ad validation guide vol2 2016 en
DESCRIPTION
Graitec Advance Design Validation Guide Vol.2 2016 ENTRANSCRIPT
VALIDATION GUIDE
Advance Design
Validation Guide
Version: 2016
Tests passed on: 17 June 2015
Number of tests: 606
INTRODUCTION Before being officially released, each version of GRAITEC software, including Advance Design, undergoes a series of validation tests. This validation is performed in parallel and in addition to manual testing and beta testing, in order to obtain the "operational version" status. This document contains a description of the automatic tests, highlighting the theoretical background and the results we have obtained using the current software release.
Usually, a test is made of a reference (independent from the specific software version tested), a transformation (a calculation or a data processing scenario), a result (given by the specific software version tested) and a difference usually measured in percentage as a drift from a set of reference values. Depending on the cases, the used reference is either a theoretical calculation done manually, a sample taken from the technical literature, or the result of a previous version considered as good by experience.
Starting with version 2012, Graitec Advance has made significant steps forward in term of quality management by extending the scope and automating the testing process. While in previous versions, the tests were always about the calculation results which were compared to a reference set, starting with version 2012, tests have been extended to user interface behavior, import/export procedures, etc. The next major improvement is the capacity to pass the tests automatically. These current tests have obviously been passed on the “operational version”, but they are actually passed on a daily basis during the development process, which helps improve the daily quality by solving potential issues, immediately after they have been introduced in the code.
In the field of structural analysis and design, software users must keep in mind that the results highly depend on the modeling (especially when dealing with finite elements) and on the settings of the numerous assumptions and options available in the software. A software package cannot replace engineers experience and analysis. Despite all our efforts in term of quality management, we cannot guaranty the correct behavior and the validity of the results issued by Advance Design in any situation. With this validation guide, we are providing a set of concrete test cases showing the behavior of Advance Design in various areas and various conditions. The tests cover a wide field of expertise: modeling, climatic load generation according to Eurocode 1, combinations management, meshing, finite element calculation, reinforced concrete design according to Eurocode 2, steel member design according to Eurocode 3, steel connection design according to Eurocode 3, timber member design according to Eurocode 5, seismic analysis according to Eurocode 8, report generation, import / export procedures and user interface behavior.
We hope that this guide will highly contribute to the knowledge and the confidence you are placing in Advance Design.
Manuel LIEDOT
Chief Product Office
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Contents
13 STEEL DESIGN ....................................................................................................................... 9
13.1 Verifying the calculation results for steel cables (TTAD #11623) ................................................................ 10
13.2 Generating the shape sheet by system (TTAD #11471) ............................................................................. 10
13.3 Verifying shape sheet on S275 beam (TTAD #11731) ................................................................................ 10
13.4 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770) ......................... 10
13.5 Verifying the cross section optimization according to EC3 (TTAD #11516) ................................................ 10
13.6 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522) ................................. 11
13.7 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545) ........................................... 11
13.8 Verifying the shape sheet results for a column (TTAD #11550) .................................................................. 11
13.9 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method .................................. 11
13.10 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method ........................... 11
13.11 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method ............................. 12
13.12 Verifying the buckling coefficient Xy on a class 2 section ......................................................................... 12
13.13 EC3 Test 7: Class section classification and compression resistance for an IPE600 column ................... 13
13.14 Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873) . 19
13.15 EC3 Test 28: Verifying an user defined I section class 1, column fixed on base and without any other restraint ... 20
13.16 EC3 Test 22: Verifying the lateral torsional buckling of a IPE300 beam ................................................... 50
13.17 EC3 Test 23: Verifying a IPE400 column for compression, shear, bending moment, buckling, lateral torsional buckling and bending and axial compression ......................................................................................... 56
13.18 EC3 Test 20: Verifying the buckling resistance of a RC3020100 column ................................................. 92
13.19 EC3 Test 13: Verifying the resistance of a rectangular hollow section column subjected to bending and shear efforts ......................................................................................................................................................... 100
13.20 EC3 test 10: Verifying the classification and the bending resistance of a welded built-up beam ............ 107
13.21 EC3 test 11: Cross section classification and compression resistance verification of a rectangular hollow section column ..................................................................................................................................................... 113
13.22 EC3 Test 25: Verifying an user defined I section class 4 column fixed on the bottom and with a displacement restraint at 2.81m from the bottom ................................................................................................. 118
13.23 EC3 Test 19: Verifying the buckling resistance for a IPE300 column ...................................................... 152
13.24 EC3 Test 12: Verifying the design plastic shear resistance of a rectangular hollow section beam ......... 159
13.25 EC3 Test 27: Verifying an user defined I section class 3 beam simply supported with a displacement restraint .............................................................................................................................................. 162
13.26 EC3 Test 26: Verifying an user defined I section class 3 column fixed on the bottom ............................ 186
13.27 EC3 test 9: Verifying the classification and the compression resistance of a welded built-up column .... 209
13.28 EC3 Test 21: Verifying the buckling resistance of a CHS219.1x6.3H column ......................................... 216
13.29 EC3 Test 24: Verifying an user defined I section class 4 column fixed on the bottom and without any other restraint ............................................................................................................................... 222
13.30 EC3 Test 29: Verifying an user defined I section class 1, column hinged on base and restrained on top for the X, Y translation and Z rotation ........................................................................................................................ 248
13.31 EC3 Test 31: Verifying IPE450 column fixed on base subjected to axial compression and bending moment, both applied on top ................................................................................................................................ 289
13.32 EC3 Test 15: Verifying a rectangular hollow section column subjected to bending and axial efforts ....... 290
13.33 EC3 Test 33: Verifying UPN300 simple supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and punctual vertical force by Z axis ................................................................................. 297
13.34 EC3 Test 35: Verifying C310x30.8 class 4 cantilever, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load applied on the free end ......................................................... 297
13.35 EC3 Test 37: Verifying RHS350x150x8.5H class 3 column, loaded with centric compression, punctual lateral load and bending moment, all applied to the top of the column ................................................................. 297
13.36 EC3 Test 30: Verifying IPE300 beam, simply supported, loaded with centric compression and uniform linear efforts by Y and Z axis ................................................................................................................................ 297
13.37 EC3 Test 34: Verifying C310x30.8 class 3beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle ......................................................................... 298
13.38 EC3 Test 36: Verifying RHS300x150x9H class 1 simply supported beam, loaded with centric compression, uniform linear horizontal efforts and a vertical punctual load in the middle ..................................... 298
13.39 EC3 Test 14: Verifying the bending resistance of a rectangular hollow section column made of S235 steel ........................................................................................................................................................... 299
13.40 EC3 Test 32: Verifying IPE600 simple supported beam, loaded with centric compression and uniform linear efforts by Y and Z axis ................................................................................................................................ 305
