AD Validation Guide Vol2 2015 En

8/11/2019 AD Validation Guide Vol2 2015 En

http://slidepdf.com/reader/full/ad-validation-guide-vol2-2015-en 1/457





Advance Design

Validation Guide

Part 2

Version: 2015

Tests passed on: 16 April 2014

Number of tests: 519









ADVANCE VALIDATION GUIDE

7

Table of Contents

1 GENERAL APPLICATIONS.................................................................................................. 13

1.1 Verifying element creation using commas for coordinates (TTAD #11141) ..................................................14

1.2 Defining the reinforced concrete design assumptions (TTAD #12354).........................................................14

1.3 Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724)...........................................14

1.4 Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238) ..................................14

1.5 Verifying the precision of linear and planar concrete covers (TTAD #12525)...............................................14

1.6 Changing the default material (TTAD #11870) ............................................................................................. 14

1.7 Verifying the objects rename function (TTAD #12162) ................................................................................. 15

1.8 Importing a cross section from the Advance Steel profiles library (TTAD #11487) .......................................15

1.9 Creating a new Advance Design file using the "New" command from the "Standard" toolbar (TTAD #12102)... ...... 15

1.10 Verifying the appearance of the local x orientation legend (TTAD #11737)................................................15

1.11 Creating system trees using the copy/paste commands (DEV2012 #1.5) ..................................................15

1.12 Creating system trees using the copy/paste commands (DEV2012 #1.5) ..................................................16

1.13 Launching the verification of a model containing steel connections (TTAD #12100) ..................................16

1.14 Creating and updating model views and post-processing views (TTAD #11552) .......................................16

1.15 Verifying mesh, CAD and climatic forces - LPM meeting............................................................................16

1.16 Verifying material properties for C25/30 (TTAD #11617) ............................................................................16

1.17 Verifying geometry properties of elements with compound cross sections (TTAD #11601) .......................16

1.18 Verifying the synthetic table by type of connection (TTAD #11422)............................................................17

1.19 Verifying the overlapping when a planar element is built in a hole.(TTAD #13772) ....................................17

2 IMPORT / EXPORT................................................................................................................ 19

2.1 Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952) ..........................................20

2.2 Exporting an Advance Design model to DO4 format (DEV2012 #1.10) ........................................................20

2.3 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) ..........................................................20

2.4 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) ..........................................................20

2.5 Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172)....................20

2.6 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197).......... 21

2.7 Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9) .............21

2.8 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197).......... 21

2.9 System stability when importing AE files with invalid geometry (TTAD #12232) ..........................................21

2.10 Verifying the releases option of the planar elements edges after the model was exported and imported viaGTC format (TTAD #12137) ................................................................................................................................... 21

2.11 Verifying the load case properties from models imported as GTC files (TTAD #12306) .............................22

2.12 Exporting linear elements to IFC format (TTAD #10561)............................................................................22 2.13 Importing IFC files containing continuous foundations (TTAD #12410) ......................................................22




8

2.14 Exporting a meshed model to GTC (TTAD #12550)................................................................................... 22

2.15 Importing GTC files containing "PH.RDC" system (TTAD #12055) ............................................................ 22

3 JOINT DESIGN...................................................................................................................... 23

3.1 Deleting a welded tube connection - 1 gusset bar (TTAD #12630).............................................................. 24

3.2 Creating connections groups (TTAD #11797) .............................................................................................. 24

4 MESH..................................................................................................................................... 25

4.1 Verifying the mesh for a model with generalized buckling (TTAD #11519) .................................................. 26

4.2 Verifying the options to take into account loads in linear and planar elements mesh.(TTAD #15251) .........26

4.3 Creating triangular mesh for planar elements (TTAD #11727)..................................................................... 26

4.4 Verifying mesh points (TTAD #11748) ......................................................................................................... 26

4.5 Verifying the mesh of a planar element influenced by peak smoothing........................................................ 26

5 OLD CLIMATIC GENERATOR ............................................................................................. 27 5.1 NV2009: Verifying wind on a protruding canopy. (TTAD #13880)................................................................ 28

5.2 NV2009: Verifying wind and snow reports for a protruding roof (TTAD #11318).......................................... 28

5.3 NV2009: Generating wind loads on a 2 slopes 3D portal frame at 15m height (TTAD #12604)................... 28

5.4 NV2009: generating wind loads and snow loads on a simple structure with planar support (TTAD #11380)28

6 REPORTS GENERATOR...................................................................................................... 29

6.1 Generating a report with modal analysis results (TTAD #10849)................................................................. 30

6.2 System stability when the column releases interfere with support restraints (TTAD #10557) ...................... 30

6.3 Generating the critical magnification factors report (TTAD #11379)............................................................. 30

6.4 Modal analysis: eigen modes results for a structure with one level .............................................................. 30

6.5 Verifying the model geometry report (TTAD #12201)................................................................................... 30

6.6 Verifying the shape sheet for a steel beam (TTAD #12455)......................................................................... 31

6.7 Verifying the steel shape sheet display (TTAD #12657)............................................................................... 31

6.8 Verifying the shape sheet strings display (TTAD #12622)............................................................................ 31

6.9 Verifying the shape sheet report (TTAD #12353)......................................................................................... 31

6.10 Verifying the EC2 calculation assumptions report (TTAD #11838)............................................................. 31

6.11 Verifying the Max row on the user table report (TTAD #12512).................................................................. 31

6.12 Verifying the global envelope of linear elements stresses (on all quarters of super element) (TTAD #12230)........ 32

6.13 Verifying the global envelope of linear elements stresses (on the end point of super element) (TTAD#12230, TTAD #12261).......................................................................................................................................... 32

6.14 Verifying the Min/Max values from the user reports (TTAD# 12231) .......................................................... 32

6.15 Verifying the global envelope of linear elements forces result (on end points and middle of super element)(TTAD #12230) ...................................................................................................................................................... 32

6.16 Verifying the global envelope of linear elements displacements (on all quarters of super element) (TTAD #12230)....33

6.17 Verifying the global envelope of linear elements displacements (on end points and middle of superelement) (TTAD #12230)........................................................................................................................................ 33

6.18 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element) (TTAD #12230) ...... 33




9

6.19 Creating the rules table (TTAD #11802) ..................................................................................................... 33

6.20 Creating the steel materials description report (TTAD #11954)..................................................................34

6.21 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element) (TTAD #12230)..........34

6.22 Verifying the global envelope of linear elements forces result (on start and end of super element) (TTAD #12230)....34

6.23 Verifying the global envelope of linear elements forces result (on the start point of super element) (TTAD #12230)...34

6.24 Verifying the global envelope of linear elements forces result (on all quarters of super element) (TTAD #12230).......35

6.25 Verifying the global envelope of linear elements forces result (on the end point of super element) (TTAD#12230, #12261) .................................................................................................................................................... 35

6.26 Verifying the global envelope of linear elements stresses (on each 1/4 of mesh element) (TTAD #12230)35

6.27 Verifying the global envelope of linear elements displacements (on the end point of super element) (TTAD#12230, TTAD #12261) .......................................................................................................................................... 35

6.28 Verifying the global envelope of linear elements displacements (on the start point of super element) (TTAD #12230)36

6.29 Verifying the global envelope of linear elements displacements (on start and end of super element) (TTAD #12230) 36

6.30 Verifying the global envelope of linear elements stresses (on start and end of super element) (TTAD #12230).........36

6.31 Verifying the global envelope of linear elements stresses (on the start point of super element) (TTAD #12230).........36

6.32 Verifying the global envelope of linear elements stresses (on end points and middle of super element)(TTAD #12230)....................................................................................................................................................... 37

6.33 Verifying the modal analysis report (TTAD #12718) ................................................................................... 37

7 SEISMIC ANALYSIS ............................................................................................................. 39

7.1 EC8 Fr Annex: Generating forces results per modes on linear and planar elements (TTAD #13797)..........40

7.2 Verifying the displacements results of a linear element for an envelope spectrum according to RomanianEC8 appendix (DEV2013 #8.2) .............................................................................................................................. 40

7.3 Verifying the displacements results of a linear element for an envelope spectrum according to FrenchPS92/100 standard (DEV2013 #8.2) ...................................................................................................................... 40

7.4 Verifying the displacements results of a linear element according to Marocco seismic standards (RPS 2011)(DEV2013 #3.6)...................................................................................................................................................... 40

7.5 Verifying the displacements results of a linear element for an envelope spectrum according to Maroccoseismic standards (RPS 2011) (DEV2013 #8.2)..................................................................................................... 41

7.6 Verifying the displacements results of a linear element for an envelope spectrum according to EurocodeEC8 standard (DEV2013 #8.2)............................................................................................................................... 41

7.7 EC8 Romanian Annex: verifying action results and torsors per modes on point, linear and planar supports(TTAD #14840)....................................................................................................................................................... 41

7.8 EC8 French Annex: verifying torsors on walls, elastic linear supports and user-defined section cuts (TTAD #14460) ..41

7.9 EC 8 General Annex: verifying torsors on a 6 storey single concrete core subjected to horizontal forces andseismic action ......................................................................................................................................................... 42

7.10 Verifying the displacements results of a linear element for spectrum with renewed building option,according to Eurocode EC8 standard (TTAD #14161) ........................................................................................... 42

7.11 Verifying the spectrum results for EC8 seism (TTAD #12472)....................................................................42

7.12 EC8 French Annex: verifying torsors on walls............................................................................................. 42

7.13 EC8: verifying the sum of actions on supports and nodes restraints (TTAD #12706) .................................42

7.14 Seismic norm PS92: verifying efforts and torsors on planar elements (TTAD #12974) ..............................43

7.15 Verifying seismic efforts on planar elements with Q4 and T3-Q4 mesh type (TTAD #14244) ....................43

7.16 Verifying the damping correction influence over the efforts in supports (TTAD #13011). ...........................43




10

7.17 EC8 French Annex: verifying seismic results when a design spectrum is used (TTAD #13778) ................ 43

7.18 Verifying the displacements results of a linear element according to Algerian seismic standards (RPA99/2003) (DEV2013 #3.5) ...................................................................................................................................... 43

7.19 EC8 French Annex: verifying torsors on grouped walls from a multi-storey concrete structure..................44

7.20 Verifying the spectrum results for EC8 seism (TTAD #11478) ................................................................... 44

7.21 EC8 : Verifying the displacements results of a linear element according to Czech seismic standards (CSNEN 1998-1) (DEV2012 #3.18) ................................................................................................................................ 44

7.22 Verifying signed concomitant linear elements envelopes on Fx report (TTAD #11517) ............................. 44

7.23 Verifying the combinations description report (TTAD #11632) ................................................................... 44

7.24 EC8: Verifying earthquake description report in analysis with Z axis down. (TTAD #15095) ..................... 45

8 STEEL DESIGN..................................................................................................................... 47

8.1 Verifying shape sheet on S275 beam (TTAD #11731)................................................................................. 48

8.2 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770) .......................... 48

8.3 Generating the shape sheet by system (TTAD #11471) .............................................................................. 48

8.4 Verifying the calculation results for steel cables (TTAD #11623) ................................................................. 48

8.5 Verifying the cross section optimization according to EC3 (TTAD #11516) ................................................. 49

8.6 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522) .................................. 49

8.7 Verifying the shape sheet results for a column (TTAD #11550)................................................................... 49

8.8 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method ................................... 49

8.9 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method................................. 50

8.10 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method ............................ 50

8.11 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545)........................................... 50

8.12 Verifying the buckling coefficient Xy on a class 2 section........................................................................... 50

8.13 Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873)51

8.14 EC3 Test 22: Verifying the lateral torsional buckling of a IPE300 beam..................................................... 52

8.15 EC3 Test 28: Verifying an user defined I section class 1, column fixed on base and without any other restraint.... 58

8.16 EC3 Test 23: Verifying a IPE400 column for compression, shear, bending moment, buckling, lateraltorsional buckling and bending and axial compression ......................................................................................... 88

8.17 EC3 Test 7: Class section classification and compression resistance for an IPE600 column .................. 124

8.18 EC3 test 4: Class section classification and bending moment verification of an IPE300 column.............. 130 8.19 EC3 Test 2: Class section classification and shear verification of an IPE300 beam subjected to linearuniform loading..................................................................................................................................................... 136

8.20 EC3 Test 3: Class section classification, shear and bending moment verification of an IPE300 column .143

8.21 EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 beam . 151

8.22 EC3 Test 1: Class section classification and compression verification of an IPE300 column................... 159

8.23 EC3 Test 5: Class section classification and combined axial force with bending moment verification of anIPE300 column..................................................................................................................................................... 166

8.24 Changing the steel design template for a linear element (TTAD #12491) ................................................ 172

8.25 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation(TTAD #12389) .................................................................................................................................................... 172






12

8.55 EC3 Test 44: Determining lateral torsional buckling parameters for a I-shaped welded built-up beamconsidering the load applied on the upper flange................................................................................................. 363

8.56 EC3 Test 43: Determining lateral torsional buckling parameters for a I-shaped laminated beam consideringthe load applied on the lower flange..................................................................................................................... 369

8.57 EC3 test 8: Verifying the classification and the resistance of a column subjected to bending and axial load370

8.58 EC3 Test 20: Verifying the buckling resistance of a RC3020100 column................................................. 376

8.59 EC3 test 10: Verifying the classification and the bending resistance of a welded built-up beam.............. 383

8.60 EC3 test 11: Cross section classification and compression resistance verification of a rectangular hollowsection column ..................................................................................................................................................... 389

8.61 EC3 Test 19: Verifying the buckling resistance for a IPE300 column....................................................... 394

8.62 EC3 Test 12: Verifying the design plastic shear resistance of a rectangular hollow section beam........... 401

8.63 EC3 Test 13: Verifying the resistance of a rectangular hollow section column subjected to bending andshear efforts ......................................................................................................................................................... 404

9 TIMBER DESIGN................................................................................................................. 411

9.1 Modifying the "Design experts" properties for timber linear elements (TTAD #12259) ............................... 412

9.2 Verifying the units display in the timber shape sheet (TTAD #12445)........................................................ 412

9.3 Verifying the timber elements shape sheet (TTAD #12337)....................................................................... 412

9.4 EC5: Verifying a timber purlin subjected to oblique bending ...................................................................... 413

9.5 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes .............................. 417

9.6 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending ...................................... 420

9.7 EC5: Shear verification for a simply supported timber beam...................................................................... 425

9.8 EC5: Verifying a timber beam subjected to simple bending ....................................................................... 426

9.9 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression.......430

9.10 EC5: Verifying a C24 timber beam subjected to shear force .................................................................... 435

9.11 EC5: Verifying a timber column subjected to tensile forces...................................................................... 439

9.12 EC5: Verifying a timber column subjected to compression forces............................................................ 442

9.13 EC5: Verifying a timber beam subjected to combined bending and axial tension..................................... 446

9.14 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression ................................. 452



1 General applications




14

1.1 Verifying element creation using commas for coordinates (TTAD #11141)

Test ID: 4554

Test status: Passed

1.1.1 Description

Verifies element creation using commas for coordinates in the command line.

1.2 Defining the reinforced concrete design assumptions (TTAD #12354)

Test ID: 4528

Test status: Passed

1.2.1 Description

Verifies the definition of the reinforced concrete design assumptions.

1.3 Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724)

Test ID: 4361

Test status: Passed

1.3.1 Description

Generates liquid pressure on a concrete structure with horizontal and vertical surfaces. Generates the loadingsdescription report.

1.4 Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238)

Test ID: 4480Test status: Passed

1.4.1 Description

Performs the finite elements calculation on a model with 2 joined vertical elements with the clipping option enabled.

1.5 Verifying the precision of linear and planar concrete covers (TTAD #12525)

Test ID: 4547

Test status: Passed

1.5.1 DescriptionVerifies the linear and planar concrete covers precision.

1.6 Changing the default material (TTAD #11870)

Test ID: 4436

Test status: Passed

1.6.1 Description

Selects a different default material for concrete elements.




15

1.7 Verifying the objects rename function (TTAD #12162)

Test ID: 4229

Test status: Passed

1.7.1 Description

Verifies the renaming function from the properties list of a linear element. The name contains the "_" character.

1.8 Importing a cross section from the Advance Steel profiles library (TTAD #11487)

Test ID: 3719

Test status: Passed

1.8.1 Description

Imports the Canam Z 203x10.0 section from the Advance Steel Profiles library in the Advance Design list of available

cross sections.

1.9 Creating a new Advance Design file using the "New" command from the "Standard" toolbar(TTAD #12102)

Test ID: 4095

Test status: Passed

1.9.1 Description

Creates a new .fto file using the "New" command from the "Standard" toolbar. Creates a rigid punctual support andtests the CAD coordinates.

1.10 Verifying the appearance of the local x orientation legend (TTAD #11737)

Test ID: 4109

Test status: Passed

1.10.1Description

Verifies the color legend display. On a model with a planar element, the color legend is enabled; it displays theelements by the local x orientation color.

1.11 Creating system trees using the copy/paste commands (DEV2012 #1.5)

Test ID: 4162

Test status: Passed

1.11.1Description

Creates system trees using the copy/paste commands in the Pilot. The source system consists of severalsubsystems. The target system is on the same level as the source system.




16

1.12 Creating system trees using the copy/paste commands (DEV2012 #1.5)

Test ID: 4164

Test status: Passed

1.12.1Description

Creates system trees using the copy/paste commands in the Pilot. The source system consists of severalsubsystems. The target system is in the source system.

1.13 Launching the verification of a model containing steel connections (TTAD #12100)

Test ID: 4096

Test status: Passed

1.13.1Description

Launches the verification function for a model containing a beam-column steel connection.

1.14 Creating and updating model views and post-processing views (TTAD #11552)

Test ID: 3820

Test status: Passed

1.14.1Description

Creates and updates the model view and the post-processing views for a simple steel frame.

1.15 Verifying mesh, CAD and climatic forces - LPM meeting

Test ID: 4079

Test status: Passed

1.15.1Description

Generates the "Description of climatic loads" report for a model consisting of a concrete structure with dead loads,wind loads, snow loads and seism loads. Performs the model verification, meshing and finite elements calculation.Generates the "Displacements of linear elements by element" report.

1.16 Verifying material properties for C25/30 (TTAD #11617)

Test ID: 3571

Test status: Passed

1.16.1Description

Verifies the material properties on a model with a concrete (C25/30) bar. Generates the material description report.

1.17 Verifying geometry properties of elements with compound cross sections (TTAD #11601)

Test ID: 3546

Test status: Passed

1.17.1Description

Verifies the geometry properties of a steel structure with elements which have compound cross sections (CS4IPE330 UPN240). Generates the geometrical data report.




17

1.18 Verifying the synthetic table by type of connection (TTAD #11422)

Test ID: 3646

Test status: Passed

1.18.1Description

Performs the finite elements calculation and the steel calculation on a structure with four types of connections.Generates the "Synthetic table by type of connection" report.

The structure consists of linear steel elements (S275) with CE505, IPE450, IPE140 and IPE500 cross section. Themodel connections: columns base plates, beam - column fixed connections, beam - beam fixed connections andgusset plate. Live loads, snow loads and wind loads are applied.

1.19 Verifying the overlapping when a planar element is built in a hole.(TTAD #13772)

Test ID: 6214

Test status: Passed

1.19.1Description

Verifies that when a planar element is built in a hole of another planar element by checking the mesh.





2 Import / Export




20

2.1 Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952)

Test ID: 3628

Test status: Passed

2.1.1 Description

Exports a linear element to GTC.

2.2 Exporting an Advance Design model to DO4 format (DEV2012 #1.10)

Test ID: 4101

Test status: Passed

2.2.1 Description

Launches the "Export > Text file" command and saves the current project as a .do4 archive file.

The model contains all types of structural elements, loads and geometric objects.

2.3 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3)

Test ID: 4195

Test status: Passed

2.3.1 Description

Exports the analysis model to ADA (through GTC) with:

- Export results: disabled

- Export meshed model: enabled

2.4 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3)

Test ID: 4193

Test status: Passed

2.4.1 Description

Exports the analysis model to ADA (through GTC) with:

- Export results: enabled

- Export meshed model: disabled

2.5 Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172)

Test ID: 4297

Test status: Passed

2.5.1 Description

Imports a GTC file from SuperSTRESS. The file contains steel linear elements with haunch sections.




21

2.6 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS(TTAD #12197)

Test ID: 4388

Test status: Passed

2.6.1 Description

Imports a GTC file from SuperSTRESS. The file contains elements with circular hollow section. Verifies if the crosssections are imported from the attached "UK Steel Sections" database.

2.7 Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9)

Test ID: 4445

Test status: Passed

2.7.1 DescriptionVerifies the GTC files exchange (import/export) between Advance Design and SuperSTRESS.

2.8 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS(TTAD #12197)

Test ID: 4389

Test status: Passed

2.8.1 Description

Imports a GTC file from SuperSTRESS when the "UK Steel Sections" database is not attached. The file contains

elements with circular hollow section. Verifies the cross sections definition.

2.9 System stability when importing AE files with invalid geometry (TTAD #12232)

Test ID: 4479

Test status: Passed

2.9.1 Description

Imports a complex model containing elements with invalid geometry.

2.10 Verifying the releases option of the planar elements edges after the model was exported and

imported via GTC format (TTAD #12137)

Test ID: 4506

Test status: Passed

2.10.1Description

Exports to GTC a model with planar elements on which the edges releases were defined. Imports the GTC file toverify the planar elements releases option.




22

2.11 Verifying the load case properties from models imported as GTC files (TTAD #12306)

Test ID: 4515

Test status: Passed

2.11.1Description

Performs the finite elements calculation on a model with dead load cases and exports the model to GTC. Imports theGTC file to verify the load case properties.

2.12 Exporting linear elements to IFC format (TTAD #10561)

Test ID: 4530

Test status: Passed

2.12.1Description

Exports to IFC format a model containing linear elements having sections of type "I symmetric" and "I asymmetric".

2.13 Importing IFC files containing continuous foundations (TTAD #12410)

Test ID: 4531

Test status: Passed

2.13.1Description

Imports an IFC file containing a continuous foundation (linear support) and verifies the element display.

2.14 Exporting a meshed model to GTC (TTAD #12550)

Test ID: 4552

Test status: Passed

2.14.1Description

Exports a meshed model to GTC. The meshed planar element from the model contains a triangular mesh.

2.15 Importing GTC files containing "PH.RDC" system (TTAD #12055)

Test ID: 4548

Test status: Passed

2.15.1Description

Imports a GTC file exported from Advance Design. The file contains the automatically created system "PH.RDC".



3 Joint Design




24

3.1 Deleting a welded tube connection - 1 gusset bar (TTAD #12630)

Test ID: 4561

Test status: Passed

3.1.1 Description

Deletes a welded tube connection - 1 gusset bar after the joint was exported to ADSC.

3.2 Creating connections groups (TTAD #11797)

Test ID: 4250

Test status: Passed

3.2.1 Description

Verifies the connections groups function.



4 Mesh




26

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

Test ID: 3649

Test status: Passed

4.1.1 Description

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

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

Test ID: 6215

Test status: Passed

4.2.1 Description

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

4.3 Creating triangular mesh for planar elements (TTAD #11727)

Test ID: 3423

Test status: Passed

4.3.1 Description

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

4.4 Verifying mesh points (TTAD #11748)

Test ID: 3458

Test status: Passed

4.4.1 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 thestructure: dead loads, live loads, wind loads and snow loads, according to Eurocodes.

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

Test ID: 6190

Test status: Passed

4.5.1 Description

Verifies the mesh of a planar concrete element influenced by peak smoothing.

The model consists in a c25/30 concrete planar element and four concrete columns (2 x r20/30, 1 x IPE400, 1 x D40)and is subjected to self weight and 2 live loads of -20KN respectively -100 KN.



5 Old Climatic Generator




28

5.1 NV2009: Verifying wind on a protruding canopy. (TTAD #13880)

Test ID: 6173

Test status: Passed

5.1.1 Description

Generates wind on a protruding canopy according to NV2009 - French climatic standards.

5.2 NV2009: Verifying wind and snow reports for a protruding roof (TTAD #11318)

Test ID: 4536

Test status: Passed

5.2.1 Description

Generates wind loads and snow loads according to NV2009 - French climatic standards. Verifies wind and snowreports for a protruding roof.

5.3 NV2009: Generating wind loads on a 2 slopes 3D portal frame at 15m height (TTAD #12604)

Test ID: 4567

Test status: Passed

5.3.1 Description

Generates wind loads on a 2 slopes 3D portal frame at 15m height, according to the French standard (NV2009).

5.4 NV2009: generating wind loads and snow loads on a simple structure with planar support(TTAD #11380)

Test ID: 4091

Test status: Passed

5.4.1 Description

Generates wind loads and snow loads on the windwalls of a concrete structure with a planar support, according toNV2009 French standards.

The structure has concrete columns and beams (R20*30 cross section and C20/25 material).



6 Reports Generator




30

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

Test ID: 3734

Test status: Passed

6.1.1 Description

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

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

Test ID: 3717

Test status: Passed

6.2.1 Description

Performs the finite elements calculation and generates the systems description report for a structure which hascolumn releases that interfere with the supports restraints.

The structure consists of steel beams and steel columns (S235 material, HEA550 cross section) with rigid fixedsupports.

6.3 Generating the critical magnification factors report (TTAD #11379)

Test ID: 3647

Test status: Passed

6.3.1 Description

Performs the finite elements calculation and the steel calculation on a structure with four types of connections.Generates the "Synthetic table by type of connection" report.

The structure consists of linear steel elements (S275) with CE505, IPE450, IPE140 and IPE500 cross section. Themodel connections: columns base plates, beam - column fixed connections, beam - beam fixed connections andgusset plate. Live loads, snow loads and wind loads are applied.

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

Test ID: 3668

Test status: Passed

6.4.1 Description

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 isdefined.

6.5 Verifying the model geometry report (TTAD #12201)

Test ID: 4467

Test status: Passed

6.5.1 Description

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




31

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

Test ID: 4535

Test status: Passed

6.6.1 Description

Verifies the shape sheet for a steel beam.

6.7 Verifying the steel shape sheet display (TTAD #12657)

Test ID: 4562

Test status: Passed

6.7.1 Description

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

6.8 Verifying the shape sheet strings display (TTAD #12622)

Test ID: 4559

Test status: Passed

6.8.1 Description

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

6.9 Verifying the shape sheet report (TTAD #12353)

Test ID: 4545

Test status: Passed

6.9.1 Description

Generates and verifies the shape sheet report.

6.10 Verifying the EC2 calculation assumptions report (TTAD #11838)

Test ID: 4544

Test status: Passed

6.10.1Description

Verifies the EC2 calculation assumptions report.

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

Test ID: 4558

Test status: Passed

6.11.1Description

Verifies the Max row on the user table report.




32

6.12 Verifying the global envelope of linear elements stresses (on all quarters of super element)(TTAD #12230)

Test ID: 4499

Test status: Passed

6.12.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements stresses on all quarters ofsuper element by generating the "Global envelope of linear elements stresses report".

The model consists of a concrete portal frame with rigid fixed supports.

6.13 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

6.13.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements stresses on the end pointof super element by generating the "Global envelope of linear elements stresses report".


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

Test ID: 4505

Test status: Passed

6.14.1DescriptionPerforms the finite elements calculation and generates a user report containing the results of Min/Max values.

6.15 Verifying the global envelope of linear elements forces result (on end points and middle ofsuper element) (TTAD #12230)

Test ID: 4488

Test status: Passed

6.15.1Description

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.




33

6.16 Verifying the global envelope of linear elements displacements (on all quarters of superelement) (TTAD #12230)

Test ID: 4493

Test status: Passed

6.16.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements displacements on allquarters of super element by generating the "Global envelope of linear elements displacements report".


6.17 Verifying the global envelope of linear elements displacements (on end points and middle ofsuper element) (TTAD #12230)

Test ID: 4494

Test status: Passed

6.17.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements displacements on endpoints and middle of super element by generating the "Global envelope of linear elements displacements report".


6.18 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element)(TTAD #12230)

Test ID: 4492

Test status: Passed

6.18.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements displacements on each1/4 of mesh element by generating the "Global envelope of linear elements displacements report".


6.19 Creating the rules table (TTAD #11802)

Test ID: 4099

Test status: Passed

6.19.1Description

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 thesteel calculation.




34

6.20 Creating the steel materials description report (TTAD #11954)

Test ID: 4100

Test status: Passed

6.20.1Description

Generates the "Steel materials" report as a .txt file.

