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Advance Design
Validation Guide
Part 2
Version: 2015
Tests passed on: 16 April 2014
Number of tests: 519
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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
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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
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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
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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
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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
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1 General applications
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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.
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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.
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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.
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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.
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2 Import / Export
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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.
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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.
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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".
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3 Joint Design
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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.
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4 Mesh
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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.
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5 Old Climatic Generator
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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).
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6 Reports Generator
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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.
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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.
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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".
The model consists of a concrete portal frame with rigid fixed supports.
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.
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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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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.
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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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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".
The model consists of a concrete portal frame with rigid fixed supports.
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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".
The model consists of a concrete portal frame with rigid fixed supports.
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.
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7 Seismic analysis
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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.
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.
7.4 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).
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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.
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.
7.7 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).
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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.
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.
7.11 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.
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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).
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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.
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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.
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8 Steel design
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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.
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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.
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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.
Generates the "Buckling and lateral-torsional buckling lengths" report.
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.
Generates the "Buckling and lateral-torsional buckling lengths" report.
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.
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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.
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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
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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
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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
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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:
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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).
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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 %
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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
8.15.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Fz=--328000N N; My=50000Nm; Mx=10000Nm;
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (mm).
Units
Metric System
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Geometrical properties
■ 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
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength f y = 275 MPa,
■ Longitudinal elastic modulus: E = 210000 MPa.
■ Shear modulus of rigidity: G=80800MPa
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8.15.2.2 Cross section Class
According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
1253.006.179
30.45
sup
inf
Mpa
Mpa
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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
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-for beam flange:
316.8924.0968.6
924.0
68.67.10
2
1.7150
t
c
t
c
therefore the haunch is considered to be Class1
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
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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
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b) over the weak axis of the section, z-z:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
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
Cross section area:206.4904 mm A
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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
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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
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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,
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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:
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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
Torsional moment of inertia: 497.149294 mm I t
Working inertial moment: 688.19351706542 mm I w
- 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,,
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kN N
mm
mmmm N
l
I E N
fz
z z cr 43.39563.395426
²5620
46.6025866/210000²
²
² 42
,
(previously calculated)
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:
For the top part of the column:
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
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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,
(previously calculated)
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
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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
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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,
(previously calculated)
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
(previously calculated)
kN N l
I E N
fz
z z cr 43.39563.395426
²
²,
(previously calculated)
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)
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Nm M Ed y 50000,
904.01
1 0,0,
LT y
LT y
mymymya
aC C C
(previously calculated)
1564.0²958.0²145.1145.1
1²²
1
LT LT LT
LT
(previously calculated)
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,
(previously calculated)
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
(previously calculated)
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
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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
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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
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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
Finite elements modeling
■ Linear element: S beam,
■ 7 nodes,
■ 1 linear element.
y coefficient corresponding to non-dimensional slenderness y
Column subjected to axial and shear force to the top
y
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z coefficient corresponding to non-dimensional slenderness z
Column subjected to axial and shear force to the top
z
Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
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Internal factor, zz k
Column subjected to axial and shear force to the top
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
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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
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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
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The elastic moment for lateral-torsional buckling calculation
The elastic moment for lateral-torsional buckling calculation
Mcr
The appropriate non-dimensional slenderness
The appropriate non-dimensional slenderness
LT
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8.15.2.1 Reference results
Result name Result description Reference value
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 %
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8.15.3Calculated results
Result name Result description Value Error
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 %
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■ Cross section dimensions are in millimeters (mm).
Units
Metric System
Geometrical properties
■ Column length: L=9000mm
■ Cross section area:28446mm A
■ Overall breadth: mmb 180
■ Flange thickness: mmt f 5.13
■ Root radius: mmr 21
■ Web thickness: mmt w 6.8
■ Depth of the web: mmhw 400
■ Elastic modulus after the Y axis, 33
, 101156 mmW yel
■ Plastic modulus after the Y axis, 33101307 mmW y
■ Elastic modulus after the Z axis, 33
, 1040.146 mmW z el
■ Plastic modulus after the Z axis, 33
, 10229 mmW z pl
■ Flexion inertia moment around the Y axis: Iy=23130.00x104mm
4
■ Flexion inertia moment around the Z axis: Iz=1318.00x104mm
4
■ Torsional moment of inertia: It=51.08x104mm
4
■ Working inertial moment: Iw=490000x106mm6
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Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength f y = 275 MPa,
■ Longitudinal elastic modulus: E = 210000 MPa.
■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation along X, Y and Z axis,
► Support at the end point (z = 9.00) restrained in translation along Y and Z axis and restrained rotationalong X axis.
■ Inner: lateral (xoz) restraint at z=3m
Loading
The column is subjected to the following loadings:
■ External: Point load From X=0.00m and Y=9.00m: FZ =-125000N and Fx=28330N■ Internal: None.
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CTICM model
The model is presented in the CTICM 2006-4-Resistance barre comprimee selon
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8.16.2.2 Cross section Class
According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-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
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 1
According to CTICM document:
The cross section is considered to be Class 1. The column strength will be determined considering the plasticcharacteristics of the cross-section. Below can be seen the CTICM conclusion, extracted from CTICM 2006-4:
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8.16.2.3 Compression verification
According to Advance Design calculations:
The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element 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
According to CTICM document:
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
According to Advance Design calculations:
The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element 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)
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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
y f is the yielding strength:2/275 mm N f y
0 M is the partial safety factor: 10 M
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
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According to CTICM document:
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
According to Advance Design calculations:
The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element 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
y f is the yielding strength:2/275 mm N f y
0 M is the partial safety factor: 10 M
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
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According to CTICM document:
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:
8.16.2.6 Buckling verification
According to Advance Design calculations:
a) over the strong axis of the section, y-y:
The cross section buckling curve will be chosen according to Table 6.2:
The imperfection factor α will be chosen according to Table 6.1:
21.0
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:
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
,
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The imperfection factor α will be chosen according to Table 6.1:
34.0
z coefficient corresponding to non-dimensional slenderness z will be determined from the relevant buckling curve
according to:
11
22
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
,
*
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:
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
According to CTICM document:
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:
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8.16.2.7 Lateral torsional buckling verification
According to Advance Design calculations:
a) for the 3m part of the column:
The elastic moment for lateral-torsional buckling calculation, Mcr :
-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
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)
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
■ Flexion inertia moment around the Y axis: Iy=23130.00x104mm
4
■ Flexion inertia moment around the Z axis: Iz=1318.00x104mm
4
■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
■ Torsional moment of inertia: It=51.08x104mm
4
■ Working inertial moment: Iw=490000x106mm6
■ Length of the column part: L=3000mm
■ Shear modulus of rigidity: G=80800MPa
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
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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
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 chosen according to Table 6.4:
The imperfection factor α will be chosen according to Table 6.1:
34.0
80228.0²66754.02.066754.034.015.0 LT
180173.0²66754.0²80228.080228.0
1
²²
1
LT LT LT
LT
According to CTICM document:
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:
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b) for the 6m part of the column:
The elastic moment for lateral-torsional buckling calculation, Mcr :
-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
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)
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
■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
■ Torsional moment of inertia: It=51.08x104mm
4
■ Working inertial moment: Iw=490000x106mm
6
■ Length of the column part: L=6000mm
■ Shear modulus of rigidity: G=80800MPa
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
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According to CTICM document:
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:
8.16.2.8 Bending and axial compression verification
According to Advance Design calculations:
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
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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:
87968.0 y (previously calculated)
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
According to CTICM document:
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The myC will be calculated according to Table A.1:
Calculation of the 0 term:
LT
C 10
-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)
is the fraction of the bending moment from the column extremities: 0
77.11 C
66754.02.806585210
/275101307 233
Nmm
mm N mm M
f W cr
y y LT
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88811.066754.077.1110 LT LT C C
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 ,,
²,
2
0
,
T cr
wt T cr
L
I E I G
I
A N
The mass moment of inertia 0 I
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
Shear modulus of rigidity: G=80800MPa
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
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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
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
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 0mC coefficient is defined according to the Table A.2:
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
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According to CTICM document:
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
a) for the 3m part of the column:
LT y
LT y
mymymmya
aC C C
11 0,0,3.
90119.14 y (previously calculated)
99779.0 LT a (previously calculated)
The 0mC coefficient is defined according to the Table A.2:
is the fraction of the bending moment from the column part extremities: 66667.097.254
98.169
kNm
kNm
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ycr
Ed my
N
N C
,
0, 33.036.021.079.0
Where:
N N ycr 773.5918472, (previously calculated)
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 z cr 153.1142396, (previously calculated)
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
a) for the 6m part of the column:
LT y
LT y
mymymmya
aC C C
11 0,0,6.
yel Ed
Ed y
yW
A
N
M
,
,
Elastic modulus after the Y axis, 33
, 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
The 0mC coefficient is defined according to the Table A.2:
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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:
N N ycr 773.5918472, (previously calculated)
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 z cr 153.1142396, (previously calculated)
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
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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
■ Elastic modulus after the Y axis, 33
, 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
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■ 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:
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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:
37096.0 z (previously calculated)
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
According to CTICM document:
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The myC will be calculated according to Table A.1:
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
(previously calculated)
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)
The 0mC coefficient is defined according to the Table A.2:
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The bending moment is null at the end of the column, therefore: 0
78749.033.036.021.079.0,
0, ycr
Ed my
N
N C
(previously calculated)
95619.01
1 0,0,
LT y
LT y
mymymya
aC C C
(previously calculated)
1
05596.1
116,
3,
,,
2
mmLT
mmLT
T cr
Ed
z cr
Ed
LT mymLT
C
C
N
N
N
N
aC C
(previously calculated)
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
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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
y (previously calculated)
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
(previously calculated)
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
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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
According to CTICM 2006-4:
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
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-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
According to CTICM 2006-4:
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Ratio of the design normal force to design compresion resistance
Column subjected to axial and shear force to the top
Work ratio - Fx
Ratio of the design share force to design share resistance
Column subjected to axial and shear force to the top
Work ratio - Fz
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Ratio of the design share force to design share resistance
Column subjected to axial and shear force to the top
Work ratio - oblique
y coefficient corresponding to non-dimensional slenderness y
Column subjected to axial and shear force to the top
y
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z coefficient corresponding to non-dimensional slenderness z
Column subjected to axial and shear force to the top
z
Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
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Internal factor, yz k
Column subjected to axial and shear force to the top
yz k
8.16.2.11 Reference results
Result name Result description Reference value
Work ratio - Fx Ratio of the design normal force to design compression resistance 5.38
Work ratio - Fz Ratio of the design share force to design share resistance 4.18
Work ratio - Oblique Ratio of the design moment resistance to design bending resistanceone the principal axis
70.94
y y coefficient corresponding to non-dimensional slenderness y
0.88
z z coefficient corresponding to non-dimensional slenderness z
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
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8.16.3Calculated results
Result name Result description Value Error
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 %
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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.