13.41 EC3 Test 16: Verifying a simply supported rectangular hollow section beam subjected to biaxial bending ................................................................................................................................................................ 306
13.42 EC3 Test 18: Verifying a simply supported circular hollow section element subjected to torsional efforts 313
13.43 EC3 Test 39: Verifying CHS323.9x6.3H class 2 beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle ......................................................................... 313
13.44 EC3 Test 38: Verifying RHS350x150x5H class 4 column, loaded with centric compression, punctual horizontal force by Y and a bending moment, all applied to the top ..................................................................... 313
13.45 EC3 Test 45: Comparing the shear resistance of a welded built-up beam made from different steel materials ............................................................................................................................................................... 314
13.46 EC3 Test 43: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange .................................................................................................. 318
13.47 EC3 Test 41: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange .................................................................................................. 318
13.48 EC3 test 8: Verifying the classification and the resistance of a column subjected to bending and axial load ............................................................................................................................................................. 319
13.49 EC3 Test 17: Verifying a simply supported rectangular hollow section beam subjected to torsional efforts ........ 325
13.50 EC3 Test 44: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange ................................................................................................. 326
13.51 EC3 Test 40: Verifying CHS508x8H class 3, simply supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle ................................................... 332
13.52 EC3 Test 42: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange ................................................................................................. 332
13.53 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533) .......................... 332
13.54 Changing the steel design template for a linear element (TTAD #12491) ............................................... 332
13.55 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389) ..................................................................................................................................................... 332
13.56 EC3: Verifying the buckling length results (TTAD #11550)...................................................................... 333
13.57 EC3 fire verification: Verifying the work ratios after performing an optimization for steel profiles (TTAD #11975) 333
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13.58 EC3 test 4: Class section classification and bending moment verification of an IPE300 column ............ 334
13.59 EC3 Test 2: Class section classification and shear verification of an IPE300 beam subjected to linear uniform loading ..................................................................................................................................................... 340
13.60 EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 beam 347
13.61 EC3 Test 1: Class section classification and compression verification of an IPE300 column ................. 354
13.62 EC3 Test 5: Class section classification and combined axial force with bending moment verification of an IPE300 column ..................................................................................................................................................... 360
13.63 EC3 Test 3: Class section classification, shear and bending moment verification of an IPE300 column 366
14 TIMBER DESIGN ................................................................................................................. 375
14.1 EC5: Verifying a timber purlin subjected to oblique bending ..................................................................... 376
14.2 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes ............................. 380
14.3 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending ..................................... 383
14.4 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression ......................................................................................................................................................... 388
14.5 EC5: Shear verification for a simply supported timber beam .................................................................... 393
14.6 Modifying the "Design experts" properties for timber linear elements (TTAD #12259) .............................. 393
14.7 Verifying the timber elements shape sheet (TTAD #12337) ...................................................................... 393
14.8 Verifying the units display in the timber shape sheet (TTAD #12445) ....................................................... 393
14.9 EC5: Verifying a timber beam subjected to simple bending ...................................................................... 394
14.10 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression ................................ 398
14.11 EC5: Verifying a timber column subjected to tensile forces .................................................................... 403
14.12 EC5: Verifying a timber column subjected to compression forces........................................................... 406
14.13 EC5: Verifying a timber beam subjected to combined bending and axial tension ................................... 410
14.14 EC5: Verifying a C24 timber beam subjected to shear force ................................................................... 415
13 Steel design
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13.1 Verifying the calculation results for steel cables (TTAD #11623)
Test ID: 3560
Test status: Passed
13.1.1 Description Performs the finite elements calculation and the steel calculation for a steel model with cables (D4) and a static nonlinear case. Generates the steel analysis report: data and results.
13.2 Generating the shape sheet by system (TTAD #11471)
Test ID: 3575
Test status: Passed
13.2.1 Description Generates shape sheets by system, on a model with 2 systems.
13.3 Verifying shape sheet on S275 beam (TTAD #11731)
Test ID: 3434
Test status: Passed
13.3.1 Description Performs the steel calculation for two horizontal bars and generates the shape sheets report.
The bars have cross sections from different catalogues (1016x305x487 UKB and UKB1016x305x487). They are made of the same material (S275); each is subjected to a -500.00 kN linear horizontal dead load and has two rigid supports at both ends.
13.4 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770)
Test ID: 3406
Test status: Passed
13.4.1 Description Performs the steel calculation for two vertical bars with different thicknesses and generates the shape sheets report.
The bars are made of the same material (S275 H - EN 10210-1), have rectangular hollow cross sections, but with different thicknesses (R80*40/4.1 and R80*40/3.9). Each is subjected to a -150.00 kN vertical live load and has a rigid support.
13.5 Verifying the cross section optimization according to EC3 (TTAD #11516)
Test ID: 3620
Test status: Passed
13.5.1 Description Verifies the cross section optimization of a steel element, according to Eurocodes 3.
Performs the finite elements calculation and the steel calculation. Generates the "Envelopes and shapes optimization" report.
The steel bar has a IPE360 cross section, a rigid hinge support at one end and a rigid support with translation restraints on X, Y and Z and rotation restraint on X. Two loads are applied: a punctual dead load of -1.00 kN on FZ and a punctual live load of -40.00 kN on FZ.
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13.6 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522)
Test ID: 3612
Test status: Passed
13.6.1 Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet results report.
The structure is a simple vault consisting of three steel elements (S235 material, IPEA240 cross section) with two rigid fixed supports. The loads applied on the structure: self weight and a linear live load of -10.00kN on FZ.
13.7 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545)
Test ID: 3641
Test status: Passed
13.7.1 Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a horizontal steel element, verifies the cross section class.
The steel element (S235 material, IPE300 cross section) has a rigid fixed support at one end and a rigid support with translation restraints on Y and Z and rotation restraint on X at the other end. A linear live load of -10.00 kN on FX and a punctual live load of 1000 kN on FX are applied.
13.8 Verifying the shape sheet results for a column (TTAD #11550)
Test ID: 3640
Test status: Passed
13.8.1 Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a vertical steel element.
The steel element (S235 material, IPE300 cross section) has a rigid fixed support. A vertical live load of -200.00 kN is applied.
13.9 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method
Test ID: 3819
Test status: Passed
13.9.1 Description Verifies the buckling length for a steel portal frame, using the kA kB method, according to Eurocodes 3.
Generates the "Buckling and lateral-torsional buckling lengths" report.
13.10 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method
Test ID: 3814
Test status: Passed
13.10.1Description Verifies the buckling length for a steel portal frame with one level, using the roA roB method, according to CM66.
Generates the "Buckling and lateral-torsional buckling lengths" report.
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13.11 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method
Test ID: 3813
Test status: Passed
13.11.1Description Verifies the buckling length for a steel portal frame with one level, using the kA kB method, according to CM66.
