 ×

0.991 = 0.944 328000 N 1− 95035371.44 N

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

12.29.2.10Internal factor, k zz , calculation a) for the top part of the column:

k zz = C mz ×

µz 1−

N Ed N cr , z

= 0.784 ×

0.991 = 0.815 328000 N 1− 6989347.62 N

111

ADVANCE DESIGN VALIDATION GUIDE

b) for the bottom part of the column:

k zz = C mz ×

µz

N 1 − Ed N cr , z

= 0.784 ×

0.991 = 0.815 328000 N 1− 6989347.62 N

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

12.29.2.11Bending and axial compression verification

M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k yz × z , Ed + k yy ×  N M M z , Rk  χ y × Rk χ LT × y , Rk γ M1 γ M1 γ M1   M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k zz × z , Ed + k zy ×  N Rk M M z , Rk χ LT × y , Rk χz × γ M1 γ M1 γ M1  N Rk = f y × Ai

a) for the top part of the column:

328000 N 1274 × 10 6 Nmm + × + 0 . 952 7815.2mm 2 × 275 N / mm 2 3179229.533mm 3 × 275 N / mm 2 1× 0.803 × 1 1 6 127.4 × 10 Nmm + 0.823 × = 0.15 + 1.73 + 1.58 = 3.46 242068.10mm 3 × 275 N / mm 2 1 328000 N 1274 × 10 6 Nmm + 0.944 × + 7815.2mm 2 × 275 N / mm 2 3179229.533mm 3 × 275 N / mm 2 0.812 × 0.803 × 1 1 6 127.4 × 10 Nmm + 0.815 × = 0.19 + 1.71 + 1.56 = 3.56 242068.10mm 3 × 275 N / mm 2 1

112

ADVANCE DESIGN VALIDATION GUIDE

b) for the bottom part of the column:

328000 N 1274 × 10 6 Nmm + 0 . 952 × + 7815.2mm 2 × 275 N / mm 2 3179229.533mm 3 × 275 N / mm 2 0.812 × 0.803 × 1 1 6 127.4 × 10 Nmm + 0.823 × = 0.19 + 1.73 + 1.58 = 3.46 242068.10mm 3 × 275 N / mm 2 1 328000 N 1274 × 10 6 Nmm + 0.944 × + 7815.2mm 2 × 275 N / mm 2 3179229.533mm 3 × 275 N / mm 2 0.812 × 0.803 × 1 1 6 127.4 × 10 Nmm + 0.815 × = 0.19 + 1.71 + 1.56 = 3.56 242068.10mm 3 × 275 N / mm 2 1 Note: The software does not give the results of the lower section because it is not the most solicited segment. Finite elements modeling ■ ■ ■

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

χ y coefficient corresponding to non-dimensional slenderness λ y Column subjected to axial and shear force to the top

χy

113

ADVANCE DESIGN VALIDATION GUIDE

χ z coefficient corresponding to non-dimensional slenderness λ z Column subjected to axial and shear force to the top

χz

Internal factor, k yy Column subjected to axial and shear force to the top

k yy

114

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yz Column subjected to axial and shear force to the top

k yz

Internal factor, k zy Column subjected to axial and shear force to the top

k zy

115

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k zz Column subjected to axial and shear force to the top

k zz

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

116

ADVANCE DESIGN VALIDATION GUIDE

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

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

117

ADVANCE DESIGN VALIDATION GUIDE

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

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

118

ADVANCE DESIGN VALIDATION GUIDE

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

12.29.2.1 Reference results Result name

Result description

Reference value

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

1

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.81

k yy

Internal factor, k yy

0.95

k yz

Internal factor, k yz

0.82

k zy

Internal factor, k zy

0.94

k zy

Internal factor, k zy

0.82

SNy

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

0.15

SMyy SMyz SNz SMzy SMzz

1.72 1.58 0.19 1.71 1.56

119

ADVANCE DESIGN VALIDATION GUIDE

12.29.3Calculated results

120

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 depending of the compression effort over the Y axis

term

0.152455 adim

1.6367 %

SMyy

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

term

1.72357 adim

0.2076 %

SMyz

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

term

1.57508 adim

-0.3114 %

SNz

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

term

0.187789 adim

-1.1637 %

SMzy

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

term

1.70775 adim

-0.1316 %

SMzy

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

term

1.70775 adim

-0.1316 %

ADVANCE DESIGN VALIDATION GUIDE

12.30 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 3 column fixed on the bottom (evaluated by SOCOTEC France - ref. Test 26) Test ID: 5714 Test status: Passed

12.30.1Description The test verifies a user defined cross section column. 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. The calculations are made according to Eurocode 3 French Annex.

12.30.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.30.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■ ■

Exploitation loadings (category A): Fz=-1000000N N; My=200000Nm The ultimate limit state (ULS) combination is: Cmax = 1 x Q Cross section dimensions are in millimeters (mm).

121

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Geometrical properties ■

Column length: L=2000mm

■ ■

A = 9108.72mm 2 Overall breadth: b = 190mm Flange thickness: t f = 14.6mm

■

Root radius:

■

Web thickness: t w

■

Depth of the web:

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, W y

■

Elastic modulus after the Z axis, Wel , z

= 175962.65mm 3

■

Plastic modulus after the Z axis, W pl , z

= 271897.69

■ ■ ■ ■

Flexion inertia moment around the Y axis: Iy=257332751mm4 4 Flexion inertia moment around the Z axis: Iz=16716452.10 mm Torsional moment of inertia: It=492581.13 mm4 6 Working inertial moment: Iw=645759981974.33mm

■

Cross section area:

r = 0mm

= 9.4mm hw = 408mm = 1261435.06mm 3

= 1428491.78mm 3

Materials properties S275 steel material is used. The following characteristics are used: ■ ■ ■

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

Boundary conditions The boundary conditions are described below: ■

Outer: ►

122

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

ADVANCE DESIGN VALIDATION GUIDE

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

12.30.2.2 Cross section Class According to Advance Design calculations: Cross-class classification is made according to Table 5.2

- for beam web: The web dimensions are 378.8x9.4mm.

ψ = 2⋅

N Ed 1.000 −1 = 2 ⋅ − 1 = −0.20 > −1 A⋅ f y 0.0091× 275

123

ADVANCE DESIGN VALIDATION GUIDE

1 

N



1 

1



Ed  = ⋅ 1 + α = ⋅ 1 +  = 1.01 > 0.5 f y × t × d  2  275 × 0.3788 × 0.0094  2 

ε=

235 = fy

235 = 0.924 275

c 408mm − 2 ×14.6mm  = = 40.30 c 42 × ε 42 × 0.924 = = 64.25 t 9.4mm  ⇒ = 170 > ψ t 0 . 67 + 0 . 33 × 0 . 67 + 0 . 33 × ( − 0 . 20 )  ε = 0.924 therefore the beam web is considered to be Class 3. - for beam flange:

c 90.30  = = 6.18 c t 14.6  ⇒ = 6.18 ≤ 9 × 0.924 = 8.316 t  ε = 0.924 In conclusion, the section is considered to be Class 3.

124

therefore the haunch is considered to be Class1

ADVANCE DESIGN VALIDATION GUIDE

12.30.2.3 Buckling verification a) over the strong axis of the section, y-y: - the imperfection factor α will be selected according to Tables 6.1 and 6.2:

α = 0.34 Coefficient corresponding to non-dimensional slenderness after Y-Y axis:

χ y coefficient

corresponding to non-dimensional slenderness

λy

will be determined from the relevant buckling

curve according to:

χy =

1 2

Φ y + Φ y − λy

2

≤1

(6.49)

λ y the non-dimensional slenderness corresponding to Class 4 cross-sections:

λy =

A * fy N cr , y

Cross section area:

A = 9108.72mm 2

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

4

125

ADVANCE DESIGN VALIDATION GUIDE

N cr , y =

π²× E × Iy l fy ²

A× f y

λy =

N cr , y

=

π ² × 210000 N / mm 2 × 257332751mm 4

(2000mm)²

= 133338053.7 N = 133338.05kN

9108.72mm 2 × 275 N / mm 2 = 0.137 133338053.7 N

=

[

]

φ y = 0.5 × 1 + α × (λ y − 0.2) + λ y ² = 0.5 × [1 + 0.34 × (0.137 − 0.2 ) + 0.137 2 ] = 0.499

χy =

1 2

Φy + Φy −λy

χy ≤1

2

=

1

 = 1.022 0.499 + 0.499 − 0.137  ⇒ χy =1   2

2

b) over the weak axis of the section, z-z: - the imperfection factor α will be selected according to Tables 6.1 and 6.2:

α = 0.49 Coefficient corresponding to non-dimensional slenderness after Z-Z axis:

χ z coefficient corresponding to non-dimensional slenderness according to: 126

λz

will be determined from the relevant buckling curve

ADVANCE DESIGN VALIDATION GUIDE

χz =

1 2

Φz + Φz − λ z

2

≤1

(6.49)

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

A* f y

λz =

N cr , z

l fz = 2.00m Flexion inertia moment around the Z axis: Iz=16716452.10 mm4 Cross section area:

N cr , z =

π ²× E × Iz l fz ²

A× fy

λz =

A = 9108.72mm 2

N cr , z

=

=

π ² × 210000 N / mm 2 ×16716452.10mm 4

(2000mm )²

= 8661700.38 N = 8661.70kN

9108.72mm 2 × 275 N / mm 2 = 0.538 8661700.38 N

[

]

φ z = 0.5 × 1 + α × (λ z − 0.2) + λ z ² = 0.5 × [1 + 0.49 × (0.538 − 0.2 ) + 0.538 2 ] = 0.728

χz =

1 2

Φz + Φz − λ z

χz ≤ 1

2

=

1

 = 0.821 0.728 + 0.728 2 − 0.538 2  ⇒ χ z = 0.821  

12.30.2.4 Lateral 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:

M cr = C1 ×

L² × G × It I π ² × E × Iz × w+ Iz π ² × E × Iz L² 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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1-AN France; Chapter 3; …(6)

127

ADVANCE DESIGN VALIDATION GUIDE

- in order to simplify the calculation, we will consider

C1 = 1

Flexion inertia moment around the Y axis: Iy=257332751mm4 Flexion inertia moment around the Z axis: Iz=16716452.10 mm4 Longitudinal elastic modulus: E = 210000 N/mm2. Torsional moment of inertia: It=492581.13 mm4 Warping inertial moment: IW is the warping inertia (deformation inertia moment):

Iw =

I z × (h − t f

)

2

4

h cross section height; h=408mm

tf

flange thickness;

Iw =

t f = 14.6mm

16716452.10mm 4 × (408mm − 14.6mm ) = 6.46774 × 1011 mm 6 4 2

According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column: L=2000mm Shear modulus of rigidity: G=80800N/mm2

M cr = C1 ×

π ²× E × Iz L²

×

I w L² × G × I t π 2 × 210000 N / mm 2 ×16716452.10mm 4 = 1× × + Iz π ²× E × Iz (2000mm )2

6.46774 ×1011 mm 6 (2000mm ) × 80800 N / mm 2 × 492581.13mm 4 + = 1× 8661700.384 N × 208.052mm = 16716452.10mm 4 π 2 × 210000 N / mm 2 ×16716452.10mm 4 = 1802088994 Nmm = 1802.089kNm

×

128

2

ADVANCE DESIGN VALIDATION GUIDE

The elastic modulus : Wel , y

λ LT =

Weff , y f y M cr

Calculation of the

χ LT =

=

=

Iy z max

257332751mm 4 = = 1261434.172mm 3 204mm

1261434.172mm 3 ×275 N / mm 2 = 0.439 1802088994 Nmm

χ LT for appropriate non-dimensional slenderness λ LT

1

φLT + φLT ² − λ LT ²

≤1

will be determined with formula:

(6.56)

[

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

]

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

h 408mm = = 2.147 > 2 b 190mm

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

[

(

)

]

α = 0.76

φ LT = 0.5 × 1 + α LT × λ LT − 0.2 + λ LT ² = 0.5 × [1 + 0.76 × (0.439 − 0.2 ) + 0.439² ] = 0.687

χ LT =

1

φ LT + φ LT ² − λ LT ²

=

1 = 0.813 ≤ 1 0.687 + 0.687² − 0.439²

129

ADVANCE DESIGN VALIDATION GUIDE

12.30.2.5 Internal factor, k yy , calculation The internal factor k yy 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:

k yy = C my × C mLT ×

µy 1−

N Ed N cr , y

N Ed N cr , y µy = N 1 − χ y × Ed N cr , y 1−

χ y = 1.022

(previously calculated)

N Ed = 1000kN N cr , y =

π²× E × Iy l fy ²

= 133338053.7 N = 133338.05kN

N Ed 1000000 N 1− N cr , y 133338053 .7 N = 1 µy = = 1000000 N N Ed 1 − 1× 1− χ y × 133338053.7 N N cr , y 1−

130

(previously calculated)

ADVANCE DESIGN VALIDATION GUIDE

The

Cmy will be calculated according to Table A.1:

Calculation of the λ 0 term:

λ0 =

Weff , y × f y M cr 0 According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2

Wel , y =

Iy z max

=

257332751mm 4 = 1261434.172mm 3 204mm

The calculation the

M cr = C1 ×

M cr 0

π ²× E × Iz L²

will be calculated using

×

C1 = 1 and C2 = 0 , therefore:

I w L² × G × I t π 2 × 210000 N / mm 2 ×16716452.10mm 4 + = 1× × Iz π ²× E × Iz (2000mm )2

6.46774 ×1011 mm 6 (2000mm ) × 80800 N / mm 2 × 492581.13mm 4 + = 1× 8661700.384 N × 208.052mm = 16716452.10mm 4 π 2 × 210000 N / mm 2 ×16716452.10mm 4 = 1802088994 Nmm = 1802.089kNm 2

×

λ0 =

Weff , y f y M cr

1261434.172mm 3 ×275 N / mm 2 = = 0.439 1802088994 Nmm

131

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed N cr , z   N cr ,TF 

   

term:

Where: - for a symmetrical section for the both axis,

N cr ,TF = N cr ,T

π 2 × E × I w  1  × G × I + t 2 Lcr ,T ²  i0 

N cr ,T =

The mass moment of inertia I 0 2

2

2

i0 = i y + i z + y02 + z 02 = 0.0301 Torsional moment of inertia: It=492581.13 mm4 Working inertial moment: Iw=645759981974.33mm6 - the buckling length,

Lcr ,T , Lcr ,T = 2.00m

N cr ,T

1mm 2 = 0.0301mm 4

 π 2 × 210000 N / mm 2 × 645759981974.33mm 6  2 4  = ×  80800 N / mm × 492581.13mm + (2000mm )²  

= 1.244 ×1013 N N Ed = 1000000 N N cr ,TF = N cr ,T = 1.244 ×1013 N

N cr , z =

π ²× E × Iz l fz ²

=

π ² × 210000 N / mm 2 × 16716452.10mm 4

(2000mm)²

= 8661700.38 N

(previously calculated)

C1=1 for the top part of the column For the top part of the column:

 N 0.20 × C1 × 4 1 − Ed  N cr , z = 0.172

132

   × 1 − N Ed   N cr ,TF  

  = 0.20 × 1 × 4  

1000000 N   1000000 N    × 1 − = 1 − 13  8661700.38 N   1.244 ×10 N 

ADVANCE DESIGN VALIDATION GUIDE

Therefore: For the top part of the column:

    ε y × a LT  C my = C my , 0 + (1 − C my , 0 )× λ 0 = 0.469   1 + ε y × a LT       N N   0.20 × C1 × 4 1 − Ed  × 1 − Ed  = 0.172  ⇒ C mz = C mz , 0  N cr , z   N cr ,TF    a LT 2  C mLT = C my × ≥1       N N     Ed Ed N N  ⋅ 1 −  = 0.172  λ0 = 0.469 > 0.20 ⋅ C1 ⋅ 4 1 − 1 − Ed  × 1 − Ed    N  N  N   N   , cr , TF cr z       cr ,T  cr , z    The

Cmy 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.

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξy =

M y , Ed N Ed

×

ε y × aLT 1 + ε y × aLT

Aeff Weff , y

Elastic modulus after the Y axis,

ξy =

M y , Ed N Ed

×

Aeff Weff , y

=

Wel , y = Weff , y = 1261435.06mm 3

200 ×10 6 Nmm 9108.72mm 2 × = 1.444 1000000 N 1261435.06mm 3

a LT

It 492581.13 mm 4 = 1− = 1− = 0.998 ≈ 1 Iy 257332751mm 4

The

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

133

ADVANCE DESIGN VALIDATION GUIDE

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

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) ×

=1

N Ed N cr , y

Where:

N cr , y =

π²× E× Iy l fy ²

=

π ² × 210000 N / mm 2 × 257332751mm 4

= 133338053.7 N

(2000mm)²

(previously calculated)

N Ed = 1000000 N C my , 0 = 0.79 + 0.21×1 + 0.36 × (1 − 0.33)×

C my = C my , 0 + (1 − C my , 0 )×

ξ y × a LT 1 + ξ y × a LT

Equivalent uniform moment factor,

 1 − N Ed  N cr , z 

= 1.002 + (1 − 1.002 )×

   × 1 − N Ed   N cr ,T  

   

= 1.0012 ×

C mLT ≥ 1 Therefore the

1.444 ×1 = 1.001 1 + 1.444 ×1

CmLT , calculation

a LT

2 × C mLT = C my

1000000 N = 1.002 133338053.7 N

1

 = 1.065 1000000 N  1000000 N       × 1 − 1 − 13  ⇒ C mLT = 1.065 38 . 8661700 N 10 . 244 1 × N       

k yy term corresponding to the top part of the column will be:

k yy = C my × C mLT ×

µy 1−

N Ed N cr , y

= 1.001 × 1.065 ×

1 = 1.074 1000000 N 1− 133338053.7 N

12.30.2.6 Internal factor, k yz , calculation

k yz = C mz ×

Cmz

µy 1−

N Ed N cr , z

term must be calculated for the whole column length

N cr , z =

π ²× E × Iz l fz ²

=

π ² × 210000 N / mm 2 ×16716452.10mm 4

(2000mm )²

C mz = C mz , 0 = 0.79 + 0.36 × (− 0.33) × k yz = C mz ×

134

µy

N 1 − Ed N cr , z

  M ψ = Ed ,inf = 200kNm = 1  M Ed ,sup 200kNm  

= 0.776 ×

= 8661700.38 N = 8661.70kN

N Ed 1000000 N = 0.776 = 0.79 + 0.36 × (− 0.33) × 8661700.38 N N cr , z

1 = 0.878 1000000 N 1− 8661700.38 N

ADVANCE DESIGN VALIDATION GUIDE

12.30.2.7 Internal factor, k zy , calculation

k zy = C my × C mLT ×

µz 1−

N Ed N cr , y

N Ed 1000000 N 1− N cr , z 8661700.38 N µz = = = 0.799 N Ed 1000000 N 1 − 0 . 821 × 1− χz × 8661700.38 N N cr , z 1−

χ z = 0.821 (previously calculated) k zy = C my × C mLT ×

µz

N 1 − Ed N cr , y

= 1.001 × 1.065 ×

0.977 = 1.050 1000000 N 1− 133338053.7 N

12.30.2.8 Internal factor, k zz , calculation

k zz = C mz ×

µz

N 1 − Ed N cr , z

= 0.776 ×

0.977 = 0.857 1000000 N 1− 8661700.38 N

12.30.2.9 Bending and axial compression verification

M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k yy × + k yz × z , Ed  N M M z , Rk  χ y × Rk χ LT × y , Rk γ M1 γ M1 γ M1   M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k zy × + k zz × z , Ed  N Rk M M z , Rk χ LT × y , Rk χz × γ M1 γ M1 γ M1  N Rk = f y × Ai

135

ADVANCE DESIGN VALIDATION GUIDE

1000000 N 200 ×10 6 Nmm + 1 . 074 × + 9108.72mm 2 × 275 N / mm 2 1261435.06mm 3 × 275 N / mm 2 1× 0.813 × 1 1 6 10 ×10 Nmm + 0.878 × = 0.40 + 0.76 + 0.18 = 1.34 175962.5mm 3 × 275 N / mm 2 1 1000000 N 200 ×10 6 Nmm + 1.050 × + 9108.72mm 2 mm 2 × 275 N / mm 2 1261435.06mm 3 × 275 N / mm 2 0.821× 0.813 × 1 1 6 10 ×10 Nmm + 0.857 × = 0.49 + 0.74 + 0.18 = 1.41 175962.5mm 3 × 275 N / mm 2 1 Finite elements modeling ■ ■ ■

136

Linear element: S beam, 5 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

χ y coefficient corresponding to non-dimensional slenderness λ y Column subjected to axial and shear force to the top

χy

χ z coefficient corresponding to non-dimensional slenderness λ z Column subjected to axial and shear force to the top

χz

137

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yy Column subjected to axial and shear force to the top

k yy

Internal factor, k yz Column subjected to axial and shear force to the top

k yz

138

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k zy Column subjected to axial and shear force to the top

k zy

Internal factor, k zz Column subjected to axial and shear force to the top

k zz

139

ADVANCE DESIGN VALIDATION GUIDE

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

140

ADVANCE DESIGN VALIDATION GUIDE

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

141

ADVANCE DESIGN VALIDATION GUIDE

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

142

ADVANCE DESIGN VALIDATION GUIDE

12.30.2.10Reference results Result name

Result description

Reference value

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

1

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.821

k yy

Internal factor, k yy

1.074

k yz

Internal factor, k yz

0.878

k zy

Internal factor, k zy

1.050

k zy

Internal factor, k zy

0.857

SNy

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

0.40

SMyy SMyz SNz SMzy SMzz

0.76 0.18 0.49 0.74 0.18

12.30.3Calculated 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 %

Kyy

Internal factor kyy

1.11767 adim

4.0661 %

Kyz

Internal factor kyz

0.877605 adim

-0.0450 %

Kzy

Internal factor kzy

1.09224 adim

4.0229 %

Kzz

Internal factor kzz

0.857639 adim

0.0746 %

#SNy

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

term

0.399218 adim

-0.1955 %

SMyy

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

term

0.783306 adim

3.0666 %

SMyz

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

term

0.181362 adim

0.7567 %

SNz

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

term

0.485883 adim

-0.8402 %

SMzy

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

term

0.765485 adim

3.4439 %

SMzz

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

term

0.177236 adim

-1.5356 %

143

ADVANCE DESIGN VALIDATION GUIDE

12.31 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 3 beam simply supported with a displacement restraint (evaluated by SOCOTEC France - ref. Test 27) Test ID: 5717 Test status: Passed

12.31.1Description 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 cross section has an “I symmetric” shape with: 530mm height; 190mm width; 12mm center thickness; 19mm flange thickness; 0mm fillet radius and 0mm rounding radius. 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. The calculations are made according to Eurocode 3 French Annex.

12.31.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.31.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■ ■

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).

Units Metric System

144

ADVANCE DESIGN VALIDATION GUIDE

Geometrical properties ■

Column length: L=5000mm

■

Cross section area:

■ ■

A = 14264mm 2 Overall breadth: b = 220mm Flange thickness: t f = 19mm r = 0mm

■

Root radius:

■

Web thickness: t w

■

Depth of the web:

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, W y

■

Elastic modulus after the Z axis, W el , z

= 307177.41mm 3

■

Plastic modulus after the Z axis, W pl , z

= 477512.00mm 3

■

Flexion inertia moment around the Y axis:

I y = 665089874.67mm 4

■

Flexion inertia moment around the Z axis:

I z = 33789514.67mm 4

■

Torsional moment of inertia:

■

Working inertial moment:

= 12mm hw = 530mm = 2509773.11mm 3

= 2862172.00mm 3

I t = 1269555.73mm 4

I w = 2201162989666.67mm 6

Materials properties S235 steel material is used. The following characteristics are used: ■ ■ ■

Yield strength fy = 235 MPa, Longitudinal elastic modulus: E = 210000 MPa. Shear modulus of rigidity: G=80800MPa

Boundary conditions The boundary conditions are described below: ■

Outer: ► ►

■

Support at start point (x = 0) restrained in translation along X, Y and Z axis, 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). 145

ADVANCE DESIGN VALIDATION GUIDE

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

146

ADVANCE DESIGN VALIDATION GUIDE

12.31.2.2 Cross 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.

ψ=

σ inf 246.94Mpa = =1 σ sup 246.94Mpa

147

ADVANCE DESIGN VALIDATION GUIDE

ε=

148

235 = fy

235 =1 235

ADVANCE DESIGN VALIDATION GUIDE

c 530mm − 2 × 19mm  = = 41 c t 12mm  ⇒ 38 × ε = 38 < = 41 ≤ 42 × ε = 42 t ε =1 

therefore

the

beam

web

is

considered to be Class 3 -for beam flange:

220 − 12   c c 2 = = 5.47  ⇒ = 4.57 ≤ 9 × 1 = 9 t 19 t   ε =1

therefore the haunch is considered to be Class1

In conclusion, the section is considered to be Class 3.

12.31.2.3 Buckling 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:

α = 0.34

149

ADVANCE DESIGN VALIDATION GUIDE

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

χ y coefficient

corresponding to non-dimensional slenderness

λy

will be determined from the relevant buckling

curve according to:

χy =

1 2

Φ y + Φ y − λy

2

≤1

(6.49)

λ y the non-dimensional slenderness corresponding to Class 3 cross-sections: A * fy

λy =

N cr , y A = 14264mm 2

Cross section area:

Flexion inertia moment around the Y axis:

N cr , y =

π²× E× Iy l fy ²

A× f y

λy =

=

I y = 665089874.67 mm 4

π ² × 210000 N / mm 2 × 665089874.67mm 4

(5000mm )²

N cr , y

14264mm 2 × 235 N / mm 2 = = 0.247 55139061.21N

[

]

= 55139061.21N = 55139.06kN

φ y = 0.5 × 1 + α × (λ y − 0.2) + λ y ² = 0.5 × [1 + 0.34 × (0.247 − 0.2 ) + 0.247 2 ] = 0.538

χy =

1 Φy + Φy −λy

χy ≤1

150

2

2

=

1

 = 0.984 0.538 + 0.538 − 0.247  ⇒ χ y = 0.984   2

2

ADVANCE DESIGN VALIDATION GUIDE

b) over the weak axis of the section, z-z: -the imperfection factor α will be selected according to the Table 6.1 and 6.2:

α = 0.49 Coefficient corresponding to non-dimensional slenderness after Z-Z axis:

χ z coefficient corresponding to non-dimensional slenderness according to:

χz =

1 2

Φz + Φz − λ z

2

λz

will be determined from the relevant buckling curve

≤1

(6.49)

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

λz =

A* f y N cr , z

l fz = 2.50m Flexion inertia moment around the Z axis: Cross section area:

I z = 33789514.67mm 4

A = 14264mm 2 151

ADVANCE DESIGN VALIDATION GUIDE

N cr , z =

π²× E × Iz l fz ²

A× f y

λz =

N cr , z

=

=

π ² × 210000 N / mm 2 × 33789514.67mm 4

(2500mm)²

= 11205235.19 N = 11205.235kN

14264mm 2 × 235 N / mm 2 = 0.547 11205235.19 N

[

]

φ z = 0.5 × 1 + α × (λ z − 0.2) + λ z ² = 0.5 × [1 + 0.49 × (0.547 − 0.2 ) + 0.547 2 ] = 0.735

χz =

1 2

Φz + Φz − λ z

χz ≤1

2

=

1

 = 0.819 0.735 + 0.735 2 − 0.547 2  ⇒ χ z = 0.816  

12.31.2.4 Lateral torsional buckling verification

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

µ=

q × L ² 10000 N / m × (2500mm )² = = 0.25 8× M 8 × 31.25 × 10 3 Nm

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

M cr = C1 ×

I L² × G × It π ² × E × Iz × w+ Iz π ² × E × Iz L² 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 According to EN 1993-1-1-AN France; Chapter 3; …(6)

ψ =0

152

therefore:

ADVANCE DESIGN VALIDATION GUIDE

C1 = 1.31 Flexion inertia moment around the Y axis:

I y = 665089874.67mm4

Flexion inertia moment around the Z axis:

I z = 33789514.67mm 4

Longitudinal elastic modulus:

E = 210000 N / mm 2

Torsional moment of inertia:

I t = 1269555.73mm 4

Working inertial moment:

I w = 2201162989666.67mm 6

Shear modulus of rigidity:

G = 80800 N / mm 2

Buckling length of the beam

L = 2500mm

Elastic modulus after the Y axis, Wel , y

M cr = C1 ×

π ²× E × Iz L²

= 2509773.11mm 3

I w L² × G × I t π 2 × 210000 N / mm 2 × 33789514.67mm 4 = 1.31× × × + Iz π ²× E × Iz (2500mm)2

22.01163 × 1011 mm 6 (2500mm ) × 80800 N / mm 2 × 1269555.73mm 4 × + = 1.31× 11205235.19 N × 272.58mm = 33789514.67 mm 4 π 2 × 210000 N / mm 2 × 33789514.67mm 4 = 4001163141Nmm 2

153

ADVANCE DESIGN VALIDATION GUIDE

λ LT =

Weff , y f y M cr

Calculation of the

=

2509773.11mm 3 ×235 N / mm 2 = 0.384 4001163141Nmm

χ LT for appropriate non-dimensional slenderness λ LT

will be determined below:

Note:

M Ed 31.25 × 10 6 Nmm = = 7.81× 10 −3 M cr 4001163141   2   M Ed ≤ λ LT , 0 ⇒ −3  M 2 M cr = 7.81× 10 ≤ 0.04 Ed = 0.00781 ≤ λ LT , 0 = 0.0156 ⇒  M cr  b 220 2 = 0.3 × = 0.3 × = 0.125 ⇒ λ LT , 0 = 0.0156  h 530

λ LT = 0.384 > 0.20 M Ed M cr

λ LT ,0

For slendernesses apply.

M Ed ≤λ M cr

According to EN 1993-1-1-AN France; AN.3; Chapter 6.3.2.2(4) 2 LT , 0

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

-therefore:

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

12.31.2.5 Internal factor, k yy , calculation The internal factor k yy 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:

154

ADVANCE DESIGN VALIDATION GUIDE

k yy = C my × C mLT ×

µy 1−

N Ed N cr , y

N Ed N cr , y µy = N 1 − χ y × Ed N cr , y 1−

χ y = 0.984

(previously calculated)

N Ed = 3700kN N cr , y =

π²× E× Iy l fy ²

=

π ² × 210000 N / mm 2 × 665089874.67mm 4

(5000mm )²

= 55139061.21N

(previously

calculated)

N Ed 3700000 N 1− N cr , y 55130961.21N = 0.999 = µy = 3700000 N N Ed 1 − 0.984 × 1− χy × 55130961.21N N cr , y 1−

- 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.

155

ADVANCE DESIGN VALIDATION GUIDE

The

Cmy will be calculated according to Table A.1:

Calculation of the λ 0 term:

λ0 =

Weff , y × f y M cr 0 -according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2

Wel , y =

Iy z max

=

665089874.67mm 4 = 2509773.11mm 3 265mm

The calculation the

M cr , 0 = C1 ×

M cr 0

will be calculated using

π²× E × Iz L²

×

C1 = 1 and C2 = 0 , therefore:

I w L² × G × I t π 2 × 210000 N / mm 2 × 33789514.67mm 4 + = 1× × Iz π²× E × Iz (2500mm)2

22.01163 × 1011 mm 6 (2500mm ) × 80800 N / mm 2 × 1269555.73mm 4 + = 1 × 11205235.19 N × 272.58mm = 33789514.67 mm 4 π 2 × 210000 N / mm 2 × 33789514.67mm 4 = 3054323008 Nmm 2

×

λ0 =

156

Weff , y f y M cr , 0

=

2509773.11mm 3 × 235 N / mm 2 = 0.439 3054323008 Nmm

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed N cr , z   N cr ,TF 

   

term:

Where: -for a symmetrical section for the both axis,

π 2 × E × I w  1  × G × I + t 2 Lcr ,T ²  i0 

N cr ,T = 2

N cr ,TF = N cr ,T

2

2

i0 = i y + i z + y 02 + z 02 = 0.0491 Torsional moment of inertia: Working inertial moment: - the buckling length,

I t = 1269555.73mm 4

I w = 2201162989666.67mm 6

Lcr ,T , Lcr ,T = 2.50m

 1mm 2 π 2 × 210000 N / mm 2 × 2201162989666.67 mm 6  2 4   = × × + N mm mm 80800 / 1269555 . 73 (2500mm)² 0.0491mm 4   = 1.696 × 1013 N N Ed = 3700000 N N cr ,T =

N cr ,TF = N cr ,T = 1.696 × 1013 N

N cr , z =

π ²× E × Iz l fz ²

=

π ² × 210000 N / mm 2 × 33789514.67mm 4

(2500mm)²

= 11205235.19 N

(previously calculated)

C1=1

 N 0.20 × C1 × 4 1 − Ed  N cr , z = 0.181

  N  × 1 − Ed   N cr ,TF  

  = 0.20 × 1 × 4  

3700000 N   3700000 N   =  × 1 − 1 − 13  11205235.19 N   1.696 × 10 N 

Therefore:

λ 0 = 0.439  N 0.20 × C1 × 4 1 − Ed  N cr , z

   × 1 − N Ed   N cr ,TF  

  = 0.181  

 N   N  λ0 = 0.439 > 0.20 ⋅ C1 ⋅ 4 1 − Ed  ⋅ 1 − Ed  N N 

The

cr , z

 

cr ,TF



    ε y × a LT  C my = C my , 0 + (1 − C my , 0 )×   1 + ε y × a LT    ⇒ C mz = C mz , 0   a LT 2  C mLT = C my × ≥1       N N = 0.181  1 − Ed  × 1 − Ed   N   N    cr , z   cr ,T   

Cmy 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.

157

ADVANCE DESIGN VALIDATION GUIDE

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξy =

M y , Ed N Ed

×

ε y × aLT 1 + ε y × aLT

Aeff Weff , y

Elastic modulus after the Y axis,

ξy =

M y , Ed N Ed

×

Aeff Weff , y

=

Wel , y = Weff , y = 2509773.11mm 3

31.25 × 10 6 Nmm 14264mm 2 × = 0.048 3700000 N 2509773.11mm 3

a LT

It 1269555.73mm 4 = 1− = 1− = 0.998 Iy 665089874.67 mm 4

The

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

C my , 0 = 1 + 0.03 ×

N Ed N cr , y

Where:

N cr , y =

π²× E× Iy l fy ²

=

π ² × 210000 N / mm 2 × 665089874.67mm 4

(5000mm )²

= 55139061.21N (previously calculated)

N Ed = 3700000 N C my , 0 = 1 + 0.03 ×

3700000 N = 1.002 55139061.21N

C my = C my , 0 + (1 − C my , 0 )×

ξ y × a LT 1 + ξ y × a LT

Equivalent uniform moment factor,

= 1.002 + (1 − 1.002 ) ×

CmLT , calculation

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

158

0.048 × 1 = 1.002 1 + 0048 × 1

ADVANCE DESIGN VALIDATION GUIDE

2 CmLT = Cmy ×

aLT    1 − N Ed  × 1 − N Ed  N   N cr , z   cr ,T 

   

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

δ z = −0.0001396m π 2 × E × Iy × δ  N  × Ed = Cmy , 0 = 1 +  − 1  L2 × M y , Ed  N cr , y    π 2 × 210000 N / mm 2 × 665089874.67mm 4 × 0.0001396  3700000 N = 1 +  − 1 × = 0.999 2 6 (2500mm) × 31.25 × 10 Nmm   55139061.21N

159

ADVANCE DESIGN VALIDATION GUIDE

C my = C my , 0 + (1 − C my , 0 )×

1 + ε y × a LT

a LT

2 × C mLT = C my

Therefore the

ε y × a LT

 N 1 − Ed  N cr , z 

  N  × 1 − Ed   N cr ,T  

   

= 0.999 + (1 − 0.999 ) × = 0.999² ×

0.219 × 0.998 = 0.999 1 + 0.219 × 0.998 0.998

3700000 N  3700000 N      × 1 − 1 − 13  11205235.19 N   1.696 × 10 N 

k yy term corresponding to the top part of the column will be:

k yy = C my × C mLT ×

µy 1 = 1.002 × 1.377 × = 1.476 N Ed 3700000 N 1− 1− 55130961.21N N cr , y

12.31.2.6 Internal factor, k yz , calculation

k yz = C mz ×

µy 1−

N Ed N cr , z

Cmz coefficient must be calculated considering the beam along its length

C mz = C mz , 0 = 1 + 0.03 × k yz = C mz ×

160

µy 1−

N Ed N cr , z

N Ed 3700000 N = 1 + 0.03 × = 1.010 N cr , z 11205235.19 N

= 0.776 ×

1 = 1.506 3700000 N 1− 11205235.19 N

= 1.377

ADVANCE DESIGN VALIDATION GUIDE

12.31.2.7 Internal factor, k zy , calculation

k zy = C my × C mLT ×

µz 1−

N Ed N cr , y

N Ed 3700000 N 1− N cr , z 11205235.19 N = µz = = 0.917 N Ed 3700000 N 1 − 0.816 × 1− χz × 11205235.19 N N cr , z 1−

χ z = 0.816

(previously calculated)

k zy = C my C mLT

µz

N 1 − Ed N cr , y

= 1.002 × 1.377 ×

0.917 = 1.355 3700000 N 1− 55130961.21N

12.31.2.8 Internal factor, k zz , calculation

k zz = C mz ×

µz 1−

N Ed N cr , z

= 0.776 ×

0.917 = 1.383 3700000 N 1− 11205235.19 N

12.31.2.9 Bending and axial compression verification

M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k yz × z , Ed + k yy ×  N M M z , Rk  χ y × Rk χ LT × y , Rk γ M1 γ M1 γ M1   M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k zz × z , Ed + k zy ×  N Rk M M z , Rk χ LT × y , Rk χz × γ M1 γ M1 γ M1  N Rk = f y × Ai

161

ADVANCE DESIGN VALIDATION GUIDE

3700000 N 31.25 × 10 6 Nmm + × + 1 . 476 14264mm 2 × 235 N / mm 2 2509773.11mm 3 × 235 N / mm 2 0.984 × 1× 1 1 6 15.62 × 10 Nmm + 1.506 × = 0.12 + 0.08 + 0.33 = 0.53 307177.41mm 3 × 235 N / mm 2 1 3700000 N 31.25 × 10 6 Nmm + 1.355 × + 14264mm 2 × 235 N / mm 2 2509773.11mm 3 × 235 N / mm 2 0.816 × 1× 1 1 6 15.62 × 10 Nmm + 1.383 × = 1.35 + 0.07 + 0.30 = 1.72 307177.41mm 3 × 235 N / mm 2 1 Finite elements modeling ■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

χ y coefficient corresponding to non-dimensional slenderness λ y Column subjected to axial and shear force to the top

χy

χz

coefficient corresponding to non-dimensional slenderness

Column subjected to axial and shear force to the top

χz

162

λz

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yy Column subjected to axial and shear force to the top

k yy

Internal factor, k yz Column subjected to axial and shear force to the top

k yz

Internal factor, k zy Column subjected to axial and shear force to the top

k zy

163

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k zz Column subjected to axial and shear force to the top

k zz

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

164

ADVANCE DESIGN VALIDATION GUIDE

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

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

165

ADVANCE DESIGN VALIDATION GUIDE

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

12.31.2.10Reference results Result name

Result description

Reference value

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

0.98

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.82

k yy

Internal factor, k yy

1.47

k yz

Internal factor, k yz

1.51

k zy

Internal factor, k zy

1.35

k zz

Internal factor, k zz

1.38

SNy

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

1.12

SMyy SMyz SNz SMzy SMzz

166

0.08 0.33 1.35 0.07 0.30

ADVANCE DESIGN VALIDATION GUIDE

12.31.3Calculated 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 depending of the compression effort over the Y

term

1.12239 adim

0.2134 %

SMyy

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

term

0.078033 adim

-2.4587 %

SMyz

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

term

0.325974 adim

-1.2200 %

SNz

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

term

1.35209 adim

0.1548 %

SMzy

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

term

0.0716405 adim

2.3436 %

SMzz

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

term

0.29927 adim

-0.2433 %

167

ADVANCE DESIGN VALIDATION GUIDE

12.32 EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling resistance for a IPE300 column (evaluated by SOCOTEC France - ref. Test 19) Test ID: 5699 Test status: Passed

12.32.1Description The test verifies the buckling resistance for a IPE300 column made of S235 steel. The verifications are made according to Eurocode3 French Annex.

12.32.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.32.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■

Exploitation loadings (category A): Q1 = -200 000 N, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in millimeters (mm).

Units Metric System Geometrical properties ■ ■ ■

168

Cross section area: A=5380mm2 Flexion inertia moment around the Y axis: Iy=603.80x104mm4 4 4 Flexion inertia moment around the Z axis: Iz=8356x10 mm

ADVANCE DESIGN VALIDATION GUIDE

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

Loading The column is subjected to the following loadings: ■ ■

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

169

ADVANCE DESIGN VALIDATION GUIDE

12.32.2.2 Buckling 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.

N Ed × 100 ≤ 100% N b , Rd

(6.46)

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

χ × A× fy γ M1

N b , Rd =

(6.47)

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

χ coefficient corresponding to non-dimensional slenderness λ

will be determined from the relevant buckling curve

according to:

χ=

1 Φ + Φ2 − λ

2

≤1

(6.49)

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

λ=

A* f y N cr

Where: A is the cross section area; A=5380mm2; fy is the yielding strength of the material; fy=235N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

N cr =

λ=

π ² × E × Iz L fy ²

A× fy N cr

[

=

=

π ² × 210000MPa × 8356 × 104 mm 4

(10000mm )2

5380mm 2 × 235 N / mm 2 = 0.854 1731878.70 N

φ = 0.5 1 + α (λ − 0.2) + λ ²

]

It will be used the following buckling curve:

170

= 1731878.70 N

ADVANCE DESIGN VALIDATION GUIDE

α

The imperfection factor

corresponding to the appropriate buckling curve will be 0.21:

[

]

φ = 0.5 × 1 + α × (λ − 0.2) + λ ² = 0.5 × [1 + 0.21 × (0.854 − 0.2) + 0.8542 ] = 0.933 Therefore:

χ= γ M1

1

φ + φ −λ 2

=

2

1 0.933 + 0.9332 − 0.8542

is a safety coefficient,

N b , Rd =

= 0.764 ≤ 1

γ M1 = 1

0.764 × 5380mm2 × 235 N / mm2 = 966051.089 N 1

N Ed = 200000 N N Ed 200000 N × 100 = × 100 = 20.703% N b , Rd 966051.089 N 12.32.2.3 Buckling 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.