8.17.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
► Fz = -100 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength f y = 235 MPa,
■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
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Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (x = 5.00).■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 5.0: N = Fz = -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.
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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.
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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
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Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
Finite elements results
Work ratio of the design resistance for uniform compression
Column subjected to bending and axial force
Work ratio - Fx
8.17.2.4 Reference results
Result name Result description Reference value
Work ratio - Fx Compression resistance work ratio [%] 2.79518 %
8.17.3Calculated results
Result name Result description Value Error
Work ratio - Fx Compression resistance work ratio 2.79495 % -0.0083 %
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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.
8.18.2.1 Model description
■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load cases and load combination are used:
■ Exploitation loadings (category A): Q = 50kN,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
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Units
Metric System
Materials properties
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x=0) restrained in translation and rotation along X, Y and Z axis,
► Support at end point (z = 5.00) free.
■ Inner: None.
8.18.2.2 Reference results for calculating the cross section class
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In this case, the column is subjected to a lateral load, therefore the stresses distribution on the most stressed point(the column base) is like in the picture below.
The Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.
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Therefore:
This means that the column web is Class 1.
Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.
The section geometrical properties are described in the picture below:
According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:
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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
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.5(1)
0
,
*
M
y pl
Rd c
f w M
for Class1 cross sections
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.5(2)
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
Finite elements modeling
■ Linear element: S beam,
■ 7 nodes,
■ 1 linear element.
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Finite elements results
Combined oblique bending
Combined oblique bending
Work ratio - Oblique
8.18.2.4 Reference results
Result name Result description Reference value
Combined obliquebending
Work ratio - Oblique 85 %
8.18.3Calculated results
Result name Result description Value Error
Work ratio -Oblique
Work ratio - Oblique 84.6459 % -0.4166 %
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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.
8.19.2.1 Model description
■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load cases and load combination are used:
■ Exploitation loadings (category A): Q = -50kN/m,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
Units
Metric System
Materials properties
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Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x=0) restrained in translation along X, Y and Z axis,
► Support at end point (x = 5.00) restrained in translation along Y and Z axis and rotation restrained on X
axis■ Inner: None.
8.19.2.2 Reference results for calculating the cross section class
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In this case the stresses distribution along the section is like in the picture below:
■ compression for the top flange
■ compression and tension for the web
■ tension for the bottom flange
To determine the web class it will be used the Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.
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The section geometrical properties are described in the picture below:
According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:
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Therefore:
This means that the column web is Class 1.
To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter5.5.2
The section geometrical properties are described in the picture below:
According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:
276.57.10
45.56
mm
mm
t
c
92.0
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Therefore:
28.892.0*9*9276.57.10
45.56
mm
mm
t
c this means that the column flanges are Class 1.
A cross-section is classified by quoting the heist (least favorable) class of its compression elements.
According to the calculation above, the beam section have a Class 1 web and Class 1 flanges; therefore the classsection for the entire beam section will be considered Class 1.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)
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
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(2)
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
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3)
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
,
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For more:
Verification of the shear buckling resistance for webs without stiffeners:
*72
w
w
t
h
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(7)
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)
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
Finite elements results
Shear resistance work ratio
Work ratio - Fz
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8.19.2.4 Reference results
Result name Result description Reference value
Fz Shear force 125 kN
Work ratio Work ratio - Fz 31 %
8.19.3Calculated results
Result name Result description Value Error
Fz Fz -125 kN -0.0000 %
Work ratio - Fz Work ratio Fz 30.6579 % -1.1034 %
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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
Classification and verification of an IPE 300 column made of S235 steel.
The column is connected to the ground by a fixed connection and is free on the top part.
In the middle, the column is subjected to a 200 kN force applied on the web direction, defined as a live load.
The dead load will be neglected.
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.
8.20.2.1 Model description
■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load cases and load combination are used:
■ Exploitation loadings (category A): Q = -200kN,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
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Units
Metric System
Materials properties
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,
► Support at end point (z = 5.00) free.
■ Inner: None.
8.20.2.2 Reference results for calculating the cross section class
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In case the column is subjected to a lateral load, the stresses distribution on the most stressed point (the columnbase) is like in the picture below:
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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.
The section geometrical properties are described in the picture below:
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According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:
Therefore:
This means that the column web is Class 1.
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.
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The section geometrical properties are described in the picture below:
According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:
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).
8.20.2.3 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
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(2)
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
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According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3)
f y nominal yielding strength for S235 f y=235MPa
0 M partial safety coefficient 10 M
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
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(3)
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
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3)
f y nominal yielding strength for S235 f y=235MPa
0 M partial safety coefficient 10 M
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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
Finite elements modeling
■ Linear element: S beam,
■ 7 nodes,
■ 1 linear element.
Finite elements results
Shear z direction work ratio
Shear z direction work ratio
Work ratio - Fz
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Combined oblique bending
Combined oblique bending
Work ratio - Oblique
8.20.2.5 Reference results
Result name Result description Reference value
Shear z directionwork ratio Work ratio - Fz 57 %
Combined obliquebending
Work ratio - Oblique 341 %
8.20.3Calculated results
Result name Result description Value Error
Work ratio - Fz Work ratio Fz 57.4021 % 0.7054 %
Work ratio -
Oblique
Work ratio - Oblique 341.348 % 0.1021 %
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Metric System
Materials properties
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,
► Support at end point (x = 5.00) restrained in translation and rotation along Y, Z axis and rotationblocked along X axis.
■ Inner: None.
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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:
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Therefore:
This means that the beam web is Class 1.
Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.
The section geometrical properties are described in the picture below:
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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
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)
8.21.2.3 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 ,,
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(4)
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 ,,
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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 ,,
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(4)
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 %
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
Finite elements results
Work ratio – oblique bending
Beam subjected to combined bending
Work ratio – Oblique [%]
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8.21.2.4 Reference results
Result name Result description Reference value
Combined obliquebending
Combined oblique bending [%] 110.86 %
8.21.3Calculated results
Result name Result description Value Error
Work ratio -Oblique
Work ratio-Oblique 110.691 % -0.2783 %
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8.22 EC3 Test 1: Class section classification and compression verification of an IPE300 column
Test ID: 5383
Test status: Passed
8.22.1Description
Classification and verification of an IPE 300 column made of S235 steel.
The column is connected to the ground by a fixed connection and is free on the top part.
On top, the column is subjected to a 100 kN force applied gravitationally, defined as a live load.
The dead load will be neglected.
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.
8.22.2.1 Model description
■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load cases and load combination are used:
■ Exploitation loadings (category A): Q = -100kN,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
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Units
Metric System
Materials properties
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x=0) restrained in translation and rotation along X, Y and Z axis,
► Support at end point (z = 5.00) free.
■ Inner: None.
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8.22.2.2 Reference results for calculating the cross section class
In this case, the column is subjected only to compression, therefore the distribution of stresses along the section 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
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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:
1
Therefore:
This means that the column web is Class 2.
To determine the flanges class, we will use Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
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The section geometrical properties are described in the picture below:
According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:
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.
A cross-section is classified by quoting the heist (least favorable) class of its compression elements.
According to the calculation above, the column section have a Class 2 web and Class 1 flanges; therefore the classsection for the entire column section will be considered Class 2.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)
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Finite elements results
Compressive resistance work ratio
Column subjected to compressive load
Work ratio [%]
8.22.2.4 Reference results
Result name Result description Reference value
Work ratio Compressive resistance work ratio [%] 8 %
8.22.3Calculated results
Result name Result description Value Error
Work ratio - Fx Work ratio Fx 7.90805 % -1.1494 %
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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
Classification and verification of an IPE 300 column made of S235 steel.
The column is connected to the ground by a fixed connection and is free on the top part.
The column is subjected to a 500 kN compressive force applied on top and a 5 kN/m uniform linear load applied onall the length of the column, on the web direction, both defined as live loads.
The dead load will be neglected.
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.
8.23.2.1 Model description
■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load cases and load combination are used:
■ Exploitation loadings (category A): Q1 = 500kN, Q2 = 5kN/m,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
Units
Metric System
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Materials properties
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x=0) restrained in translation and rotation along X, Y and Z axis,
► Support at end point (z = 5.00) free.
■ Inner: None.
8.23.2.2 Reference results for calculating the cross section class
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In this case the column is subjected to compression and lateral load, therefore the stresses distribution on the 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.
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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:
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Therefore:
Therefore:
This means that the column web is Class 3.
Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.
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The section geometrical properties are described in the picture below:
According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:
Therefore:
This means that the column flanges are Class 1.
A cross-section is classified by quoting the heist (least favorable) class of its compression elements.
According to the calculation above, the column section have a Class 3 web and Class 1 flanges; therefore the classsection for the entire column section will be considered Class 3.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)
Cross sections for class 3, the maximum longitudinal stress should check:
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.4(3)
In this case:
In absence of shear force, for Class 3 cross-sections the maximum longitudinal stress shall satisfy the criterion:
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.2(1)
This means:
The corresponding work ratio is: WR = 0.8728 x 100 = 87.28 %
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Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
Finite elements results
Work ratio – bending and axial compression
Column subjected to bending and axial compression
Work ratio – Oblique [%]
8.23.2.3 Reference results
Result name Result description Reference value
Bending and axialcompression
Work ratio – oblique [%] 87.28 %
8.23.3Calculated results
Result name Result description Value Error
Work ratio -
Oblique
Work ratio- Oblique 87.2799 % 0.3217 %
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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.
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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.
8.29.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear;
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
► Fz = - 20 000 N/ml,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
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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 .
Materials properties
500 MPa yield strength user-defined material and S275 steel are used. The following characteristics are used:
■ Yield strength f y = 500 MPa,
■ 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.
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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
Finite elements modeling
■ Linear element: S beam,
■ 7 nodes,■ 1 linear element.
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Finite elements results
Work ratio of the design shear resistance (beam made of 500 MPa yield strength user-defined material)
Beam subjected to uniformly distributed load applied over its length
Work ratio – Fz
Work ratio of the design shear resistance (beam made of S275 steel)
Beam subjected to uniformly distributed load applied over its length
Work ratio – Fz
8.29.2.3 Reference results
Result name Result description Reference value
Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 275 MPa) 13.27 %
Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 500 MPa) 8.76 %
8.29.3Calculated results
Result name Result description Value Error
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 %
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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.
The calculations are made according to Eurocode 3 French Annex.
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.
The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.
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Geometry
Below are described the column cross section characteristics:
■ Height: h = 630 mm,
■ Flange width: b = 500 mm,
■ Flange thickness: tf = 18 mm,
■ Web thickness: tw = 8 mm,
■ Column length: L = 5000 mm,
■ Section area: A = 22752 mm2 ,
■ Partial factor for resistance of cross sections: 0.10 M .
Materials properties
S355 steel material is used. The following characteristics are used:
■ Yield strength f y = 355 MPa,
■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditionsThe boundary conditions are described below:
■ Outer:
► Support at start point (z = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (z = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External:
► Point load at Z = 5.0: N = FZ = -100 000 N,
■ Internal: None.