Generates the "Buckling and lateral-torsional buckling lengths" report.
13.12 Verifying the buckling coefficient Xy on a class 2 section
Test ID: 4443
Test status: Passed
13.12.1Description Performs the finite elements calculation and the steel elements calculation. Verifies the buckling coefficient Xy on a class 2 section and generates the shape sheets report.
The model consists of a vertical linear element (I26*0.71+15*1.07 cross section and S275 material) with a rigid hinge support at the base and a rigid support with translation restraints on X and Y and rotation restraint on Z, at the top. A punctual live load is applied.
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13.13 EC3 Test 7: Class section classification and compression resistance for an IPE600 column
Test ID: 5620
Test status: Passed
13.13.1Description Verifies the classification and the compression resistance for an IPE 600 column made of S235 steel. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.13.2Background Classification and verification under compression efforts for an IPE 600 column made of S235 steel. The column is fixed at its base and free on the top. The column is subjected to a compression force (100 000 N) applied at its top. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.13.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = -100 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (x = 5.00).
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■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 5.0: N = Fz = -100 000 N, ■ Internal: None.
13.13.2.2Reference results for calculating the cross section class The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.
In this case, the column is subjected to a punctual compression load, therefore the stresses distribution is like in the picture below:
Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the web class.
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The section geometrical properties are described in the picture below:
The web class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1:
83.4212
224219600=
×−×−=
mmmmmmmm
tc
0.1235==
yfε
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Therefore:
424283.42 =>= εtc
This means that the column web is Class 4.
Table 5.2 - sheet 2, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the flanges class.
The top flange class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 2:
21.419
2/)22412220(=
×−−=
mmmmmmmm
tc
0.1235==
yfε
Therefore:
9921.4 =≤= εtc
This means that the top column flange is Class 1. Having the same dimensions, the bottom column flange is also Class 1.
A cross-section is classified according to the least favorable classification of its compression elements (chapter 5.5.2(6) from EN 1993-1-1: 2001).
According to the calculation above, the column section have Class 4 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 4.
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13.13.2.3Reference results for calculating the compression resistance of the cross-section The compression resistance for Class 4 cross-section is determined with the formula (6.11) from EN 1993-1-1:2001.
In order to verify the compression resistance for Class 4 cross-section, it is necessary to determine the effective area of the cross-section.
The effective area of the cross section takes into account the reduction factor, ρ, which is applying in this case only for the web of the IPE 600 cross-section.
The following parameters have to be determined in order to calculate the reduction factor: the buckling factor and the stress ratio, and the plate modified slenderness. They will be calculated considering only the cross-section web.
The buckling factor (kσ) and the stress ratio(Ψ)
Taking into account that the stress distribution on web is linear, the stress ratio becomes:
0.11
2 ==σσψ
0.4=σk
The plate modified slenderness (λp)
The formula used to determine the plate modified slenderness is:
( ) 754.00.40.14.28
12/2421926004.28
/=
×××−×−
==mmmmmmmm
ktb
pσε
λ
The reduction factor (ρ)
Because λp > 0.673, the reduction factor has the following formula:
( )0.1
3055.02 ≤
+×−=
p
p
λψλ
ρ
Effective area
The effective area is determined considering the following:
( ) ( ) ( ) 275.1522312242192600939.01156001 mmtbAA weff =××−×−×−−=××−−= ρ
Compression resistance of the cross section
For Class 4 cross-section, EN 1993-1-1: 2011 provides the following formula in order to calculate the compression resistance of the cross-section:
NMPammfAN
M
yeffRdc 3577581
0.123575.15223 2
0, =
×=
×=
γ
Work ratio
Work ratio = %79518.21003577581100000100
,
=×=×NN
NN
Rdc
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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
13.13.2.4Reference results
Result name Result description Reference value Work ratio - Fx Compression resistance work ratio [%] 2.79518 %
13.13.3Calculated results Result name Result description Value Error Work ratio - Fx Compression resistance work ratio 2.79495 % -0.0082 %
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13.14 Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873)
Test ID: 4289
Test status: Passed
13.14.1Description Verifies the steel shape optimization when using sections from Advance Steel Profiles database. Performs the finite elements calculation and the steel elements calculation and generates the steel shapes report.
The structure consists of columns with UKC152x152x23 cross section, beams with UKB254x102x22 cross section and roof beams with UKB127x76x13 cross section. Dead loads and live loads are applied on the structure.
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13.15 EC3 Test 28: Verifying an user defined I section class 1, column fixed on base and without any other restraint
Test ID: 5720
Test status: Passed
13.15.1Description The test verifies a user defined cross section column.
The cross section has an “I symmetric” shape with: 260mm height; 150mm width; 7.1mm center thickness; 10.7mm flange thickness; 0mm fillet radius and 0mm rounding radius. The column is made of S275 steel.
The column is subjected to 328 kN compression axial force, 10 kNm bending moment over the X axis and 50 kNm bending moment over the Y axis. All the efforts are applied on the top of the column.
The calculations are made according to Eurocode 3 French Annex.
13.15.2Background An I260*7.1+150*10.7 shaped column subjected to compression and bending, made from S275 steel. The column has a 260x7.1mm web and 150x10.7mm flanges. The column is fixed at its base. The column is subjected to an axial compression load -328000 N, a 10000Nm bending moment after the X axis and a 50000Nm bending moment after the Y axis.
This test was evaluated by the French control office SOCOTEC.
13.15.2.1Model 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: 206.4904 mmA =
■ Overall breadth: mmb 150=
■ Flange thickness: mmt f 70.10=
■ Root radius: mmr 0=
■ Web thickness: mmtw 10.7=
■ Depth of the web: mmhw 260=
■ Elastic modulus after the Y axis, 3, 63.445717 mmW yel =
■ Plastic modulus after the Y axis, 318.501177 mmWy =
■ Elastic modulus after the Z axis, 3, 89.80344 mmW zel =
■ Plastic modulus after the Z axis, 3, 96.123381 mmW zpl =
■ Flexion inertia moment around the Y axis: 464.57943291 mmI y =
■ Flexion inertia moment around the Z axis: 446.6025866 mmIz =
■ Torsional moment of inertia: 497.149294 mmIt =
■ Working inertial moment: 688.19351706542 mmIw =
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength fy = 275 MPa, ■ Longitudinal elastic modulus: E = 210000 MPa. ■ Shear modulus of rigidity: G=80800MPa
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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 start point (x = 0) restrained in translation along X and Y axis, and restrained inrotation along Z
axis,
Loading
The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =-328000N; Mx=10000Nm and My=50000Nm
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13.15.2.2Cross section Class According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
1253.006.179
30.45
sup
inf −>−=−
==Mpa
Mpaσσψ
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=×
=
=×
=⇒=
++
==73.190
36.22406.1796.238
175.4836.224
3.456.238
36.2246.238
06.17930.4506.17930.45 y
xyxyx
5.080.06.238
73.1906.238
>===xα
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924.0275235235
===yf
ε
93.3818.013
924.0396113
3966.33924.0
61.331.7
7.102260=
−××
=−×
×≤=⇒
=
=×−
=α
ε
ε tc
mmmmmm
tc
therefore the beam
web is considered to be Class 1
-for beam flange:
316.8924.0968.6
924.0
68.67.10
21.7150
=×≤=⇒
=
=
−
=tc
tc
ε
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 1.