N Ed × 100 ≤ 100% N b , Rd

(6.46)

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

N b , Rd =

χ × A× fy γ M1

(6.47)

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

χ coefficient corresponding to non-dimensional slenderness according to:

χ=

1 Φ+ Φ −λ 2

2

λ

will be determined from the relevant buckling curve

≤1

(6.49)

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

λ=

A* f y N cr

Where: A is the cross section area; A=5380mm2; fy is the yielding strength of the material; fy=235N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

N cr =

π ² × E × Iz L fy ²

=

π ² × 210000MPa × 603.80 × 104 mm4

(10000mm)2

= 125144.610 N

171

ADVANCE DESIGN VALIDATION GUIDE

λ=

A× fy N cr

=

5380mm2 × 235 N / mm2 = 3.178 125144.610 N

[

φ = 0.5 1 + α (λ − 0.2) + λ ²

]

It will be used the following buckling curve:

The imperfection factor

α

corresponding to the appropriate buckling curve will be 0.34:

[

]

φ = 0.5 × 1 + α × (λ − 0.2) + λ ² = 0.5 × [1 + 0.34 × (3.178 − 0.2) + 3.1782 ] = 6.056 Therefore:

χ= γ M1

1

φ + φ2 − λ

2

=

1 6.056 + 6.0562 − 3.1782

is a safety coefficient,

N b , Rd =

γ M1 = 1

0.089 × 5380mm2 × 235 N / mm2 = 112771.78 N 1

N Ed = 200000 N 200000 N N Ed × 100 = 177.349% × 100 = 112771.78 N N b , Rd Finite elements modeling ■ ■ ■

172

= 0.089 ≤ 1

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

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Coefficient corresponding to non-dimensional slenderness after Y-Y axis 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)

173

ADVANCE DESIGN VALIDATION GUIDE

12.32.2.4 Reference results Result name

Result description

Reference value

χy

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

0.764

χz

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 %

12.32.3Calculated results

174

Result name

Result description

Value

Error

Xy

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

0.763129 adim

-0.1140 %

Xz

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

0.0891543 adim

0.1734 %

SNy

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

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 %

ADVANCE DESIGN VALIDATION GUIDE

12.33 EC3 / NF EN 1993-1-1/NA - France: Verifying a IPE400 column for compression, shear, bending moment, buckling, lateral torsional buckling and bending and axial compression (evaluated by SOCOTEC France - ref. Test 23) Test ID: 5703 Test status: Passed

12.33.1Description 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.

12.33.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.33.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■

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

175

ADVANCE DESIGN VALIDATION GUIDE

■

Cross section dimensions are in millimeters (mm).

Units Metric System Geometrical properties ■

Column length: L=9000mm

■ ■

A = 8446mm 2 Overall breadth: b = 180mm Flange thickness: t f = 13.5mm

■

Root radius:

■

Web thickness: t w

= 8.6mm

■

Depth of the web:

hw = 400mm

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, Wy

■

Elastic modulus after the Z axis, Wel , z

= 146.40 ×10 3 mm 3

■

Plastic modulus after the Z axis, W pl , z

= 229 × 103 mm3

■ ■ ■ ■

4 4 Flexion inertia moment around the Y axis: Iy=23130.00x10 mm 4 4 Flexion inertia moment around the Z axis: Iz=1318.00x10 mm 4 4 Torsional moment of inertia: It=51.08x10 mm Working inertial moment: Iw=490000x106mm6

■

176

Cross section area:

r = 21mm

= 1156 × 103 mm3

= 1307 × 103 mm3

ADVANCE DESIGN VALIDATION GUIDE

Materials properties S235 steel material is used. The following characteristics are used: ■ ■ ■

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

Boundary conditions The boundary conditions are described below: ■

Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, ► Support at the end point (z = 9.00) restrained in translation along Y and Z axis and restrained rotation 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.

177

ADVANCE DESIGN VALIDATION GUIDE

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

178

ADVANCE DESIGN VALIDATION GUIDE

12.33.2.2 Cross section Class According to Advance Design calculations: Cross-class classification is made according to Table 5.2 -for beam web:

c 331mm  = = 38.49 c t 8.6mm  ⇒ = 38.49 ≤ 72 × ε = 72 t  ε =1 

therefore the beam web is considered to be Class 1

-for beam flange:

c 67.47mm  = = 4.50 c t 13.5mm  ⇒ = 4.50 ≤ 9 × ε = 9 t  ε =1

therefore the haunch is considered to be Class1

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

179

ADVANCE DESIGN VALIDATION GUIDE

12.33.2.3 Compression 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.

N Ed × 100 ≤ 100% N c ,Rd

(6.9)

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

N c , Rd =

A× f y

(6.10)

γ M0

-where: A is the section area:

A = 8446mm 2

f y is the yielding strength: f y = 275 N / mm 2

γ M0

is the partial safety factor:

N c , Rd =

A× f y

γ M0

γ M0 =1

8446mm 2 × 275 N / mm 2 = = 2322650 N = 2322.65kN 1

N Ed = 125000 N N Ed 125000 N = × 100 = 5.38% ≤ 100% N c ,Rd 2322650 N According to CTICM document: The compression resistance of the column is

N c , Rd = 2324kN

as it can be seen from conclusion extracted from

CTCIM 2006-4:

12.33.2.4 Shear 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.

V Ed × 100 ≤ 100% Vc , Rd

180

(6.17)

ADVANCE DESIGN VALIDATION GUIDE

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

 f  AV ×  y   3 =

Vc , Rd

(6.18)

γM0

-where: AV is the shear area:

AV = A − 2 × b × t f + (t w + 2 × r ) × t f ≥ η × hw × t w -where: A is the cross section area: b is the overall breadth:

b = 180mm

tf is the flange thickness: r is the root radius:

A = 8446mm 2

t f = 13.5mm

r = 21mm

tw is the web thickness: t w

= 8.6mm

hw is the depth of the web:

hw = 400mm

η =1 AV = A − 2 × b × t f + (t w + 2 × r ) × t f = = 8446mm 2 − 2 × 180mm × 13.5mm + (8.6mm + 2 × 21mm ) × 13.5mm = 4269.1mm 2

η × hw × t w = 1 × 8.6mm × 400mm = 3440mm 2 AV = 4269.1mm 2 ≥ 3440mm 2 f y is the yielding strength: f y = 275 N / mm 2

γ M0

Vc , Rd

is the partial safety factor:

γ M0 =1

 275 N / mm 2   fy  2   4269.1mm ×  AV ×  3 3    = 677810.66 N  = = γM0 1

VEd = 28330 N 28330 N V Ed × 100 = × 100 = 0.04179 × 100 = 4.180% ≤ 100% Vc , Rd 677810.66 N

181

ADVANCE DESIGN VALIDATION GUIDE

According to CTICM document: The shear resistance of the column is

V pl , z , Rd = 677.8kN

as it can be seen from conclusion extracted from CTCIM

2006-4:

12.33.2.5 Bending 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.

M Ed × 100 ≤ 100% M pl , Rd

(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:

M pl , Rd =

w pl × f y

(6.13)

γM0

-where: wpl is the plastic modulus:

wpl = 1307000mm3

f y is the yielding strength: f y = 275 N / mm 2

γ M0

is the partial safety factor:

M pl , Rd =

wpl × f y

γM0

=

γ M0 =1

1307000mm3 × 275 N / mm 2 = 359425000 Nmm 1

M Ed = 254970000 Nmm M Ed 254970000 Nmm × 100 = × 100 = 0.70938 × 100 = 70.938% ≤ 100% M pl , Rd 359425000 Nmm

182

ADVANCE DESIGN VALIDATION GUIDE

According to CTICM document: The bending moment resistance of the column is

M pl , y , Rd = 359.7 kNm as

it can be seen from the conclusion

extracted from CTCIM 2006-4:

12.33.2.6 Buckling 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:

α = 0.21

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

χ y coefficient

corresponding to non-dimensional slenderness

λy

will be determined from the relevant buckling

curve according to:

χy =

1 2

Φ y + Φ y − λy

2

≤1

(6.49)

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

λy =

A* fy N cr , y

2 2 Where: A is the cross section area; A=8446mm ; fy is the yielding strength of the material; fy=275N/mm and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

N cr , y =

π²× E × Iy L fz ²

=

π ² × 210000 MPa × 23130 × 10 4 mm 4

(9000mm)2

= 5918472.77 N

183

ADVANCE DESIGN VALIDATION GUIDE

λy =

A× fy N cr , y

=

8446mm 2 × 275 N / mm 2 = 0.62645 5918472.77 N

[

]

φ y = 0.5 × 1 + α × (λ y − 0.2) + λ y ² = 0.5 × [1 + 0.21 × (0.62645 − 0.2 ) + 0.626452 ] = 0.740997

χy =

1 2

Φy + Φy − λ y

2

=

1 0.740997 + 0.740997 2 − 0.626452

= 0.87968 ≤ 1

According to CTICM document: The determined value for the coefficient corresponding to non-dimensional slenderness for the strong section, y-y axis, λ y is:

χ y = 0.8796

as it can be observed from the conclusion extracted from CTCIM 2006-4:

b) over the strong axis of the section, z-z: The cross section buckling curve will be chosen according to Table 6.2:

184

ADVANCE DESIGN VALIDATION GUIDE

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

α = 0.34

χ z coefficient corresponding to non-dimensional slenderness λ z

will be determined from the relevant buckling curve

according to:

χz =

1 2

Φz + Φz − λ z

2

≤1

(6.49)

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

A* fy

λz =

N cr , z

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: 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 eigenmode of instability corresponds to an amplification factor equal to critical α cr

N cr , z =

λz =

= 9.15 . The normal critical force can be directly calculated:

π ² × E × Iz l fy ²

A× fy N cr , z

=

=

π ² × 210000 N / mm2 × 1318 × 104 mm4

(4890mm)²

= 1142396.153N

8446mm 2 × 275 N / mm 2 = 1.42588 1142396.153

[

]

φz = 0.5 × 1 + α × (λ z − 0.2) + λ z ² = 0.5 × [1 + 0.34 × (1.42588 − 0.2 ) + 1.425882 ] = 1.72497

χz =

1 2

Φz + Φz − λ z

2

=

1 1.72497 + 1.72497 2 − 1.425882

= 0.37096 ≤ 1

According to CTICM document: The determined value for the coefficient corresponding to non-dimensional slenderness for the strong section, z-z axis, λ z is:

χ z = 0.3711

as it can be observed from the conclusion extracted from CTCIM 2006-4:

185

ADVANCE DESIGN VALIDATION GUIDE

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

M cr = C1 ×

π ² × E × Iz I L² × G × It × w+ L² Iz π ² × E × Iz 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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1-AN France; Chapter 3; …(6)

ψ

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

C1 =

169.98kNm = 0.66667 254.97 kNm

1 1 = = 1.17932 0.325 + 0.423ψ + 0.252ψ ² 0.325 + 0.423 × 0.66667 + 0.252 × 0.66667² Flexion inertia moment around the Y axis: Iy=23130.00x104mm4 4 4 Flexion inertia moment around the Z axis: Iz=1318.00x10 mm 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa. 4 4 Torsional moment of inertia: It=51.08x10 mm 6 6 Working inertial moment: Iw=490000x10 mm Length of the column part: L=3000mm Shear modulus of rigidity: G=80800MPa

■ ■ ■ ■ ■ ■ ■

M cr = C1 ×

=

π ² × E × Iz L²

×

I w L² × G × It π ² × 210000 N / mm 2 × 1318 × 104 mm 4 = 1.17932 × × + Iz π ² × E × Iz (3000mm )2

49 × 1010 mm6 (3000mm ) × 80800 N / mm 2 × 51.08 × 104 mm 4 = 1.17932 × 3035232.34 N × 225.33396mm = × + 1318 × 104 mm 4 π ² × 210000 N / mm 2 × 1318 × 104 mm 4 = 806585210.2 Nmm 2

186

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the non-dimensional slenderness factor,

λ LT = Plastic modulus, Wy

λ LT =

Wy × f y

Calculation of the

= 1307 × 10 mm

=

M cr

3

λ LT :

Wy f y M cr

3

1307 × 103 mm3 × 275 N / mm2 = 0.66754 806585210.2 Nmm

χ LT for appropriate non-dimensional slenderness λ LT

χ LT =

[

1

φLT + φLT ² − λ LT ²

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

≤1

will be determined with formula: (6.56)

]

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

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

α = 0.34

φLT = 0.5 × [1 + 0.34 × (0.66754 − 0.2 ) + 0.66754² ] = 0.80228

χ LT =

1

φLT + φLT ² − λ LT ²

=

1 = 0.80173 ≤ 1 0.80228 + 0.80228² − 0.66754²

According to CTICM document: The determined value for the coefficient corresponding to non-dimensional slenderness, it can be observed from the conclusion extracted from CTCIM 2006-4:

λ LT

is:

χ LT = 0.7877

as

187

ADVANCE DESIGN VALIDATION GUIDE

b) for the 6m part of the column: 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:

M cr = C1 ×

π ² × E × Iz I L² × G × It × w+ L² Iz π ² × E × Iz 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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1-AN France; Chapter 3; …(6)

ψ

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

■ ■ ■ ■ ■ ■ ■

M cr = C1 ×

=0

C1 = 1.77

Flexion inertia moment around the Y axis: Iy=23130.00x104mm4 4 4 Flexion inertia moment around the Z axis: Iz=1318.00x10 mm 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa. Torsional moment of inertia: It=51.08x104 mm4 6 6 Working inertial moment: Iw=490000x10 mm Length of the column part: L=6000mm Shear modulus of rigidity: G=80800MPa

π ² × E × Iz L²

π ² × 210000 N / mm 2 × 1318 × 104 mm 4 I w L² × G × It × + = 1.77 × × 6000² mm 2 Iz π ² × E × Iz

49 × 1010 mm6 6000² mm 2 × 80800 N / mm 2 × 51.08 × 104 mm 4 × + = 1.77 × 758808.085 N × 302.604mm = π ² × 210000 N / mm 2 × 1318 × 104 mm 4 1318 × 104 mm 4 = 406423987 Nmm 188

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the non-dimensional slenderness factor,

λ LT :

λ LT = Plastic modulus, Wy

λ LT =

Wy × f y M cr

Calculation of the

M cr

= 1307 × 103 mm3

1307 × 103 mm3 × 275 N / mm2 = 0.94040 = 406423987 Nmm

χ LT for appropriate non-dimensional slenderness λ LT

χ LT =

[

Wy f y

1

φLT + φLT ² − λ LT ²

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

will be determined with formula:

≤1

(6.56)

]

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

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

α = 0.34

φLT = 0.5 × [1 + 0.34 × (0.94040 − 0.2 ) + 0.94040² ] = 1.06804 1 1 χ LT = = = 0.63518 ≤ 1 φLT + φLT ² − λ LT ² 1.06804 + 1.06804² − 0.94040²

189

ADVANCE DESIGN VALIDATION GUIDE

According to CTICM document: The determined value for the coefficient corresponding to non-dimensional slenderness, λ LT is: can be observed from the conclusion extracted from CTCIM 2006-4:

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

M + ∆M y , Rd M + ∆M z , Ed N Ed + k yy × y , Ed + k yz × z , Ed ≤1 N Rk M y , Rd M z , Rk χy × χ LT ×

(6.61)

+ ∆M y , Rd M + ∆M z , Ed M N Ed ≤1 + k zz × z , Ed + k zy × y , Ed M z , Rk M y , Rd N Rk χz × χ LT ×

(6.62)

γ M1

γ M1

γ M1

γ M1

There is no bending on the small inertia axis: The section is considered to be a Class1:

γ M1

γ M1

The formulae can be simplified because:

M z , Ed = 0

∆M y , Rd = 0 and ∆M z , Rd = 0

Therefore the formulae are:

M y , Ed N Ed + k yy × ≤ 1.00 N Rk M y , Rk χy × χ LT ×

(6.61)

M y , Ed N Ed + k zy × ≤ 1.00 N Rk M y , Rk χz × χ LT ×

(6.62)

γ M1

γ M1

190

γ M1

γ M1

χ LT = 0.694

as it

ADVANCE DESIGN VALIDATION GUIDE

12.33.2.9 Internal factor, k yy , calculation: The internal factor k yy corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:

k yy = Cmy × CmLT ×

µy

N 1 − Ed N cr , y

×

1 C yy

Auxiliary terms:

N Ed N cr , y µy = N 1 − χ y × Ed N cr , y 1−

Where:

χ y = 0.87968 N cr , y =

(previously calculated)

π ² × E × Iy l fy ²

=

π ² × 210000 N / mm 2 × 23130 × 104 mm4

(9000mm)²

= 5918472.773 N

125000 N 5918472.773 N µy = = 0.99741 125000 N 1 − 0.87968 × 5918472.773 1−

According to CTICM document:

191

ADVANCE DESIGN VALIDATION GUIDE

The

Cmy will be calculated according to Table A.1:

Calculation of the λ0 term:

λ0 = C1 × λLT -where:

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

C1 =

ψ

1 0.325 + 0.423ψ + 0.252ψ ²

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

C1 = 1.77

λ LT =

192

Wy × f y M cr

=

1307 × 103 mm3 × 275 N / mm2 = 0.66754 806585210.2 Nmm

ADVANCE DESIGN VALIDATION GUIDE

λ0 = C1 × λLT = C1 × λLT = 1.77 × 0.66754 = 0.88811

Calculation of the

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed N cr , z   N cr ,TF 

   

term:

Where: -for a symmetrical section for the both axis,

N cr ,T =

N cr ,TF = N cr ,T

A  π 2 × E × I w  ×  G × It + I0  Lcr ,T ² 

The mass moment of inertia I 0

I 0 = I y + I z + A ⋅ z g2

4 4 Flexion inertia moment around the Y axis: Iy=23130.00x10 mm

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

A = 8446mm 2

Distance between the section neutral axis and the section geometrical center: z g

=0

I 0 = I y + I z + A ⋅ z g2 = I y + I z = 23130 × 104 mm 4 + 1318 × 104 mm 4 = 24448 × 104 mm 4 -for simplification, it will be considered the same buckling length,

4

Torsional moment of inertia: It=51.08x10 mm

4

Lcr ,T , for all the column parts:

Lcr ,T = 6m

Working inertial moment: Iw=490000x106mm6 Longitudinal elastic modulus: E = 210000 MPa Shear modulus of rigidity: G=80800MPa

N cr ,T =

8446mm 2 24448 × 104 mm 4

 π 2 × 210000 N / mm 2 × 49 × 1010 mm6   = ×  80800 N / mm 2 × 51.08 × 104 mm 4 + (6000mm )²  

= 2400423.788 N N cr , TF = N cr , T = 2400423.788 N N cr , z = 1142396.153 N

(previously calculated)

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed  N   N cr , z   cr ,TF  = 0.25505

  = 0.20 × 1.77 × 4 1 − 125000 N  × 1 − 125000 N  =   1142396.153 N   2400423.788 N  

193

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

 N  N   λ0 = 0.88811 > 0.20 × C1 × 4 1 − Ed  × 1 − Ed  N N 

The

cr , z

 

cr ,TF



  C = C + (1 − C )× ε y × aLT my , 0 my , 0  my 1 + ε y × aLT   = 0.25505 ⇒ Cmz = Cmz , 0  aLT 2 CmLT = Cmy ≥1 ×      1 − N Ed  × 1 − N Ed    N   N   cr , z   cr ,T  

Cmy 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.

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξy =

M y , Ed N Ed

×

M y , Ed N Ed

×

1 + ε y × aLT

A Wel , y

Elastic modulus after the Y axis, Wel , y

ξy =

ε y × aLT

= 1156 × 103 mm3

8446mm2 254.94 × 106 Nmm A = 14.90119 × = 1156 × 103 mm3 125000 N Wel , y

a LT

It 51.08 ⋅10 4 mm 4 = 1− = 1− = 0.99779 Iy 23130 ×10 4 mm 4

The

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

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

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) ×

=0

N Ed N cr , y

Where:

N cr , y = 5918472.773 N

(previously calculated)

C my , 0 = 0.79 + 0.36 × (0 − 0.33)×

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

194

125000 N = 0.78749 5918472.773N

ξ y × aLT 1 + ξ y × aLT

= 0.78749 + (1 − 0.78749 ) ×

14.90119 × 0.99779 = 0.95619 1 + 14.90119 × 0.99779

ADVANCE DESIGN VALIDATION GUIDE

According to CTICM document:

The

CmLT coefficient takes into account the laterally restrained parts of the column. The CmLT coefficient must be calculated

individually for each of the column parts.

2 × CmLT = Cmy

a LT  1 − N Ed  N cr , z 

   × 1 − N Ed   N cr ,T  

   

≥1

a) for the 3m part of the column:

ξ y a LT C my.3m = C my , 0 + (1 − C my , 0 ) 1 + ξ y a LT ξ y = 14.90119 (previously calculated) a LT = 0.99779 (previously calculated) The

ψ

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

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

=

169.98kNm = 0.66667 254.97 kNm

195

ADVANCE DESIGN VALIDATION GUIDE

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) ×

N Ed N cr , y

Where:

N cr , y = 5918472.773 N

(previously calculated)

C my , 0 = 0.79 + 0.21× 0.66667 + 0.36 × (0.66667 − 0.33)×

125000 N = 0.93256 5918472.773N

ξ y × a LT 14.90119 × 0.99779 = 0.98609 C my ,3m = C my , 0 + (1 − C my , 0 ) = 0.93256 + (1 − 0.93256 ) × 1 + ξ y × a LT 1 + 14.90119 × 0.99779 N cr , z = 1142396.153 N

(previously calculated)

N cr ,T = 2400423.788 N (previously calculated) a LT = 0.99779 (previously calculated)

a LT

       1 − N Ed  × 1 − N Ed   N   N   cr , z   cr ,T     0.99779 = 0.98609² × = 1.05596 ⇒ CmLT ,3m = 1.05596 125000 N   125000 N    1 −  × 1 −   1142396 . 153 2400423 . 788 N N      CmLT ,3m ≥ 1   

2 CmLT ,3m = Cmy ,3m ×

=

a) for the 6m part of the column:

ξ y a LT C my.6 m = C my , 0 + (1 − C my , 0 ) 1 + ξ y a LT

ξy =

M y , Ed N Ed

×

A Wel , y

Elastic modulus after the Y axis, Wel , y

ξy =

M y , Ed N Ed

×

= 1156 × 103 mm3

A 169.98 × 10 6 Nmm 8446mm 2 = × = 9.9353 Wel , y 125000 N 1156 × 10 3 mm 3

a LT

It 51.08 ⋅10 4 mm 4 = 1− = 1− = 0.99779 Iy 23130 ×10 4 mm 4

The

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

196

ADVANCE DESIGN VALIDATION GUIDE

ψ

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

=0

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) ×

N Ed N cr , y

Where:

N cr , y = 5918472.773 N

(previously calculated)

C my , 0 = 0.79 + 0.21× 0 + 0.36 × (0 − 0.33) ×

125000 N = 0.78749 5918472.773 N

ξ y × a LT 9.9353 × 0.99779 C my , 6 m = C my , 0 + (1 − C my , 0 ) = 0.78749 + (1 − 0.78749 ) × = 0.94873 1 + ξ y × a LT 1 + 9.9353 × 0.99779 N cr , z = 1142396.153 N

(previously calculated)

N cr ,T = 2400423.788 N (previously calculated)

a LT

      N N  Ed Ed 1 −     N  × 1 − N   cr , z   cr ,T     0.99779 = 0.94873² × = 0.97746 ⇒ C mLT , 6 m = 1 125000 N   125000 N     1 −  × 1 −   1142396.153 N   2400423.788 N   C mLT , 6 m ≥ 1    2 C mLT , 6 m = C my ,6 m ×

=

In conclusion:

C mLT ,3m = 1.05596 C mLT =  C mLT , 6 m = 1

197

ADVANCE DESIGN VALIDATION GUIDE

The

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

 1.6  Wel , y  1.6 2 2 × C my × λmax − × C my × λ2max  × n pl − bLT  ≥ C yy = 1 + ( w y − 1) ×  2 −   wy wy   W pl , y  Where:

M y , Ed

M y , Ed M z , Ed = 0.5 × aLT × λ20 × ×0 = 0 M pl , Rd χ LT × M pl , y , Rd

■

bLT = 0.5 × aLT × λ20 ×

■

wy =

■

Elastic modulus after the Y axis, Wel , y

= 1156 × 103 mm3

■

Plastic modulus after the Y axis, W pl , y

= 1307 × 10 3 mm 3

■

wy =

■

n pl =

W pl , y Wel , y

W pl , y Wel , y

χ LT × M pl , y , Rd

×

≤ 1.5

=

1307 × 103 mm3 = 1.13062 ≤ 1.5 1156 × 103 mm3

N Ed N Rk

γ M1

198

■

N Rk = N c , Rd =

■

n pl =

A× fy

γM0

=

8446mm 2 × 275 N / mm 2 = 2322650 N 1

N Ed 125000 N = = 0.05382 N Rk 2322650 N γ M1 1

ADVANCE DESIGN VALIDATION GUIDE

(

)

■

λ max = max λ y ; λ z

■

λ y = 0.62645

■

λ z = 1.42504 (previously calculated)

■

λ max = max λ y ; λ z = max(0.62645;1.42504 ) = 1.42504

■

C my = 0.95619 (previously calculated)

(

(previously calculated)

)

  1.6 1.6  C yy = 1 + (1.13062 − 1) ×  2 − × 0.95619² × 1.42504 − × 0.95619² × 1.42504²  × 0.05382 − 0 = 1.13062   1.13062  = 0.98511

C yy = 0.98511 ≥

Wel , y W pl , y

=

1156 × 10 3 mm 3 = 0.88447 1307 × 10 3 mm 3

In conclusion:

µy  1 0.99741 1 × = 0.95619 ×1.05569 × × = 1.04409 k yy `,3m = Cmy × CmLT ,3m × N Ed C yy 125000 N 0 . 98511  1− 1− 5918472.773 N N cr , y  k yy =  µy 1 0.99741 1 k × = 0.95619 ×1× × = 0.98902 yy , 6 m = C my × C mLT , 6 m ×  N Ed C yy 125000 N 0 . 98511 1− 1−  5918472.773 N N cr , y  According to CTICM 2006-4:

199

ADVANCE DESIGN VALIDATION GUIDE

12.33.2.10Internal factor, k zy , calculation: The internal factor k zy corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:

k zy = C my × C mLT ×

µz 1−

N Ed N cr , y

×

wy 1 × 0.6 × C zy wz

Auxiliary terms:

N Ed N cr , z µz = N 1 − χ z × Ed N cr , z 1−

Where:

χ z = 0.37096 N cr , z =

(previously calculated)

π ² × E × Iz l fy ²

=

π ² × 210000 N / mm2 × 1318 × 104 mm4

(4890mm)²

125000 N 1142396 .1533 N µz = = 0.92826 125000 N 1 − 0.37096 × 1142396.153 N 1−

According to CTICM document:

200

= 1142396.153N

ADVANCE DESIGN VALIDATION GUIDE

The

Cmy will be calculated according to Table A.1:

   × 1 − N Ed N cr , z   N cr ,TF

 N λ0 = 0.88811 > 0.20 × C1 × 4 1 − Ed 

  C = C + (1 − C )× ε y × aLT my , 0 my , 0  my 1 + ε y × aLT    = 0.208 ⇒ Cmz = Cmz , 0    aLT 2 CmLT = Cmy ≥1 ×      1 − N Ed  × 1 − N Ed    N   N   cr ,T  cr , z   

(previously calculated)

ξy =

M y , Ed N Ed

aLT = 1 − The

×

6

2

A 254.94 × 10 Nmm 8446mm = × = 14.90119 Wel , y 125000 N 1156 × 103 mm3

(previously calculated)

51.08 ⋅ 104 mm 4 It = 0.99779 (previously calculated) =1− 23130 × 104 mm 4 Iy

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

201

ADVANCE DESIGN VALIDATION GUIDE

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

Cmy , 0 = 0.79 + 0.21×ψ + 0.36 × (ψ − 0.33) ×

=0

N Ed = 0.78749 N cr , y (previously calculated)

Cmy = Cmy , 0 + (1 − Cmy , 0 )

ξ y a LT 1 + ξ y a LT

= 0.95619 (previously calculated)

2 × CmLT = Cmy

a LT  1 − N Ed  N cr , z 

   × 1 − N Ed   N cr ,T  

CmLT ,3m = 1.05596 =  CmLT , 6 m = 1    (previously calculated)

The

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

  C 2 × λ2  wy Wel , y × Czy = 1 + ( wy − 1) ×  2 − 14 × my 5 max  × n pl − d LT  ≥ 0.6 ×  wy w W pl , y   z  

202

ADVANCE DESIGN VALIDATION GUIDE

Where:

d LT = 2 × aLT ×

wy =

wz =

W pl , y Wel , y

W pl , z Wel , z

=

λ0 0.1 + λ

4 z

×

M y , Ed Cmy × χ LT × M pl , y , Rd

1307 × 103 mm3 = 1.13062 ≤ 1.5 1156 × 103 mm3

×

M z , Ed Cmz × M pl , z , Rd

≤ 1.5 = 146.40 ×10 3 mm 3

Plastic modulus after the Z axis, W pl , z

= 229 × 103 mm3

W pl , z Wel , z

=

wz ≤ 1.5 n pl =

M y , Ed λ0 × ×0 = 0 4 0.1 + λz Cmy × χ LT × M pl , y , Rd

(previously calculated)

Elastic modulus after the Z axis, Wel , z

wz =

= 2 × aLT ×

 229 × 103 mm3 = 1.564 3 3 146.40 × 10 mm  ⇒ wz = 1.5  

N Ed N Rk

γ M1 n pl =

N Ed 125000 N = = 0.05382 N Rk 2322650 N γ M1 1

(

(previously calculated)

)

λ max = max λ y ; λ z = max(0.62645;1.42504 ) = 1.42504 (previously calculated)   0.986092 × 1.425042   × − 0 . 05382 0 C zy = 1 + (1.13062 − 1) ×  2 − 14 ×  = 0.90887  1.130625    0.6 ×

wy wz

×

Wel , y W pl , y

1.13062 1156 × 103 mm3 = 0.46073 = 0.6 × × 1307 × 103 mm3 1.5

  C 2 × λ2  wy Wel , y C zy = 1 + ( wy − 1) ×  2 − 14 × my 5 max  × n pl − d LT  = 0.90887 ≥ 0.6 × × = 0.46073  wy w W   z pl y ,   In conclusion:

k zy = C my × C mLT ×

µz 1−

N Ed N cr , y

×

wy 1.13062 1 0.923830 1 × × 0 .6 × = × 0 .6 × = 0.98609 ×1.05569 × 125000 N 1 .5 0.90887 C zy wz 1− 5918472.773 N

= 0.56307

203

ADVANCE DESIGN VALIDATION GUIDE

 µy wy 1 × × 0.6 × = k zy `,3m = C my × C mLT ,3m × N Ed C zy wz  1−  N cr , z   0.99741 1 1.13062 × × 0.6 × = 0.52306 = 0.95619 ×1.05569 × 125000 N 0.90887 1.5 1−   5918472.773 N k zy =  µy wy 1  × × 0.6 × = k zy , 6 m = C my × C mLT , 6 m × N Ed C zy wz 1−  N cr , z   0.99741 1 1.13062 = 0.95619 ×1× × × 0.6 × = 0.50752  125000 N 0.98511 1.5 1−  5918472.773 N According to CTICM 2006-4:

The bending and axial compression verifications are: -for the 3m column part:

M y ,Ed N Ed + k yy × = N Rk M y ,Rk χy × χ LT ×

γ M1

γ M1

125000 N 254970000 Nmm + 1.04409 × = 2322650 N 359425000 Nmm 0.87968 × 0.80173 × 1 1 = 0.06119 + 0.92383 = 0.98501 ≤ 1 =

M y ,Ed N Ed + k zy × = N Rk M y ,Rk χz × χ LT ×

γ M1

γ M1

125000 N 254970000 Nmm + 0.52306 × = 2322650 N 359425000 Nmm 0.37096 × 0.80173 × 1 1 = 0.14508 + 0.46281 = 0.60789 ≤ 1 =

204

ADVANCE DESIGN VALIDATION GUIDE

-for the 6m column part:

M y ,Ed N Ed + k yy × = N M y ,Rk χ y × Rk χ LT ×

γ M1

γ M1

125000 N 168980000 Nmm + 0.98902 × = 2322650 N 359425000 Nmm 0.87968 × 0.63518 × 1 1 0 . 06118 0 . 73204 0 . 79322 1 = + = ≤ =

M y ,Ed N Ed + k zy × = N Rk M y ,Rk χz × χ LT ×

γ M1

γ M1

125000 N 168980000 Nmm + 0.50752 × = 2322650 N 359425000 Nmm 0.37096 × 0.63518 × 1 1 = 0.14508 + 0.37565 = 0.52073 ≤ 1 =

According to CTICM 2006-4:

Finite elements modeling ■ ■ ■

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

205

ADVANCE DESIGN VALIDATION GUIDE

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

206

ADVANCE DESIGN VALIDATION GUIDE

Ratio of the design share force to design share resistance Column subjected to axial and shear force to the top Work ratio - oblique

χ y coefficient corresponding to non-dimensional slenderness λ y Column subjected to axial and shear force to the top

χy

207

ADVANCE DESIGN VALIDATION GUIDE

χ z coefficient corresponding to non-dimensional slenderness λ z Column subjected to axial and shear force to the top

χz

Internal factor, k yy Column subjected to axial and shear force to the top

k yy

208

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yz Column subjected to axial and shear force to the top

k yz

12.33.2.11Reference results 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

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

0.88

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.37

k yy

Internal factor, k yy for the 3m segment

1.04

k yy

Internal factor, k yy for the 6m segment

0.99

k zy

Internal factor, k zy for the 3m segment

0.52

k zy

Internal factor, k zy for the 6m segment

0.51

209

ADVANCE DESIGN VALIDATION GUIDE

12.33.3Calculated 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 %

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 %

Work ratio Oblique

210

-

ADVANCE DESIGN VALIDATION GUIDE

12.34 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 4 column fixed on the bottom and without any other restraint (evaluated by SOCOTEC France - ref. Test 24) Test ID: 5709 Test status: Passed

12.34.1Description The test verifies an user defined cross section column. 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 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. The calculations are made according to Eurocode 3 French Annex.

12.34.2Background 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. This test was evaluated by the French control office SOCOTEC.

211

ADVANCE DESIGN VALIDATION GUIDE

12.34.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■ ■

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 Geometrical properties ■

Column length: L=5620mm

■ ■

A = 10850mm 2 Overall breadth: b = 220mm Flange thickness: t f = 15mm

■

Root radius:

■

Web thickness: t w

= 5mm

■

Depth of the web:

hw = 880mm

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, W y

■

Elastic modulus after the Z axis, Wel , z

= 242.08 × 103 mm 3

■

Plastic modulus after the Z axis, W pl , z

= 368.31 × 103 mm 3

■ ■ ■ ■

4 4 Flexion inertia moment around the Y axis: Iy=149058.04x10 mm 4 4 Flexion inertia moment around the Z axis: Iz=2662.89x10 mm Torsional moment of inertia: It=51.46x104 mm4 6 6 Working inertial moment: Iw=4979437.37x10 mm

■

212

Cross section area:

r = 0mm

= 3387.66 × 103 mm 3

= 3757.62 × 103 mm 3

ADVANCE DESIGN VALIDATION GUIDE

Materials properties S275 steel material is used. The following characteristics are used: ■ ■ ■

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

Boundary conditions The boundary conditions are described below: ■

Outer: ► ►

Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at the end point (z = 5.62) restrained in translation along Y and Z axis and restrained rotation along X axis.

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

213

ADVANCE DESIGN VALIDATION GUIDE

12.34.2.2Cross 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.

ψ = 2⋅

N Ed 0.328 −1 = 2 ⋅ − 1 = −0.78 > −1 A⋅ fy 0.0109 × 275

1 

N



1 

0.328



Ed  = ⋅ 1 + α = ⋅ 1 +  = 0.64 > 0.5 f y × t × d  2  275 × 0.850 × 0.005  2 

214

ADVANCE DESIGN VALIDATION GUIDE

ε=

235 = fy

235 = 0.924 275

c 880mm − 2 × 15mm  = = 170 42 × 0.924 c 42 × ε = = 94.057 t 5mm  ⇒ = 170 > 0.67 + 0.33 ×ψ 0.67 + 0.33 × (−0.78) t  ε =1  therefore the beam web is considered to be Class 4 -for beam flange:

c 107.5  = = 7.61 c t 15  ⇒ = 7.61 ≤ 9 × ε = 9 t  ε =1

therefore the haunch is considered to be Class1

In conclusion, the section is considered to be Class 4

12.34.2.3 Effective 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

c 880mm − 2 × 15mm = = 170 : t 5mm

ψ = 1 ⇒ kσ = 4 -according EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1

bw t λp = 28.4 × ε × kσ

bw

is the width of the web;

bw = 850mm

t is the web thickness; t=5mm

ε=

235 235 = = 0.9244 fy 275

850mm 5mm λp = = 3.261 28.4 × 0.9244 × 4 -according EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4 -the web is considered to be an internal compression element, therefore:

λ p = 3.261 > 0.673 3 +ψ = 4 ≥ 0

⇒ ρ = 

λ p − 0.055 × (3 +ψ ) 3.261 − 0.055 × 4 λp

2

=

3.2612

= 0.286

215

ADVANCE DESIGN VALIDATION GUIDE

beff = ρ × bw  beff = 0.286 × 850mm = 243.1mm   be1 = 0.5 × beff  ⇒ be1 = 0.5 × 243.1mm = 121.55mm   be 2 = 0.5 × beff  be 2 = 0.5 × 243.1mm = 121.55mm Aeff ,web = t w × be1 + t w × be 2 = 5mm ×121.55mm + 5mm ×121.55mm = 1215.5mm 2 Aeff , flange = t f × b f = 15mm × 220mm = 3300mm 2

Aeff = Aeff , web + 2 × Aeff , flange = 1215.5mm 2 + 2 × 3300mm 2 = 7815.2mm 2 12.34.2.4 Effective elastic section modulus of Class4 cross-sections -In order to simplify the calculation the section will be considered in pure bending

ψ =

σ inf 850mm = 425mm = −1 ⇒ bc = bt = 2 σ sup

ψ = −1 ⇒ kσ = 23.9 -according EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1

c 880mm − 2 × 15mm = = 170 t 5mm

bw t λp = 28.4 × ε × kσ

bw

is the width of the web;

bw = 850mm

t is the web thickness; t=5mm

ε=

235 235 = = 0.9244 fy 275

850mm 5mm = 1.325 λp = 28.4 × 0.9244 × 23.9 -according EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4

λ p = 1.325 > 0.673 3 +ψ = 2 ≥ 0 beff = ρ × bc = ρ × be1 = 0.4 × beff be 2 = 0.6 × beff

216

⇒ ρ = 

λ p − 0.055 × (3 +ψ ) 1.325 − 0.055 × 2 λp

2

=

1.325 2

bw   850mm beff = 0.692 × = 294.1mm   1 − (− 1) 1 −ψ    ⇒ be1 = 0.4 × 294.1mm = 117.64mm  b = 0.6 × 294.1mm = 176.46mm   e2  

= 0.692

ADVANCE DESIGN VALIDATION GUIDE

-the weight center coordinate is:

yG = =

(220 ×15)× 432.5 + (117.64 × 5)× 366.18 − (220 ×15)× 432.5 − (601.46 × 5)×124.27 = 220 × 15 + 117.64 × 5 + 601.46 × 5 + 220 × 15

− 158330.095 = −15.53mm 10195.5

-the inertial moment along the strong axis is:

15 3 × 220 117.64 3 × 5 + 220 × 15 × 448.03 2 + + 117.64 × 5 × 381.712 + 12 12 3 3 601.46 × 5 15 × 220 + + 601.46 × 5 × 107.74 2 + + 220 × 15 × 416.97 2 = 12 12 = 748854356.6 + 699380072.5 = 1448234429mm 4 Iy =

220 3 ×15 53 ×117.64 + 220 ×15 × 0 2 + + 117.64 × 5 × 0 2 + 12 12 3 5 × 601.46 220 3 ×15 2 + + 601.46 × 5 × 0 + + 220 ×15 × 0 2 = 26627490.63mm 4 12 12 Iz =

Wel , y = Wel , z

Iy z max

1448234429mm 4 = = 3179229.533mm 3 455.53mm

Iz 26627490.63mm 4 = = = 242068.10mm 3 y max 110mm

217

ADVANCE DESIGN VALIDATION GUIDE

12.34.2.5 Buckling 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:

α = 0.34 Coefficient corresponding to non-dimensional slenderness after Y-Y axis:

χ y coefficient

corresponding to non-dimensional slenderness

λy

will be determined from the relevant buckling

curve according to:

χy =

1 2

Φ y + Φ y − λy

2

≤1

λ y the non-dimensional slenderness corresponding to Class 4 cross-sections:

λy =

218

Aeff * f y N cr , y

(6.49)

ADVANCE DESIGN VALIDATION GUIDE

Where:

A is the effective cross section area; 2

Aeff = 7815.2mm 2 ;

fy is the yielding strength of the material;

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

N cr , y =

π²× E× Iy l fy ²

Aeff × f y

λy =

N cr , y

=

=

π ² × 210000 N / mm 2 × 1448234429mm 4

(5620mm)²

= 95035371.44 N = 95035.37kN

7815.2mm 2 × 275 N / mm 2 = 0.15 95035371.44 N

[

]

φ y = 0.5 × 1 + α × (λ y − 0.2) + λ y ² = 0.5 × [1 + 0.34 × (0.15 − 0.2 ) + 0.15 2 ] = 0.503

χy =

1 2

Φy + Φy −λy

χy ≥1

2

=

1

 = 1.017  0.503 + 0.5032 − 0.15 2  ⇒ χy =1  

b) over the weak axis of the section, z-z: -the imperfection factor α will be selected according to the Table 6.1 and 6.2:

α = 0.49

219

ADVANCE DESIGN VALIDATION GUIDE

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

χ z coefficient corresponding to non-dimensional slenderness according to:

χz =

1 2

Φz + Φz − λ z

2

λz

will be determined from the relevant buckling curve

≤1

(6.49)

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

Aeff * f y

λz =

N cr , z

Where:

A is the effective cross section area; 2

Aeff = 7815.2mm 2 ;

fy is the yielding strength of the material;

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

N cr , z =

π ²× E × Iz l fz ²

Aeff × f y

λz =

=

π ² × 210000 N / mm 2 × 26627490.63mm 4

(5620mm )²

= 1747336.905 N = 1747.34kN

7815.2mm 2 × 275 N / mm 2 = = 1.109 1747336.905 N

N cr , z

[

]

φ z = 0.5 × 1 + α × (λ z − 0.2) + λ z ² = 0.5 × [1 + 0.49 × (1.109 − 0.2) + 1.109 2 ] = 1.338

χz =

1 2

Φz + Φz − λ z

χz ≤1

2

=

1

 = 0.479 1.338 + 1.338 − 1.109  ⇒ χ z = 0.479   2

2

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

M cr = C1 ×

I L² × G × It π ² × E × Iz × w+ Iz π ² × E × Iz L² 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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1-AN France; Chapter 3; …(6)

ψ

220

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

=

0kNm =0 1274kNm

ADVANCE DESIGN VALIDATION GUIDE

ψ=

0kNm = 0 ⇒ C1 = 1.77 1274kNm

Flexion inertia moment around the Y axis:

I y = 1448234429mm 4

Flexion inertia moment around the Z axis:

I z = 26627490.63mm 4

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

Longitudinal elastic modulus: E = 210000 N/mm2. Torsional moment of inertia: It=514614.75mm4 Warping inertial moment: IW is the warping inertia (deformation inertia moment):

Iw =

I z × (h − t f

)

2

4

h cross section height; h=880mm

tf

Iw =

flange thickness;

t f = 15mm

26627490.63 mm 4 × (880mm − 15mm ) = 49.808 × 1011 mm 6 4 2

-according to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column part: L=5620mm Shear modulus of rigidity: G=80800N/mm2

M cr = C1 ×

π ²× E × Iz L²

×

I w L² × G × I t π 2 × 210000 N / mm 2 × 26627490.63mm 4 + = 1.77 × × Iz π ²× E × Iz (5620mm )2

49.808 ×1011 mm 6 (5620mm ) × 80800 N / mm 2 × 514614mm 4 × + = 1.77 ×1747336.905 N × 459.185mm = 26627490.63mm 4 π 2 × 210000 N / mm 2 × 26627490.63mm 4 = 1420163158 Nmm = 1420.163kNm Iy 1448234429mm 4 = = 3179229.533mm 3 The elastic modulus : Wel , y = z max 455.53mm 2

λ LT =

Weff , y f y M cr

Calculation of the

=

3179229.533mm 3 ×275 N / mm 2 = 0.785 1420163158 Nmm

χ LT for appropriate non-dimensional slenderness λ LT

χ LT =

[

1

φLT + φLT ² − λ LT ²

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

will be determined with formula:

≤1

(6.56)

]

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

h 880mm = =4>2 b 220mm

221

ADVANCE DESIGN VALIDATION GUIDE

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

α = 0.76

[

(

)

]

φ LT = 0.5 1 + α LT × λ LT − 0.2 + λ LT ² = 0.5[1 + 0.76 × (0.785 − 0.2 ) + 0.785² ] = 1.030

χ LT =

1

φLT + φLT ² − λ LT ²

=

1 = 0.589 ≤ 1 1.030 + 1.030² − 0.785²

12.34.2.7 Internal factor, k yy , calculation The internal factor k yy corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:

k yy = C my × C mLT ×

222

µy 1−

N Ed N cr , y

ADVANCE DESIGN VALIDATION GUIDE

N Ed N cr , y µy = N 1 − χ y × Ed N cr , y 1−

χy =1

(previously calculated)

N Ed = 328kN N cr , y =

π ²× E× Iy l fy ²

= 95035371.44 N = 95035.37 kN (previously calculated)

N Ed 328000 N 1− N cr , y 95035371.44 N = 1 = µy = 328000 N N Ed 1 − 1× 1− χ y × 95035371.44 N N cr , y 1−

The

Cmy will be calculated according to Table A.1:

Calculation of the λ 0 term:

λ0 =

Weff , y × f y M cr 0 -according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2 223

ADVANCE DESIGN VALIDATION GUIDE

Wel , y = 3387.66 × 103 mm 3

M cr = 1420163158 Nmm = 1420.163kNm (previously calculated) Weff , y × f y

λ0 =

M cr 0

Calculation of the

3387660mm 3 × 275 N / mm 2 = 0.810 1420163158 Nmm

=

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed N cr , z   N cr ,TF 

   

term:

Where: -for a symmetrical section for the both axis,

N cr ,T =

N cr ,TF = N cr ,T

A  π 2 × E × I w  ×  G × It + I0  Lcr ,T ² 

The mass moment of inertia I 0

I 0 = I y + I z + A ⋅ z g2 Flexion inertia moment around the Y axis:

I y = 1490580416.67 mm 4

Flexion inertia moment around the Z axis:

I z = 26628854.17 mm 4

Cross section area:

A = 10850mm 2

Distance between the section neutral axis and the section geometrical center: z g

=0

I 0 = I y + I z + A ⋅ z g2 = I y + I z = 1490580416.67 mm 4 + 26628854.17 mm 4 = 1517209270mm 4 - the buckling length,

Lcr ,T , Lcr ,T = 5.62m

Torsional moment of inertia: Working inertial moment:

I t = 514614.75mm 4

I w = 49.808 × 1011 mm 6

(previously calculated)

Longitudinal elastic modulus: E = 210000 MPa Shear modulus of rigidity: G=80800MPa

N cr ,T =

10850mm 2 1517209270mm 4

 π 2 × 210000 N / mm 2 × 49.808 ×1011 mm 6   = ×  80800 N / mm 2 × 514614.75mm 4 + (5620mm )²  

= 2634739.14 N N Ed = 328000 N N cr ,TF = N cr ,T = 2634739.14 N

N cr , z =

224

π ² × E × Iz l fz ²

= 1747336.905 N

(previously calculated)

ADVANCE DESIGN VALIDATION GUIDE

 N 0.20 × C1 × 4 1 − Ed  N cr , z = 0.244

  N  × 1 − Ed   N cr ,TF  

  = 0.20 × 1.77 × 4  

328000 N  328000 N    1 −  × 1 − =  1747336.905 N   2634739.14 N 

Therefore:

λ 0 = 0.810  N 0.20 × C1 × 4 1 − Ed N cr , z 

   × 1 − N Ed   N cr ,TF  

 N λ0 = 0.810 > 0.20 ⋅ C1 ⋅ 4 1 − Ed 

The

N cr , z

  = 0.244  

   ⋅ 1 − N Ed   N cr ,TF  

   

    ε y × a LT  Cmy = Cmy , 0 + (1 − Cmy , 0 ) ×   1 + ε y × a LT    ⇒ Cmz = Cmz , 0   a LT 2  CmLT = Cmy × ≥1       = 0.244  1 − N Ed  × 1 − N Ed     N cr , z   N cr ,T   

Cmy 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.