The model consists of a steel structure with supports and a base plate connection.

6.21 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element)(TTAD #12230)

Test ID: 4486

Test status: Passed

6.21.1DescriptionPerforms the finite elements calculation and verifies the global envelope of linear elements forces results on each 1/4of mesh element by generating the "Global envelope of linear elements forces result report".


6.22 Verifying the global envelope of linear elements forces result (on start and end of superelement) (TTAD #12230)

Test ID: 4489

Test status: Passed

6.22.1DescriptionPerforms the finite elements calculation and verifies the global envelope of linear elements forces results on the startand end of super element by generating the "Global envelope of linear elements forces result report".


6.23 Verifying the global envelope of linear elements forces result (on the start point of superelement) (TTAD #12230)

Test ID: 4490

Test status: Passed

6.23.1DescriptionPerforms the finite elements calculation and verifies the global envelope of linear elements forces results on the startpoint of super element by generating the "Global envelope of linear elements forces result report".





35

6.24 Verifying the global envelope of linear elements forces result (on all quarters of super element)(TTAD #12230)

Test ID: 4487

Test status: Passed

6.24.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements forces results on allquarters of super element by generating the "Global envelope of linear elements forces result report".


6.25 Verifying the global envelope of linear elements forces result (on the end point of superelement) (TTAD #12230, #12261)

Test ID: 4491

Test status: Passed

6.25.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the endpoint of super element by generating the "Global envelope of linear elements forces result report".


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

Test ID: 4498

Test status: Passed

6.26.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements stresses on each 1/4 ofmesh element by generating the "Global envelope of linear elements stresses report".


6.27 Verifying the global envelope of linear elements displacements (on the end point of superelement) (TTAD #12230, TTAD #12261)

Test ID: 4497

Test status: Passed

6.27.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements displacements on the endpoint of super element by generating the "Global envelope of linear elements displacements report".





36

6.28 Verifying the global envelope of linear elements displacements (on the start point of superelement) (TTAD #12230)

Test ID: 4496

Test status: Passed

6.28.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements displacements on the startpoint of super element by generating the "Global envelope of linear elements displacements report".


6.29 Verifying the global envelope of linear elements displacements (on start and end of superelement) (TTAD #12230)

Test ID: 4495

Test status: Passed

6.29.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements displacements on startand end of super element by generating the "Global envelope of linear elements displacements report".


6.30 Verifying the global envelope of linear elements stresses (on start and end of super element)(TTAD #12230)

Test ID: 4501

Test status: Passed

6.30.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements stresses on start and endof super element by generating the "Global envelope of linear elements stresses report".


6.31 Verifying the global envelope of linear elements stresses (on the start point of super element)(TTAD #12230)

Test ID: 4502

Test status: Passed

6.31.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements stresses on the start pointof super element by generating the "Global envelope of linear elements stresses report".





37

6.32 Verifying the global envelope of linear elements stresses (on end points and middle of superelement) (TTAD #12230)

Test ID: 4500

Test status: Passed

6.32.1Description

Performs the finite elements calculation and verifies the global envelope of linear elements stresses on end pointsand middle of super element by generating the "Global envelope of linear elements stresses report".


6.33 Verifying the modal analysis report (TTAD #12718)

Test ID: 4576

Test status: Passed

6.33.1Description

Generates and verifies the modal analysis report.





7 Seismic analysis




40

7.1 EC8 Fr Annex: Generating forces results per modes on linear and planar elements (TTAD#13797)

Test ID: 5455

Test status: Passed

7.1.1 Description

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

7.2 Verifying the displacements results of a linear element for an envelope spectrum according toRomanian EC8 appendix (DEV2013 #8.2)

Test ID: 5601

Test status: Passed

7.2.1 Description

Verifies the displacements results of a vertical linear element according to Romanian EC8 appendix. Performs thefinite 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 theelement: self weight and seismic loads for an envelope spectrum.

7.3 Verifying the displacements results of a linear element for an envelope spectrum according toFrench PS92/100 standard (DEV2013 #8.2)

Test ID: 5599

Test status: Passed

7.3.1 Description

Verifies the displacements results of a vertical linear element according to French PS92/100 standard. Performs thefinite elements calculation and generates the displacements of linear elements by load case and by element reports.


7.4 Verifying the displacements results of a linear element according to Marocco seismicstandards (RPS 2011) (DEV2013 #3.6)

Test ID: 5597

Test status: Passed

7.4.1 Description

Verifies the displacements results of a vertical linear element according to Marocco seismic standard. Performs thefinite 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 theelement: self weight and seismic loads (RPS 2011).




41

7.5 Verifying the displacements results of a linear element for an envelope spectrum according toMarocco seismic standards (RPS 2011) (DEV2013 #8.2)

Test ID: 5598

Test status: Passed

7.5.1 Description

Verifies the displacements results of a vertical linear element according to Marocco seismic standard. Performs thefinite 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 theelement: self weight and seismic loads for an envelope spectrum (RPS 2011).

7.6 Verifying the displacements results of a linear element for an envelope spectrum according toEurocode EC8 standard (DEV2013 #8.2)

Test ID: 5600

Test status: Passed

7.6.1 Description

Verifies the displacements results of a vertical linear element according to Eurocode EC8 standard. Performs thefinite elements calculation and generates the displacements of linear elements by load case and by element reports.


7.7 EC8 Romanian Annex: verifying action results and torsors per modes on point, linear andplanar supports (TTAD #14840)

Test ID: 5860

Test status: Passed

7.7.1 Description

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

7.8 EC8 French Annex: verifying torsors on walls, elastic linear supports and user-defined sectioncuts (TTAD #14460)

Test ID: 5861

Test status: Passed

7.8.1 Description

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




42

7.9 EC 8 General Annex: verifying torsors on a 6 storey single concrete core subjected tohorizontal forces and seismic action

Test ID: 6088

Test status: Passed

7.9.1 Description

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

7.10 Verifying the displacements results of a linear element for spectrum with renewed buildingoption, according to Eurocode EC8 standard (TTAD #14161)

Test ID: 6192

Test status: Passed

7.10.1Description

Verifies the displacements results of a linear element for spectrum with renewed building option, according toEurocode EC8 standard. Performs the finite elements calculation and generates the displacements of linear elementsby load case and by element reports.


7.11 Verifying the spectrum results for EC8 seism (TTAD #12472)

Test ID: 4537

Test status: Passed

7.11.1Description

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

7.12 EC8 French Annex: verifying torsors on walls

Test ID: 4803

Test status: Passed

7.12.1Description

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

7.13 EC8: verifying the sum of actions on supports and nodes restraints (TTAD #12706)

Test ID: 4859

Test status: Passed

7.13.1Description

Verifies the sum of actions on supports and nodes restraints for a simple structure subjected to seismic action

according to EC8 French annex.




43

7.14 Seismic norm PS92: verifying efforts and torsors on planar elements (TTAD #12974)

Test ID: 4858

Test status: Passed

7.14.1Description

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

7.15 Verifying seismic efforts on planar elements with Q4 and T3-Q4 mesh type (TTAD #14244)

Test ID: 5492

Test status: Passed

7.15.1Description

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

7.16 Verifying the damping correction influence over the efforts in supports (TTAD #13011).

Test ID: 4853

Test status: Passed

7.16.1Description

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

7.17 EC8 French Annex: verifying seismic results when a design spectrum is used (TTAD #13778)

Test ID: 5425

Test status: Passed

7.17.1Description

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

7.18 Verifying the displacements results of a linear element according to Algerian seismicstandards (RPA 99/2003) (DEV2013 #3.5)

Test ID: 5559

Test status: Passed

7.18.1Description

Verifies the displacements results of a vertical linear element according to Algerian seismic standard. Performs thefinite 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 theelement: self weight and seismic loads (RPA 99/2003).




44

7.19 EC8 French Annex: verifying torsors on grouped walls from a multi-storey concrete structure

Test ID: 4810

Test status: Passed

7.19.1Description

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

7.20 Verifying the spectrum results for EC8 seism (TTAD #11478)

Test ID: 3703

Test status: Passed

7.20.1Description

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

7.21 EC8 : Verifying the displacements results of a linear element according to Czech seismicstandards (CSN EN 1998-1) (DEV2012 #3.18)

Test ID: 3626

Test status: Passed

7.21.1Description

Verifies the displacements results of an inclined linear element according to Eurocodes 8 Czech standard. Performsthe 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 theelement: self weight and seismic loads (CSN EN 1998-1).

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

Test ID: 3576

Test status: Passed

7.22.1Description

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.

7.23 Verifying the combinations description report (TTAD #11632)

Test ID: 3544

Test status: Passed

7.23.1Description

Generates a dead load case, two live load cases and a snow load case, defines the concomitance between the

generated load cases and generates the combinations description report.



7.24 EC8: Verifying earthquake description report in analysis with Z axis down. (TTAD #15095)

Test ID: 6207

Test status: Passed

7.24.1Description

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





8 Steel design




48

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

Test ID: 3434

Test status: Passed

8.1.1 Description

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

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

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

Test ID: 3406

Test status: Passed

8.2.1 Description

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

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

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

Test ID: 3575

Test status: Passed

8.3.1 Description

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

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

Test ID: 3560

Test status: Passed

8.4.1 Description

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




49

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

Test ID: 3620

Test status: Passed

8.5.1 Description

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

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

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

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

Test ID: 3612

Test status: Passed

8.6.1 Description

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

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

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

Test ID: 3640

Test status: Passed

8.7.1 Description

Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a verticalsteel element.

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

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

Test ID: 3819

Test status: Passed

8.8.1 Description

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

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




50

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

Test ID: 3813

Test status: Passed

8.9.1 Description

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


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

Test ID: 3814

Test status: Passed

8.10.1Description

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


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

Test ID: 3641

Test status: Passed

8.11.1Description

Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a horizontalsteel 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 withtranslation restraints on Y and Z and rotation restraint on X at the other end. A linear live load of -10.00 kN on FX anda punctual live load of 1000 kN on FX are applied.

8.12 Verifying the buckling coefficient Xy on a class 2 section

Test ID: 4443

Test status: Passed

8.12.1Description

Performs the finite elements calculation and the steel elements calculation. Verifies the buckling coefficient Xy on a

class 2 section and generates the shape sheets report.The model consists of a vertical linear element (I26*0.71+15*1.07 cross section and S275 material) with a rigid hingesupport at the base and a rigid support with translation restraints on X and Y and rotation restraint on Z, at the top. Apunctual live load is applied.




51

8.13 Verifying the steel shape optimization when using sections from Advance Steel Profilesdatabase (TTAD #11873)

Test ID: 4289

Test status: Passed

8.13.1Description

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

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




52

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

Test ID: 5702

Test status: Passed

8.14.1Description

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.

8.14.2Background

Lateral torsional buckling verification for an unrestrained IPE300 beam subjected to axis bending efforts, made ofS235 steel. The beam is simply supported. The beam is subjected to a uniform vertical load (10 000 N) appliedconstantly on the entire length. The dead load will be neglected.

8.14.2.1 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

■ Flexion inertia moment around the Y axis: Iy=8356.00x104mm

4

■ Flexion inertia moment around the Z axis: Iz=603.80x104mm

4




53

Materials properties

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

■ Yield strength f y = 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 along X, Y and Z axis,

► Support at the end point (z = 3.00) restrained in translation along Y and Z axis and restrained rotationalong 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.

8.14.2.2 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 iscalculated using the percentage of the design buckling moment resistance of the bended element (Mb,Rd) from thedesigned value moment (MEd) produced by the linear force applied to the element (NEd). The design bucklingresistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.

%100100,

Rd b

Ed

M

M (6.46)

Cross-class classification is made according to Table 5.2

■ for beam web:

7272014.35

1

014.351.7

6.248

t

cmm

mm

t

c

therefore the beam web is considered to be

Class 1

■ for beam flange:

99276.5

1

276.57.10

45.56

t

cmm

mm

t

c

therefore the haunch is considered to be Class1




54

In conclusion, the section is considered to be Class 1

The buckling curve will be determined corresponding to Table 6.2:

22150

300

mm

mm

b

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

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

1

,

M

y y LT

Rd b

f W M

(6.55)

Where:

LT reduction factor for lateral-torsional buckling:

11

22

LT LT LT

LT

(6.56)

Where:

2

)2.0(15.0 LT LT LT LT

LT represents the imperfection factor; 21.0 LT




55

LT the non-dimensional slenderness corresponding:

cr

y y LT

M

f W

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

g g

Z

t z

z

w

w

z

z

z cr z C z C

I E

I G Lk

I

I

k

k

Lk

I E C M 2

2

22

2

2

2

1

(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 x104mm

4

It is the torsional inertia: It=20.12x104mm

4

IW is the warping inertia (deformation inertia moment): Iw=12.59x1010

mm6

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 ofgravity)

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 tothe center, then C2xxg=0 and the Mcr formula become:

z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1

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




56

kNm Nmmmm

mmmm N

mmmm N mm

mm

mm

mm

mmmm N M cr

61.1308.13060942480.23044.50057813.1

1080.603/210000²

1012.20/80770)²5000(

1080.603

1059.12

)²5000(

1080.603/210000²13.1

442

442

44

610

442

therefore:

063.18.130609424

/235104.628 233

Nmm

mm N mm

M

f W

cr

y y LT

156.1063.1)2.0063.1(21.015.0)2.0(15.0 22

LT LT LT LT

1621.0²063.1²156.1156.1

1

²²

1

LT LT LT

LT

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 thecross section and the effective characteristics (not applicable in this case, as the cross section is class 1).




57

Lateral torsional buckling coefficient

Simply supported beam subjected to bending efforts

Lateral torsional buckling coefficient

Elastic critical moment for lateral-torsional buckling

Simply supported beam subjected to bending efforts

Mcr

8.14.2.3 Reference results

Result name Result description Reference value

LT Lateral-torsional buckling coefficient [adim.] 0.621

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

8.14.3Calculated results

Result name Result description Value Error

XLT Lateral-torsional buckling coefficient 0.621588adim

0.0947 %

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

0.0678 %




58

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

Test ID: 5720

Test status: Passed

8.15.1Description

The test verifies a user defined cross section column.

The cross section has an “I symmetric” shape with: 260mm height; 150mm width; 7.1mm center thickness; 10.7mmflange 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 kNmbending 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.

8.15.2Background

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






■ Exploitation loadings (category A): Fz=--328000N N; My=50000Nm; Mx=10000Nm;



Units

Metric System




59


■ Column length: L=5620mm

■ Cross section area:206.4904 mm A

■ Overall breadth: mmb 150

■ Flange thickness: mmt f 70.10

■ Root radius: mmr 0

■ Web thickness: mmt w 10.7

■ Depth of the web: mmhw 260

■ Elastic modulus after the Y axis, 3

, 63.445717 mmW yel

■ Plastic modulus after the Y axis, 318.501177 mmW y

■ Elastic modulus after the Z axis, 3

, 89.80344 mmW z el

■ Plastic modulus after the Z axis, 3, 96.123381 mmW z pl

■ Flexion inertia moment around the Y axis:464.57943291 mm I y

■ Flexion inertia moment around the Z axis: 446.6025866 mm I z

■ Torsional moment of inertia: 497.149294 mm I t

■ Working inertial moment: 688.19351706542 mm I w




■ Longitudinal elastic modulus: E = 210000 MPa.

■ Shear modulus of rigidity: G=80800MPa






61

8.15.2.2 Cross section Class

According to Advance Design calculations:


-for beam web:

The web dimensions are 850x5mm.

1253.006.179

30.45

sup

inf

Mpa

Mpa




62

73.19036.224

06.1796.238

175.4836.224

3.456.238

36.224

6.238

06.17930.4506.17930.45

y

x y x y x

5.080.06.238

73.190

6.238

x

924.0275

235235

y f

93.3818.013

924.0396

113

3966.33

924.0

61.331.7

7.102260

t

cmm

mmmm

t

c

therefore the beam

web is considered to be Class 1




63

-for beam flange:

316.8924.0968.6

924.0

68.67.10

2

1.7150

t

c

t

c


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

8.15.2.3 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:

34.0




64

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

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

curve according to:

11

22

y y y

y

(6.49)

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

ycr

y

y N

f A

,

*

Cross section area:206.4904 mm A

Flexion inertia moment around the Y axis: 464.57943291 mm I y

kN N

mm

mmmm N

l

I E N

fy

y

ycr 33.380295.3802327²5620

64.57943291/210000²

²

² 42

,

5956.095.3802327

/27506.4904 22

,

N

mm N mm

N

f A

ycr

y y

7446.05956.02.05956.034.015.0²)2.0(15.0 2 y y y

839.0

1

839.05956.07446.07446.0

11

2222

y

y

y y y

y




65

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


49.0

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

z coefficient corresponding to non-dimensional slenderness z will be determined from the relevant buckling curve

according to:

1122

z z z

z

(6.49)

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

z cr

y z

N

f A

,

*

Flexion inertia moment around the Z axis:446.6025866 mm I z





66

kN N

mm

mmmm N

l

I E N

fz

z z cr 43.39563.395426

²5620

46.6025866/210000²

²

² 42

,

847.1

63.395426

/27506.4904 22

,

N

mm N mm

N

f A

z cr

y z

609.2847.12.0847.149.015.0²)2.0(15.0 2 z z z

225.0

1

225.0847.1609.2609.2

112222

z

z

z z z

z

8.15.2.4 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:

z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1

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

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

²252.0423.0325.0

11

C

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

77.1050

01

,

, C M

M

top y

botom y




67

Flexion inertia moment around the Y axis:464.57943291 mm I y

Flexion inertia moment around the Z axis: 446.6025866 mm I z

Torsional moment of inertia: 497.149294 mm I t

Working inertial moment: 688.19351706542 mm I w

Yield strength f y = 275 MPa,

Longitudinal elastic modulus: E = 210000 MPa.

Shear modulus of rigidity: G=80800MPa

Warping inertial moment (recalculated):

IW is the warping inertia (deformation inertia moment):

4

2

f z

w

t h I I

h cross section height; mmh 260

f t flange thickness; mmt f 7.10

6

24

093627638294

7.1026046.6025866mm

mmmmmm I w

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

Length of the column: L=5620mm

kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mmmmmm N

I E I G L

I I

L I E C M

z

t

z

w z cr

150181.150184702

58.21463.39542677.146.6025866/210000

97.149294/808005620

46.6025866

09362763829

562046.6025866/21000077.1

²²

²²

422

422

4

6

2

422

1

958.01.150184702

/27518.501177 23,

Nmm

mm N mm

M

f W

cr

y y pl LT

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

1²²

1

LT LT LT

LT

(6.56)

²2.015.0 LT LT LT LT

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

2733.1150

260

mm

mm

b

h




68

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

49.0

145.1²958.02.0958.049.015.0²2.015.0 LT LT LT LT

1564.0²958.0²145.1145.1

1

²²

1

LT LT LT

LT

8.15.2.5 Internal factor, yyk , calculation

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

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

yy

ycr

Ed

y

mLT my yyC

N

N C C k

1̀

1,




69

ycr

Ed y

ycr

Ed

y

N

N

N

N

,

,

1

1

839.0 y (previously calculated)

kN N Ed 328

kN N l

I E N

fy

y

ycr 33.380295.3802327²

²,

(previously calculated)

985.0

95.3802327

328000839.01

95.3802327

3280001

1

1

,

,

N

N N

N

N

N

N

N

ycr

Ed y

ycr

Ed

y

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




70

Calculation of the 0 term:

0

,0

cr

y y pl

M

f W

-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2318.501177 mmW y

The calculation the 0cr M will be calculated using 11 C and 02 C , therefore:

kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

85.8427.84850646

58.21463.395426146.6025866/210000

97.149294/808005620

46.6025866

88.19351706542

5620

46.6025866/2100001

²

²

²

²

422

422

4

6

2

422

1

274.127.84850646

/27518.501177 23,

0

Nmm

mm N mm

M

f W

cr

y y pl

Calculation of the 4

,,

1 1120.0

TF cr

Ed

z cr

Ed

N

N

N

N C term:

Where:

-for a symmetrical section for the both axis, T cr TF cr N N ,,

²1

,

2

0

,

T cr

wt T cr

L I E I G

I N

The mass moment of inertia 0 I

4442

0 1.6396915846.602586664.57943291 mmmmmm I I z A I I I z y g z y



- the buckling length, T cr L , ,

m L T cr 62.5,

mm

mmmm N mmmm N

mm

mm N T cr

607.1395246

²5620

88.19351706542/21000097.149294/80800

1.63969158

06.4904 62242

4

2

,

N N Ed 328000

N N N T cr TF cr 607.1395246,,




71

kN N

mm

mmmm N

l

I E N

fz

z z cr 43.39563.395426

²5620

46.6025866/210000²

²

² 42

,


C1=1 for the top part of the column

For the top part of the column:

120.0

607.1395246

3280001

63.395426

3280001120.01120.0 44

,,

1

N

N

N

N

N

N

N

N C

TF cr

Ed

z cr

Ed

Therefore:


1

11

11

120.01120.0274.1

120.01120.0

274.1

,,

2

0,

0,0,

4

,,

10

4

,,

1

0

T cr

Ed

z cr

Ed

LT mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C

a

a

C C C

N

N

N

N C

N

N

N

N C

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

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

LT y

LT y

mymymya

aC C C

1

1 0,0,

yeff

eff

Ed

Ed y

yW

A

N

M

,

,

Elastic modulus after the Y axis,3

,, 63.445717 mmW W yeff yel

677.163.445717

06.4904

328000

10503

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

1997.064.57943291

97.14929411

4

4

mm

mm

I

I a

y

t LT




72

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

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

ycr

Ed

ycr

Ed my

N

N

N

N C

,,

0, 33.036.079.033.036.021.079.0

Where:

kN N l

I E N

fy

y

ycr 33.380295.3802327²

²,


N N Ed 328000

780.095.3802327

32800033.036.079.00,

N

N C my

904.01677.11

1677.1780.01780.0

11 0,0,

LT y

LT y

mymymya

aC C C

Equivalent uniform moment factor, mLT C , calculation

256.2

1

256.2

607.1395246

3280001

63.395426

3280001

997.0904.0

11

2

,,

2

mLT

mLT

T cr

Ed

z cr

Ed

LT mymLT

C

C

N

N

N

N

N

N

N

N

aC C




73

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

y pl

yel

LT pl my

y

my

y

y yyW

W bnC

wC

wwC

,

,2

max2

max2 6.16.1

2)1(1

1534.0

/27596.123381

10000000

/27518.501177564.0

50000000274.1997.05.0

5.05.0

2323

2

,

,

,

,

2

0

,,

,

,,

,

2

0

mm N mm

Nmm

mm N mm

Nmm

f W

M

f W

M a

M

M

M

M ab

y z pl

Ed z

y y pl LT

Ed y LT

Rd z pl

Ed z

Rd y pl LT

Ed y LT LT

5.1124.163.445717

18.5011773

3

,

, mm

mm

W

W w

yel

y pl

y

243.0

1

/27506.4904

32800022

1

mm N mm

N

N

N n

M

Rk

Ed pl

847.1847.1;5956.0max;maxmax

z y

857.01534.0243.0²847.1²904.0124.1

6.1847.1²904.0

124.1

6.12)1124.1(1

yyC




74

889.0889.018.501177

63.445717

857.0

,

,

3

3

,

,

yy

y pl

yel

yy

y pl

yel

yy

C

W

W C

mm

mm

W

W

C

kN N l

I E N

fy

y

ycr 33.380295.3802327²

²,


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

47.2889.0

1

95.3802327

3280001

985.0256.2904.0

1̀

1,

N

N C

N

N C C k

yy

ycr

Ed

y

mLT my yy

8.15.2.6 Internal factor, yz k , calculation

y

z

yz

z cr

Ed

y

mz yz w

w

C

N

N C k

6.0

1

1,

691.063.395426

32800033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

985.0

95.3802327

328000839.01

95.3802327

328000

1

1

1

,

,

N

N N

N

N

N

N

N

ycr

Ed y

ycr

Ed

y


kN N l

I E N

fz

z z cr 43.39563.395426

²

²,


LT pl

z

mz z yz cn

w

C wC

5

2

max2

142)1(1

Rd y pl lt my

Ed y

z

LT LT M C

M ac

,,

,

4

2

0

5

10

997.064.57943291

97.14929411

4

4

mm

mm

I

I a

y

t LT (previously calculated)

274.127.84850646

/27518.501177 23,

0

Nmm

mm N mm

M

f W

cr

y y pl (previously calculated)

847.163.395426

/27506.4904

22

, N

mm N mm N

f A z cr

y z (previously calculated)




75

Nm M Ed y 50000,

904.01

1 0,0,

LT y

LT y

mymymya

aC C C


1564.0²958.0²145.1145.1

1²²

1

LT LT LT

LT


Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23

,,

664.05.137823724564.0904.0

50000000

847.15

²274.1997.010

5

10

4

,,

,

4

2

0

Nmm

Nmm

M C

M ac

Rd y pl lt my

Ed y

z

LT LT

LT pl

z

mz z yz cn

w

C wC

5

2

max2

142)1(1

5.1

5.1

5.1536.189.80344

96.1233813

3

,

,

z

z

z el

z pl

z w

w

mm

mm

W

W w

kN N l

I E N

fz

z z cr 43.39563.395426

²

²,


691.063.395426

32800033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

847.1847.1;5956.0max;maxmax

z y

243.0

1

M

Rk

Ed pl N

N n


532.0691.0243.05.1

847.1691.0142)15.1(1

142)1(1

5

22

5

2

max

2

LT pl

z

mz z yz cn

wC wC

20.5124.1

5.16.0

532.0

1

63.395426

3280001

985.0691.06.0

1

1,

N

N w

w

C

N

N C k

y

z

yz

z cr

Ed

y

mz yz




76

8.15.2.7 Internal factor, zyk , calculation

y

z

zy

ycr

Ed

z mLT my zy

w

w

C

N

N C C k

6.0

1

1,

210.0

63.395426

328000225.01

63.395426

3280001

1

1

,

,

N

N N

N

N

N

N

N

z cr

Ed z

z cr

Ed

z

225.0 z (previously calculated)

y pl

yel

z

y

LT pl

y

my

y zyW

W

w

wd n

w

C wC

,

,

5

2

max2

6.0142)1(1

Nmmmm N mm f W M z z Rd z pl 33930039/27596.123381 23

,,

771.0

33930039691.0

10000000

5.137823724564.0904.0

50000000

847.11.0

274.1997.02

1.0

2

4

,,

,

,,

,

4

0

Nmm

Nmm

Nmm

Nmm

M C

M

M C

M ad

Rd z pl mz

Ed z

Rd y pl LT my

Ed y

z

LT LT

310.0771.0243.0124.1

847.1904.0142)1124.1(1

142)1(1

5

22

5

2

max

2

LT pl

y

my y zy d n

wC wC

462.018.501177

63.445717

5.1

124.16.06.0

3

3

,

, mm

mm

W

W

w

w

y pl

yel

z

y

462.0

462.06.0

310.0

6.0

,

,

,

,

zy

y pl

yel

z

y

zy

y pl

yel

z

y

zy

C

W

W

w

w

C

W

W

w

wC

256.2mLT C (previously calculated)

529.05.1

124.16.0

462.0

1

95.3802327

3280001

210.0256.2904.06.0

1

1,

N

N w

w

C

N

N C C k

z

y

zy

ycr

Ed

z mLT my zy




77

8.15.2.8 Internal factor, zz k , calculation

zz

z cr

Ed

z mz zz

C

N

N C k

1

1,

z pl

z el

LT pl mz

z

mz

z

z zz W

W enC

wC

wwC

,

,max

2max

2 6.16.12)1(1

131.0/27518.501177564.0904.0

50000000

847.11.0

274.1997.07.1

1.0

7.1

234

,,

,

4

0

mmmm

Nmm

M C

M ae

Rd y pl lt my

Ed y

z

LT LT

926.0131.0243.0847.1691.05.1

6.1847.1691.0

5.1

6.12)15.1(1

6.16.12)1(1

222

2

max2

max2

LT pl mz

z

mz

z

z zz enC wC wwC

919.0926.0

1

63.395426

3280001

210.0691.0

1

1,

N

N C

N

N C k

zz

z cr

Ed

z mz zz

8.15.2.9 Bending and axial compression verification

1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rk z

Rd z Ed z

zz

M

Rk y

LT

Rd y Ed y

zy

M

Rk z

Ed

M

Rk z

Rd z Ed z

yz

M

Rk y

LT

Rd y Ed y

yy

M

Rk y

Ed

M

M M k

M

M M k

N

N

M

M M k

M

M M k

N

N

i y Rk A f N




78

69.127.034.008.1

1

/27596.123381

1010919.0

1

/27518.501177564.0

1050529.0

1

/27506.4904225.0

328000

41.353.159.129.0

1/27596.123381

101020.5

1

/27518.501177564.0

1050479.2

1

/27506.4904839.0

328000

23

6

23

6

22

23

6

23

6

22

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N



■ 7 nodes,


y coefficient corresponding to non-dimensional slenderness y

Column subjected to axial and shear force to the top

y




79

z coefficient corresponding to non-dimensional slenderness z


z

Internal factor, yyk


yyk






81

Internal factor, zz k


zz k

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 effortover the Y axis

SNy




82

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 momentover 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 momentover the Y axis

SMyz






84

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 momentover the Z axis

SMzz

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

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

C1




85

The elastic moment for lateral-torsional buckling calculation


Mcr

The appropriate non-dimensional slenderness


LT




86



y y coefficient corresponding to non-dimensional slenderness y

0.839

z z coefficient corresponding to non-dimensional slenderness z

0.225

yyk Internal factor, yyk 2.47

yz k Internal factor, yz k 5.20

zyk Internal factor, zyk 0.529

zz k Internal factor, zyk 0.919

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

0.29

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

1.59

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

1.53

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

1.08

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

0.34

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

0.27

C1 Coefficient that depends of several parameters as: sectionro erties su ort conditions moment dia ram allure

1.77

Mcr The elastic moment for lateral-torsional buckling calculation 150.18

LT The appropriate non-dimensional slenderness 0.564

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




87



Xy Coefficient corresponding to non-dimensional slendernessafter Y-Y axis

0.839285adim

0.0340 %

Xz Coefficient corresponding to non-dimensional slendernessafter Z-Z axis

0.224656adim

-0.1530 %

Kyy Internal factor kyy 2.47232 adim 0.0938 %

Kyz Internal factor kyz 5.20929 adim 0.1787 %

Kzy Internal factor kzy 0.525982adim

-0.5704 %

Kzz Internal factor kzz 0.942941adim

2.6051 %

Work ratio Stability work ratio 341.352 % 0.0000 %






89


Units

Metric System



■ Cross section area:28446mm A







, 101156 mmW yel

■ Plastic modulus after the Y axis, 33101307 mmW y


, 1040.146 mmW z el

■ Plastic modulus after the Z axis, 33

, 10229 mmW z pl


4


4

■ Torsional moment of inertia: It=51.08x104mm

4

■ Working inertial moment: Iw=490000x106mm6




90






Boundary conditions


■ Outer:



■ Inner: lateral (xoz) restraint at z=3m

Loading


■ External: Point load From X=0.00m and Y=9.00m: FZ =-125000N and Fx=28330N■ Internal: None.