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8.33.2.2 Cross-section classification
Before calculating the compression resistance, the cross-section class has to be determined.
The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General 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 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
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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
This means that the column web is Class 4.
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.
8.33.2.3 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 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.
The buckling factor (k
) 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.
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Taking into account that the stress distribution on flanges is linear, the stress ratio becomes:
43.00.11
2
k
The buckling factor (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,
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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
Compression resistance of the cross section
For Class 4 cross-section, EN 1993-1-1: 2001 provides (6.11) formula in order to calculate the 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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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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.
8.34.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Q1 = -100 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (mm).
Units
Metric System
Geometrical properties
■ Tube wall thickness: t=6.3mm
■ Tube diameter: d=219.1mm
■ Cross section area: A=4210mm2
■ 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
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Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength f y = 355 MPa,
■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditionsThe boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (z = 3.00).
■ Inner: None.
■ Buckling lengths Lfy and Lfz are both imposed with 6m value
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 3.0: FZ = N = -100 000 N,
■ Internal: None.
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8.34.2.2 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 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)
Cross-class classification is made according to Table 5.2
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:
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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
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Finite elements modeling
■ Linear element: S beam,
■ 4 nodes,
■ 1 linear element.
Finite elements results
The appropriate non-dimensional slenderness
The appropriate non-dimensional slenderness
LT
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Ratio of the design normal force to design buckling resistance in the strong inertia of the profile
Column subjected to axial force
Adimensional - SNy
8.34.2.3 Reference results
Result name Result description Reference value
y y coefficient corresponding to non-dimensional slenderness y
0.636
ySN Ratio of the design normal force to design buckling resistance in thestrong inertia of the profile
0.1052
8.34.3Calculated results
Result name Result description Value Error
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 %
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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 test verifies a user defined cross section column.
The cross section has an “I symmetric” shape with: 408mm height; 190mm width; 9.4mm center thickness; 14.6mmflange 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.
The calculations are made according to Eurocode 3 French Annex.
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.
8.35.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Fz=-1000000N N; My=200000Nm
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (mm).
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Units
Metric System
Geometrical properties
■ Column length: L=2000mm
■ Cross section area:2
72.9108 mm A
■ Overall breadth: mmb 190
■ Flange thickness: mmt f 6.14
■ Root radius: mmr 0
■ Web thickness: mmt w 4.9
■ Depth of the web: mmhw 408
■ Elastic modulus after the Y axis, 3
, 06.1261435 mmW yel
■ Plastic modulus after the Y axis, 378.1428491 mmW y
■ Elastic modulus after the Z axis, 3
, 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
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength f y = 275 MPa,
■ Longitudinal elastic modulus: E = 210000 MPa.
■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,
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Loading
The column is subjected to the following loadings:
■ External: Point load From X=0.00m and z=2.00m: FZ =-1000000N; Mx=200000Nm and Fy=5000N
8.35.2.2 Cross section Class
According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
118.083.268
76.48
sup
inf
Mpa
Mpa
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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
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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
therefore the beam web is considered to be Class 3
-for beam flange:
316.8924.0918.6
924.0
18.66.14
30.90
t
ct
c
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 3.
8.35.2.3 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
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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 4 cross-sections:
ycr
y
y N
f A
,
*
Cross section area:272.9108 mm 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
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b) over the weak axis of the section, z-z:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
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
,
*
ml fz 00.2
Flexion inertia moment around the Z axis: Iz=16716452.10 mm
4
Cross section area:272.9108 mm A
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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
8.35.2.4 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)
-in order to simplify the calculation, it will be considered 11 C
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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:
IW is the warping inertia (deformation inertia moment):
4
2
f z
w
t h I I
h cross section height; h=408mm
f t flange thickness; mmt f 6.14
611
24
1046774.64
6.144080mm16716452.1mm
mmmm I w
-according to EN1993-1-1-AN France; Chapter 2 (…4)
Length of the column: L=2000mm
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
Calculation of the LT for appropriate non-dimensional slenderness LT will be determined with formula:
1
²²
1
LT LT LT
LT
(6.56)
²2.015.0 LT LT LT LT
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The cross section buckling curve will be chose according to Table 6.4:
2147.2190
408
mm
mm
b
h
The imperfection factor α will be chose according to Table 6.3:
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
8.35.2.5 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:
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Calculation of the 0 term:
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
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
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
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 ,,
²
1
,
2
2
0
,
T cr
wt T cr
L
I E I G
i N
The mass moment of inertia 0 I
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)
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C1=1 for the top part of the column
For the top part of the column:
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:
For the top part of the column:
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
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
11 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
Elastic modulus after the Y axis,3
,, 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
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8.35.2.6 Internal factor, yz k , calculation
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
8.35.2.7 Internal factor, zyk , calculation
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
821.0 z (previously calculated)
050.1
7.133338053
10000001
977.0065.1001.1
1,
N
N
N
N C C k
ycr
Ed
z mLT my zy
8.35.2.8 Internal factor, zz k , calculation
857.0
38.8661700
10000001
977.0776.0
1,
N
N
N
N C k
z cr
Ed
z
mz zz
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8.35.2.9 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
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
Finite elements modeling
■ Linear element: S beam,
■ 5 nodes,■ 1 linear element.
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y coefficient corresponding to non-dimensional slenderness y
Column subjected to axial and shear force to the top
y
z coefficient corresponding to non-dimensional slenderness z
Column subjected to axial and shear force to the top
z
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Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yz k
Column subjected to axial and shear force to the top
yz k
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Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zz k
Column subjected to axial and shear force to the top
zz k
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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
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
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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
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
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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
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
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8.35.2.10 Reference results
Result name Result description Reference value
y y coefficient corresponding to non-dimensional slenderness y
1
z z coefficient corresponding to non-dimensional slenderness z
0.821
yyk Internal factor, yyk 1.074
yz k Internal factor, yz k 0.878
zyk Internal factor, zyk 1.050
zyk Internal factor, zyk 0.857
SNy Bending and axial compression verification term depending of thecom ression effort over the Y axis
0.40
SMyy Bending and axial compression verification term depending of the Ybendin moment over the Y axis
0.76
SMyz Bending and axial compression verification term depending of the Zbendin moment over the Y axis
0.18
SNz Bending and axial compression verification term depending of thecom ression effort over the z axis
0.49
SMzy Bending and axial compression verification term depending of the Ybendin moment over the Z axis
0.74
SMzz Bending and axial compression verification term depending of the Zbendin moment over the Z axis
0.18
8.35.3Calculated results
Result name Result description Value Error
Xy coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 %
Xz coefficient corresponding to non-dimensional slenderness 0.821634 adim -0.0001 %
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 %
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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.
The calculations are made according to Eurocode 3 French Annex.
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.
8.36.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Fx=-3700000N; Fy=-5000N/m; Fz=-10000N/m
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (mm).
Units
Metric System
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Geometrical properties
■ Column length: L=5000mm
■ Cross section area:2
14264mm A
■ Overall breadth: mmb 220
■ Flange thickness: mmt f 19
■ Root radius: mmr 0
■ Web thickness: mmt w 12
■ Depth of the web: mmhw 530
■ Elastic modulus after the Y axis, 3
, 11.2509773 mmW yel
■ Plastic modulus after the Y axis, 300.2862172 mmW y
■ Elastic modulus after the Z axis, 3
, 41.307177 mmW z el
■ Plastic modulus after the Z axis, 3, 00.477512 mmW z pl
■ Flexion inertia moment around the Y axis:467.665089874 mm I y
■ 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
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength f y = 235 MPa,
■ Longitudinal elastic modulus: E = 210000 MPa.
■ Shear modulus of rigidity: G=80800MPa
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Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation along X, Y and Z axis,
► Support at the end point (x = 5) restrained in translation along Y and Z axis and restrained rotationalong X axis.
■ Inner:
► Lateral buckling restraint in the middle of the column (x=2.50).
Loading
The beam is subjected to the following loadings:
■ External: Point load from X=f.00m and z=.00m: Fx =-3700000N;
■ External: vertical uniform distributed linear load from X=0.00 to X=5.00: Fz=-10000N/m
■ External: horizontal uniform distributed linear load from X=0.00 to X=5.00: Fy=-5000N/m
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8.36.2.2 Cross section Class
According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
194.246
94.246
sup
inf Mpa
Mpa
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1235
235235
y f
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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
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 3.
8.36.2.3 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
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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 3 cross-sections:
ycr
y
y N
f A
,
*
Cross section area:214264mm A
Flexion inertia moment around the Y axis:467.665089874 mm I y
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
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b) over the weak axis of the section, z-z:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
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
,
*
ml fz 50.2
Flexion inertia moment around the Z axis:467.33789514 mm I z
Cross section area:214264mm A
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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
8.36.2.4 Lateral torsional buckling verification
The elastic moment for lateral-torsional buckling calculation, Mcr :-it must be studied separately for each beam segment
-however, the two sections are symmetrical, the same result will be obtained
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
-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
According to EN 1993-1-1-AN France; Chapter 3; …(6)
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0 therefore:
31.11 C
Flexion inertia moment around the Y axis:467.665089874 mm I y
Flexion inertia moment around the Z axis:467.33789514 mm I z
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
Elastic modulus after the Y axis, 3
, 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
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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)
8.36.2.5 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 calculatedseparately for the two column parts separate by the middle torsional lateral restraint:
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ycr
Ed
y
mLT my yy
N
N C C k
,
1
ycr
Ed y
ycr
Ed
y
N
N N
N
,
,
1
1
984.0 y (previously calculated)
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.
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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 ,,
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
Torsional moment of inertia:473.1269555 mm I t
Working inertial moment:667.6662201162989 mm I w
- the buckling length, T cr L , ,
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
,
(previously calculated)
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.
The coefficient must be calculated considering the column over the entire height.
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LT y
LT y
mymymya
aC C C
11 0,0,
yeff
eff
Ed
Ed y
y
W
A
N
M
,
,
Elastic modulus after the Y axis,3
,, 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
The 0mC coefficient is defined according to the Table A.2:
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
,
(previously calculated)
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
Equivalent uniform moment factor, mLT C , calculation
- It must be calculated for each of the two sections.