13.15.2.3Buckling verification a) over the strong axis of the section, y-y:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
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34.0=α
Coefficient corresponding to non-dimensional slenderness after Y-Y axis:
yχ coefficient corresponding to non-dimensional slenderness yλ will be determined from the relevant buckling
curve according to:
1122
≤−Φ+Φ
=yyy
yλ
χ (6.49)
yλ the non-dimensional slenderness corresponding to Class 1 cross-sections:
ycr
yy N
fA
,
*=λ
Cross section area: 206.4904 mmA =
Flexion inertia moment around the Y axis: 464.57943291 mmI y =
( ) kNNmm
mmmmNl
IEN
fy
yycr 33.380295.3802327
²562064.57943291/210000²
²² 42
, ==××
=××
=ππ
5956.095.3802327
/27506.4904 22
,
=×
=×
=N
mmNmmN
fA
ycr
yyλ
[ ] ( )[ ] 7446.05956.02.05956.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ
839.0
1
839.05956.07446.07446.0
112222 =⇒
≤
=−+
=−Φ+Φ
=y
y
yyy
yχ
χ
λχ
b) over the weak axis of the section, z-z:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
ADVANCE DESIGN VALIDATION GUIDE
27
49.0=α
Coefficient corresponding to non-dimensional slenderness after Z-Z axis:
zχ coefficient corresponding to non-dimensional slenderness zλ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=zzz
zλ
χ (6.49)
zλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
zcr
yz
NfA
,
*=λ
Flexion inertia moment around the Z axis: 446.6025866 mmI z =
Cross section area: 206.4904 mmA =
( ) kNNmm
mmmmNl
IENfz
zzcr 43.39563.395426
²562046.6025866/210000²
²² 42
, ==××
=××
=ππ
ADVANCE DESIGN VALIDATION GUIDE
28
847.163.395426
/27506.4904 22
,
=×
=×
=N
mmNmmN
fA
zcr
yzλ
[ ] ( )[ ] 609.2847.12.0847.149.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ
225.01
225.0847.1609.2609.2
112222 =⇒
≤
=−+
=−Φ+Φ
=z
z
zzz
zχ
χλ
χ
13.15.2.4Lateral torsional buckling verification The elastic moment for lateral-torsional buckling calculation, Mcr:
-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
z
t
z
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3) -where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
77.10500
1,
, =⇒=== CM
M
topy
botomyψ
ADVANCE DESIGN VALIDATION GUIDE
29
Flexion inertia moment around the Y axis: 464.57943291 mmI y =
Flexion inertia moment around the Z axis: 446.6025866 mmI z =
Torsional moment of inertia: 497.149294 mmIt =
Working inertial moment: 688.19351706542 mmI w =
Yield strength fy = 275 MPa,
Longitudinal elastic modulus: E = 210000 MPa.
Shear modulus of rigidity: G=80800MPa
Warping inertial moment (recalculated):
IW is the warping inertia (deformation inertia moment):
( )4
2fz
w
thII
−×=
h cross section height; mmh 260=
ft flange thickness; mmt f 7.10=
( ) 624
093627638294
7.1026046.6025866 mmmmmmmmI w =−×
=
-according to EN1993-1-1-AN France; Chapter 2 (…4)
Length of the column: L=5620mm
( )( )
kNmNmm
mmNmmmmN
mmmmNmmmmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
150181.150184702
58.21463.39542677.146.6025866/210000
97.149294/80800562046.6025866
09362763829
562046.6025866/21000077.1
²²
²²
422
422
4
6
2
422
1
==
=××=××
××+×
×××
×=××××
+×××
×=
π
ππ
π
958.01.150184702
/27518.501177 23, =
×==
NmmmmNmm
MfW
cr
yyplLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=
The cross section buckling curve will be chose according to Table 6.4:
2733.1150260
≤==mmmm
bh
ADVANCE DESIGN VALIDATION GUIDE
30
The imperfection factor α will be chose according to Table 6.3:
49.0=α
( )[ ] ( )[ ] 145.1²958.02.0958.049.015.0²2.015.0 =+−×+×=+−×+×= LTLTLTLT λλαφ
1564.0²958.0²145.1145.1
1²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
13.15.2.5Internal factor, yyk , calculation
The internal factor yyk corresponding to a Class 4 section will be calculated according to Annex A, Table a.1, and will be
calculated separately for the two column parts separate by the middle torsional lateral restraint:
yy
ycr
Ed
ymLTmyyy C
NNCCk 1̀
1,
×−
××=µ
ADVANCE DESIGN VALIDATION GUIDE
31
ycr
Edy
ycr
Ed
y
NN
NN
,
,
1
1
×−
−
=χ
µ
839.0=yχ (previously calculated)
kNNEd 328=
kNNl
IEN
fy
yycr 33.380295.3802327
²²
, ==××
=π
(previously calculated)
985.0
95.3802327328000839.01
95.38023273280001
1
1
,
, =×−
−=
×−
−
=
NN
NN
NN
NN
ycr
Edy
ycr
Ed
y
χµ
The myC will be calculated according to Table A.1:
ADVANCE DESIGN VALIDATION GUIDE
32
Calculation of the 0λ term:
0
,0
cr
yypl
MfW ×
=λ
-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2 318.501177 mmWy =
The calculation the 0crM will be calculated using 11 =C and 02 =C , therefore:
( )( )
kNmNmm
mmNmmmmN
mmmmNmmmm
mm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
85.8427.84850646
58.21463.395426146.6025866/210000
97.149294/80800562046.6025866
88.19351706542
562046.6025866/2100001
²²
²²
422
422
4
6
2
422
1
==
=××=××
××+×
×××
×=××××
+×××
×=
π
ππ
π
274.127.84850646
/27518.501177 23,
0 =×
==Nmm
mmNmmM
fW
cr
yyplλ
Calculation of the 4
,,1 1120.0
−×
−××
TFcr
Ed
zcr
Ed
NN
NNC term:
Where:
-for a symmetrical section for the both axis, TcrTFcr NN ,, =
××+××=
²1
,
2
0,
Tcr
wtTcr L
IEIGI
N π
The mass moment of inertia 0I
44420 1.6396915846.602586664.57943291 mmmmmmIIzAIII zygzy =+=+=×++=
Torsional moment of inertia: 497.149294 mmIt =
Working inertial moment: 688.19351706542 mmIw =
- the buckling length, TcrL , ,
mL Tcr 62.5, =
( )N
mmmmmmNmmmmN
mmmmN Tcr
607.1395246²5620
88.19351706542/21000097.149294/808001.63969158
06.4904 62242
4
2
,
=
=
××+××=
π
NNEd 328000=
NNN TcrTFcr 607.1395246,, ==
ADVANCE DESIGN VALIDATION GUIDE
33
( ) kNNmm
mmmmNl
IENfz
zzcr 43.39563.395426
²562046.6025866/210000²
²² 42
, ==××
=××
=ππ
(previously calculated)
C1=1 for the top part of the column
For the top part of the column:
120.0
607.13952463280001
63.3954263280001120.01120.0 44
,,1
=
=
−×
−××=
−×
−××
NN
NN
NN
NNC
TFcr
Ed
zcr
Ed
Therefore:
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
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
a
aCCC
NN
NNC
NN
NNC
ε
ε
λ
λ
The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.