91

CTICM model

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




92




-for beam web:

727249.38

1

49.386.8

331

t

cmmmm

t c

therefore the beam web is considered to be Class 1

-for beam flange:

9950.4

1

50.45.13

47.67

t

cmm

mm

t

c



According to CTICM document:

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




93

8.16.2.3 Compression verification


The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element iscalculated using the percentage of the design axial compression resistance of the element (Nc,Rd) from thecompression force applied to the element (NEd). The compressed resistance of the member, Nc,Rd, is calculated

according to Eurocode 3 1993-1-1-2005, Chapter 6.2.4.

%100100,

Rd c

Ed

N

N (6.9)

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

0

,

M

y

Rd c

f A N

(6.10)

-where:

A is the section area:28446mm A

y f is the yielding strength:2/275 mm N f y

0 M is the partial safety factor: 10 M

kN N mm N mm f A

N M

y

Rd c 65.232223226501

/2758446 22

0

,

N N Ed 125000

%100%38.51002322650

125000

,

N

N

N

N

Rd c

Ed


The compression resistance of the column is kN N Rd c 2324, as it can be seen from conclusion extracted from

CTCIM 2006-4:

8.16.2.4 Shear verification


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

%100100,

Rd c

Ed

V

V (6.17)




94

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

0

,

3

M

y

V

Rd c

f A

V

(6.18)

-where:

AV is the shear area:

ww f w f V t ht r t t b A A 22

-where:

A is the cross section area:2

8446mm A

b is the overall breadth: mmb 180

tf is the flange thickness: mmt f 5.13

r is the root radius: mmr 21

tw is the web thickness: mmt w 6.8

hw is the depth of the web: mmhw 400

1

22 1.42695.132126.85.1318028446

22

mmmmmmmmmmmmmm

t r t t b A A f w f V

234404006.81 mmmmmmt h ww

22 34401.4269 mmmm AV



N

mm N mm

f A

V M

y

V

Rd c 66.6778101

3

/2751.4269

3

22

0

,

N V Ed 28330

%100%180.410004179.010066.677810

28330100

,

N

N

V

V

Rd c

Ed




95


The shear resistance of the column is kN V Rd z pl 8.677,, as it can be seen from conclusion extracted from CTCIM

2006-4:

8.16.2.5 Bending moment verification


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

%100100,

Rd pl

Ed

M

M (6.12)

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

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

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

0

,

M

y pl

Rd pl

f w M

(6.13)

-where:

wpl is the plastic modulus:31307000mmw pl



Nmmmm N mm f w

M M

y pl

Rd pl 3594250001

/2751307000 23

0

,

Nmm M Ed

254970000

%100%938.7010070938.0100359425000

254970000100

,

Nmm

Nmm

M

M

Rd pl

Ed




96


The bending moment resistance of the column is kNm M Rd y pl 7.359,, as it can be seen from the conclusion

extracted from CTCIM 2006-4:




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

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

21.0

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


curve according to:

1122

y y y

y

(6.49)

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

ycr

y

y N

f A

,

*

Where: A is the cross section area; A=8446mm2; f y is the yielding strength of the material; f y=275N/mm

2 and Ncr is

the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

N

mmmm MPa

L I E N

fz

y

ycr 77.59184729000

1023130210000²²

²2

44

,






98


34.0


according to:

11

22

z z z

z

(6.49)


z cr

y z

N

f A

,

*


2 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 equalto the grater length of the two beam sections, 6m. A more accurate calculation is to perform a modal analysis of thecolumn buckling outside the frame. The first eigenmode of instability corresponds to an amplification factor equal to

critical 15.9cr . The normal critical force can be directly calculated:

N

mm

mmmm N

l

I E N

fy

z z cr 153.1142396

²4890

101318/210000²

²

² 442

,

42588.1153.1142396

/2758446 22

,

mm N mm

N

f A

z cr

y z

72497.142588.12.042588.134.015.0²)2.0(15.0 2 z z z

137096.042588.172497.172497.1

11

2222

z z z

z


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

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




99



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:

z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1


-where:


²252.0423.0325.0

11

C


is the fraction of the bending moment from the column extremities: 66667.097.254

98.169

kNm

kNm

17932.1²66667.0252.066667.0423.0325.0

1

²252.0423.0325.0

11

C


4


4



4

■ Working inertial moment: Iw=490000x106mm6

■ Length of the column part: L=3000mm


Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

2.806585210

33396.22534.303523217932.1101318/210000²

1008.51/808003000

101318

1049

3000

101318/210000²17932.1

²

²

²

²

442

4422

44

610

2

442

1




100

Calculation of the non-dimensional slenderness factor, LT :

cr

y y LT

M

f W

Plastic modulus, 33101307 mmW

y

66754.02.806585210

/275101307 233

Nmm

mm N mm

M

f W

cr

y y LT


1²²

1

LT LT LT

LT

(6.56)


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


34.0

80228.0²66754.02.066754.034.015.0 LT

180173.0²66754.0²80228.080228.0

1

²²

1

LT LT LT

LT


The determined value for the coefficient corresponding to non-dimensional slenderness, LT is: 7877.0 LT as

it can be observed from the conclusion extracted from CTCIM 2006-4:




101

b) for the 6m part of the column:



z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1



²252.0423.0325.0

11

C


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

77.11 C

■ Flexion inertia moment around the Y axis: Iy=23130.00x10

4

mm

4

■ Flexion inertia moment around the Z axis: Iz=1318.00x10

4mm

4



4

■ Working inertial moment: Iw=490000x106mm

6

■ Length of the column part: L=6000mm


Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

406423987

604.302085.75880877.1101318/210000²

1008.51/80800²6000

101318

1049

²6000

101318/210000²77.1

²

²

²

²

442

4422

44

610

2

442

1






103


The determined value for the coefficient corresponding to non-dimensional slenderness, LT is: 694.0 LT as it

can be observed from the conclusion extracted from CTCIM 2006-4:



1

1

,

,,

1

,

,,

1

M

Rk z

Ed z Ed z

yz

M

Rd y

LT

Rd y Ed y

yy

M

Rk y

Ed

M

M M k

M

M M k

N

N

(6.61)

1

1

,

,,

1

,

,,

1

M

Rk z

Ed z Ed z

zz

M

Rd y

LT

Rd y Ed y

zy

M

Rk z

Ed

M

M M k

M

M M k

N

N

(6.62)

The formulae can be simplified because:

There is no bending on the small inertia axis: 0, Ed z M

The section is considered to be a Class1: 0, Rd y M and 0, Rd z M

Therefore the formulae are:

62.600.1

61.600.1

1

,

,

1

1

,

,

1

M

Rk y

LT

Ed y

zy

M

Rk z

Ed

M

Rk y

LT

Ed y

yy

M

Rk y

Ed

M M k

N N

M

M k

N

N




104

8.16.2.9 Internal factor, yyk , calculation:

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

yy

ycr

Ed

y

mLT my yyC

N

N C C k

1

1,

Auxiliary terms:

ycr

Ed y

ycr

Ed

y

N N

N

N

,

,

1

1

Where:


N

mm

mmmm N

l

I E N

fy

y

ycr 773.5918472²9000

41023130/210000²

²

² 42

,

99741.0

773.5918472

12500087968.01

773.5918472

1250001

N N

N

y





105



LT

C 10

-where:


²252.0423.0325.0

11

C



77.11 C

66754.02.806585210

/275101307 233

Nmm

mm N mm M

f W cr

y y LT




106

88811.066754.077.1110 LT LT C C


,,

1 1120.0

TF cr

Ed

z cr

Ed

N

N

N

N C term:

Where:


²,

2

0

,

T cr

wt T cr

L

I E I G

I

A N


2

0 g z y z A I I I

Flexion inertia moment around the Y axis: Iy=23130.00x104mm

4

Flexion inertia moment around the Z axis: Iz=1318.00x10

4

mm

4

Cross section area:28446mm A

Distance between the section neutral axis and the section geometrical center: 0 g z

4444442

0 10244481013181023130 mmmmmm I I z A I I I z y g z y

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

m L T cr 6,

Torsional moment of inertia: It=51.08x104mm

4

Working inertial moment: Iw=490000x106

mm6

Longitudinal elastic modulus: E = 210000 MPa


N

mm

mmmm N mmmm N

mm

mm N T cr

788.2400423

²6000

1049/2100001008.51/80800

1024448

8446 61022442

44

2

,

N N N T cr TF cr 788.2400423,,

N N z cr 153.1142396, (previously calculated)

25505.0

788.2400423

1250001

153.1142396

125000177.120.01120.0 44

,,

1

N

N

N

N

N

N

N

N C

TF cr

Ed

z cr

Ed




107

Therefore:

1

11

11

25505.01120.088811.0

,,

2

0,

0,0,

4

,,

10

T cr

Ed

z cr

Ed

LT mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C

a

aC C C

N N

N N C



LT y

LT y

mymymya

aC C C

11 0,0,

yel Ed

Ed y

yW

A

N

M

,

,

Elastic modulus after the Y axis, 33

, 101156 mmW yel

90119.14101156

8446

125000

1094.25433

26

,

,

mm

mm

N

Nmm

W

A

N

M

yel Ed

Ed y

y

99779.01023130

1008.511144

44

mm

mm

Iy

It a LT


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

ycr

Ed

my N

N

C ,0, 33.036.021.079.0

Where:

N N ycr 773.5918472, (previously calculated)

78749.0773.5918472

12500033.0036.079.00,

N

N C my

95619.099779.090119.141

99779.090119.1478749.0178749.0

11 0,0,

LT y

LT y

mymymya

aC C C




108


The mLT C coefficient takes into account the laterally restrained parts of the column. The mLT C coefficient must be calculated

individually for each of the column parts.

1

11,,

2

T cr

Ed

z cr

Ed

LT mymLT

N

N

N

N

aC C


LT y

LT y

mymymmya

aC C C

11 0,0,3.


99779.0 LT a (previously calculated)


is the fraction of the bending moment from the column part extremities: 66667.097.254

98.169

kNm

kNm




109

ycr

Ed my

N

N C

,

0, 33.036.021.079.0

Where:


93256.0773.5918472

12500033.066667.036.066667.021.079.00,

N C my

98609.099779.090119.141

99779.090119.1493256.0193256.0

11 0,0,3,

LT y

LT y

mymymmya

aC C C


N N T cr 788.2400423, (previously calculated)

99779.0 LT a (previously calculated)

05596.1

1

05596.1

788.2400423

1250001

153.1142396

1250001

99779.0²98609.0

11

3,

3,

,,

2

3,3,

mmLT

mmLT

T cr

Ed

z cr

Ed

LT mmymmLT

C

C

N

N

N

N

N

N

N

N

aC C


LT y

LT y

mymymmya

aC C C

11 0,0,6.

yel Ed

Ed y

yW

A

N

M

,

,


, 101156 mmW yel

9353.9101156

8446125000

1098.169 33

26

,

, mmmm

N Nmm

W A

N

M

yel Ed

Ed y y

99779.01023130

1008.5111

44

44

mm

mm

Iy

It a LT





110

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

ycr

Ed my

N

N C

,

0, 33.036.021.079.0

Where:


78749.0773.5918472

12500033.0036.0021.079.00,

N

N C my

94873.099779.09353.91

99779.09353.9

78749.0178749.011 0,0,6,

LT y

LT y

mymymmya

a

C C C


N N T cr 788.2400423, (previously calculated)

1

1

97746.0

788.2400423

1250001

153.1142396

1250001

99779.0

²94873.0

11

6,

6,

,,

2

6,6,

mmLT

mmLT

T cr

Ed

z cr

Ed

LT mmymmLT

C

C

N

N

N

N

N

N

N

N

aC C

In conclusion:

1

05596.1

6,

3,

mmLT

mmLT

mLT C

C C




111

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

y pl

yel

LT pl my

y

my

y

y yyW

W bnC

wC

wwC

,

,2

max

2

max

2 6.16.12)1(1

Where:

■ 005.05.0,,

,2

0

,

,

,,

,2

0

Rd y pl LT

Ed y

LT

Rd pl

Ed z

Rd y pl LT

Ed y

LT LT M

M a

M

M

M

M ab

■ 5.1,

, yel

y pl

yW

W w


, 101156 mmW yel

■ Plastic modulus after the Y axis, 33

, 101307 mmW y pl

■ 5.113062.1101156101307 33

33

,

, mmmm

W

W w

yel

y pl y

■

1 M

Rk

Ed pl N

N n

■ N mm N mm f A

N N M

y

Rd c Rk 23226501

/2758446 22

0

,

■ 05382.0

1

2322650

125000

1

N

N

N

N n

M

Rk

Ed pl




112

■ z y ;maxmax

■ 62645.0 y (previously calculated)

■ 42504.1 z (previously calculated)

■ 42504.142504.1;62645.0max;maxmax z y

■ 95619.0myC (previously calculated)

98511.0

005382.0²42504.1²95619.013062.1

6.142504.1²95619.0

13062.1

6.12)113062.1(1

yyC

88447.0101307

10115698511.0

33

33

,

,

mm

mm

W

W C

y pl

yel

yy

In conclusion:

98902.098511.0

1

773.5918472

1250001

99741.0195619.0

1

1

04409.198511.0

1

773.5918472

1250001

99741.005569.195619.0

1

1

,

6,6,

,

3,3`,

N

N C

N

N C C k

N

N C

N

N C C k

k

yy

ycr

Ed

y

mmLT mym yy

yy

ycr

Ed

y

mmLT mym yy

yy

According to CTICM 2006-4:




113

8.16.2.10 Internal factor, zyk , calculation:

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

z

y

zy

ycr

Ed

z mLT my zy

w

w

C

N

N C C k

6.0

1

1,

Auxiliary terms:

z cr

Ed z

z cr

Ed

z

N N

N

N

,

,

1

1

Where:


N

mm

mmmm N

l

I E N

fy

z z cr 153.1142396

²4890

101318/210000²

²

² 442

,

92826.0

153.114239612500037096.01

1533.1142396

1250001

N N

N

N

z





114


1

11

11

208.01120.088811.0

,,

2

0,

0,0,

4

,,

10

T cr

Ed

z cr

Ed

LT mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C

a

aC C C

N

N

N

N C


90119.14101156

8446

125000

1094.25433

26

,

,

mm

mm

N

Nmm

W

A

N

M

yel Ed

Ed y

y (previously calculated)

99779.01023130

1008.5111

44

44

mm

mm

Iy

It a LT (previously calculated)





115


78749.033.036.021.079.0,

0, ycr

Ed my

N

N C


95619.01

1 0,0,

LT y

LT y

mymymya

aC C C


1

05596.1

116,

3,

,,

2

mmLT

mmLT

T cr

Ed

z cr

Ed

LT mymLT

C

C

N

N

N

N

aC C


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

y pl

yel

z

y

LT pl

y

my

y zyW

W

w

wd n

w

C wC

,

,

5

2

max

2

6.0142)1(1




116

Where:

001.0

21.0

2,,

,

4

0

,,

,

,,

,

4

0

Rd y pl LT my

Ed y

z

LT

Rd z pl mz

Ed z

Rd y pl LT my

Ed y

z

LT LT M C

M a

M C

M

M C

M ad

5.113062.1101156101307

33

33

,

,

mmmm

W W w

yel

y pl


5.1,

, z el

z pl

z W

W w

Elastic modulus after the Z axis, 33

, 1040.146 mmW z el

Plastic modulus after the Z axis, 33

, 10229 mmW z pl

5.15.1

564.1

1040.146

1022933

33

,

,

z

z

z el

z pl

z

ww

mm

mm

W

W w

1 M

Rk

Ed pl N

N n

05382.0

1

2322650

125000

1

N

N

N

N n

M

Rk

Ed pl


42504.142504.1;62645.0max;maxmax z y (previously calculated)

90887.0005382.013062.1

42504.198609.0142)113062.1(1

5

22

zyC

46073.0101307

101156

5.1

13062.16.06.0

33

33

,

,

mm

mm

W

W

w

w

y pl

yel

z

y

46073.06.090887.0142)1(1,

,

5

2

max

2

y pl

yel

z

y

LT pl

y

my

y zyW

W

w

wd n

w

C wC

In conclusion:

56307.0

5.1

13062.16.0

90887.0

1

773.5918472

1250001

923830.005569.198609.06.0

1

1,

N

N w

w

C

N

N C C k

z

y

zy

ycr

Ed

z mLT my zy




117

50752.05.1

13062.16.0

98511.0

1

773.5918472

1250001

99741.0195619.0

6.01

1

52306.0

5.1

13062.16.0

90887.0

1

773.59184721250001

99741.005569.195619.0

6.01

1

,

6,6,

,

3,3`,

N

N

w

w

C

N

N C C k

N N

w

w

C

N

N C C k

k

z

y

zy

z cr

Ed

y

mmLT mym zy

z

y

zy

z cr

Ed

y

mmLT mym zy

zy


The bending and axial compression verifications are:

-for the 3m column part:

160789.046281.014508.0

1

35942500080173.0

25497000052306.0

1

232265037096.0

125000

198501.092383.006119.0

1

35942500080173.0

25497000004409.1

1

232265087968.0

125000

1

,

,

1

1

,

,

1

Nmm

Nmm

N

N

M M k

N N

Nmm

Nmm

N

N

M

M k

N

N

M

Rk y

LT

Ed y zy

M

Rk z

Ed

M

Rk y

LT

Ed y

yy

M

Rk y

Ed




118

-for the 6m column part:

152073.037565.014508.0

135942500063518.0

16898000050752.0

1232265037096.0

125000

179322.073204.006118.0

1

35942500063518.0

16898000098902.0

1

232265087968.0

125000

1

,

,

1

1

,

,

1

Nmm

Nmm

N

N

M

M k

N

N

Nmm

Nmm

N

N

M

M k

N

N

M

Rk y

LT

Ed y

zy

M

Rk z

Ed

M

Rk y

LT

Ed y

yy

M

Rk y

Ed




■ 6 nodes,





119

Ratio of the design normal force to design compresion resistance


Work ratio - Fx

Ratio of the design share force to design share resistance


Work ratio - Fz




120

Ratio of the design share force to design share resistance


Work ratio - oblique



y




121



z



yyk




122

Internal factor, yz k


yz k



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 resistanceone the principal axis

70.94


0.88


0.37

yyk Internal factor, yyk for the 3m segment 1.04

yy

k Internal factor, yy

k for the 6m segment 0.99

zyk Internal factor, zyk for the 3m segment 0.52

zyk Internal factor, zyk for the 6m segment 0.51




123



Work ratio - Fx Ratio of the design normal force to design compression

resistance

5.38178 % 0.0332 %

Work ratio - Fz Ratio of the design share force to design share resistance 4.17973 % -0.0064 %

Work ratio -Oblique

Ratio of the design moment resistance to design bendingresistance 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 -0.8089 %

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 %




124

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

Test ID: 5620

Test status: Passed

8.17.1Description

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

8.17.2Background

Classification and verification under compression efforts for an IPE 600 column made of S235 steel. The column isfixed 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.






■ Exploitation loadings (category A), Q:

► Fz = -100 000 N,


■ Cross section dimensions are in milimeters (mm).

Units

Metric System








125

Boundary conditions


■ 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


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

■ Internal: None.

8.17.2.2 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 rulesand 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 thepicture below:

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






127

Therefore:

424283.42 t

c

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 inTable 5.2 - sheet 2:

21.419

2/)22412220(

mm

mmmmmm

t

c

0.1235

y f

Therefore:

9921.4 t

c

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

A cross-section is classified according to the least favorable classification of its compression elements (chapter5.5.2(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 classsection for the entire column section will be considered Class 4.




128

8.17.2.3 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 areaof the cross-section.

The effective area of the cross section takes into account the reduction factor, , which is applying in this case onlyfor 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 thestress ratio, and the plate modified slenderness. They will be calculated considering only the cross-section web.

The buckling factor (k

) and the stress ratio( )

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

0.11

2

0.4 k

The plate modified slenderness ( p )

The formula used to determine the plate modified slenderness is:

754.0

0.40.14.28

12/242192600

4.28

/

mmmmmmmm

k

t b p

The reduction factor ( )

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

0.1

3055.02

p

p

Effective area

The effective area is determined considering the following:

275.1522312242192600939.01156001 mmt b A A weff

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 compressionresistance of the cross-section:

N MPamm f A

N M

yeff

Rd c 35775810.1

23575.15223 2

0

,

Work ratio

Work ratio = %79518.21003577581

100000100

,

N

N

N

N

Rd c




129



■ 6 nodes,



Work ratio of the design resistance for uniform compression

Column subjected to bending and axial force

Work ratio - Fx



Work ratio - Fx Compression resistance work ratio [%] 2.79518 %



Work ratio - Fx Compression resistance work ratio 2.79495 % -0.0083 %




130

8.18 EC3 test 4: Class section classification and bending moment verification of an IPE300 column

Test ID: 5412

Test status: Passed

8.18.1Description

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.

8.18.2Background

Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected tothe ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 50kN forceapplied on the web direction, defined as a live load. The dead load will be neglected.


■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001;



The following load cases and load combination are used:

■ Exploitation loadings (category A): Q = 50kN,





131

Units

Metric System


Boundary conditions


■ Outer:

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

► Support at end point (z = 5.00) free.

■ Inner: None.


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.






133

Therefore:


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 column section geometrical properties, the next conclusions can be found:




134

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 classsection 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)

8.18.2.3 Reference results in calculating the bending moment resistance

1,

Rd c

Ed M

M


0

,

*

M

y pl

Rd c

f w M

for Class1 cross sections


Where:

340.628 cmw pl

f y nominal yielding strength for S235 f y=235MPa

0 M partial safety coefficient 10 M

Therefore:

MNm f w

M M

y pl

Rd V y 147674.01

235*10*40.628* 6

0

,,

MNmkNmkN m M Ed 125.012550*5.2

%84148.0

125.0

,,

Rd V y

Ed

M

M



■ 7 nodes,





135


Combined oblique bending


Work ratio - Oblique



Combined obliquebending

Work ratio - Oblique 85 %



Work ratio -Oblique

Work ratio - Oblique 84.6459 % -0.4166 %




136

8.19 EC3 Test 2: Class section classification and shear verification of an IPE300 beam subjected tolinear uniform loading

Test ID: 5410

Test status: Passed

8.19.1Description

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.

8.19.2Background

Classification and verification of sections for an IPE 300 beam made from S235 steel. The beam is subjected to a50 kN/m linear uniform load applied gravitationally. The force is considered to be a live load and the dead load isneglected.






■ Exploitation loadings (category A): Q = -50kN/m,


Units

Metric System





137

Boundary conditions


■ Outer:

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

► Support at end point (x = 5.00) restrained in translation along Y and Z axis and rotation restrained on X

axis■ Inner: None.



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

■ compression for the top flange

■ compression and tension for the web

■ tension for the bottom flange

To determine the web class it will be used the Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.




138


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




139

Therefore:


To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter5.5.2


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

276.57.10

45.56

mm

mm

t

c

92.0




140

Therefore:

28.892.0*9*9276.57.10

45.56

mm

mm

t

c this means that the column flanges are Class 1.


According to the calculation above, the beam section have a Class 1 web and Class 1 flanges; therefore the classsection for the entire beam section will be considered Class 1.


8.19.2.3 Reference results in calculating the shear resistance V pl,Rd

The design resistance of the cross-section Vpl,Rd shall be determined as follows:

0

,

3*

M

y

v

Rd pl

f A

V


Where:

Av: section shear area for rolled profiles f w f v t r t t b A A *)*2(**2

A: cross-section area A=53.81cm2

b: overall breadth b=150mm

h: overall depth h=300mm

hw: depth of the web hw=248.6mm

r: root radius r=15mm

tf : flange thickness tf =10.7mm

tw: web thickness tw=7.1mm

268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmt r t t b A A f w f v


f y: nominal yielding strength for S275 f y=275MPa

0 M : partial safety coefficient 10 M

Therefore:

kN MN

f A

V M

y

v

Rd pl 7.4074077.01

3

275*10*68.25

3*

4

0

,




141

For more:

Verification of the shear buckling resistance for webs without stiffeners:

*72

w

w

t

h


20.11

1*20.1*20.1

0

1 M

M

92.0275

235235

y f

91.9392.0

20.1*72*7201.35

1.7

6.248

w

w

t

h

There is no need for shear buckling resistance verification

According to: EC3 Part 1,5 EN 1993-1-5-2004 Chapter 5.1(2)



■ 6 nodes,



Shear resistance work ratio

Work ratio - Fz




142



Fz Shear force 125 kN

Work ratio Work ratio - Fz 31 %



Fz Fz -125 kN -0.0000 %

Work ratio - Fz Work ratio Fz 30.6579 % -1.1034 %




143

8.20 EC3 Test 3: Class section classification, shear and bending moment verification of an IPE300column

Test ID: 5411

Test status: Passed

8.20.1Description



In the middle, the column is subjected to a 200 kN force applied on the web direction, defined as a live load.


8.20.2Background

Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected tothe ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 200kN force

applied on the web direction, defined as a live load. The dead load will be neglected.






■ Exploitation loadings (category A): Q = -200kN,





144

Units

Metric System


Boundary conditions


■ Outer:



■ Inner: None.



In case the column is subjected to a lateral load, the stresses distribution on the most stressed point (the columnbase) is like in the picture below:




145

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.





146

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

Therefore:


To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter5.5.2.




147



Therefore:




According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6).