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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
Therefore the yyk term corresponding to the top part of the column will be:
476.1
21.55130961
37000001
1377.1002.1
1,
N
N
N
N C C k
ycr
Ed
y
mLT my yy
8.36.2.6 Internal factor, yz k , calculation
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
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8.36.2.7 Internal factor, zyk , calculation
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
816.0 z (previously calculated)
355.1
21.55130961
37000001
917.0377.1002.1
1,
N
N
N
N C C k
ycr
Ed
z mLT my zy
8.36.2.8 Internal factor, zz k , calculation
383.1
19.11205235
37000001
917.0776.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
8.36.2.9 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
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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
Finite elements modeling
■ Linear element: S beam,
■ 7 nodes,
■ 1 linear element.
y coefficient corresponding to non-dimensional slenderness y
Column subjected to axial and shear force to the top
y
z coefficient corresponding to non-dimensional slenderness z
Column subjected to axial and shear force to the top
z
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Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yz k
Column subjected to axial and shear force to the top
yz k
Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
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Internal factor, zz k
Column subjected to axial and shear force to the top
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
Bending and axial compression verification term depending of the Y bending moment over the Y axis: SMyy
Bending and axial compression verification term depending of the Y bending moment overthe Y axis
SMyy
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Bending and axial compression verification term depending of the Z bending moment over the Y axis: SMyz
Bending and axial compression verification term depending of the Z bending moment overthe Y axis
SMyz
Bending and axial compression verification term depending of the compression effort over the Z axis: SNz
Bending and axial compression verification term depending of the compression effortover the Z axis
SNz
Bending and axial compression verification term depending of the Y bending moment over the Z axis: SMzy
Bending and axial compression verification term depending of the Y bending momentover the Z axis
SMzy
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8.36.3Calculated results
Result name Result description Value Error
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 %
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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 test verifies a user defined cross section column.
The cross section has an “I symmetric” shape with: 880mm height; 220mm width; 5mm center thickness; 15mmflange thickness; 0mm fillet radius and 0mm rounding radius.
The column is subjected to a -328kN axial compression force; 1274 kNm bending moment after the Y axis and 127.4kNm 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.
The calculations are made according to Eurocode 3 French Annex.
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)
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8.37.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Fz=-328000N N, Mx= 127400Nm; My=1274000Nm
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (mm).
Units
Metric System
Geometrical properties
■ Column length: L=5620mm
■ Cross section area:210850mm A
■ Overall breadth: mmb 220
■ Flange thickness: mmt f 15
■ Root radius: mmr 0
■ Web thickness: mmt w 5
■ Depth of the web: mmhw 880
■ Elastic modulus after the Y axis, 33
, 1066.3387 mmW yel
■ Plastic modulus after the Y axis, 331062.3757 mmW y
■ Elastic modulus after the Z axis, 33
, 1008.242 mmW z el
■ Plastic modulus after the Z axis, 33
, 1031.368 mmW z pl
■ Flexion inertia moment around the Y axis: Iy=149058.04x104mm
4
■ Flexion inertia moment around the Z axis: Iz=2662.89x104mm
4
■ Torsional moment of inertia: It=51.46x104mm
4
■ Working inertial moment: Iw=4979437.37x106mm
6
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Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength f y = 275 MPa,
■ Longitudinal elastic modulus: E = 210000 MPa.
■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation along X, Y and Z axis,
► Support at the end point (z = 5.62) restrained in translation along Y and Z axis and restrained rotationalong X axis.
■ Inner:
► Lateral buckling restraint in the middle of the column (z=2.81).
Loading
The column is subjected to the following loadings:
■ External: Point load From X=0.00m and z=5.62m: FZ =--328000N; Mx=127400Nm and My=1274000Nm
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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
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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
therefore the beam web is considered to be Class 4
-for beam flange:
316.8924.0961.7
924
61.715
5.107
t
ct
c
therefore the haunch is considered to
be Class1
In conclusion, the section is considered to be Class 4
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
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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
-according EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1
170
5
152880
mm
mmmm
t
c
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
325.19.239244.04.28
5
850
mm
mm
p
-according EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4
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
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-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
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8.37.2.5 Buckling verification
a) over the strong axis of the section, y-y:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
34.0
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 4 cross-sections:
ycr
yeff
y N
f A
,
*
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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
b) over the weak axis of the section, z-z:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
49.0
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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:
11
22
z z z
z
(6.49)
z the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
z cr
yeff z
N
f A
,
*
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:
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
8.37.2.6 Lateral torsional buckling verification
a) for the top part of the column:
The elastic moment for lateral-torsional buckling calculation, Mcr :
-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
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)
is the fraction of the bending moment from the column extremities: 50.01274
637
kNm
kNm
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The elastic modulus : 3
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
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:
24220
880
mm
mm
b
h
The imperfection factor α will be chose according to Table 6.3:
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:
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)
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is the fraction of the bending moment from the column extremities: 0637
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
Flexion inertia moment around the Z axis:463.26627490 mm I z
Longitudinal elastic modulus: E = 210000 N/mm2.
Torsional moment of inertia: It=514614.75mm4
Warping inertial moment:
IW is the warping inertia (deformation inertia moment):
4
2
f z
w
t h I I
h cross section height; h=880mm
f t flange thickness;
mmt f 15
611
24
10808.494
15880mm326627490.6mm
mmmm I w
-according to EN1993-1-1-AN France; Chapter 2 (…4)
Length of the column part: L=2810mm
Shear modulus of rigidity: G=80800N/mm2
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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
The elastic modulus : 3
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
Calculation of the LT for appropriate non-dimensional slenderness LT will be determined with formula:
1
²²
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:
24220
880
mm
mm
b
h
The imperfection factor α will be chose according to Table 6.3:
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
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8.37.2.7 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:
a) for the top part of the column:
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,
(previously calculated)
1
44.95035371
32800011
44.95035371
3280001
1
1
,
,
N
N N
N
N
N
N
N
ycr
Ed y
ycr
Ed
y
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The myC will be calculated according to Table A.1:
Calculation of the 0 term:
0
,0
cr
y yeff
M
f W
-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2
34
max
, 533.317922953.455
1448234429mm
mm
mm
z
I W
y
yel
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
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
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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 ,,
²,
2
0
,
T cr
wt T cr
L
I E I G
I
A N
The mass moment of inertia 0 I
2
0 g z y z A I I I
Flexion inertia moment around the Y axis:467.1490580416 mm I y
Flexion inertia moment around the Z axis:417.26628854 mm I z
Cross section area:2
10850mm A
Distance between the section neutral axis and the section geometrical center: 0 g z
4442
0 151720927017.2662885467.1490580416 mmmmmm I I z A I I I z y g z y
- the buckling length, T cr L , ,
m L T cr 81.2,
Torsional moment of inertia:475.514614 mm I t
Working inertial moment:61110808.49 mm I w (previously calculated)
Longitudinal elastic modulus: E = 210000 MPa
Shear modulus of rigidity: G=80800MPa
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
For the top 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
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Therefore:
For the top part of the column:
1
11
11
224.01120.0534.0
224.01120.0
534.0
,,
2
0,
0,0,
4
,,
10
4
,,
1
0
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
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
11 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
Elastic modulus after the Y axis,3
,, 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
The 0mC coefficient is defined according to the Table A.2:
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
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b) for the bottom part of the column:
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
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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
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
mymymy a
a
C C C
11 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
Elastic modulus after the Y axis,3
,, 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
The 0mC coefficient is defined according to the Table A.2:
The bending moment is null at the end of the column, therefore: 01274
0
,sup
,inf kNm M
M
Ed
Ed
ycr
Ed my
N N C
,
0, 33.036.021.079.0
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Where:
kN N l
I E N
fy
y
ycr 37.9503544.95035371²
²,
(previously calculated)
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
Equivalent uniform moment factor, mLT C , calculation
- 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.
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8.37.2.8 Internal factor, yz k , calculation
a) for the top part of the column:
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
b) for the bottom part of the column:
823.0
62.6989347
3280001
1784.0
1,
N
N
N
N C k
z cr
Ed
y
mz yz
Note:
The software does not give the results of the lower section because it is not the most solicited segment.
8.37.2.9 Internal factor, zyk , calculation
a) for the top part of the column:
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
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b) for the bottom part of the column:
944.0
44.95035371
3280001
991.01949.0
1,
N
N
N
N C C k
ycr
Ed
z mLT my zy
Note:
The software does not give the results of the lower section because it is not the most solicited segment.
8.37.2.10 Internal factor, zz k , calculation
a) for the top part of the column:
815.0
62.6989347
3280001
991.0784.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
b) for the bottom part of the column:
815.0
62.6989347
3280001
991.0784.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
Note:
The software does not give the results of the lower section because it is not the most solicited segment.
8.37.2.11 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
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a) for the top part of the column:
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
b) for the bottom part of the column:
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:
The software does not give the results of the lower section because it is not the most solicited segment.
Finite elements modeling
■ Linear element: S beam,■ 6 nodes,
■ 1 linear element.
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y coefficient corresponding to non-dimensional slenderness y
Column subjected to axial and shear force to the top
y
z coefficient corresponding to non-dimensional slenderness z
Column subjected to axial and shear force to the top
z
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Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yz k
Column subjected to axial and shear force to the top
yz k
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Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zz k
Column subjected to axial and shear force to the top
zz k
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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
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
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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
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
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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
Bending and axial compression verification term depending of the Z bending moment over the Z axis: SMzz
Bending and axial compression verification term depending of the Z bending moment
over the Z axisSMzz
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8.37.2.12 Reference results
Result name Result description Reference value
y y coefficient corresponding to non-dimensional slenderness y
1
z z coefficient corresponding to non-dimensional slenderness z
0.81
yyk Internal factor, yyk 0.95
yz k Internal factor, yz k 0.82
zyk Internal factor, zyk 0.94
zyk Internal factor, zyk 0.82
SNy Bending and axial compression verification term depending of thecom ression effort over the Y axis
0.15
SMyy Bending and axial compression verification term depending of the Ybendin moment over the Y axis
1.72
SMyz Bending and axial compression verification term depending of the Zbendin moment over the Y axis
1.58
SNz Bending and axial compression verification term depending of thecom ression effort over the z axis
0.19
SMzy Bending and axial compression verification term depending of the Ybendin moment over the Z axis
1.71
SMzz Bending and axial compression verification term depending of the Zbendin moment over the Z axis
1.56
8.37.3Calculated results
Result name Result description Value Error
Xy Coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 %
Xz Coefficient corresponding to non-dimensional slenderness 0.811841adim
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 %
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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 test verifies a user defined cross section column.
The column is an “I symmetric” shape with: 260mm height; 150mm width; 7.1mm web thickness; 10.7mm 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.
The calculations are made according to Eurocode 3 French Annex.
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
8.38.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Fz=-328000N N; My=50000Nm; Mx=10000Nm;
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (mm).
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Units
Metric System
Geometrical properties
■ Column length: L=5620mm
■ Cross section area:2
06.4904 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
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength f y = 275 MPa,
■ Longitudinal elastic modulus: E = 210000 MPa.
■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation along X, Y and Z axis,
► Restraint of translation over the Y axis at half (z=2.81)
► Support at start point (z = 5.62) restrained in translation along X and Y axis, and restrained in rotation
along Z axis,
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Loading
The column is subjected to the following loadings:
■ External: Point load From X=0.00m and z=5.62m: FZ =-328000N; Mx=10000Nm and My=50000Nm
8.38.2.2 Cross section Class
According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
1253.006.179
30.45
sup
inf
Mpa
Mpa
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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
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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
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 1.