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξy =

M y , Ed N Ed

×

ε y × aLT 1 + ε y × aLT

Aeff Weff , y

Elastic modulus after the Y axis,

ξy =

M y , Ed N Ed

×

Aeff Weff , y

=

Wel , y = Weff , y = 3179229.533mm 3

1274 ×10 6 Nmm 7815.2mm 2 × = 9.55 328000 N 3179229.533mm 3

a LT

It 514614.75mm 4 = 1− = 1− = 0.9996 ≈ 1 Iy 1448234429mm 4

The

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

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

=0

225

ADVANCE DESIGN VALIDATION GUIDE

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) ×

N Ed N cr , y

Where:

N cr , y =

π ²× E× Iy

= 95035371.44 N = 95035.37 kN (previously calculated)

l fy ²

C my , 0 = 0.79 + 0.36 × (0 − 0.33)×

C my = C my , 0 + (1 − C my , 0 )×

328000 N = 0.79 95035371.44 N

ξ y × a LT

= 0.79 + (1 − 0.79 )×

1 + ξ y × a LT

aLT

2 CmLT = Cmy ×

 1 − N Ed  N cr , z 

   

≥1

1

CmLT = 0.949 2 ×

k yy = Cmy × CmLT ×

   × 1 − N Ed   N cr ,T  

9.55 ×1 = 0.949 1 + 9.55 ×1

328000 N  328000 N    1 −  × 1 −   1747336.905 N   2634739.14 N  µy 1−

N Ed N cr , y

= 0.949 ×1.068 ×

= 1.068

1 = 1.0161 328000 N 1− 95035371.44 N

12.34.2.8 Internal factor, k yz , calculation

k yz = C mz ×

µy 1−

N Ed N cr , z

C mz = C mz , 0 = 0.79 + 0.36 × (− 0.33)× k yz = C mz ×

µy 1−

N Ed N cr , z

= 0.7677 ×

N Ed 328000 N = 0.7677 = 0.79 + 0.36 × (− 0.33)× 1747336.905 N N cr , z

1 = 0.945 328000 N 1− 1747336.905 N

12.34.2.9 Internal factor, k zy , calculation

k zy = C my × C mLT ×

µz 1−

N Ed N cr , y

N Ed 328000 N 1− N cr , z 1747336.905 N = µz = = 0.893 N Ed 328000 N 1 − 0.479 × 1− χz × 1747336.905 N N cr , z 1−

226

ADVANCE DESIGN VALIDATION GUIDE

k zy = Cmy × CmLT ×

µz

N 1 − Ed N cr , y

= 0.949 × 1.068 ×

0.893 = 0.908 328000 N 1− 95035371.44 N

12.34.2.10Internal factor, k zz , calculation

k zz = C mz ×

µz

N 1 − Ed N cr , z

= 0.7677 ×

0.893 = 0.844 328000 N 1− 1747336.905 N

12.34.2.11Bending and axial compression verification

M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k yz × z , Ed + k yy ×  N M M z , Rk  χ y × Rk χ LT × y , Rk γ M1 γ M1 γ M1   M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k zz × z , Ed + k zy ×  N Rk M M z , Rk χ LT × y , Rk χz × γ M1 γ M1 γ M1  N Rk = f y × Ai

1274 × 10 6 Nmm 328000 N . 0161 1 × + + 3179229.533mm 3 × 275 N / mm 2 7815.2mm 2 × 275 N / mm 2 0.589 × 1× 1 1 6 127.4 × 10 Nmm = 0.15 + 2.51 + 1.808 = 4.47 + 0.945 × 242068.10mm 3 × 275 N / mm 2 1 1274 × 10 6 Nmm 328000 N + + 0.908 × 3179229.533mm 3 × 275 N / mm 2 7815.2mm 2 × 275 N / mm 2 0.589 × 0.479 × 1 1 6 127.4 × 10 Nmm + 0.844 × = 0.32 + 2.24 + 1.61 = 4.14 242068.10mm 3 × 275 N / mm 2 1 Finite elements modeling ■ ■ ■

Linear element: S beam, 6 nodes, 1 linear element. 227

ADVANCE DESIGN VALIDATION GUIDE

χ y coefficient corresponding to non-dimensional slenderness λ y Column subjected to axial and shear force to the top

χy

χ z coefficient corresponding to non-dimensional slenderness λ z Column subjected to axial and shear force to the top

χz

228

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yy Column subjected to axial and shear force to the top

k yy

Internal factor, k yz Column subjected to axial and shear force to the top

k yz

229

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k zy Column subjected to axial and shear force to the top

k zy

Internal factor, k zz Column subjected to axial and shear force to the top

k zz

230

ADVANCE DESIGN VALIDATION GUIDE

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

231

ADVANCE DESIGN VALIDATION GUIDE

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

232

ADVANCE DESIGN VALIDATION GUIDE

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

233

ADVANCE DESIGN VALIDATION GUIDE

12.34.2.12Reference results Result name

Result description

Reference value

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

1

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.48

k yy

Internal factor, k yy

1.011

k yz

Internal factor, k yz

0.95

k zy

Internal factor, k zy

0.902

k zy

Internal factor, k zy

0.84

SNy

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

0.15

SMyy SMyz SMzy SMzz

2.50 1.81 2.23 1.61

12.34.3Calculated results

234

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 %

Kyy

Internal factor kyy

1.01066 adim

-0.0336 %

Kyz

Internal factor kyz

0.9451 adim

-0.5158 %

Kzy

Internal factor kzy

0.902094 adim

0.0104 %

Kzz

Internal factor kzz

0.843573 adim

0.4254 %

#SNy

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

term

0.152455 adim

1.6367 %

#SMyy

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

term

2.49874 adim

-0.0504 %

#SMyz

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

term

1.80865 adim

-0.0746 %

#SMzy

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

term

2.23031 adim

0.0139 %

#SMzz

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

term

1.61436 adim

0.2708 %

ADVANCE DESIGN VALIDATION GUIDE

12.35 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 1, column hinged on base and restrained on top for the X, Y translation and Z rotation (evaluated by SOCOTEC France - ref. Test 29) Test ID: 5729 Test status: Passed

12.35.1Description The test verifies a user defined cross section column. 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. The calculations are made according to Eurocode 3 French Annex.

12.35.2Background 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 This test was evaluated by the French control office SOCOTEC.

235

ADVANCE DESIGN VALIDATION GUIDE

12.35.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■ ■

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).

Units Metric System Geometrical properties ■

Column length: L=5620mm

■ ■

A = 4904.06mm 2 Overall breadth: b = 150mm Flange thickness: t f = 10.70mm

■

Root radius:

■

Web thickness: t w

■

Depth of the web:

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, Wy

■

Elastic modulus after the Z axis, Wel , z

= 80344.89mm 3

■

Plastic modulus after the Z axis, W pl , z

= 123381.96mm3

■

Flexion inertia moment around the Y axis:

■

Flexion inertia moment around the Z axis: I z

■

Torsional moment of inertia: I t

■

Working inertial moment: I w

■

236

Cross section area:

r = 0mm

= 7.10mm hw = 260mm = 445717.63mm3

= 501177.18mm3

I y = 57943291.64mm 4

= 6025866.46mm 4

= 149294.97 mm 4

= 93517065421.88mm 6

ADVANCE DESIGN VALIDATION GUIDE

Materials properties S275 steel material is used. The following characteristics are used: ■ ■ ■

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

Boundary conditions The boundary conditions are described below: ■

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

237

ADVANCE DESIGN VALIDATION GUIDE

12.35.2.2 Cross section Class According to Advance Design calculations: Cross-class classification is made according to Table 5.2

- for beam web: The web dimensions are 238.6x7.1mm.

ψ = 2⋅ 1 

N Ed 0.328 −1 = 2 ⋅ − 1 = −0.78 > −1 0.0109 × 275 A⋅ fy N



1 

0.328



Ed  = ⋅ 1 + α = ⋅ 1 +  = 0.85 > 0.5 f y × t × d  2  275 × 0.2386 × 0.0071  2 

238

ADVANCE DESIGN VALIDATION GUIDE

ε=

235 = fy

235 = 0.924 275

c 260mm − 2 × 10.7 mm  = = 33.61 c 396 × ε 396 × 0.924 = = 36.41 t 7.1mm  ⇒ = 33.6 ≤ t 13 × α − 1 13 × 0.85 − 1  ε = 0.924 

therefore the beam

web is considered to be Class 1 -for beam flange:

150 − 7.1   c c 2 = = 6.68 ⇒ = 6.68 ≤ 9 × 0.924 = 8.316 t 10.7  t  ε = 0.924

therefore the haunch is considered to be Class1

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

12.35.2.3 Buckling verification a) over the strong axis of the section, y-y: - the imperfection factor α will be selected according to Tables 6.1 and 6.2:

α = 0.34 Coefficient corresponding to non-dimensional slenderness after Y-Y axis:

239

ADVANCE DESIGN VALIDATION GUIDE

χ y coefficient

corresponding to non-dimensional slenderness

λy

will be determined from the relevant buckling

curve according to:

χy =

1 2

Φ y + Φ y − λy

2

≤1

(6.49)

λ y the non-dimensional slenderness corresponding to Class 1 cross-sections: A * fy

λy =

N cr , y A = 4904.06mm 2

Cross section area:

Flexion inertia moment around the Y axis: I y

N cr , y =

π²× E× Iy l fy ²

A× f y

λy =

N cr , y

=

= 57943291.64mm 4

π ² × 210000 N / mm 2 × 57943291.64mm 4

(5620mm)²

= 3802327.95 N = 3802.33kN

4904.06mm 2 × 275 N / mm 2 = = 0.5956 3802327.95 N

[

]

φ y = 0.5 × 1 + α × (λ y − 0.2) + λ y ² = 0.5 × [1 + 0.34 × (0.5956 − 0.2 ) + 0.5956 2 ] = 0.7446

χy =

1

χy ≤1

240

2

Φy + Φy −λy

2

=

1

 = 0.839 0.7446 + 0.7446 2 − 0.5956 2  ⇒ χ y = 0.839  

ADVANCE DESIGN VALIDATION GUIDE

b) over the weak axis of the section, z-z: - the imperfection factor α will be selected according to Tables 6.1 and 6.2:

α = 0.49 Coefficient corresponding to non-dimensional slenderness after Z-Z axis:

χ z coefficient corresponding to non-dimensional slenderness according to:

χz =

1 2

Φz + Φz − λ z

2

λz

will be determined from the relevant buckling curve

≤1

(6.49)

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

λz =

A* f y N cr , z

Flexion inertia moment around the Z axis: Cross section area:

N cr , z =

λz =

A = 4904.06mm 2

π ² × E × Iz l fz ²

A× fy N cr , z

=

I z = 6025866.46mm 4

=

π ² × 210000 N / mm 2 × 6025866.46mm 4

(2810mm )²

= 1581706.51N = 31581.71kN

4904.06mm 2 × 275 N / mm 2 = 0.923 1581706.51N 241

ADVANCE DESIGN VALIDATION GUIDE

[

]

φ z = 0.5 × 1 + α × (λ z − 0.2) + λ z ² = 0.5 × [1 + 0.49 × (0.923 − 0.2) + 0.9232 ] = 1.103

χz =

1 2

Φz + Φz − λ z

χz ≤ 1

2

=

1

 = 0.586 1.103 + 1.103 − 0.923  ⇒ χ z = 0.586   2

2

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

M cr = C1 ×

π ² × E × Iz I L² × G × It × w+ L² Iz π ² × E × Iz 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

C1 =

ψ=

242

1 0.325 + 0.423ψ + 0.252ψ ²

M y ,botom M y ,top

=

25kN = 0.5 ⇒ C1 = 1.31 25kN

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

ADVANCE DESIGN VALIDATION GUIDE

Flexion inertia moment around the Y axis:

I y = 57943291.64mm 4

Flexion inertia moment around the Z axis: I z Torsional moment of inertia: I t Working inertial moment: I w

= 6025866.46mm 4

= 149294.97 mm 4

= 93517065421.88mm 6

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):

Iw =

I z × (h − t f

)

2

4

h cross section height; h

= 260mm

tf

t f = 10.7 mm

Iw =

flange thickness;

6025866.46mm 4 × (260mm − 10.7mm ) = 93627638290mm 6 4 2

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

M cr = C1 ×

π ² × E × Iz L²

×

π 2 × 210000 N / mm 2 × 6025866.46mm 4 I w L² × G × It + = 1.31 × × Iz π ² × E × Iz (2810mm )2

93627638290mm6 (2810mm ) × 80800 N / mm 2 × 149294.97mm 4 × + = 1.31 × 1581706.51N × 152.20mm = π 2 × 210000 N / mm 2 × 6025866.46mm 4 6025866.46mm 4 = 315363380.74 Nmm = 315.36kNm 2

λ LT =

W pl , y f y M cr

Calculation of the

=

χ LT for appropriate non-dimensional slenderness λ LT

χ LT =

[

501177.18mm3 × 275 N / mm 2 = 0.661 315363380.74 Nmm

1

φLT + φLT ² − λ LT ²

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

≤1

will be determined with formula:

(6.56)

]

243

ADVANCE DESIGN VALIDATION GUIDE

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

h 260mm = = 1.733 ≤ 2 b 150mm

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

α LT = 0.49

[

(

)

]

φLT = 0.5 × 1 + α LT × λ LT − 0.2 + λ LT ² = 0.5 × [1 + 0.49 × (0.661 − 0.2) + 0.661² ] = 0.831

χ LT =

1

φLT + φLT ² − λ LT ²

=

1 = 0.749 ≤ 1 0.831 + 0.831² − 0.661²

b) for the bottom part of the column: 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:

M cr = C1 ×

L² × G × It I π ² × E × Iz × w+ L² Iz π ² × E × Iz According to EN 1993-1-1-AN France; Chapter 2; …(3)

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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1-AN France; Chapter 3; …(6)

244

ADVANCE DESIGN VALIDATION GUIDE

ψ

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

=

0 =0 637 kNm

ψ = 0 ⇒ C1 = 1.77

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

Flexion inertia moment around the Y axis:

I y = 57943291.64mm 4

Flexion inertia moment around the Z axis: I z Torsional moment of inertia: I t Working inertial moment: I w

= 6025866.46mm 4

= 149294.97 mm 4

= 93517065421.88mm 6

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):

Iw =

I z × (h − t f

)

2

4

h cross section height; h

tf

Iw =

flange thickness;

= 260mm

t f = 10.7 mm

6025866.46mm 4 × (260mm − 10.7mm ) = 93627638290mm 6 4 2

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

245

ADVANCE DESIGN VALIDATION GUIDE

M cr = C1 ×

π ² × E × Iz L²

×

π 2 × 210000 N / mm 2 × 6025866.46mm 4 I w L² × G × It + = 1.77 × × Iz π ² × E × Iz (2810mm )2

93627638290mm6 (2810mm ) × 80800 N / mm 2 × 149294.97mm 4 × + = 1.77 × 1581706.51N × 152.20mm = π 2 × 210000 N / mm 2 × 6025866.46mm 4 6025866.46mm 4 = 426102243.6 Nmm = 426.10kNm 2

λ LT =

W pl , y f y M cr

Calculation of the

=

501177.18mm3 × 275 N / mm 2 = 0.569 426102243.6 Nmm

χ LT for appropriate non-dimensional slenderness λ LT

χ LT =

[

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

]

1

φLT + φLT ² − λ LT ²

will be determined with formula:

≤1

(6.56)

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

h 260mm = = 1.733 ≤ 2 b 150mm

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

α LT = 0.49

[

]

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ² = 0.5 × [1 + 0.49 × (0.569 − 0.2 ) + 0.569² ] = 0.752

χ LT =

246

1

φLT + φLT ² − λ LT ²

=

1 = 0.804 ≤ 1 0.752 + 0.752² − 0.569²

ADVANCE DESIGN VALIDATION GUIDE

12.35.2.5 Internal factor, k yy , calculation The internal factor k yy 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: a) for the top part of the column:

k yy = Cmy × CmLT ×

µy `1 × N Ed C yy 1− N cr , y

N Ed N cr , y µy = N 1 − χ y × Ed N cr , y 1−

χ y = 0.839

(previously calculated)

N Ed = 328kN N cr , y =

π²× E × Iy l fy ²

= 3802327.95 N = 3802.33kN

(previously calculated)

N Ed 328000 N 1− N cr , y 3802327.95 N µy = = = 0.985 N Ed 328000 N 1 − 0.839 × 1− χ y × 3802327.95 N N cr , y 1−

247

ADVANCE DESIGN VALIDATION GUIDE

The

Cmy will be calculated according to Table A.1:

Calculation of the λ 0 term:

λ0 =

W pl , y × f y M cr 0 According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2

W y = 501177.18mm The calculation the

M cr = C1 ×

3

M cr 0

π ² × E × Iz L²

will be calculated using

×

C1 = 1 and C2 = 0 , therefore:

π 2 × 210000 N / mm 2 × 6025866.46mm 4 I w L² × G × It + = 1× × Iz π ² × E × Iz (2810mm )2

93517065421.88mm6 (2810mm ) × 80800 N / mm 2 × 149294.97 mm 4 + = 1 × 1581706.51N × 152.20mm = π 2 × 210000 N / mm 2 × 6025866.46mm 4 6025866.46mm 4 = 240735730.8 Nmm = 240.73kNm 2

×

λ0 =

248

W pl , y f y M cr

=

501177.18mm3 × 275 N / mm 2 = 0.757 240735730.8 Nmm

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed N cr , z   N cr ,TF 

   

term:

Where: - for a symmetrical section for the both axis,

N cr ,T =

N cr ,TF = N cr ,T

π 2 × E × I w  1  ×  G × It + I0  Lcr ,T ² 

The mass moment of inertia I 0

I 0 = I y + I z + A × z g2 = I y + I z = 57943291.64mm 4 + 6025866.46mm 4 = 63969158.1mm 4 Torsional moment of inertia: I t Working inertial moment: I w - the buckling length,

= 149294.97 mm 4

= 93517065421.88mm 6

Lcr ,T , Lcr ,T = 2.81m

 π 2 × 210000 N / mm 2 × 93517065421.88mm6   = ×  80800 N / mm 2 × 149294.97mm 4 + (2810mm )²   = 2806625.68 N = 2806.63kN N cr ,T =

4904.06mm 2 63969158.1mm 4

N Ed = 328000 N N cr ,TF = N cr ,T = 2806625.68 N

N cr , z =

π ² × E × Iz l fz ²

= 1581706.51N = 31581.71kN

(previously calculated)

C1=1 for the top part of the column For the top part of the column:

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed  N cr , z   N cr ,TF = 0.183

 328000 N   328000 N    4  = 0.20 × 1 × 1 − 1581706.51N  × 1 − 2806625.68 N  = 

249

ADVANCE DESIGN VALIDATION GUIDE

Therefore: For the top part of the column:

λ 0 = 0.757  N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed  N cr , z   N cr ,TF

  = 0.183  

   ⋅ 1 − N Ed N cr , z   N cr ,TF

 N λ0 = 0.757 > 0.20 ⋅ C1 ⋅ 4 1 − Ed 

The

   

    ε y × aLT  Cmy = Cmy , 0 + (1 − Cmy , 0 )×   1 + ε y × aLT    ⇒ Cmz = Cmz , 0   aLT 2  CmLT = Cmy ≥1 ×       = 0.183  1 − N Ed  × 1 − N Ed   N   N    cr , z   cr ,T   

Cmy 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.

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξy =

M y , Ed N Ed

×

ε y × aLT 1 + ε y × aLT

Aeff Weff , y

Elastic modulus after the Y axis,

ξy =

M y ,Ed N Ed

aLT = 1 − The

250

×

Aeff Weff , y

=

Wel , y = Weff , y = 445717.63mm3

50 × 106 Nmm 4904.06mm 2 × = 1.677 328000 N 445717.63mm3

It 149294.97 mm 4 =1− = 0.997 ≈ 1 57943291.64mm 4 Iy

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

ADVANCE DESIGN VALIDATION GUIDE

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

Cmy , 0 = 0.79 + 0.21×ψ + 0.36 × (ψ − 0.33)×

=0

N Ed N = 0.79 − 0.36 × 0.33 × Ed N cr , y N cr , y

Where:

N cr , y =

π²× E × Iy l fy ²

= 3802327.95 N = 3802.33kN

(previously calculated)

N Ed = 328000 N Cmy , 0 = 0.79 − 0.36 × 0.33 ×

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

328000 N = 0.780 3802327.95 N

ξ y × aLT 1 + ξ y × aLT

= 0.780 + (1 − 0.780 ) ×

Equivalent uniform moment factor,

CmLT , calculation

Equivalent uniform moment factor,

CmLT , calculation

-

1.677 × 1 = 0.904 1 + 1.677 × 1

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

2 CmLT = Cmy ×

- the

C my

a LT  1 − N Ed  N cr , z 

term used for

   × 1 − N Ed   N cr ,T  

CmLT calculation,

   

≥1

must be recalculated for the corresponding column part (in this case the top column

part) - this being the case,

C my

will be calculated using ψ

Cmy = Cmy , 0 + (1 − Cmy , 0 )

ξ y a LT 1 + ξ y a LT

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) ×

ξy =

M y ,Ed N Ed

×

Aeff Weff , y

=

= 0.5 :

238000 N N Ed = 0.900 = 0.79 + 0.21 × 0.5 + 0.36 × (0.5 − 0.33) × 3802327.95 N N cr , y

50 × 106 Nmm 4904.06mm 2 × = 1.677 328000 N 445717.63mm3

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξ y × aLT 1 + ξ y × aLT

= 0.900 + (1 − 0.900) ×

(previously calculated)

1.677 × 1 = 0.956 1 + 1.677 × 1

251

ADVANCE DESIGN VALIDATION GUIDE

aLT

      N N  Ed Ed    1 −  N  × 1 − N   cr , z   cr ,T     0 . 997 = 1.089 ⇒ CmLT = 1.089 = 0.9562 × 328000 N   328000 N      × 1 − 1 −   1581706.51N   2806625.68 N   CmLT ≥ 1    The C yy coefficient is defined according to the Table A.1, Auxiliary terms: 2 CmLT = Cmy ×

=

− − 2  1.6  W  1.6 2 2  C yy = 1 + ( wy − 1) ×  2 − × Cmy × λ max − × Cmy × λ max  × n pl − bLT  ≥ el , y   wy wy    W pl , y

−2

bLT = 0.5 × aLT × λ 0 × -

M y , Ed

χ LT × M pl , y , Rd

×

M z , Ed M pl , z , Rd −

bLT must be calculated separately for each of the two column parts, depending on λ 0 −

−

λ 0 = C1 × λ LT = 1.31 × 0.661 = 0.757

252

(for the top part of the column)

and

χ LT :

ADVANCE DESIGN VALIDATION GUIDE

M y , Ed

−2

bLT = 0.5 × aLT × λ 0 ×

χ LT × M pl , y , Rd

= 0.5 × 0.997 × 0.757 2 × = 0.041

wy = n pl =

W pl , y Wel , y

=

×

−2 M y , Ed M z , Ed M z , Ed = 0.5 × aLT × λ 0 × × χ LT × W pl , y × f y W pl , z × f y M pl , z , Rd

50000000 Nmm 10000000 Nmm × = 3 2 0.749 × 501177.18mm × 275 N / mm 123381.96mm3 × 275 N / mm 2

501177.18mm 3 = 1.124 ≤ 1.5 445717.63mm 3

N Ed 328000 N = 0.243 = N Rk 4904.06mm 2 × 275 N / mm 2 γ M1 1

λmax = max λ y ; λ z  = max(0.5956;0.923) = 0.923 −

−





  1.6 1.6  C yy = 1 + (1.124 − 1) ×  2 − × 0.904² × 0.923 − × 0.904² × 0.923²  × 0.243 − 0.041 = 0.993 1.124   1.124    C yy = 0.993   3 Wel , y 445717.63mm  0 . 889 = =  ⇒ C yy = 0.993 3 W pl , y 501177.18mm   Wel , y  C yy ≥ W pl , y 

N cr , y =

π²× E × Iy l fy ²

Therefore, the

= 3802327.95 N = 3802.33kN

(previously calculated)

k yy term corresponding to the top part of the column will be:

k yy = Cmy × CmLT ×

µy 1−

N Ed N cr , y

×

`1 = 0.904 × 1.089 × C yy 1−

0.985 1 × = 1.069 328000 N 0.993 3802327.95 N

253

ADVANCE DESIGN VALIDATION GUIDE

b) for the bottom part of the column:

k yy = Cmy × CmLT ×

- the terms:

N 1 − Ed N cr , y

CmLT ; bLT

2 CmLT = Cmy ×

- the

µy

×

`1 C yy

C yy

and

must be recalculated:

aLT    1 − N Ed  × 1 − N Ed  N   N cr , z   cr ,T 

   

≥1

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

Cmy , 0 = 0.79 + 0.21 ×ψ + 0.36 × (ψ − 0.33) ×

ξy =

M y , Ed N Ed

×

Aeff Weff , y

=

238000 N N Ed = 0.780 = 0.79 + 0.36 × (− 0.33) × 3802327.95 N N cr , y

25 × 106 Nmm 4904.06mm 2 × = 0.839 328000 N 445717.63mm3

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξ y × aLT 1 + ξ y × aLT

aLT

= 0.780 + (1 − 0.780 ) ×

0.839 × 1 = 0.885 1 + 0.839 × 1

       1 − N Ed  × 1 − N Ed   N   N   cr , z   cr ,T     0.997 = 0.8852 × = 0.933 ⇒ CmLT = 1 328000 N   328000 N    1 −  × 1 −   N N 1581706 . 51 2806625 . 68      CmLT ≥ 1    2 CmLT = Cmy ×

254

=

ADVANCE DESIGN VALIDATION GUIDE

− − 2  W  1.6  1.6 2 2 C yy = 1 + ( wy − 1) ×  2 − × Cmy × λ max − × Cmy × λ max  × n pl − bLT  ≥ el , y   wy wy  W pl , y  

M y , Ed

−2

bLT = 0.5 × aLT × λ 0 × -the

χ LT × M pl , y , Rd

×

M z , Ed M pl , z , Rd −

bLT must be calculated separately for each of the two column parts, depending of λ 0

W pl , y f y

λ0 =

M cr

χ LT =

501177.18mm3 × 275 N / mm 2 = = 0.757 240735730.8 Nmm

1

φLT + φLT ² − λ LT ²

χ LT × M pl , y , Rd

= 0.5 × 0.997 × 0.757 2 × = 0.0095

wy = n pl =

W pl , y Wel , y

χ LT :

(for the bottom part of the column)

1 = 0.804 ≤ 1 (for the bottom part of the column) 0.752 + 0.752² − 0.569²

M y , Ed

−2

bLT = 0.5 × aLT × λ 0 ×

=

and

×

−2 M y , Ed M z , Ed M z , Ed = 0.5 × aLT × λ 0 × × M pl , z , Rd χ LT × W pl , y × f y W pl , z × f y

25000000 Nmm 5000000 Nmm × = 3 2 0.804 × 501177.18mm × 275 N / mm 123381.96mm3 × 275 N / mm 2

501177.18mm 3 = = 1.124 ≤ 1.5 445717.63mm 3

N Ed 328000 N = = 0.243 N Rk 4904.06mm 2 × 275 N / mm 2 γ M1 1

λmax = max λ y ; λ z  = max(0.5956;0.923) = 0.923 −

−





  1.6 1.6  C yy = 1 + (1.124 − 1) ×  2 − × 0.904² × 0.923 − × 0.904² × 0.923²  × 0.243 − 0.0095 = 0.997 1.124   1.124    C yy = 0.997   3 Wel , y 445717.63mm  = = 0 . 889  ⇒ C yy = 0.997 3 W pl , y 501177.18mm   W  C yy ≥ el , y W pl , y 

N cr , y =

π ²× E × Iy l fy ²

= 3802327.95 N = 3802.33kN

(previously calculated)

255

ADVANCE DESIGN VALIDATION GUIDE

Therefore the

k yy term corresponding to the bottom part of the column will be:

k yy = Cmy × CmLT ×

µy

N 1 − Ed N cr , y

×

`1 = 0.904 × 1 × C yy 1−

0.985 1 × = 0.977 328000 N 0.997 3802327.95 N

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

12.35.2.6 Internal factor, k yz , calculation

k yz = Cmz ×

-the

µy

×

N 1 − Ed N cr , z

1 wz × 0.6 × C yz wy

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

Cmz = Cmz , 0 = 0.79 + 0.36 × (− 0.33) ×

328000 N N Ed = 0.79 + 0.36 × (− 0.33) × = 0.765 1581706.51N N cr , z

N Ed 328000 N 1− N cr , y 3802327.95 N = 0.985 = µy = 328000 N N Ed 1 0 . 839 − × 1− χy × 3802327.95 N N cr , y 1−

N cr , z =

π ² × E × Iz l fz ²

= 1581706.51N

(previously calculated) (previously calculated)

−2    2  × C λ max    mz C yz = 1 + ( wz − 1) ×  2 − 14 ×  × n pl − cLT  5 w z      −2

cLT = 10 × aLT ×

aLT = 1 −

λ0 =

λz =

λ0

−4

5+ λz

×

M y , Ed Cmy × χ lt × M pl , y , Rd

It 149294.97 mm 4 =1− = 0.997 Iy 57943291.64mm 4

W pl , y f y M cr

A× fy N cr , z

=

=

501177.18mm3 × 275 N / mm 2 = 0.757 240735730.8 Nmm

4904.06mm 2 × 275 N / mm 2 = 0.923 1581706.51N

M y , Ed = 50000 Nm

wz =

W pl , z Wel , z

w z ≤ 1.5

256

=

 123381.96mm 3 = 1.536 ≤ 1.5 3 80344.89mm  ⇒ w z = 1 .5  

(previously calculated)

(previously calculated)

(previously calculated)

ADVANCE DESIGN VALIDATION GUIDE

wy = - the

W pl , y Wel , y

Cmy

=

501177.18mm 3 = 1.124 ≤ 1.5 445717.63mm 3

term will be considered separately for each column part:

a) for the top part of the column:

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

χ LT =

ξ y × aLT 1 + ξ y × aLT

1

φLT + φLT ² − λ LT ²

=

= 0.900 + (1 − 0.900 ) ×

1.677 × 1 = 0.956 1 + 1.677 × 1

1 = 0.749 ≤ 1 0.831 + 0.831² − 0.661²

(previously calculated)

(previously calculated)

M pl , y , Rd = Wy × f y = 501177.18mm3 × 275 N / mm 2 = 137823724.5 Nmm −2

λ0

cLT = 10 × aLT ×

−4

×

M y , Ed Cmy × χ lt × M pl , y , Rd

=

5 + λz 0.757² 50000000 Nmm = 10 × 0.997 × × = 0.506 4 5 + 0.923 0.956 × 0.749 × 137823724.5 Nmm N cr , z =

π ² × E × Iz

= 1581706.51N

l fz ²

(previously calculated

Cmz = Cmz , 0 = 0.79 + 0.36 × (− 0.33) ×

328000 N N Ed = 0.765 = 0.79 + 0.36 × (− 0.33) × 1581706.51N N cr , z

λmax = max λ y ; λ z  = max(0.5956;0.923) = 0.923 −

−



n pl =



N Ed = 0.243 N Rk

(previously calculated)

γ M1

- Therefore: −2    2  × λ max  Cmz   C yz = 1 + ( wz − 1) ×  2 − 14 ×  × n pl − cLT  = 5 wz    

  0.7652 × 0.9232   × 0.243 − 0.506 = 0.878 = 1 + (1.5 − 1) ×  2 − 14 × 5 1.5    k yz = Cmz ×

µy

N 1 − Ed N cr , z

×

1 wz × 0.6 × = 0.765 × C yz wy 1−

0.985 1 1. 5 × × 0.6 × = 0.750 328000 N 0.878 1.124 1581706.51N

257

ADVANCE DESIGN VALIDATION GUIDE

b) for the bottom part of the column:

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

χ LT =

1

φLT + φLT ² − λ LT ²

ξ y × aLT 1 + ξ y × aLT

=

= 0.780 + (1 − 0.780 ) ×

0.839 × 1 = 0.885 1 + 0.839 × 1

1 = 0.804 ≤ 1 0.752 + 0.752² − 0.569²

(previously calculated)

(previously calculated)

M pl , y , Rd = Wy × f y = 501177.18mm3 × 275 N / mm 2 = 137823724.5 Nmm −2

λ0

cLT = 10 × aLT ×

−4

×

M y , Ed Cmy × χ lt × M pl , y , Rd

=

5 + λz 0.757² 25000000 Nmm = 10 × 0.997 × × = 0.254 4 5 + 0.923 0.885 × 0.804 × 137823724.5 Nmm π ² × E × Iz N cr , z = = 1581706.51N (previously calculated l fz ²

Cmz = Cmz , 0 = 0.79 + 0.36 × (− 0.33) ×

N Ed 328000 N = 0.79 + 0.36 × (− 0.33) × = 0.765 N cr , z 1581706.51N

λmax = max λ y ; λ z  = max(0.5956;0.923) = 0.923 −

−





n pl =

N Ed = 0.243 N Rk

(previously calculated)

γ M1

-Therefore: −2    2  Cmz × λ max    C yz = 1 + ( wz − 1) ×  2 − 14 ×  × n pl − cLT  = wz5     

  0.7652 × 0.9232   × 0.243 − 0.254 = 1.0043 = 1 + (1.5 − 1) ×  2 − 14 × 5 1.5    k yz = Cmz ×

µy 1−

N Ed N cr , z

×

1 wz × 0.6 × = 0.765 × C yz wy 1−

0.985 1 1.5 × × 0.6 × = 0.656 328000 N 1.0043 1.124 1581706.51N

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

258

ADVANCE DESIGN VALIDATION GUIDE

12.35.2.7 Internal factor, k zy , calculation a) for the top part of the column:

k zy = Cmy × CmLT ×

µz

N 1 − Ed N cr , y

×

1 wz × 0.6 × C zy wy

N Ed 328000 N 1− N cr , z 1581706.51N µz = = = 0.902 328000 N N Ed 1 − 0.586 × 1− χz × 1581706.51N N cr , z 1−

χ z = 0.586

(previously calculated)

−2    2  Cmy × λ max  wy Wel , y   C zy = 1 + ( wy − 1) ×  2 − 14 × ×  × n pl − d LT  ≥ 0.6 × 5 wy wz W pl , y       

M pl , y , Rd = Wy × f y = 501177.18mm3 × 275 N / mm 2 = 137823724.5 Nmm M pl , z , Rd = Wz × f z = 123381.96mm3 × 275 N / mm2 = 33930039 Nmm -in order to calculate the -the term

d LT term, the terms Cmy and Cmz

must be recalculated for each column part;

Cmz must be recalculated for the top column part only, using ψ = 0.5 :

Cmz = Cmz , 0 = 0.79 + 0.21 × 0.5 + 0.36 × (0.5 − 0.33) × = 0.79 + 0.21 × 0.5 + 0.36 × (0.5 − 0.33) × −

d LT = 2 × aLT ×

λ0 −4

×

M y ,Ed

N Ed = N cr , z

328000 N = 0.908 1581706.51N ×

M z ,Ed = Cmz × M pl , z ,Rd

Cmy × χ LT × M pl , y ,Rd 0.1 + λ z 0.757 50000000 Nmm 10000000 Nmm = 2 × 0.997 × × × = 4 0.1 + 0.923 0.956 × 0.749 × 137823724.5 Nmm 0.908 × 33930039 Nmm = 0.301

259

ADVANCE DESIGN VALIDATION GUIDE

     2 2    0.904 × 1.847    = 1 + (1.124 − 1) ×  2 − 14 × 0 . 243 0 . 301 0 . 859 × − =   1.1245   ⇒ C zy = 0.859    wy Wel , y 1.5 445717.63mm 3  0 .6 × × = 0.6 × × = 0.616 3  wz W pl , y 1.124 501177.18mm   wy Wel , y C zy ≥ 0.6 × ×  wz W pl , y 

−2    2  C λ × max    my C zy = 1 + ( wy − 1) ×  2 − 14 ×  × n pl − d LT  = 5 wy      

- for the calculation of the C zy term,

k zy = Cmy × CmLT ×

µz

N 1 − Ed N cr , y

×

Cmy

will be used for the entire column and

wy 1 × 0 .6 × = 0.904 × 1.089 × C zy wz 1−

d LT

will be used for the top column part:

0.902 1 1.124 × × 0 .6 × = 0.588 328000 N 0.859 1 .5 3802327.95 N

b) for the bottom part of the column:

k zy = Cmy × CmLT ×

µz 1−

N Ed N cr , y

×

wz 1 × 0.6 × C zy wy

N Ed 328000 N 1− N cr , z 1581706.51N µz = = = 0.902 328000 N N Ed 1 − 0.586 × 1− χz × 1581706.51N N cr , z 1−

χ z = 0.586

(previously calculated)

−2    2  λ C wy Wel , y × max    my × C zy = 1 + ( wy − 1) ×  2 − 14 ×  × n pl − d LT  ≥ 0.6 × 5 w w W pl , y y z      

M pl , y , Rd = Wy × f y = 501177.18mm3 × 275 N / mm 2 = 137823724.5 Nmm M pl , z , Rd = Wz × f z = 123381.96mm3 × 275 N / mm2 = 33930039 Nmm - in order to calculate the - the term

d LT term, the terms Cmy and Cmz

Cmz must be recalculated for the top column part only, using ψ = 0 :

Cmz = Cmz , 0 = 0.79 + 0.36 × (− 0.33) ×

260

must be recalculated for each column part;