8.20.2.3 Reference results in calculating the shear resistance V pl,Rd

The design resistance of the cross-section Vpl,Rd , is determined as follows:

0

,

3*

M

y

v

Rd pl

f A

V


Where:

Av section shear area for rolled profiles f w f v t r t t b A A *)*2(**2

A cross-section area A=53.81cm2

b overall breadth b=150mm

h overall depth h=300mm

hw depth of the web hw=248.6mm

r root radius r=15mm

tf flange thickness tf =10.7mm

tw web thickness tw=7.1mm268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmt r t t b A A f w f v




148




Therefore:

kN MN

f A

V M

y

v

Rd pl 42.3483484.01

3

235*10*68.253

* 4

0

,

8.20.2.4 Reference results in calculating the bending moment resistance

%50%4.5742.348

200

,

Rd pl

Ed

V

V

The shear force is greater than half of the plastic shear resistance. Its effect on the moment resistance must be takeninto account.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(1)(2)

Where:

0223.01

348.0

200.0*21

2 22

,

Rd pl

Ed

V

V


f w f v t r t t b A A *)*2(**2

Av section shear area for rolled profiles

A cross-section area A=53.81cm2

b overall breadth b=150mm

h overall depth h=300mm

hw depth of the web hw=248.6mm

r root radius r=15mm

tf flange thickness tf =10.7mm

tw web thickness tw=7.1mm

268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmt r t t b A A f w f v







149

Therefore:

MNm

f t

Aw

M M

y

w

v pl

Rd V y 146.01

235*0071.0*4

)²10*68.25(*0223.010*40.628*

4

²* 46

0

,,

MNmkNmkN m M Ed 5.0500200*5.2

%342146.0

500.0

,,

Rd V y

Ed

M

M



■ 7 nodes,



Shear z direction work ratio

Shear z direction work ratio

Work ratio - Fz




150






Shear z directionwork ratio Work ratio - Fz 57 %


Work ratio - Oblique 341 %



Work ratio - Fz Work ratio Fz 57.4021 % 0.7054 %

Work ratio -

Oblique

Work ratio - Oblique 341.348 % 0.1021 %






152

Units

Metric System


Boundary conditions


■ Outer:


► Support at end point (x = 5.00) restrained in translation and rotation along Y, Z axis and rotationblocked along X axis.

■ Inner: None.






154






155

Therefore:

This means that the beam web is Class 1.






156

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

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




8.21.2.3 Reference results for calculating the combined biaxial bending

1,

,

,

,

Ed Nz

Ed z

Rd Ny

Ed Y

M

M

M

M

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(5)

In which α and β are constants, which may conservatively be taken as unity, otherwise as follows:

For I and H sections:

2

)1;max(n

00

,,

Rd pl Rd pl

Ed

N N

N n therefore 1

Bending around Y:

For cross-sections without bolts holes, the following approximations may be used for standard rolled I or H sectionsand for welded I or H sections with equal flanges:

a

n M M

Rd y pl

Rd Ny*5.01

)1(*,,

,

but Rd ply Rd Ny M M ,,


00

,,

Rd pl Rd pl

Ed

N N

N n

5.0403.010*81.53

)0107.0*15.0*210*81.53()**2(4

4

A

t b Aa

f

8.0)403.0*5.01(

,,,,

,,

Rd y pl Rd y pl

Rd y N

M M M

Rd y pl Rd y N Rd y pl Rd y N M M M M ,,,,,,,,*8.0 but Rd ply Rd Ny M M ,,




157

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 sectionsand for welded I or H sections with equal flanges:

a

n M M

Rd z pl

Rd Nz *5.01

)1(*,,

,

but

Rd plz Rd Nz M M ,,


Rd z pl Rd z N Rd z pl Rd z N M M M M ,,,,,,,,*8.0 but Rd plz Rd Nz M M ,, therefore it will be considered:

MNm f w

M M M

y z pl

Rd z pl Rd z N 030.01

235*10*20.125* 6

0

,

,,,,

In conclusion:

11086.1029375.0

03125.0

148.0

03125.0 12

,

,

,

,

Ed Nz

Ed z

Rd Ny

Ed Y

M

M

M

M

The coresponding work ratio is:

WR = 1.1086 x 100 = 110.86 %



■ 6 nodes,



Work ratio – oblique bending

Beam subjected to combined bending

Work ratio – Oblique [%]




158




Combined oblique bending [%] 110.86 %



Work ratio -Oblique

Work ratio-Oblique 110.691 % -0.2783 %




159

8.22 EC3 Test 1: Class section classification and compression verification of an IPE300 column

Test ID: 5383

Test status: Passed

8.22.1Description



On top, the column is subjected to a 100 kN force applied gravitationally, defined as a live load.


8.22.2Background

Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected tothe ground by a fixed connection and is free on the top part. On top, the column is subjected to a 100kN force appliedgravitationally, defined as a live load. The dead load will be neglected.






■ Exploitation loadings (category A): Q = -100kN,





160

Units

Metric System


Boundary conditions


■ Outer:



■ Inner: None.




161


In this case, the column is subjected only to compression, therefore the distribution of stresses along the section islike 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




162



1

Therefore:


To determine the flanges class, we will use Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2




163



276.57.10

45.56

mm

mm

t

c

1

Therefore:

9*9276.57.10

45.56

mm

mm

t

c this means that the column flanges are Class 1.









165


Compressive resistance work ratio

Column subjected to compressive load

Work ratio [%]



Work ratio Compressive resistance work ratio [%] 8 %



Work ratio - Fx Work ratio Fx 7.90805 % -1.1494 %




166

8.23 EC3 Test 5: Class section classification and combined axial force with bending momentverification of an IPE300 column

Test ID: 5421

Test status: Passed

8.23.1Description



The column is subjected to a 500 kN compressive force applied on top and a 5 kN/m uniform linear load applied onall the length of the column, on the web direction, both defined as live loads.


8.23.2Background

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 forceapplied on top and a 5kN/m uniform linear load applied for all the length of the column, on the web direction, bothdefined as live loads. The dead load will be neglected.






■ Exploitation loadings (category A): Q1 = 500kN, Q2 = 5kN/m,


Units

Metric System




167


Boundary conditions


■ Outer:



■ Inner: None.



In this case the column is subjected to compression and lateral load, therefore the stresses distribution on the moststressed 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.




168






169

Therefore:

Therefore:






170



Therefore:





Cross sections for class 3, the maximum longitudinal stress should check:


In this case:

In absence of shear force, for Class 3 cross-sections the maximum longitudinal stress shall satisfy the criterion:


This means:

The corresponding work ratio is: WR = 0.8728 x 100 = 87.28 %




171



■ 6 nodes,



Work ratio – bending and axial compression

Column subjected to bending and axial compression

Work ratio – Oblique [%]



Bending and axialcompression

Work ratio – oblique [%] 87.28 %



Work ratio -

Oblique

Work ratio- Oblique 87.2799 % 0.3217 %




172

8.24 Changing the steel design template for a linear element (TTAD #12491)

Test ID: 4540

Test status: Passed

8.24.1Description

Selects a different design template for steel linear elements.

8.25 Verifying the "Shape sheet" command for elements which were excluded from the specializedcalculation (TTAD #12389)

Test ID: 4529

Test status: Passed

8.25.1Description

Verifies the program behavior when the "Shape sheet" command is used for elements which were excluded from thespecialized calculation (chords).

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

Test ID: 4549

Test status: Passed

8.26.1Description

Verifies the shape sheet for a steel beam with circular cross-section when the lateral torsional buckling is computedand when it is not.

8.27 EC3 fire verification: Verifying the work ratios after performing an optimization for steelprofiles (TTAD #11975)

Test ID: 4484

Test status: Passed

8.27.1Description

Runs the Steel elements verification and generates the "Envelopes and optimizing profiles" report in order to verifythe work ratios. The verification is performed using the EC3 norm with Romanian annex.

8.28 EC3: Verifying the buckling length results (TTAD #11550)

Test ID: 4481

Test status: Passed

8.28.1Description

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 thebase. A punctual live load of 200.00 kN is applied.




173

8.29 EC3 Test 45: Comparing the shear resistance of a welded built-up beam made from differentsteel materials

Test ID: 5745

Test status: Passed

8.29.1Description

The shear resistance of a welded built-up beam made of S275 steel is compared with the shear resistance of thesame built-up beam made of a user-defined steel material.

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

8.29.2Background

Verifies the shear resistance of a welded built-up beam made of 500 MPa yield strength user-defined steel. Thebeam is simply supported and it is subjected to a uniformly distributed load (20 000 N/ml) applied over its length. Thedead load will be neglected.

Verifies also the shear resistance of the same welded built-up beam made of S275 steel. The loading and supportconditions are the same.



■ Analysis type: static linear;




► Fz = - 20 000 N/ml,



Units

Metric System




174

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,

■ Section area: A = 5188 mm2,

■ Partial factor for resistance of cross sections: 0.10 M .


500 MPa yield strength user-defined material and S275 steel are used. The following characteristics are used:


■ Yield strength (for S275 steel) f y = 275 MPa.

Boundary conditionsThe 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.

8.29.2.2 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 beused:

0.1,

Rd pl

Ed

V

V

■ VEd represents the design value of the shear force:

N mmml N Lq

V Ed 500002

5000/20000

2

■ Vpl,Rd represents the design plastic shear resistance. The design plastic shear resistance of the cross-sectionis determined with formula (6.18) from EN 1993-1-1:2001. Before using it, the shear area (A v) has to bedetermined.




175

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 () 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 isparallel to web, the shear area is:

206.1978)1.76.278(0.1 mmmmmmt h A wwv

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 () is 1.2.

As the load is parallel to web, the shear area becomes:

267.2373)1.76.278(2.1 mmmmmmt h A wwv

Design plastic shear resistance of the cross section made of 500 MPa yield strength user-defined material

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

N

MPamm

f A

V M

y

v

Rd pl 7.5710160.1

3

50006.1978

3

2

0

,

The verification of the design plastic shear resistance of the cross section is done with criterion (6.18) from EN 1993-1-1:

0.10876.07.571016

50000

,

N

N

V

V

Rd pl

Ed

The corresponding work ratio is:

Work ratio = %76.81007.571016

50000100,

N

N

V

V

Rd pl

Ed

Design plastic shear resistance of the cross section made of S275 steel

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

N

MPamm

f A

V M

y

v

Rd pl 7.3768700.1

3

27567.2373

3

2

0

,

The verification of the design plastic shear resistance of the cross section is done with criterion (6.18) from EN 1993-

1-1:

0.1133.07.376870

50000

,

N

N

V

V

Rd pl

Ed

The corresponding work ratio is:

Work ratio = %27.131007.376870

50000100

,

N

N

V

V

Rd pl

Ed



■ 7 nodes,■ 1 linear element.




176


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



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 %



Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 275MPa)

13.2671 % -0.0216 %

Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 500MPa)

8.75631 % -0.0421 %




177

8.30 EC3 Test 41: Determining lateral torsional buckling parameters for a I-shaped laminated beamconsidering the load applied on the lower flange

Test ID: 5753

Test status: Passed

8.30.1Description

Determines the lateral torsional buckling parameters for a I-shaped laminated beam made of S235 steel, consideringthe load applied on the lower flange. The loadings applied on the beam are: a uniformly distributed load and 2punctual 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.

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

Test ID: 5744

Test status: Passed

8.31.1Description

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 thebeam and 7 kN/m linear uniform horizontal load.


8.32 EC3 Test 42: Determining lateral torsional buckling parameters for a I-shaped welded built-upbeam considering the load applied on the upper flange

Test ID: 5752

Test status: Passed

8.32.1Description

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 loadand 2 punctual negative bending moments applied at beam extremities.







179

Geometry

Below are described the column cross section characteristics:



■ Flange thickness: tf = 18 mm,

■ Web thickness: tw = 8 mm,

■ Column length: L = 5000 mm,

■ Section area: A = 22752 mm2 ,







■ Outer:

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

► Free at end point (z = 5.00).

■ Inner: None.

Loading


■ External:

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

■ Internal: None.




180

8.33.2.2 Cross-section classification

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


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. Thepicture below shows an extract from this table.

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

67.1318

2/)8500(

mm

mmmm

t

c

81.0235

y f

Therefore:

34.111467.13 t c




181

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

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

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

25.748

218630

mm

mmmm

t

c

81.0235

y f

Therefore:

02.344225.74 t

c


A cross-section is classified according to the least favorable classification of its compression elements (chapter5.5.2(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 classsection for the entire column section will be considered Class 4.


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 areaof the cross-section.

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

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


) and the stress ratio( ) - for flanges

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




182

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

43.00.11

2

k


) and the stress ratio( ) - for web

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

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

0.40.11

2

k

The plate modified slenderness ( p ) – for flanges

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

906.0

43.081.04.28

18/2/8500

4.28

/

mmmmmm

k

t c p

The plate modified slenderness ( p) – for web

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

614.1

0.481.04.28

8/182630

4.28

/

mmmmmm

k

t b p

The reduction factor ( ) – for flanges

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

0.1188.0

2

p

p

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

mm

mmmmcb f eff 25.215

2

8500875.0,




183

The reduction factor ( ) – for web

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

0.1

3055.02

p

p

The effective width of the web can now be calculated:

mmmmmmbb weff 8.317182630535.0,

Effective area

The effective area is determined considering the following:

weff ww f eff f eff f eff bt t bbt A ,,, )(2


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

N MPamm f A

N M

yeff

Rd c 65065820.1

3554.18328 2

0

,

Work ratio

Work ratio = %54.11006506582

100000100

,

N

N

N

N

Rd c



■ 6 nodes,







185

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

Test ID: 5701

Test status: Passed

8.34.1Description

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

The tests are made according to Eurocode 3 French Annex.

8.34.2Background

Buckling verification under compression efforts for an CHS219.1x6.3H column made of S355 steel. The column isfixed 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.


■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;■ Analysis type: static linear (plane problem);






Units

Metric System


■ Tube wall thickness: t=6.3mm

■ Tube diameter: d=219.1mm


■ Radius of gyration about the relevant axis: i=75.283mm

■ Flexion inertia moment around the Y axis: Iy=2366x104mm

4

■ Flexion inertia moment around the Z axis: Iz=2366x104mm

4




186






■ Outer:



■ Inner: None.

■ Buckling lengths Lfy and Lfz are both imposed with 6m value

Loading


■ External: Point load at Z = 3.0: FZ = N = -100 000 N,

■ Internal: None.




187


The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element iscalculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from thecompression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, iscalculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.

%100100,

Rd b

Ed

N N (6.46)


778.343.6

1.219

mm

mm

t

d

814.0355

235235

y f

381.46814.05080778.34 2

t

d therefore the section is considered to be Class 2

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




188

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

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

1

,

M

y

Rd b

f A N

(6.47)

Where:

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

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

according to:

11

22

y y y

y

(6.49)

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

1

*

i

L

N

f Acr

cr

y

y (6.50)

41.76355

2100001

y f

E

mmmm

mm

A

I i

y

283.754210

238600002

4

043.141.76283.75

6000

1

mm

mm

i

Lcr y

132.1²043.1)2.0043.1(21.015.0)2.0(15.0 2 y y y

1636.0²043.1²132.1132.1

1

y

A is the cross section area; A=4210mm2; f y is the yielding strength of the material; f y=355N/mm

2 and 1 M is a safety

coefficient, 11 M

The design buckling resistance of the compression member will be:

N mm N mm f A

N M

y y

Rd b 8.9505331

/3554210636.0 22

1

,

N N Ed 100000

%520.101008.950533

100000100

,

N

N

N

N

Rd b

Ed




189



■ 4 nodes,





LT




190

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

Column subjected to axial force

Adimensional - SNy




0.636

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

0.1052



Xy coefficient corresponding to non-dimensional slenderness 0.635463adim

0.0001 %

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

0.105293adim

-0.0001 %




191

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

Test ID: 5714

Test status: Passed

8.35.1Description


The cross section has an “I symmetric” shape with: 408mm height; 190mm width; 9.4mm center thickness; 14.6mmflange 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 theefforts are applied on the top of the column.


8.35.2Background

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 axialcompression load -1000000 N, a 200000Nm bending moment after the Y axis and a 5000N lateral force after the Yaxis.






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






192

Units

Metric System



■ Cross section area:2

72.9108 mm A







, 06.1261435 mmW yel



, 65.175962 mmW z el

■ Plastic modulus after the Z axis, 69.271897, z pl W

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

■ Flexion inertia moment around the Z axis: Iz=16716452.10 mm

4

■ Torsional moment of inertia: It=492581.13 mm

4

■ Working inertial moment: Iw=645759981974.33mm6






Boundary conditions


■ Outer:





193

Loading


■ External: Point load From X=0.00m and z=2.00m: FZ =-1000000N; Mx=200000Nm and Fy=5000N




-for beam web:


118.083.268

76.48

sup

inf

Mpa

Mpa




194

909.320

59.317

83.26876.378

152.5859.317

76.4876.378

59.317

76.378

83.26876.4883.26876.48 y

x y x y x

5.0153.076.378

152.5876.378

x




195

924.0275

235235

y f

56.63)18.0(33.067.0

924.042

33.067.0

42170

924.0

30.404.9

6.142408

t

cmm

mmmm

t

c


-for beam flange:

316.8924.0918.6

924.0

18.66.14

30.90

t

ct

c






34.0




196



curve according to:

11

22

y y y

y

(6.49)


ycr

y

y N

f A

,

*


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

kN N mm

mmmm N

l

I E N

fy

y ycr 05.1333387.133338053²2000

257332751/210000²

²

² 42

,

137.07.133338053

/27572.9108 22

,

N

mm N mm

N

f A

ycr

y y

499.0137.02.0137.034.015.0²)2.0(15.0 2 y y y

1

1

022.1137.0499.0499.0

11

2222

y

y

y y y

y




197



49.0



according to:

1122

z z z

z

(6.49)


z cr

y z

N

f A

,

*

ml fz 00.2

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

4





198

kN N

mm

mmmm N

l

I E N

fz

z z cr 70.866138.8661700

²2000

10.16716452/210000²

²

² 42

,

538.0

38.8661700

/27572.9108 22

,

N

mm N mm

N

f A

z cr

y z

728.0538.02.0538.049.015.0²)2.0(15.0 2 z z z

821.0

1

821.0538.0728.0728.0

11

2222

z

z

z z z

z



z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1



²252.0423.0325.0

11

C


-in order to simplify the calculation, it will be considered 11 C




199

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

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

4

Longitudinal elastic modulus: E = 210000 N/mm2.

Torsional moment of inertia: It=492581.13 mm

4

Warping inertial moment:


4

2

f z

w

t h I I

h cross section height; h=408mm


611

24

1046774.64

6.144080mm16716452.1mm

mmmm I w



Shear modulus of rigidity: G=80800N/mm2

kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

089.18021802088994

052.208384.8661700110.16716452/210000

13.492581/808002000

10.16716452

1046774.6

2000

10.16716452/2100001

²

²

²

²

422

422

4

611

2

422

1

The elastic modulus : 3

4

max

, 172.1261434204

257332751mm

mm

mm

z

I W

y

yel

439.01802088994

/275172.1261434 23,

Nmm

mm N mm

M

f W

cr

y yeff LT


1

²²

1

LT LT LT

LT

(6.56)





200


2147.2190

408

mm

mm

b

h


76.0

687.0²439.02.0439.076.015.0²2.015.0 LT LT LT LT

1813.0²439.0²687.0687.0

1

²²

1

LT LT LT

LT


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

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






202


0

,0

cr

y yeff

M

f W

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

34

max

, 172.1261434204

257332751mm

mm

mm

z

I W

y

yel


kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

089.18021802088994

052.208384.8661700110.16716452/210000

13.492581/808002000

10.16716452

1046774.6

2000

10.16716452/2100001

²

²

²

²

422

422

4

611

2

422

1

439.01802088994

/275172.1261434 23,

0

Nmm

mm N mm

M

f W

cr

y yeff


,,

1 1120.0

TF cr

Ed

z cr

Ed

N

N

N

N C term:

Where:


²

1

,

2

2

0

,

T cr

wt T cr

L

I E I G

i N


0301.02

0

2

0

222

0 z yiii z y

Torsional moment of inertia: It=492581.13 mm

4

Working inertial moment: Iw=645759981974.33mm6

- the buckling length,T cr

L,

,

m L T cr 00.2,

mm

mmmm N mmmm N

mm

mm N T cr

13

62242

4

2

,

10244.1

²2000

33.746457599819/21000013.492581/80800

0301.0

1

N N Ed 1000000

N N N T cr TF cr

13

,, 10244.1

N

mmmmmm N

l I E N

fz

z z cr 38.8661700

²200010.16716452/210000²

²²

42

, (previously calculated)




203



172.010244.1

10000001

38.8661700

10000001120.01120.0 4

134

,,

1

N

N

N

N

N

N

N

N C

TF cr

Ed

z cr

Ed

Therefore:


1

11

11

172.01120.0469.0

172.01120.0

469.0

,,

2

0,

0,0,

4

,,

10

4

,,

1

0

T cr

Ed

z cr

Ed

LT mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C

a

aC C C

N

N

N

N C

N

N

N

N C



LT y

LT y

mymymya

aC C C

11 0,0,

yeff

eff

Ed

Ed y

yW

A

N

M

,

,


,, 06.1261435 mmW W yeff yel

444.106.1261435

72.9108

1000000

102003

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

1998.0257332751

492581.1311

4

4

mm

mm

I

I a

y

t LT






205


z cr

Ed

y

mz yz

N

N C k

,

1

mz C term must be calculated for the hole column length

1

200

200

,sup

,inf

kNm

kNm

M

M

Ed

Ed

kN N

mm

mmmm N

l

I E N

fz

z z cr 70.866138.8661700

²2000

10.16716452/210000²

²

² 42

,

776.038.8661700

100000033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

878.0

38.8661700

10000001

1776.0

1,

N

N

N

N C k

z cr

Ed

y

mz yz


ycr

Ed

z mLT my zy

N

N C C k

,

1

799.0

38.8661700

1000000821.01

38.8661700

1000000

1

1

1

,

,

N

N N

N

N

N

N

N

z cr

Ed

z

z cr

Ed

z


050.1

7.133338053

10000001

977.0065.1001.1

1,

N

N

N

N C C k

ycr

Ed

z mLT my zy


857.0

38.8661700

10000001

977.0776.0

1,

N

N

N

N C k

z cr

Ed

z

mz zz




206


1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rk z

Rd z Ed z

zz

M

Rk y

LT

Rd y Ed y

zy

M

Rk z

Ed

M

Rk z

Rd z Ed z

yz

M

Rk y

LT

Rd y Ed y

yy

M

Rk

y

Ed

M

M M k

M

M M k

N

N

M

M M k

M

M M k

N

N

i y Rk A f N

41.118.074.049.0

1

/2755.175962

1010857.0

1

/27506.1261435813.0

10200050.1

1

/27572.9108821.0

1000000

34.118.076.040.0

1

/2755.175962

1010878.0

1

/27506.1261435813.0

10200074.1

1

/27572.91081

1000000

23

6

23

6

222

23

6

23

6

22

mm N mm

Nmm

mm N mm

Nmm

mm N mmmm

N

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N







207



y



z




208



yyk



yz k




209

Internal factor, zyk


zyk



zz k




210



SNy



SMyy




211



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 effortover the Z axis

SNz




212

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 momentover the Z axis

SMzy



SMzz




213




1


0.821






0.40


0.76


0.18


0.49


0.74


0.18



Xy coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 %

Xz coefficient corresponding to non-dimensional slenderness 0.821634 adim -0.0001 %

Kyy Internal factor kyy 1.11767 adim -0.0003 %

Kyz Internal factor kyz 0.877605 adim -0.0000 %

Kzy Internal factor kzy 1.09224 adim -0.0001 %

Kzz Internal factor kzz 0.857639 adim -0.0001 %

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

0.399218 adim -0.0000 %

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

0.783306 adim -0.0000 %

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

0.181362 adim -0.0001 %

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

0.485883 adim 0.0000 %

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

0.765485 adim 0.0000 %

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

0.177236 adim -0.0002 %




214

8.36 EC3 Test 27: Verifying an user defined I section class 3 beam simply supported with adisplacement restraint

Test ID: 5717

Test status: Passed

8.36.1Description

The test verifies a user defined cross section beam. The beam is hinged at one end and the translations over the Yand 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; 19mmflange 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 kNpunctual force applied on the end of the beam.


8.36.2Background An I53*1.2+22*1.9 beam column subjected to axial compression, uniform distributed vertical force and uniformdistributed horizontal force, made from S235 steel. The beam has a 53x12mm web and 220x19mm flanges. Thebeam is simply supported. The beam is subjected to an axial compression load 3700000 N, 10000 N/m uniformdistributed load over the Z axis and 5000 N/m horizontal uniform distributed force after the Y axis.






■ Exploitation loadings (category A): Fx=-3700000N; Fy=-5000N/m; Fz=-10000N/m



Units

Metric System




215




14264mm A


■ Flange thickness: mmt f 19


■ Web thickness: mmt w 12



, 11.2509773 mmW yel



, 41.307177 mmW z el



■ Flexion inertia moment around the Z axis:

467.33789514 mm I z

■ Torsional moment of inertia:

473.1269555 mm I t

■ Working inertial moment:

667.6662201162989 mm I w









216

Boundary conditions


■ Outer:


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

■ Inner:

► Lateral buckling restraint in the middle of the column (x=2.50).

Loading


■ 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




217




218




-for beam web:


194.246

94.246

sup

inf Mpa

Mpa




219

1235

235235

y f




220

4242413838

1

4112

192530

t

cmm

mmmm

t

c

therefore the beam web is

considered to be Class 3

-for beam flange:

91957.4

1

47.519

2

12220

t

c

t

c






34.0




221



curve according to:

11

22

y y y

y

(6.49)


ycr

y

y N

f A

,

*



kN N

mm

mmmm N

l

I E N

fy

y

ycr 06.5513921.55139061²5000

67665089874./210000²

²

² 42

,

247.021.55139061

/23514264 22

,

N

mm N mm

N

f A

ycr

y y

538.0247.02.0247.034.015.0²)2.0(15.0 2 y y y

984.0

1

984.0247.0538.0538.0

11

2222

y

y

y y y

y




222



49.0



according to:

1122

z z z

z

(6.49)


z cr

y z

N

f A

,

*

ml fz 50.2






223

kN N

mm

mmmm N

l

I E N

fz

z z cr 235.1120519.11205235

²2500

67.33789514/210000²

²

² 42

,

547.019.11205235

/23514264 22

,

N

mm N mm

N

f A

z cr

y z

735.0547.02.0547.049.015.0²)2.0(15.0 2 z z z

816.0

1

819.0547.0735.0735.0

11

2222

z

z

z z z

z


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

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

25.0

1025.318

²2500/10000

8

²3

Nm

mmm N

M

Lq


z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1


-where:






224

0 therefore:

31.11 C



Longitudinal elastic modulus:2

/210000 mm N E

Torsional moment of inertia:473.1269555 mm I t

Working inertial moment:667.6662201162989 mm I w

Shear modulus of rigidity:

2

/80800 mm N G

Buckling length of the beam mm L 2500


, 11.2509773 mmW yel

Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

4001163141

58.27219.1120523531.167.33789514/210000

73.1269555/808002500

67.33789514

1001163.22

2500

67.33789514/21000031.1

²

²

²

²

422

422

4

611

2

422

1




225

384.04001163141

/23511.2509773 23,

Nmm

mm N mm

M

f W

cr

y yeff LT

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

Note:

36

1081.74001163141

1025.31

Nmm

M

M

cr

Ed

0156.000781.0

0156.0125.0530

2203.03.0

04.01081.7

20.0384.0

2

0,

20,0,

2

0,3

LT

cr

Ed

LT LT

LT

cr

Ed

cr

Ed

LT

M

M

h

b

M

M

M

M

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

For slendernesses2

0, LT

cr

Ed

M

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

apply.

-therefore:

1 LT

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


The internal factor yyk corresponding to a Class 4 section will be calculated according to Annex A, Table a.1, and will be calculatedseparately for the two column parts separate by the middle torsional lateral restraint:




226

ycr

Ed

y

mLT my yy

N

N C C k

,

1

ycr

Ed y

ycr

Ed

y

N

N N

N

,

,

1

1


kN N Ed 3700

N

mm

mmmm N

l

I E N

fy

y

ycr 21.55139061²5000

67665089874./210000²

²

² 42

,

(previously

calculated)

999.0

21.55130961

3700000984.01

21.55130961

37000001

1

1

,

,

N

N N

N

N

N

N

N

ycr

Ed y

ycr

Ed

y

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

- Must be calculated considering the beam along its length.