8.38.2.3 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:
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34.0
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:
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
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b) over the weak axis of the section, z-z:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
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
Cross section area:206.4904 mm A
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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
8.38.2.4 Lateral torsional buckling verification
a) for the top part of the column:
The elastic moment for lateral-torsional buckling calculation, Mcr :-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
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 diagramallure
²252.0423.0325.0
11
C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
31.15.025
251
,
, C kN
kN
M
M
top y
botom y
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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: 4
97.149294 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=2810mm
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
Calculation of the LT for appropriate non-dimensional slenderness LT will be determined with formula:
1²²
1
LT LT LT
LT
(6.56)
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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
The imperfection factor α will be chose according to Table 6.3:
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
b) for the bottom part of the column:
The elastic moment for lateral-torsional buckling calculation, Mcr :
-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
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)
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is the fraction of the bending moment from the column extremities: 0637
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: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=2810mm
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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
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
The imperfection factor α will be chose according to Table 6.3:
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
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8.38.2.5 Internal factor, yyk , calculation
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:
a) for the top part of the column:
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
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
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The myC will be calculated according to Table A.1:
Calculation of the 0 term:
0
,0
cr
y y pl
M
f W
-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2
318.501177 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
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
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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 ,,
²
1
,
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
Torsional moment of inertia: 497.149294 mm I t
Working inertial moment: 6
88.19351706542 mm I w
- the buckling length, T cr L , ,
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,
(previously calculated)
C1=1 for the top part of the column
For the top part of the column:
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
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Therefore:
For the top part of the column:
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
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 ymymymy
aaC 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
The 0mC coefficient is defined according to the Table A.2:
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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,
(previously calculated)
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
Equivalent uniform moment factor, mLT C , calculation
- 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
y (previously calculated)
956.01677.11
1677.1900.01900.0
11 0,0,
LT y
LT y
mymymya
aC C C
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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)
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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,
(previously calculated)
Therefore the yyk term corresponding to the top part of the column will be:
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
b) for the bottom part of the column:
yy
ycr
Ed
y
mLT my yyC
N
N C C k
1̀
1,
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-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
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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,
(previously calculated)
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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.
8.38.2.6 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,
-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
(previously calculated)
N l
I E N
fz
z z cr 51.1581706
²
²,
(previously calculated)
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
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-the myC term will be considered separately for each column part:
a) for the top part of the column:
956.0
1677.11
1677.1900.01900.0
1
1 0,0,
LT y
LT y
mymymy
a
aC C C
(previously calculated)
1749.0²661.0²831.0831.0
1
²²
1
LT LT LT
LT
(previously calculated)
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
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
(previously calculated)
-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
b) for the bottom part of the column:
885.01839.01
1839.0780.01780.0
11 0,0,
LT y
LT y
mymymya
aC C C
(previously calculated)
1804.0²569.0²752.0752.0
1
²²
1
LT
LT LT
LT
(previously calculated)
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Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
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
(previously calculated)
-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
Note: The software does not give the results of the lower section because it is not the most requested segment.
8.38.2.7 Internal factor, zyk , calculation
a) for the top part of the column:
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
586.0 z (previously calculated)
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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
,,
Nmmmm N mm f W M z z Rd z pl 33930039/27596.123381 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
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b) for the bottom part of the column:
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
586.0 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 y y Rd y pl 5.137823724/27518.501177 23
,,
Nmmmm N mm f W M z z Rd z pl 33930039/27596.123381 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 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
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-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
Note: The software does not give the results of the lower section because it is not the most requested segment.
8.38.2.8 Internal factor, zz k , calculation
a) for the top part of the column:
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
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
-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
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b) for the bottom part of the column:
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
Nmmmm N mm f W M y y Rd y pl 5.137823724/27518.501177 23
,,
-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
Note: The software does not give the results of the lower section because it is not the most requested segment.
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8.38.2.9 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
a) for the top part of the column:
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
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b) for the bottom part of the column:
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
Finite elements modeling
■ Linear element: S beam,
■ 7 nodes,
■ 1 linear element.
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y coefficient corresponding to non-dimensional slenderness y
Column subjected to axial and shear force to the top
y
z coefficient corresponding to non-dimensional slenderness z
Column subjected to axial and shear force to the top
z
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Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zz k
Column subjected to axial and shear force to the top
zz k
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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
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
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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
Bending and axial compression verification term depending of the compression effort over the Z axis: SNz
Bending and axial compression verification term depending of the compression effort
over the Z axisSNz
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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
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
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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
The elastic moment for lateral-torsional buckling calculation
The elastic moment for lateral-torsional buckling calculation
Mcr
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The appropriate non-dimensional slenderness
The appropriate non-dimensional slenderness
LT
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8.38.2.10 Reference results
a) for the top part of the column:
Result name Result description Reference value
y y coefficient corresponding to non-dimensional slenderness y
0.839
z z coefficient corresponding to non-dimensional slenderness z 0.586
yyk Internal factor, yyk 1.069
yz k Internal factor, yz k 0.750
zyk Internal factor, zyk 0.588
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
Mcr The elastic moment for lateral-torsional buckling calculation 315.36
LT The appropriate non-dimensional slenderness 0.749
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b) for the bottom part of the column:
Result name Result description Reference value
y y coefficient corresponding to non-dimensional slenderness y
0.839
z z coefficient corresponding to non-dimensional slenderness z 0.586
yyk Internal factor, yyk 0.977
yz k Internal factor, yz k 0.656
zyk Internal factor, zyk 0.566
zz k Internal factor, zz k 0.821
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
0.22
SMyz Bending and axial compression verification term depending of the Z
bendin moment over the Y axis
0.97
SNz Bending and axial compression verification term depending of thecom ression effort over the z axis
0.42
SMzy Bending and axial compression verification term depending of the Ybendin moment over the Z axis
0.13
SMzz Bending and axial compression verification term depending of the Zbendin moment over the Z axis
0.12
C1 Coefficient that depends of several parameters as: section properties;su ort conditions moment dia ram allure
1.77
Mcr The elastic moment for lateral-torsional buckling calculation 426.10
LT The appropriate non-dimensional slenderness 0.804
Note: The software does not give the results of the lower section because it is not the most requested segment.
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8.38.3Calculated results
Result name Result description Value Error
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 %
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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.
The calculations are made according to Eurocode 3 French Annex.
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.
The calculations are made according to Eurocode 3 French Annex.
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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.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
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.
8.41.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
► Fz = - 500 000 N,
► Fx = 5 000 N/ml,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
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Units
Metric System
Geometry
Below are described the column cross section characteristics:
■ Height: h = 300 mm,■ Width: b = 200 mm,
■ Thickness: t = 10 mm,
■ Outer radius: r = 15 mm,
■ Column height: L = 5000 mm,
■ Section area:A = 9490 mm2,
■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,
■ Partial factor for resistance of cross sections: 0.10 M .
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 (z = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (z = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External:
► Point load at z = 5.0: Fz = - 500 000 N,
► Uniformly distributed load: q = Fx = 5 000 N/ml
■ Internal: None.
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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
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 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.
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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.
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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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Finite elements results
Work ratio of the design resistance for uniform compression
Column subjected to bending and axial efforts
Work ratio – Fx
Work ratio of the design resistance for bending
Column subjected to bending and axial efforts
Work ratio – Oblique
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8.41.2.3 Reference results
Result name Result description Reference value
Work ratio – Fx Compression resistance work ratio [%] 22.42 %
Work ratio – Oblique Work ratio of the design resistance for bending 27.82 %
8.41.3Calculated results
Result name Result description Value Error
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 %
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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.
The calculations are made according to Eurocode 3 French Annex.
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.
The calculations are made according to Eurocode 3 French Annex.
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 cross section has an “I symmetric” shape with: 880mm height; 220mm width; 5mm center thickness; 15mmflange thickness; 0mm fillet radius and 0mm rounding radius.
The column is subjected to 328 kN axial compression force; 1274 kNm bending moment after the Y axis and 127.4kNm 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.
The calculations are made according to Eurocode 3 French Annex.
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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.
8.44.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Fz=-328000N N, Mx= 127400Nm; My=1274000Nm■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (mm).
Units
Metric System
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Geometrical properties
■ Column length: L=5620mm
■ Cross section area:2
10850mm A
■ Overall breadth: mmb 220
■ Flange thickness: mmt f 15
■ Root radius: mmr 0
■ Web thickness: mmt w 5
■ Depth of the web: mmhw 880
■ Elastic modulus after the Y axis, 33
, 1066.3387 mmW yel
■ Plastic modulus after the Y axis, 331062.3757 mmW y
■ Elastic modulus after the Z axis, 33
, 1008.242 mmW z el
■ Plastic modulus after the Z axis, 33, 1031.368 mmW z pl
■ Flexion inertia moment around the Y axis: Iy=149058.04x104mm
4
■ Flexion inertia moment around the Z axis: Iz=2662.89x104mm
4
■ Torsional moment of inertia: It=51.46x104mm
4
■ Working inertial moment: Iw=4979437.37x106mm
6
Materials properties
S275 steel material is used. The following characteristics are used:■ Yield strength f y = 275 MPa,
■ Longitudinal elastic modulus: E = 210000 MPa.
■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation along X, Y and Z axis,
► Support at the end point (z = 5.62) restrained in translation along Y and Z axis and restrained rotationalong X axis.
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Loading
The column is subjected to the following loadings:
■ External: Point load From X=0.00m and z=5.62m: FZ =--328000N; Mx=127400Nm and My=1274000Nm
8.44.2.2 Cross section Class
According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
1873.030.406
84.354
sup
inf
Mpa
Mpa
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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
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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
therefore the beam web is considered to be Class 4
-for beam flange:
9961.7
1
61.715
5.107
t
ct
c
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 4
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
-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
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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
-according EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1
1705
152880
mm
mmmm
t
c
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
325.19.239244.04.28
5
850
mm
mm
p
-according EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4
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
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-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
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8.44.2.5 Buckling verification
a) over the strong axis of the section, y-y:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
34.0
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 4 cross-sections:
ycr
yeff
y N
f A
,
*
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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
b) over the weak axis of the section, z-z:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
49.0
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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:
11
22
z z z
z
(6.49)
z the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
z cr
yeff z
N
f A
,
*
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 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
8.44.2.6 Lateral torsional buckling verification
-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 diagramallure
²252.0423.0325.0
11
C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
is the fraction of the bending moment from the column extremities: 01274
0
kNm
kNm
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77.101274
01 C
kNm
kNm
According to EN 1993-1-1-AN France; Chapter 3.2; Table 1
Flexion inertia moment around the Y axis:41448234429mm I y
Flexion inertia moment around the Z axis: 463.26627490 mm I z
Longitudinal elastic modulus: E = 210000 N/mm2.