N Ed 328000 N = 0.79 + 0.36 × (− 0.33) × = 0.765 1581706.51N N cr , z

ADVANCE DESIGN VALIDATION GUIDE

−

λ0

d LT = 2 × a LT ×

−4

M y ,Ed

×

×

M z ,Ed = Cmz × M pl , z ,Rd

Cmy × χ LT × M pl , y ,Rd 0.1 + λ z 0.757 25000000 Nmm 5000000 Nmm = 2 × 0.997 × × × = 4 0.1 + 0.923 0.885 × 0.804 × 137823724.5 Nmm 0.765 × 33930039 Nmm = 0.090

λmax = max λ y ; λ z  = max(0.5956;0.923) = 0.923 −

−





     2 2    0.885 × 0.923    0 . 243 0 . 090 0 . 892 = × − = 1 + (1.124 − 1) ×  2 − 14 ×   1.1245   ⇒ C zy = 0.892    wy Wel , y 1.124 445717.63mm3  0.6 × × = 0.6 × × = 0.462 3  wz W pl , y 1.5 501177.18mm   wy Wel , y C zy ≥ 0.6 × ×  wz W pl , y 

−2    2  C λ × max    my C zy = 1 + ( wy − 1) ×  2 − 14 ×  × n pl − d LT  = 5 wy      

- for the calculation of the

k zy = Cmy × CmLT ×

C zy

µz

N 1 − Ed N cr , y

term,

×

Cmy

will be used for the entire column and

wy 1 × 0.6 × = 0.904 × 1.089 × C zy wz 1−

d LT

will be used for the top column part:

0.902 1 1.124 × × 0.6 × = 0.566 328000 N 0.892 1.5 3802327.95 N

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

12.35.2.8 Internal factor, k zz , calculation a) for the top part of the column:

k zz = Cmz ×

µz

N 1 − Ed N cr , z

×

1 C zz

 1.6  W  1.6 2 2 × Cmz × λ max − × Cmz × λ max  × n pl − eLT  ≥ el , z C zz = 1 + ( wz − 1) ×  2 − wz wz    W pl , z M pl , y , Rd = Wy × f y = 501177.18mm3 × 275 N / mm 2 = 137823724.5 Nmm - in calculating the

eLT , the Cmy −

eLT = 1.7 × a LT ×

λ0 −4

×

term must be used accordingly with the corresponding column part

M y ,Ed Cmy × χ lt × M pl , y ,Rd

=

0.1 + λ z 0.757 50000000 Nmm = 1.7 × 0.997 × × = 0.787 4 0.1 + 0.923 0.956 × 0.749 × 501177.18mm 3 × 275 N / mm 2 261

ADVANCE DESIGN VALIDATION GUIDE

- for the calculation of the

C zz

term, C mz will be used for the entire column and

eLT will be used for the top column part:

      1.6 1.6 2 2 2 = 1 + (1.5 − 1) ×  2 − × 0.765 × 1.847 − × 0.765 × 1.847  − 0.787  × 0.243 = 1.013  1 . 5 1 . 5     ⇒ C zz = 1.013 Wel , z 80344.89mm3  = = 0.651  W pl , z 123381.96mm3   Wel , z C zz ≥  W pl , z    2  1.6 1.6 2 2 C zz = 1 + ( wz − 1) ×  2 − × Cmz × λ max − × Cmz × λ max  − eLT  × n pl = wz wz   

k zz = Cmz ×

µz

N 1 − Ed N cr , z

×

1 = 0.765 × C zz 1−

0.902 1 × = 0.860 328000 N 1.013 1581706.51N

b) for the bottom part of the column:

k zz = Cmz ×

µz 1−

N Ed N cr , z

1 C zz

×

 1.6  W  1.6 2 2 × Cmz × λ max − × Cmz × λ max  × n pl − eLT  ≥ el , z C zz = 1 + ( wz − 1) ×  2 − wz wz    W pl , z M pl , y , Rd = Wy × f y = 501177.18mm3 × 275 N / mm 2 = 137823724.5 Nmm - in calculating the

eLT , the Cmy −

λ0

eLT = 1.7 × aLT ×

−4

×

term must be used according to the corresponding column part

M y , Ed Cmy × χ lt × M pl , y , Rd

=

0.1 + λ z 25000000 Nmm 0.757 = 1.7 × 0.997 × × = 0.396 4 0.1 + 0.923 0.885 × 0.804 × 501177.18mm3 × 275 N / mm 2 - for the calculation of the C zz term, C mz will be used for the entire column and eLT will be used for the top column part:

λmax = max λ y ; λ z  = max(0.5956;0.923) = 0.923 −



−



      1.6 1.6  = 1 + (1.5 − 1) ×  2 − × 0.7652 × 0.923 − × 0.7652 × 0.9232  − 0.396 × 0.243 = 1.060  1.5 1.5     ⇒ C zz = 1.060 3 Wel , z 80344.89mm  = = 0.651 3  W pl , z 123381.96mm   Wel , z C zz ≥  W pl , z 

  2  1.6 1.6 2 2 × Cmz × λ max − × Cmz × λ max  − eLT  × n pl = C zz = 1 + ( wz − 1) ×  2 − wz wz   

262

ADVANCE DESIGN VALIDATION GUIDE

k zz = Cmz ×

µz

N 1 − Ed N cr , z

×

1 = 0.765 × C zz 1−

0.902 1 × = 0.821 328000 N 1.060 1581706.51N

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

12.35.2.9 Bending and axial compression verification

M y , Ed + ∆M y , Rd + ∆M z , Rd M  N Ed + k yz × z , Ed + k yy ×  M z , Rk M N  χ y × Rk χ LT × y , Rk γ M1 γ M1 γ M1   M y , Ed + ∆M y , Rd + ∆M z , Rd M  N Ed + k zz × z , Ed + k zy ×  M z , Rk M N Rk χ LT × y , Rk χz × γ M1 γ M1 γ M1  N Rk = f y × Ai

a) for the top part of the column:

328000 N 50 × 106 Nmm + × + 1 . 069 4904.06mm 2 × 275 N / mm 2 501177.18mm3 × 275 N / mm 2 0.839 × 0.749 × 1 1 6 10 × 10 Nmm + 0.750 × = 0.29 + 0.52 + 0.21 = 1.02 123381.96mm3 × 275 N / mm 2 1 328000 N 50 × 106 Nmm + × + 0 . 588 4904.06mm 2 × 275 N / mm 2 501177.18mm3 × 275 N / mm 2 0.586 × 0.804 × 1 1 6 10 × 10 Nmm + 0.860 × = 0.41 + 0.27 + 0.25 = 0.93 123381.96mm3 × 275 N / mm 2 1

263

ADVANCE DESIGN VALIDATION GUIDE

b) for the bottom part of the column:

328000 N 25 × 106 Nmm 0 . 977 + × + 4904.06mm 2 × 275 N / mm 2 501177.18mm3 × 275 N / mm 2 0.839 × 0.804 × 1 1 6 5 × 10 Nmm + 0.656 × = 0.29 + 0.22 + 0.97 = 1.30 123381.96mm3 × 275 N / mm 2 1 328000 N 25 × 106 Nmm 0 . 566 + × + 4904.06mm 2 × 275 N / mm 2 501177.18mm3 × 275 N / mm 2 0.586 × 0.804 × 1 1 6 5 × 10 Nmm + 0.821 × = 0.42 + 0.13 + 0.12 = 0.67 123381.96mm3 × 275 N / mm 2 1 Finite elements modeling ■ ■ ■

264

Linear element: S beam, 7 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

χ y coefficient corresponding to non-dimensional slenderness λ y Column subjected to axial and shear force to the top

χy

χ z coefficient corresponding to non-dimensional slenderness λ z Column subjected to axial and shear force to the top

χz

265

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yy Column subjected to axial and shear force to the top

k yy

Internal factor, k yz Column subjected to axial and shear force to the top

k yz

266

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k zy Column subjected to axial and shear force to the top

k zy

Internal factor, k zz Column subjected to axial and shear force to the top

k zz

Bending and axial compression verification term depending of the compression effort over the Y axis: SNy 267

ADVANCE DESIGN VALIDATION GUIDE

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

268

ADVANCE DESIGN VALIDATION GUIDE

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

269

ADVANCE DESIGN VALIDATION GUIDE

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

270

ADVANCE DESIGN VALIDATION GUIDE

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

271

ADVANCE DESIGN VALIDATION GUIDE

The appropriate non-dimensional slenderness The appropriate non-dimensional slenderness

χ LT

272

ADVANCE DESIGN VALIDATION GUIDE

12.35.2.10Reference results a) for the top part of the column: Result name

Result description

Reference value

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

0.839

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.586

k yy

Internal factor, k yy

1.069

k yz

Internal factor, k yz

0.750

k zy

Internal factor, k zy

0.588

k zz

Internal factor, k zz

0.860

SNy

Bending and axial compression verification term depending of the compression effort over the Y axis Bending and axial compression verification term depending of the Y bending moment over the Y axis Bending and axial compression verification term depending of the Z bending moment over the Y axis Bending and axial compression verification term depending of the compression effort over the z axis Bending and axial compression verification term depending of the Y bending moment over the Z axis Bending and axial compression verification term depending of the Z bending moment over the Z axis Coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure The elastic moment for lateral-torsional buckling calculation

0.29

The appropriate non-dimensional slenderness

0.749

SMyy SMyz SNz SMzy SMzz C1 Mcr

χ LT

0.52 0.21 0.41 0.27 0.25 1.77 315.36

273

ADVANCE DESIGN VALIDATION GUIDE

b) for the bottom part of the column: Result name

Result description

Reference value

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

0.839

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.586

k yy

Internal factor, k yy

0.977

k yz

Internal factor, k yz

0.656

k zy

Internal factor, k zy

0.566

k zz

Internal factor, k zz

0.821

SNy

Bending and axial compression verification term depending of the compression effort over the Y axis Bending and axial compression verification term depending of the Y bending moment over the Y axis Bending and axial compression verification term depending of the Z bending moment over the Y axis Bending and axial compression verification term depending of the compression effort over the z axis Bending and axial compression verification term depending of the Y bending moment over the Z axis Bending and axial compression verification term depending of the Z bending moment over the Z axis Coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure The elastic moment for lateral-torsional buckling calculation

0.29

The appropriate non-dimensional slenderness

0.804

SMyy SMyz SNz SMzy SMzz C1 Mcr

χ LT

0.22 0.97 0.42 0.13 0.12 1.77 426.10

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

12.35.3Calculated results

274

Result name

Result description

Value

Error

Xy

Coefficient corresponding slenderness by Y axis

to

non-dimensional

0.839285 adim

0.0340 %

Xz

Coefficient corresponding slenderness by the Z axis

to

non-dimensional

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 %

ADVANCE DESIGN VALIDATION GUIDE

12.36 EC3 / NF EN 1993-1-1/NA - France: Verifying a rectangular hollow section column subjected to bending and axial efforts (evaluated by SOCOTEC France - ref. Test 15) Test ID: 5735 Test status: Passed

12.36.1Description Verifies a rectangular hollow section column made of S235 steel subjected to bending and axial efforts. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

12.36.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.36.2.1 Model 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 = - 500 000 N, ► Fx = 5 000 N/ml, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

►

275

ADVANCE DESIGN VALIDATION GUIDE

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, 2 Section area:A = 9490 mm , 3 Plastic section modulus about y-y axis: Wpl,y = 956000 mm ,

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

Materials properties S235 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 235 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa.

Boundary conditions The boundary conditions are described below: ■

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 The column is subjected to the following loadings: ■

External: Point load at z = 5.0: Fz = - 500 000 N, ► Uniformly distributed load: q = Fx = 5 000 N/ml Internal: None. ►

■

276

ADVANCE DESIGN VALIDATION GUIDE

12.36.2.2 Reference 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 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 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 class for compressed parts. The picture below shows an extract from this table.

277

ADVANCE DESIGN VALIDATION GUIDE

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).

c b − 2 × r − 2 × t 200mm − 2 × 15mm − 2 × 10mm = = = 15 t t 10mm

ε=

235 = 1.0 fy

Therefore:

c = 15 ≤ 33ε = 33 t 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 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.

α 132 MPa = → α = 0.832 > 0.5 1 − α 26.63MPa c h − 2 × r − 2 × t 300mm − 2 × 15mm − 2 × 10mm = = = 25 t t 10mm

ε=

235 = 1.0 fy

Therefore:

c 396ε = 25 ≤ = 40.34 t 13α − 1 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. 278

ADVANCE DESIGN VALIDATION GUIDE

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.

N c , Rd =

A× f

γM0

y

=

9490mm 2 × 235MPa = 2230150 N 1.0

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 =

F N Ed 500000 N ×100 = z ×100 = ×100 = 22.42% N c , Rd N c , Rd 2230150 N

Verifying the design plastic moment resistance The design plastic moment resistance, for Class 1 cross-section, is determined with formula (6.13) from EN 1993-11:2001.

M c , Rd = M pl , Rd =

W pl , y × f

γM0

y

=

956000mm 3 × 235MPa = 224660000 Nmm 1.0

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 =

M Ed × 100 = M c , Rd

L 5000mm 5 N / mm × 5000mm × 2 × 100 = 2 × 100 = 27.82% M c , Rd 224660000 Nmm

q× L×

Finite elements modeling ■ ■ ■

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

279

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Work ratio of the design resistance for uniform compression Column subjected to bending and axial efforts Work ratio – Fx

Work ratio of the design resistance for bending Column subjected to bending and axial efforts Work ratio – Oblique

280

ADVANCE DESIGN VALIDATION GUIDE

12.36.2.3 Reference results 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 %

12.36.3Calculated results Result name Work ratio - Fx Work ratio Oblique

-

Result description

Value

Error

Compression resistance work ratio

22.42 %

0.0000 %

Work ratio of the design resistance for bending

27.8198 %

-0.0007 %

281

ADVANCE DESIGN VALIDATION GUIDE

12.37 EC3 / NF EN 1993-1-1/NA - France: Determining lateral torsional buckling parameters for a Ishaped welded built-up beam considering the load applied on the upper flange (evaluated by SOCOTEC France - ref. Test 44) Test ID: 5749 Test status: Passed

12.37.1Description 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. The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.

12.37.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.37.2.1 Model 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 = - 50 000 N/ml, ► My,1 = 142 x 106 Nmm, ► My,2 = - 113.6 x 106 Nmm, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

►

282

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

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, 2 Section area: A = 5188 mm , 4 Flexion inertia moment about the z axis: Iz = 6025866.46 mm , 4 Torsional moment of inertia: It = 149294.97 mm , 6 Warping constant: Iw = 93517065421.88 mm , Plastic modulus about the y axis: Wy = 501177.18 mm3

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

Materials properties S235 steel material is used. The following characteristics are used: ■ ■ ■

Yield strength fy = 235 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa; Shear modulus of rigidity: G=80800MPa.

Boundary conditions The boundary conditions are described below: ■

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 The beam is subjected to the following loadings: ■

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. ►

■

283

ADVANCE DESIGN VALIDATION GUIDE

12.37.2.2 Reference 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:

ψ= ■

− 113600 Nm = 0.8 − 142000 Nm

Ratio between the moment given by uniformly distributed load and the biggest bending moment from extremity:

µ=

q × L2 50 N / mm × (5000mm ) = = 1.1 8× M 8 × 142 × 10 6 Nmm 2

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: ■

β = ψ + 4 × µ − 1 = 0.8 + 4 × 1.1 − 1 = 4.2

■

γ = β 2 − 8 × µ = 4.2 2 − 8 × 1.1 = 8.84

■

a = 0.5 × (1 + β ) + 0.1413364γ − 0.6960364 βµ + 0.9126223µ 2 = 1.738

■

b = 0.5 × (1 + β ) + 0.1603341γ − 0.9240091βµ + 1.4281556 µ 2 = 1.4765

■

c = −0.1801266 β − 0.0900633γ + 0.5940757 βµ − 0.9352904 µ 2 = 0.0602

■

A = a × b − c 2 = 2.5625 b B = 2 × a + = 4.2143 2

■

284

■

d1 = µ + 0.52 × (1 + ψ ) = 2.036

■

e1 = 0.3

■

As d1 > e1, the factor r1 is equal to 1.0

■

ξ = 0.5 −

■

m = 1 − ξ × (1 −ψ ) + 4 × µ × ξ × (1 − ξ ) = 2.002 ≥ 1.0

■

C10 = r1 ×

■

d 2 = 0.425 + µ + 0.675 ×ψ = 2.065

■

e2 = 0.65 − 0.35 ×ψ = 0.37

1 −ψ = 0.4773 ≤ 1.0 8µ

B − B2 − 4× A = 0.5363 2× A

ADVANCE DESIGN VALIDATION GUIDE

■ As d2 > e2, the factor r2 is equal to 1.0 Having the above factors, C1 and C2 coefficients become: ■

C1 = m × C10 = 1.074

■

C 2 = 0.398 × r2 × µ × C10 = 0.235

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: ■

z g = +130mm

The French Annex of EN 1993-1-1 provides the analytical relationship used to determine the value of the elastic critical moment: ■

M cr = C1 ×

π 2 × E × Iz L2

 I  L2 × G × I t 2 × w + 2 + (C 2 × z g ) − C 2 × z g  = 91.71772 × 10 6 Nmm  I z π × E × I z 

12.37.2.3 Reference 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:

λLT ■

, imperfection factor αLT, φLT. Non dimensional slenderness for lateral torsional buckling,

λ LT = ■

W pl , y × f y

λLT

:

501177.18mm 3 × 235 N / mm 2 = = 1.133 91.71772 × 10 6 Nmm

M cr

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:

α LT = 0.49 ■

The value used to determine the reduction factor χLT, φLT, becomes:

■

The reduction factor for lateral torsional buckling is calculated using the formula (6.56) from EN 1993-1-1:

[

(

)

]

φ LT = 0.5 × 1 + α LT × λ LT − 0.2 + λ LT 2 = 0.5 × [1 + 0.49 × (1.133 − 0.2 ) + 1.133 2 ] = 1.370

χ LT =

1 2

φ LT + φ LT − λ LT

2

=

1 1.37 + 1.37 2 − 1.133 2

= 0.467 ≤ 1.0

Finite elements modeling ■ ■ ■

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

285

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results 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

286

ADVANCE DESIGN VALIDATION GUIDE

12.37.2.4 Reference results 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

12.37.3Calculated 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 %

287

ADVANCE DESIGN VALIDATION GUIDE

12.38 EC3 / NF EN 1993-1-1/NA - France: Class section classification and compression verification of an IPE300 column (evaluated by SOCOTEC France - ref. Test 1) Test ID: 5383 Test status: Passed

12.38.1Description 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. On top, the column is subjected to a 100 kN force applied gravitationally, defined as a live load. The dead load will be neglected.

12.38.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.38.2.1 Model 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: ■ ■

288

Exploitation loadings (category A): Q = -100kN, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Materials properties

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, Support at end point (z = 5.00) free. Inner: None.

► ►

■

12.38.2.2 Reference 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

289

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

ε =1 Therefore:

This means that the column web is Class 2. 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

290

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

c 56.45mm = = 5.276 t 10.7 mm

ε =1 Therefore:

c 56.45mm = = 5.276 ≤ 9 * ε = 9 t 10.7 mm

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 2 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 2. According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6) 291

ADVANCE DESIGN VALIDATION GUIDE

12.38.2.3 Reference 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

N c , Rd =

A* fy

γM0

Where: A

section area A=53.81cm2

Fy

nominal yielding strength for S235 fy=235MPa

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

N c , Rd =

A* fy

γM0

=

53.81 * 10−4 * 235 = 1.264535MN = 1264.54kN 1 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.4(2)

Finite elements modeling ■ ■ ■

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

Finite elements results Compressive resistance work ratio Column subjected to compressive load Work ratio [%]

292

ADVANCE DESIGN VALIDATION GUIDE

12.38.2.4 Reference results Result name

Result description

Reference value

Work ratio

Compressive resistance work ratio [%]

8%

12.38.3Calculated results Result name

Result description

Value

Error

Work ratio - Fx

Work ratio Fx

7.90805 %

-1.1494 %

293

ADVANCE DESIGN VALIDATION GUIDE

12.39 EC3 / NF EN 1993-1-1/NA - France: Class section classification and shear verification of an IPE300 beam subjected to linear uniform loading (evaluated by SOCOTEC France - ref. Test 2) Test ID: 5410 Test status: Passed

12.39.1Description 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.

12.39.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.39.2.1 Model 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: ■ ■

294

Exploitation loadings (category A): Q = -50kN/m, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Materials properties

Boundary conditions The boundary conditions are described below: ■

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.

► ►

■

12.39.2.2 Reference 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 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. 295

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:

296

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

This means that the column web is Class 1. 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

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:

c 56.45mm = = 5.276 t 10.7 mm

ε = 0.92 297

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

c 56.45mm = = 5.276 ≤ 9 * ε = 9 * 0.92 = 8.28 t 10.7mm

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 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. According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)

12.39.2.3 Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd shall be determined as follows:

Av *

V pl , Rd =

fy

γM0

3 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(2)

Where: Av: section shear area for rolled profiles

Av = A − 2 * b * t f + (t w + 2 * r ) * t f

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

Av = A − 2 * b * t f + (t w + 2 * r ) * t f = 53.81 − 2 * 15 * 17 + (0.71 + 2 * 1.5) * 1.07 = 25.68cm 2 According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3) fy: nominal yielding strength for S275 fy=275MPa

γ M 0 : partial safety coefficient γ M 0 = 1 Therefore:

V pl , Rd =

298

Av *

fy

γM0

3 =

25.68 * 104 * 1

275 3 = 0.4077 MN = 407.7kN

ADVANCE DESIGN VALIDATION GUIDE

For more: Verification of the shear buckling resistance for webs without stiffeners:

hw η ≤ 72 * tw ε According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(7)

η = 1.20 *

ε=

γ M1 1 = 1.20 * = 1.20 γM0 1

235 = fy

235 = 0.92 275

1.20 η hw 248.6 = = 35.01 ≤ 72 * = 72 * = 93.91 7.1 ε 0.92 tw There is no need for shear buckling resistance verification According to: EC3 Part 1,5 EN 1993-1-5-2004 Chapter 5.1(2) Finite elements modeling ■ ■ ■

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

Finite elements results Shear resistance work ratio Work ratio - Fz

299

ADVANCE DESIGN VALIDATION GUIDE

12.39.2.4 Reference results Result name

Result description

Reference value

Fz

Shear force

125 kN

Work ratio

Work ratio - Fz

31 %

12.39.3Calculated results

300

Result name

Result description

Value

Error

Fz

Fz

-125 kN

0.0000 %

Work ratio - Fz

Work ratio Fz

30.6579 %

-1.1035 %

ADVANCE DESIGN VALIDATION GUIDE

12.40 EC3 / NF EN 1993-1-1/NA - France: Class section classification and combined axial force with bending moment verification of an IPE300 column (evaluated by SOCOTEC France - ref. Test 5) Test ID: 5421 Test status: Passed

12.40.1Description 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. 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. The dead load will be neglected.

12.40.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.40.2.1 Model 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): Q1 = 500kN, Q2 = 5kN/m, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

301

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Materials properties

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, Support at end point (z = 5.00) free. Inner: None.

► ►

■

12.40.2.2 Reference 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 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.

302

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

ψ = 2⋅

N Ed 0.500 −1 = 2 ⋅ − 1 = −0.21 > −1 A⋅ f y 0.005381× 235

1 

N



1 

0.5



Ed  = ⋅ 1 + α = ⋅ 1 +  = 1.10 > 0.5 2  f y × t × d  2  235 × 0.2486 × 0.0071 

303

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

42ε 42 * 1 = = 69.92 0.67 + 0.33ψ 0.67 + 0.33 * (−0.21) Therefore:

c 42ε = 35.014 ≤ = 69.92 t 0.67 + 0.33ψ This means that the column web is Class 3. Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.

304

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

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 3 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 3. According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6) Cross sections for class 3, the maximum longitudinal stress should check:

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.4(3) In this case:

In absence of shear force, for Class 3 cross-sections the maximum longitudinal stress shall satisfy the criterion:

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.2(1) This means:

The corresponding work ratio is: WR = 0.8728 x 100 = 87.28 % 305

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling ■ ■ ■

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

Finite elements results Work ratio – bending and axial compression Column subjected to bending and axial compression Work ratio – Oblique [%]

12.40.2.3 Reference results Result name

Result description

Reference value

Bending and axial compression

Work ratio – oblique [%]

87.28 %

12.40.3Calculated results Result name Work ratio Oblique

306

-

Result description

Value

Error

Work ratio- Oblique

87.2799 %

0.3217 %

ADVANCE DESIGN VALIDATION GUIDE

12.41 EC3 / NF EN 1993-1-1/NA - France: Class section classification and combined biaxial bending verification of an IPE300 beam (evaluated by SOCOTEC France - ref. Test 6) Test ID: 5424 Test status: Passed

12.41.1Description 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. The dead load will be neglected.

12.41.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.41.2.1 Model 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): Q1 = -10kN/m, Q2 = 10kN/m, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

307

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Materials properties

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, ► Support at end point (x = 5.00) restrained in translation and rotation along Y, Z axis and rotation blocked along X axis. Inner: None. ►

■

12.41.2.2 Reference 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 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. 308

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

309

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

This means that the beam web is Class 1. Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.

The section geometrical properties are described in the picture below:

According to Table 5.2 and the beam section geometrical properties, the next conclusions can be found:

310

ADVANCE DESIGN VALIDATION GUIDE

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 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)

12.41.2.3 Reference results for calculating the combined biaxial bending α

β

 M Y , Ed   M z , Ed   +   ≤1  M Ny , Rd   M Nz , Ed  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:

α =2

β = max(n;1) n=

N Ed 0 = = 0 therefore β = 1 N pl , Rd N pl , Rd

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:

M Ny , Rd =

M pl , y , Rd * (1 − n) 1 − 0.5 * a

but

M Ny , Rd ≤ M ply , Rd According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(4)

n=

a=

0 N Ed = =0 N pl , Rd N pl , Rd

(A − 2*b *tf )

M N , y , Rd =

A

=

(53.81 * 10−4 − 2 * 0.15 * 0.0107) = 0.403 ≤ 0.5 53.81 * 10− 4

M pl , y , Rd (1 − 0.5 * 0.403)

=

M pl , y , Rd 0.8

0.8 * M N , y , Rd = M pl , y , Rd ⇒ M N , y , Rd > M pl , y , Rd

but

M Ny , Rd ≤ M ply , Rd

311

ADVANCE DESIGN VALIDATION GUIDE

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:

M Nz , Rd =

M pl , z , Rd * (1 − n) 1 − 0.5 * a

but

M Nz , Rd ≤ M plz , Rd

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(4)

0.8 * M N , z , Rd = M pl , z , Rd ⇒ M N , z , Rd > M pl , z , Rd

M N , z , Rd = M pl , z , Rd =

wpl , z * f y

γM0

but

M Nz , Rd ≤ M plz , Rd therefore it will be considered:

125.20 * 10−6 * 235 = = 0.030MNm 1

In conclusion:

 M Y , Ed   M Ny , Rd

α

  M z , Ed  +  M Nz , Ed 

β

2 1   0.03125   0.03125  = +   0.148   0.029375  = 1.1086 > 1 

The coresponding work ratio is: WR = 1.1086 x 100 = 110.86 % Finite elements modeling ■ ■ ■

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

Finite elements results Work ratio – oblique bending Beam subjected to combined bending Work ratio – Oblique [%]

312

ADVANCE DESIGN VALIDATION GUIDE

12.41.2.4 Reference results Result name Combined bending

oblique

Result description

Reference value

Combined oblique bending [%]

110.86 %

12.41.3Calculated results Result name Work ratio Oblique

-

Result description

Value

Error

Work ratio-Oblique

110.691 %

-0.2784 %

313

ADVANCE DESIGN VALIDATION GUIDE

12.42 EC3 / NF EN 1993-1-1/NA - France: Verifying the bending resistance of a rectangular hollow section column made of S235 steel (evaluated by SOCOTEC France - ref. Test 14) Test ID: 5728 Test status: Passed

12.42.1Description 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.

12.42.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.42.2.1 Model 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: ■

314

■

Exploitation loadings (category A), Q:  Fx = 50 000 N, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

ADVANCE DESIGN VALIDATION GUIDE

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, 3 Plastic section modulus about y-y axis: Wpl,y = 956000 mm ,

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

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 (z = 0) restrained in translation and rotation along X, Y and Z axis, ► Free at end point (z = 5.00). Inner: None. ►

Loading The column is subjected to the following loadings: ■

External: Point load at z = 2.5: V= Fx = 50 000 N, Internal: None. ►

■

315

ADVANCE DESIGN VALIDATION GUIDE

12.42.2.2 Reference 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:

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.

316

ADVANCE DESIGN VALIDATION GUIDE

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).

c b − 2 × r − 2 × t 200mm − 2 × 15mm − 2 × 10mm = = = 15 t t 10mm

ε=

235 = 1.0 fy

Therefore:

c = 15 ≤ 33ε = 33 t 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):

c h − 2 × r − 2 × t 300mm − 2 × 15mm − 2 × 10mm = = = 25 t t 10mm

ε=

235 = 1.0 fy

Therefore:

c = 25 ≤ 72ε = 72 t 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.

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ADVANCE DESIGN VALIDATION GUIDE

Design resistance for bending The design resistance for bending, for Class 1 cross-section, is determined with formula (6.13) from EN 1993-11:2001.

M c , Rd =

W pl , y × f

γ M0

y

=

956000mm 3 × 235MPa = 224660000 Nmm 1.0

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 =

L 5000mm V× 50000 N × M Ed 2 × 100 = 2 × 100 = × 100 = 55.64% M c , Rd M c , Rd 224660000 Nmm

Finite elements modeling ■ ■ ■

318

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

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Work ratio of the design resistance for bending Column subjected to a punctual horizontal load applied to the middle height Work ratio – Oblique

12.42.2.3 Reference results Result name

Result description

Reference value

Work ratio - Oblique

Work ratio of the design resistance for bending [%]

55.64 %

12.42.3Calculated results Result name Work ratio Oblique

-

Result description

Value

Error

Work ratio of the design resistance for bending

55.6396 %

-0.0007 %

319

ADVANCE DESIGN VALIDATION GUIDE

12.43 EC3 / NF EN 1993-1-1/NA - France: Comparing the shear resistance of a welded built-up beam made from different steel materials (evaluated by SOCOTEC France - ref. Test 45) Test ID: 5745 Test status: Passed

12.43.1Description 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. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

12.43.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.43.2.1 Model 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 = - 20 000 N/ml, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

Units Metric System

320

ADVANCE DESIGN VALIDATION GUIDE

Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■

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, 2 Section area: A = 5188 mm ,

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

Materials properties 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 The boundary conditions are described below: ■

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 The beam is subjected to the following loadings: ■

External: Uniformly distributed load: q = Fz = -20 000 N/ml Internal: None.

►

■

12.43.2.2 Reference 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:

VEd ≤ 1.0 V pl , Rd ■

VEd represents the design value of the shear force:

VEd = ■

q × L 20000 N / ml × 5000mm = = 50000 N 2 2

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.

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ADVANCE DESIGN VALIDATION GUIDE

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:

Av = η × ∑ (hw × t w ) = 1.0 × (278.6mm × 7.1mm) = 1978.06mm 2 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:

Av = η × ∑ (hw × t w ) = 1.2 × (278.6mm × 7.1mm) = 2373.67mm 2 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:

V pl , Rd =

Av ×

f

γ M0

y

3 =

1978.06mm 2 × 1.0

500 MPa 3 = 571016.7 N

The verification of the design plastic shear resistance of the cross section is done with criterion (6.18) from EN 19931-1:

VEd 50000 N = = 0.0876 ≤ 1.0 V pl , Rd 571016.7 N The corresponding work ratio is: Work ratio =

VEd 50000 N × 100 = × 100 = 8.76% V pl , Rd 571016.7 N

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:

V pl , Rd =

Av ×

f

γ M0

y

3 =

2373.67mm 2 × 1.0

275MPa 3 = 376870.7 N

The verification of the design plastic shear resistance of the cross section is done with criterion (6.18) from EN 19931-1:

VEd 50000 N = 0.133 ≤ 1.0 = V pl , Rd 376870.7 N The corresponding work ratio is: Work ratio =

VEd 50000 N × 100 = 13.27% × 100 = 376870.7 N V pl , Rd

Finite elements modeling ■ ■ ■

322

Linear element: S beam, 7 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results 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

12.43.2.3 Reference results 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 %

12.43.3Calculated 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 %

323

ADVANCE DESIGN VALIDATION GUIDE

12.44 EC3 / NF EN 1993-1-1/NA - France: Class section classification and bending moment verification of an IPE300 column (evaluated by SOCOTEC France - ref. Test 4) Test ID: 5412 Test status: Passed

12.44.1Description 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.

12.44.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.44.2.1 Model 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: ■ ■

324

Exploitation loadings (category A): Q = 50kN, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Materials properties

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, ► Support at end point (z = 5.00) free. Inner: None. ►

■

12.44.2.2 Reference 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.

325

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

326

ADVANCE DESIGN VALIDATION GUIDE

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

Therefore:

This means that the column web is Class 1. Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.

The section geometrical properties are described in the picture below:

327

ADVANCE DESIGN VALIDATION GUIDE

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

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)

12.44.2.3 Reference results in calculating the bending moment resistance

M Ed ≤1 M c , Rd According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.5(1)

M c , Rd =

wpl * f y

γM0

for Class1 cross sections According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.5(2)

Where:

w pl = 628.40cm 3 fy

nominal yielding strength for S235 fy=235MPa

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

M y ,V , Rd =

wpl * f y

γM0

=

628.40 * 10−6 * 235 = 0.147674MNm 1

M Ed = 2.5m * 50kN = 125kNm = 0.125MNm M Ed 0.125 = = 84% M y ,V , Rd 0.148 Finite elements modeling ■ ■ ■

328

Linear element: S beam, 7 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Combined oblique bending Combined oblique bending Work ratio - Oblique

12.44.2.4 Reference results Result name Combined bending

oblique

Result description

Reference value

Work ratio - Oblique

85 %

12.44.3Calculated results Result name Work ratio Oblique

-

Result description

Value

Error

Work ratio - Oblique

84.6459 %

-0.4166 %

329

ADVANCE DESIGN VALIDATION GUIDE

12.45 EC3 / NF EN 1993-1-1/NA - France: Verifying the classification and the resistance of a column subjected to bending and axial load (evaluated by SOCOTEC France - ref. Test 8) Test ID: 5632 Test status: Passed

12.45.1Description 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.

12.45.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.45.2.1 Model 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

■

Cross section dimensions are in milimeters (mm).

►

Units Metric System Geometry Below are described the column cross section characteristics:

330

■ ■ ■ ■ ■

Height: h = 600 mm, Flange width: b = 220 mm, Flange thickness: tf = 19 mm, Column length: L = 5000 mm, 2 Section area: A = 15600 mm ,

■

Plastic section modulus about the strong y-y axis:

W pl , y = 3512000mm 3 ,

ADVANCE DESIGN VALIDATION GUIDE

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

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). Inner: None. ►

Loading The column is subjected to the following loadings: ■

External: Point load at Z = 5.0: N = FZ = -1 000 000 N, ► Uniformly distributed load: q = Fx = 50 000 N/ml Internal: None. ►

■

331

ADVANCE DESIGN VALIDATION GUIDE

12.45.2.2 Reference 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:

N c , Rd =

A× fy

γM0

=

15600mm 2 × 235MPa = 3666000 N 1.0

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:

N Ed N 1000000 N = = = 0.273 ≤ 1.0 → 27.3% ≤ 100% N c , Rd N c , Rd 3666000 N 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.

N Ed ≤ 0.25 × N pl , Rd → 1000000 N > 0.25 × 3666000 N = 916500 N 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). ■

332

Design plastic moment resistance:

ADVANCE DESIGN VALIDATION GUIDE

►

■

■

=

γ M0

3512000mm 3 × 235MPa = 825320000 Nmm 1.0

n=

N

=

N pl , Rd

1000000 N = 0.273 3666000 N

Ratio of web area to gross area: ►

■

W pl , y × f y

Ratio of design normal force to design plastic resistance to normal force of the gross cross-section: ►

■

M pl , y , Rd =

a=

A − 2×b×t f A

=

15600mm 2 − 2 × 220mm × 19mm = 0.464 15600mm 2

Design plastic moment resistance reduced due to the axial force is determined according to expression (6.36) from EN 1993-1-1:

1− n 1 − 0 .5 × a

►

M N , y , Rd = M pl , y , Rd ×

►

M N , y , Rd = 825320000 Nmm ×

but

M N , y , Rd ≤ M pl , y , Rd

1 − 0.273 = 781259948 Nmm ≤ 825320000 Nmm 1 − 0.5 × 0.464

The column subjected to bending and axial force is verified with criterion (6.41) from EN 1993-1-1: α

β

►

 M y , Ed  M  N , y , Rd

►

Because the column doesn’t have bending moment about z axis, the second term from criterion (6.41)

  M z , Ed  +  M   N , z , Rd

  ≤ 1.0  

is neglected. The verification becomes:

►

 M y , Ed  M  N , y , Rd

2     =  q × L / 2  ≤ 1.0  M    N , y , Rd 

 q × L2 / 2   50 N / mm × (5000mm )2 / 2  =   = 0.799 ≤ 1.0 → 79.9% < 100%  M   781259948 Nmm   N , y , Rd  

Finite elements modeling ■ ■ ■

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

333

ADVANCE DESIGN VALIDATION GUIDE

Work ratio of the design resistance for uniform compression Column subjected to bending and axial force Work ratio - Fx

Work ratio of the design resistance for oblique bending Column subjected to bending and axial force Work ratio - Oblique

334

ADVANCE DESIGN VALIDATION GUIDE

12.45.2.3 Reference results Result name

Result description

Reference value

Work ratio - Fx Work ratio - Oblique

Work ratio of the design resistance for uniform compression [%] Work ratio of the design resistance for oblique bending [%]

27.3 % 79.9 %

12.45.3Calculated results Result name

Result description

Work ratio - Fx

Work ratio of compression

Work ratio Oblique

-

the

design

resistance

for

uniform

Work ratio of the design resistance for oblique bending

Value

Error

27.2777 %

-0.0084 %

79.9691 %

-0.0386 %

335

ADVANCE DESIGN VALIDATION GUIDE

12.46 EC3 / NF EN 1993-1-1/NA - France: Verifying the classification and the compression resistance of a welded built-up column (evaluated by SOCOTEC France - ref. Test 9) Test ID: 5674 Test status: Passed

12.46.1Description 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.

12.46.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.46.2.1 Model 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 Geometry Below are described the column cross section characteristics: ■ ■ ■ 336

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

ADVANCE DESIGN VALIDATION GUIDE

■ ■ ■

Web thickness: tw = 8 mm, Column length: L = 5000 mm, 2 Section area: A = 22752 mm ,

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

Materials properties S355 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 355 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa.

Boundary conditions The boundary conditions are described below: ■

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 The column is subjected to the following loadings: ■

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

►

■

337

ADVANCE DESIGN VALIDATION GUIDE

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

c (500mm − 8mm) / 2 = = 13.67 t 18mm

ε=

235 = 0.81 fy

Therefore:

c = 13.67 > 14ε = 11.34 t 338

ADVANCE DESIGN VALIDATION GUIDE

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:

c 630mm − 18mm × 2 = = 74.25 t 8mm

ε=

235 = 0.81 fy

Therefore:

c = 74.25 > 42ε = 34.02 t This means that the column web is Class 4. 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 4 flanges; therefore the class section for the entire column section will be considered Class 4.

12.46.2.3 Reference 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.

339

ADVANCE DESIGN VALIDATION GUIDE

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

ψ=

σ2 = 1.0 → kσ = 0.43 σ1

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:

ψ=

σ2 = 1.0 → kσ = 4.0 σ1

The plate modified slenderness (λp) – for flanges The formula used to determine the plate modified slenderness for flanges is:

λp =

((500mm − 8mm ) / 2) / 18mm = 0.906 c/t = 28.4ε kσ 28.4 × 0.81× 0.43

The plate modified slenderness (λp) – for web The formula used to determine the plate modified slenderness for web is:

λp =

(630mm − 2 × 18mm ) / 8mm = 1.614 b/t = 28.4ε kσ 28.4 × 0.81× 4.0

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:

ρ=

λ p − 0.188 ≤ 1.0 λ2p

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

beff , f = ρ × c = 0.875 ×

(500mm − 8mm ) = 215.25mm 2

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:

340

ADVANCE DESIGN VALIDATION GUIDE

ρ=

λ p − 0.055 × (3 + ψ ) ≤ 1.0 λ2p

The effective width of the web can now be calculated:

beff ,w = ρ × b = 0.535 × (630mm − 2 × 18mm ) = 317.8mm Effective area The effective area is determined considering the following:

Aeff = 2 × t f × (beff , f + beff , f + t w ) + t w × beff ,w Compression resistance of the cross section 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:

N c , Rd =

Aeff × f

γ M0

y

=

18328.4mm 2 × 355MPa = 6506582 N 1.0

Work ratio Work ratio =

N 100000 N × 100 = × 100 = 1.54% N c , Rd 6506582 N

Finite elements modeling ■ ■ ■

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

341

ADVANCE DESIGN VALIDATION GUIDE

Work ratio of the design resistance for uniform compression Column subjected to compression axial force Work ratio - Fx

12.46.2.4 Reference results Result name

Result description

Reference value

Work ratio - Fx

Compression resistance work ratio [%]

1.53 %

12.46.3Calculated results

342

Result name

Result description

Value

Error

Work ratio - Fx

Compression resistance work ratio

1.5322 %

0.1438 %

ADVANCE DESIGN VALIDATION GUIDE

12.47 EC3 / NF EN 1993-1-1/NA - France: Verifying the classification and the bending resistance of a welded built-up beam (evaluated by SOCOTEC France - ref. Test 10) Test ID: 5692 Test status: Passed

12.47.1Description 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.

12.47.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.47.2.1 Model 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 = -15 000 N/ml, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

►

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, 2 Section area: A = 22752 mm ,

343

ADVANCE DESIGN VALIDATION GUIDE

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

Materials properties S355 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 355 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 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 column is subjected to the following loadings: ■

External: Uniformly distributed load: q = Fz = -15 000 N/ml, Internal: None.

►

■

344

ADVANCE DESIGN VALIDATION GUIDE

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

c (500mm − 8mm) / 2 = = 13.67 t 18mm

ε=

235 = 0.8136 fy

Therefore:

c = 13.67 > 14ε = 11.39 t 345

ADVANCE DESIGN VALIDATION GUIDE

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):

c 630mm − 18mm × 2 = = 74.25 8mm t

ε=

235 = 0.8136 fy

Therefore:

c = 74.25 ≤ 124ε = 100.89 t This means that the beam web is Class 3. 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 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.

12.47.2.3 Reference 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 1993-11:2001. 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 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.

346

ADVANCE DESIGN VALIDATION GUIDE

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

ψ=

σ2 = 1.0 → kσ = 0.43 σ1

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

λp =

((500mm − 8mm ) / 2) / 18mm = 0.902 c/t = 28.4ε kσ 28.4 × 0.8136 × 0.43

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:

ρ=

λ p − 0.188 ≤ 1.0 λ2p

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

beff , f = ρ × c = 0.8776 ×

(500mm − 8mm) = 215.89mm 2

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.