228


,,

1 1120.0

TF cr

Ed

z cr

Ed

N

N

N

N C term:

Where:


²

1

,

2

2

0

,

T cr

wt T cr

L

I E I G

i N

0491.02

0

2

0

222

0 z yiii z y


Working inertial moment:667.6662201162989 mm I w


m L T cr 50.2,

mm

mmmm N mmmm N

mm

mm N T cr

13

62242

4

2

,

10696.1

²2500

67.6662201162989/21000073.1269555/80800

0491.0

1

N N Ed 3700000

N N N T cr TF cr

13

,, 10696.1

N

mm

mmmm N

l

I E N

fz

z z cr 19.11205235

²2500

67.33789514/210000²

²

² 42

,


C1=1

181.0

10696.1

37000001

19.11205235

37000001120.01120.0 4

134

,,

1

N

N

N

N

N

N

N

N C

TF cr

Ed

z cr

Ed

Therefore:

1

11

11

181.01120.0439.0

181.01120.0

439.0

,,

2

0,

0,0,

4

,,

10

4

,,

1

0

T cr

Ed

z cr

Ed

LT mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C

a

aC C C

N

N

N

N C

N

N

N

N C

ThemyC

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





229

LT y

LT y

mymymya

aC C C

11 0,0,

yeff

eff

Ed

Ed y

y

W

A

N

M

,

,


,, 11.2509773 mmW W yeff yel

048.011.2509773

14264

3700000

1025.313

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

998.067.665089874

73.126955511

4

4

mm

mm

I

I a

y

t LT


ycr

Ed my

N

N C

,

0, 03.01

Where:

N

mm

mmmm N

l

I E N

fy

y

ycr 21.55139061²5000

67665089874./210000²

²

² 42

,


N N Ed 3700000

002.121.55139061

370000003.010,

N

N C my

002.1100481

1048.0002.11002.1

11 0,0,

LT y

LT y

mymymya

aC C C


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






231

999.0998.0219.01

998.0219.0999.01999.0

11 0,0,

LT y

LT y

mymymya

aC C C

377.1

10696.137000001

19.1120523537000001

998.0²999.0

11 13

,,

2

N N

N N

N

N

N

N

aC C

T cr

Ed

z cr

Ed

LT

mymLT


476.1

21.55130961

37000001

1377.1002.1

1,

N

N

N

N C C k

ycr

Ed

y

mLT my yy


z cr

Ed

y

mz yz

N

N C k

,

1

Cmz coefficient must be calculated considering the beam along its length

010.119.11205235

370000003.0103.01

,0,

N

N

N

N C C

z cr

Ed

mz mz

506.1

19.11205235

37000001

1776.0

1,

N

N

N

N C k

z cr

Ed

y

mz yz




232


ycr

Ed

z mLT my zy

N

N C C k

,1

917.0

19.11205235

3700000816.01

19.11205235

37000001

1

1

,

,

N

N N

N

N

N

N

N

z cr

Ed z

z cr

Ed

z


355.1

21.55130961

37000001

917.0377.1002.1

1,

N

N

N

N C C k

ycr

Ed

z mLT my zy


383.1

19.11205235

37000001

917.0776.0

1,

N

N

N

N C k

z cr

Ed

z mz zz


1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rk z

Rd z Ed z

zz

M

Rk y

LT

Rd y Ed y

zy

M

Rk z

Ed

M

Rk z

Rd z Ed z

yz

M

Rk y

LT

Rd y Ed y

yy

M

Rk y

Ed

M

M M k

M

M M k

N

N

M

M M k

M

M M k

N

N

i y Rk A f N




233

72.130.007.035.1

1

/23541.307177

1062.15383.1

1

/23511.25097731

1025.31355.1

1

/23514264816.0

3700000

53.033.008.012.0

1/23541.307177

1062.15506.1

1

/23511.25097731

1025.31476.1

1

/23514264984.0

3700000

23

6

23

6

22

23

6

23

6

22

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N



■ 7 nodes,




y



z




234



yyk



yz k



zyk




235



zz k



SNy


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

SMyy




236


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

SMyz



SNz



SMzy






238



Xy coefficient corresponding to non-dimensional slenderness 0.983441 adim 0.0000 %

Xz coefficient corresponding to non-dimensional slenderness 0.816369 adim -0.0000 %Kyy Internal factor, kyy 1.47276 adim -0.0003 %

Kyz Internal factor, kyz 1.50599 adim -0.0003 %

Kzy Internal factor, kzy 1.35211 adim -0.0003 %

Kzz Internal factor, kzz 1.38261 adim 0.0002 %

SNy Bending and axial compression verification term dependingof the compression effort over the Y

1.12239 adim 0.0001 %

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

0.078033 adim -0.0000 %

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

0.325974 adim 0.0001 %

SNz Bending and axial compression verification term dependingof the compression effort over the z axis 1.35209 adim 0.0001 %

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

0.0716405 adim -0.0001 %

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

0.29927 adim 0.0001 %




239

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

Test ID: 5712

Test status: Passed

8.37.1Description



The column is subjected to a -328kN axial compression force; 1274 kNm bending moment after the Y axis and 127.4kNm bending moment after the Z axis. All the efforts are applied on the top of the column. The column height is5.62m and has a restraint of displacement at 2.81m from the bottom over the weak axis.


8.37.2Background An I880*5+220*15 shaped column subjected to compression and bending, made from S275 steel. The column has a880x5mm web and 220x15mm flanges. The column is hinged at it’s base and at his top the end is translation ispermitted only on vertical direction and the rotation is blocked for the long axis of the column. The column issubjected to an axial compression load -328000 N, a 127400Nm bending moment after the X axis and a1274000Nmbending 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)




240






■ Exploitation loadings (category A): Fz=-328000N N, Mx= 127400Nm; My=1274000Nm



Units

Metric System



■ Cross section area:210850mm A







, 1066.3387 mmW yel



, 1008.242 mmW z el


, 1031.368 mmW z pl


4


4


4

■ Working inertial moment: Iw=4979437.37x106mm

6




241






Boundary conditions


■ Outer:



■ Inner:

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

Loading


■ External: Point load From X=0.00m and z=5.62m: FZ =--328000N; Mx=127400Nm and My=1274000Nm






243

27.396

14.761

84.354850

73.45314.761

3.406850

14.761

850

84.3543.40684.3543.406 y

x y x y x

5.05338.0850

73.453

850

x




244

924.0275

235235

y f

61.101)873.0(33.067.0

924.042

33.067.0

42170

924

1705

152880

t

cmm

mmmm

t

c


-for beam flange:

316.8924.0961.7

924

61.715

5.107

t

ct

c

therefore the haunch is considered to

be Class1


8.37.2.3 Effective cross-sections of Class4 cross-sections

-the section is composed from Class 4 web and Class 1 flanges, therefore will start the webcalculation:

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

1705

152880

mm

mmmm

t

c:

41 k

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

k

t

bw

p

4.28

wb is the width of the web; mmbw 850

t is the web thickness; t=5mm

9244.0275

235235

y f

261.349244.04.28

5

850

mm

mm

p

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

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

286.0

261.3

4055.0261.33055.0

043

673.0261.322

p

p p

mmmmb

mmmmb

mmmmb

bb

bb

bb

e

e

eff

eff e

eff e

weff

55.1211.2435.0

55.1211.2435.0

1.243850286.0

5.0

5.0

2

1

2

1




245

2

21, 5.121555.121555.1215 mmmmmmmmmmbt bt A ewewwebeff

2

, 330022015 mmmmmmbt A f f flangeeff

222

,, 2.7815330025.12152 mmmmmm A A A flangeeff webeff eff

8.37.2.4 Effective elastic section modulus of Class4 cross-sections

-In order to simplify the calculation the section will be considered in pure bending

mmmm

bb t c 4252

8501

sup

inf

9.231 k


170

5

152880

mm

mmmm

t

c

k

t

bw

p

4.28



9244.0275

235235

y f

325.19.239244.04.28

5

850

mm

mm

p


692.0

325.1

2055.0325.13055.0

023

673.0325.122

p

p p

mmmmb

mmmmb

mmmm

b

bb

bb

bbb

e

e

eff

eff e

eff e

wceff

46.1761.2946.0

64.1171.2944.0

1.294

11

850692.0

6.0

4.01

2

1

2

1




246

-the weight center coordinate is:

mm

yG

53.155.10195

095.158330

15220546.601564.11715220

27.124546.6015.4321522018.366564.1175.43215220

-the inertial moment along the strong axis is:

4

23

23

23

23

14482344295.6993800726.748854356

97.4161522012

2201574.107546.601

12

546.601

71.381564.11712

564.11703.44815220

12

22015

mm

I y

423

23

23

23

63.2662749001522012

152200546.601

12

46.6015

0564.11712

64.1175015220

12

15220

mm

I z

34

max,

533.317922953.455

1448234429mm

mm

mm

z

I W

y

yel

34

max

, 10.242068110

63.26627490mm

mm

mm

y

I W z

z el




247




34.0



curve according to:

11

22

y y y

y

(6.49)


ycr

yeff

y N

f A

,

*




248

Where: A is the effective cross section area;22.7815 mm Aeff ; f y is the yielding strength of the material;

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

properties:

kN N

mm

mmmm N

l

I E N

fy

y

ycr 37.9503544.95035371

²5620

1448234429/210000²

²

² 42

,

15.044.95035371

/2752.7815 22

,

N

mm N mm

N

f A

ycr

yeff y

503.015.02.015.034.015.0²)2.0(15.0 2 y y y

1

1

017.115.0503.0503.0

11

2222

y

y

y y y

y



49.0




249



according to:

11

22

z z z

z

(6.49)


z cr

yeff z

N

f A

,

*



properties:

ml fz

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

kN N

mm

mmmm N

l

I E N

fz

z z cr 35.698962.6989347

²2810

63.26627490/210000²

²

² 42

,

555.062.6989347

/2752.7815 22

,

N

mm N mm

N

f A

z cr

yeff z

741.0555.02.0555.049.015.0²)2.0(15.0 2 z z z

812.0

1

812.0555.0741.0741.0

11

2222

z

z

z z z

z


a) for the top part of the column:



z

t

z

w z cr

I E

I G L

I

I

L

I E

C M

²

²

²

²1



²252.0423.0325.0

11

C


is the fraction of the bending moment from the column extremities: 50.01274

637

kNm

kNm






251


4

max

, 533.317922953.455

1448234429mm

mm

mm

z

I W

y

yel

466.04022433856

/275533.3179229 23,

Nmm

mm N mm

M

f W

cr

y yeff LT


1²²

1

LT LT LT

LT

(6.56)



24220

880

mm

mm

b

h


76.0

710.0²466.02.0466.076.015.0²2.015.0 LT LT LT LT

1803.0²466.0²710.0710.0

1

²²

1

LT LT LT

LT

b) for the bottom part of the column:



z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1


-where:


²252.0423.0325.0

11

C





252


0

kNm

77.10 1 C

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

Flexion inertia moment around the Y axis:41448234429mm I y



Torsional moment of inertia: It=514614.75mm4



4

2

f z

w

t h I I


f t flange thickness;

mmt f 15

611

24

10808.494

15880mm326627490.6mm

mmmm I w


Length of the column part: L=2810mm





253

kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

89.54345434891269

32.439626.698934777.163.26627490/210000

514614/808002810

63.26627490

10808.49

2810

63.26627490/21000077.1

²

²

²

²

422

422

4

611

2

422

1


4

max

, 533.317922953.455

1448234429mm

mm

mm

z

I W

y

yel

401.05434891269

/275533.3179229 23,

Nmm

mm N mm

M

f W

cr

y yeff LT


1

²²

1

LT LT LT

LT

(6.56)



24220

880

mm

mm

b

h


76.0

657.0²401.02.0401.076.015.0²2.015.0 LT LT LT LT

1849.0²401.0²657.0657.0

1

²²

1

LT LT LT

LT




254


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

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


ycr

Ed

ymLT my yy

N

N C C k

,

1

ycr

Ed y

ycr

Ed

y

N

N

N

N

,

,

1

1

1 y (previously calculated)

kN N Ed 328

kN N l

I E N

fy

y

ycr 37.9503544.95035371²

²,


1

44.95035371

32800011

44.95035371

3280001

1

1

,

,

N

N N

N

N

N

N

N

ycr

Ed y

ycr

Ed

y




255



0

,0

cr

y yeff

M

f W


34

max

, 533.317922953.455

1448234429mm

mm

mm

z

I W

y

yel


kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

56.30703070560199

32.439626.6989347163.26627490/210000

514614/808002810

63.26627490

10808.49

2810

63.26627490/2100001

²

²

²

²

422

422

4

611

2

422

10

534.03070560199

/275533.3179229 23

0

,0

Nmm

mm N mm

M

f W

cr

y yeff




256


,,

1 1120.0

TF cr

Ed

z cr

Ed

N

N

N

N C term:

Where:


²,

2

0

,

T cr

wt T cr

L

I E I G

I

A N


2

0 g z y z A I I I



Cross section area:2

10850mm A


4442

0 151720927017.2662885467.1490580416 mmmmmm I I z A I I I z y g z y


m L T cr 81.2,


Working inertial moment:61110808.49 mm I w (previously calculated)



mm

mmmm N mmmm N

mm

mm N T cr

24.9646886

²2810

10808.49/21000075.514614/80800

1517209270

10850 6112242

4

2

,

N N Ed 328000

N N N T cr TF cr 24.9646886,,

kN N l

I E N fz

z z cr 35.698962.6989347

²

², (previously calculated)

C1=1.31 for the top part of the column

C1=1.77 for the bottom part of the column


224.0

24.9646886

3280001

62.6989347

328000131.120.01120.0 44

,,

1

N

N

N

N

N

N

N

N C

TF cr

Ed

z cr

Ed




257

Therefore:


1

11

11

224.01120.0534.0

224.01120.0

534.0

,,

2

0,

0,0,

4

,,

10

4

,,

1

0

T cr

Ed

z cr

Ed

LT mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C

a

aC C C

N

N

N

N C

N

N

N

N C



LT y

LT y

mymymya

aC C C

11 0,0,

yeff

eff

Ed

Ed y

yW

A

N

M

,

,


,, 533.3179229 mmW W yeff yel

55.9533.3179229

2.7815

328000

1012743

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

19996.01448234429

75.51461411

4

4

mm

mm

I

I a

y

t LT



ycr

Ed my N

N C

,

0, 33.036.021.079.0






259


ycr

Ed

y

mLT my yy

N

N C C k

,

1

534.03070560199

/275533.3179229 23

0

,0

Nmm

mm N mm

M

f W

cr

y yeff

For the botom part of the column:

260.024.9646886

3280001

62.6989347

328000177.120.01120.0 44

,,

1

N

N

N

N

N

N

N

N C

TF cr

Ed

z cr

Ed




260

Therefore:

For the bottom part of the column:

1

11

1

1

260.01120.0534.0

260.01120.0

534.0

,,

2

0,

0,0,

4

,,

10

4

,,

1

0

T cr

Ed

z cr

Ed

LT mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C a

aC C C

N

N

N

N C

N

N

N

N C



LT y

LT y

mymymy a

a

C C C

11 0,0,

yeff

eff

Ed

Ed y

yW

A

N

M

,

,


,, 533.3179229 mmW W yeff yel

55.9533.3179229

2.7815

328000

1012743

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

19996.01448234429

75.51461411 4

4

mmmm

I I a y

t LT



0

,sup

,inf kNm M

M

Ed

Ed

ycr

Ed my

N N C

,

0, 33.036.021.079.0




261

Where:

kN N l

I E N

fy

y

ycr 37.9503544.95035371²

²,


79.044.95035371

328000

33.0036.079.00, N

N

C my

949.0155.91

155.979.0179.0

11

0,0,

LT y

LT y

mymymya

aC C C


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

1

11 ,,

2

T cr

Ed

z cr

Ed

LT mymLT

N

N

N

N

aC C

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

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

LT y

LT y

mymymya

aC C C

11 0,0,

895.0

44.95035371

32800033.05.036.05.021.079.033.036.021.079.0

,

0,

N

N

N

N C

ycr

Ed my

77.4533.3179229

2.7815

328000

106373

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

967.0177.41

177.4895.01895.0

11 0,0,

LT y

LT y

mymymya

aC C C

1

1

974.0

24.9646886

3280001

62.6989347

3280001

1967.0

11

2

,,

2

mLT

mLT

T cr

Ed

z cr

Ed

LT mymLT

C

C

N

N

N

N

N

N

N

N

aC C

952.0

44.95035371

3280001

11949.0

1,

N

N

N

N C C k

ycr

Ed

y

mLT my yy

Note:

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




262



z cr

Ed

y

mz yz

N

N C k

,

1

The

mz C term must be calculated for the hole column length

0

1274

0

,sup

,inf

kNm M

M

Ed

Ed

784.062.6989347

32800033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

823.0

62.6989347

3280001

1784.0

1,

N

N

N

N C k

z cr

Ed

y

mz yz


823.0

62.6989347

3280001

1784.0

1,

N

N

N

N C k

z cr

Ed

y

mz yz

Note:




ycr

Ed

z mLT my zy

N

N C C k

,

1

991.0

62.6989347

328000812.01

62.6989347

3280001

1

1

,

,

N

N N

N

N

N

N

N

z cr

Ed z

z cr

Ed

z

944.0

44.95035371

3280001

991.01949.0

1,

N

N

N

N C C k

ycr

Ed

z mLT my zy




263


944.0

44.95035371

3280001

991.01949.0

1,

N

N

N

N C C k

ycr

Ed

z mLT my zy

Note:




815.0

62.6989347

3280001

991.0784.0

1,

N

N

N

N C k

z cr

Ed

z mz zz


815.0

62.6989347

3280001

991.0784.0

1,

N

N

N

N C k

z cr

Ed

z mz zz

Note:



1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rk z

Rd z Ed z

zz

M

Rk y

LT

Rd y Ed y

zy

M

Rk z

Ed

M

Rk z

Rd z Ed z yz

M

Rk y

LT

Rd y Ed y yy

M

Rk y

Ed

M

M M k

M

M M k

N

N

M

M M k M

M M k N N

i y Rk A f N




264


56.356.171.119.0

1

/27510.242068

104.127815.0

1

/275533.3179229803.0

101274944.0

1

/2752.7815812.0

328000

46.358.173.115.0

1

/27510.242068104.127823.0

1

/275533.3179229803.0

101274952.0

1

/2752.78151

328000

23

6

23

6

22

23

6

23

6

22

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N

mm N mm Nmm

mm N mm

Nmm

mm N mm

N


56.356.171.119.0

1

/27510.242068

104.127815.0

1

/275533.3179229803.0

101274944.0

1

/2752.7815812.0

328000

46.358.173.119.0

1

/27510.242068

104.127823.0

1

/275533.3179229803.0

101274952.0

1

/2752.7815812.0

328000

23

6

23

6

22

23

6

23

6

22

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N

Note:



■ Linear element: S beam,■ 6 nodes,





265



y



z




266



yyk



yz k




267



zyk



zz k




268



SNy



SMyy




269



SMyz



SNz




270



SMzy


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

over the Z axisSMzz




271




1


0.81






0.15


1.72


1.58


0.19


1.71


1.56



Xy Coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 %

Xz Coefficient corresponding to non-dimensional slenderness 0.811841adim

0.0000 %

Kyy Internal factor, kyy 0.950358adim

0.0000 %

Kyz Internal factor, kyz 0.823048adim

-0.0000 %

Kzy Internal factor, kzy 0.941635adim

-0.0000 %

Kzz Internal factor, kzz 0.815493adim

-0.0000 %

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

0.152455adim

-0.0001 %

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

1.72357 adim 0.0000 %

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

1.57508 adim -0.0002 %

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

0.187789adim

0.0001 %

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

1.70775 adim -0.0000 %

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

1.70775 adim -0.0000 %




272

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

Test ID: 5729

Test status: Passed

8.38.1Description


The column is an “I symmetric” shape with: 260mm height; 150mm width; 7.1mm web thickness; 10.7mm flangethickness; 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 10kNmbending moment after the Z axis. All the efforts are applied to the top of the column.


8.38.2Background

An I260*7.1+150*10.7 shaped column subjected to compression and bending, made from S275 steel. The columnhas a 260x7.1mm web and 150x10.7mm flanges. The column is fixed for all translations and free for all rotations, atit’s base, and on the top end, the translations over the X and Y axes and the rotation over the Z axis are notpermitted. In the middle of the column there is a restraint over the Y axis, therefore the bucking length for the XYplane is equal to half of the column length. The column is subjected to an axial compression load -328000 N, a10000Nm bending moment after the X axis and a 50000Nm bending moment after the Y axis






■ Exploitation loadings (category A): Fz=-328000N N; My=50000Nm; Mx=10000Nm;






273

Units

Metric System




06.4904 mm A







, 63.445717 mmW yel



, 89.80344 mmW z el


, 96.123381 mmW z pl


■ Flexion inertia moment around the Z axis: 446.6025866 mm I z

■ Torsional moment of inertia: 497.149294 mm I t

■ Working inertial moment: 688.19351706542 mm I w






Boundary conditions


■ Outer:


► 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,




274

Loading


■ External: Point load From X=0.00m and z=5.62m: FZ =-328000N; Mx=10000Nm and My=50000Nm




-for beam web:


1253.006.179

30.45

sup

inf

Mpa

Mpa




275

73.190

36.224

06.1796.238

175.4836.224

3.456.238

36.224

6.238

06.17930.4506.17930.45 y

x y x y x

5.080.06.23873.190

6.238 x




276

924.0275

235235

y f

93.3818.013

924.0396

113

3966.33

924.0

61.331.7

7.102260

t

cmm

mmmm

t

c

therefore the beam

web is considered to be Class 1

-for beam flange:

316.8924.0968.6

924.0

68.67.10

2

1.7150

t

c

t

c









277

34.0



curve according to:

11

22

y y y

y

(6.49)


ycr

y

y N

f A

,

*


Flexion inertia moment around the Y axis:

4

64.57943291 mm I y

kN N

mm

mmmm N

l

I E N

fy

y

ycr 33.380295.3802327²5620

64.57943291/210000²

²

² 42

,

5956.095.3802327

/27506.4904 22

,

N

mm N mm

N

f A

ycr

y y

7446.05956.02.05956.034.015.0²)2.0(15.0 2 y y y

839.0

1

839.05956.07446.07446.0

11

2222

y

y

y y y

y




278



49.0



according to:

1122

z z z

z

(6.49)


z cr

y z

N

f A

,

*






279

kN N

mm

mmmm N

l

I E N

fz

z z cr 71.3158151.1581706

²2810

46.6025866/210000²

²

² 42

,

923.051.1581706

/27506.4904 22

,

N

mm N mm

N

f A

z cr

y z

103.1923.02.0923.049.015.0²)2.0(15.0 2 z z z

586.0

1

586.0923.0103.1103.1

112222

z

z

z z z

z




z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1


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

²252.0423.0325.0

11

C


31.15.025

251

,

, C kN

kN

M

M

top y

botom y




280



Torsional moment of inertia: 4

97.149294 mm I t







4

2

f z

w

t h I

I



6

24

093627638294

7.1026046.6025866mm

mmmmmm I w



kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mmmmmm N

I E I G L

I I

L I E C M

z

t

z

w z cr

36.31574.315363380

20.15251.158170631.146.6025866/210000

97.149294/808002810

46.6025866

09362763829

281046.6025866/21000031.1

²²

²²

422

422

4

6

2

422

1

661.074.315363380

/27518.501177 23,

Nmm

mm N mm

M

f W

cr

y y pl LT


1²²

1

LT LT LT

LT

(6.56)




281



2733.1150

260

mm

mm

b

h


49.0 LT

831.0²661.02.0661.049.015.0²2.015.0 LT LT LT LT

1749.0²661.0²831.0831.0

1

²²

1

LT LT LT

LT




z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1


-where:

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

allure

²252.0423.0325.0

11

C





282


0

kNm

77.10 1 C











4

2

f z

w

t h I I



6

24

093627638294

7.1026046.6025866mm

mmmmmm I w






283

kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

10.4266.426102243

20.15251.158170677.146.6025866/210000

97.149294/808002810

46.6025866

09362763829

2810

46.6025866/21000077.1

²

²

²

²

422

422

4

6

2

422

1

569.06.426102243

/27518.501177 23,

Nmm

mm N mm

M

f W

cr

y y pl LT


1²²

1

LT LT LT

LT

(6.56)

²2.015.0 LT LT

LT LT


2733.1150

260

mm

mm

b

h


49.0 LT

752.0²569.02.0569.049.015.0²2.015.0 LT LT LT LT

1804.0²569.0²752.0752.0

1

²²

1

LT LT LT

LT




284


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

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


yy

ycr

Ed

y

mLT my yy C

N

N C C k

1̀

1,

ycr

Ed y

ycr

Ed

y

N

N

N

N

,

,

1

1


kN N Ed 328

kN N l

I E N

fy

y

ycr 33.380295.3802327²

²,


985.0

95.3802327

328000839.01

95.3802327

3280001

1

1

,

,

N

N N

N

N

N

N

N

ycr

Ed y

ycr

Ed

y




285



0

,0

cr

y y pl

M

f W


318.501177 mmW y


kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

73.2408.240735730

20.15251.1581706146.6025866/210000

97.149294/808002810

46.6025866

88.19351706542

2810

46.6025866/2100001

²

²

²

²

422

422

4

6

2

422

1

757.08.240735730

/27518.501177 23,

0

Nmm

mm N mm

M

f W

cr

y y pl




286


,,

1 1120.0

TF cr

Ed

z cr

Ed

N

N

N

N C term:

Where:


²

1

,

2

0

,

T cr

wt T cr

L

I E I G

I N


4442

0 1.6396915846.602586664.57943291 mmmmmm I I z A I I I z y g z y


Working inertial moment: 6

88.19351706542 mm I w


m L T cr 81.2,

k N

mm

mmmm N mmmm N

mm

mm N T cr

63.280668.2806625

²2810

88.19351706542/21000097.149294/80800

1.63969158

06.4904 62242

4

2

,

N N Ed 328000

N N N T cr TF cr 68.2806625,,

kN N l

I E N

fz

z z cr 71.3158151.1581706

²

²,




183.0

68.2806625

3280001

51.1581706

3280001120.01120.0 44

,,

1

N

N

N

N

N

N

N

N C

TF cr

Ed

z cr

Ed




287

Therefore:


1

11

11

183.01120.0757.0

183.01120.0

757.0

,,

2

0,

0,0,

4

,,

10

4

,,

1

0

T cr

Ed

z cr

Ed

LT mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C

a

aC C C

N

N

N

N C

N

N

N

N C



LT y

LT ymymymy

aaC C C

1

1 0,0,

yeff

eff

Ed

Ed y

yW

A

N

M

,

,


,, 63.445717 mmW W yeff yel

677.163.445717

06.4904

328000

10503

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

1997.064.57943291

97.14929411

4

4

mm

mm

I

I a

y

t LT





288

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

ycr

Ed

ycr

Ed my

N

N

N

N C

,,

0, 33.036.079.033.036.021.079.0

Where:

kN N l

I E N

fy

y

ycr 33.380295.3802327²

²,


N N Ed 328000

780.095.3802327

32800033.036.079.00,

N

N C my

904.01677.11

1677.1780.01780.0

11 0,0,

LT y

LT y

mymymya

aC C C



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

1

11,,

2

T cr

Ed

z cr

Ed

LT mymLT

N

N

N

N

aC C

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

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

LT y

LT y

mymymya

aC C C

11 0,0,

900.095.3802327

23800033.05.036.05.021.079.033.036.021.079.0

,

0, N

N

N

N C

ycr

Ed my

677.1

63.445717

06.4904

328000

10503

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y


956.01677.11

1677.1900.01900.0

11 0,0,

LT y

LT y

mymymya

aC C C




289

089.1

1

089.1

68.2806625

3280001

51.1581706

3280001

997.0

956.0

11

2

,,

2

mLT

mLT

T cr

Ed

z cr

Ed

LT mymLT

C

C

N

N

N

N

N

N

N

N

aC C

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

y pl

yel

LT pl my

y

my

y

y yyW

W bnC

wC

wwC

,

,2

max2

max2 6.16.1

2)1(1

Rd z pl

Ed z

Rd y pl LT

Ed y

LT LT M M

M M ab

,,

,

,,

,2

05.0

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

and LT :

757.0661.031.110

LT C (for the top part of the column)