The coefficient must be calculated considering the column over the entire height.
( )LTy
LTymymymy a
aCCC
×+
××−+=
ε
ε
11 0,0,
yeff
eff
Ed
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 3,, 63.445717 mmWW yeffyel ==
677.163.445717
06.4904328000
10503
26
,
, =××
=×=mm
mmN
NmmWA
NM
yeff
eff
Ed
Edyyξ
1997.064.57943291
97.14929411 4
4
≈=−=−=mm
mmIIa
y
tLT
ADVANCE DESIGN VALIDATION GUIDE
34
The 0mC coefficient is defined according to the Table A.2:
The bending moment in null at one end of the column, therefore: 0=ψ
( )ycr
Ed
ycr
Edmy N
NNNC
,,0, 33.036.079.033.036.021.079.0 ××−=×−×+×+= ψψ
Where:
kNNl
IEN
fy
yycr 33.380295.3802327
²²
, ==××
=π
(previously calculated)
NNEd 328000=
780.095.3802327
32800033.036.079.00, =××−=N
NCmy
( ) ( ) 904.01677.11
1677.1780.01780.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
Equivalent uniform moment factor, mLTC , calculation
256.2
1
256.2
607.13952463280001
63.3954263280001
997.0904.0
11
2
,,
2
=⇒
≥
=
−×
−
×=
=
−×
−
×=
mLT
mLT
Tcr
Ed
zcr
Ed
LTmymLT
C
CN
NN
N
NN
NN
aCC
ADVANCE DESIGN VALIDATION GUIDE
35
The yyC coefficient is defined according to the Table A.1, Auxiliary terms:
ypl
yelLTplmy
ymy
yyyy W
WbnC
wC
wwC
,
,2
max2
max2 6.16.12)1(1 ≥
−×
××−××−×−+=
−−
λλ
1534.0/27596.123381
10000000/27518.501177564.0
50000000274.1997.05.0
5.05.0
23232
,
,
,
,2
0
,,
,
,,
,2
0
=
=×
×××
×××=
××
×××××=×
××××=
−−
mmNmmNmm
mmNmmNmm
fWM
fWM
aMM
MM
abyzpl
Edz
yyplLT
EdyLT
Rdzpl
Edz
RdyplLT
EdyLTLT χ
λχ
λ
5.1124.163.44571718.501177
3
3
,
, ≤===mmmm
WW
wyel
yply
243.0
1/27506.4904
32800022
1
=×
==mmNmm
NNNn
M
Rk
Edpl
γ
( ) 847.1847.1;5956.0max;maxmax ==
=
−−
zy λλλ
857.01534.0243.0²847.1²904.0124.16.1847.1²904.0
124.16.12)1124.1(1 =
−×
××−××−×−+=yyC
ADVANCE DESIGN VALIDATION GUIDE
36
889.0889.018.50117763.445717
857.0
,
,
3
3
,
, =⇒
≥
==
=
yy
ypl
yelyy
ypl
yel
yy
C
WW
C
mmmm
WWC
kNNl
IEN
fy
yycr 33.380295.3802327
²²
, ==××
=π
(previously calculated)
Therefore the yyk term corresponding to the top part of the column will be:
47.2889.01
95.38023273280001
985.0256.2904.01̀
1,
=×−
××=×−
××=
NNC
NNCCk
yy
ycr
Ed
ymLTmyyy
µ
13.15.2.6Internal factor, yzk , calculation
y
z
yz
zcr
Ed
ymzyz w
wC
NNCk ×××
−×= 6.01
1,
µ
( ) ( ) 691.063.395426
32800033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
985.0
95.3802327328000839.01
95.38023273280001
1
1
,
, =×−
−=
×−
−
=
NN
NN
NN
NN
ycr
Edy
ycr
Ed
y
χµ
(previously calculated)
kNNl
IENfz
zzcr 43.39563.395426
²²
, ==××
=π
(previously calculated)
−×
×
×−×−+=
−
LTplz
mzzyz cn
wCwC 5
2
max2
142)1(1 λ
Rdyplltmy
Edy
z
LTLT MCM
ac,,
,4
2
0
510
×××
+××=
−
−
χλ
λ
997.064.57943291
97.14929411 4
4
=−=−=mm
mmIIa
y
tLT (previously calculated)
274.127.84850646
/27518.501177 23,
0 =×
==Nmm
mmNmmM
fW
cr
yyplλ (previously calculated)
847.163.395426
/27506.4904 22
,
=×
=×
=N
mmNmmN
fA
zcr
yzλ (previously calculated)
ADVANCE DESIGN VALIDATION GUIDE
37
NmM Edy 50000, =
( ) 904.01
1 0,0, =×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ (previously calculated)
1564.0²958.0²145.1145.1
1²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ (previously calculated)
NmmmmNmmfWM yyRdypl 5.137823724/27518.501177 23,, =×=×=
664.05.137823724564.0904.0
50000000847.15
²274.1997.010
510
4
,,
,4
2
0
=××
×+
××=
=××
×+
××=−
−
NmmNmm
MCM
acRdyplltmy
Edy
z
LTLT χλ
λ
−×
×
×−×−+=
−
LTplz
mzzyz cn
wCwC 5
2
max2
142)1(1 λ
5.15.1
5.1536.189.8034496.123381
3
3
,
,
=⇒
≤
≤===z
z
zel
zplz w
wmmmm
WW
w
kNNl
IENfz
zzcr 43.39563.395426
²²
, ==××
=π
(previously calculated)
( ) ( ) 691.063.395426
32800033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
( ) 847.1847.1;5956.0max;maxmax ==
=
−−
zy λλλ
243.0
1
==
M
Rk
Edpl N
Nn
γ
(previously calculated)
532.0691.0243.05.1
847.1691.0142)15.1(1
142)1(1
5
22
5
2
max2
=
−×
××−×−+=
=
−×
×
×−×−+=−
LTplz
mzzyz cn
wCwC λ
20.5124.1
5.16.0532.01
63.3954263280001
985.0691.06.01
1,
=×××−
×=×××−
×=
NNw
wC
NNCk
y
z
yz
zcr
Ed
ymzyz
µ
ADVANCE DESIGN VALIDATION GUIDE
38
13.15.2.7Internal factor, zyk , calculation
y
z
zy
ycr
Ed
zmLTmyzy w
wC
NNCCk ×××
−××= 6.01
1,
µ
210.0
63.395426328000225.01
63.3954263280001
1
1
,
, =×−
−=
×−
−=
NN
NN
NN
NN
zcr
Edz
zcr
Ed
z
χµ
225.0=zχ (previously calculated)
ypl
yel
z
yLTpl
y
myyzy W
Www
dnw
CwC
,
,5
2
max2
6.0142)1(1 ××≥
−×
×
×−×−+=
−
λ
NmmmmNmmfWM zzRdzpl 33930039/27596.123381 23,, =×=×=
771.033930039691.0
100000005.137823724564.0904.0
50000000847.11.0
274.1997.02
1.02
4
,,
,
,,
,4
0
=
=×
×××
×+
××=
=×
×××
×+
××=−
−
NmmNmm
NmmNmm
MCM
MCM
adRdzplmz
Edz
RdyplLTmy
Edy
z
LTLT χλ
λ
310.0771.0243.0124.1
847.1904.0142)1124.1(1
142)1(1
5
22
5
2
max2
=−×
××−×−+=
=
−×
×
×−×−+=
−
LTply
myyzy dn
wC
wCλ
462.018.50117763.445717
5.1124.16.06.0 3
3
,
, =××=××mmmm
WW
ww
ypl
yel
z
y
462.0
462.06.0
310.0
6.0
,
,
,
,
=⇒
=××
=
××≥
zy
ypl
yel
z
y
zy
ypl
yel
z
yzy
C
WW
ww
CWW
ww
C
256.2=mLTC (previously calculated)
529.05.1
124.16.0462.01
95.38023273280001
210.0256.2904.06.01
1,
=×××−
××=×××−
××=
NNw
wC
NNCCk
z
y
zy
ycr
Ed
zmLTmyzy
µ
ADVANCE DESIGN VALIDATION GUIDE
39
13.15.2.8Internal factor, zzk , calculation
zz
zcr
Ed
zmzzz C
NNCk 1
1,
×−
×=µ
zpl
zelLTplmz
zmz
zzzz W
WenC
wC
wwC
,
,max
2max
2 6.16.12)1(1 ≥
−×
××−××−×−+= λλ
131.0/27518.501177564.0904.0