347

ADVANCE DESIGN VALIDATION GUIDE

Weff , y ,sup = 5204392.91mm 3

Weff , y ,inf = 5736064.4mm 3 Weff , y ,min = min (Weff , y ,sup ,Weff , y ,inf ) = 5204392.91mm 3 Design resistance for bending For Class 4 cross-section, EN 1993-1-1: 2001 provides (6.15) formula in order to calculate the design resistance for bending:

M c , Rd =

Weff ,min × f

y

γM0

=

5204392.91mm 3 × 355MPa = 1847559483 Nmm 1.0

Work ratio

15 N / mm × (5000mm ) / 8 q × L2 / 8 M ×100 = 2.54% ×100 = ×100 = 1847559483 Nmm M c , Rd M c , Rd 2

Work ratio =

Finite elements modeling ■ ■ ■

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

Work ratio of the design resistance for bending Beam subjected to uniformly distributed load Work ratio - Oblique

12.47.2.4 Reference results Result name

Result description

Reference value

Work ratio - Oblique

Design resistance for bending work ratio [%]

2.54 %

12.47.3Calculated results Result name Work ratio Oblique

348

-

Result description

Value

Error

Design resistance for bending work ratio

2.53713 %

-0.1130 %

ADVANCE DESIGN VALIDATION GUIDE

12.48 NTC 2008 - Italy: Stability check for steel column hinged base Test ID: 6270 Test status: Passed

12.48.1Description Verification of a steel column with a point load applied on the top and a linear load applied along the linear element, the column is made of S235 material and is hinged on the bottom extremity and restrained in translation on x, y axis and restrained in rotation on z axis.

12.48.2Background 12.48.2.1 Model description ■ ■

Analysis type: static linear (plane problem); Element type: linear.

Units Metric Geometry ■ ■ ■

Length: L = 5.00 m, Area: A = 31.40 cm² Inertia: I = 1033.00 cm4

349

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■ ■

E = 210000 MPa S235

Boundary conditions ■ ■

Hinged at end x = 0, Restrained in translation X,Y at the top (x = 5.00 m) and restrained in rotation along Z.

Loading Point load on the top of the column, linear load on the left of the column and self-weight.

12.48.2.2 Reference results Calculation method used to obtain the reference solution

350

ADVANCE DESIGN VALIDATION GUIDE

The result for My is considered in the middle point of the column, so the work ratio will be 0.551 + 0.350 = 0.90 = 90%

351

ADVANCE DESIGN VALIDATION GUIDE

12.48.2.3 Theoretical results Result description

Reference value

SNz

Result name

Work ratio of the Normal force for the stability check

0.551

SMy

Work ratio of the bending moment for the stability check calculated in the middle Stability work ratio

0.350

Work ratio

0.900

12.48.3Calculated results

352

Result name

Result description

Value

Error

SNz

Effect of the normal force

0.549572 adim

-0.2592 %

SMzy

Effect of My

0.348335 adim

0.0000 %

Work ratio

Work ratio

91.1238 %

-1.5091 %

ADVANCE DESIGN VALIDATION GUIDE

12.49 EC3 / NF EN 1993-1-1 - France: Buckling resistance of a steel column Test ID: 6271 Test status: Passed

12.49.1Description Compares the buckling resistance of a compresed column with the value from a CTICM example.

12.49.2Background Compares the buckling resistance of a compressed column with the value from a CTICM example.

12.49.2.1 Model description One windwall (dimensions: 10m x 5m) Punctual support (fixed) Snow parameters ■ ■ ■

Section: IPE200 Material: S235 Buckling lengths: section is prevented from buckling in its secondary plane at x = 3.5m from the bottom

353

ADVANCE DESIGN VALIDATION GUIDE

Geometrical properties ■

Column length: L=6m

■

Cross section area:

■

Overall breadth:

■

Flange thickness:

■

Root radius:

■

Web thickness: t w

■

Depth of the web:

■

Plastic modulus after the Y axis, W pl , y

= 220.60cm 6

■

Plastic modulus after the Z axis, W pl , z

= 44.61cm 6

■ ■ ■ ■

Flexion inertia moment around the Y axis: Iy=1943 cm4 4 Flexion inertia moment around the Z axis: Iz=142.4 cm 4 4 Torsional moment of inertia: It=6.98x10 mm Working inertial moment: Iw=12990x106cm6

A = 28.48cm² b = 100mm

t f = 8.5mm

r = 12mm

= 5.6mm hw = 159mm

Materials properties S235 steel material is used. The following characteristics are used: ■ ■ ■

Yield strength fy = 235 MPa, Longitudinal elastic modulus: E = 210000 MPa. Shear modulus of rigidity: G=80800MPa

Boundary conditions The boundary conditions are described below: ■

Outer: ► ►

Hinged support at start point (x = 0m) restrained in translation along X, Y and Z axis, Support at the end point (z = 6m) restrained in translation along X and Y axis and restrained rotation around Z axis.

Loading The column is subjected to the following loadings: ■ External: Point load on column top : FZ =-176kN

354

ADVANCE DESIGN VALIDATION GUIDE

12.49.2.2 Cross section Class Cross-class classification is made according to Table 5.2

ε=

235 =1 fy

Compressed flange:

(b − t w − 2r ) / 2 = 4.1 < 9ε = 9 tf Flange is Class 1. Web:

(h − 2t f − 2r ) / 2 tw

= 28.4 < 33ε = 33

Web is Class 1. Section class is Class 1. Plastic characteristics are to be used.

12.49.2.3 Buckling lengths Section is prevented from buckling in its secondary plane at x = 3.5m from the bottom Lfz (strong inertia): 6m Lfy (weak inertia): 3.5m

12.49.2.4 Slenderness Strong axis:

i y = 8.26cm

λ y = 72.6 −

λy =

λy 93.9ε

= 0.773

Weak axis:

iz = 2.24cm

λz = 156.25 − λ λ z = z = 1.664 93.9ε 12.49.2.5 Buckling curves Strong axis: Curve a

(α = 0.21)

Weak axis: Curve b

(α = 0.34)

12.49.2.6 Buckling factors Strong axis: Weak axis:

χ y = 0.810 χ y = 0.287

We keep the minimal value:

χ min = 0.287

355

ADVANCE DESIGN VALIDATION GUIDE

12.49.2.7 Buckling resistance

N b , Rd = χ min × A × f y / γ M 1 = (0.287 × 2848 × 235 / 1) × 10−3 = 192kN 12.49.2.8 Work ratio

N Ed / N b , Rd = 176 / 192 = 91.66% 12.49.2.9 Reference results Result name

Result description

Reference value

Stability

Stability work ratio

91.66%

12.49.1Calculated results

356

Result name

Result description

Value

Error

Work ratio

Buckling work ratio

91.456 %

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

12.50 EC3 / NF EN 1993-1-2 - France: Verifying the bending resistance of a purlin for a 15min duration Test ID: 6259 Test status: Passed

12.50.1Description Verifies bending resistance of a purlin for 15min duration according to Eurocode 3 - part 2 - French standards (EN 1993-1-2).

12.50.2Background Simple Bending Design for in fire condition Verifies the adequacy of an IPE220 purlin made from S235 to resist simple bending for a 15min duration.

12.50.2.1 Model description ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■ ■ ■ ■

Loadings from the structure: G = 0.875 kN/m Snow loads : S = 1.1kN/m, Frequent combination of actions: CCQ = 1.0 x G + 0.2 x Q Fire duration : 15mins Exposed faces : 4 faces

Units Metric System Geometrical properties ■

Beam length: L=10000mm

■ ■

A = 3337mm 2 Overall breadth: b = 110mm Flange thickness: t f = 9.20mm

■

Web thickness: t w

= 5.90mm

■

Depth of the web:

hw = 201.6mm

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, W y

■

Cross section area:

= 252 × 103 mm 3

= 285.4 × 103 mm 3 357

ADVANCE DESIGN VALIDATION GUIDE

■

Elastic modulus after the Z axis, Wel , z

= 37.250 × 103 mm 3

■

Plastic modulus after the Z axis, W pl , z

= 58.11 × 103 mm 3

■ ■

Flexion inertia moment around the Y axis: Iy=2772x104mm4 4 4 Flexion inertia moment around the Z axis: Iz=2049x10 mm

Materials properties S235 steel material is used. The following characteristics are used: ■ ■ ■

Yield strength fy = 235 MPa, Longitudinal elastic modulus: E = 210000 MPa. Shear modulus of rigidity: G=80800MPa

Boundary conditions The boundary conditions are described below: ■

Plane workplane Support at start point (x = 0) : hinged support ► Support at the end point (x = 10) : hinged support ►

Loading The beam is subjected to the following load combinations: Load combinations: ■

SLS Frequent combination of actions:

N ser , fr = G + 0.2Q = 135 + 150 = 1.095 kN / ml 12.50.2.2 Critical temperature For a IPE220 with t=15mins and 4 exposed faces,

358

θ a = 702°C

ADVANCE DESIGN VALIDATION GUIDE

12.50.2.3 k y ,θ coefficient

k y ,θ

is then read from Table 3.1 :

k y ,θ = 0.2276 12.50.2.4 Bending resistance in fire conditions

M fi ,θ ,Rd = k y ,θ ×

γM0 × M Rd γ M , fi

M fi ,θ , Rd = k y ,θ ×

γ M 0 w pl × f y × γ M , fi γM0

1 285.4 × 10 −6 × 235 M fi ,θ , Rd = 0.2276 × × 1 1 M fi ,θ , Rd = 15.26kN .m 12.50.2.5 Bending moment My (design value)

M y , fi ,Ed =

ql ² 1.095 × 10² = = 13.69kN .m 8 8

12.50.2.6 Work ratio

M y , fi , Ed M fi ,θ , Rd

=

13.69 = 89.71% 15.26

Finite elements modeling ■ ■ ■

Linear element: S beam, 11 nodes, 1 linear element. 359

ADVANCE DESIGN VALIDATION GUIDE

Bending moment My in fire conditions

Work ratio for bending My in fire conditions

12.50.2.7 Reference results Result name

Result description

Reference value

Work ratio - My

Work ratio for bending moment My in fire conditions

0.8970

12.50.1Calculated results

360

Result name

Result description

Value

Error

Work ratio - My

My bending Fire

89.6982 %

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

12.51 EC3 / NF EN 1993-1-1/NA - France: Verifying an user defined I section class 1, column fixed on base and without any other restraint (evaluated by SOCOTEC France - ref. Test 28) Test ID: 5720 Test status: Passed

12.51.1Description 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.

12.51.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.51.2.1 Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■ ■

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).

361

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Geometrical properties ■

Column length: L=5620mm

■

■

A = 4904.06mm 2 Overall breadth: b = 150mm Flange thickness: t f = 10.70mm Root radius: r = 0mm

■

Web thickness: t w

■

Depth of the web:

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, Wy

■

Elastic modulus after the Z axis, Wel , z

= 80344.89mm 3

■

Plastic modulus after the Z axis, W pl , z

= 123381.96mm3

■

Flexion inertia moment around the Y axis:

■

Flexion inertia moment around the Z axis: I z

■

Torsional moment of inertia: I t

■

Working inertial moment: I w

■ ■

Cross section area:

= 7.10mm hw = 260mm = 445717.63mm3

= 501177.18mm3

I y = 57943291.64mm 4

= 6025866.46mm 4

= 149294.97 mm 4

= 93517065421.88mm 6

Materials properties S275 steel material is used. The following characteristics are used: ■ ■ ■

362

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

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions The boundary conditions are described below: ■

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

363

ADVANCE DESIGN VALIDATION GUIDE

12.51.2.2 Cross 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.

ψ=

364

σ inf 45.30Mpa = = −0.253 > −1 σ sup − 179.06Mpa

ADVANCE DESIGN VALIDATION GUIDE

238.6 × 45.3  x= = 48.175  x y x+ y 238.6  224.36 = = = ⇒ 238.6 × 179.06 45.30 179.06 45.30 + 179.06 224.36  y= = 190.73 224.36 

α=

x 190.73 = = 0.80 > 0.5 238.6 238.6 365

ADVANCE DESIGN VALIDATION GUIDE

ε=

235 = fy

235 = 0.924 275

c 260mm − 2 × 10.7 mm  = = 33.61 c 396 × ε 396 × 0.924 = 38.93 = t 7.1mm  ⇒ = 33.6 ≤ t 13 × α − 1 13 × 0.8 − 1  ε = 0.924 

therefore the beam

web is considered to be Class 1 -for beam flange:

150 − 7.1   c c 2 = = 6.68 ⇒ = 6.68 ≤ 9 × 0.924 = 8.316 t 10.7  t  ε = 0.924

therefore the haunch is considered to be Class1

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

12.51.2.3 Buckling 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:

366

ADVANCE DESIGN VALIDATION GUIDE

α = 0.34 Coefficient corresponding to non-dimensional slenderness after Y-Y axis:

χ y coefficient

corresponding to non-dimensional slenderness

λy

will be determined from the relevant buckling

curve according to:

χy =

1 2

Φ y + Φ y − λy

2

≤1

(6.49)

λ y the non-dimensional slenderness corresponding to Class 1 cross-sections: A * fy

λy =

N cr , y A = 4904.06mm 2

Cross section area:

Flexion inertia moment around the Y axis: I y

N cr , y =

π²× E× Iy l fy ²

A× f y

λy =

N cr , y

=

= 57943291.64mm 4

π ² × 210000 N / mm 2 × 57943291.64mm 4

(5620mm)²

= 3802327.95 N = 3802.33kN

4904.06mm 2 × 275 N / mm 2 = 0.5956 = 3802327.95 N

[

]

φ y = 0.5 × 1 + α × (λ y − 0.2) + λ y ² = 0.5 × [1 + 0.34 × (0.5956 − 0.2 ) + 0.5956 2 ] = 0.7446

χy =

1 2

Φy + Φy −λy

χy ≤1

2

=

1

 = 0.839 0.7446 + 0.7446 − 0.5956  ⇒ χ y = 0.839   2

2

367

ADVANCE DESIGN VALIDATION GUIDE

b) over the weak axis of the section, z-z: -the imperfection factor α will be selected according to the Table 6.1 and 6.2:

α = 0.49 Coefficient corresponding to non-dimensional slenderness after Z-Z axis:

χ z coefficient corresponding to non-dimensional slenderness according to:

χz =

1 2

Φz + Φz − λ z

2

λz

will be determined from the relevant buckling curve

≤1

(6.49)

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

λz =

A* f y N cr , z

Flexion inertia moment around the Z axis: Cross section area:

368

A = 4904.06mm 2

I z = 6025866.46mm 4

ADVANCE DESIGN VALIDATION GUIDE

N cr , z =

π ²× E × Iz l fz ²

A× f y

λz =

N cr , z

=

=

π ² × 210000 N / mm 2 × 6025866.46mm 4

(5620mm )²

= 395426.63 N = 395.43kN

4904.06mm 2 × 275 N / mm 2 = 1.847 395426.63 N

[

]

φ z = 0.5 × 1 + α × (λ z − 0.2) + λ z ² = 0.5 × [1 + 0.49 × (1.847 − 0.2 ) + 1.847 2 ] = 2.609

χz =

1 2

Φz + Φz − λ z

χz ≤ 1

2

=

1

 = 0.225 2.609 + 2.609 − 1.847  ⇒ χ z = 0.225   2

2

12.51.2.4 Lateral 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:

M cr = C1 ×

π ² × E × Iz I L² × G × It × w+ L² Iz π ² × E × Iz 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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1-AN France; Chapter 3; …(6)

ψ=

M y ,botom M y ,top

=

0 = 0 ⇒ C1 = 1.77 50

369

ADVANCE DESIGN VALIDATION GUIDE

Flexion inertia moment around the Y axis:

I y = 57943291.64mm 4

Flexion inertia moment around the Z axis: I z Torsional moment of inertia: I t Working inertial moment: I w

= 6025866.46mm 4

= 149294.97 mm 4

= 93517065421.88mm 6

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):

Iw =

I z × (h − t f

)

2

4

h cross section height; h

= 260mm

tf

t f = 10.7 mm

flange thickness;

Iw =

6025866.46mm 4 × (260mm − 10.7mm ) = 93627638290mm 6 4 2

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

M cr = C1 ×

π ² × E × Iz L²

×

I w L² × G × I t π 2 × 210000 N / mm 2 × 6025866.46mm 4 + = 1.77 × × Iz π ² × E × Iz (5620mm )2

93627638290mm 6 (5620mm ) × 80800 N / mm 2 × 149294.97 mm 4 + = 1.77 × 395426.63N × 214.58mm = 6025866.46mm 4 π 2 × 210000 N / mm 2 × 6025866.46mm 4 = 150184702.1Nmm = 15018kNm 2

×

λ LT =

W pl , y f y M cr

=

Calculation of the

501177.18mm3 × 275 N / mm 2 = 0.958 150184702.1Nmm

χ LT for appropriate non-dimensional slenderness λ LT

χ LT =

[

1

φLT + φLT ² − λ LT ²

φLT = 0.5 × 1 + α LT × (λ LT − 0.2 ) + λ LT ²

≤1

]

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

h 260mm = = 1.733 ≤ 2 b 150mm

370

will be determined with formula: (6.56)

ADVANCE DESIGN VALIDATION GUIDE

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

[

(

)

]

α = 0.49

φLT = 0.5 × 1 + α LT × λ LT − 0.2 + λ LT ² = 0.5 × [1 + 0.49 × (0.958 − 0.2 ) + 0.958² ] = 1.145

χ LT =

1

φLT + φLT ² − λ LT ²

=

1 = 0.564 ≤ 1 1.145 + 1.145² − 0.958²

12.51.2.5 Internal factor, k yy , calculation The internal factor k yy 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:

k yy = Cmy × CmLT ×

µy `1 × N Ed C yy 1− N cr , y

371

ADVANCE DESIGN VALIDATION GUIDE

N Ed N cr , y µy = N 1 − χ y × Ed N cr , y 1−

χ y = 0.839

(previously calculated)

N Ed = 328kN N cr , y =

π²× E × Iy l fy ²

= 3802327.95 N = 3802.33kN

(previously calculated)

N Ed 328000 N 1− N cr , y 3802327.95 N µy = = = 0.985 N Ed 328000 N 1 − 0.839 × 1− χy × 3802327.95 N N cr , y 1−

The

Cmy will be calculated according to Table A.1:

Calculation of the λ 0 term:

λ0 =

W pl , y × f y M cr 0 -according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2

372

ADVANCE DESIGN VALIDATION GUIDE

W y = 501177.18mm3 The calculation the

M cr = C1 ×

M cr 0

π ² × E × Iz L²

will be calculated using

C1 = 1 and C2 = 0 , therefore:

I w L² × G × I t π 2 × 210000 N / mm 2 × 6025866.46mm 4 + = 1× × Iz π ² × E × Iz (5620mm)2

×

93517065421.88mm 6 (5620mm ) × 80800 N / mm 2 × 149294.97 mm 4 + = 1 × 395426.63N × 214.58mm = 6025866.46mm 4 π 2 × 210000 N / mm 2 × 6025866.46mm 4 = 84850646.27 Nmm = 84.85kNm 2

×

λ0 =

W pl , y f y M cr

Calculation of the

=

501177.18mm3 × 275 N / mm 2 = 1.274 84850646.27 Nmm

 N   N 0.20 × C1 × 4 1 − Ed  × 1 − Ed N cr , z   N cr ,TF 

   

term:

Where: -for a symmetrical section for the both axis,

N cr ,T =

N cr ,TF = N cr ,T

π 2 × E × I w  1  ×  G × It + Lcr ,T ²  I0 

The mass moment of inertia I 0

I 0 = I y + I z + A × z g2 = I y + I z = 57943291.64mm 4 + 6025866.46mm 4 = 63969158.1mm 4 Torsional moment of inertia: I t Working inertial moment: I w - the buckling length,

= 149294.97 mm 4

= 93517065421.88mm 6

Lcr ,T , Lcr ,T = 5.62m

N cr ,T =

4904.06mm 2 63969158.1mm 4

 π 2 × 210000 N / mm 2 × 93517065421.88mm 6   = ×  80800 N / mm 2 × 149294.97 mm 4 + (5620mm)²  

= 1395246.607 N N Ed = 328000 N N cr ,TF = N cr ,T = 1395246.607 N

N cr , z =

π ² × E × Iz l fz ²

=

π ² × 210000 N / mm 2 × 6025866.46mm 4

(5620mm)²

= 395426.63 N = 395.43kN

(previously calculated)

C1=1 for the top part of the column

373

ADVANCE DESIGN VALIDATION GUIDE

For the top part of the column:

 N 0.20 × C1 × 4 1 − Ed  N cr , z = 0.120

   × 1 − N Ed   N cr ,TF  

  = 0.20 × 1 × 4  

328000 N  328000 N    =  × 1 − 1 −  395426.63 N   1395246.607 N 

Therefore: For the top part of the column:

λ 0 = 1.274  N 0.20 × C1 × 4 1 − Ed  N cr , z

   × 1 − N Ed   N cr ,TF  

  = 0.120  

 N   N  λ0 = 1.274 > 0.20 ⋅ C1 ⋅ 4 1 − Ed  ⋅ 1 − Ed  N N 

The

cr , z

 

cr ,TF



    ε y × aLT  Cmy = Cmy , 0 + (1 − Cmy , 0 )×   1 + ε y × aLT    ⇒ Cmz = Cmz , 0   aLT 2  CmLT = Cmy × ≥1       N N = 0.120  1 − Ed  × 1 − Ed   N   N    cr , z   cr ,T   

Cmy 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.

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

ξy =

M y , Ed N Ed

×

ε y × aLT 1 + ε y × aLT

Aeff Weff , y

Elastic modulus after the Y axis,

ξy =

M y ,Ed N Ed

aLT = 1 −

374

×

Aeff Weff , y

Wel , y = Weff , y = 445717.63mm3

50 × 106 Nmm 4904.06mm 2 × = 1.677 = 445717.63mm3 328000 N

It 149294.97 mm 4 =1− = 0.997 ≈ 1 Iy 57943291.64mm 4

ADVANCE DESIGN VALIDATION GUIDE

The

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

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

Cmy , 0 = 0.79 + 0.21×ψ + 0.36 × (ψ − 0.33)×

=0

N Ed N = 0.79 − 0.36 × 0.33 × Ed N cr , y N cr , y

Where:

N cr , y =

π²× E × Iy l fy ²

= 3802327.95 N = 3802.33kN

(previously calculated)

N Ed = 328000 N

Cmy , 0 = 0.79 − 0.36 × 0.33 × Cmy = Cmy , 0 + (1 − Cmy , 0 )×

328000 N = 0.780 3802327.95 N

ξ y × aLT 1 + ξ y × aLT

Equivalent uniform moment factor,

aLT

= 0.780 + (1 − 0.780 ) ×

1.677 × 1 = 0.904 1 + 1.677 × 1

CmLT , calculation

      N N  Ed Ed    1 −  N  × 1 − N   cr , z   cr ,T     0.997 2 = 2.256 ⇒ CmLT = 2.256 = 0.904 × 328000 N   328000 N      × 1 − 1 −   395426.63 N   1395246.607 N   CmLT ≥ 1    2 CmLT = Cmy ×

=

375

ADVANCE DESIGN VALIDATION GUIDE

The

C yy coefficient is defined according to the Table A.1, Auxiliary terms:

− − 2  1.6  W  1.6 2 2  × Cmy × λ max − × Cmy × λ max  × n pl − bLT  ≥ el , y C yy = 1 + ( wy − 1) ×  2 −   wy wy    W pl , y

M y ,Ed

−2

bLT = 0.5 × aLT × λ 0 ×

χ LT × M pl , y ,Rd

= 0.5 × 0.997 × 1.274 2 × = 0.1534

wy = n pl =

W pl , y Wel , y

=

×

−2 M y ,Ed M z ,Ed M z ,Ed = 0.5 × aLT × λ 0 × × χ LT × W pl , y × f y W pl , z × f y M pl , z ,Rd

50000000 Nmm 10000000 Nmm × = 3 2 0.564 × 501177.18mm × 275 N / mm 123381.96mm3 × 275 N / mm 2

501177.18mm 3 = 1.124 ≤ 1.5 445717.63mm 3

N Ed 328000 N = = 0.243 N Rk 4904.06mm 2 × 275 N / mm 2 γ M1 1

λmax = max λ y ; λ z  = max(0.5956;1.847 ) = 1.847 −



−



  1.6 1.6  C yy = 1 + (1.124 − 1) ×  2 − × 0.904² ×1.847 − × 0.904² ×1.847²  × 0.243 − 0.1534 = 0.857 1.124   1.124 

376

ADVANCE DESIGN VALIDATION GUIDE

  C yy = 0.857   3 Wel , y 445717.63mm  = = 0 . 889  ⇒ C yy = 0.889 3 W pl , y 501177.18mm   W  C yy ≥ el , y W pl , y  N cr , y =

π²× E × Iy

= 3802327.95 N = 3802.33kN

l fy ²

Therefore the

(previously calculated)

k yy term corresponding to the top part of the column will be:

k yy = Cmy × CmLT ×

µy

N Ed N cr , y

1−

×

`1 = 0.904 × 2.256 × C yy 1−

1 0.985 = 2.47 × 328000 N 0.889 3802327.95 N

12.51.2.6 Internal factor, k yz , calculation

k yz = Cmz ×

µy 1−

1 wz × 0.6 × C yz wy

×

N Ed N cr , z

Cmz = Cmz , 0 = 0.79 + 0.36 × (− 0.33) ×

N Ed 328000 N = 0.79 + 0.36 × (− 0.33) × = 0.691 N cr , z 395426.63 N

N Ed 328000 N 1− N cr , y 3802327.95 N µy = = = 0.985 328000 N N Ed 1 − 0.839 × 1− χy × 3802327.95 N N cr , y 1−

N cr , z =

π ² × E × Iz l fz ²

= 395426.63 N = 395.43kN

(previously calculated)

(previously calculated)

−2    2  Cmz × λ max    C yz = 1 + ( wz − 1) ×  2 − 14 ×  × n pl − cLT  5 wz     −2

cLT = 10 × aLT ×

aLT = 1 −

λ0 =

λz =

λ0

−4

5+ λz

×

M y , Ed Cmy × χ lt × M pl , y , Rd

It 149294.97 mm 4 =1− = 0.997 Iy 57943291.64mm 4

W pl , y f y M cr

A× f y N cr , z

=

501177.18mm3 × 275 N / mm 2 = 1.274 84850646.27 Nmm

4904.06mm 2 × 275 N / mm 2 = = 1.847 395426.63 N

(previously calculated)

(previously calculated)

(previously calculated)

377

ADVANCE DESIGN VALIDATION GUIDE

M y , Ed = 50000 Nm

Cmy = Cmy , 0 + (1 − Cmy , 0 )×

χ LT =

ξ y × aLT 1 + ξ y × aLT

1

φLT + φLT ² − λ LT ²

=

= 0.904

(previously calculated)

1 = 0.564 ≤ 1 1.145 + 1.145² − 0.958²

(previously calculated)

M pl , y , Rd = Wy × f y = 501177.18mm3 × 275 N / mm 2 = 137823724.5 Nmm −2

λ0

cLT = 10 × aLT ×

−4

×

M y , Ed Cmy × χ lt × M pl , y , Rd

=

5+ λz 1.274² 50000000 Nmm = 10 × 0.997 × × = 0.664 4 5 + 1.847 0.904 × 0.564 × 137823724.5 Nmm

−2    2  × λ max  Cmz   C yz = 1 + ( wz − 1) ×  2 − 14 ×  × n pl − cLT  5 wz    

wz =

W pl , z Wel , z

=

wz ≤ 1.5

N cr , z =

 123381.96mm3 = 1.536 ≤ 1.5 3 80344.89mm  ⇒ wz = 1.5  

π ² × E × Iz

= 395426.63 N = 395.43kN

l fz ²

Cmz = Cmz , 0 = 0.79 + 0.36 × (− 0.33) ×

(previously calculated)

328000 N N Ed = 0.691 = 0.79 + 0.36 × (− 0.33) × 395426.63N N cr , z

λmax = max λ y ; λ z  = max(0.5956;1.847 ) = 1.847 −

−





n pl =

N Ed = 0.243 N Rk

(previously calculated)

γ M1

−2    2  Cmz × λ max    C yz = 1 + ( wz − 1) ×  2 − 14 ×  × n pl − cLT  = 5 wz       0.6912 × 1.847 2   × 0.243 − 0.691 = 0.532 = 1 + (1.5 − 1) ×  2 − 14 × 5 1.5   

k yz = Cmz ×

378

µy

N 1 − Ed N cr , z

×

1 0.985 1 1.5 wz × 0.6 × = 0.691 × × × 0.6 × = 5.20 328000 N 0.532 1.124 C yz wy 1− 395426.63 N

ADVANCE DESIGN VALIDATION GUIDE

12.51.2.7 Internal factor, k zy , calculation

k zy = Cmy × CmLT ×

µz

N 1 − Ed N cr , y

×

wz 1 × 0.6 × C zy wy

N Ed 328000 N 1− N cr , z 395426.63 N µz = = = 0.210 N Ed 328000 N 1 − 0.225 × 1− χz × 395426.63 N N cr , z 1−

χ z = 0.225

(previously calculated)

−2    2  C × λ wy Wel , y max    my C zy = 1 + ( wy − 1) ×  2 − 14 × ×  × n pl − d LT  ≥ 0.6 × 5 wy wz W pl , y      

M pl , z , Rd = Wz × f z = 123381.96mm3 × 275 N / mm2 = 33930039 Nmm −

λ0

d LT = 2 × aLT ×

−4

×

M y , Ed Cmy × χ LT × M pl , y , Rd

×

M z , Ed Cmz × M pl , z , Rd

=

0.1 + λ z 1.274 50000000 Nmm 10000000 Nmm = 2 × 0.997 × × × = 4 0.1 + 1.847 0.904 × 0.564 × 137823724.5 Nmm 0.691× 33930039 Nmm = 0.771

−2    2  C λ × max    my Czy = 1 + ( wy − 1) ×  2 − 14 ×  × n pl − d LT  = 5 wy      

 0.9042 × 1.847 2  = 1 + (1.124 − 1) ×  2 − 14 ×  × 0.243 − 0.771 = 0.310 1.1245   0.6 ×

wy wz

×

Wel , y W pl , y

wy

C zy ≥ 0.6 ×

wz

×

C zy = 0.310 0 .6 ×

wy wz

×

Wel , y W pl , y

CmLT = 2.256 k zy = Cmy × CmLT ×

1.124 445717.63mm3 = 0.6 × × = 0.462 501177.18mm3 1.5

  W pl , y    ⇒ C zy = 0.462  = 0.462  Wel , y

(previously calculated)

µz

N 1 − Ed N cr , y

×

wy 1 × 0.6 × = 0.904 × 2.256 × C zy wz 1−

0.210 1 1.124 = 0.529 × × 0.6 × 328000 N 0.462 1.5 3802327.95 N

379

ADVANCE DESIGN VALIDATION GUIDE

12.51.2.8 Internal factor, k zz , calculation

k zz = Cmz ×

µz

N 1 − Ed N cr , z

1 C zz

×

 1.6  W  1.6 2 2 C zz = 1 + ( wz − 1) ×  2 − × Cmz × λ max − × Cmz × λ max  × n pl − eLT  ≥ el , z wz wz    W pl , z −

eLT = 1.7 × aLT ×

λ0 −4

×

M y , Ed Cmy × χ lt × M pl , y , Rd

=

0.1 + λ z 1.274 50000000 Nmm = 1.7 × 0.997 × × = 0.131 4 0.1 + 1.847 0.904 × 0.564 × 501177.18mm3 × 275 N / mm 2

  2  1.6 1.6 2 2 C zz = 1 + ( wz − 1) ×  2 − × Cmz × λ max − × Cmz × λ max  × n pl − eLT  = wz wz      1.6 1.6  = 1 + (1.5 − 1) ×  2 − × 0.6912 × 1.847 − × 0.6912 × 1.847 2  × 0.243 − 0.131 = 0.926 1.5 1.5    k zz = Cmz ×

µz

N 1 − Ed N cr , z

×

1 0.210 1 = 0.691 × × = 0.919 328000 N 926 0 . C zz 1− 395426.63 N

12.51.2.9 Bending and axial compression verification

M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k yz × z , Ed + k yy ×  N M M z , Rk  χ y × Rk χ LT × y , Rk γ M1 γ M1 γ M1   M y , Ed + ∆M y , Rd M + ∆M z , Rd  N Ed + k zz × z , Ed + k zy ×  N Rk M M z , Rk χ LT × y , Rk χz × γ M1 γ M1 γ M1  N Rk = f y × Ai

380

ADVANCE DESIGN VALIDATION GUIDE

328000 N 50 × 106 Nmm + 2 . 479 × + 4904.06mm 2 × 275 N / mm 2 501177.18mm3 × 275 N / mm 2 0.839 × 0.564 × 1 1 6 10 × 10 Nmm + 5.20 × = 0.29 + 1.59 + 1.53 = 3.41 123381.96mm3 × 275 N / mm 2 1 328000 N 50 × 106 Nmm + 0 . 529 × + 4904.06mm 2 × 275 N / mm 2 501177.18mm3 × 275 N / mm 2 0.225 × 0.564 × 1 1 6 10 × 10 Nmm + 0.919 × = 1.08 + 0.34 + 0.27 = 1.69 123381.96mm3 × 275 N / mm 2 1 Finite elements modeling ■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

381

ADVANCE DESIGN VALIDATION GUIDE

χ y coefficient corresponding to non-dimensional slenderness λ y Column subjected to axial and shear force to the top

χy

χ z coefficient corresponding to non-dimensional slenderness λ z Column subjected to axial and shear force to the top

χz

382

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yy Column subjected to axial and shear force to the top

k yy

Internal factor, k yz Column subjected to axial and shear force to the top

k yz

383

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k zy Column subjected to axial and shear force to the top

k zy

Internal factor, k zz Column subjected to axial and shear force to the top

k zz

384

ADVANCE DESIGN VALIDATION GUIDE

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

385

ADVANCE DESIGN VALIDATION GUIDE

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

386

ADVANCE DESIGN VALIDATION GUIDE

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

387

ADVANCE DESIGN VALIDATION GUIDE

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

388

ADVANCE DESIGN VALIDATION GUIDE

The appropriate non-dimensional slenderness The appropriate non-dimensional slenderness

χ LT

389

ADVANCE DESIGN VALIDATION GUIDE

12.51.2.10Reference results Result name

Result description

Reference value

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

0.839

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.225

k yy

Internal factor, k yy

2.47

k yz

Internal factor, k yz

5.20

k zy

Internal factor, k zy

0.529

k zz

Internal factor, k zy

0.919

SNy

Bending and axial compression verification term depending of the compression effort over the Y axis Bending and axial compression verification term depending of the Y bending moment over the Y axis Bending and axial compression verification term depending of the Z bending moment over the Y axis Bending and axial compression verification term depending of the compression effort over the z axis Bending and axial compression verification term depending of the Y bending moment over the Z axis Bending and axial compression verification term depending of the Z bending moment over the Z axis Coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure The elastic moment for lateral-torsional buckling calculation

0.29

χ LT

The appropriate non-dimensional slenderness

0.564

Work ratio

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

341 %

SMyy SMyz SNz SMzy SMzz C1 Mcr

1.59 1.53 1.08 0.34 0.27 1.77 150.18

12.51.3Calculated results

390

Result name

Result description

Value

Error

Xy

Coefficient corresponding slenderness after Y-Y axis

to

non-dimensional

0.839285 adim

0.0340 %

Xz

Coefficient corresponding slenderness after Z-Z axis

to

non-dimensional

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 %

ADVANCE DESIGN VALIDATION GUIDE

12.52 EC3 / NF EN 1993-1-1/NA - France: Verifying a simply supported rectangular hollow section beam subjected to biaxial bending (evaluated by SOCOTEC France - ref. Test 16) Test ID: 5736 Test status: Passed

12.52.1Description Verifies a simply supported rectangular hollow section beam made of S235 steel subjected to biaxial bending. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

12.52.2Background 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. This test was evaluated by the French control office SOCOTEC.

12.52.2.1 Model 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, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in milimeters (mm).

►

Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■

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

391

ADVANCE DESIGN VALIDATION GUIDE

■ ■ ■ ■

Beam length: L = 5000 mm, Section area: A = 9490 mm2, 3 Plastic section modulus about y-y axis: Wpl,y = 956000 mm , Plastic section modulus about z-z axis: Wpl,z = 721000 mm3,

■

Partial factor for resistance of cross sections:

γ M 0 = 1.0 .

Materials properties S235 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 235 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa.

Boundary conditions The boundary conditions are described below: ■

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 The beam is subjected to the following loadings: ■

External: Uniformly distributed load: q1 = Fz = -10 000 N/ml ► Uniformly distributed load: q2 = Fy = 10 000 N/ml Internal: None. ►

■

392

ADVANCE DESIGN VALIDATION GUIDE

12.52.2.2 Reference 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 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 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 class for compressed parts. The picture below shows an extract from this table.

393

ADVANCE DESIGN VALIDATION GUIDE

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).

c b − 2 × r − 2 × t 300mm − 2 ×15mm − 2 ×10mm = = = 25 t t 10mm

ε=

235 = 1.0 fy

Therefore:

c = 25 ≤ 72ε = 72 t 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. 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. 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:

n= ■

394

N Ed → as the beam is not subjected to axial efforts n = 0. N pl , Rd

Determination of aw for hollow section:

ADVANCE DESIGN VALIDATION GUIDE

A − 2×b×t ≤ 0.5 → a w = 0.5 A

aw = ■

Determination of af for hollow section:

af = ■

Determination of aw for hollow section:

A − 2×b×t ≤ 0.5 → a w = 0.5 A

aw = ■

A − 2× h×t ≤ 0.5 → a f = 0.3678 A

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

M N , y , Rd =

M pl , y , Rd × (1 − n )

(1 − 0.5 × aw )

≤ M pl , y , Rd

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

M N , y , Rd = M pl , y , Rd = ■

W pl , y × f y

γM0

=

956000mm 3 × 235MPa = 22466 ×10 4 Nmm 1.0

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

M N , z , Rd =

M pl , z , Rd × (1 − n )

(1 − 0.5 × a )

≤ M pl , z , Rd

f

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

M N , z , Rd = M pl , z , Rd =

W pl , z × f y

γ M0

721000mm 3 × 235MPa = = 16943.5 ×10 4 Nmm 1.0

Verifying for bi-axial bending Criterion (6.41) from EN 1993-1-1 has to be fulfilled:

 M y , Ed  M  N , y , Rd ■

α

  M z , Ed  +  M   N , z , Rd

Determination of constants α and β for rectangular hollow section:

α =β = ■

1.66 ≤ 6 → α = β = 1.66 1 − 1.13 × n 2

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

M y , Ed =

M z , Ed = ■

β

  ≤ 1.0  

q1 × L2 8

q 2 × L2 8

=

10 N / mm × 5000 2 mm 2 = 3125 × 10 4 Nmm 8

=

10 N / mm × 5000 2 mm 2 = 3125 ×10 4 Nmm 8

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

395

ADVANCE DESIGN VALIDATION GUIDE

1.66

 3125 × 10 4 Nmm    4  22466 × 10 Nmm  Work ratio =

1.66

 3125 × 10 4 Nmm   +  4  16943.5 × 10 Nmm 

= 0.0378 + 0.0604 = 0.098 ≤ 1.0

0.098 × 100 = 9.8% 1.0

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:

M y , Ed M pl , y , Rd

3125 ×10 4 Nmm = = 0.139 ≤ 1.0 22466 ×10 4 Nmm

Work ratio =

0.139 × 100 = 13.9% 1.0

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:

M z , Ed M pl , z , Rd

=

Work ratio =

3125 ×10 4 Nmm = 0.1844 ≤ 1.0 16943.5 ×10 4 Nmm 0.1844 × 100 = 18.44% 1.0

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

396

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

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Work ratio of the design resistance for bending Beam subjected to bending efforts Work ratio – Oblique

12.52.2.3 Reference results Result name

Result description

Reference value

Work ratio - Oblique

Work ratio of the design resistance for biaxial bending

18.44 %

12.52.3Calculated results Result name Work ratio Oblique

-

Result description

Value

Error

Work ratio of the design resistance for biaxial bending

18.4437 %

0.2375 %

397

ADVANCE DESIGN VALIDATION GUIDE

12.53 EC3 / CSN EN 1993-1-1 - Czech Republic: Compressed IPE300 Test ID: 6275 Test status: Passed

12.53.1Description Class section classification and compression verification of an IPE300 column

12.53.2Background 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.

12.53.2.1 Model 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: ■ ■

398

Exploitation loadings (category A): Q = -100kN, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

ADVANCE DESIGN VALIDATION GUIDE

Units Metric System Materials properties

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, Support at end point (z = 5.00) free. Inner: None.

► ►

■

12.53.2.2 Reference 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 CSN EN 1993-1-1 Chapter 5.5.2

399

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:

ε =1 Therefore:

This means that the column web is Class 2. To determine the flanges class, we will use Table 5.2, sheet 2, from CSN EN 1993-1-1 Chapter 5.5.2

400

ADVANCE DESIGN VALIDATION GUIDE

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:

c 56.45mm = = 5.276 t 10.7 mm

ε =1 Therefore:

c 56.45mm = = 5.276 ≤ 9 * ε = 9 t 10.7 mm

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 2 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 2. According to: CSN EN 1993-1-1 Chapter 5.5.2(6) 401

ADVANCE DESIGN VALIDATION GUIDE

12.53.2.3 Reference 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

N c , Rd =

A* fy

γM0

Where: A

section area A=53.81cm2

Fy

nominal yielding strength for S235 fy=235MPa

γ M 0 partial safety coefficient γ M 0 = 1 Therefore:

N c , Rd =

A* fy

γM0

53.81 * 10−4 * 235 = = 1.264535MN = 1264.54kN 1 According to: CSN EN 1993-1-1 Chapter 6.2.4(2)

Finite elements modeling ■ ■ ■

402

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

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Compressive resistance work ratio Column subjected to compressive load Work ratio [%]

12.53.2.4 Reference results Result name

Result description

Reference value

Work ratio

Compressive resistance work ratio [%]

8%

12.53.3Calculated results Result name

Result description

Value

Error

Max Work ratio

Max work ratio

7.90805 %

-1.1494 %

Work ratio - Fx

Work ratio Fx

7.90805 %

-1.1494 %

403

ADVANCE DESIGN VALIDATION GUIDE

12.54 NTC 2008 - Italy: Deflection check on simply supported beam Test ID: 6266 Test status: Passed

12.54.1Description Deflection check on simply supported beam made of steel S355, the cross section used is HEA140, the span of the beam is 6.00 meter, the verification is made according to NTC 2008

12.54.2Background 12.54.2.1 Model description ■ ■

Analysis type: static linear (plane problem); Element type: linear.

Units Metric Geometry ■ ■ ■

Length: L = 6.00 m, Area: A = 31.42 cm² Inertia: I = 1033.00 cm4

Materials properties ■ ■

E = 210000 MPa S355

Boundary conditions ■ ■

Hinge at end x = 0, Hinge at end x = 6.00 m.