Torsional moment of inertia: It=514614.75mm4
Warping inertial moment:
IW is the warping inertia (deformation inertia moment):
4
2
f z
w
t h I I
h cross section height; h=880mm
f t flange thickness; mmt f 15
611
24
10808.494
15880mm326627490.6mm
mmmm I w
-according to EN1993-1-1-AN France; Chapter 2 (…4)
Length of the column part: L=5620mm
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
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
The elastic modulus : 3
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
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
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The cross section buckling curve will be chose according to Table 6.4:
24220
880
mm
mm
b
h
The imperfection factor α will be chose according to Table 6.3:
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
8.44.2.7 Internal factor, yyk , calculation
The internal factor yyk corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:
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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,
(previously calculated)
1
44.95035371
32800011
44.950353713280001
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:
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Calculation of the 0 term:
0
,0
cr
y yeff
M
f W
-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2
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
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 ,,
²,
2
0
,
T cr
wt T cr
L
I E I G
I
A N
The mass moment of inertia 0 I
2
0 g z y z A I I I
Flexion inertia moment around the Y axis:467.1490580416 mm I
y
Flexion inertia moment around the Z axis:417.26628854 mm I z
Cross section area:210850mm A
Distance between the section neutral axis and the section geometrical center: 0 g z
4442
0 151720927017.2662885467.1490580416 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,
Torsional moment of inertia: 475.514614 mm I t
Working inertial moment:61110808.49 mm I w (previously calculated)
Longitudinal elastic modulus: E = 210000 MPa
Shear modulus of rigidity: G=80800MPa
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,,
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N l
I E N
fz
z z cr 905.1747336
²
²,
(previously calculated)
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
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
11 0,0,
yeff
eff
Ed
Ed y
yW
A
N
M
,
,
Elastic modulus after the Y axis,3
,, 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
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The 0mC coefficient is defined according to the Table A.2:
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:
kN N l
I E N
fy
y
ycr 37.9503544.95035371²
²,
(previously calculated)
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
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8.44.2.8 Internal factor, yz k , calculation
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
8.44.2.9 Internal factor, zyk , calculation
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
8.44.2.10 Internal factor, zz k , calculation
844.0
905.1747336
3280001
893.07677.0
1,
N
N
N
N C k
z cr
Ed
z mz zz
8.44.2.11 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
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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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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y coefficient corresponding to non-dimensional slenderness y
Column subjected to axial and shear force to the top
y
z coefficient corresponding to non-dimensional slenderness z
Column subjected to axial and shear force to the top
z
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Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yz k
Column subjected to axial and shear force to the top
yz k
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Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zz k
Column subjected to axial and shear force to the top
zz k
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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
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
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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
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
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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
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
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8.44.2.12 Reference results
Result name Result description Reference value
y y coefficient corresponding to non-dimensional slenderness y
1
z z coefficient corresponding to non-dimensional slenderness z 0.48
yyk Internal factor, yyk 1.011
yz k Internal factor, yz k 0.95
zyk Internal factor, zyk 0.902
zyk Internal factor, zyk 0.84
SNy Bending and axial compression verification term depending of thecom ression effort over the Y axis
0.15
SMyy Bending and axial compression verification term depending of the Ybendin moment over the Y axis
2.50
SMyz Bending and axial compression verification term depending of the Zbendin moment over the Y axis
1.81
SMzy Bending and axial compression verification term depending of the Ybendin moment over the Z axis
2.23
SMzz Bending and axial compression verification term depending of the Zbendin moment over the Z axis
1.61
8.44.3Calculated results
Result name Result description Value Error
Xy Coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 %
Xz Coefficient corresponding to non-dimensional slenderness 0.479165 adim 0.0000 %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 %
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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.
8.45.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
Fx = 50 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
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Units
Metric System
Geometry
Below are described the column cross section characteristics:
■ Height: h = 300 mm,■ Width: b = 200 mm,
■ Thickness: t = 10 mm,
■ Outer radius: r = 15 mm,
■ Column height: L = 5000 mm,
■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,
■ Partial factor for resistance of cross sections: 0.10 M .
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 (z = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (z = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External:
► Point load at z = 2.5: V= Fx = 50 000 N,
■ Internal: None.
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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
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 bending efforts, therefore the stresses distribution is like in the picture below:
Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class forcompressed parts. The picture below shows an extract from this table.
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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 tobending):
2510
10215230022
mm
mmmmmm
t
t r h
t
c
0.1235 y f
Therefore:
727225 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.
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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
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 = %64.55100224660000
2
500050000
1002100,,
Nmm
mm N
M
LV
M
M
Rd c Rd c
Ed
Finite elements modeling
■ Linear element: S beam,■ 6 nodes,
■ 1 linear element.
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Finite elements results
Work ratio of the design resistance for bending
Column subjected to a punctual horizontal load applied to the middle height
Work ratio – Oblique
8.45.2.3 Reference results
Result name Result description Reference value
Work ratio - Oblique Work ratio of the design resistance for bending [%] 55.64 %
8.45.3Calculated results
Result name Result description Value Error
Work ratio -Oblique
Work ratio of the design resistance for bending 55.6396 % -0.0007 %
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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.
The calculations are made according to Eurocode 3 French Annex.
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Geometry
Below are described the beam cross section characteristics:
■ Height: h = 300 mm,
■ Width: b = 200 mm,
■ Thickness: t = 10 mm,
■ Outer radius: r = 15 mm,
■ Beam length: L = 5000 mm,
■ Section area: A = 9490 mm2,
■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,
■ Plastic section modulus about z-z axis: Wpl,z = 721000 mm3,
■ Partial factor for resistance of cross sections: 0.10 M .
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 (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.
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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
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 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.
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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.
Because a cross-section is classified according to the least favorable classification of its compression elements(chapter 5.5.2(6) from EN 1993-1-1: 2001), this means that the cross-section is Class 1.
Determining the design plastic moment resistance
Before verifying for bi-axial bending a rectangular structural hollow section of uniform thickness, the design 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:
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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:
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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.
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Finite elements results
Work ratio of the design resistance for bending
Beam subjected to bending efforts
Work ratio – Oblique
8.47.2.3 Reference results
Result name Result description Reference value
Work ratio - Oblique Work ratio of the design resistance for biaxial bending 18.44 %
8.47.3Calculated results
Result name Result description Value Error
Work ratio -Oblique
Work ratio of the design resistance for biaxial bending 18.4437 % 0.2372 %
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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.
The beam is subjected to 12 kN axial compression force, 7 kN punctual vertical load applied to the middle of thebeam and 3 kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
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.
The calculations are made according to Eurocode 3 French Annex.
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.
The calculations are made according to Eurocode 3 French Annex.
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.
The calculations are made according to Eurocode 3 French Annex.
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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.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
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.
The beam is subjected to 20 kN axial compression force, 50 kN punctual vertical load applied to the middle of thebeam and 4 kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
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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.
The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.
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.
8.55.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear;
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
► Fz = - 50 000 N/ml,
► My,1 = 142 x 106Nmm,
► My,2 = - 113.6 x 106Nmm,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
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Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 260 mm,■ Flange width: b = 150 mm,
■ Flange thickness: tf = 10.7 mm,
■ Web thickness: tw = 7.1 mm,
■ Beam length: L = 5000 mm,
■ Section area: A = 5188 mm2,
■ Flexion inertia moment about the z axis: Iz = 6025866.46 mm4,
■ Torsional moment of inertia: It = 149294.97 mm4,
■ Warping constant: Iw = 93517065421.88 mm6,
■ Plastic modulus about the y axis: Wy = 501177.18 mm3
■ Partial factor for resistance of cross sections: 0.10 M .
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;
■ Shear modulus of rigidity: G=80800MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation along X, Y and Z axis,
► Support at end point (x = 5.00) restrained in translation along Y and Z axis and restrained rotationalong X axis.
■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ External:
► Uniformly distributed load over its length: q = Fz = -50 000 N/ml
► Bending moment at x=0: My,1 = 142 x 106
Nmm
► Bending moment at x=5: My,2 = - 113.6 x 106
Nmm,
■ Internal: None.
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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
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■ 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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Finite elements results
C1 parameter
Simply supported beam
C1
C2 parameter
Simply supported beam
C2
Elastic critical moment
Simply supported beam
Mcr
Reduction factor for lateral torsional buckling
Simply supported beam
XLT
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8.55.2.4 Reference results
Result name Result description Reference value
C1 C1 parameter [adim.] 1.074
C2 C2 parameter [adim.] 0.235
Mcr Elastic critical moment [kNm] 91.72 kNm
XLT Reduction factor for lateral torsional buckling [adim.] 0.467
8.55.3Calculated results
Result name Result description Value Error
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 %
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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.
The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.
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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.
8.57.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case (Q1) and load combination are used:
■ Exploitation loadings (category A), Q1:
► Fz = -1 000 000 N,
► Fx = 50 000 N/ml,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q1
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
Geometry
Below are described the column cross section characteristics:
■ Height: h = 600 mm,
■ Flange width: b = 220 mm,
■ Flange thickness: tf = 19 mm,
■ Column length: L = 5000 mm,
■ Section area: A = 15600 mm2 ,
■ Plastic section modulus about the strong y-y axis:3
,
3512000mmW y pl
,
■ Partial factor for resistance of cross sections: 0.10 M .
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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 conditionsThe boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (x = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External:
► Point load at Z = 5.0: N = FZ = -1 000 000 N,
► Uniformly distributed load: q = Fx = 50 000 N/ml
■ Internal: None.
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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:
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► 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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Work ratio of the design resistance for uniform compression
Column subjected to bending and axial force
Work ratio - Fx
Work ratio of the design resistance for oblique bending
Column subjected to bending and axial force
Work ratio - Oblique
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8.57.2.3 Reference results
Result name Result description Reference value
Work ratio - Fx Work ratio of the design resistance for uniform compression [%] 27.3 %
Work ratio - Oblique Work ratio of the design resistance for oblique bending [%] 79.9 %
8.57.3Calculated results
Result name Result description Value Error
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 %
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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.
8.58.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Q1 = -200 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
Geometrical properties
■ Cross section area: A=9490mm2
■ Flexion inertia moment around the Y axis: Iy=11819x104mm
4
■ Flexion inertia moment around the Z axis: Iz=6278x104mm
4
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.
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Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (x = 5.00).
■ Inner: None.■ Buckling lengths Lfy and Lfz are both imposed (10m)
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 5.0: FZ = N = -200 000 N,
■ Internal: None.
8.58.2.2 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 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)
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
(6.49)
the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
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
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It will be used the following buckling curve:
The imperfection factor corresponding to the appropriate buckling curve will be 0.21:
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)
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)
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 Z-Z axis
coefficient corresponding to non-dimensional slenderness will be determined from the relevant buckling curve
according to:
11
22
(6.49)
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the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
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
It will be used the following buckling curve:
The imperfection factor corresponding to the appropriate buckling curve will be 0.21:
473.1309.12.0309.121.015.0²)2.0(15.0 2
Therefore:
1465.0309.1473.1473.1
11
2222
1 M is a safety coefficient, 11 M
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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Finite elements results
Coefficient corresponding to non-dimensional slenderness after Y-Y axis
Buckling of a column subjected to compression force
Non-dimensional slenderness after Y-Y axis
Coefficient corresponding to non-dimensional slenderness after Z-Z axis
Buckling of a column subjected to compression force
Non-dimensional slenderness after Z-Z axis
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Ratio of the design normal force to design buckling resistance (strong inertia)
Buckling of a column subjected to compression force
Work ratio (y-y)
Ratio of the design normal force to design buckling resistance (weak inertia)
Buckling of a column subjected to compression force
Work ratio (z-z)
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8.58.2.4 Reference results
Result name Result description Reference value
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%
8.58.3Calculated results
Result name Result description Value Error
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 %
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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.