290

041.0

/27596.123381

10000000

/27518.501177749.0

50000000757.0997.05.0

5.05.0

2323

2

,

,

,

,2

0

,,

,

,,

,2

0

mm N mm

Nmm

mm N mm

Nmm

f W

M

f W

M a

M

M

M

M ab

y z pl

Ed z

y y pl LT

Ed y

LT

Rd z pl

Ed z

Rd y pl LT

Ed y

LT LT

5.1124.163.445717

18.5011773

3

,

, mm

mm

W

W w

yel

y pl

y

243.0

1

/27506.4904

32800022

1

mm N mm

N

N

N n

M

Rk

Ed pl

923.0923.0;5956.0max;maxmax

z y

993.0041.0243.0²923.0²904.0124.1

6.1923.0²904.0

124.1

6.12)1124.1(1

yyC

993.0889.018.501177

63.445717

993.0

,

,

3

3

,

,

yy

y pl

yel

yy

y pl

yel

yy

C

W

W C

mm

mm

W

W

C

kN N l

I E N

fy

y

ycr 33.380295.3802327²

²,



069.1993.0

1

95.3802327

3280001

985.0089.1904.0

1̀

1,

N

N C

N

N C C k

yy

ycr

Ed

y

mLT my yy


yy

ycr

Ed

y

mLT my yyC

N

N C C k

1̀

1,




291

-the terms: mLT C ; LT b and yyC must be recalculated:

1

11

,,

2

T cr

Ed

z cr

Ed

LT mymLT

N

N

N

N

aC C

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

780.095.3802327

23800033.036.079.033.036.021.079.0

,

0, N

N

N

N C

ycr

Ed my

839.063.445717

06.4904

328000

10253

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

885.01839.01

1839.0780.01780.0

11 0,0,

LT y

LT y

mymymya

aC C C

1

1

933.0

68.2806625

3280001

51.1581706

3280001

997.0885.0

11

2

,,

2

mLT

mLT

T cr

Ed

z cr

Ed

LT mymLT

C

C

N

N

N

N

N

N

N

N

aC C




292

y pl

yel

LT pl my

y

my

y

y yyW

W bnC

wC

wwC

,

,2

max2

max2 6.16.1

2)1(1

Rd z pl

Ed z

Rd y pl LT

Ed y

LT LT

M

M

M

M ab

,,

,

,,

,2

05.0

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

and LT :

757.08.240735730

/27518.501177 23,

0

Nmm

mm N mm

M

f W

cr

y y pl (for the bottom part of the column)

1804.0²569.0²752.0752.0

1

²²

1

LT LT LT

LT

(for the bottom part of the column)

0095.0

/27596.123381

5000000

/27518.501177804.0

25000000757.0997.05.0

5.05.0

2323

2

,

,

,

,

2

0

,,

,

,,

,

2

0

mm N mm

Nmm

mm N mm

Nmm

f W M

f W

M a

M M

M

M ab

y z pl

Ed z

y y pl LT

Ed y LT

Rd z pl

Ed z

Rd y pl LT

Ed y LT LT

5.1124.163.445717

18.5011773

3

,

, mm

mm

W

W w

yel

y pl

y

243.0

1

/27506.4904328000 22

1

mm N mm N

N N n

M

Rk

Ed pl

923.0923.0;5956.0max;maxmax

z y

997.00095.0243.0²923.0²904.0124.1

6.1923.0²904.0

124.1

6.12)1124.1(1

yyC

997.0889.018.501177

63.445717

997.0

,

,

3

3

,

,

yy

y pl

yel

yy

y pl

yel

yy

C

W

W C

mm

mm

W

W

C

kN N l

I E N

fy

y

ycr 33.380295.3802327²

²,





293

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

977.0997.0

1

95.3802327

3280001

985.01904.0

1̀

1,

N

N C

N

N C C k

yy

ycr

Ed

y

mLT my yy

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


y

z

yz

z cr

Ed

y

mz yz w

w

C

N

N C k

6.0

1

1,

-the mz C ter will be considered for the entire column length (with )0 :

765.051.1581706

32800033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

985.0

95.3802327

328000839.01

95.3802327

3280001

1

1

,

,

N

N N

N

N

N

N

N

ycr

Ed y

ycr

Ed

y


N l

I E N

fz

z z cr 51.1581706

²

²,


LT pl

z

mz z yz cn

w

C wC

5

2

max2

142)1(1

Rd y pl lt my

Ed y

z

LT LT M C

M ac

,,

,

4

2

0

5

10

997.064.57943291

97.14929411

4

4

mm

mm

I

I a

y

t LT (previously calculated)

757.0

8.240735730

/27518.501177 23,

0

Nmm

mm N mm

M

f W

cr

y y pl (previously calculated)

923.051.1581706

/27506.4904 22

,

N

mm N mm

N

f A

z cr

y z (previously calculated)

Nm M Ed y 50000,

5.1

5.1

5.1536.189.80344

96.1233813

3

,

,

z

z

z el

z pl

z w

w

mm

mm

W

W w

5.1124.163.44571718.501177

3

3

,

, mmmm

W W w

yel

y pl y




294

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


956.0

1677.11

1677.1900.01900.0

1

1 0,0,

LT y

LT y

mymymy

a

aC C C


1749.0²661.0²831.0831.0

1

²²

1

LT LT LT

LT



,,

506.05.137823724749.0956.0

50000000

923.05

²757.0

997.010

5

10

4

,,

,

4

2

0

Nmm

Nmm

M C

M ac

Rd y pl lt my

Ed y

z

LT LT

N l

I E N

fz

z z cr 51.1581706

²

²,

(previously calculated

765.051.1581706

32800033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

923.0923.0;5956.0max;maxmax

z y

243.0

1

M

Rk

Ed

pl N

N

n


-Therfor:

878.0506.0243.05.1

923.0765.014215.11

142)1(1

5

22

5

2

max2

LT pl

z

mz z yz cn

w

C wC

750.0124.1

5.16.0

878.0

1

51.1581706

3280001

985.0765.06.0

1

1,

N

N w

w

C

N

N C k

y

z

yz

z cr

Ed

y

mz yz


885.01839.01

1839.0780.01780.0

11 0,0,

LT y

LT y

mymymya

aC C C


1804.0²569.0²752.0752.0

1

²²

1

LT

LT LT

LT





295


,,

254.05.137823724804.0885.0

25000000

923.05

²757.0997.010

5

10

4

,,

,

4

2

0

Nmm

Nmm

M C

M ac

Rd y pl lt my

Ed y

z

LT LT

N l

I E N

fz

z z cr 51.1581706

²

²,

(previously calculated

765.051.1581706

32800033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

923.0923.0;5956.0max;maxmax

z y

243.0

1

M

Rk

Ed pl N

N n


-Therfor:

0043.1254.0243.0

5.1

923.0765.014215.11

142)1(1

5

22

5

2

max2

LT pl

z

mz z yz cn

w

C wC

656.0124.1

5.16.0

0043.1

1

51.1581706

3280001

985.0765.06.0

1

1,

N

N w

w

C

N

N C k

y

z

yz

z cr

Ed

y

mz yz




y

z

zy

ycr

Ed

z mLT my zy

ww

C

N

N C C k

6.011

,

902.0

51.1581706

328000586.01

51.1581706

3280001

1

1

,

,

N

N N

N

N

N

N

N

z cr

Ed z

z cr

Ed

z





296

y pl

yel

z

y

LT pl

y

my

y zyW

W

w

wd n

w

C wC

,

,

5

2

max2

6.0142)1(1

Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177

23

,,


,,

-in order to calculate the LT d term, the terms myC and mz C must be recalculated for each column part;

-the term mz C must be recalculated for the top column part only, using 5.0 :

908.0

51.1581706

32800033.05.036.05.021.079.0

33.05.036.05.021.079.0,

0,

N

N

N

N C C

z cr

Ed mz mz

301.0

33930039908.0

10000000

5.137823724749.0956.0

50000000

923.01.0

757.0997.02

1.0

2

4

,,

,

,,

,

4

0

Nmm

Nmm

Nmm

Nmm

M C

M

M C

M ad

Rd z pl mz

Ed z

Rd y pl LT my

Ed y

z

LT LT

859.0

6.0

616.018.501177

63.445717

124.1

5.16.06.0

859.0301.0243.0124.1

847.1904.0142)1124.1(1

142)1(1

,

,

3

3

,

,

5

22

5

2

max2

zy

y pl

yel

z

y

zy

y pl

yel

z

y

LT pl y

my

y zy

C

W

W

w

wC

mm

mm

W

W

w

w

d nw

C

wC

-fot the calculation of zyC term , it will be used the myC for the entire column and the LT d for the top column part:

588.05.1

124.16.0

859.0

1

95.3802327

3280001

902.0089.1904.06.0

1

1,

N

N w

w

C

N

N C C k

z

y

zy

ycr

Ed

z mLT my zy




297


y

z

zy

ycr

Ed

z mLT my zy

w

w

C

N

N C C k

6.0

1

1,

902.0

51.1581706

328000586.01

51.1581706

3280001

1

1

,

,

N

N N

N

N

N

N N

z cr

Ed z

z cr

Ed

z


y pl

yel

z

y

LT pl

y

my

y zyW

W

w

wd n

w

C wC

,

,

5

2

max2

6.0142)1(1


,,


,,

-in order to calculate the LT d term, the terms myC and mz C must be recalculated for each column part;

-the term mz C must be recalculated for the top column part only, using 0 :

765.051.1581706

32800033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

090.0

33930039765.0

5000000

5.137823724804.0885.0

25000000

923.01.0

757.0997.02

1.0

2

4

,,

,

,,

,

4

0

Nmm

Nmm

Nmm

Nmm

M C

M

M C

M ad

Rd z pl mz

Ed z

Rd y pl LT my

Ed y

z

LT LT

923.0923.0;5956.0max;maxmax

z y

892.0

6.0

462.018.501177

63.445717

5.1

124.16.06.0

892.0090.0243.0124.1

923.0885.0142)1124.1(1

142)1(1

,

,

3

3

,

,

5

22

5

2

max2

zy

y pl

yel

z

y

zy

y pl

yel

z

y

LT pl

y

my

y zy

C

W

W

w

wC

mm

mm

W

W

w

w

d nw

C wC




298

-fot the calculation of

zyC term , it will be used the myC for the entire column and the LT d for the top column part:

566.05.1

124.16.0

892.0

1

95.3802327

328000

1

902.0089.1904.06.0

1

1,

N

N w

w

C

N

N C C k

z

y

zy

ycr

Ed

z mLT my zy




zz

z cr

Ed

z mz zz

C

N

N C k

1

1,

z pl

z el

LT pl mz

z

mz

z

z zz W

W enC

wC

wwC

,

,max

2max

2 6.16.12)1(1


,,

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

787.0/27518.501177749.0956.0

50000000923.01.0

757.0997.07.1

1.0

7.1

234

,,

,

4

0

mm N mm Nmm

M C

M ae

Rd y pl lt my

Ed y

z

LT LT

-fot the calculation of

zz C term , it will be used the mz C for the entire column and the LT e for the top column part:

013.1

651.0

96.123381

89.80344

013.1243.0787.0847.1765.05.1

6.1847.1765.0

5.1

6.12)15.1(1

6.16.12)1(1

,

,

3

3

,

,

222

2

max2

max2

zz

z pl

z el

zz

z pl

z el

pl LT mz

z

mz

z

z zz

C

W

W C

mm

mm

W

W

neC w

C w

wC

860.0013.1

1

51.1581706

3280001

902.0765.0

1

1,

N

N C

N

N C k

zz

z cr

Ed

z mz zz




299


zz

z cr

Ed

z mz zz

C

N

N C k

1

1,

z pl

z el LT pl mz

z

mz

z

z zz W W enC

wC

wwC

,

,max

2max

2 6.16.12)1(1


,,

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

396.0/27518.501177804.0885.0

25000000

923.01.0

757.0

997.07.1

1.0

7.1

234

,,

,

4

0

mmmm

Nmm

M C

M ae

Rd y pl lt my

Ed y

z

LT LT

-for the calculation of zz C term , it will be used the mz C for the entire column and the LT e for the top column part:

923.0923.0;5956.0max;maxmax

z y

060.1651.0

96.123381

89.80344

060.1243.0396.0923.0765.05.1

6.1923.0765.0

5.1

6.12)15.1(1

6.16.12)1(1

,

,

3

3

,

,

222

2

max2

max2

zz

z pl

z el

zz

z pl

z el

pl LT mz

z

mz

z

z zz

C

W

W C

mm

mm

W

W

neC w

C w

wC

821.0060.1

1

51.1581706

3280001

902.0765.0

1

1,

N

N C

N

N C k

zz

z cr

Ed

z mz zz





300


1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rk z

Rd z Ed z

zz

M

Rk y

LT

Rd y Ed y

zy

M

Rk z

Ed

M

Rk z

Rd z Ed z

yz

M

Rk y

LT

Rd y Ed y

yy

M

Rk

y

Ed

M

M M k

M

M M k

N

N

M

M M k

M

M M k

N

N

i y Rk A f N


93.025.027.041.0

1

/27596.123381

1010860.0

1

/27518.501177804.0

1050588.0

1

/27506.4904586.0

328000

02.121.052.029.0

1

/27596.123381

1010750.0

1

/27518.501177749.0

1050069.1

1

/27506.4904839.0

328000

23

6

23

6

22

23

6

23

6

22

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N




301


67.012.013.042.0

1

/27596.123381

105821.0

1

/27518.501177804.0

1025566.0

1

/27506.4904586.0

328000

30.197.022.029.0

1

/27596.123381

105656.0

1

/27518.501177804.0

1025977.0

1

/27506.4904839.0

328000

23

6

23

6

22

23

6

23

6

22

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N



■ 7 nodes,





302



y



z






304



zyk



zz k




305



SNy



SMyy




306



SMyz


Bending and axial compression verification term depending of the compression effort

over the Z axisSNz




307



SMzy



SMzz




308

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

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

C1



Mcr




309



LT




310





0.839

z z coefficient corresponding to non-dimensional slenderness z 0.586




zz k Internal factor, zz k 0.860

SNy Bending and axial compression verification term depending of the compression

effort over the Y axis

0.29

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

0.52

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

0.21

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

0.41

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

0.27

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

0.25

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

conditions; moment diagram allure

1.77






311




0.839





zz k Internal factor, zz k 0.821


0.29


0.22

SMyz Bending and axial compression verification term depending of the Z

bendin moment over the Y axis

0.97


0.42


0.13


0.12

C1 Coefficient that depends of several parameters as: section properties;su ort conditions moment dia ram allure

1.77







312



Xy Coefficient corresponding to non-dimensionalslenderness by Y axis

0.839285 adim 0.0000 %

Xz Coefficient corresponding to non-dimensionalslenderness by the Z axis

0.585533 adim 0.0000 %

Kyy Internal coefficient kyy 1.07027 adim -0.0001 %

Kyy Internal coefficient kyy 1.04152 adim -0.0001 %

Kyz Internal coefficient kyz 0.750217 adim -0.0000 %

Kyz Internal coefficient kyz 0.685239 adim -0.0000 %

Kzy Internal coefficient kzy 0.593445 adim 0.0000 %

Kzy Internal coefficient kzy 0.561245 adim 0.0001 %

Kzz Internal coefficient kzz 0.860237 adim 0.0000 %

Kzz Internal coefficient kzz 0.834265 adim -0.0000 %




313

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

Test ID: 5730

Test status: Passed

8.39.1Description

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/mlinear uniform load applied horizontal.


8.40 EC3 Test 31: Verifying IPE450 column fixed on base subjected to axial compression andbending moment, both applied on top

Test ID: 5731

Test status: Passed

8.40.1Description

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.





314

8.41 EC3 Test 15: Verifying a rectangular hollow section column subjected to bending and axialefforts

Test ID: 5735

Test status: Passed

8.41.1Description

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


8.41.2Background

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 isfixed at its base and it is subjected to a uniformly distributed load over its height and a punctual axial load applied onthe top. The dead load will be neglected.







► Fz = - 500 000 N,

► Fx = 5 000 N/ml,






315

Units

Metric System

Geometry


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

■ Thickness: t = 10 mm,

■ Outer radius: r = 15 mm,

■ Column height: L = 5000 mm,

■ Section area:A = 9490 mm2,

■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,






Boundary conditions


■ Outer:



■ Inner: None.

Loading


■ External:

► Point load at z = 5.0: Fz = - 500 000 N,

► Uniformly distributed load: q = Fx = 5 000 N/ml

■ Internal: None.




316

8.41.2.2 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 uniformcompression (Nc,Rd) and also the design plastic moment resistance (Mpl,Rd) have to be compared with the designvalues of the corresponding efforts.

The design resistance for uniform compression is verified considering the relationship (6.9) from chapter 6.2.4 (EN1993-1-1), while the design plastic moment resistance is verified considering the criterion (6.12) from chapter 6.2.5

(EN 1993-1-1).

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

Cross section class


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

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




317

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

1510

10215220022

mm

mmmmmm

t

t r b

t

c

0.1235

y f

Therefore:

333315 t

c

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 geometricalproperties 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 (). is determinedconsidering the stresses distribution on the web.

5.0832.063.26

132

1

MPa

MPa

2510

10215230022

mm

mmmmmm

t

t r h

t

c

0.1235

y f

Therefore:

34.40113

39625

t

c

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.




318

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 EN1993-1-1:2001.

N MPamm f A

N M

y

Rd c 22301500.1

2359490 2

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 = %42.221002230150

500000100100

,,

N

N

N

F

N

N

Rd c

z

Rd c

Ed

Verifying the design plastic moment resistance

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

Nmm

MPamm f W

M M M

y y pl

Rd pl Rd c 2246600000.1

235956000 3

0

,

,,

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

Work ratio = %82.27100224660000

2

50005000/5

1002100,,

Nmm

mmmmmm N

M

L Lq

M

M

Rd c Rd c

Ed



■ 6 nodes,





319



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




320



Work ratio – Fx Compression resistance work ratio [%] 22.42 %

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



Work ratio - Fx Compression resistance work ratio 22.42 % 0.0001 %

Work ratio -Oblique

Work ratio of the design resistance for bending 27.8198 % -0.0007 %




321

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

Test ID: 5734

Test status: Passed

8.42.1Description

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 thebeam and 1.5 kN/m linear uniform horizontal load.


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

Test ID: 5733

Test status: Passed

8.43.1Description

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 thebeam and 5kN/m linear uniform horizontal load.


8.44 EC3 Test 24: Verifying an user defined I section class 4 column fixed on the bottom and

without any other restraintTest ID: 5709

Test status: Passed

8.44.1Description

The test verifies an user defined cross section column.


The column is subjected to 328 kN axial compression force; 1274 kNm bending moment after the Y axis and 127.4kNm bending moment after the Z axis. All the efforts are applied on the top of the column. The column height is5.62m and has no restraints over its length.





322

8.44.2Background

An I880*5+220*15 shaped column subjected to compression and bending, made from S275 steel. The column has a880x5mm web and 220x15mm flanges. The column is hinged at its base and, at his top end, translation is permittedonly on vertical direction and the rotation is blocked for the long axis of the column. The column is subjected to anaxial compression load 328000 N, 127400Nm bending moment after the X axis and 1274000Nm bending momentafter the Y axis.






■ Exploitation loadings (category A): Fz=-328000N N, Mx= 127400Nm; My=1274000Nm■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q


Units

Metric System




323




10850mm A







, 1066.3387 mmW yel



, 1008.242 mmW z el



4


4


4

■ Working inertial moment: Iw=4979437.37x106mm

6


S275 steel material is used. The following characteristics are used:■ Yield strength f y = 275 MPa,



Boundary conditions


■ Outer:






324

Loading


■ External: Point load From X=0.00m and z=5.62m: FZ =--328000N; Mx=127400Nm and My=1274000Nm




-for beam web:


1873.030.406

84.354

sup

inf

Mpa

Mpa




325

27.396

14.761

84.354850

73.45314.761

3.406850

14.761

850

84.3543.40684.3543.406 y

x y x y x

5.05338.0850

73.453

850

x




326

924.0275

235235

y f

61.101)873.0(33.067.0

924.042

33.067.0

42170

1

1705

152880

t

cmm

mmmm

t

c


-for beam flange:

9961.7

1

61.715

5.107

t

ct

c



8.44.2.3 Effective cross-sections of Class4 cross-sections

-the section is composed from Class 4 web and Class 1 flanges, therefore will start the webcalculation:

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

1705

152880

mm

mmmm

t

c:

41 k


k t

bw

p

4.28



9244.0275

235235

y f

261.349244.04.28

5

850

mm

mm

p


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

286.0

261.3

4055.0261.33055.0

043

673.0261.322

p

p p




327

mmmmb

mmmmb

mmmmb

bb

bb

bb

e

e

eff

eff e

eff e

weff

55.1211.2435.0

55.1211.2435.0

1.243850286.0

5.0

5.0

2

1

2

1

2

21, 5.121555.121555.1215 mmmmmmmmmmbt bt A ewewwebeff

2

, 330022015 mmmmmmbt A f f flangeeff

222

,, 2.7815330025.12152 mmmmmm A A A flangeeff webeff eff

8.44.2.4 Effective elastic section modulus of Class4 cross-sections

-In order to simplify the calculation the section will be considered in pure bending

mmmm

bb t c 4252

8501

sup

inf

9.231 k


1705

152880

mm

mmmm

t

c

k

t

bw

p

4.28



9244.0275

235235

y f

325.19.239244.04.28

5

850

mm

mm

p


692.0325.1

2055.0325.13055.0

023

673.0325.122

p

p p

mmmmb

mmmmb

mmmm

b

bb

bb

bbb

e

e

eff

eff e

eff e

wceff

46.1761.2946.0

64.1171.2944.0

1.29411

850692.0

6.0

4.0

1

2

1

2

1




328

-the weight center coordinate is:

mm

yG

53.155.10195

095.158330

15220546.601564.11715220

27.124546.6015.4321522018.366564.1175.43215220

-the inertial moment along the strong axis is:

4

23

23

23

23

14482344295.6993800726.748854356

97.4161522012

2201574.107546.601

12

546.601

71.381564.11712

564.11703.44815220

12

22015

mm

I y

423

23

23

23

63.2662749001522012

152200546.601

12

46.6015

0564.11712

64.1175015220

12

15220

mm

I z

34

max,

533.317922953.455

1448234429mm

mm

mm

z

I W

y

yel

34

max

, 10.242068110

63.26627490mm

mm

mm

y

I W z

z el




329




34.0



curve according to:

11

22

y y y

y

(6.49)


ycr

yeff

y N

f A

,

*




330



properties:

kN N

mm

mmmm N

l

I E N

fy

y

ycr 37.9503544.95035371

²5620

1448234429/210000²

²

² 42

,

15.044.95035371

/2752.7815 22

,

N

mm N mm

N

f A

ycr

yeff y

503.015.02.015.034.015.0²)2.0(15.0 2 y y y

1

1

017.115.0503.0503.0

11

2222

y

y

y y y

y



49.0




331



according to:

11

22

z z z

z

(6.49)


z cr

yeff z

N

f A

,

*



properties:

kN N mm

mmmm N l

I E N fz

z z cr 34.1747905.1747336

²562063.26627490/210000²

²²

42

,

109.1905.1747336

/2752.7815 22

,

N

mm N mm

N

f A

z cr

yeff z

338.1109.12.0109.149.015.0²)2.0(15.0 2 z z z

479.0

1

479.0109.1338.1338.1

11

2222

z

z

z z z

z



z

t

z

w z cr

I E

I G L

I

I

L

I E C M

²

²

²

²1


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

²252.0423.0325.0

11

C



0

kNm

kNm




332

77.101274

01 C

kNm

kNm


Flexion inertia moment around the Y axis:41448234429mm I y



Torsional moment of inertia: It=514614.75mm4



4

2

f z

w

t h I I


f t flange thickness; mmt f 15

611

24

10808.494

15880mm326627490.6mm

mmmm I w


Length of the column part: L=5620mm


kNm Nmm

mm N mmmm N

mmmm N mm

mm

mm

mm

mmmm N

I E

I G L

I

I

L

I E C M

z

t

z

w z cr

163.14201420163158

185.459905.174733677.163.26627490/210000

514614/808005620

63.26627490

10808.49

5620

63.26627490/21000077.1

²

²

²

²

422

422

4

611

2

422

1


4

max

, 533.317922953.455

1448234429mm

mm

mm

z

I W

y

yel

785.01420163158

/275533.3179229 23,

Nmm

mm N mm

M

f W

cr

y yeff LT


1²²

1

LT LT LT

LT

(6.56)





333


24220

880

mm

mm

b

h


76.0

030.1²785.02.0785.076.015.0²2.015.0 LT LT LT LT

1589.0²785.0²030.1030.1

1

²²

1

LT LT LT

LT


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




334

ycr

Ed

y

mLT my yy

N

N C C k

,

1

ycr

Ed y

ycr

Ed

y

N

N N

N

,

,

1

1

1 y (previously calculated)

kN N Ed 328

kN N l

I E N

fy

y

ycr 37.9503544.95035371²

²,


1

44.95035371

32800011

44.950353713280001

1

1

,

,

N

N N

N

N

N

N

N

ycr

Ed y

ycr

Ed

y





335


0

,0

cr

y yeff

M

f W


33

, 1066.3387 mmW yel

kNm Nmm M cr 163.14201420163158 (previously calculated)

810.01420163158

/2753387660 23

0

,0

Nmm

mm N mm

M

f W

cr

y yeff


,,

1 1120.0

TF cr

Ed

z cr

Ed

N

N

N

N C term:

Where:


²,

2

0

,

T cr

wt T cr

L

I E I G

I

A N


2

0 g z y z A I I I

Flexion inertia moment around the Y axis:467.1490580416 mm I

y




4442

0 151720927017.2662885467.1490580416 mmmmmm I I z A I I I z y g z y


m L T cr 62.5,


Working inertial moment:61110808.49 mm I w (previously calculated)



mm

mmmm N mmmm N

mm

mm N T cr

14.2634739

²5620

10808.49/21000075.514614/80800

1517209270

10850 6112242

4

2

,

N N Ed 328000

N N N T cr TF cr 14.2634739,,




336

N l

I E N

fz

z z cr 905.1747336

²

²,


244.0

14.2634739

3280001

905.1747336

328000177.120.01120.0 44

,,

1

N

N

N

N

N

N

N

N C

TF cr

Ed

z cr

Ed

Therefore:

1

11

11

244.01120.0810.0

244.01120.0

810.0

,,

2

0,

0,0,

4

,,

10

4

,,

1

0

T cr

Ed

z cr

Ed

LT

mymLT

mz mz

LT y

LT y

mymymy

TF cr

Ed

z cr

Ed

TF cr

Ed

z cr

Ed

N

N

N

N

aC C

C C

a

aC C C

N

N

N

N C

N

N

N

N C



LT y

LT y

mymymya

aC C C

11 0,0,

yeff

eff

Ed

Ed y

yW

A

N

M

,

,


,, 533.3179229 mmW W yeff yel

55.9533.3179229

2.7815

328000

1012743

26

,

,

mm

mm

N

Nmm

W

A

N

M

yeff

eff

Ed

Ed y

y

19996.01448234429

75.51461411

4

4

mm

mm

I

I a

y

t LT




337



ycr

Ed my

N

N C

,

0, 33.036.021.079.0

Where:

kN N l

I E N

fy

y

ycr 37.9503544.95035371²

²,


79.044.95035371

32800033.0036.079.00,

N

N C my

949.0155.91

155.979.0179.0

11 0,0,

LT y

LT y

mymymya

a

C C C

1

11,,

2

T cr

Ed

z cr

Ed

LT mymLT

N

N

N

N

aC C

068.1

14.2634739

3280001

905.1747336

3280001

1949.0 2

N

N

N

N C mLT

0161.1

44.95035371

3280001

1068.1949.0

1,

N

N

N

N C C k

ycr

Ed

y

mLT my yy




338


z cr

Ed

y

mz yz

N

N C k

,1

7677.0905.1747336

32800033.036.079.033.036.079.0

,

0, N

N

N

N C C

z cr

Ed mz mz

945.0

905.1747336

3280001

17677.0

1,

N

N

N

N C k

z cr

Ed

y

mz yz


ycr

Ed

z mLT my zy

N

N C C k

,

1

893.0

905.1747336

328000479.01

905.1747336

3280001

1

1

,

,

N

N N

N

N

N

N

N

z cr

Ed z

z cr

Ed

z

908.0

44.950353713280001

893.0068.1949.0

1,

N N

N N

C C k

ycr

Ed

z mLT my zy


844.0

905.1747336

3280001

893.07677.0

1,

N

N

N

N C k

z cr

Ed

z mz zz


1

,

,,

1

,

,,

1

1

,

,,

1

,

,,

1

M

Rk z

Rd z Ed z

zz

M

Rk y

LT

Rd y Ed y

zy

M

Rk z

Ed

M

Rk z

Rd z Ed z

yz

M

Rk y

LT

Rd y Ed y

yy

M

Rk y

Ed

M

M M k

M

M M k

N

N

M

M M k

M

M M k

N

N

i y Rk A f N




339

14.461.124.232.0

1

/27510.242068

104.127844.0

1

/275533.3179229589.0

101274908.0

1

/2752.7815479.0

328000

47.4808.151.215.0

1

/27510.242068

104.127945.0

1

/275533.3179229589.0

1012740161.1

1

/2752.78151

328000

23

6

23

6

22

23

6

23

6

22

mm N mm

Nmm

mm N mm Nmm

mm N mm N

mm N mm

Nmm

mm N mm

Nmm

mm N mm

N



■ 6 nodes,





340



y



z




341



yyk



yz k




342



zyk



zz k




343



SNy



SMyy




344



SMyz



SNz




345



SMzy



SMzz




346




1







0.15


2.50


1.81


2.23


1.61



Xy Coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 %

Xz Coefficient corresponding to non-dimensional slenderness 0.479165 adim 0.0000 %Kyy Internal factor kyy 1.01066 adim 0.0005 %

Kyz Internal factor kyz 0.9451 adim 0.0000 %

Kzy Internal factor kzy 0.902094 adim -0.0000 %

Kzz Internal factor kzz 0.843573 adim -0.0000 %

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

0.152455 adim -0.0001 %

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

2.49874 adim 0.0001 %

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

1.80865 adim 0.0001 %

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

2.23031 adim 0.0002 %

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

1.61436 adim -0.0002 %




347

8.45 EC3 Test 14: Verifying the bending resistance of a rectangular hollow section column made ofS235 steel

Test ID: 5728

Test status: Passed

8.45.1Description

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.