Loading Uniformly distributed force of q = -10.00 kN/m on beam AB.

404

ADVANCE DESIGN VALIDATION GUIDE

12.54.2.2 Reference results Calculation method used to obtain the reference solution

f = (5/384) X (10 X 60004 ) / (2100000 X 10330000) = 777.9 mm = L / 75

12.54.2.3 Theoretical result Result name Deflection

Result description

Reference value

Maximum deflection calculated in the middle of the steel beam

777.9 mm = L / 75

12.54.3Calculated results Result name

Result description

Value

Error

Max. deflection

Max deflection

74.7801 adim

0.0000 %

405

ADVANCE DESIGN VALIDATION GUIDE

12.55 AISC 360 2010 - United States of America: Deflection and Strength for a W-Shape Continuously Braced Flexural Member in Strong-Axis Bending (LRFD) Test ID: 6273 Test status: Passed

12.55.1Description Verifies the available flexural strength of a W18x50 in ASTM A992 simple supported beam with a span of 35 ft. The deflection is limited to a ratio of L/360. The nominal loads are a uniform dead load of 0.45 kip/ft and a uniform live load of 0.75 kip/ft. The beam is continuously braced. (LRFD)

12.55.2Background Verifies the deflections, the deflection constraints, the bending moment and it related resistance to a simply supported W18x50. The primary objective of this test is to validate the resistance of member exposed to uniformly distributed load which would cause bending moments in the beam. With this the deflection will be verified against a max deflection criterion.

12.55.2.1 Model Description Reference: AISC Design Example Version 14.1 Example as taken from “Example F.1-1A&C W-Shape Flexural Member Design in Strong-Axis Bending, Continuously Braced” Select an ASTM A992 W-shape beam with a simple span of 35 ft. Limit the member to a maximum nominal depth of 18 in. Limit the live load deflection to L/360. The nominal loads are a uniform dead load of 0.45 kip/ft and a uniform live load of 0.75 kip/ft. Assume the beam is continuously braced. Try a W18x50.

■

Analysis Type: Static

■

Element Type: Linear

The following load cases and load combination are used ■

Dead load: 0.450 kip/ft uniformly distributed load (Self-weight included in this value – Therefore not needed for AD to calculate)

■

Live Load: 0.750 kip/ft uniformly distributed load

■

Combination used for this example is: ULS Combination #102 - 1.2 × 𝐷𝑒𝑎𝑑 + 1.6 × 𝐿𝑖𝑣𝑒

1.2 × (0.45 𝑘𝑖𝑝/𝑓𝑡) + 1.6 × (0.750 𝑘𝑖𝑝/𝑓𝑡) 0.54 𝑘𝑖𝑝/𝑓𝑡 + 1.2 𝑘𝑖𝑝/𝑓𝑡

SLS Combination #103 - 1.0 × 𝐿𝑖𝑣𝑒

1.74 𝑘𝑖𝑝/𝑓𝑡

1.0 × (0.750 𝑘𝑖𝑝/𝑓𝑡) 0.750 𝑘𝑖𝑝/𝑓𝑡

406

ADVANCE DESIGN VALIDATION GUIDE

■

Steel cross-section used is an AISC W18x50

■

Unit System: IMPERIAL

■

Conditions: Continuously braced the full length of the beam

■

Steel Type: ASTM A992

■

Fy = 50 ksi

■

Fu = 65 ksi

12.55.2.2 Calculation Deflection Calculations: Required Moment of Inertia for the Live-Load Deflection Criteria of L/360

𝐼𝑥(𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑) =

Or for W18x50 where Ix = 800.04 in4

𝛥𝑚𝑎𝑥 =

𝛥𝑚𝑎𝑥 =

𝐿 35𝑓𝑡(12 𝑖𝑛⁄𝑓𝑡 ) = = 1.17 𝑖𝑛 360 360

3

5𝑤𝑙 𝑙 4 5(0.750 𝑘𝑖𝑝⁄𝑓𝑡)(35.0 𝑓𝑡)4 (12 𝑖𝑛⁄𝑓𝑡) = = 746 𝑖𝑛4 384𝐸𝛥𝑚𝑎𝑥 384(29,000 𝑘𝑠𝑖)(1.17 𝑖𝑛) 3

5𝑤𝑙 𝑙 4 5(0.750 𝑘𝑖𝑝⁄𝑓𝑡)(35.0 𝑓𝑡)4 (12 𝑖𝑛⁄𝑓𝑡) = = 1.09 𝑖𝑛 384𝐸𝐼𝑥 384(29,000 𝑘𝑠𝑖)(800.04 𝑖𝑛4 )

𝑀𝑎𝑥 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 =

𝐿

𝛥𝑚𝑎𝑥

=

35𝑓𝑡(12 𝑖𝑛⁄𝑓𝑡 ) 𝐿 = 1.0915 384.80

Moment Calculations: The required flexural strength must surpass: 𝑀𝑢 =

1.74 𝑘𝑖𝑝⁄𝑓𝑡 (35 𝑓𝑡)2 = 266𝑘𝑖𝑝 · 𝑓𝑡 8

407

ADVANCE DESIGN VALIDATION GUIDE

From the AISC Manual Table 1-1 for a W18x50 Zx=101 in3 Calculating the Nominal Flexural Strength, Mn 𝑀𝑛 = 𝑀𝑝 = 𝐹𝑦 𝑍𝑥

𝑀𝑛 = 50 𝑘𝑠𝑖 × 101𝑖𝑛3 = 5050 𝑘𝑖𝑝 ∙ 𝑖𝑛 = 421 𝑘𝑖𝑝 ∙ 𝑓𝑡

Finally, the available flexural strength is

𝛷𝑏 𝑀𝑛 = 0.90 × 421 𝑘𝑖𝑝 ∙ 𝑖𝑛 = 379 𝑘𝑖𝑝 ∙ 𝑓𝑡

Work Ratio is therefore

266 𝑘𝑖𝑝𝑠 ∙ 𝑓𝑡 𝑀𝑢 × 100% = × 100% = 70% 𝛷𝑏 𝑀𝑛 379 𝑘𝑖𝑝𝑠 ∙ 𝑓𝑡

12.55.2.3 Reference Results Result name

Result description

Reference value

Max Deflection

L/? for Deflection Criteria

L / 384.80

Actual Deflection

1.09 in

Required Flexural Strength

266 kips·ft

Available Flexural Strength

379 kips·ft

Design Load Ratio due to Strong Axis Bending

70%

Mu ΦbMn

𝛥𝑚𝑎𝑥

Work Ratio, My

12.55.3Calculated results

408

Result name

Result description

Value

Error

Max. deflection

Deflection Criteria

378.242 adim

-1.7043 %

ADVANCE DESIGN VALIDATION GUIDE

12.56 EC3 / NF EN 1993-1-1/NA - France: Verifying critical moment Mcr on a beam with intermediate restraints Test ID: 6279 Test status: Passed

12.56.1Description The test verifies the critical bending moment Mcr of a IPE220 beam made of S235 steel. It ensures Advance Design uses the intermediate restraint definition from the upper flange when the beam is subject to forces acting downwards. The calculations are made according to Eurocode 3, French Annex.

12.56.2Background Critical bending moment verification for a restrained IPE220 beam subjected to axis bending efforts, made of S235 steel. The beam is simply supported.

12.56.2.1 Model description ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■

Exploitation loadings (category A): Q1 = -20 000 N, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in millimeters (mm).

Units Metric System Geometrical properties ■ ■ ■ ■

Beam length: 5m 2 Cross section area: A=3337mm Flexion inertia moment around the Y axis: Iy=2772x104mm4 4 4 Flexion inertia moment around the Z axis: Iz=2049x10 mm

Materials properties S235 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 235 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa.

409

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions The boundary conditions are described below: ■

Outer: Hinged support at start point (x = 0) Hinged support at the end point (z = 5.00) ► Elastic support at start point (x = 0) : KRY = 45 kN.m/°. ► Elastic support at end point (x = 5) : KRY = 45 kN.m/°. Inner: None. ► ►

■

Intermediate restraints The boundary conditions are described below: ■ ■

Top flange: intermediate restraints at x = 1.25m, x = 2.5m, x = 3.75m Bottom flange: intermediate restraint at x = 2. 5m

Loading The column is subjected to the following loadings: ■ ■

410

External: Linear load From X=0.00m to X=5.00m: FZ = N = -20 000 N, Internal: None.

ADVANCE DESIGN VALIDATION GUIDE

12.56.2.2 Critical moment Mcr calculation 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.

M cr = C1 ×

π 2 × E × I z   k z  I w

(k z × L )2

×

 × +   k w  I z 

(k z × L )2 × G × I t + (C × z )2 − C × z  2 2 g g 2  π × E × IZ

(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=2049 x104mm4 It is the torsional inertia: It=90700mm4 IW is the warping inertia (deformation inertia moment): Iw=2.267x1010mm6 L is the distance between intermediate restraints on the compressed flange (in this case : top flange) : L = 1.25m 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 C1 = 1.114 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:

M cr = C1 ×

π ²× E × Iz

M cr = 1.4815 ×

L²

I w L² × G × I t + Iz π ²× E × Iz

×

π ² × 210000 × 204.90 ×10 −8 (1.250)²

×

22670 ×10 −12 (1.25)² × 80800 × 9.04 ×108 + 204.9 ×10 −8 π ² × 210000 × 204.9 ×10 −8

M cr = 355.06kNm Finite elements modeling ■ ■ ■

Linear element: S beam, 21 nodes, 1 linear element.

411

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Elastic critical moment for lateral-torsional buckling Simply supported beam subjected to bending efforts Mcr

12.56.2.3 Reference results Result name

Result description

Reference value

Mcr

Elastic critical moment for lateral-torsional buckling [kNm]

355.06

12.56.3Calculated results Result name

Result description

Value

Error

Mcr

Mcr

355.838 kN*m

0.0000 %

412

ADVANCE DESIGN VALIDATION GUIDE

12.57 EC3 / CSN EN1993-1-1 - Czech Republic: Bended beam without stability failure Test ID: 6285 Test status: Passed

12.57.1Description Classification and verification on bending of simply supported IPE240 beam made of S235 steel. Considered loading consists of dead load and snow load. Self weigth is also considered. Test verifies finite element results for deflections and design forces and steel design results for work ratios in shear and bending and allowable deflections.

12.57.2Background Verification of simple supported steel profile Verifies the simple supported roof beam designed of IPE240 profile without stability failure. During this test is verified bearing capacity for ultimate limit state and deflections for serviceability limit state.

12.57.2.1 Model description ■

Reference: "Martina Eliášková, Zdeněk Sokol: Ocelové konstrukce - Príklady"; Chapter "1.1 Ohyb nosníku bez ztráty stability" ■ Analysis type: static linear (plane problem); ■ Element type: linear. Load cases and load combinations: ■ ■ ■ ■

Dead load: g = 2.20 kN/m, γG = 1.2 Snow load: s = 4.80 kN/m, γS = 1.4 ULS combination: CULS = 1.2*G + 1.4*S SLS combination: CSLS = 1.0*G + 1.0*S

Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■ ■

Height: h = 240 mm, Width: b = 120 mm, Length: L = 7200 mm, 2 Section area: A = 3912 mm , Bending moment of inertia around y-axis: Iy = 38.97*106 mm4, 2 Reduced cross-section: Avz = 1914,0 mm , IPE 2 Weight: g = 30.7 kg/m ,

413

ADVANCE DESIGN VALIDATION GUIDE

■ ■ ■ ■

Compressed part of web: d = 194 mm Web thickness: tw = 6.2 mm Compressed part of flange: c = 60.0 mm Glange thickness: tf = 9.8 mm

Materials properties Steel: S235 Material coefficient: γM0 = 1.15 Boundary conditions The boundary conditions are described below: ■

Outer: Support at start point (x=0) , hinged connection, Support at end point (x = 7.20) restrained in translation along Y and Z, in rotation around X. Inner: None.

► ►

■

Loading The beam is subjected to the following load combinations: Load combinations: ■

Ultimate Limit State: NULS = 1.2*(g + gIPE) + 1.4*s = 1.2*(2.2 + 0.307) + 1.4*4.80 = 9.73 kN/m Serviceability Limit State: IPE NSLS = 1.0*(G + G ) + 1.0*S = 1.0*(2.2 + 0.307) + 1.0*4.80 = 7.31 kN/m

■

12.57.2.2 Cross-section classification Web d 194.0 235 = = 31.29 < 72 ∗ ε = 72 ∗ � = 72.0 tw 6.2 fy

According to: CSN EN1993-1-1 Chapter 5.6, Table 5.2 (Sheet 1)

Flange cf tf

=

60.0 9.8

= 6.1 < 10 ∗ ε = 10.0 Table 5.2 (sheet 2)

235

𝜀 = � 𝑓 = 1.0 𝑦

Cross-section belongs to class 1

414

According to: CSN EN1993-1-1 Chapter 5.6, Table 5.2 (Sheet 2)

ADVANCE DESIGN VALIDATION GUIDE

12.57.2.3 Verification of bearing capacity for Ultimate Limit State Bending resistance verification Design value of bending moment from loading:

Bending resistance of IPE240:

𝑀𝐸𝑑 = 𝑀𝑝𝑙,𝑅𝑑 =

Verification:

63.,

74.9

1 1 𝑁 ∗ 𝐿2 = 9.73 ∗ 7.22 = 63.0 𝑘𝑁𝑚 8 𝑈𝐿𝑆 8

𝑊𝑝𝑙 ∗ 𝑓𝑦 366.6 ∗ 103 ∗ 235 = = 74.9 𝑘𝑁𝑚 𝛾𝑀0 1.15

According to: CSN EN1993-1-1 Chapter 5.2.5(2), formula (6.13)

𝑀𝐸𝑑 ≤ 1.0 𝑀𝑝𝑙,𝑅𝑑

= 0.84 < 1.0

=> Work ratio for bending is 84% According to: CSN EN1993-1-1 Chapter 5.2.5(1), formula (6.12) Shear resistance verification Design value of shear force from loading:

Shear resistance of IPE240:

𝑉𝐸𝑑 =

1 1 𝑁 ∗ 𝐿 = 9.73 ∗ 7.2 = 35.0 𝑘𝑁 2 𝑈𝐿𝑆 2

𝑉𝑝𝑙,𝑅𝑑 = Verification:

𝐴𝑣𝑧 ∗ 𝑓𝑦

𝛾𝑀0 ∗ √3

=

1914 ∗ 235 1.15 ∗ √3

= 225.8 𝑘𝑁

According to: CSN EN1993-1-1 Chapter 5.2.6(2), formula (6.18)

𝑉𝐸𝑑 ≤ 1.0 𝑉𝑝𝑙,𝑅𝑑

=> Work ratio for shear is 16 %

35.0 = 0.16 < 1.0 225.8

According to: CSN EN1993-1-1 Chapter 5.2.6(1), formula (6.17)

Finite elements modeling ■ Linear element: S beam, ■ 8 nodes, ■ 1 linear element.

415

ADVANCE DESIGN VALIDATION GUIDE

Finite element results (Ultimate Limit State) Design value of bending moment My (reference value 63.0 kNm)

Design value of shear force Fz (reference value 35.0 kN)

Steel design results Work ratio in bending (reference value 84%)

Work ratio in shear (reference value 16%)

12.57.2.4 Verification of deflections for Serviceability Limit State Deflection from live load (snow):

Total deflection:

𝛿2 = 𝛿𝑚𝑎𝑥 =

5 ∗ 𝑠 ∗ 𝐿4 5 ∗ 4.80 ∗ 7.24 = = 20.6 𝑚𝑚 384 ∗ 𝐸 ∗ 𝐼𝑦 384 ∗ 210 ∗ 38.92 5 ∗ 𝑁𝑆𝐿𝑆 ∗ 𝐿4 5 ∗ 7.37 ∗ 7.24 = = 31.3 𝑚𝑚 384 ∗ 𝐸 ∗ 𝐼𝑦 384 ∗ 210 ∗ 38.92

Deflections of roof beams must fulfill following conditions:

■ Deflection from live loads: δ2 ≤ L/250 20.6 mm < 28.8 mm => Work ratio for live load deflections is 72% ■ Total deflection: δmax ≤ L/200 31.3 mm < 36.0 mm => Work ratio for total deflections is 87% ■

416

According to: CSN EN1993-1-1 National Annex NA.2.22, Table NA.1

ADVANCE DESIGN VALIDATION GUIDE

Finite element results (Serviceability Limit State) Live load deflection (reference value 20.6 mm)

Total deflection (reference value 31.3 mm)

Steel design results Live load deflection work ratio (reference value 72%)

Total deflection work ratio (reference value 87%)

12.57.2.5 Reference results Result name

Result description

Reference value

My

Design value of bending moment for ULS load combination [kNm]

63.0 kNm

Fz

Design value of shear force for ULS load combination [kN]

35.0 kNm

Work ratio - Oblique

Maximal work ratio for bending resistance [%]

84 %

Work ratio - Fx

Maximal work ratio for shear resistance [%]

16 %

D

Live load deflection for ULS load combination [mm]

20.6 mm

D

Total deflection for ULS load combination [mm]

31.3 mm

Deviat.all.defl.

Deviation of allowed live load deflections [%]

72 %

Deviat.all.defl.

Deviation of allowed total deflections [%]

87 %

12.57.3Calculated results Result name

Result description

Value

Error

D

Deflection total

31.5644 mm

0.8447 %

D

Deflection snow

20.7514 mm

0.7350 %

My

Design moment

-62.9946 kN*m

0.0086 %

Fz

Design shear force

34.997 kN

-0.0086 %

Deviat. all. defl.

WR Deflection total

87.679 %

0.7805 %

WR Deflection snow

72.0535 %

0.0743 %

WR Oblique

84.0893 %

0.1063 %

WR Fz

15.492 %

-0.0516 %

Deviat. all. defl. Work ratio Oblique Work ratio - Fz

-

417

ADVANCE DESIGN VALIDATION GUIDE

12.58 AISC 360 2010 - United States of America: Example G.1 W-shape in strong axis shear (LRFD) Test ID: 6287 Test status: Passed

12.58.1Description Determines the available shear strength and adequacy of a W24x62 ASTM A992 beam using the AISC Manual with end shears of 48 kips from dead load and 145 kips from live load. Obtains the available shear strength, which is determined by the tabulated values of the AISC Manual.

12.58.2Background Determines the shear applied to the cross-section, its resistance to shear and finally the work ratio due to shear alone. The objective of this test is to check the calculation of steel resistance due to shearing forces.

12.58.2.1 Model Description ■ Reference: AISC Design Example Version 14.1 ■ Example as taken from “Example E.1A W-Shape Column Design with Pinned Ends” Example G.1A Determine the available shear strength and adequacy of a W24x62 ASTM A992 beam using the AISC Manual with end shears of 48 kips from dead load and 145 kips from live load Example G.1B The available shear strength, which can be easily determined by the tabulated values of the AISC Manual, can be verified by directly applying the provisions of the AISC Specification. Determine the available shear strength for the W-shape in Example G.1A by applying the provisions of the AISC Specification. ■ ■ ■

Analysis Type: static Element Type: linear Steel cross-section used is and AISC W24x62

The following load cases and load combination are used: ■ ■ ■ ■

Dead load: 48 kips (Self-weight included in this value – Therefore not needed for AD to calculate) Live Load: 145 kips axial force Unit System: IMPERIAL Steel Type: ASTM A992 Fy = 50 ksi Fu = 65 ksi

418

ADVANCE DESIGN VALIDATION GUIDE

12.58.2.2 Results Combination used for this example is (ULS), Vu: 1.2 × 𝐷𝑒𝑎𝑑 + 1.6 × 𝐿𝑖𝑣𝑒

1.2 × (48 𝑘𝑖𝑝𝑠) + 1.6 × (145 𝑘𝑖𝑝𝑠) 57.6 𝑘𝑖𝑝𝑠 + 232 𝑘𝑖𝑝𝑠 289.6 𝑘𝑖𝑝𝑠

“G2.1 Members with Unstiffened or Stiffened webs” is applicable to webs for the shear strength for singly or doubly symmetric members and channels subject to shear in the plane of the web The nominal shear strength, Vn, of unstiffened or stiffened webs according to the, limit states of shear yield and shear buckling, is

Part (a) for webs of rolled I-shaped member with

𝑉𝑛 = 0.6𝐹𝑦 𝐴𝑤 𝐶𝑣 ℎ ≤ 2.24 �𝐸 ⁄𝐹𝑦 𝑡𝑤

Where Φv = 1 and Cv = 1 and

Therefore,

And finally

Work Ratio

21.06 𝑖𝑛 ≤ 2.24 �29,000 𝑘𝑠𝑖⁄50 𝑘𝑠𝑖 0.43 𝑖𝑛 48.98 ≤ 53.95

𝐴𝑤 = 𝑑𝑡𝑤 = 23.7 𝑖𝑛 × 0.430 𝑖𝑛 = 10.2 𝑖𝑛2 𝑉𝑛 = 0.6 × 50 𝑘𝑠𝑖 × 10.2 𝑖𝑛2 × 1.0 = 306 𝑘𝑖𝑝𝑠 𝛷𝑣 𝑉𝑛 = 1.0 × 306 𝑘𝑖𝑝𝑠 289.6 𝑘𝑖𝑝𝑠 𝑉𝑢 × 100% = × 100% = 94.65% 𝛷𝑣 𝑉𝑛 306 𝑘𝑖𝑝𝑠

419

ADVANCE DESIGN VALIDATION GUIDE

12.58.2.3 Theoretical result Result Name

Result Description

Reference

Vu

Required Shear Strength

289.6 kips

ΦvVn

Available Shear Strength

306 kips

Work Ratio

Design Load Rate

94.65 %

12.58.3Calculated results

420

Result name

Result description

Value

Error

Fz

Shear Verify (not forces)

-289.993 kip

0.0024 %

ADVANCE DESIGN VALIDATION GUIDE

12.59 EC3 / EN 1993-1-1 - United Kingdom: Simply supported laterally restrained (from P374 Hollow Sections Example 3) Test ID: 6345 Test status: Passed

12.59.1Description The 250x150x16 RHS in S355 beam is fully restrained, loading including a UDL and mid-span point load. These are the steel design results for example 3 from P374 from the SCI.

12.59.2Background The 250x150x16 RHS beam in S355 steel is fully restrained laterally along its length. The design aspects covered in this example are: ■ ■ ■

■ ■

Calculation of design values of actions for ULS and SLS Cross section classification Cross-sectional resistance: ► Shear buckling ► Shear ► Bending moment Resistance of web to transverse forces Vertical deflection of beam at SLS.

12.59.2.1 Model description ■ ■ ■

Reference: SCI Publication P374; Analysis type: static linear (3D problem); Element type: linear.

Units Metric System Geometry Below are described the column cross section characteristics: ■ ■ ■ ■ ■ ■ ■ ■

Depth: h = 250.0 mm Width: b = 150.0 mm Wall thickness: t = 16.0 mm Second moment of area about the y-axis: Iy = 8880 cm4 Radius of gyration about the y-axis iy = 8.79 cm Radius of gyration about the z-axis iz = 5.80 cm Plastic modulus about the y-axis Wpl,y = 906 cm3 Cross sectional area: A = 115 cm2

421

ADVANCE DESIGN VALIDATION GUIDE

For buildings that will be built in the UK the nominal values of the yield strength (fy) and the ultimate strength (fu) for structural steel should be those obtained from the product standard. Where a range is given the lowest nominal value should be used. Materials properties For S355 steel and t ≤ 16 mm: ■

Yield strength: fy = ReH = 355 N/mm

2

Loading Permanent actions ■

Uniformly distributed load: 𝑔1 = 3 𝑘𝑁

■ Concentrated load: 𝐺2 = 40 𝑘𝑁 Variable actions ■

Uniformly distributed load: 𝑞1 = 3 𝑘𝑁

■ Concentrated load: 𝑄2 = 50 𝑘𝑁 The variable actions given above are not due to storage and are not independent of each other. There are no other variable actions to be considered. Partial factors for actions For the design of structural members not involving geotechnical actions, the partial factors for actions to be used for ultimate limit state strength verifications should be obtained from Table A1.2(B). Note 2 to Table A1.2(B) allows the National Annex to specify different values for the partial factors. ■ ■ ■

Partial factor for permanent actions 𝛾𝐺 = 1.35 Partial factor for variable actions 𝛾𝑄 = 1.50

Reduction factor 𝜉 = 0.925

For this example the factor for the combination value of a variable action is: 𝜓0 = 0.7 Design values of combined actions for ultimate limit state

BS EN 1990 presents two methods for determining the effects due to combination of actions for the ultimate limit state verification for the resistance of a structural member. The methods are to use expression (6.10) on its own or to determine the less favourable of the values from expressions (6.10a) and (6.10b). Note 1 to Table NA.A1.2(B) in the UK National Annex to BS EN 1990 allows either method to be used. The second method using expressions (6.10a) and (6.10b) is used here. Therefore the design values are taken as the most onerous values obtained from the following expressions: 𝛾𝐺𝑗,𝑠𝑢𝑝 𝐺𝑗,𝑠𝑢𝑝 + 𝛾𝑄,1 𝜓0,1 𝑄1 + 𝛾𝑄,𝑖 𝜓0,𝑖 𝑄𝑖 𝜉𝛾𝐺𝑗,𝑠𝑢𝑝 𝐺𝑗,𝑠𝑢𝑝 + 𝛾𝑄,1 𝑄1 + 𝛾𝑄,𝑖 𝜓0,𝑖 𝑄𝑖

(6.10a) (6.10b)

Here Qi is not required as the variable actions are not independent of each other and expression 6.10b gives the more onerous value. The design values are:

■ ■

Combination of uniformly distributed loads: 𝐹𝑑,1 = 𝜉𝛾𝐺 𝑔1 + 𝛾𝑄 𝑞1 = (0.925 × 1.35 × 3) + (1.35 × 3) = 8.2 kN/m

Combination of concentrated loads: 𝐹𝑑,2 = 𝜉𝛾𝐺 𝐺2 + 𝛾𝑄 𝑄2 = (0.925 × 1.35 × 40) + (1.5 × 50) = 125 kN/m

12.59.2.2 Design bending moments and shear forces at ultimate limit state Span of beam L = 5000 mm Maximum value of the design bending moment occurs at the mid-span: 𝑀𝐸𝑑 =

𝐹𝑑,1 𝐿2 𝐹𝑑,2 𝐿 8.2 × 52 125 × 5 + = + = 182.0 𝐾𝑁𝑚 8 8 4 4

Maximum design value of shear occurs at the supports: 422

ADVANCE DESIGN VALIDATION GUIDE

𝑉𝐸𝑑 =

𝐹𝑑,1 𝐿 𝐹𝑑,2 8.2 × 5 125 + = + = 83.0 𝐾𝑁𝑚 2 2 2 2

Design value of shear force at the mid-span:

𝑉𝐸𝑑,𝐶 = VEd −

𝐹𝑑,1 𝐿 8.2 × 5 = 83 − = 62.5 𝐾𝑁𝑚 2 2

The design shear force and bending moment diagrams are shown in Figure 3.2.

Figure 3.2

12.59.2.3 Cross section classification 𝜀=�

235 235 =� = 0.81 𝑓𝑦 355

Internal part subject to bending (web) c = h - 3t = 250 - 3 x 16 = 202 mm 𝑐 202 = = 12.63 𝑡 16

The limiting value for Class 1 is

𝑐 𝑡

≤ 72𝜀 = 72 × 0.81 = 58.32

12.63 < 58.32 therefore the internal part in bending is Class 1. Internal part subject to compression (flange) c = b - 3t = 150 - 3 x 16 = 102 mm 𝑐 102 = = 6.38 16 𝑡

The limiting value for Class 1 is

𝑐 𝑡

≤ 33𝜀 = 33 × 0.81 = 26.73

6.38 < 26.73 therefore the internal part in compression is Class 1. Therefore the section is Class 1 for bending about the y-y axis.

12.59.2.4 Partial factors for resistance 𝛾𝑀0 = 1.0 𝛾𝑀1 = 1.0

423

ADVANCE DESIGN VALIDATION GUIDE

12.59.2.5 Cross-sectional resistance Shear buckling The shear buckling resistance for webs should be verified according to Section 5 of BS EN 1993-1-5 if:

ℎ𝑤 72𝜀 = 𝑡𝑤 𝜂

𝜂 = 1.0 (conservative)

ℎ𝑤 = ℎ − 2𝑡 = 250 − 2 × 16 = 218 mm

ℎ𝑤 218 = = 13.6 𝑡𝑤 16

72𝜀 72 × 0.81 = = 58.3 𝜂 1.0

13.6 < 58.3

Therefore the shear buckling resistance of the web does not need to be verified. Shear resistance

Verify that: 𝑉𝐸𝑑

𝑉𝑐,𝑅𝑑

≤ 1.0

For plastic design, 𝑉𝑐,𝑅𝑑 is the design plastic shear resistance (𝑉𝑝𝑙,𝑅𝑑 ). 𝑉𝑐,𝑅𝑑 = 𝑉𝑝𝑙,𝑅𝑑 =

𝐴𝑣 �𝑓𝑦 /√3� 𝛾𝑀0

𝐴𝑣 is the shear area and is determined as follows for rolled RHS sections with the load applied parallel to the depth. 𝐴𝑣 =

𝐴ℎ 11500 × 250 = = 7187.5 𝑚𝑚2 𝑏+ℎ 150 + 250

𝑉𝑐,𝑅𝑑 =

𝐴𝑣 �𝑓𝑦 /√3� 7187.5 × �355/√3� = × 10−3 = 1473 kN 1.0 𝛾𝑀0

Maximum design shear 𝑉𝐸𝑑 = 83.0 𝑘𝑁

𝑉𝐸𝑑 83 = = 0.06 < 1.0 𝑉𝑐,𝑅𝑑 1473

Therefore the shear resistance of the RHS is adequate. Resistance to bending Verify that: 𝑀𝐸𝑑

𝑀𝑐,𝑅𝑑

≤ 1.0

At the point of maximum bending moment (mid-span) check whether the shear force will reduce the bending resistance of the section.

𝑉𝑐,𝑅𝑑 2

=

1473 2

= 736.5 𝑘𝑁

𝑉𝐸𝑑,𝑐 = 62.5 𝑘𝑁 < 736.5 𝑘𝑁

Therefore no reduction in bending resistance due to shear is required. The design resistance for bending for Class 1 and 2 cross sections is:

𝑀𝑐,𝑅𝑑 = 𝑀𝑝𝑙,𝑅𝑑 = 424

𝑊𝑝𝑙,𝑦 𝑓𝑦 906 × 103 × 355 = × 10−6 = 322 kNm 𝛾𝑀0 1.0

ADVANCE DESIGN VALIDATION GUIDE

𝑀𝐸𝑑

𝑀𝑐,𝑅𝑑

=

182 322

= 0.57 < 1.0

Therefore the bending resistance is adequate.

12.59.2.6 Reference results Result name

Result description

Reference value

Steel Strength – Work ratio - Fz

Shear resistance

5.6%

Steel Strength – Work ratio - Oblique

Resistance to bending

56.5%

12.59.3Calculated results Result name

Result description

Value

Error

Work ratio - Fz

Fz

5.63971 %

0.7091 %

Work ratio - Oblique

Oblique

56.5734 %

0.1299 %

Work ratio

Work Ratio

56.5734 %

0.1299 %

425

ADVANCE DESIGN VALIDATION GUIDE

12.60 EC3 / EN 1993-1-1 - United Kingdom: Pin-ended column (from P374 Hollow Sections Example 2) Test ID: 6344 Test status: Passed

12.60.1Description The pin-ended 200x200x6.3 SHS in S355 column is subject to compression. These are the steel design results for example 2 from P374 from the SCI.

12.60.2Background A pin-ended, hot finished 200x200x6.3 SHS column in S355 steel is subjected to compression. The design aspects covered in this example are: ► ► ►

Cross section classification Cross-sectional resistance to axial compression Flexural buckling resistance.

12.60.2.1 Model description ■ ■ ■

Reference: SCI Publication P374; Analysis type: static linear (3D problem); Element type: linear.

Units Metric System Geometry Below are described the column cross section characteristics: ■ ■ ■ ■ ■

426

Depth: h = 200 mm Width: b = 200 mm Wall thickness: t = 6.3 mm Radius of gyration: i = 7.89 cm 2 Cross sectional area: A = 48.40 cm

ADVANCE DESIGN VALIDATION GUIDE

Materials properties S355 steel is used. The following characteristics are used in relation to this material: 2

■ Yield strength fy = ReH = 355 N/mm For buildings that will be built in the UK the nominal values of the yield strength (fy) and the ultimate strength (fu) for structural steel should be those obtained from the product standard. Where a range is given the lowest nominal value should be used.

12.60.2.2 Cross section classification 𝜀=�

235 235 =� = 0.81 𝑓2 355

Internal compression parts 𝑐 = ℎ − 3𝑡 = 200 − 3𝑥6.3 = 181.1𝑚𝑚

𝑐 181.1 = = 28.75 6.3 𝑡

The limiting value for Class 1 is The limiting value for Class 2 is

𝑐 𝑡

𝑐 𝑡

≤ 33𝜀 = 33𝑥0.81 = 26.73 ≤ 38𝜀 = 38𝑥0.81 = 30.78

26.73 < 28.75 < 30.78, therefore the internal compression parts are Class 2. As b = h only one check is required; therefore the cross section is Class 2.

12.60.2.3 Partial factors for resistance 𝛾𝑀0 = 1.0 𝛾𝑀1 = 1.0

12.60.2.4 Cross-sectional resistance Compression resistance Verify that

𝑁𝐸𝑑

𝑁𝑐,𝑅𝑑

= 1.0 ,

the design resistance of the cross section to compression, is: 𝑁𝑐,𝑅𝑑 =

𝑁𝑐,𝑅𝑑 = 𝑁𝐸𝑑

𝑁𝑐,𝑅𝑑

=

𝐴×𝑓𝑦 𝛾𝑀0

(For Class 1, 2 and 3 cross sections)

4840 × 355 × 10−3 = 1718 𝐾𝑛 1.0 920

1718

= 0.54 < 1.0

Therefore the compressive resistance of the SHS cross section is adequate.

427

ADVANCE DESIGN VALIDATION GUIDE

12.60.2.5 Flexural buckling resistance Verify that

𝑁𝐸𝑑

𝑁𝑐,𝑅𝑑

= 1.0 ,

the design buckling resistance, is determined from: 𝑁𝑐,𝑅𝑑 =

𝜒𝐴𝑓𝑦

(For Class 1 and 2 cross sections)

𝛾𝑀1

𝜒 is the reduction factor for the relevant buckling mode and is determined from:

𝜒=

1

𝜙 + �𝜙 − 𝜆̅2

but 𝜒 ≤ 1.0

Where: 𝜙 = 0.5�1 + 𝛼�𝜆̅ − 0.2� + 𝜆̅2 �

For hot finished SHS in S355 steel, use buckling curve ‘a’ For buckling curve ‘a’ the imperfection factor 𝛼 = 0.21

For flexural buckling the slenderness is determined from: 𝐴𝑓𝑦 𝐿 1 𝜆̅ = �𝑁 = � 𝑖𝑐𝑟 � �𝜆 � (For Class 1, 2 and 3 cross sections) 𝑐𝑟

1

𝜆̅1 = 93.9𝜀 = 93.9 × 0.81 = 76.1

The buckling length about both axes is 𝐿𝑐𝑟 = 𝐿 = 6000𝑚𝑚

As the cross section is square 𝜆̅y = 𝜆̅z

6000 1 Lcr 1 �� � = 1.00 𝜆̅ = � � � � = � i 𝜆1 78.9 76.1

𝜙 = 0.5�1 + 𝛼�𝜆̅ − 0.2� + 𝜆̅2 � = 0.5(1 + 0.21 × (1.0 − 0.2) + 1. 02 ) = 1.08

𝜒=

1

ϕ + ��ϕ2 − 𝜆̅2 �

Therefore, 𝜒 = 0.67 𝑁𝐸𝑑 = 𝑁𝐸𝑑

𝑁𝑏,𝑅𝑑

=

=

1

1.08 + �(1.082 − 1.02 )

= 0.67

𝜒𝐴𝑓𝑦 0.67 × 4840 × 355 = × 10−3 = 1151 𝑘𝑁 𝛾𝑀1 1.0 920

1151

= 0.80 < 1.0

Therefore, the flexural buckling resistance of SHS is adequate.

428

ADVANCE DESIGN VALIDATION GUIDE

12.60.2.6 Reference results Result name

Result description

Reference value

Steel Strength – Work ratio - Fx

Compression resistance

53%

Steel Stability – Work ratio

Flexural buckling

80%

12.60.3Calculated results Result name Work ratio - Fx Work ratio

Result description Fx Work Ratio

Value 53.5593 %

Error -0.8161 %

80.0974 %

0.1217 %

429

ADVANCE DESIGN VALIDATION GUIDE

12.61 EC3 / EN 1993-1-1 - United Kingdom: Laterally unrestrained beam (from P374 Hollow Sections Example 4) Test ID: 6346 Test status: Passed

12.61.1Description The 250x150x16 RHS in S355 beam is fully restrained, loading including a UDL and mid-span point load. These are the steel design results for example 3 from P374 from the SCI.

12.61.2Background The 250x150x16 RHS in S355 beam is fully restrained, loading including a UDL and mid-span point load. These are the steel design results for example 3 from P374 from the SCI. The design aspect covered in this example is: ■

Lateral torsional buckling

12.61.2.1 Model description ■ ■ ■

Reference: SCI Publication P374; Analysis type: static linear (3D problem); Element type: linear.

Units Metric System Geometry Below are described the column cross section characteristics: ■ Depth: h = 250.0 mm ■ Width: b = 150.0 mm ■ Wall thickness: t = 16.0 mm ■ Second moment of area about the y-axis: Iy = 8880 cm4 ■ Second moment of area about the z-axis: Iz = 3870 cm4 ■ Plastic modulus about the y-axis Wpl,y = 906 cm3 ■ St Venant torsional constant: IT = 8870 cm2 For buildings that will be built in the UK the nominal values of the yield strength (fy) and the ultimate strength (fu) for structural steel should be those obtained from the product standard. Where a range is given the lowest nominal value should be used. Materials properties For S355 steel and t ≤ 16 mm: ■

430

Yield strength: fy = ReH = 355 N/mm2

ADVANCE DESIGN VALIDATION GUIDE

Design shear force and moments at ultimate limit state The design shear force and bending moment diagrams at ULS are shown in the figure below:

12.61.2.2 Lateral torsional buckling If the slenderness for lateral torsional buckling (𝜆̅𝐿𝑇 ) is less than or equal to 𝜆̅𝐿𝑇,0 the effects of lateral torsional buckling may be neglected, and only cross-sectional verifications apply. In the UK National Annex the value of 𝜆̅𝐿𝑇,0 for rolled sections is given as 𝜆̅𝐿𝑇,0 = 0.4 𝜆̅𝐿𝑇 = �

𝑊𝑦 × 𝑓𝑦 𝑀𝑐𝑟

𝑊𝑦 = 𝑊𝑝𝑙,𝑦 For class 1 or 2 cross sections.

BS EN 1993-1-1 does not give an expression for the determination of the elastic critical buckling moment for lateral torsional buckling (𝑀𝑐𝑟 ). Access Steel document SN003 presents expressions that may be used to determine 𝑀𝑐𝑟 . The general expression for doubly symmetrical sections is: 𝑀𝑐𝑟 = 𝐶1 where:

𝜋 2 𝐸𝐼𝑧 𝑘 2 𝐼𝑤 (𝑘𝐿)2 𝐺𝐼𝑇 2 � �� � + + �𝐶2 𝑧𝑔 � − 𝐶2 𝑧𝑔 � 2 (𝑘𝐿) 𝜋 2 𝐸𝐼𝑧 𝑘𝑤 𝐼𝑧

𝐿 is the member length between points of lateral restraint

𝑧𝑔 is the distance between the shear centre and the point of application of the transverse loading

𝑘 and 𝑘𝑤 are effective length factors

𝐶1 & 𝐶2 are coefficients which depend on the shape of the bending moment and the end restraint conditions

𝑘𝑤 = 1.0 (unless advised otherwise)

𝑘 = 1.0 (for simply supported members). 𝐼

For RHS the effects of warping � 𝐼𝑤 � are negligible compared with the effects of torsion �

when determining 𝑀𝑐𝑟 .

𝑧

(𝑘𝐿)2 𝐺𝐼𝑇 𝜋2 𝐸𝐼𝑧

�, so may be neglected

Also, the effect of applying of the actions above or below the shear centre of the section may be neglected for RHS sections (i.e. take 𝑧𝑔 = 0). 𝑘

𝐼

Therefore, the terms �𝑘 � 𝐼𝑤 and 𝐶2 𝑧𝑔 may be removed from the expression and the following simplified expression used for RHS sections: 𝑀𝑐𝑟 = 𝐶1

𝑤

𝑧

𝜋 2 𝐸𝐼𝑧 (𝑘𝐿)2 𝐺𝐼𝑇 � 2 (𝑘𝐿)2 𝜋 𝐸𝐼𝑧 431

ADVANCE DESIGN VALIDATION GUIDE

For the design bending moment shown in Figure 4.2, 𝐶1 is: 𝐶1 = 1.127

𝑀𝑐𝑟 = 1.127 × 𝜆̅𝐿𝑇 = �

(1 × 5000)2 × 81000 × 8870 × 104 𝜋 2 × 210000 × 3870 × 104 × �� � = 5500 × 106 𝑁𝑚𝑚 2 (1 × 5000) 𝜋 2 × 210000 × 3870 × 104

𝑊𝑦 × 𝑓𝑦 906 × 103 × 355 =� = 0.24 𝑀𝑐𝑟 5500 × 106

Since 𝜆̅𝐿𝑇 < 𝜆̅𝐿𝑇,0 (0.24 < 0.4) the lateral torsional buckling effects are neglected and only cross-sectional verifications apply.

Since the cross-sectional verifications are satisfied in Example 3, the 250 X 150 X 16 RHS in S355 steel is adequate to be unrestrained between the supports shown in Figure 4.1.

12.61.2.3 Reference results Result name

Result description

Reference value

Mcr 𝜆̅𝐿𝑇

Steel Stability Mcr Steel Stability 𝜆̅𝐿𝑇

5500 kNm 0.24

12.61.3Calculated results

432

Result name

Result description

Value

Error

Mcr

Mcr

5400.35 kN*m

-1.8118 %

ADVANCE DESIGN VALIDATION GUIDE

12.62 NTC 2008 - Italy: Verifying the lateral torsional buckling of a IPE300 beam Test ID: 6348 Test status: Passed

12.62.1Description The test verifies the lateral-torsional buckling of a simply supported IPE300 beam made of S235 steel. The calculations are made according Italian standard NTC 2008.

12.62.2Background 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.

12.62.2.1 Model description ■ ■ ■ ■ ■

Analysis type: static linear (plane problem); Element type: linear. The following load case and load combination are used: Exploitation loadings (category A): Q1 = -10kN/m; Cross section dimensions are in millimeters (mm).