50000000847.11.0
274.1997.07.1
1.07.1
234
,,
,4
0
=×××
×+
××=
=××
×+
××=−
−
mmNmmNmm
MCM
aeRdyplltmy
Edy
z
LTLT χλ
λ
926.0131.0243.0847.1691.05.16.1847.1691.0
5.16.12)15.1(1
6.16.12)1(1
222
2max
2max
2
=
−×
××−××−×−+=
=
−×
××−××−×−+= LTplmz
zmz
zzzz enC
wC
wwC λλ
919.0926.01
63.3954263280001
210.0691.01
1,
=×−
×=×−
×=
NNC
NNCk
zz
zcr
Ed
zmzzz
µ
13.15.2.9Bending and axial compression verification
∆+×+
×
∆+×+
×
∆+×+
×
∆+×+
×
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rkz
RdzEdzzz
M
RkyLT
RdyEdyzy
M
Rkz
Ed
M
Rkz
RdzEdzyz
M
RkyLT
RdyEdyyy
M
Rky
Ed
MMM
kMMM
kNN
MMM
kMMM
kNN
γγχ
γχ
γγχ
γχ
iyRk AfN ×=
ADVANCE DESIGN VALIDATION GUIDE
40
69.127.034.008.1
1/27596.123381
1010919.0
1/27518.501177564.0
1050529.0
1/27506.4904225.0
328000
41.353.159.129.0
1/27596.123381
101020.5
1/27518.501177564.0
1050479.2
1/27506.4904839.0
328000
23
6
23
6
22
23
6
23
6
22
=++=×
××+
+×
×
××+
××
=++=×
××+
+×
×
××+
××
mmNmmNmm
mmNmmNmm
mmNmmN
mmNmmNmm
mmNmmNmm
mmNmmN
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
41
yχ coefficient corresponding to non-dimensional slenderness yλ
Column subjected to axial and shear force to the top
yχ
zχ coefficient corresponding to non-dimensional slenderness zλ
Column subjected to axial and shear force to the top
zχ
ADVANCE DESIGN VALIDATION GUIDE
42
Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yzk
Column subjected to axial and shear force to the top
yzk
ADVANCE DESIGN VALIDATION GUIDE
43
Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zzk
Column subjected to axial and shear force to the top
zzk
ADVANCE DESIGN VALIDATION GUIDE
44
Bending and axial compression verification term depending of the compression effort over the Y axis: SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
SNy
Bending and axial compression verification term depending of the Y bending moment over the Y axis: SMyy
Bending and axial compression verification term depending of the Y bending moment over the Y axis
SMyy
ADVANCE DESIGN VALIDATION GUIDE
45
Bending and axial compression verification term depending of the Z bending moment over the Y axis: SMyz
Bending and axial compression verification term depending of the Z bending moment over the Y axis
SMyz
Bending and axial compression verification term depending of the compression effort over the Z axis: SNz
Bending and axial compression verification term depending of the compression effort over the Z axis
SNz
ADVANCE DESIGN VALIDATION GUIDE
46
Bending and axial compression verification term depending of the Y bending moment over the Z axis: SMzy
Bending and axial compression verification term depending of the Y bending moment over the Z axis
SMzy
Bending and axial compression verification term depending of the Z bending moment over the Z axis: SMzz
Bending and axial compression verification term depending of the Z bending moment over the Z axis
SMzz
ADVANCE DESIGN VALIDATION GUIDE
47
Coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
Coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
C1
The elastic moment for lateral-torsional buckling calculation
The elastic moment for lateral-torsional buckling calculation Mcr
ADVANCE DESIGN VALIDATION GUIDE
48
The appropriate non-dimensional slenderness
The appropriate non-dimensional slenderness
LTχ
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13.15.2.10Reference results
Result name Result description Reference value
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
0.839
zχ zχ coefficient corresponding to non-dimensional slenderness zλ
0.225
yyk Internal factor, yyk
2.47
yzk Internal factor, yzk
5.20
zyk Internal factor, zyk
0.529
zzk Internal factor, zyk
0.919
SNy
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 bending moment over the Y axis
1.59
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
1.53
SNz
Bending and axial compression verification term depending of the compression effort over the z axis
1.08
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
0.34
SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
0.27 C1 Coefficient that depends of several parameters as: section properties;
support conditions; moment diagram allure 1.77
Mcr The elastic moment for lateral-torsional buckling calculation 150.18
LTχ The appropriate non-dimensional slenderness 0.564
Work ratio Stability work ratio (bending and axial compression verification) [%] 341 %
13.15.3Calculated results Result name Result description Value Error Xy Coefficient corresponding to non-dimensional slenderness
after Y-Y axis 0.839285 adim
0.0340 %
Xz Coefficient corresponding to non-dimensional slenderness after Z-Z axis
0.224656 adim
-0.1529 %
Kyy Internal factor kyy 2.47232 adim
0.0939 %
Kyz Internal factor kyz 5.20929 adim
0.1787 %
Kzy Internal factor kzy 0.525982 adim
-0.5705 %
Kzz Internal factor kzz 0.942941 adim
2.6051 %
Work ratio Stability work ratio 341.352 % 0.0000 %
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13.16 EC3 Test 22: Verifying the lateral torsional buckling of a IPE300 beam
Test ID: 5702
Test status: Passed
13.16.1Description The test verifies the lateral torsional buckling of a IPE300 beam made of S235 steel.