8.45.2Background

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 isRC3020100 and can be found in the Advance Design OTUA library. The column is fixed at its base and it issubjected to a punctual horizontal load applied to the middle height (50 000 N). The dead load will be neglected.







Fx = 50 000 N,






348

Units

Metric System

Geometry


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



■ Column height: L = 5000 mm,







Boundary conditions


■ Outer:



■ Inner: None.

Loading


■ External:

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

■ Internal: None.




349

8.45.2.2 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


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





350

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

1510

10215220022

mm

mmmmmm

t

t r b

t

c

0.1235

y f

Therefore:

333315 t

c


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

2510

10215230022

mm

mmmmmm

t

t r h

t

c

0.1235 y f

Therefore:

727225 t

c






351

Design resistance for bending

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

Nmm MPamm f W

M M

y y pl

Rd c 224660000

0.1

235956000 3

0

,

,

Work ratio


Work ratio = %64.55100224660000

2

500050000

1002100,,

Nmm

mm N

M

LV

M

M

Rd c Rd c

Ed


■ Linear element: S beam,■ 6 nodes,





352



Column subjected to a punctual horizontal load applied to the middle height




Work ratio - Oblique Work ratio of the design resistance for bending [%] 55.64 %



Work ratio -Oblique





353

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

Test ID: 5732

Test status: Passed

8.46.1Description

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 linearuniform horizontal load.







355

Geometry



■ Width: b = 200 mm,






■ Plastic section modulus about z-z axis: Wpl,z = 721000 mm3,






Boundary conditions


■ Outer:

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

► Support at end point (z = 5.00) restrained in translation along Y and Z axis, and restrained in rotationabout 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.




356

8.47.2.2 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


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





357

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

2510

10215230022

mm

mmmmmm

t

t r b

t

c

0.1235

y f

Therefore:

727225 t

c

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


Determining the design plastic moment resistance

Before verifying for bi-axial bending a rectangular structural hollow section of uniform thickness, the design plasticmoment resistance reduced due to the axial force (MN,Rd) needs to be determined. Its determination has to be madeabout 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:

Rd pl

Ed

N

N n

,

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

■ Determination of aw for hollow section:




358

5.05.02

ww a A

t b Aa

■ Determination of af for hollow section:

3678.05.02

f f a

A

t h Aa

■ Determination of aw for hollow section:

5.05.02

ww a A

t b Aa

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

Rd y pl

w

Rd y pl

Rd y N M a

n M M ,,

,,

,,5.01

1

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

Nmm MPamm f W

M M M

y y pl

Rd y pl Rd y N

43

0

,

,,,, 10224660.1

235956000

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

Rd z pl

f

Rd z pl

Rd z N M a

n M M ,,

,,

,,5.01

1

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

Nmm MPamm f W

M M M

y z pl

Rd z pl Rd z N

43

0

,

,,,, 105.169430.1

235721000

Verifying for bi-axial bending

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

0.1,,

,

,,

,

Rd z N

Ed z

Rd y N

Ed y

M

M

M

M

■ Determination of constants and for rectangular hollow section:

66.1613.11

66.12

n

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

Nmmmmmm N Lq

M Ed y

4222

, 1031258

5000/10

8

1

Nmmmmmm N Lq

M Ed z

4222

, 1031258

5000/10

8

2

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




359

0.1098.00604.00378.0105.16943

103125

1022466

10312566.1

4

466.1

4

4

Nmm

Nmm

Nmm

Nmm

Work ratio =

%8.91000.1

098.0

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:

0.1139.01022466

1031254

4

,,

,

Nmm

Nmm

M

M

Rd y pl

Ed y

Work ratio =

%9.131000.1

139.0

Verifying for simple bending about z-z axis

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

0.11844.0105.16943

1031254

4

,,

,

Nmm

Nmm

M

M

Rd z pl

Ed z

Work ratio =

%44.181000.1

1844.0

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



■ 6 nodes,





360



Beam subjected to bending efforts




Work ratio - Oblique Work ratio of the design resistance for biaxial bending 18.44 %



Work ratio -Oblique

Work ratio of the design resistance for biaxial bending 18.4437 % 0.2372 %




361

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

Test ID: 5738

Test status: Passed

8.48.1Description

The test verifies an RHS300x150x9H beam made of S275 steel.



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

Test ID: 5739

Test status: Passed

8.49.1Description

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 kNmbending moment after the Y axis. All loads are applied on the top of the column.


8.50 EC3 Test 38: Verifying RHS350x150x5H class 4 column, loaded with centric compression,

punctual horizontal force by Y and a bending moment, all applied to the topTest ID: 5740

Test status: Passed

8.50.1Description

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.


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

Test ID: 5737

Test status: Passed

8.51.1Description

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 thebeam and 1.2.kN/m linear uniform horizontal load.





362

8.52 EC3 Test 17: Verifying a simply supported rectangular hollow section beam subjected totorsional efforts

Test ID: 5742

Test status: Passed

8.52.1Description

Verifies a simply supported rectangular hollow section beam made of S235 steel subjected to torsional efforts.


8.53 EC3 Test 18: Verifying a simply supported circular hollow section element subjected totorsional efforts

Test ID: 5743

Test status: Passed

8.53.1Description

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.

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

Test ID: 5741

Test status: Passed

8.54.1Description

The test verifies a CHS323.9x6.3H beam made of S275 steel.






363

8.55 EC3 Test 44: Determining lateral torsional buckling parameters for a I-shaped welded built-upbeam considering the load applied on the upper flange

Test ID: 5749

Test status: Passed

8.55.1Description

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 loadand 2 punctual bending moments, acting opposite to each other, applied at beam extremities.


8.55.2Background

Determines the elastic critical moment (Mcr ) and factors (C1, C2, LT) involved in the torsional buckling verification fora simply supported steel beam. The beam is made of S235 steel and it is subjected to a uniformly distributed load (50000 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.



■ Analysis type: static linear;




► Fz = - 50 000 N/ml,

► My,1 = 142 x 106Nmm,

► My,2 = - 113.6 x 106Nmm,






364

Units

Metric System

Geometry


■ Height: h = 260 mm,■ Flange width: b = 150 mm,

■ Flange thickness: tf = 10.7 mm,

■ Web thickness: tw = 7.1 mm,



■ Flexion inertia moment about the z axis: Iz = 6025866.46 mm4,

■ Torsional moment of inertia: It = 149294.97 mm4,

■ Warping constant: Iw = 93517065421.88 mm6,

■ Plastic modulus about the y axis: Wy = 501177.18 mm3





■ Longitudinal elastic modulus: E = 2.1 x 105 MPa;

■ Shear modulus of rigidity: G=80800MPa.

Boundary conditions


■ Outer:


► Support at end point (x = 5.00) restrained in translation along Y and Z axis and restrained rotationalong X axis.

■ Inner: None.

Loading


■ External:

► Uniformly distributed load over its length: q = Fz = -50 000 N/ml

► Bending moment at x=0: My,1 = 142 x 106

Nmm

► Bending moment at x=5: My,2 = - 113.6 x 106

Nmm,

■ Internal: None.




365

8.55.2.2 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 C2coefficients 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, themoments distribution is defined considering two parameters:

■ Ratio between the moments at extremities:

8.0142000

113600

Nm

Nm

■ Ratio between the moment given by uniformly distributed load and the biggest bending moment fromextremity:

1.1

101428

5000/50

8 6

22

Nmm

mmmm N

M

Lq

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 , , a, b, c, A, B, d1, e1, r 1, , m, C10, d2, e2, r 2 need to becalculated considering the analytical relationships provided in chapter 3.5 from French Annex of EN 1993-1-1:

■ 2.411.148.014

■

84.81.182.48

22

■ 738.19126223.06960364.01413364.015.0 2 a

■ 4765.14281556.19240091.01603341.015.0 2 b

■ 0602.09352904.05940757.00900633.01801266.0 2 c

■ 5625.22 cba A

■ 2143.42

2 ba B

■ 036.2152.01 d

■ 3.01 e

■ As d1 > e1, the factor r 1 is equal to 1.0

■ 0.14773.08

15.0

■ 0.1002.21411 m

■ 5363.02

42

110

A

A B Br C

■ 065.2675.0425.02 d

■ 37.035.065.02 e




366

■ As d2 > e2, the factor r 2 is equal to 1.0

Having the above factors, C1 and C2 coefficients become:

■ 074.1101 C mC

■ 235.0398.0 1022 C r C

The load being applied at the top fibre it tends to accentuate the lateral torsional buckling, so it will reduce the valueof elastic critical moment. In this case, the distance from the shear centre to the point of load application (zg) will bepositive:

■ mm z g 130

The French Annex of EN 1993-1-1 provides the analytical relationship used to determine the value of the elasticcritical moment:

■ Nmm z C z C I E

I G L

I

I

L

I E C M g g

z

t

z

w z cr

6

2

2

22

2

2

2

1 1071772.91

8.55.2.3 Reference results for calculating the reduction factor for lateral torsional bucklingThe calculation of the reduction factor for lateral torsional buckling (LT) is done using the formula (6.56) from chapter6.3.2.2 (EN 1993-1-1).

Before determining the reduction factor for lateral torsional buckling (LT), the following terms should be determined:

LT , imperfection factor LT, LT.

■ Non dimensional slenderness for lateral torsional buckling, LT :

133.11071772.91

/23518.5011776

23,

Nmm

mm N mm

M

f W

cr

y y pl

LT

■ In order to determine the imperfection factor LT, 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 torsionalbuckling curve is “c”. In this case, table 6.3 from EN 1993-1-1 recommends the value for imperfection factor

LT:

49.0 LT

■ The value used to determine the reduction factor LT, LT, becomes:

370.1133.12.0133.149.015.02.015.0 22 LT LT LT LT

■ The reduction factor for lateral torsional buckling is calculated using the formula (6.56) from EN 1993-1-1:

0.1467.0133.137.137.1

11

2222

LT LT LT

LT



■ 6 nodes,





367


C1 parameter

Simply supported beam

C1

C2 parameter


C2

Elastic critical moment


Mcr

Reduction factor for lateral torsional buckling


XLT




368



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



C1 C1 parameter 1.07375 adim 0.0004 %

C2 C2 parameter 0.234812 adim -0.0000 %

Mcr Elastic critical moment 91.72 kN*m 0.0000 %

XLT Reduction factor for lateral torsional buckling 0.466895 adim -0.0001 %




369

8.56 EC3 Test 43: Determining lateral torsional buckling parameters for a I-shaped laminated beamconsidering the load applied on the lower flange

Test ID: 5750

Test status: Passed

8.56.1Description

Determines the lateral torsional buckling parameters for a I-shaped laminated beam made of S235 steel, consideringthe load applied on the lower flange.





370

8.57 EC3 test 8: Verifying the classification and the resistance of a column subjected to bendingand axial load

Test ID: 5632

Test status: Passed

8.57.1Description

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

8.57.2Background

Classification and verification of an IPE 600 column, made of S235 steel, subjected to bending and axial force. Thecolumn is fixed at its base and free on the top. The column is loaded by a compression force (1 000 000 N), appliedat its top, and a uniformly distributed load (50 000 N/ml). The dead load will be neglected.





The following load case (Q1) and load combination are used:

■ Exploitation loadings (category A), Q1:

► Fz = -1 000 000 N,

► Fx = 50 000 N/ml,

■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q1


Units

Metric System

Geometry







■ Plastic section modulus about the strong y-y axis:3

,

3512000mmW y pl

,





371






■ Outer:


► Free at end point (x = 5.00).

■ Inner: None.

Loading


■ External:

► Point load at Z = 5.0: N = FZ = -1 000 000 N,

► Uniformly distributed load: q = Fx = 50 000 N/ml

■ Internal: None.




372

8.57.2.2 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 uniformcompression (Nc,Rd) and also the design plastic moment resistance reduced due to the axial force (M N,Rd) have to becompared 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 (EN1993-1-1), 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 isbigger 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 MPamm f A

N M

y

Rd c 36660000.1

23515600 2

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:

%100%3.270.1273.03666000

1000000

,,

N

N

N

N

N

N

Rd c Rd c

Ed

Verifying the column subjected to bending and axial force

According to paragraph 6.2.9.1 (4) from EN 1993-1-1, allowance will not be made for the effect of the axial force onthe plastic resistance moment about the y-y axis if relationship (6.33) is fulfilled.

N N N N N Rd pl Ed 916500366600025.0100000025.0 ,

In this case, because the above verification is not fulfilled, the axial force has an impact on the plastic resistancemoment 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 beused. Supplementary terms need to be determined: design resistance for bending (Mpl,Rd), ratio of design normalforce 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).

■ Design plastic moment resistance:




373

► Nmm MPamm f W

M M

y y pl

Rd y pl 8253200000.1

2353512000 3

0

,

,,

■ Ratio of design normal force to design plastic resistance to normal force of the gross cross-section:

► 273.0

3666000

1000000

,

N

N

N

N n

Rd pl

■ Ratio of web area to gross area:

► 464.015600

1922021560022

2

mm

mmmmmm

A

t b Aa

f

■ Design plastic moment resistance reduced due to the axial force is determined according to expression (6.36)from EN 1993-1-1:

► a

n M M Rd y pl Rd y N

5.01

1,,,,

but Rd y pl Rd y N M M ,,,,

► Nmm Nmm Nmm M Rd y N 825320000781259948

464.05.01

273.01825320000,,

■ The column subjected to bending and axial force is verified with criterion (6.41) from EN 1993-1-1:

► 0.1,,

,

,,

,

Rd z N

Ed z

Rd y N

Ed y

M

M

M

M

► Because the column doesn’t have bending moment about z axis, the second term from criterion (6.41)

is neglected. The verification becomes: 0.12/

,,

2

,,

,

Rd y N Rd y N

Ed y

M

Lq

M

M

►

%100%9.790.1799.0

781259948

2/5000/502/ 2

,,

2

Nmm

mmmm N

M

Lq

Rd y N



■ 6 nodes,





374



Work ratio - Fx

Work ratio of the design resistance for oblique bending






375



Work ratio - Fx Work ratio of the design resistance for uniform compression [%] 27.3 %

Work ratio - Oblique Work ratio of the design resistance for oblique bending [%] 79.9 %



Work ratio - Fx Work ratio of the design resistance for uniform compression 27.2777 % -0.0085 %

Work ratio -Oblique

Work ratio of the design resistance for oblique bending 79.9691 % -0.0386 %




376

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

Test ID: 5700

Test status: Passed

8.58.1Description

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

The verifications are made according to Eurocode3 French Annex.

8.58.2Background

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








Units

Metric System



■ Flexion inertia moment around the Y axis: Iy=11819x104mm

4


4








377

Boundary conditions


■ Outer:


► Free at end point (x = 5.00).

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

Loading



■ Internal: None.



%100100,

Rd b

Ed

N

N (6.46)


1

,

M

y

Rd b

f A N

(6.47)

Where:

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


according to:

11

22

(6.49)


cr

y

N

f A*

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

N

mm

mm MPa

L

I E N

fy

z cr 943.2449625

10000

1011819210000²

²

²2

44

954.0943.2449625

/2359490 22

N

mm N mm

N

f A

cr

y

²)2.0(15.0




378

It will be used the following buckling curve:


034.1954.02.0954.021.015.0²)2.0(15.0 2

Therefore:

1698.0954.0034.1034.1

112222

1 M is a safety coefficient, 11 M

N mm N mm

N Rd b 7.15566441

/2359490698.0 22

,

N N Ed 200000

%848.121007.1556644

200000100

, N

N

N

N

Rd b

Ed

8.58.2.3 Buckling in the weak inertia of the profile (along Z-Z)


%100100,

Rd b

Ed

N

N (6.46)


1

,

M

y

Rd b

f A N

(6.47)

Where:

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


according to:

11

22

(6.49)




379


cr

y

N

f A*

Where: A is the cross section area; A=5380mm

2

; f y is the yielding strength of the material; f y=235N/mm

2

and Ncr is theelastic critical force for the relevant buckling mode based on the gross cross sectional properties:

N

mm

mm MPa

L

I E N

fy

z cr 905.1301188

10000

106278210000²

²

²2

44

309.1905.1301188

/2359490 22

N

mm N mm

N

f A

cr

y

²)2.0(15.0



473.1309.12.0309.121.015.0²)2.0(15.0 2

Therefore:

1465.0309.1473.1473.1

11

2222


N

mm N mm

N Rd b 75.10370191

/2359490465.0 22

,

N N Ed 200000

%286.1910075.1037019

200000100

,

N

N

N

N

Rd b

Ed



■ 6 nodes,





380



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


Non-dimensional slenderness after Z-Z axis




381

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


Work ratio (y-y)

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


Work ratio (z-z)




382



y coefficient corresponding to non-dimensional slenderness after Y-Yaxis

0.698

z

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 (weakinertia) [%]

19.29%



Xy coefficient corresponding to non-dimensional slenderness

after Y-Y axis

0.697433 adim -0.0001 %

Xz coefficient corresponding to non-dimensional slendernessafter Z-Z axis

0.465226 adim 0.0000 %

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

0.128586 adim 0.0000 %

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

0.192767 adim -0.0002 %




383

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

Test ID: 5692

Test status: Passed

8.59.1Description

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

8.59.2Background

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







► Fz = -15 000 N/ml,



UnitsMetric System

Geometry





■ Web thickness: tw = 8 mm,







384






■ Outer:


► Support at end point (x = 5.00) restrained in translation along Y, Z axis and restrained in rotation alongX axis.

■ Inner: None.

Loading


■ External:

► Uniformly distributed load: q = Fz = -15 000 N/ml,■ Internal: None.






386

Therefore:

39.111467.13 t

c

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. Thepicture 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 conditionsdescribed in Table 5.2 - sheet 1 (above extract):

25.748

218630

mm

mmmm

t

c

8136.0235

y f

Therefore:

89.10012425.74 t

c

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 forbottom flange; therefore the class section for the entire beam section will be considered Class 4.

8.59.2.3 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 1993-1-1:2001.

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

The effective section modulus of the cross section takes into account the reduction factor, , which is applying only toparts 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.




387


) and the stress ratio( ) - for flanges

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

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

43.00.11

2

k

The plate modified slenderness ( p )

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

902.0

43.08136.04.28

18/2/8500

4.28

/

mmmmmm

k

t c p

The reduction factor ( )

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

0.1188.0

2

p

p

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

mm

mmmmcb f eff 89.215

2

85008776.0,

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.

3

,sup, 91.5204392 mmW yeff

3

,inf , 4.5736064 mmW yeff

3

,inf ,,sup,min,, 91.5204392,min mmW W W yeff yeff yeff




388


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

Nmm MPamm f W

M M

yeff

Rd c 18475594830.1

35591.5204392 3

0

min,

,

Work ratio

Work ratio =

%54.21001847559483

8/5000/15100

8/100

2

,

2

,

Nmm

mmmm N

M

Lq

M

M

Rd c Rd c



■ 6 nodes,



Beam subjected to uniformly distributed load




Work ratio - Oblique Design resistance for bending work ratio [%] 2.54 %



Work ratio -Oblique

Design resistance for bending work ratio 2.53713 % -0.1129 %




389

8.60 EC3 test 11: Cross section classification and compression resistance verification of arectangular hollow section column

Test ID: 5705

Test status: Passed

8.60.1Description

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


8.60.2Background

Classification and verification under compression efforts of a hot rolled rectangular hollow section column made ofS235 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 itstop. The dead load will be neglected.







► Fz = -100 000 N,



Units

Metric System




390

Geometry








■ Partial factor for resistance of cross sections: .






■ Outer:



■ Inner: None.

Loading


■ External: Point load at Z = 5.0:

► N = Fz = -100 000 N,

■ Internal: None.




391



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

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

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

2510

10215230022

mm

mmmmmm

t

t r h

t

c

0.1235 y f




392

Therefore:

333325 t

c



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


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

N MPamm f A

N M

y

Rd c 22301500.1

2359490 2

0

,

Work ratio

Work ratio = %48.41002230150

100000100

,

N

N

N

N

Rd c



■ 6 nodes,





Work ratio - Fx






394

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

Test ID: 5699

Test status: Passed

8.61.1Description

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

The verifications are made according to Eurocode3 French Annex.

8.61.2Background

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








Units

Metric System




4


4








395

Boundary conditions


■ Outer:


► Free at end point (x = 5.00).■ Inner: None.

■ Buckling lengths Lfy and Lfz are doth imposed with 10m value

Loading



■ Internal: None.






397

933.0854.02.0854.021.015.0²)2.0(15.0 2

Therefore:

1764.0854.0933.0933.0

112222


N mm N mm

N Rd b 089.9660511

/2355380764.0 22

,

N N Ed 200000

%703.20100089.966051

200000100,

N

N N N

Rd b

Ed

8.61.2.3 Buckling in the weak inertia of the profile (along Z-Z)


%100100,

Rd b

Ed

N

N (6.46)


1

,

M

y

Rd b

f A N

(6.47)

Where:

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


according to:

1

122

(6.49)


cr

y

N

f A*


2 and Ncr is the

elastic critical force for the relevant buckling mode based on the gross cross sectional properties:

N

mm

mm MPa

L

I E N

fy

z cr 610.125144

10000

1080.603210000²

²

²2

44




398

178.3610.125144

/2355380 22

N

mm N mm

N

f A

cr

y

²)2.0(15.0



056.6178.32.0178.334.015.0²)2.0(15.0 2

Therefore:

1089.0178.3056.6056.6

112222


N mm N mm

N Rd b 78.1127711

/2355380089.0 22

,

N N Ed 200000

%349.177100

78.112771

200000100

,

N

N

N

N

Rd b

Ed



■ 6 nodes,





399




Non-dimensional slenderness after Y-Y axis

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


Work ratio (y-y)






401

8.62 EC3 Test 12: Verifying the design plastic shear resistance of a rectangular hollow sectionbeam

Test ID: 5706

Test status: Passed

8.62.1Description

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


8.62.2Background

Verifies the adequacy of a rectangular hollow section beam made of S275 steel to resist shear. Verification of theshear resistance at ultimate limit state is realised. The name of the cross-section is RC3020100 and can be found inthe 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.







► Fz = -50 000 N/ml,



Units

Metric System

Geometry











402






Boundary conditions


■ Outer:


► Support at end point (x = 5.00) restrained in translation along Y, Z axis and restrained in rotation alongX axis.

■ Inner: None.

Loading


■ External:

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

■ Internal: None.

8.62.2.2 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) fromEN 1993-1-1. As the load is parallel to depth, the shear area is:

22

5694200300

3009490mm

mmmm

mmmm

hb

h A Av

Design plastic shear resistance of the cross section

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

N

MPamm

f A

V M

y

v

Rd pl 9040440.1

3

2755694

3

2

0

,

Work ratio

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

Work ratio = %83.13100904044

125000100

904044

2

5000/50

1002100,,

N

N

N

mmmm N

V

Lq

V

V

Rd pl Rd pl

Ed







403


Work ratio of the design shear resistance

Beam subjected to uniformly distributed load

Work ratio - Fz



Work ratio - Fz Work ratio of the design shear resistance [%] 13.83 %



Work ratio - Fz Shear resistance work ratio 13.8219 % 0.1587 %




404

8.63 EC3 Test 13: Verifying the resistance of a rectangular hollow section column subjected tobending and shear efforts

Test ID: 5707

Test status: Passed

8.63.1Description

Verifies the resistance of a rectangular hollow section column (made of S235 steel) subjected to bending and shearefforts.


8.63.2Background

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 isfixed 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.







► Fx = 200 000 N,












408

1510

10215220022

mm

mmmmmm

t

t r b

t

c

0.1235

y f

Therefore:

333315 t

c


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

2510

10215230022

mm

mmmmmm

t

t r h

t

c

0.1235

y f

Therefore:

727225 t

c




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

Nmm MPamm f W

M M

y y pl

Rd c 2246600000.1

235956000 3

0

,

,

Work ratio


Work ratio = %56.222100224660000

2

5000

2000001002100

,,

Nmm

mm

N

M

L

V

M

M

Rd c Rd c

Ed



■ 6 nodes,





Work ratio of the design shear resistance


Work ratio - Fz







410



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 %



Work ratio - Fz Work ratio of the design plastic shear resistance 25.8793 % -0.0413 %

Work ratio -Oblique









413

9.4 EC5: Verifying a timber purlin subjected to oblique bending

Test ID: 4878

Test status: Passed

9.4.1 Description

Verifies a rectangular timber purlin made from solid timber C24 to resist oblique bending. The verification is madefollowing the rules from Eurocode 5 French annex.

9.4.2 Background

Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to oblique bending. Theverification of the bending stresses at ultimate limit state is performed.


■ Reference: Guide de validation Eurocode 5, test E.3;


■ Element type: linear;

■ Distance between adjacent purlins (span): d = 1.8 m.


■ Loadings from the structure: G = 550 N/m2;

■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2;

■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2;

All loads will be projected on the purlin direction since the roof slope is 17°.

Simply supported purlin subjected to loadings

Units

Metric System

Geometry

Purlin cross section characteristics:

■ Height: h = 0.20 m,

■ Width: b = 0.10 m,

■ Length: L = 3.5 m,

■ Section area: A = 0.02 m2 ,

■ Elastic section modulus about the strong axis, y:3

22

000666.06

20.01.0

6m

hbW y

,

■ Elastic section modulus about the strong axis, z:3

22

000333.06

20.01.0

6 m

hb

W z

.




414


Rectangular solid timber C24 is used. The following characteristics are used in relation to this material:

■ Characteristic bending strength: f m,k = 24 x 106 Pa,

■ Service class 2.


■ 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/m

2 x 1.8 m x

cos17° = 3601.92 N/m,

■ Internal: None.