Units Metric System

Geometrical properties ■ ■ ■ ■ ■ ■

Column length: L = 5000 mm Cross section area: A = 5310 mm2 4 Flexion inertia moment around the Y axis: Iy = 8356.00x104 mm 4 Flexion inertia moment around the Z axis: Iz = 603.80x104 mm 4 Torsional moment of inertia: It = 20.12x104 mm 6 Warping inertial moment: Iw = 12.59x1010 mm

433

ADVANCE DESIGN VALIDATION GUIDE

Materials properties S235 steel material is used. The following characteristics are used: ■ ■ ■

Yield strength fyk = 235 MPa, Longitudinal elastic modulus: E = 210000 MPa. Shear modulus of rigidity: G = 80800MPa

Boundary conditions The boundary conditions are described below: Outer: ■ ■

Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at start point (x = 5m) restrained in translation along X and Y axis, and restrained in rotation along X axis

Loadings The linear load applied along the beam is equal to 10 kN/m

12.62.2.2 Reference calculation Cross section Class Cross-class classification is made according to table 4.2.I from NTC 2008

434

■

for beam web:

■

for beam flange:

ADVANCE DESIGN VALIDATION GUIDE

435

ADVANCE DESIGN VALIDATION GUIDE

Lateral torsional buckling verification The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the elements 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. The design buckling resistance of the compressed member, Mb,Rd, is calculated according to NTC 2008 Chapter 4.2.4.1.3.2.

Where: ■ MEd is the designed value moment ■ Mb,Rd is the design buckling moment resistance The design buckling moment resistance against lateral-torsional buckling is calculated with the formula 4.2.50 from NTC 2008:

Where: Wy is the resistance module, which is equal to plastic module Wpl,y for section in class 1 and 2, or equal to Wel,y for section in class 3 and it could be taken as Weff,y for section in class 4. ΧLT is the reduction factor for lateral – torsional buckling defined below “fyk” is the steel yield strength γM1 is the safety coefficient taken from table 4.2.V below for elements stability

Where:

436

ADVANCE DESIGN VALIDATION GUIDE

The coefficient αLT represents the imperfection factor according the indications in the Table 4.2.VII from Chapter 4.2.4.1.3 from NTC 2008.

The coefficient β can be taken as 1 (not less than 0.75 value). The coefficient λLT,0 can be taken as 0.2 or its value doesn’t exceed 0.4. The factor f consider the real distribution of the bending moment between 2 torsional restraint and it is calculated:

Where the coefficient kc depends on the load distribution according Table 4.2.VII from NTC 2008

The dimensional slenderness coefficient λLT is given by the formula 4.2.52 from NTC 2008.

The elastic moment for lateral-torsional buckling calculation, Mcr is calculated as follows: The transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:

M cr = C1 ×

π ² × E × Iz L²

×

I w L² × G × It + Iz π ² × E × Iz

According to EN 1993-1-1; Chapter 2

437

ADVANCE DESIGN VALIDATION GUIDE

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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ²

According to EN 1993-1-1 ; Chapter 3

𝑀𝑐𝑟

= 1.77 ×

𝑁 × 603.80 × 104 𝑚𝑚2 (5000 𝑚𝑚)2

𝜋 2 × 210000

𝑁 4 4 (5000)2 × 80800 12.59 × 1010 𝑚𝑚6 2 × 20.12 × 10 𝑚𝑚 𝑚𝑚 × � + 𝑁 603.80 × 104 𝑚𝑚4 𝜋 2 × 210000 × 603.80 × 104 𝑚𝑚4 𝑚𝑚2 = 1.77 𝑥 500578.44𝑁 × 230.92𝑚𝑚 = 204607478 𝑁𝑚𝑚 = 204.60 𝑘𝑁𝑚

Note:

The formula 4.2.52 defines the slenderness λLT

𝜆𝐿𝑇 = �

𝑊𝑦 × 𝑓𝑦 628.4 × 103 𝑚𝑚3 × 235 𝑁/𝑚𝑚2 = � = 0.849 𝑀𝑐𝑟 204607478 𝑁/𝑚𝑚

The distribution coefficient f is calculated as follows, where the coefficients kc = 0.94 from the table 4.2.VIII and λLT = 0.849 2 f = 1 – 0.5 x (1 – 0.94) x [1 – 2.0 x (0.849 - 0.8) ] = 0.971

So the factor is defined as follows: ΦLT = 0.5 x [1 + 0.34 x (0.849 – 0.2) + 1 x 0.8492] = 0.970

ΧLT =

1

0.971

×

1

0.970+√0.9702 −0.8492

= 0.715 < 1

The design buckling moment resistance is calculated according the formula 4.2.50 in NTC 2008: 235

𝑀𝑏,𝑅𝑑 = 0.71 × 628400 × 1.05 = 99.85 𝑘𝑁𝑚 > 𝑀𝐸𝑑 = 46.88 𝑘𝑁𝑚 → 𝑂𝐾 The work ratio is calculated with formula 4.2.29:

𝑀𝐸𝑑 46.88 = = 0.469 < 1 (47%) 𝑀𝑏,𝑅𝑑 99.85

438

ADVANCE DESIGN VALIDATION GUIDE

12.62.2.3 Reference results Result name

Result description The appropriate non-dimensional slenderness

Reference value 0.715

Work Ratio

Stability work ratio

0.469

χ LT

12.62.3Calculated results Result name

Result description

Value

Error

XLT

Reduction factor for lateral-torsional buckling

0.71492 adim

-0.0112 %

Work ratio

Stability work ratio

46.6197 %

-0.5977 %

439

ADVANCE DESIGN VALIDATION GUIDE

12.63 EC3 / CSN EN 1993-1-1 - Czech Republic: Compressed U_profile Test ID: 6265 Test status: Passed

12.63.1Description Verifies the classification and the compression resistance for an UPE240 vertical beam made of steel S235.

12.63.2Background Verification of compressed steel profile Verifies the steel centrically compressed vertical beam made of UPE240, steel S235. The beam is supported by hinges at both extremities and blocked to swerve perpendicularly to local z-axis of the profile at all quarters of its length.

12.63.2.1 Model description ■ Reference: Manually calculated example ■ Analysis type: static linear (plane problem); ■ Element type: linear. Load cases and load combinations: ■

Dead load: g = 500 kN

Units Metric System Geometry Below are described the beam cross section characteristics: ■ Height: h = 240 mm, ■ Width: b = 90 mm, ■ Length: L = 7200 mm,

440

ADVANCE DESIGN VALIDATION GUIDE

■ Section area: A = 3850 mm2, ■ Bending moment of inertia around y-axis: Iy = 35.99*106 mm4, 6 4 ■ Bending moment of inertia around z-axis: Iz = 3.109*10 mm , 𝐼

𝑦 ■ Radius of inertia around y-axis: 𝑖𝑦 = � 𝐴 = 96.7 𝑚𝑚,

𝐼

■ Radius of inertia around z-axis: 𝑖𝑧 = � 𝐴𝑧 = 28.4 𝑚𝑚,

■ ■ ■ ■

Compressed part of web: d = 185.0 mm, Web thickness: tw = 7.0 mm, Compressed part of flange: c = 68.0 mm, Flange thickness: tf = 12.5 mm

Materials properties Steel: S235, fy = 235 N/mm2 Material coefficient: γM0 = γM1 = 1.15 Boundary conditions The boundary conditions are described below: ■ Outer: ► ►

Support at start point (Z=0), hinged connection restrained to rotate around Z, Support at end point (Z = 7.20) hinged connection,

■ Inner: ►

Support at each quarter of beam length (Z = 1.80, 3.60, 5.40), restrained to translate along Y

Loading The beam is subjected only to axial force: Load combinations: ■ Ultimate Limit State: NEd = 1*g = 500.0 kN

12.63.2.2 Cross-section classification Web d 185.0 = = 26.43 < 72 ∗ ε = 72.0 tw 7

According to: CSN EN1993-1-1 Chapter 5.6, Table 5.2 (Sheet 1)

Flange cf tf

=

68.0 12.5

= 5.44 < 10 ∗ ε = 10.0

235

𝜀 = � 𝑓 = 1.0 𝑦

According to: CSN EN1993-1-1 Chapter 5.6, Table 5.2 (Sheet 2)

Cross-section belongs to class 1

441

ADVANCE DESIGN VALIDATION GUIDE

12.63.2.3 Verification of resistance in compression for Ultimate Limit State Buckling lengths Ly = L = 7200 mm, Lz = L/4 = 1800 mm Slenderness 𝜆𝑧 =

𝜆𝑦 =

λy >λz

𝐼𝑧 1800 = = 63.4 𝑖𝑧 28.4

𝐼𝑦 7200 = = 74.5 𝑖𝑦 96.7

=> to consider λy, corresponding to buckling curve "C"

According to: CSN EN1993-1-1 Chapter 6.3.1.2(2), Table 6.2 Buckling coefficient 𝜆=� λ1 = 93.9*ε = 93.9

𝐴 ∗ 𝑓𝑦 𝐿𝑐𝑟 1 7200 1 = ∗ = ∗ = 0.7934 𝑁𝑐𝑟 𝑖 𝜆1 96.7 93.9

𝜀=�

235 =1 𝑓𝑦

According to: CSN EN1993-1-1 Chapter 6.3.1.3(1), Formula (6.50) For buckling curve "C" and 𝜆 = 0.7934 the corresponding value is κ = 0.67

According to: CSN EN1993-1-1 Chapter 6.3.1.2(3), Picture 6.4

Verification of resistance in compression 𝑁𝑏,𝑅𝑑 =

𝜅∗𝐴∗𝑓𝑦 𝛾𝑀1

=

0.67∗3850∗235 1.15

= 527.1 𝑘𝑁

According to: CSN EN1993-1-1 Chapter 6.3.1.1(3), Formula (6.47) 𝑁𝐸𝑑 ≤ 1.0 𝑁𝑏,𝑅𝑑

500.0 = 0.95 < 1.0 527.1

=> Work ratio in compression is 95% Finite elements modeling ■ ■ ■

442

Linear element: S beam, 5 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

Steel design results Buckling coefficient κ (reference value 0.67)

Work ratio in compression (reference value 95%)

12.63.2.4 Reference results Result name

Result description

Reference value

κy

Buckling coefficient corresponding to local y-axis [-]

0.67

Work ratio

Work ratio in compression [%]

95 %

12.63.3Calculated results Result name

Result description

Value

Error

Xy

Buckling coefficient

0.666583 adim

-0.5100 %

Work ratio

Work ratio

95.7491 %

0.7885 %

443

ADVANCE DESIGN VALIDATION GUIDE

12.64 EC3 / NF EN 1993-1-1/NA - France: Verifying critical bending moment on a IPE200 beam with complex loading Test ID: 6276 Test status: Passed

12.64.1Description Verifies critical moment for lateral-torsional buckling on a IPE200 beam subject to linear loading and end moments according to EN 1993-1-1 (French standards).

12.64.2Background Verifies critical moment for lateral-torsional buckling on a IPE200 beam subject to linear loading and end moments according to EN 1993-1-1 (French standards).

12.64.2.1 Model description ■ ■

Section : IPE200 Material : S235

Geometrical properties ■ ■ ■ ■

Beam length: L=4m Flexion inertia moment around the Z axis: Iz=142.4 cm4 4 Torsional moment of inertia: It=7.02mm 6 Working inertial moment: Iw=13052cm

Materials properties S235 steel material is used. The following characteristics are used: ■ ■ ■

444

Yield strength fy = 235 MPa, Longitudinal elastic modulus: E = 210000 MPa. Shear modulus of rigidity: G=80800MPa

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions Beam is simply supported. Loading The column is subjected to the following loadings: ■ Linear loads Fz = -9 kN/ml applied on the upper flange. ■ End moments : My = -30kN.m at x = 0m and My = 6kN.m at x = 4m

12.64.2.2 Bending moment diagram (kN.m)

According to M.3.4 from EN 1993-1-1 (French appendix) :

M = 30kN .m

ψ=

6 = 0.2 30

q = 9kN / m

µ=−

ql ² 9 × 4² =− = −0.6 8M 8 × 30

445

ADVANCE DESIGN VALIDATION GUIDE

12.64.2.3 C1 calculation C1 can be read from the curves given in Figure M.4

12.64.2.4 C2 calculation C2 can be read from the curves given in Figure M.6

446

ADVANCE DESIGN VALIDATION GUIDE

12.64.2.5 Mcr calculation

M cr = C1

 π ² EI z  I w L ²GI t 2 + + (C 2 z g ) − C 2 z g   L ²  I z π ² EI z 

C1 = 3.75 C2 = 0.7

z g = +0.1m Note:

the linear applied on the top flange tends to increase the lateral torsional buckling effect on the critical position (x=0m) => Zg is taken with a positive sign.

E = 210000 MPa

I z = 142.4cm4 I w = 13052cm 6 I t = 7.02cm 4 L = 4m G = 80800 MPa Soit :

M cr = 98.kN .m 12.64.2.6 Reference results Result name

Result description

Reference value

Mcr

Critical bending moment

98kN.m

12.64.1Calculated results Result name

Result description

Value

Error

Mcr

Mcr

94.7469 kN*m

0.0000 %

447

ADVANCE DESIGN VALIDATION GUIDE

12.65 EC3 / NF EN 1993-1-1/NA - France: Verifying critical moment Mcr of a beam with intermediate restraints subject to upwards loading Test ID: 6281 Test status: Passed

12.65.1Description The test verifies the critical bending moment Mcr of a IPE220 beam made of S235 steel. It ensures Advance Design uses the intermediate restraint definition from the bottom flange when the beam is subject to forces acting upwards. The calculations are made according to Eurocode 3, French Annex.

12.65.2Background Critical bending moment verification for a restrained IPE220 beam subjected to axis bending efforts, made of S235 steel. The beam is simply supported.

12.65.2.1 Model description ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■

Exploitation loadings (category A): Q1 = -20 000 N, The ultimate limit state (ULS) combination is: Cmax = 1 x Q

■

Cross section dimensions are in millimeters (mm).

Units Metric System Geometrical properties ■ ■ ■ ■

448

Beam length: 5m 2 Cross section area: A=3337mm Flexion inertia moment around the Y axis: Iy=2772x104mm4 4 4 Flexion inertia moment around the Z axis: Iz=2049x10 mm

ADVANCE DESIGN VALIDATION GUIDE

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: Hinged support at start point (x = 0) Hinged support at the end point (z = 5.00) ► Elastic support at start point (x = 0) : KRY = 45 kN.m/°. ► Elastic support at end point (x = 5) : KRY = 45 kN.m/°. Inner: None. ► ►

■

Intermediate restraints The boundary conditions are described below: ■ ■

Top flange: intermediate restraints at x = 1.25m, x = 2.5m, x = 3.75m Bottom flange: intermediate restraint at x = 2. 5m

Loading The column is subjected to the following loadings: ■ ■

External: Linear load From X=0.00m to X=5.00m: FZ = N = +20 000 N, Internal: None.

449

ADVANCE DESIGN VALIDATION GUIDE

12.65.2.2 Critical moment Mcr calculation 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.

M cr = C1 ×

π 2 × E × I z   k z  I w

(k z × L )2

×

 × +   k w  I z 

(k z × L )2 × G × I t + (C × z )2 − C × z  2 2 g g 2  π × E × IZ



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=2049 x104mm4 It is the torsional inertia: It=90700mm4 IW is the warping inertia (deformation inertia moment): Iw=2.267x1010mm6 L is the distance between intermediate restraints on the compressed flange (in this case : top flange) : L = 1.25m 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 C1 = 1.4815 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:

M cr = C1 ×

π ²× E × Iz

M cr = 1.4815 ×

L²

I w L² × G × I t + Iz π ²× E × Iz

×

π ² × 210000 × 204.90 ×10 −8 (2.50)²

M cr = 148.68kNm Finite elements modeling ■ ■ ■

450

Linear element: S beam, 21 nodes, 1 linear element.

×

22670 ×10 −12 (2.50)² × 80800 × 9.04 ×108 + 204.9 ×10 −8 π ² × 210000 × 204.9 ×10 −8

ADVANCE DESIGN VALIDATION GUIDE

Finite elements results Elastic critical moment for lateral-torsional buckling Simply supported beam subjected to bending efforts Mcr

12.65.2.3 Reference results Result name

Result description

Reference value

Mcr

Elastic critical moment for lateral-torsional buckling [kNm]

148.68

12.65.3Calculated results Result name

Result description

Value

Error

Mcr

Mcr

148.937 kN*m

0.0000 %

451

ADVANCE DESIGN VALIDATION GUIDE

12.66 EC3 / CSN EN1993-1-1 - Czech Republic: Tensioned diagonal Test ID: 6289 Test status: Passed

12.66.1Description Example according to "Eliášková, Sokol: Ocelové konstrukce - príklady" (example 1.2). Verifies WR in compression

12.66.2Background Verification of tensioned steel profile Verifies the centrically tensioned horizontal steel beam made of L90*8, steel S235. Beam is simply supported.

12.66.2.1 Model description ■ Reference: Martina Eliášová, Zdeněk Sokol: Ocelové konstrukce - Příklady (2005); Chapter 1.2 ■ Analysis type: static linear (plane problem); ■ Element type: linear. Load cases and load combinations: ■

Dead load: g = 251.6 kN

Units Metric System Geometry Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 90 mm, Width: b = 90 mm, Length: L = 2000 mm, 2 Section area: A = 1390 mm ,

Materials properties Steel: S235, fy = 235 N/mm2 Material coefficient: γM0 = 1.15

452

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions The boundary conditions are described below: ■

Outer: Support at start point (Z=0), hinged connection, ► Support at end point (Z = 2.00) restrained to translate along Y and Z, restrained to rotate around X, Inner: none ►

■

Loading The beam is subjected only to normal force: Load combinations: ■ Ultimate Limit State: NEd = 1*g = 251.6 kN

12.66.2.2 Verification of resistance in tension 𝑁𝑡,𝑅𝑑 =

𝐴 ∗ 𝑓𝑦 1390 ∗ 235 = = 284.0 𝑘𝑁 𝛾𝑀0 1.15

Profile is not weakened by holes.

According to: CSN EN1993-1-1 Chapter 6.2.3(2), Formula (6.6) 𝑁𝐸𝑑 251.6 = = 0.89 ≤ 1.0 𝑁𝑡,𝑅𝑑 284.0

=> Work ratio in compression is 89 %

According to: CSN EN1993-1-1 Chapter 6.2.3(1)P, Formula (6.5) Finite elements modeling ■ ■ ■

Linear element: beam, 3 nodes, 1 linear element.

Steel design results Work ratio in tension (reference value 89%)

12.66.2.3 Reference results Result name

Result description

Reference value

Work ratio - Fx

Work ratio in tension [%]

89 %

12.66.3Calculated results Result name

Result description

Value

Error

Work ratio - Fx

WR Tension

88.6418 %

-0.4025 %

453

ADVANCE DESIGN VALIDATION GUIDE

12.67 EC3 / EN 1993-1-1 - United Kingdom: Simply supported laterally restrained (from P364 Open Sections Example 2) Test ID: 6328 Test status: Passed

12.67.1Description The 533x210x92 UKB in S275 beam is fully laterally restrained along its length and pinned supports, includes a UDL and point load at the centre. These are just the steel design results for example 2 from P364 from the SCI.

12.67.2Background The 533x210x92 UKB in S275 beam is fully laterally restrained along its length and pinned supports, includes a UDL and point load at the centre. These are just the steel design results for example 2 from P364 from the SCI.

12.67.2.1 Model description Reference: SCI PUBLICATION P364, Steel Building Design: Worked Examples - Open Sections. In accordance with Eurocodes and the UK National Annexes, Example 2 - Simply supported laterally restrained beam ■ ■

Analysis type: static linear (plane problem) Element type: linear The beam shown in the figure below is fully laterally restrained along its length and has bearing lengths of 50 mm at the unstiffened supports and 75 mm under the point load. Design the beam in S275 steel for the loading shown below.

The design aspects covered in this example are: ■ ■ ■

■ ■

Calculation of design values of actions for ULS and SLS Cross section classification Cross sectional resistance: ► Shear buckling ► Shear ► Bending moment Resistance of web to transverse forces Vertical deflection of beam at SLS.

Units Metric System Geometry Below are described the column cross section characteristics: ■ ■

Depth: h = 533.1 mm Width: b = 209.3 mm

■

Flange thickness: 𝑡𝑓 = 15.6 mm

■ ■ ■ ■

454

Root radius: r = 12.7 mm Depth between flange fillets: d = 476.5 mm Second moment of area, y-y axis: 𝐼𝑦 = 55200 cm4

Plastic modulus, y-y ax: 𝑊𝑝𝑙,𝑦 = 2360 cm4

ADVANCE DESIGN VALIDATION GUIDE

Section area: A = 117 cm2

■

Modulus of elasticity: 𝐸 = 210000 N/mm2

■

Materials properties

For buildings that will be built in the UK, the nominal values of the yield strength (fy) and the ultimate strength (fu) for structural steel should be those obtained from the product standard. Where a range is given, the lowest nominal value should be used. For S275 steel and t ≤ 16 mm Yield strength: 𝑓𝑦 = 𝑅𝑒𝐻 = 275 𝑁/𝑚𝑚2

■

12.67.2.2 Resistance to bending Verify that: 𝑀𝐸𝑑

𝑀𝑐,𝑅𝑑

≤ 1.0

At the point of maximum bending moment (mid-span) check whether the shear force will reduce the bending resistance of the section.

𝑉𝑐,𝑅𝑑 2

=

909 2

= 454.5 𝑘𝑁

Shear force at maximum bending moment 𝑉𝑐,𝐸𝑑 = 62.5 𝑘𝑁

𝑉𝐸𝑑,𝑐 = 62.5 𝑘𝑁 < 454.4 𝑘𝑁

Therefore no reduction in bending resistance due to shear is required. The design resistance for bending for Class 1 and 2 cross sections is:

𝑀𝑐,𝑅𝑑 = 𝑀𝑝𝑙,𝑅𝑑 = 𝑀𝐸𝑑

𝑀𝑐,𝑅𝑑

=

539.5 649

𝑊𝑝𝑙,𝑦 𝑓𝑦 2360 × 103 × 275 = × 10−6 = 649 kNm 𝛾𝑀0 1.0

= 0.83 < 1.0

Therefore the bending resistance is adequate.

12.67.2.3 Reference results Result name

Result description

Reference value

Steel Strength - Work ratio - Oblique

Flexural buckling resistance

83%

12.67.3Calculated results Result name Work ratio - Fz Work ratio Oblique

-

Result description

Value

Error

Fz work ratio

29.6673 %

-1.1090 %

Oblique work ratio

83.1471 %

0.1772 %

455

ADVANCE DESIGN VALIDATION GUIDE

12.68 NTC 2008 - Italy: Strenght verification of a steel linear hollow section Test ID: 6282 Test status: Passed

12.68.1Description Verifies the strengh of a steel linear hollow section with a point load applied on the top and a linear load applied along the linear element, the element is fixed on the base

12.68.2Background 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 European 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.

12.68.2.1 Model description ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■

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

■

Cross section dimensions are in millimeters (mm).

Units Metric System Geometry Below are described the column cross section characteristics: ■ ■ 456

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

ADVANCE DESIGN VALIDATION GUIDE

■ ■ ■ ■ ■

Thickness: t = 10 mm, Outer radius: r = 15 mm, Column height: L = 5000 mm, 2 Section area: A = 9260 mm , 3 Plastic section modulus about y-y axis: Wpl,y = 921000 mm ,

■

Partial factor for resistance of cross sections:

.

Materials properties S235 steel material is used. The following characteristics are used: ■ ■

Yield strength fy = 235 MPa, 5 Longitudinal elastic modulus: E = 2.1 x 10 MPa.

Boundary conditions The boundary conditions are described below: ■

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 The column is subjected to the following loadings: ■

■

External: ► Point load at z = 5.0: Fz = - 500 kN, ► Uniformly distributed load: q = Fx = 5 kN/ml Internal: None.

457

ADVANCE DESIGN VALIDATION GUIDE

12.68.2.2 Reference 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 (4.2.4) from chapter 4.2.4.1.1 (NTC 2008), while the design plastic moment resistance is verified considering the criterion (4.2.12) from chapter 4.2.4.1.2 (NTC 2008). Before starting the above verifications, the cross-section class has to be determined. Cross section class The following results are determined according to NTC 2008: “Costruzioni civili ed industriali” – Chapter 4.2.4. In this case, the stresses distribution is like in the picture below:

Table 4.2.I - from Chapter 4.2.3.1 (NTC 2008), establish the rules to determine the class for compressed parts. The picture below shows an extract from this table.

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 4.2.I

458

ADVANCE DESIGN VALIDATION GUIDE

c b − 2 × r − 2 × t 200mm − 2 × 15mm − 2 × 10mm = = = 15 t t 10mm

ε=

235 = 1.0 fy

Therefore:

c = 15 ≤ 33ε = 33 t 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 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.

α 132 MPa = → α = 0.832 > 0.5 1 − α 26.63MPa c h − 2 × r − 2 × t 300mm − 2 × 15mm − 2 × 10mm = = = 25 t 10mm t

ε=

235 = 1.0 fy

Therefore:

396ε c = 40.34 = 25 ≤ 13α − 1 t This means that the left/right web is Class 1. Verifying the design resistance for uniform compression The design resistance for uniform compression, for Class 1 cross-section, is determined with formula (4.2.11) from NTC 2008.

Nc, Rd = A x fyk / gamma m0 = 9260 mm2 x 235 kN/mm2 / 1.05 = 2072.47 kN The verification of the design resistance for uniform compression is done with relationship (4.2.10) from NTC 2008. The corresponding work ratio is: Work ratio =

Ned / Nc,Rd = 500 kN / 2072.47 kN = 0.24 < 1

459

ADVANCE DESIGN VALIDATION GUIDE

Verifying the design plastic moment resistance The design plastic moment resistance, for Class 1 cross-section, is determined with formula (4.2.13) from NTC 2008.

921000 mm3 x 235 kN / mm2 / 1.05 = 206125.57 kNm The verification of the design resistance for bending is done with relationship (6.12) from EN 1993-1-1. The corresponding work ratio is:

62500 KNm / 206125.57 kNm = 0.3032 < 1 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 efforts Work ratio – Fx

460

ADVANCE DESIGN VALIDATION GUIDE

Work ratio of the design resistance for bending Column subjected to bending and axial efforts Work ratio – Oblique

12.68.2.3 Reference results Result name

Result description

Reference value

Work ratio – Fx

Compression resistance work ratio [%]

24.13 %

Work ratio – Oblique

Work ratio of the design resistance for bending

30.32 %

12.68.3Calculated results Result name Work ratio - Fx Work ratio Oblique

-

Result description

Value

Error

Strenght work ratio Fx

24.1257 %

0.5237 %

Strenght work ratio - oblique

30.3209 %

1.0697 %

461

ADVANCE DESIGN VALIDATION GUIDE

12.69 NTC 2008 - Italy: Lateral torsional buckling verification of a steel column Test ID: 6293 Test status: Passed

12.69.1Description Verifies the lateral torsional buckling of a steel column with a point load applied on the top and a bending moment on y direction and another bending moment on the x direction both applied on the top. The column is hinged on the base and restrained in traslation on x,y direction and restrained in rotation on x axis.

12.69.2Background 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.

12.69.2.1 Model description ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used: ■ ■ ■

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).

Units Metric System

462

ADVANCE DESIGN VALIDATION GUIDE

Geometrical properties ■

Column length: L=5620mm

■

Cross section area:

■

Overall breadth:

■

Flange thickness:

■

Root radius:

■

Web thickness:

■

Depth of the web:

■

Elastic modulus after the Y axis, Wel , y

■

Plastic modulus after the Y axis, Wy

■

Elastic modulus after the Z axis, Wel , z

= 80344.89mm 3

■

Plastic modulus after the Z axis, W pl , z

= 123381.96mm3

■

Flexion inertia moment around the Y axis:

■

Flexion inertia moment around the Z axis: I z

■

Torsional moment of inertia: I t

■

Working inertial moment: I w

A = 4904.06mm 2 b = 150mm

t f = 10.70mm

r = 0mm

t w = 7.10mm

hw = 260mm

= 445717.63mm3

= 501177.18mm3

I y = 57943291.64mm 4 = 6025866.46mm 4

= 149294.97mm 4

= 93517065421.88mm 6

Materials properties S275 steel material is used. The following characteristics are used: ■ ■ ■

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

Boundary conditions The boundary conditions are described below: ■

Outer: ► ►

Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at start point (x = 5.62) restrained in translation along X and Y axis, and restrained in rotation along Z axis,

463

ADVANCE DESIGN VALIDATION GUIDE

Loading The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =-328 kN; Mx=10 kNm and My=50 kNm

12.69.2.2 Cross section Class According to Advance Design calculations: Cross-class classification is made according to Table 4.2 from NTC 2008 standard

464

ADVANCE DESIGN VALIDATION GUIDE

- for beam web: The web dimensions are 850x5mm.

ψ=

σ inf 45.30Mpa = = −0.253 > −1 σ sup − 179.06Mpa

465

ADVANCE DESIGN VALIDATION GUIDE

238.6 × 45.3  = 48.175 x=  238.6 x y x+ y  224.36 = ⇒ = = 238.6 × 179.06 45.30 179.06 45.30 + 179.06 224.36  = 190.73 y=  224.36

α= ε=

x 190.73 = = 0.80 > 0.5 238.6 238.6 235 = fy

235 = 0.924 275

c 260mm − 2 × 10.7 mm  = = 33.61 c 396 × ε 396 × 0.924 = 38.93 = t 7.1mm  ⇒ = 33.6 ≤ t α 13 1 13 0 . 8 1 × − × −  ε = 0.924

therefore the beam

web is considered to be Class 1 -for beam flange:

150 − 7.1   c c  2 = = 6.68 ⇒ = 6.68 ≤ 9 × 0.924 = 8.316 t 10.7 t   ε = 0.924 In conclusion, the section is considered to be Class 1.

466

therefore the haunch is considered to be Class1

ADVANCE DESIGN VALIDATION GUIDE

12.69.2.3 Buckling verification a) over the strong axis of the section, y-y: -the imperfection factor α will be selected according to the Table 4.2.VI from NTC 2008 standard:

α = 0.34 Coefficient corresponding to non-dimensional slenderness after Y-Y axis:

χ y coefficient

corresponding to non-dimensional slenderness

λy

will be determined from the relevant buckling

curve according to the formula taken from NTC 2008 standard:

χy =

1 2

Φ y + Φ y − λy

2

≤1

(4.2.45)

467

ADVANCE DESIGN VALIDATION GUIDE

λ y the non-dimensional slenderness corresponding to Class 1 cross-sections: A * fy

λy =

N cr , y A = 4904.06mm 2

Cross section area:

Flexion inertia moment around the Y axis: I y

N cr , y =

π²× E× Iy l fy ²

A× f y

λy =

N cr , y

=

= 57943291.64mm 4

π ² × 210000 N / mm 2 × 57943291.64mm 4

(5620mm)²

= 3802327.95 N = 3802.33kN

4904.06mm 2 × 275 N / mm 2 = = 0.5956 3802327.95 N

[

]

φ y = 0.5 × 1 + α × (λ y − 0.2) + λ y ² = 0.5 × [1 + 0.34 × (0.5956 − 0.2 ) + 0.5956 2 ] = 0.7446

χy =

1

χy ≤1

468

2

Φy + Φy −λy

2

=

1

 = 0.839 0.7446 + 0.7446 2 − 0.5956 2  ⇒ χ y = 0.839  

ADVANCE DESIGN VALIDATION GUIDE

b) over the weak axis of the section, z-z: -the imperfection factor α will be selected according to the Table 4.2.VI from NTC 2008 standard:

α = 0.49 Coefficient corresponding to non-dimensional slenderness after Z-Z axis:

χ z coefficient corresponding to non-dimensional slenderness according to the formula from NTC 2008 standard:

λz

will be determined from the relevant buckling curve

469

ADVANCE DESIGN VALIDATION GUIDE

χz =

1 2

Φz + Φz − λ z

2

≤1

(4.2.45)

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

A* f y

λz =

N cr , z

Flexion inertia moment around the Z axis: Cross section area:

N cr , z =

A = 4904.06mm 2

π ² × E × I z π ² × 210000 N / mm 2 × 6025866.46mm 4 = = 395426.63 N = 395.43kN (5620mm )² l fz ² A× f y

λz =

I z = 6025866.46mm 4

N cr , z

=

4904.06mm 2 × 275 N / mm 2 = 1.847 395426.63 N

[

]

φ z = 0.5 × 1 + α × (λ z − 0.2) + λ z ² = 0.5 × [1 + 0.49 × (1.847 − 0.2) + 1.847 2 ] = 2.609 χz =

1 2

Φz + Φz − λ z

2

=

χz ≤ 1

1

 = 0.225 2.609 + 2.609 − 1.847  ⇒ χ z = 0.225   2

2

12.69.2.4 Lateral 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:

M cr = C1 ×

π ² × E × Iz L²

×

I w L² × G × It + Iz π ² × E × Iz According to EN 1993-1-1; Chapter 2

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

C1 =

1 0.325 + 0.423ψ + 0.252ψ ² According to EN 1993-1-1 ; Chapter 3

470

ADVANCE DESIGN VALIDATION GUIDE

ψ=

M y ,botom M y ,top

=

0 = 0 ⇒ C1 = 1.77 50

Flexion inertia moment around the Y axis:

I y = 57943291.64mm 4

Flexion inertia moment around the Y axis:

I y = 57943291.64mm 4

Flexion inertia moment around the Z axis: I z Torsional moment of inertia: I t Working inertial moment: I w

= 6025866.46mm 4

= 149294.97mm 4

= 93517065421.88mm 6

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):

Iw =

I z × (h − t f

)

2

4

h cross section height; h

tf

flange thickness;

= 260mm

t f = 10.7mm

6025866.46mm 4 × (260mm − 10.7mm ) = 93627638290mm 6 Iw = 4 2

Length of the column: L=5620mm

471

ADVANCE DESIGN VALIDATION GUIDE

The calculation of the Mcr

NOTE: For the calculation of the elastic moment, I used the formula from the Eurocode 3 and not the formula from the “Circolare Ministeriale n. 917/2009”, because of the distribution of the moment along the column. Knowing the Mcr we can calculate the non-dimensional slenderness for lateral – torsional buckling:

λ LT =

W pl , y f y M cr

Calculation of the with this formula:

=

501177.18mm3 × 275 N / mm 2 = 0.958 150184702.1Nmm

χ LT for appropriate non-dimensional slenderness λ LT

according NTC 2008 standard will be determined

Where

The cross section buckling curve will be chose according to Table 4.2.VII from NTC 2008 standard:

h 260mm = = 1.733 ≤ 2 b 150mm

The imperfection factor α will be chose according to Table 4.2.VI:

α = 0.49 472

ADVANCE DESIGN VALIDATION GUIDE

f = 1 -0.5 x (1 – 0.75) x [1 – 2.0 x (0.958 – 0.8)2] = 0.879

[

(

)

]

φLT = 0.5 × 1 + α LT × λ LT − 0.2 + λ LT ² = 0.5 × [1 + 0.49 × (0.958 − 0.2) + 0.958² ] = 1.145 χ LT =

1

φLT + φLT ² − λ LT ²

=

1 = 0.564 ≤ 1 1.145 + 1.145² − 0.958²

The result is get without consider the effect of the reduction factor (with reduction factor

= 0,639

The stability check for section class 1, 2 or 3 is performed checking:

Finite elements modeling ■ ■ ■

Linear element: S beam, 7 nodes, 1 linear element.

473

ADVANCE DESIGN VALIDATION GUIDE

χ y coefficient corresponding to non-dimensional slenderness λ y Column subjected to axial and shear force to the top

χy

474

ADVANCE DESIGN VALIDATION GUIDE

χ z coefficient corresponding to non-dimensional slenderness λ z Column subjected to axial and shear force to the top

χz

475

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yy Column subjected to axial and shear force to the top

k yy

476

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k yz Column subjected to axial and shear force to the top

k yz

477

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k zy Column subjected to axial and shear force to the top

k zy

478

ADVANCE DESIGN VALIDATION GUIDE

Internal factor, k zz Column subjected to axial and shear force to the top

k zz

479

ADVANCE DESIGN VALIDATION GUIDE

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

480

ADVANCE DESIGN VALIDATION GUIDE

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

481

ADVANCE DESIGN VALIDATION GUIDE

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

482

ADVANCE DESIGN VALIDATION GUIDE

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

483

ADVANCE DESIGN VALIDATION GUIDE

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

484

ADVANCE DESIGN VALIDATION GUIDE

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

485

ADVANCE DESIGN VALIDATION GUIDE

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

486

ADVANCE DESIGN VALIDATION GUIDE

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

487

ADVANCE DESIGN VALIDATION GUIDE

The appropriate non-dimensional slenderness The appropriate non-dimensional slenderness

χ LT

488

ADVANCE DESIGN VALIDATION GUIDE

12.69.2.5 Reference results Result name

Result description

Reference value

χy

χ y coefficient corresponding to non-dimensional slenderness λ y

0.839

χz

χ z coefficient corresponding to non-dimensional slenderness λ z

0.225

k yy

Internal factor, k yy

2.47

k yz

Internal factor, k yz

5.20

k zy

Internal factor, k zy

0.529

k zz

Internal factor, k zy

0.919

SNy

Bending and axial compression verification term depending of the compression effort over the Y axis Bending and axial compression verification term depending of the Y bending moment over the Y axis Bending and axial compression verification term depending of the Z bending moment over the Y axis Bending and axial compression verification term depending of the compression effort over the z axis Bending and axial compression verification term depending of the Y bending moment over the Z axis Bending and axial compression verification term depending of the Z bending moment over the Z axis Coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure The elastic moment for lateral-torsional buckling calculation

0.297

χ LT

The appropriate non-dimensional slenderness

0.564

Work Ratio

Stability work ratio

0.357

SMyy SMyz SNz SMzy SMzz

Mcr

1.668 1.609 1.13 0.35 0.28 1.75 150.17

489

ADVANCE DESIGN VALIDATION GUIDE

12.69.3Calculated results

490

Result name

Result description

Value

Error

Xy

Xy coefficent

0.839285 adim

0.0000 %

Xz

Xz coefficent

0.224656 adim

-0.1529 %

Kyy

Kyy coefficent

2.47232 adim

0.0939 %

Kyz

Kyz coefficent

5.26997 adim

1.3456 %

Kzy

Kzy coefficent

0.525982 adim

0.0000 %

Kzz

Kzz coefficent

0.948136 adim

4.1908 %

SNy

SNy coefficent

0.304274 adim

4.9221 %

SMyy

SMyy coefficent

1.66785 adim

-0.0090 %

SMyz

SMyz coefficent

1.63085 adim

1.3580 %

SNz

SNy coefficent

1.13673 adim

0.5956 %

SMzy

SMzy coefficent

0.354833 adim

1.3809 %

SMzz

SMzz coefficent

0.29341 adim

4.7893 %

C1

C1 coefficent

1.77 adim

0.0000 %

Mcr

Critical bending moment

150.166 kN*m

0.0000 %

XLT

X LT coefficent

0.564654 adim

0.0000 %

Work ratio

Work ratio

360.297 %

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

12.70 EC3 / NF EN 1993-1-1/NA France: Verifying the stability of an asymmetric I-shape subjected to axial force and bending moment Test ID: 6303 Test status: Passed

12.70.1Description Verifies the stability of a welded column subjected to compression and bending, made from S355 steel. Section is an asymmetric I-shape. The results will be compared with the ones obtained by the CTIM in their training guide “EC3/1 Résistance ultime des sections” from April 2006.

12.70.2Background Unrestrained welded column subjected to compression and bending, made from S355 steel. Section is an asymmetric I-shape. The results will be compared with the ones obtained by the CTIM in their training guide “EC3/1 Résistance ultime des sections” from April 2006.

12.70.2.1 Model description ■ ■

Analysis type: static linear (plane problem); Element type: linear.

■

Cross section dimensions are in millimeters (mm).

Units Metric System Geometrical properties ■

Column length: L=5m

■

Cross section area:

■

Elastic modulus around the y axis, Wel , y ,sup

A = 118cm²

= 3701cm 4

and

Wel , y ,inf = 3129cm 4

491

ADVANCE DESIGN VALIDATION GUIDE

Materials properties S355 steel material is used. The following characteristics are used: ■ ■ ■

Yield strength fy = 255 MPa, Longitudinal elastic modulus: E = 210000 MPa. Shear modulus of rigidity: G=80800MPa

Boundary conditions The boundary conditions are described below: ■

Outer: ► ►

Hinged support at column bottom (x = 0) restrained in translation along X, Y and Z axis, Support at the end point (z = 5.00) restrained in translation along X and Y axis and restrained rotation around Z axis.

Loading The column is subjected to the following loadings: ■ ■

492

External: Point load at column top: FZ =-390kN and My=630kN.m Internal: None.

ADVANCE DESIGN VALIDATION GUIDE

CTICM model The model is presented in the CTICM training guide “Stage EC3/1 : Résistance ultime des sections”.

12.70.2.2 Cross section Class According to Advance Design calculations: Cross-class classification is made according to Table 5.2 -for beam web:

ψ =2

N Ed 0.39 −1 = 2 − 1 = −0.814 A× fy 0.0118 × 355

c (820 − 2 × 10 − 2 × 6)  = = 131  c 42ε t 6 ⇒ >  42ε 42 × 0.81 t 0.67 + 0.33ψ = = 84.5  0.67 + 0.33ψ 0.67 + 0.33 × (−0.81)

therefore

the

beam

web

is

considered to be Class 4

493

ADVANCE DESIGN VALIDATION GUIDE

In conclusion, the section is considered to be Class 4. According to CTICM document:

494

ADVANCE DESIGN VALIDATION GUIDE

12.70.2.3 Aeff calculation (in pure compression) Top flange:

ψ=

σ2 = 1 ⇒ kσ = 0.43 σ1

beff = ρc

λp = For

c/t 28.4ε kσ

=

19.1 28.4 × 0.81 × 0.43

λ p > 0.748 , ρ =

= 1.266 > 0.748

λ p − 0.188 1.266 − 0.188 = = 0.673 1.266² λp ²

(4.3)

beff = ρc = 0.673 × (400 − 2 × 6 − 6) / 2 = 129mm

becomes According to CTICM document:

495

ADVANCE DESIGN VALIDATION GUIDE

Web:

ψ=

σ2 = 1 ⇒ kσ = 4 σ1

beff = ρbw

λp = For

c/t 28.4ε kσ

=

131 28.4 × 0.81 × 4

λ p > 0.748 , ρ =

= 2.847 > 0.748

λ p − 0.188 2.847 − 0.188 = = 0.328 2.847² λp ²

beff = ρbw = 0.328 × (800 − 2 × 6) = 258mm

which means

becomes

496

(4.3)

129mm

for top and bottom parts

ADVANCE DESIGN VALIDATION GUIDE

According to CTICM document:

Bottom flange:

ψ=

σ2 = 1 ⇒ kσ = 0.43 σ1

beff = ρc

λp = For

c/t 28.4ε kσ

=

14.1 28.4 × 0.81 × 0.43

λ p > 0.748 , ρ =

= 0.935 > 0.748

λ p − 0.188 0.935 − 0.188 = = 0.854 0.935² λp ²

(4.3)

beff = ρc = 0.854 × (300 − 2 × 6 − 6) / 2 = 120mm

becomes

497

ADVANCE DESIGN VALIDATION GUIDE

According to CTICM document:

Total:

Aeff = A − (62 × 10) × 2 − (21 × 10) × 2 − 530 × 6 = 118 − (6.2 × 1.0) × 2 − (2.1 × 1.0) × 2 − 53.0 × 0.6 = 69.60cm² Eccentricity calculation:

becomes Which means eccentricity: eN = 25mm. According to CTICM document:

498

ADVANCE DESIGN VALIDATION GUIDE

12.70.2.4 Weff calculation (in pure bending) Top flange:

ψ =

σ2 = 1 ⇒ kσ = 0.43 σ1

beff = ρc

λp = For

c/t 28.4ε kσ

=

19.1 28.4 × 0.81 × 0.43

λ p > 0.748 , ρ =

= 1.266 > 0.748

λ p − 0.188 1.266 − 0.188 = = 0.673 1.266² λp ²

(4.3)

beff = ρc = 0.673 × 191 = 129mm

becomes According to CTICM document:

499

ADVANCE DESIGN VALIDATION GUIDE

Web:

ψ =

σ2 1 =− = −1.19 ⇒ kσ = 5.98 × (1 + 1.19)² = 28.680 σ1 0.84

beff = ρbw

λp = For

c/t 28.4ε kσ

=

131 28.4 × 0.81 × 22.68

λ p > 0.748 , ρ =

λ p − 0.055 × (3 + ψ ) 1.066 − 0.055 × (3 − 1.19) = = 0.735 1.066² λp ²

beff = ρbw = 0.735 × 365 = 268mm

500

= 1.066 > 0.748

(4.3)

ADVANCE DESIGN VALIDATION GUIDE

becomes According to CTICM document:

501

ADVANCE DESIGN VALIDATION GUIDE

Bottom flange: Bottom flange is not compressed.