The calculations are made according to Eurocode 3, French Annex.
13.16.2Background Lateral torsional buckling verification for an unrestrained IPE300 beam subjected to axis bending efforts, made of S235 steel. The beam is simply supported. The beam is subjected to a uniform vertical load (10 000 N) applied constantly on the entire length. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.16.2.1Model 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=8356.00x104mm4 ■ Flexion inertia moment around the Z axis: Iz=603.80x104 mm4
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Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at the end point (z = 3.00) restrained in translation along Y and Z axis and restrained rotation
along X axis. ■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Linear load From X=0.00m to X=5.00m: FZ = N = -10 000 N, ■ Internal: None.
13.16.2.2Buckling in the strong inertia of the profile (along Y-Y) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling moment resistance of the bended element (Mb,Rd) from the designed value moment (MEd) produced by the linear force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.
%100100,
≤×Rdb
Ed
MM
(6.46)
Cross-class classification is made according to Table 5.2
■ for beam web:
7272014.351
014.351.76.248
=×≤=⇒
=
==ε
ε tc
mmmm
tc
therefore the beam web is considered to be
Class 1
■ for beam flange:
99276.51
276.57.1045.56
=×≤=⇒
=
==ε
ε tc
mmmm
tc
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:
22150300
≤==mmmm
bh
the buckling curve about Y-Y will be considered “a”
The design buckling resistance moment against lateral-torsional buckling is calculated according the next formula:
1,
M
yyLTRdb
fWM
γχ ××
= (6.55)
Where:
LTχ reduction factor for lateral-torsional buckling:
1122
≤−Φ+Φ
=LTLTLT
LTλ
χ (6.56)
Where:
[ ]2)2.0(15.0 LTLTLTLT λλαφ +−×+×=
LTα represents the imperfection factor; 21.0=LTα
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LTλ the non-dimensional slenderness corresponding:
cr
yyLT
MfW ×
=λ
Mcr is the elastic critical moment for lateral-torsional buckling, is based on gross cross sectional properties and takes into account the loading conditions, the real moment distribution and the lateral restraints.
( )( ) ( )
×−×+
×××××
+×
×
×××
×= ggZ
tz
z
w
w
z
z
zcr zCzC
IEIGLk
II
kk
LkIECM 2
222
2
2
2
1 ππ
(1)
according to EN 1993-1-1-AN France; AN.3 Chapter 2
Where:
E is the Young’s module: E=210000N/mm2
G is the share modulus: G=80770N/mm2
Iz is the inertia of bending about the minor axis Z: Iz=603.8 x104mm4
It is the torsional inertia: It=20.12x104mm4
IW is the warping inertia (deformation inertia moment): Iw=12.59x1010mm6
L is the beam length: L=5000mm
kz and kw are buckling coefficients
zg is the distance between the point of load application and the share center (which coincide with the center of gravity)
C1 and C2 are coefficients depending on the load variation over the beam length
If the bending moment is linear along the bar, if there are no transversal loads or if the transverse load is applied to the center, then C2xxg=0 and the Mcr formula become:
z
t
z
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
The C1 coefficient is chosen from the Table2 of the EN 1993-1-1-AN France; AN.3 Chapter 3.3:
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kNmNmmmmNmmmmN
mmmmNmmmmmm
mmmmmmNM cr
61.1308.13060942480.23044.50057813.11080.603/210000²
1012.20/80770)²5000(1080.603
1059.12
)²5000(1080.603/210000²13.1
442
442
44
610
442
==××=
=×××
×××+
××
×
××××
×=
π
π
therefore:
063.18.130609424
/235104.628 233
=××
=×
=Nmm
mmNmmM
fW
cr
yyLTλ
[ ] [ ] 156.1063.1)2.0063.1(21.015.0)2.0(15.0 22=+−×+×=+−×+×= LTLTLTLT λλαφ
1621.0²063.1²156.1156.1
1
²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
Finite elements results
The steel calculation results can be found in the Shape Sheet window. The “Class” tab shows the classification of the cross section and the effective characteristics (not applicable in this case, as the cross section is class 1).
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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
13.16.2.3Reference results
Result name Result description Reference value
LTχ Lateral-torsional buckling coefficient [adim.] 0.621
Mcr Elastic critical moment for lateral-torsional buckling [kNm] 130.61
13.16.3Calculated results Result name Result description Value Error XLT Lateral-torsional buckling coefficient 0.621588
adim 0.0947 %
Mcr Elastic critical moment for lateral-torsional buckling 130.699 kN*m
0.0681 %
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13.17 EC3 Test 23: Verifying a IPE400 column for compression, shear, bending moment, buckling, lateral torsional buckling and bending and axial compression
Test ID: 5703
Test status: Passed
13.17.1Description The test verfies a IPE400 column, made of S275 steel, subjected to compression, shear, bending moment, buckling, lateral torsional buckling and bending and axial compression
A lateral restraint is placed at 3m from the base.
The verifications are made according to Eurocode 3 French Annex.
13.17.2Background Unrestrained IPE400 column subjected to compression and bending, made from S275 steel. The column is fixed at its base and free on the top end. A lateral restraint is placed at 3m from the base. The column is subjected to an axial compression load (-125000 N) applied and to a lateral load after the X global axis (28330N). Both loads are applied on the top end of the column. The dead load will be neglected. The results will be compared with the ones obtained by the CTIM n4-2006.
This test was evaluated by the French control office SOCOTEC.