9.4.2.2 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, M, 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 andmaximum work ratio for strength verification) it is necessary to determine some parameters involved in calculations

(kmod, M, 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:

M = 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):

m,y,d = Pam

mm

N

W

Lq

W

M

y

z

y

y 6

3

222

102814.8000666.08

5.392.3601

8




415

■ Design bending stress about z axis (induced by uniformly distributed load, qy):

m,z,d = Pam

mm

N

W

Lq

W

M

z

y

z

z 6

3

222

100638.5000333.08

5.322.1101

8

■ Design bending strength:

f m,y,d = f m,z,d = Pak k k

f h sys

M

k m

66mod, 10615.160.10.1

3.1

9.01024

■ Maximum work ratio for strength verification; it represents the maximum value between the work ratiosobtained with formulae 6.17 and 6.18 from EN 1995-1-1 norm:

1max

,,

,,

,,

,,

,,

,,

,,

,,

d z m

d z m

d z m

d z m

m

d ym

d ym

m

d ym

d ym

f

f k

f k

f



■ 5 nodes,





416

Stress SMy diagram

Simply supported purlin subjected to uniformly distributed load, qz

Stress SMy [Pa]

Stress SMz diagram

Simply supported purlin subjected to uniformly distributed load, qz

Stress SMz [Pa]

Maximum work ratio for strength verification

Strength of a simply supported purlin subjected to oblique bending

Work ratio [%]



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 %

9.4.3 Calculated results


Stress SMy Design bending stress about y axis 8.47672e+006 Pa 0.0000 %

Stress SMz Design bending stress about z axis 5.18319e+006 Pa -0.0000 %

Work ratio Maximum work ratio for strength verification 71.153 % -0.0000 %




417

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

Test ID: 4896

Test status: Passed

9.5.1 Description

Verifies the residual cross section of a column exposed to fire for 60 minutes. The column is made from gluedlaminated 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.

9.5.2 Background

Verifies the adequacy of the cross sectional resistance for a rectangular cross section, which is made from gluedlaminated timber GL24, exposed to fire for 60 minutes on 3 faces. The verification is made according to chapter 4.2.2from EN 1995-1-2 norm.


■ Reference: Guide de validation Eurocode 5, test F.1;



Timber column with fixed base

Units

Metric System

Geometry


■ Depth: h = 0.60 m,

■ Width: b = 0.20 m,

■ Section area: A = 0.12 m2

■ Height: H = 5.00 m






419

Residual cross section

Column with fixed base exposed to fire for 60 minutes

Afi



Afi Residual cross section [m2

] 0.056202 m2



Afi Residual cross section 0.056202 m² -0.0000 %




420

9.6 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending

Test ID: 4901

Test status: Passed

9.6.1 Description

Verifies the fire resistance of a rectangular cross section purlin made from solid timber C24 to resist simple bending.The purlin is exposed to fire on 3 faces for 30 minutes. The verification is made according to chapter 4.2.2 (Reducedcross section method) from EN 1995-1-2 norm.

9.6.2 Background

Verifies the adequacy of the fire resistance for a rectangular cross section purlin made from solid timber C24 to resistsimple 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.


■ Reference: Guide de validation Eurocode 5, test F.2;




► Loadings from the structure: G = 500 N/m2,

► Snow load: S = 700 N/m2,

► Frequent combination of actions: CFQ = 1.0 x G + 0.5 x S = 850 N/m2

Purlin with 3 supports

Units

Metric System

Geometry


■ Height: h = 0.20 m,

■ Width: b = 0.075 m,

■ Length: L = 3.30 m,

■ Distance between adjacent purlins (span): d = 1.5 m,

■ Section area: A = 15.0 x 10-3

m2 ,




421


Rectangular solid timber C24 is used.The following characteristics are used in relation to this material:

■ Characteristic compressive strength along the grain: f c,0,k = 21 x 106 Pa,


■ Density: = 350 kg/m3,

■ Design charring rate (softwood): n = 0.8 x 10-3 m/min,


Boundary conditions


■ Outer:

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

► Support at middle point (X = 3.30) restrained in translation along X, Y and Z,

► Support at end point (X = 6.60) restrained in translation along X, Y, Z and restrained in rotation alongX.

■ Inner: None.

Loading


■ External:

► Uniformly distributed load: q = CFQ x d = 850 N/m2 x 1.5 m = 1275 N/m,

■ Internal: None.

9.6.2.2 Reference results in calculating the fire resistance of a timber purlin subjected to simplebending

In order to verify the fire resistance for a timber purlin subjected to simple bending it is necessary to determine theresidual 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, M,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 charringdepth according to chapter 4.2.2 from EN 1995-1-2. Before calculating the effective charring depth we need todetermine 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 = mm

t n 024.0min30

min

108.0 3 (according to relation 3.2 from EN 1995-1-2)

■ Effective charring depth:

def = 00, d k d nchar

■ Residual cross section:

Afi = )2()( ef ef ef ef d bd hbh




422

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, M,fi, kfi).

■ Modification factor in case of a verification done with residual section:

kmod,fi = 1.0 (according to paragraph 5 from chapter 4.2.2 from EN 1995-1-2)

■ Partial safety factor for timber in fire situations:

M,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) – accordingto paragraph 2 from chapter 6.1.6 (EN 1995-1-1):

km = 0.7

■ Design bending stress (taking into account the residual cross section):

m,d

=2

6

ef ef

y

y

y

hb

M

W

M

The picture below shows the bending moment diagram (kNm). My from the above formula represents themaximum bending moment achieved from frequent combination of actions.

■ Design bending strength (for fire situation):

f m,d,fi = Pak

f k fi M

fi

k m fi

66

,

mod,

, 10300.1

0.1102425.1

■ Work ratio according to formulae 6.17 from EN 1995-1-1 norm (considering that the axial effort, as well as thebending moment about z axis, are null):

0.1,

, d m

d m

f

■ Work ratio according to formulae 6.18 from EN 1995-1-1 norm (considering that the axial effort, as well as thebending moment about z axis, are null):

0.1,

, d m

d mm

f k

■ Maximum work ratio for bending verification for fire situation:

100100;max,

,

,

,

,

,

d m

d m

d m

d m

m

d m

d m

f f k

f WR



■ 8 nodes,







424



Afi Residual area [m2] 0.002197 m

2

Stress Design bending stress for residual cross-section [Pa] 27568524 Pa

Work ratio Maximum work ratio for fire verification [%] 91.9 %



Afi Residual area 0.002197 m² -0.0000 %

Stress Design bending stress for residual cross-section 2.75626e+007 Pa -0.0001 %

Work ratio Maximum work ratio for fire verification 91.8752 % 0.0000 %




425

9.7 EC5: Shear verification for a simply supported timber beam

Test ID: 4822

Test status: Passed

9.7.1 Description

Verifies a rectangular cross section beam made from solid timber C24 to shear efforts. The verification of the shearstresses at ultimate limit state is performed.




426

9.8 EC5: Verifying a timber beam subjected to simple bending

Test ID: 4682

Test status: Passed

9.8.1 Description

Verifies a rectangular cross section beam made from solid timber C24 to resist simple bending. Verifies the bendingstresses at ultimate limit state, as well as the deflections at serviceability limit state.

9.8.2 Background

Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending. Verificationof the bending stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit stateare performed.


■ Reference: Guide de validation Eurocode 5, test C;




■ Loadings from the structure: G = 0.5 kN/m2,

■ Exploitation loadings (category A): Q = 1.5 kN/m2,

■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2

■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q

■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q


Units

Metric System

GeometryBelow 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,


m2 ,

■ Elastic section modulus about the strong axis y:3

22

0005.06

20.0075.0

6m

hbW y




427






Boundary conditions


■ Outer:

► Support at start point (z=0) restrained in translation along X, Y and Z,


■ Inner: None.

Loading


■ External:

Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,

■ Internal: None.

9.8.2.2 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, M, kh, ksys, km, mustbe 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 theeffective values with the limiting values for deflections specified in EN 1995-1-1 norm.


Before calculating the reference solution (design bending stress, design bending strength and work ratios) it is

necessary to determine some parameters involved in calculations (kmod, M, kh, ksys, km).

■ Modification factor for duration of load (medium term) and moisture content:



M = 1.3


kh = 1.0




km = 0.7

■ Design bending stress (induced by the applied forces):

m,d = Pahb

Lq

W

M

y

y 6

2

2

2

2

104039.72.0075.08

5.44625.16

8

6


f m,d = Pak k k

f h sys

M

k m

66mod, 10769.140.10.1

3.1

8.01024




428

■ Work ratio according to formulae 6.11 from EN 1995-1-1 norm:

0.1,

, d m

d m

f


0.1,

, d m

d m

m f

k

Reference solution for serviceability limit state verification

The following limiting values for instantaneous deflection (for a base variable action), final deflection and netdeflection are considered:

300)( L

Qwinst

125

Lw fin

200,

Lw finnet

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are the followings:

■ Instantaneous deflection (for a base variable action):

8.600)(00749.0)(

LQwmQw inst inst

■ Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

45.45000999.0

Lwmd w

inst CQinst

■ In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), thedeformation factor (kdef ) has to be chosen:

6.0def k (calculated value for service class 1, according to table 3.2 from EN 1995-1-1)

95.157800285.000475.06.06.0

Lwmmd w creepQP creep

■ Final deflection:

47.35001284.000285.000999.0

Lwmmmwww fincreepinst fin

■ Net deflection:

47.35001284.0001284.0 ,,

Lwmmmwww finnet c fin finnet



■ 6 nodes,





429

Work ratio diagram

Simply supported beam subjected to bending

Strength work ratio



m,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



Stress Design bending stress 7.40391e+006 Pa -0.0001 %

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 0.0001 %

Wnet,fin Net deflection 0.0128367 m 0.0001 %




430

9.9 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending andaxial compression

Test ID: 4877

Test status: Passed

9.9.1 Description

Verifies the lateral torsional stability for a rectangular timber beam subjected to combined bending and axialcompression. The verification is made following the rules from Eurocode 5 French annex.

9.9.2 Background

Verifies the lateral torsional stability of a rectangular cross section made from solid timber C24 subjected to simplebending (about the strong axis) and axial compression.





■ Distance between adjacent rafters (span): d = 0.5 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


■ Height: h = 0.20 m,

■ Width: b = 0.05 m,

■ Length: L = 5.00 m,

■ Section area: A = 10 x 10-3

m2 ,

■ Elastic section modulus about the strong axis, y:3

22

000333.06

20.005.0

6m

hbW y

.








433

kz = 0.5 [1 + c (rel,z – 0.3) + rel,z2] (according to relation 6.28 from EN 1995-1-1)

ky = 0.5 [1 + c (rel,y – 0.3) + rel,y2] (according to relation 6.27 from EN 1995-1-1)

2

,

2,

1

z rel z z

z c

k k k

(according to relation 6.26 from EN 1995-1-1)

2

,

2,

1

yrel y y

yc

k k k



Before calculating the reference solution (the design compressive stress, the design bending stress and the workratio based on formula (6.35) from EN 1995-1-1) it is necessary to determine some parameters involved in

calculations (kmod, M, kh, ksys, km, kcrit).




M = 1.3


kh = 1.0



■ Factor which takes into account the reduced bending strength due to lateral buckling:

Kcrit = 1.56 – 0.75rel,m (because 0.75 < rel,m < 1.4)

■ Design compressive stress (induced by the compressive component, N):

c,d = Pa

m

N

A

N 219122

1010

22.219123

■ Design compressive strength:

f c,0,d = Pak

f M

k c

66mod,0, 10538.14

3.1

9.01021

■ Design bending stress (induced by uniformly distributed load, q):

m,d = Pam

mm

N

W

Lq

W

M

y y

y 6

3

222

10213.8000333.08

00.515.875

8


f m,y,d = Pak k k

f h sys

M

k m

66mod, 10615.160.10.1

3.1

9.01024

■ Work ratio according to formula 6.35 from EN 1995-1-1 norm:

1,0,,

,

2

,,

,

d c z c

d c

d ymcrit

d m

f k f k



■ 6 nodes,





434

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 [%]



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 %



Stress SFx Design compressive stress 219124 Pa 0.0002 %

Stress SMy Design bending stress 8.20519e+006 Pa -0.0000 %

Kcrit kcrit factor 0.592598 adim -0.0001 %

Work ratio Work ratio according to formula 6.35 0.759538 adim 0.0000 %




435

9.10 EC5: Verifying a C24 timber beam subjected to shear force

Test ID: 5036

Test status: Passed

9.10.1Description

Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification ofthe shear stresses at ultimate limit state is performed.

9.10.2Background

Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification ofthe shear stresses at ultimate limit state is performed.


■ Reference: Guide de validation Eurocode 5, test D;




■ Loadings from the structure: G = 0.5 kN/m2,

■ Exploitation loadings (category A): Q = 1.5 kN/m2,

■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2


Units

Metric System

Geometry

Beam cross section characteristics:

■ Height: h = 0.225 m,

■ Width: b = 0.075 m,

■ Length: L = 5.00 m,

■ Distance between adjacent beams (span): d = 0.5 m,

■ Section area: A = 16.875 x 10-3

m2 ,



■ Characteristic shear strength: f v,k = 2.5 x 106 Pa,





436

Boundary conditions


■ Outer:

► Support at start point (z=0) restrained in translation along X, Y and Z,


■ Inner: None.

Loading


■ External:

► Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,

■ Internal: None.

9.10.2.2 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 , M, 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 shearstrength and the corresponding work ratios, is calculated.


Before calculating the reference solution (design shear stress, design shear strength and work ratio) it is necessary to

determine some parameters involved in calculations (kmod, M, kcr , kf , beff , heff ).




M = 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):

d = Pa

mm

N

hb

F k

eff eff

d v f 6,10485075.0

225.005025.0

25.36562

3

■ Design shear strength:

f v,d = Pa Pak

f M

k v

66mod, 10538.1

3.1

8.0105.2


0.1,

d v

d

f






438



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 %



Fz Shear force -3.65625 kN -0.0000 %

Stress S_d Design shear stress 485075 Pa -0.0000 %

Working ratio S_d Shear strength work ratio 31.5299 % -1.4692 %




439

9.11 EC5: Verifying a timber column subjected to tensile forces

Test ID: 4693

Test status: Passed

9.11.1Description

Verifies the tensile resistance of a rectangular cross section column (fixed at base) made from solid timber C24.

9.11.2Background

Verifies the adequacy of the tension resistance for a rectangular cross section made from solid timber C24. Theverification is made according to formula (6.1) from EN 1995-1-1 norm.


■ Reference: Guide de validation Eurocode 5, test A;


■ Element type: linear.Column with fixed base

Units

Metric System

Geometry

Cross section characteristics:

■ Height: h = 0.122 m,

■ Width: b = 0.036m,


m2

■ I = 5.4475 x 10-6

m4.




440



■ Longitudinal elastic modulus: E = 1.1 x 1010

Pa,

■ Characteristic tensile strength along the grain: f t,0,k = 14 x 106 Pa,


Boundary conditions

The boundary conditions:

■ Outer:

Fixed at base (z = 0),

Free at top (z = 5),

■ Inner: None.

Loading


■ External: Point load at z = 5: Fz = N = 10000 N,

■ Internal: None.

9.11.2.2 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, M, kh). After this, the design tensile stress, the designtensile 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: M = 1.3

■ Depth factor (“h” represents the width, because the element is tensioned):

kh = min

3.1

150 2.0

h

■ Design tensile stress (induced by the ultimate limit state force, N):

t,0,d = A

N

■ Design tensile strength:

f t,0,d = h

M

k t k k

f

mod,0,

■ Work ratio:

SFx = 0.1,0,

,0, d t

d t

f




■ 6 nodes,





441

Work ratio SFx diagram

Column with fixed base, subjected to tension force

Work ratio SFx



t,0,d Design tensile stress [Pa] 2276867.03 Pa

SFx Work ratio [%] 18 %



Stress SFx Design tensile stress 2.27687e+006 Pa -0.0001 %

Work ratio Work ratio 18.0704 % 0.3910 %




442

9.12 EC5: Verifying a timber column subjected to compression forces

Test ID: 4823

Test status: Passed

9.12.1Description

Verifies the compressive resistance of a rectangular cross section column (hinged at base) made from solid timberC18.

9.12.2Background

Verifies the adequacy of the compressive resistance for a rectangular cross section made from solid timber C18. Theverification is made according to formula (6.35) from EN 1995-1-1 norm.


■ Reference: Guide de validation Eurocode 5, test B;



Simply supported column

Units

Metric System

Geometry

Column cross section characteristics:

■ Height: h = 0.15 m,

■ Width: b = 0.10 m,


m2




443




Pa,

■ 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: f c,0,k = 18 x 10

6

Pa,■ Service class 1.

Boundary conditions

The boundary conditions:

■ Outer:

► Support at base (z=0) restrained in translation along X, Y and Z,

► Support at top (z = 3.2) restrained in translation along X, Y and restrained in rotation along Z.

■ Inner: None.

Loading


■ External: Point load at z = 3.2: Fz = N = -20000 N,

■ Internal: None.

9.12.2.2 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: slendernessratios, relative slenderness ratios, instability factors. After this, the reference solution is calculated. This includes: thedesign 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:

85.1101.0

2.31121212

m

m

b

l m

b

l g c z

■ Slenderness ratio corresponding to bending about the y axis (informative):

9.7315.0

2.31121212

m

m

h

l m

h

l g c y

Relative slenderness ratios

The relative slenderness ratios are:

■ Relative slenderness ratio corresponding to bending about the z axis:

933.1106.0

1018

1.0

122.31

,

12

, 10

6

050

,0,

050

,0,

,

Pa

Pa

m

m

E

f

b

l m

E

f k c g k c z z rel

■ Relative slenderness ratio corresponding to bending about the y axis (informative):

288.1106.0

1018

15.0

122.31

,

12

, 10

6

050

,0,

050

,0,

,

Pa

Pa

m

m

E

f

h

l m

E

f k c g k c y

yrel

So there is a risk of buckling, because rel,max 0.3.




444

Instability factors

In order to determine the instability factors we need to determine the c factor. It is a factor for solid timber memberswithin the straightness limits defined in Section 10 from EN 1995-1-1:

c = 0.2 (according to relation 6.29 from EN 1995-1-1)

The instability factors are:■ kz = 0.5 [1 + c (rel,z – 0.3) + rel,z

2] (according to relation 6.28 from EN 1995-1-1)

■ ky = 0.5 [1 + c (rel,y – 0.3) + rel,y2] (informative)

■ 2

,

2,

1

z rel z z

z c

k k k


■ 2

,

2,

1

yrel y y

yc

k k k

(informative)

Reference solution

Before calculating the reference solution (design compressive stress, design compressive strength and work ratio) weneed to determine some parameters involved in calculations (kmod, M).



■ Partial factor for material properties: M = 1.3

■ Design compressive stress (induced by the applied forces):

c,0,d = A

N


f c,0,d = M

k c

k

f

mod

,0,

■ Work ratio:

Work ratio = 0.1,0,,

,0, d c z c

d c

f k




■ 4 nodes,







446

9.13 EC5: Verifying a timber beam subjected to combined bending and axial tension

Test ID: 4872

Test status: Passed

9.13.1Description

Verifies a rectangular cross section rafter made from solid timber C24 to resist combined bending and axial tension.The verification of the cross-section subjected to combined stresses at ultimate limit state, as well as the verificationof the deflections at serviceability limit state are performed.

9.13.2Background

Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending and axialtension. The verification of the deflections at serviceability limit state is also performed.


■ Reference: Guide de validation Eurocode 5, test E;



■ Distance between adjacent rafters (span): d = 0.5 m.



■ Snow load (structure is located at an altitude > 1000m above sea level): S = 900 N/m2;


■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x S;

■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.2 x S.

All loads will be projected on the rafter direction since its slope is 50% (26.6°).

Simply supported rafter subjected to projected loadings

Units

Metric System

Geometry


■ Height: h = 0.20 m,

■ Width: b = 0.05 m,

■ Length: L = 5.00 m,

■ Section area: A = 10 x 10-3

m2 ,

■ Elastic section modulus about the strong axis y:3

22

000333.06

20.005.0

6m

hbW

y

.




447



■ Characteristic tensile strength along the grain: f t,0,k = 14 x 106 Pa,



Boundary conditions


■ Outer:

► Support at start point (z=0) restrained in translation along Y, Z and restrained in rotation along X.

► Support at end point (z = 5.00) restrained in translation along X, Y, Z.

■ Inner: None.

Loading

The rafter is subjected to the following projected loadings (at ultimate limit state):

■ External:

► Uniformly distributed load: q = Cmax x d x cos26.6° = 1957.5 N/m2 x 0.5 m x cos26.6° = 875.15 N/m,

► Tensile load component: N = Cmax x d x sin26.6° x L = 1957.5 N/m2 x 0.5m x sin26.6° x 5.00m =

= 2191.22 N

■ Internal: None.

9.13.2.2 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, M,kh, ksys, km, must be determined. After this the reference solution, which consists of the design tensile stress, thedesign tensile strength and the corresponding work ratio and also the work ratios of the combined stresses, iscalculated.

A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the

effective values with the limiting values for deflections specified in EN 1995-1-1 norm.


Before calculating the reference solution (the design tensile stress, the design tensile strength and the correspondingwork ratio, and also the work ratios of the combined stresses) it is necessary to determine some parameters involved

in calculations (kmod, M, kh, ksys, km).




M = 1.3

■ Depth factor (“h” represents the width in millimeters because the element is tensioned):

kh = min 25.13.1

25.1min

3.1

50

150

min

3.1

150 2.02.0

h




km = 0.7




448

■ Design tensile stress (induced by the ultimate limit state force, N):

t,0,d = Pam

N

A

N 219122

1010

22.219123

■ Design tensile strength:

f t,0,d = Pak k

f h

M

k t 66mod,0, 10115.1225.13.1

9.01014

■ Work ratio:

SFx = 0.1,0,

,0, d t

d t

f


■ Design bending stress (induced by the applied forces):

m,y,d = Pamm

mm

N

hb

Lq

W

M

y

y 6

22

22

2

2

102045.82.005.08

00.515.8756

8

6


f m,y,d = Pak k k

f h sys

M

k m

66mod, 10615.160.10.1

3.1

9.01024


1,,

,,

,,

,,

,0,

,0, d z m

d z m

m

d ym

d ym

d t

d t

f k

f f


1,,

,,

,,

,,

,0,

,0, d z m

d z m

d ym

d ym

md t

d t

f f k

f

Reference solution for serviceability limit state verification

The following limiting values for instantaneous deflection (for a base variable action), final deflection and netdeflection are considered:

300)( L

Qwinst

125

Lw fin

200,

Lw finnet

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are:

■ Instantaneous deflection (for a base variable action):

05.547)(00914.0)(

LQwmQw inst inst

■ Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

7.36401371.0

Lwmd w inst CQinst




449

■ In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), thedeformation factor (kdef ) has to be chosen:

8.0def k (value determined for service class 2, according to table 3.2 from EN 1995-1-1)

6.976

00512.00064.08.08.0 L

wmmd w creepQP creep

■ Final deflection:

5.26501883.000512.001371.0

Lwmmmwww fincreepinst fin

■ Net deflection:

5.26501883.0001883.0 ,,

Lwmmmwww finnet c fin finnet



■ 6 nodes,


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 w inst (Q)

Simply supported beam subjected to snow loads

Instantaneous deflection winst(Q) [m]




450

Instantaneous deflection w inst (CQ)

Simply supported beam subjected to a characteristic load combination of actions

Instantaneous deflection winst(CQ) [m]

Final deflection w fin

Simply supported beam subjected to characteristic load combination of actions


Net deflection w net,fin

Simply supported beam subjected to characteristic load combination of actions







452

9.14 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression

Test ID: 4879

Test status: Passed

9.14.1Description

Verifies the stability of a rectangular timber purlin made from solid timber C24 subjected to biaxial bending and axialcompression. The verification is made following the rules from Eurocode 5 French annex.

9.14.2Background

Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to biaxial bending andaxial compression. The verification is made according to formulae (6.23) and (6.24) from EN 1995-1-1 norm.





■ Distance between adjacent purlins (span): d = 1.8 m.


■ The ultimate limit state (ULS) combination is: CULS = 1.35 x G + 1.5 x S + 0.9 x W;



■ Axial compression force due to wind effect on the supporting elements: W = 15000 N;

■ Uniformly distributed load corresponding to the ultimate limit state combination:

Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2.

All loads will be projected on the purlin direction since its slope is 30% (17°).

Simply supported purlin subjected to loadings

Units

Metric System

Geometry


■ Height: h = 0.20 m,

■ Width: b = 0.10 m,

■ Length: L = 3.50 m,

■ Section area: A = 0.02 m2 ,

■ Elastic section modulus about the strong axis, y:

322

000666.06

20.01.0

6 m

hb

W y

,




453

■ Elastic section modulus about the strong axis, z:3

22

000333.06

20.01.0

6m

hbW z

.


Rectangular solid timber C24 is used. The followings characteristics are used in relation to this material:




Pa,

■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 1010

Pa,


Boundary conditions


■ Outer:

► Support at start point (z = 0) restrained in translation along X, Y, Z;

► Support at end point (z = 3.50) restrained in translation along Y, Z and restrained in rotation along X.

■ Inner: None.

Loading

The purlin is subjected to the following projected loadings (corresponding to the ultimate limit state combination):

■ External:

► Axial compressive load: N =0.9 x W = 13500 N;

► Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x

sin17° = 1101.22 N/m,

► Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x

cos17° = 3601.92 N/m,

■ Internal: None.

9.14.2.2 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 limitstate, the formulae (6.23) and (6.24) from EN 1995-1-1 norm are used. Before applying these formulae we need todetermine some parameters involved in calculations like: slenderness ratios, relative slenderness ratios, instabilityfactors. After this, we’ll calculate the maximum work ratio for stability verification, which represents in fact thereference 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:

24.1211.0

5.31121212

m

m

b

l m

b

l g c z

■ Slenderness ratio corresponding to bending about the y axis:

62.602.0

5.31121212

m

m

h

l m

h

l g c y




454

Relative slenderness ratios

The relative slenderness ratios are:

■ Relative slenderness ratio corresponding to bending about the z axis:

056.21074.0

102124.121

10

6

05,0

,0,

,

Pa

Pa

E

f k c z

z rel

■ Relative slenderness ratio corresponding to bending about the y axis:

028.11074.0

102162.60

, 10

6

050

,0,

,

Pa

Pa

E

f k c y

yrel

■ Maximum relative slenderness ratio:

056.2,max ,,max, yrel z rel rel

So there is a risk of buckling because rel,max 0.3.

Instability factors

In order to determine the instability factors we need to determine the c factor. It is a factor for solid timber memberswithin the straightness limits defined in Section 10 from EN 1995-1-1:

c = 0.2 (according to relation 6.29 from EN 1995-1-1)

The instability factors are:

kz = 0.5 [1 + c (rel,z – 0.3) + rel,z2] (according to relation 6.28 from EN 1995-1-1)

ky = 0.5 [1 + c (rel,y – 0.3) + rel,y2] (according to relation 6.27 from EN 1995-1-1)

2

,

2,

1

z rel z z

z c

k k k


2

,

2,

1

yrel y y

yc

k k k



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 compressivestrength, the design bending stress, the design bending strength and some parameters involved in calculations (kmod,

M, kh, ksys, km).

■ Design compressive stress (induced by the axial compressive load from the corresponding ULS combination,N):

c,0,d = Pam N

A N 675000

02.013500 2

■ Design bending stress about the y axis (induced by uniformly distributed load, qz):

m,y,d = Pam

mm

N

W

Lq

W

M

y

z

y

y 6

3

222

102814.8000666.08

50.392.3601

8

■ Design bending stress about the z axis (induced by uniformly distributed load, qy):

m,z,d = Pa

m

mm

N

W

Lq

W

M

z

y

z

z 6

3

222

100638.5

000333.08

50.322.1101

8

■ Modification factor for duration of load (instantaneous action) and moisture content (service class 2):




455



M = 1.3


kh = 1.0




f c,0,d = Pak

f M

k c

66mod,0, 10769.17

3.1

1.11021


f m,y,d = f m,z,d = Pak k k

f h sys

M

k m

66mod, 10308.200.10.1

3.1

1.11024

■ Maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1:

1max

,,

,,

,,

,,

,0,,

,0,

,,

,,

,,

,,

,0,,

,0,

d z m

d z m

d ym

d ym

m

d c z c

d c

d z m

d z m

m

d ym

d ym

d c yc

d c

f f k

f k

f k

f f k



■ 6 nodes,





456

Instability factor, k cy

Simply supported purlin subjected to biaxial bending and axial compression

Kc,y

Instability factor, k cz


Kc,z

Maximum work ratio for stability verification


Work ratio [%]



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 %



Kc,y Instability factor, kc,y 0.665025 adim -0.0001 %

Kc,z Instability factor, kc,z 0.212166 adim 0.0002 %

Work ratio Maximum work ratio for stability verification 70.6586 % -0.6209 %



Documents

AD Validation Guide Vol2 2015 En