8.59.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
► Fz = -15 000 N/ml,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
UnitsMetric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 630 mm,
■ Flange width: b = 500 mm,
■ Flange thickness: tf = 18 mm,
■ Web thickness: tw = 8 mm,
■ Beam length: L = 5000 mm,
■ Section area: A = 22752 mm2,
■ Partial factor for resistance of cross sections: 0.10 M .
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Materials properties
S355 steel material is used. The following characteristics are used:
■ Yield strength f y = 355 MPa,
■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditionsThe boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation along X, Y and Z axis,
► Support at end point (x = 5.00) restrained in translation along Y, Z axis and restrained in rotation alongX axis.
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External:
► Uniformly distributed load: q = Fz = -15 000 N/ml,■ Internal: None.
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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.
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The buckling factor (k
) 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
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Design resistance for bending
For Class 4 cross-section, EN 1993-1-1: 2001 provides (6.15) formula in order to calculate the design resistance 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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
Work ratio of the design resistance for bending
Beam subjected to uniformly distributed load
Work ratio - Oblique
8.59.2.4 Reference results
Result name Result description Reference value
Work ratio - Oblique Design resistance for bending work ratio [%] 2.54 %
8.59.3Calculated results
Result name Result description Value Error
Work ratio -Oblique
Design resistance for bending work ratio 2.53713 % -0.1129 %
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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.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
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.
8.60.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
► Fz = -100 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
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Geometry
Below are described the column cross section characteristics:
■ Height: h = 300 mm,
■ Width: b = 200 mm,
■ Thickness: t = 10 mm,
■ Outer radius: r = 15 mm,
■ Column length: L = 5000 mm,
■ Section area: A = 9490 mm2 ,
■ Partial factor for resistance of cross sections: .
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength f y = 235 MPa,
■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditionsThe boundary conditions are described below:
■ Outer:
► Support at start point (z = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (z = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 5.0:
► N = Fz = -100 000 N,
■ Internal: None.
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8.60.2.2 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 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
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Therefore:
333325 t
c
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.
8.60.2.3 Reference results for calculating the compression resistance of the cross-section
The compression resistance for Class 1 cross-section is determined with formula (6.10) from EN 1993-1-1:2001.
Compression resistance of the cross section
For Class 1 cross-section, EN 1993-1-1: 2011 provides the following formula in order to calculate the 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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
Finite elements results
Work ratio of the design resistance for uniform compression
Column subjected to bending and axial force
Work ratio - Fx
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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.
8.61.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005;■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A): Q1 = -200 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (mm).
Units
Metric System
Geometrical properties
■ Cross section area: A=5380mm2
■ Flexion inertia moment around the Y axis: Iy=603.80x104mm
4
■ Flexion inertia moment around the Z axis: Iz=8356x104mm
4
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.
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Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis,
► Free at end point (x = 5.00).■ Inner: None.
■ Buckling lengths Lfy and Lfz are doth imposed with 10m value
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 5.0: FZ = N = -200 000 N,
■ Internal: None.
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933.0854.02.0854.021.015.0²)2.0(15.0 2
Therefore:
1764.0854.0933.0933.0
112222
1 M is a safety coefficient, 11 M
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)
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)
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 for Z-Z axis
coefficient corresponding to non-dimensional slenderness will be determined from the relevant buckling curve
according to:
1
122
(6.49)
the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
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/mm
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
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178.3610.125144
/2355380 22
N
mm N mm
N
f A
cr
y
²)2.0(15.0
It will be used the following buckling curve:
The imperfection factor corresponding to the appropriate buckling curve will be 0.34:
056.6178.32.0178.334.015.0²)2.0(15.0 2
Therefore:
1089.0178.3056.6056.6
112222
1 M is a safety coefficient, 11 M
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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Finite elements results
Coefficient corresponding to non-dimensional slenderness after Y-Y axis
Buckling of a column subjected to compression force
Non-dimensional slenderness after Y-Y axis
Ratio of the design normal force to design buckling resistance (strong inertia)
Buckling of a column subjected to compression force
Work ratio (y-y)
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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.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
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.
8.62.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
► Fz = -50 000 N/ml,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 300 mm,
■ Width: b = 200 mm,
■ Thickness: t = 10 mm,
■ Outer radius: r = 15 mm,
■ Beam length: L = 5000 mm,
■ Section area: A = 9490 mm2 ,
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■ Partial factor for resistance of cross sections: 0.10 M .
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength f y = 275 MPa,
■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (x = 0) restrained in translation along X, Y and Z axis,
► Support at end point (x = 5.00) restrained in translation along Y, Z axis and restrained in rotation alongX axis.
■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ 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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,■ 1 linear element.
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Finite elements results
Work ratio of the design shear resistance
Beam subjected to uniformly distributed load
Work ratio - Fz
8.62.2.3 Reference results
Result name Result description Reference value
Work ratio - Fz Work ratio of the design shear resistance [%] 13.83 %
8.62.3Calculated results
Result name Result description Value Error
Work ratio - Fz Shear resistance work ratio 13.8219 % 0.1587 %
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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.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
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.
8.63.2.1 Model description
■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load case and load combination are used:
■ Exploitation loadings (category A), Q:
► Fx = 200 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
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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 tobending):
2510
10215230022
mm
mmmmmm
t
t r h
t
c
0.1235
y f
Therefore:
727225 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.
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 2246600000.1
235956000 3
0
,
,
Work ratio
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 = %56.222100224660000
2
5000
2000001002100
,,
Nmm
mm
N
M
L
V
M
M
Rd c Rd c
Ed
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Finite elements results
Work ratio of the design shear resistance
Column subjected to a punctual horizontal load applied to the middle height
Work ratio - Fz
Work ratio of the design resistance for bending
Column subjected to a punctual horizontal load applied to the middle height
Work ratio – Oblique
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8.63.2.4 Reference results
Result name Result description Reference value
Work ratio - Fz Work ratio of the design plastic shear resistance [%] 25.89 %
Work ratio - Oblique Work ratio of the design resistance for bending [%] 222.56 %
8.63.3Calculated results
Result name Result description Value Error
Work ratio - Fz Work ratio of the design plastic shear resistance 25.8793 % -0.0413 %
Work ratio -Oblique
Work ratio of the design resistance for bending 222.559 % -0.0007 %
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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.
9.4.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test E.3;
■ Analysis type: static linear (plane problem);
■ Element type: linear;
■ Distance between adjacent purlins (span): d = 1.8 m.
The following load cases and load combination are used:
■ Loadings from the structure: G = 550 N/m2;
■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2;
■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2;
All loads will be projected on the purlin direction since the roof slope is 17°.
Simply supported purlin subjected to loadings
Units
Metric System
Geometry
Purlin cross section characteristics:
■ Height: h = 0.20 m,
■ Width: b = 0.10 m,
■ Length: L = 3.5 m,
■ Section area: A = 0.02 m2 ,
■ Elastic section modulus about the strong axis, y: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
.
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Materials properties
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.
Boundary conditionsThe boundary conditions are described below:
■ Outer:
► Support at start point (z=0) restrained in translation along X, Y, Z;
► Support at end point (z = 3.5) restrained in translation along X, Y, Z and restrained in rotation along X.
■ Inner: None.
Loading
The purlin is subjected to the following projected loadings (at ultimate limit state):
■ External:
► Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x
sin17° = 1101.22 N/m,► Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/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
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■ 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
Finite elements modeling
■ Linear element: S beam,
■ 5 nodes,
■ 1 linear element.
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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 [%]
9.4.2.3 Reference results
Result name Result description Reference value
Smy Design bending stress about y axis [Pa] 8281441 Pa
SMz Design bending stress about z axis [Pa] 5063793 Pa
Work ratio Maximum work ratio for strength verification [%] 71.2 %
9.4.3 Calculated results
Result name Result description Value Error
Stress SMy Design bending stress about y axis 8.47672e+006 Pa 0.0000 %
Stress SMz Design bending stress about z axis 5.18319e+006 Pa -0.0000 %
Work ratio Maximum work ratio for strength verification 71.153 % -0.0000 %
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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.
9.5.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test F.1;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
Timber column with fixed base
Units
Metric System
Geometry
Below are described the column cross section characteristics:
■ Depth: h = 0.60 m,
■ Width: b = 0.20 m,
■ Section area: A = 0.12 m2
■ Height: H = 5.00 m
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Residual cross section
Column with fixed base exposed to fire for 60 minutes
Afi
9.5.2.3 Reference results
Result name Result description Reference value
Afi Residual cross section [m2
] 0.056202 m2
9.5.3 Calculated results
Result name Result description Value Error
Afi Residual cross section 0.056202 m² -0.0000 %
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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.
9.6.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test F.2;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load cases and load combination are used:
► Loadings from the structure: G = 500 N/m2,
► Snow load: S = 700 N/m2,
► Frequent combination of actions: CFQ = 1.0 x G + 0.5 x S = 850 N/m2
Purlin with 3 supports
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ 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 ,
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Materials properties
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,
■ Characteristic bending strength: f m,k = 24 x 106 Pa,
■ Density: = 350 kg/m3,
■ Design charring rate (softwood): n = 0.8 x 10-3 m/min,
■ Service class 1.
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (X = 0) restrained in translation along X, Y and Z,
► Support at middle point (X = 3.30) restrained in translation along X, Y and Z,
► Support at end point (X = 6.60) restrained in translation along X, Y, Z and restrained in rotation alongX.
■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ External:
► Uniformly distributed load: q = CFQ x d = 850 N/m2 x 1.5 m = 1275 N/m,
■ Internal: None.
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
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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
Finite elements modeling
■ Linear element: S beam,
■ 8 nodes,
■ 1 linear element.
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9.6.2.3 Reference results
Result name Result description Reference value
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 %
9.6.3 Calculated results
Result name Result description Value Error
Afi Residual area 0.002197 m² -0.0000 %
Stress Design bending stress for residual cross-section 2.75626e+007 Pa -0.0001 %
Work ratio Maximum work ratio for fire verification 91.8752 % 0.0000 %
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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.
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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.
9.8.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test C;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load cases and load combination are used:
■ Loadings from the structure: G = 0.5 kN/m2,
■ Exploitation loadings (category A): Q = 1.5 kN/m2,
■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2
■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q
■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q
Simply supported beam
Units
Metric System
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,
■ Section area: A = 15.0 x 10-3
m2 ,
■ Elastic section modulus about the strong axis y:3
22
0005.06
20.0075.0
6m
hbW y
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Materials properties
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,
■ Characteristic bending strength: f m,k = 24 x 106 Pa,
■ Service class 1.