Using Cross Section Tool, we get Weff:

According to CTICM document:

502

ADVANCE DESIGN VALIDATION GUIDE

12.70.2.5 Section verification Applying formula (6.44)

M y , Ed + N Ed e Ny M z , Ed + N Ed e Nz N Ed + + ≤1 Aeff f y / γ M 0 Weff , y , min f y / γ M 0 Wef , z , min f y / γ M 0 With:

Aeff = 69.60cm²

Weff ,min = 2700cm 3 e N = 25mm N Ed = 390kN

M y , Ed = 630kN .m

M y , Ed + N Ed e Ny N Ed 0.390 0.630 + 0.39 × 0.025 + = + = 0.825 −4 Aeff f y / γ M 0 Weff , y , min f y / γ M 0 69.60 ×10 × 355 / 1 2700 ×10 − 6 × 355 / 1 According to CTICM document:

503

ADVANCE DESIGN VALIDATION GUIDE

12.70.2.6 Result in Advance Design It is to be noted that Advance Design takes into account the interaction between the axial force and the bending moment through the kyy interaction coefficient.

Therefore, the work ratio we can expect from Advance Design is:

0.390 0.630 + 0.39 × 0.025 + 0.79 × = 0.684 −4 69.60 ×10 × 355 / 1 2700 ×10 − 6 × 355 / 1 Indeed:

12.70.2.7 Reference results Result name

Result description

Reference value

Work ratio - Oblique

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

68.4%

12.70.3Calculated results

504

Result name

Result description

Value

Error

Work ratio

Work ratio

70.0434 %

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

12.71 EC3 / NF EN 1993-1-1/NA - France: Verifying IPE450 column fixed on base subjected to axial compression and bending moment, both applied on top (ref. Test 31) Test ID: 5731 Test status: Passed

12.71.1Description 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. The calculations are made according to Eurocode 3 French Annex.

12.72 EC3 / NF EN 1993-1-1/NA - France: Verifying IPE600 simple supported beam, loaded with centric compression and uniform linear efforts by Y and Z axis (ref. Test 32) Test ID: 5732 Test status: Passed

12.72.1Description 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. The calculations are made according to Eurocode 3 French Annex.

12.73 EC3 / NF EN 1993-1-1/NA - France: Verifying RHS300x150x9H class 1 simply supported beam, loaded with centric compression, uniform linear horizontal efforts and a vertical punctual load in the middle (ref. Test 36) Test ID: 5738 Test status: Passed

12.73.1Description 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. The calculations are made according to Eurocode 3 French Annex.

12.74 EC3 / NF EN 1993-1-1/NA - France: 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 (ref. Test 35) Test ID: 5737 Test status: Passed

12.74.1Description 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. The calculations are made according to Eurocode 3 French Annex.

505

ADVANCE DESIGN VALIDATION GUIDE

12.75 EC3 / NF EN 1993-1-1/NA - France: Verifying IPE300 beam, simply supported, loaded with centric compression and uniform linear efforts by Y and Z axis (ref. Test 30) Test ID: 5730 Test status: Passed

12.75.1Description 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. The calculations are made according to Eurocode 3 French Annex.

12.76 EC3 / NF EN 1993-1-1/NA - France: Verifying a simply supported rectangular hollow section beam subjected to torsional efforts (ref. Test 17) Test ID: 5742 Test status: Passed

12.76.1Description Verifies a simply supported rectangular hollow section beam made of S235 steel subjected to torsional efforts. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.

12.77 EC3 / NF EN 1993-1-1/NA - France: 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 (ref. Test 39) Test ID: 5741 Test status: Passed

12.77.1Description The test verifies a CHS323.9x6.3H beam made of S275 steel. The beam is subjected to 20 kN axial compression force, 50 kN punctual vertical load applied to the middle of the beam and 4 kN/m linear uniform horizontal load. The calculations are made according to Eurocode 3 French Annex.

12.78 EC3 / NF EN 1993-1-1/NA - France: Verifying C310x30.8 class 3beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle (ref. Test 34) Test ID: 5734 Test status: Passed

12.78.1Description 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. The calculations are made according to Eurocode 3 French Annex.

506

ADVANCE DESIGN VALIDATION GUIDE

12.79 EC3 / NF EN 1993-1-1/NA - France: Determining lateral torsional buckling parameters for a Ishaped welded built-up beam considering the load applied on the upper flange (ref. Test 42) Test ID: 5752 Test status: Passed

12.79.1Description 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. The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.

12.80 EC3 / NF EN 1993-1-1/NA - France: Determining lateral torsional buckling parameters for a Ishaped laminated beam considering the load applied on the lower flange (ref. Test 43) Test ID: 5750 Test status: Passed

12.80.1Description 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.

12.81 EC3 / NF EN 1993-1-1/NA - France: Verifying RHS350x150x5H class 4 column, loaded with centric compression, punctual horizontal force by Y and a bending moment, all applied to the top (ref. Test 38) Test ID: 5740 Test status: Passed

12.81.1Description 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. The calculations are made according to Eurocode 3 French Annex.

12.82 EC3 / NF EN 1993-1-1/NA - France: 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 (ref. Test 40) Test ID: 5744 Test status: Passed

12.82.1Description The test verifies a CHS508x8H beam made of S235 steel. The beam is subjected to 20 kN axial compression force, 30 kN punctual vertical load applied to the middle of the beam and 7 kN/m linear uniform horizontal load. The calculations are made according to Eurocode 3 French Annex.

507

ADVANCE DESIGN VALIDATION GUIDE

12.83 EC3 / NF EN 1993-1-1/NA - France: Verifying UPN300 simple supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and punctual vertical force by Z axis (ref. Test 33) Test ID: 5733 Test status: Passed

12.83.1Description 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. The calculations are made according to Eurocode 3 French Annex.

12.84 EC3 / NF EN 1993-1-1/NA - France: Verifying RHS350x150x8.5H class 3 column, loaded with centric compression, punctual lateral load and bending moment, all applied to the top of the column (ref. Test 37) Test ID: 5739 Test status: Passed

12.84.1Description 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. The calculations are made according to Eurocode 3 French Annex.

12.85 EC3 / NF EN 1993-1-1/NA - France: Verifying a simply supported circular hollow section element subjected to torsional efforts (ref. Test 18) Test ID: 5743 Test status: Passed

12.85.1Description 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.

12.86 EC3 / NF EN 1993-1-1/NA - France: Determining lateral torsional buckling parameters for a Ishaped laminated beam considering the load applied on the lower flange (ref. Test 41) Test ID: 5753 Test status: Passed

12.86.1Description 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. The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.

508

ADVANCE DESIGN VALIDATION GUIDE

12.87 EC3 / NF EN 1993-1-1/NA - France: Changing the steel design template for a linear element (TTAD #12491) Test ID: 4540 Test status: Passed

12.87.1Description Selects a different design template for steel linear elements.

12.88 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533) Test ID: 4549 Test status: Passed

12.88.1Description Verifies the shape sheet for a steel beam with circular cross-section when the lateral torsional buckling is computed and when it is not.

12.89 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389) Test ID: 4529 Test status: Passed

12.89.1Description Verifies the program behavior when the "Shape sheet" command is used for elements which were excluded from the specialized calculation (chords).

12.90 Generating the shape sheet by system (TTAD #11471) Test ID: 3575 Test status: Passed

12.90.1Description Generates shape sheets by system, on a model with 2 systems.

12.91 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522) Test ID: 3612 Test status: Passed

12.91.1Description 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.

509

ADVANCE DESIGN VALIDATION GUIDE

12.92 EC3 / NF EN 1993-1-1/NA - France: Verifying the cross section optimization of a steel element (TTAD #11516) Test ID: 3620 Test status: Passed

12.92.1Description 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.

12.93 EC3 / NF EN 1993-1-1/NA - France: Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545) Test ID: 3641 Test status: Passed

12.93.1Description 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.

12.94 Verifying the shape sheet results for a column (TTAD #11550) Test ID: 3640 Test status: Passed

12.94.1Description 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.

12.95 Verifying shape sheet on S275 beam (TTAD #11731) Test ID: 3434 Test status: Passed

12.95.1Description 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.

510

ADVANCE DESIGN VALIDATION GUIDE

12.96 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770) Test ID: 3406 Test status: Passed

12.96.1Description 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.

12.97 Verifying the calculation results for steel cables (TTAD #11623) Test ID: 3560 Test status: Passed

12.97.1Description 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.

12.98 EC3 / NF EN 1993-1-1/NA: Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873) Test ID: 4289 Test status: Passed

12.98.1Description 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 UKC152x152x37 cross section, beams with UKB254x146x37 cross section and roof beams with UKB203x133x25 cross section. Dead loads and live loads are applied on the structure.

12.99 CM66 (steel design) - France: Verifying the buckling length for a steel portal frame, using the roA roB method Test ID: 3814 Test status: Passed

12.99.1Description Verifies the buckling length for a steel portal frame with one level, using the roA roB method, according to CM66. Generates the "Buckling and lateral-torsional buckling lengths" report.

511

ADVANCE DESIGN VALIDATION GUIDE

12.100EC3 / EN 1993-1-1 - General: Verifying the buckling length for a steel portal frame, using the kA kB method Test ID: 3819 Test status: Passed

12.100.1Description 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.

12.101CM66 (steel design) - France: Verifying the buckling length for a steel portal frame, using the kA kB method Test ID: 3813 Test status: Passed

12.101.1Description Verifies the buckling length for a steel portal frame with one level, using the kA kB method, according to CM66. Generates the "Buckling and lateral-torsional buckling lengths" report.

12.102EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling coefficient Xy on a class 2 section Test ID: 4443 Test status: Passed

12.102.1Description 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.

12.103EC3 / SR EN 1993-1-1-2006 - Romania: Fire verification: verifying the work ratios after performing an optimization for steel profiles (TTAD #11975) Test ID: 4484 Test status: Passed

12.103.1Description 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.

512

ADVANCE DESIGN VALIDATION GUIDE

12.104EC3 / NF EN 1993-1-1/NA - France: Verifying the buckling length results (TTAD #11550) Test ID: 4481 Test status: Passed

12.104.1Description 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.

513

13 Timber design

ADVANCE DESIGN VALIDATION GUIDE

13.1 Modifying the "Design experts" properties for timber linear elements (TTAD #12259) Test ID: 4509 Test status: Passed

13.1.1Description Defines the "Design experts" properties for a timber linear element, in a model created with a previous version of the program.

13.2 EC5 / NF EN 1995-1-1 - France: Verifying the units display in the timber shape sheet (TTAD #12445) Test ID: 4539 Test status: Passed

13.2.1Description Verifies the Afi units display in the timber shape sheet.

13.3 EC5 / NF EN 1995-1-1 - France: Verifying the timber elements shape sheet (TTAD #12337) Test ID: 4538 Test status: Passed

13.3.1Description Verifies the timber elements shape sheet.

13.4 EC5 / NF EN 1995-1-1 - France: Shear verification for a simply supported timber beam Test ID: 4822 Test status: Passed

13.4.1Description 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.

516

ADVANCE DESIGN VALIDATION GUIDE

13.5 EC5 / NF EN 1995-1-1 - France: Verifying a timber purlin subjected to biaxial bending and axial compression Test ID: 4879 Test status: Passed

13.5.1Description 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.

13.5.2Background 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.

13.5.2.1 Model 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; 2 Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m ; Axial compression force due to wind effect on the supporting elements: W = 15000 N; Uniformly distributed load corresponding to the ultimate limit state combination: 2 Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m .

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 Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.20 m, Width: b = 0.10 m, Length: L = 3.50 m, 2 Section area: A = 0.02 m ,

■

Elastic section modulus about the strong axis, y:

Wy =

b × h 2 0.1 ⋅ 0.20 2 = = 0.000666m 3 , 6 6 517

ADVANCE DESIGN VALIDATION GUIDE

■

b 2 × h 0.12 ⋅ 0.20 = = 0.000333m3 . Elastic section modulus about the strong axis, z: Wz = 6 6

Materials properties 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, 6 Characteristic bending strength: fm,k = 24 x 10 Pa, 10 Longitudinal elastic modulus: E = 1.1 x 10 Pa, 10 Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 10 Pa, Service class 2.

Boundary conditions The boundary conditions are described below: ■

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. ►

■

13.5.2.2 Reference 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: ■

Slenderness ratio corresponding to bending about the z axis:

λ z = 12 ■

m × lg lc 1× 3.5m = 12 = 12 = 121.24 0.1m b b

Slenderness ratio corresponding to bending about the y axis:

λ y = 12

m × lg lc 1× 3.5m = 12 = 12 = 60.62 h h 0.2m

Relative slenderness ratios The relative slenderness ratios are: ■

518

Relative slenderness ratio corresponding to bending about the z axis:

ADVANCE DESIGN VALIDATION GUIDE

λz π

λrel , z = ■

E0, 05

=

121.24

π

21× 10 6 Pa = 2.056 0.74 × 1010 Pa

Relative slenderness ratio corresponding to bending about the y axis:

λrel , y = ■

f c , 0,k

λy π

f c , 0,k E0,05

=

60.62

π

21× 10 6 Pa = 1.028 0.74 × 1010 Pa

Maximum relative slenderness ratio:

λrel ,max = max(λrel , z , λrel , y ) = 2.056

So there is a risk of buckling because λrel,max ≥ 0.3. 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,z2] (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)

k c,z = kc, y =

1 k z + k z2 − λ2rel , z 1 k y + k y2 − λ2rel , y

(according to relation 6.26 from EN 1995-1-1)

(according to relation 6.25 from EN 1995-1-1)

Reference solution for ultimate limit state verification 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 =

■

Design bending stress about the y axis (induced by uniformly distributed load, qz):

σm,y,d = ■

N 13500 N = = 675000 Pa A 0.02m 2 My

q × L2 = z = 8 ×Wy Wy

N × 3.50 2 m 2 m = 8.2814 × 10 6 Pa 8 × 0.000666m 3

3601.92

Design bending stress about the z axis (induced by uniformly distributed load, qy): 2

σm,z,d =

M z qy × L = = 8 × Wz Wz

N × 3.50 2 m 2 m = 5.0638 × 10 6 Pa 8 × 0.000333m 3

1101.22

■

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)

■

Partial factor for material properties:

519

ADVANCE DESIGN VALIDATION GUIDE

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)

■

Design compressive strength: fc,0,d =

■

f c , 0,k ×

γM

= 21× 10 6 ×

1.1 = 17.769 × 10 6 Pa 1.3

Design bending strength: fm,y,d = fm,z,d =

■

k mod

f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

Maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1:

σ m , y ,d σ  σ c , 0,d + + k m ⋅ m , z ,d  f m , y ,d f m , z ,d  k c , y ⋅ f c , 0,d max   σ c , 0,d + k ⋅ σ m , y ,d + σ m , z ,d m  k c , z ⋅ f c , 0,d f m , y ,d f m , z ,d  Finite elements modeling ■ ■ ■

520

1.1 × 1.0 × 1.0 = 20.308 × 10 6 Pa 1.3

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

    ≤1   

ADVANCE DESIGN VALIDATION GUIDE

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 [%]

13.5.2.3 Reference results 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 %

13.5.3Calculated 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 %

521

ADVANCE DESIGN VALIDATION GUIDE

13.6 EC5 / NF EN 1995-1-1 - France: Verifying a timber purlin subjected to oblique bending Test ID: 4878 Test status: Passed

13.6.1Description 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.

13.6.2Background 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.

13.6.2.1 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; 2 ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m ; 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:

522

■ ■ ■ ■

Height: h = 0.20 m, Width: b = 0.10 m, Length: L = 3.5 m, 2 Section area: A = 0.02 m ,

■

Elastic section modulus about the strong axis, y:

Wy =

b × h 2 0.1 ⋅ 0.20 2 = = 0.000666m 3 , 6 6

■

Elastic section modulus about the strong axis, z:

Wz =

b 2 × h 0.12 ⋅ 0.20 = = 0.000333m3 . 6 6

ADVANCE DESIGN VALIDATION GUIDE

Materials properties 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 The boundary conditions are described below: ■

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.

13.6.2.2 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 = ■

My

q × L2 = z = Wy 8 ×Wy

N × 3.5 2 m 2 m = 8.2814 ×10 6 Pa 3 8 × 0.000666m

3601.92

Design bending stress about z axis (induced by uniformly distributed load, qy):

σm,z,d =

N 2 1101.22 × 3.5 2 m 2 M z qy × L m = = = 5.0638 ×10 6 Pa 3 8 × Wz Wz 8 × 0.000333m

523

ADVANCE DESIGN VALIDATION GUIDE

■

Design bending strength: fm,y,d = fm,z,d =

■

f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

0 .9 × 1.0 × 1.0 = 16.615 × 10 6 Pa 1.3

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:

σ σ m , y , d  + k m m , z ,d   f m , z ,d   f m , y ,d max   ≤1 k σ m , y , d + σ m , z , d   m f m , y ,d f m , z ,d    Finite elements modeling ■ ■ ■

Linear element: S beam, 5 nodes, 1 linear element.

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]

524

ADVANCE DESIGN VALIDATION GUIDE

Maximum work ratio for strength verification Strength of a simply supported purlin subjected to oblique bending Work ratio [%]

13.6.2.3 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 %

13.6.3Calculated results Result name

Result description

Value

Error

Stress SMy

Design bending stress about y axis

8.47672e+00 6 Pa

0.0000 %

Stress SMz

Design bending stress about z axis

5.18319e+00 6 Pa

0.0000 %

Work ratio

Maximum work ratio for strength verification

71.153 %

0.0000 %

525

ADVANCE DESIGN VALIDATION GUIDE

13.7 EC5 / NF EN 1995-1-1 - France: Verifying a timber column subjected to tensile forces Test ID: 4693 Test status: Passed

13.7.1Description Verifies the tensile resistance of a rectangular cross section column (fixed at base) made from solid timber C24.

13.7.2Background 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.

13.7.2.1 Model 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: ■ ■ ■ ■ 526

Height: h = 0.122 m, Width: b = 0.036m, Section area: A = 43.92 x 10-4 m2 -6 4 I = 5.4475 x 10 m .

ADVANCE DESIGN VALIDATION GUIDE

Materials properties Rectangular solid timber C24 is used. The following characteristics are used in relation to this material: ■ ■ ■

Longitudinal elastic modulus: E = 1.1 x 1010 Pa, 6 Characteristic tensile strength along the grain: ft,0,k = 14 x 10 Pa, Service class 2.

Boundary conditions The boundary conditions: ■

Outer: Fixed at base (z = 0), Free at top (z = 5),

■

Inner: None.

Loading The column is subjected to the following loadings: ■ ■

External: Point load at z = 5: Fz = N = 10000 N, Internal: None.

13.7.2.2 Reference 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):

 150  0.2   kh = min  h  1.3  ■

Design tensile stress (induced by the ultimate limit state force, N): σt,0,d =

■

Design tensile strength: ft,0,d =

■

N A f t , 0,k ×

k mod

γM

× kh

Work ratio: SFx =

σ t , 0,d f t , 0,d

≤ 1.0

(according to relation 6.1 from EN 1995-1-1)

Finite elements modeling ■ ■ ■

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

527

ADVANCE DESIGN VALIDATION GUIDE

Work ratio SFx diagram Column with fixed base, subjected to tension force Work ratio SFx

13.7.2.3 Reference results Result name

Result description

Reference value

σt,0,d SFx

Design tensile stress [Pa]

2276867.03 Pa

Work ratio [%]

18 %

13.7.3Calculated results

528

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 %

ADVANCE DESIGN VALIDATION GUIDE

13.8 EC5 / NF EN 1995-1-2 - France: Verifying the residual section of a timber column exposed to fire for 60 minutes Test ID: 4896 Test status: Passed

13.8.1Description 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.

13.8.2Background 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.

13.8.2.1 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 Below are described the column cross section characteristics: ■ ■ ■ ■

Depth: h = 0.60 m, Width: b = 0.20 m, Section area: A = 0.12 m2 Height: H = 5.00 m 529

ADVANCE DESIGN VALIDATION GUIDE

Materials properties 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 The boundary conditions are described below: ■

Outer: Fixed at base (Z = 0), ► Support at top (Z = 5.00) restrained in translation along X and Y, Inner: None. ►

■

Loading The column is subjected to the following loadings: ■ ■

External: Point load at Z = 5.00: Fz = N = - 100000 N, Internal: None.

13.8.2.2 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: −3 dchar,n = β n ⋅ t = 0.7 ⋅ 10

■

m ⋅ 60 min = 0.042m (according to relation 3.2 from EN 1995-1-2) min

Effective charring depth: def = dchar,n + k 0 ⋅ d 0

■

Residual cross section: Afi = (h − def ) × (b − 2 ⋅ def )

Finite elements modeling ■ ■ ■

530

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

ADVANCE DESIGN VALIDATION GUIDE

Residual cross section Column with fixed base exposed to fire for 60 minutes Afi

13.8.2.3 Reference results Result name

Result description

Reference value

Afi

Residual cross section [m2]

0.056202 m2

13.8.3Calculated results Result name

Result description

Value

Error

Afi

Residual cross section

0.056202 m²

0.0000 %

531

ADVANCE DESIGN VALIDATION GUIDE

13.9 EC5 / NF EN 1995-1-2 - France: Verifying the fire resistance of a timber purlin subjected to simple bending Test ID: 4901 Test status: Passed

13.9.1Description 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.

13.9.2Background 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 realised. Chapter 1.1.1.3 presents the results obtained with the theoretical background explained at chapter 1.1.1.2.

13.9.2.1 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 Below are described the beam cross section characteristics: ■ ■ ■ ■ ■

532

Height: h = 0.20 m, Width: b = 0.075 m, Length: L = 3.30 m, Distance between adjacent purlins (span): d = 1.5 m, -3 2 Section area: A = 15.0 x 10 m ,

ADVANCE DESIGN VALIDATION GUIDE

Materials properties 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, 6 Characteristic bending strength: fm,k = 24 x 10 Pa,

■

Density: ρ = 350 kg/m3,

■ ■

Design charring rate (softwood): βn = 0.8 x 10-3 m/min, Service class 1.

Boundary conditions The boundary conditions are described below: ■

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 The beam is subjected to the following loadings: ■

External: Uniformly distributed load: q = CFQ x d = 850 N/m2 x 1.5 m = 1275 N/m, Internal: None.

►

■

13.9.2.2 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 =

■

m ⋅ 30 min = 0.024m min

(according to relation 3.2 from EN 1995-1-2)

Effective charring depth: def =

■

β n ⋅ t = 0.8 ⋅10 −3

d char ,n + k 0 ⋅ d 0

Residual cross section: Afi =

hef × bef = (h − d ef ) × (b − 2 ⋅ d ef )

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).

533

ADVANCE DESIGN VALIDATION GUIDE

■

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 =

My Wy

=

6× M y bef × hef

2

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 =

■

k fi × f m ,k ×

k mod, fi

γ M , fi

= 1.25 × 24 ×10 6 ×

1.0 = 30 ×10 6 Pa 1.0

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):

σ m ,d ≤ 1.0 f m ,d ■

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):

km × ■

σ m ,d ≤ 1.0 f m ,d

Maximum work ratio for bending verification for fire situation:

σ σ WR = max m ,d ; k m × m ,d f m ,d  f m ,d Finite elements modeling ■ ■ ■

534

Linear element: S beam, 8 nodes, 1 linear element.

 σ m ,d   × 100 = f × 100 m ,d 

ADVANCE DESIGN VALIDATION GUIDE

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 [%]

535

ADVANCE DESIGN VALIDATION GUIDE

13.9.2.3 Reference results Result name

Result description

Reference value

2

Afi

Residual area [m ]

0.002197 m2

Stress

Design bending stress for residual cross-section [Pa]

27568524 Pa

Work ratio

Maximum work ratio for fire verification [%]

91.9 %

13.9.3Calculated results Result name

536

Result description

Value

Error

Afi

Residual area

0.002197 m²

0.0000 %

Stress

Design bending stress for residual cross-section

2.75626e+00 7 Pa

0.0000 %

Work ratio

Maximum work ratio for fire verification

91.8752 %

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

13.10 EC5 / NF EN 1995-1-1 - France: Verifying a timber column subjected to compression forces Test ID: 4823 Test status: Passed

13.10.1Description Verifies the compressive resistance of a rectangular cross section column (hinged at base) made from solid timber C18.

13.10.2Background 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.

13.10.2.1 Model 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

Materials properties Rectangular solid timber C18 is used. The following characteristics are used in relation to this material: ■

Longitudinal elastic modulus: E = 0.9 x 1010 Pa, 537

ADVANCE DESIGN VALIDATION GUIDE

■ ■ ■

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 The column is subjected to the following loadings: ■ ■

External: Point load at z = 3.2: Fz = N = -20000 N, Internal: None.

13.10.2.2 Reference 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. ■

Slenderness ratio corresponding to bending about the z axis:

λ z = 12 ■

m × lg lc 1× 3.2m = 12 = 12 = 110.85 0.1m b b

Slenderness ratio corresponding to bending about the y axis (informative):

λ y = 12

m × lg lc 1× 3.2m = 12 = 12 = 73.9 0.15m h h

Relative slenderness ratios The relative slenderness ratios are: ■

Relative slenderness ratio corresponding to bending about the z axis:

λrel , z = ■

λz π

f c , 0,k E0,05

=

m × l g × 12

f c , 0,k

b ×π

E0,05

=

1× 3.2m × 12 0.1m × π

18 × 10 6 Pa = 1.933 0.6 × 1010 Pa

Relative slenderness ratio corresponding to bending about the y axis (informative):

λrel , y =

λy π

f c , 0,k E0,05

=

m × l g × 12

f c , 0,k

h ×π

E0,05

=

1× 3.2m × 12 0.15m × π

18 × 10 6 Pa = 1.288 0.6 × 1010 Pa

So there is a risk of buckling, because λrel,max ≥ 0.3. 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: ■ 538

kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z2] (according to relation 6.28 from EN 1995-1-1)

ADVANCE DESIGN VALIDATION GUIDE

■

ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y2] (informative)

■

kc,z =

■

kc, y =

1 k z + k z2 − λ2rel , z

(according to relation 6.26 from EN 1995-1-1)

1

(informative)

k y + k y2 − λ2rel , y

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). ■

Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)

■ ■

Partial factor for material properties: γM = 1.3 Design compressive stress (induced by the applied forces): σc,0,d =

■

Design compressive strength: fc,0,d =

■

N A f c , 0,k

k mod

γM

Work ratio: Work ratio =

σ c , 0,d k c , z × f c , 0,d

≤ 1.0

(according to relation 6.35 from EN 1995-1-1)

Finite elements modeling ■ ■ ■

Linear element: S beam, 4 nodes, 1 linear element.

539

ADVANCE DESIGN VALIDATION GUIDE

Work ratio diagram Simply supported column subjected to compression force Work ratio

13.10.2.3 Reference results Result name

Result description

Reference value

kc,z

Instability factor

0.2400246

kc,y

Instability factor (informative)

0.488869

σc,0,d Work ratio

Design compressive stress [Pa]

1333333 Pa

Work ratio [%]

50 %

13.10.3Calculated results

540

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 %

ADVANCE DESIGN VALIDATION GUIDE

13.11 EC5 / NF EN 1995-1-1 - France: Verifying a timber beam subjected to combined bending and axial tension Test ID: 4872 Test status: Passed

13.11.1Description 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.

13.11.2Background 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.

13.11.2.1 Model 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: 2 ■ Loadings from the structure: G = 450 N/m ; 2 ■ Snow load (structure is located at an altitude > 1000m above sea level): S = 900 N/m ; ■ 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 Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.20 m, Width: b = 0.05 m, Length: L = 5.00 m, -3 2 Section area: A = 10 x 10 m ,

■

Elastic section modulus about the strong axis y:

Wy =

b × h 2 0.05 ⋅ 0.20 2 = = 0.000333m 3 . 6 6 541

ADVANCE DESIGN VALIDATION GUIDE

Materials properties Rectangular solid timber C24 is used. The following characteristics are used in relation to this material: ■ ■ ■

Characteristic tensile strength along the grain: ft,0,k = 14 x 106 Pa, 6 Characteristic bending strength: fm,k = 24 x 10 Pa, Service class 2.

Boundary conditions The boundary conditions are described below: ■

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. ►

■

13.11.2.2 Reference 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. Reference solution for ultimate limit state verification 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). ■

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 (“h” represents the width in millimeters because the element is tensioned):

 150  0.2  150  0.2 1.25     kh = min  h  = min  50  = min  = 1.25 1.3 1.3 1.3  

542

■

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 tensile stress (induced by the ultimate limit state force, N):

ADVANCE DESIGN VALIDATION GUIDE

N 2191.22 N = = 219122 Pa A 10 ×10 −3 m 2

σt,0,d = ■

Design tensile strength:

f t , 0,k ×

ft,0,d = ■

σ t , 0,d

≤ 1.0

(according to relation 6.1 from EN 1995-1-1)

My

6 × q × L2 = = Wy 8 × b × h 2

N × 5.00 2 m 2 m = 8.2045 ×10 6 Pa 2 2 8 × 0.05m × 0.2 m

6 × 875.15

f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

0.9 × 1.0 × 1.0 = 16.615 × 10 6 Pa 1.3

Work ratio according to formulae 6.17 from EN 1995-1-1 norm:

σ t , 0 ,d f t , 0 ,d ■

f t , 0,d

Design bending strength: fm,y,d =

■

0.9 ×1.25 = 12.115 ×10 6 Pa 1.3

Design bending stress (induced by the applied forces):

σm,y,d = ■

γM

× k h = 14 ×10 6 ×

Work ratio: SFx =

■

k mod

+

σ m , y ,d f m , y ,d

+ km

σ m , z ,d f m , z ,d

≤1

Work ratio according to formulae 6.18 from EN 1995-1-1 norm:

σ t , 0 ,d f t , 0 ,d

+ km

σ m , y ,d f m , y ,d

+

σ m , z ,d f m , z ,d

≤1

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:

winst (Q) ≤ w fin ≤

L 300

L 125

wnet , fin ≤

L 200

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are: ■

Instantaneous deflection (for a base variable action):

winst (Q) = 0.00914m ⇒ winst (Q) = ■

Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

winst = d CQ = 0.01371m ⇒ winst = ■

L 547.05

L 364.7

In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the deformation factor (kdef) has to be chosen:

543

ADVANCE DESIGN VALIDATION GUIDE

k def = 0.8

(value determined for service class 2, according to table 3.2 from EN 1995-1-1)

wcreep = 0.8 × d QP = 0.8 × 0.0064m = 0.00512m ⇒ wcreep = ■

L 976.6

Final deflection:

w fin = winst + wcreep = 0.01371m + 0.00512m = 0.01883m ⇒ w fin = ■

Net deflection:

wnet , fin = w fin + wc = 0.01883m + 0m = 0.01883m ⇒ wnet , fin = Finite elements modeling ■ ■ ■

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

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]

544

L 265.5

L 265.5

ADVANCE DESIGN VALIDATION GUIDE

Instantaneous deflection winst(CQ) Simply supported beam subjected to a characteristic load combination of actions Instantaneous deflection winst(CQ) [m]

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]

13.11.2.3 Reference results 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

545

ADVANCE DESIGN VALIDATION GUIDE

13.11.3Calculated results

546

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 %

ADVANCE DESIGN VALIDATION GUIDE

13.12 EC5 / NF EN 1995-1-1 - France: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression Test ID: 4877 Test status: Passed

13.12.1Description 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.

13.12.2Background 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.

13.12.2.1 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; 2 ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m ; 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 Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.20 m, Width: b = 0.05 m, Length: L = 5.00 m, -3 2 Section area: A = 10 x 10 m ,

■

Elastic section modulus about the strong axis, y:

Wy =

b × h 2 0.05 ⋅ 0.20 2 = = 0.000333m 3 . 6 6

547

ADVANCE DESIGN VALIDATION GUIDE

Materials properties 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, 6 Characteristic bending strength: fm,k = 24 x 10 Pa, 10 Longitudinal elastic modulus: E = 1.1 x 10 Pa, 10 Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 10 Pa, Service class 2.

Boundary conditions The boundary conditions are described below: ■

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.

13.12.2.2 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:

λ z = 12

m × lg lc 1× 5m = 12 = 12 = 346.4 b b 0.05m

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:

λ z = 12 ■

m × lg lc 1×1.667 m = 12 = 12 = 115.5 < 120 b b 0.05m

Slenderness ratio corresponding to bending about the y axis:

λ y = 12

m × lg lc 1 × 5m = 12 = 12 = 86.6 h h 0.2m

Relative slenderness ratios The relative slenderness ratios are: ■

548

Relative slenderness ratio corresponding to bending about the z axis:

ADVANCE DESIGN VALIDATION GUIDE

λrel , z = ■

E0,05

=

115.5

π

21×10 6 Pa = 1.958 0.74 ×1010 Pa

Relative slenderness ratio corresponding to bending about the y axis:

λrel , y = ■

f c , 0,k

λz π λy π

f c , 0 ,k E0,05

=

86.6

π

21 × 10 6 Pa = 1.468 0.74 × 1010 Pa

Maximum relative slenderness ratio:

λrel ,max = max(λrel , z , λrel , y ) = 1.958

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:

leff = 0.9 ⋅ l + 2 ⋅ h = 0.9 ⋅ 5.0m + 2 ⋅ 0.2m = 4.9m ■

Critical bending stress (determined according to formula 6.32 from EN 1995-1-1): σm,y,crit =

■

0.78 ⋅ b 2 E0.05 0.78 ⋅ 0.052 m 2 ⋅ 0.74 ⋅ 1010 Pa = = 14.724 × 106 Pa 0.2m ⋅ 4.9m h ⋅ lef

Relative slenderness for bending (determined according to formula 6.30 from EN 1995-1-1): λrel,m =

f m ,k

σ m,crit

=

24 ⋅ 106 Pa = 1.277 14.724 ⋅ 106 Pa

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,z2] (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)

kc,z = kc, y =

1 k z + k z2 − λ2rel , z 1 k y + k y2 − λ2rel , y

(according to relation 6.26 from EN 1995-1-1)

(according to relation 6.25 from EN 1995-1-1)

Reference solution for ultimate limit state verification 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). ■

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:

549

ADVANCE DESIGN VALIDATION GUIDE

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 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 =

■

Design compressive strength: fc,0,d =

■

k mod

γM

= 21×10 6 ×

My

q × L2 = = Wy 8 ×Wy f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

2

 σ c ,d  + ≤1  k c , z ⋅ f c , 0 ,d 

Finite elements modeling

550

N × 5.00 2 m 2 m = 8.213 ×10 6 Pa 3 8 × 0.000333m

875.15

0.9 × 1.0 × 1.0 = 16.615 × 10 6 Pa 1.3

Work ratio according to formula 6.35 from EN 1995-1-1 norm:

 σ m ,d  k ⋅ f  crit m , y ,d ■ ■ ■

0.9 = 14.538 ×10 6 Pa 1.3

Design bending strength: fm,y,d =

■

f c , 0,k ×

Design bending stress (induced by uniformly distributed load, q):

σm,d = ■

N 2191.22 N = = 219122 Pa A 10 ×10 −3 m 2

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

ADVANCE DESIGN VALIDATION GUIDE

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 [%]

13.12.2.3 Reference results 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 %

13.12.3Calculated results Result name

Result description

Value

Error

Stress SFx

Design compressive stress

219124 Pa

0.0000 %

Stress SMy

Design bending stress

8.20519e+00 6 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 %

551

ADVANCE DESIGN VALIDATION GUIDE

13.13 EC5 / NF EN 1995-1-1 - France: Verifying a C24 timber beam subjected to shear force Test ID: 5036 Test status: Passed

13.13.1Description 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.

13.13.2Background 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.

13.13.2.1 Model 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, 2 The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m

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, -3 2 Section area: A = 16.875 x 10 m ,

Materials properties Rectangular solid timber C24 is used. The following characteristics are used in relation to this material: ■ ■

552

Characteristic shear strength: fv,k = 2.5 x 106 Pa, Service class 1.

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions The boundary conditions are described below: ■

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 The beam is subjected to the following loadings: ■ ■

External: ► Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m, Internal: None.

13.13.2.2 Reference 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. Reference solution for ultimate limit state verification 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). ■

Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)

■

Partial factor for material properties: γM = 1.3

■

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 = ■

beff × heff

3 × 3656.25 N 2 = = 0.485075 × 10 6 Pa 0.05025m × 0.225m

Design shear strength: fv,d =

■

k f × Fv ,d

f v ,k ×

k mod

γM

= 2.5 × 10 6 Pa ×

0.8 = 1.538 × 10 6 Pa 1.3

Work ratio according to formulae 6.13 from EN 1995-1-1 norm:

τd

f v ,d

≤ 1.0

553

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling ■ ■ ■

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

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 [%]

554

ADVANCE DESIGN VALIDATION GUIDE

13.13.2.3Reference results 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 %

13.13.3Calculated 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 %

555

ADVANCE DESIGN VALIDATION GUIDE

13.14 EC5 / SR EN-1995-1-1-2004 - Romania: Timber column subjected to shear stress and torsion Test ID: 6304 Test status: Passed

13.14.1Description Verifies the stability of a timber column subjected to shear stresses and torsion, made from C40 timber. The section is a rectangular one, with 20x25 cm dimensions. The objective of this test is to validate the resistance to torsional shear stresses.

13.15 EC5 / SR EN-1995-1-1-2004 - Romania: Timber column subjected to compression Test ID: 6297 Test status: Passed

13.15.1Description Verifies the adequacy of the compressive resistance for a rectangular cross section made from solid timber C30. The verification is made according to formula (6.23) from EN 1995-1-1 norm.

13.16 EC5 / SR EN-1995-1-1-2004 - Romania: Timber beam subjected to simple bending Test ID: 6291 Test status: Passed

13.16.1Description Verifies a rectangular cross section beam made from solid timber D30 to resist simple bending. Verifies the bending stress SMy at ultimate limit state, as well as the work ratio for SMy.

13.17 EC5 / SR EN 1995-1-1 - Romania: Verifying compression strength for C14 circular column with fixed base Test ID: 6240 Test status: Passed

13.17.1Description Verifies compression strength for a circular column with fixed base made of C14 timber. Verification is made according to EN 1995-1-1 Romanian Annex.

556

ADVANCE DESIGN VALIDATION GUIDE

13.18 EC5 / NF EN 1995-1-1 - France: Verifying a timber beam subjected to simple bending Test ID: 4682 Test status: Passed

13.18.1Description 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.

13.18.2Background 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.

13.18.2.1 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: ■ ■ ■ ■ ■

2 Loadings from the structure: G = 0.5 kN/m , 2 Exploitation loadings (category A): Q = 1.5 kN/m , 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 Below are described the beam cross section characteristics: ■ ■ ■ ■ ■

Height: h = 0.20 m, Width: b = 0.075 m, Length: L = 4.50 m, Distance between adjacent beams (span): d = 0.5 m, -3 2 Section area: A = 15.0 x 10 m ,

■

Elastic section modulus about the strong axis y: W y =

b × h 2 0.075 ⋅ 0.20 2 = = 0.0005m 3 6 6

557

ADVANCE DESIGN VALIDATION GUIDE

Materials properties 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, 6 Characteristic bending strength: fm,k = 24 x 10 Pa, Service class 1.

Boundary conditions The boundary conditions are described below: ■

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 The beam is subjected to the following loadings: ■

External: Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,

■

Internal: None.

13.18.2.2 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. Reference solution for ultimate limit state verification 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)

■

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 (induced by the applied forces): σm,d =

■

558

Wy

=

6 × q × L2 6 × 1.4625 × 4.5 2 = = 7.4039 × 10 6 Pa 8× b × h2 8 × 0.075 × 0.2 2

Design bending strength: fm,d =

■

My

f m ,k ×

k mod

γM

× k sys × k h = 24 × 10 6 ×

0. 8 × 1.0 × 1.0 = 14.769 × 10 6 Pa 1.3

Work ratio according to formulae 6.11 from EN 1995-1-1 norm:

ADVANCE DESIGN VALIDATION GUIDE

σ m ,d f m ,d ■

≤ 1.0

Work ratio according to formulae 6.12 from EN 1995-1-1 norm:

km ×

σ m ,d f m ,d

≤ 1.0

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:

winst (Q) ≤ w fin ≤

L 300

L 125

wnet , fin ≤

L 200

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):

winst (Q) = 0.00749m ⇒ winst (Q) = ■

Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

winst = d CQ = 0.00999m ⇒ winst = ■

L 600.8 L 450.45

In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the deformation factor (kdef) has to be chosen:

k def = 0.6

(calculated value for service class 1, according to table 3.2 from EN 1995-1-1)

wcreep = 0.6 × d QP = 0.6 × 0.00475m = 0.00285m ⇒ wcreep = ■

L 1578.95

Final deflection:

w fin = winst + wcreep = 0.00999m + 0.00285m = 0.01284m ⇒ w fin = ■

L 350.47

Net deflection:

wnet , fin = w fin + wc = 0.01284m + 0m = 0.01284m ⇒ wnet , fin =

L 350.47

Finite elements modeling ■ ■ ■

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

559

ADVANCE DESIGN VALIDATION GUIDE

Work ratio diagram Simply supported beam subjected to bending Strength work ratio

13.18.2.3 Reference results 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

13.18.3Calculated results

560

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 %

031617-0409-0906