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (z=0) restrained in translation along X, Y and Z,
► Support at end point (z = 4.5) restrained in translation along X, Y, Z and restrained in rotation along X.
■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ External:
Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,
■ Internal: None.
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.
Reference solution for ultimate limit state verification
Before calculating the reference solution (design bending stress, design bending strength and work ratios) it is
necessary to determine some parameters involved in calculations (kmod, M, kh, ksys, km).
■ Modification factor for duration of load (medium term) and moisture content:
kmod = 0.8 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties:
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 (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
■ Design bending strength:
f m,d = Pak k k
f h sys
M
k m
66mod, 10769.140.10.1
3.1
8.01024
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■ Work ratio according to formulae 6.11 from EN 1995-1-1 norm:
0.1,
, d m
d m
f
■ Work ratio according to formulae 6.12 from EN 1995-1-1 norm:
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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Work ratio diagram
Simply supported beam subjected to bending
Strength work ratio
9.8.2.3 Reference results
Result name Result description Reference value
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
9.8.3 Calculated results
Result name Result description Value Error
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 %
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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.
9.9.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test E.2;
■ Analysis type: static linear (plane problem);
■ Element type: linear;
■ Distance between adjacent rafters (span): d = 0.5 m.
The following load cases and load combination are used:
■ Loadings from the structure: G = 450 N/m2;
■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2;
■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m2;
All loads will be projected on the rafter direction, since its slope is 50% (26.6°).
Simply supported rafter subjected to projected loadings
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ 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
.
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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
(according to relation 6.25 from EN 1995-1-1)
Reference solution for ultimate limit state verification
Before calculating the reference solution (the design compressive stress, the design bending stress and the 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).
■ 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 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
■ Design bending strength:
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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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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 [%]
9.9.2.3 Reference results
Result name Result description Reference value
SFx Design compressive stress [Pa] 219122 Pa
SMy Design bending stress [Pa] 8212744 Pa
kcrit Kcrit factor 0.602
Work ratio Lateral-torsional stability work ratio [%] 76 %
9.9.3 Calculated results
Result name Result description Value Error
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 %
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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.
9.10.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test D;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
The following load cases and load combination are used:
■ Loadings from the structure: G = 0.5 kN/m2,
■ Exploitation loadings (category A): Q = 1.5 kN/m2,
■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2
Simply supported beam
Units
Metric System
Geometry
Beam cross section characteristics:
■ Height: h = 0.225 m,
■ Width: b = 0.075 m,
■ Length: L = 5.00 m,
■ Distance between adjacent beams (span): d = 0.5 m,
■ Section area: A = 16.875 x 10-3
m2 ,
Materials properties
Rectangular solid timber C24 is used. The following characteristics are used in relation to this material:
■ Characteristic shear strength: f v,k = 2.5 x 106 Pa,
■ Service class 1.
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Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (z=0) restrained in translation along X, Y and Z,
► Support at end point (z = 5.00) restrained in translation along X, Y, Z and restrained in rotation along X.
■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ External:
► Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,
■ Internal: None.
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.
Reference solution for ultimate limit state verification
Before calculating the reference solution (design shear stress, design shear strength and work ratio) it is necessary to
determine some parameters involved in calculations (kmod, M, kcr , kf , beff , heff ).
■ Modification factor for duration of load (medium term) and moisture content:
kmod = 0.8 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties:
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
■ Work ratio according to formulae 6.13 from EN 1995-1-1 norm:
0.1,
d v
d
f
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9.10.2.3 Reference results
Result name Result description Reference value
Fz Shear force [kN] 3.65625 kN
Stress S_d Design shear stress [Pa] 485074.63 Pa
Work ratio S_d Shear work ratio (6.13) [%] 32 %
9.10.3Calculated results
Result name Result description Value Error
Fz Shear force -3.65625 kN -0.0000 %
Stress S_d Design shear stress 485075 Pa -0.0000 %
Working ratio S_d Shear strength work ratio 31.5299 % -1.4692 %
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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.
9.11.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test A;
■ Analysis type: static linear (plane problem);
■ Element type: linear.Column with fixed base
Units
Metric System
Geometry
Cross section characteristics:
■ Height: h = 0.122 m,
■ Width: b = 0.036m,
■ Section area: A = 43.92 x 10-4
m2
■ I = 5.4475 x 10-6
m4.
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Materials properties
Rectangular solid timber C24 is used. The following characteristics are used in relation to this material:
■ Longitudinal elastic modulus: E = 1.1 x 1010
Pa,
■ Characteristic tensile strength along the grain: f t,0,k = 14 x 106 Pa,
■ Service class 2.
Boundary conditions
The boundary conditions:
■ Outer:
Fixed at base (z = 0),
Free at top (z = 5),
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ 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
(according to relation 6.1 from EN 1995-1-1)
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Work ratio SFx diagram
Column with fixed base, subjected to tension force
Work ratio SFx
9.11.2.3 Reference results
Result name Result description Reference value
t,0,d Design tensile stress [Pa] 2276867.03 Pa
SFx Work ratio [%] 18 %
9.11.3Calculated results
Result name Result description Value Error
Stress SFx Design tensile stress 2.27687e+006 Pa -0.0001 %
Work ratio Work ratio 18.0704 % 0.3910 %
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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.
9.12.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test B;
■ Analysis type: static linear (plane problem);
■ Element type: linear.
Simply supported column
Units
Metric System
Geometry
Column cross section characteristics:
■ Height: h = 0.15 m,
■ Width: b = 0.10 m,
■ Section area: A = 15.0 x 10-3
m2
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Materials properties
Rectangular solid timber C18 is used. The following characteristics are used in relation to this material:
■ Longitudinal elastic modulus: E = 0.9 x 1010
Pa,
■ 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
The column is subjected to the following loadings:
■ 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.
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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
(according to relation 6.26 from EN 1995-1-1)
■ 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).
■ Modification factor for duration of load (medium term) and moisture content:
kmod = 0.8 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties: M = 1.3
■ Design compressive stress (induced by the applied forces):
c,0,d = A
N
■ Design compressive strength:
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
(according to relation 6.35 from EN 1995-1-1)
Finite elements modeling
■ Linear element: S beam,
■ 4 nodes,
■ 1 linear element.
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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.
9.13.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test E;
■ Analysis type: static linear (plane problem);
■ Element type: linear;
■ Distance between adjacent rafters (span): d = 0.5 m.
The following load cases and load combination are used:
■ Loadings from the structure: G = 450 N/m2;
■ Snow load (structure is located at an altitude > 1000m above sea level): S = 900 N/m2;
■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m2;
■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x S;
■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.2 x S.
All loads will be projected on the rafter direction since its slope is 50% (26.6°).
Simply supported rafter subjected to projected loadings
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ 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
.
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Materials properties
Rectangular solid timber C24 is used. The following characteristics are used in relation to this material:
■ Characteristic tensile strength along the grain: f t,0,k = 14 x 106 Pa,
■ Characteristic bending strength: f m,k = 24 x 106 Pa,
■ Service class 2.
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (z=0) restrained in translation along 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.
Reference solution for ultimate limit state verification
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).
■ 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 (“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
■ 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
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■ 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
(according to relation 6.1 from EN 1995-1-1)
■ 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
■ Design bending strength:
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 formulae 6.17 from EN 1995-1-1 norm:
1,,
,,
,,
,,
,0,
,0, d z m
d z m
m
d ym
d ym
d t
d t
f k
f f
■ Work ratio according to formulae 6.18 from EN 1995-1-1 norm:
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
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■ 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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
SFx work ratio diagram
Simply supported beam subjected to tensile forces
Work ratio SFx
Strength work ratio diagram
Simply supported beam subjected to combined stresses
Strength work ratio
Instantaneous deflection w inst (Q)
Simply supported beam subjected to snow loads
Instantaneous deflection winst(Q) [m]
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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
Instantaneous deflection winst(CQ) [m]
Net deflection w net,fin
Simply supported beam subjected to characteristic load combination of actions
Instantaneous deflection winst(CQ) [m]
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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.
9.14.2.1 Model description
■ Reference: Guide de validation Eurocode 5, test E.4;
■ Analysis type: static linear (plane problem);
■ Element type: linear;
■ Distance between adjacent purlins (span): d = 1.8 m.
The following load cases and load combination are used:
■ The ultimate limit state (ULS) combination is: CULS = 1.35 x G + 1.5 x S + 0.9 x W;
■ Loadings from the structure: G = 550 N/m2;
■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2;
■ Axial compression force due to wind effect on the supporting elements: W = 15000 N;
■ Uniformly distributed load corresponding to the ultimate limit state combination:
Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2.
All loads will be projected on the purlin direction since its slope is 30% (17°).
Simply supported purlin subjected to loadings
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ 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
,
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■ Elastic section modulus about the strong axis, z:3
22
000333.06
20.01.0
6m
hbW z
.
Materials properties
Rectangular solid timber C24 is used. The followings characteristics are used in relation to this material:
■ Characteristic compressive strength along the grain: f c,0,k = 21 x 106 Pa,
■ Characteristic bending strength: f m,k = 24 x 106 Pa,
■ Longitudinal elastic modulus: E = 1.1 x 1010
Pa,
■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 1010
Pa,
■ Service class 2.
Boundary conditions
The boundary conditions are described below:
■ Outer:
► Support at start point (z = 0) restrained in translation along X, Y, Z;
► Support at end point (z = 3.50) restrained in translation along Y, Z and restrained in rotation along X.
■ Inner: None.
Loading
The purlin is subjected to the following projected loadings (corresponding to the ultimate limit state combination):
■ External:
► Axial compressive load: N =0.9 x W = 13500 N;
► Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x
sin17° = 1101.22 N/m,
► Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x
cos17° = 3601.92 N/m,
■ Internal: None.
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
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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
(according to relation 6.26 from EN 1995-1-1)
2
,
2,
1
yrel y y
yc
k k k
(according to relation 6.25 from EN 1995-1-1)
Reference solution for ultimate limit state verification
Before calculating the reference solution (the maximum work ratio for stability verification based on formulae (6.23)and (6.24) from EN 1995-1-1) it is necessary to determine the design compressive stress, the design 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):
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kmod = 1.1 (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)
■ Design compressive strength:
f c,0,d = Pak
f M
k c
66mod,0, 10769.17
3.1
1.11021
■ Design bending strength:
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
Finite elements modeling
■ Linear element: S beam,
■ 6 nodes,
■ 1 linear element.
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Instability factor, k cy
Simply supported purlin subjected to biaxial bending and axial compression
Kc,y
Instability factor, k cz
Simply supported purlin subjected to biaxial bending and axial compression
Kc,z
Maximum work ratio for stability verification
Simply supported purlin subjected to biaxial bending and axial compression
Work ratio [%]
9.14.2.3 Reference results
Result name Result description Reference value
Kc,y Instability factor, kc,y 0.67
Kc,z Instability factor, kc,z 0.21
Work ratio Maximum work ratio for stability verification [%] 71.1 %
9.14.3Calculated results
Result name Result description Value Error
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 %
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