13.17.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2005; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A): Q1x= 28330 N, Q1z= -125000 N ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
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■ Cross section dimensions are in millimeters (mm).
Units
Metric System
Geometrical properties
■ Column length: L=9000mm
■ Cross section area: 28446mmA = ■ Overall breadth: mmb 180=
■ Flange thickness: mmt f 5.13=
■ Root radius: mmr 21=
■ Web thickness: mmtw 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 mmWy ×=
■ Elastic modulus after the Z axis, 33, 1040.146 mmW zel ×=
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■ Plastic modulus after the Z axis, 33, 10229 mmW zpl ×=
■ Flexion inertia moment around the Y axis: Iy=23130.00x104mm4 ■ Flexion inertia moment around the Z axis: Iz=1318.00x104 mm4 ■ Torsional moment of inertia: It=51.08x104 mm4 ■ Working inertial moment: Iw=490000x106mm6
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 275 MPa, ■ Longitudinal elastic modulus: E = 210000 MPa. ■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at the end point (z = 9.00) restrained in translation along Y and Z axis and restrained rotation
along X axis. ■ Inner: lateral (xoz) restraint at z=3m
Loading
The column is subjected to the following loadings: ■ External: Point load From X=0.00m and Y=9.00m: FZ =-125000N and Fx=28330N ■ Internal: None.
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CTICM model
The model is presented in the CTICM 2006-4-Resistance barre comprimee selon
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13.17.2.2Cross section Class According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
727249.381
49.386.8
331=×≤=⇒
=
==ε
ε tc
mmmm
tc
therefore the beam web is considered to be Class 1
-for beam flange:
9950.41
50.45.1347.67
=×≤=⇒
=
==ε
ε tc
mmmm
tc
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 1
According to CTICM document:
The cross section is considered to be Class 1. The column strength will be determined considering the plastic characteristics of the cross-section. Below can be seen the CTICM conclusion, extracted from CTICM 2006-4:
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13.17.2.3Compression verification According to Advance Design calculations:
The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design axial compression resistance of the element (Nc,Rd) from the compression force applied to the element (NEd). The compressed resistance of the member, Nc,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.2.4.
%100100,
≤×Rdc
Ed
NN
(6.9)
The design resistance of the cross-section for uniform compression Nc,Rd is determined using the formula below:
0,
M
yRdc
fAN
γ×
= (6.10)
-where:
A is the section area: 28446mmA =
yf is the yielding strength: 2/275 mmNf y =
0Mγ is the partial safety factor: 10 =Mγ
kNNmmNmmfAN
M
yRdc 65.23222322650
1/2758446 22
0, ==
×=
×=
γ
NNEd 125000=
%100%38.51002322650125000
,
≤=×=N
NNN
Rdc
Ed
According to CTICM document:
The compression resistance of the column is kNN Rdc 2324, = as it can be seen from conclusion extracted from
CTCIM 2006-4:
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13.17.2.4Shear verification According to Advance Design calculations:
The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design shear resistance of the element (Vc,Rd) from the shear force applied to the element (VEd). The shear resistance of the member, Vc,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.2.6.
%100100,
≤×Rdc
Ed
VV
(6.17)
The design shear resistance of the element, Vc,Rd is determined using the formula below:
0,
3M
yV
Rdc
fA
Vγ
×
= (6.18)
-where:
AV is the shear area:
( ) wwfwfV thtrttbAA ××≥××++××−= η22
-where:
A is the cross section area: 28446mmA =
b is the overall breadth: mmb 180=
tf is the flange thickness: mmt f 5.13=
r is the root radius: mmr 21=
tw is the web thickness: mmtw 6.8=
hw is the depth of the web: mmhw 400=
1=η
( )( ) 22 1.42695.132126.85.1318028446
22
mmmmmmmmmmmmmm
trttbAA fwfV
=××++××−=
=××++××−=
234404006.81 mmmmmmth ww =××=××η
22 34401.4269 mmmmAV ≥=
yf is the yielding strength: 2/275 mmNf y =
0Mγ is the partial safety factor: 10 =Mγ
N
mmNmmf
AV
M
yV
Rdc 66.6778101
3/2751.4269
3
22
0, =
×
=
×
=γ
NVEd 28330=
%100%180.410004179.010066.677810
28330100,
≤=×=×=×N
NVV
Rdc
Ed
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According to CTICM document:
The shear resistance of the column is kNV Rdzpl 8.677,, = as it can be seen from conclusion extracted from CTCIM
2006-4:
13.17.2.5Bending moment verification According to Advance Design calculations:
The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design bending moment resistance of the element (Mpl,Rd) from the bending moment effor applied to the element (MEd). The Bending moment resistance of the member, Mpl,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.2.5.
%100100,
≤×Rdpl
Ed
MM
(6.12)
-the shear force does not exceed 50% of the shear plastic resistance, therefore there is no influence of the shear on the composed bending;
-the axial compression force does not exceed 25% of the plastic resistance, therefore there is no influence of the compression on the composed bending
The design bending moment resistance of the element, Mpl,Rd is determined using the formula below:
0,
M
yplRdpl
fwM
γ×
= (6.13)
-where:
wpl is the plastic modulus: 31307000mmwpl =
yf is the yielding strength: 2/275 mmNf y =
0Mγ is the partial safety factor: 10 =Mγ
NmmmmNmmfwM
M
yplRdpl 359425000
1/2751307000 23
0, =
×=
×=
γ
NmmM Ed 254970000=
%100%938.7010070938.0100359425000254970000100
,
≤=×=×=×NmmNmm
MM
Rdpl
Ed
According to CTICM document:
The bending moment resistance of the column is kNmM Rdypl 7.359,, = as it can be seen from the conclusion
extracted from CTCIM 2006-4:
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13.17.2.6Buckling verification According to Advance Design calculations:
a) over the strong axis of the section, y-y:
The cross section buckling curve will be chosen according to Table 6.2:
The imperfection factor α will be chosen according to Table 6.1:
21.0=α
Coefficient corresponding to non-dimensional slenderness after Y-Y axis
yχ coefficient corresponding to non-dimensional slenderness yλ will be determined from the relevant buckling
curve according to:
1122
≤−Φ+Φ
=yyy
yλ
χ (6.49)
yλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
ycr
yy N
fA
,
*=λ
Where: A is the cross section area; A=8446mm2; fy is the yielding strength of the material; fy=275N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
( )N
mmmmMPa
LIE
Nfz
yycr 77.5918472
90001023130210000²
²²
2
44
, =×××
=××
=ππ
62645.077.5918472
/2758446 22
,
=×
=×
=N
mmNmmN
fA
ycr
yyλ
[ ] ( )[ ] 740997.062645.02.062645.021.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ
187968.062645.0740997.0740997.0
112222
≤=−+
=−Φ+Φ
=yyy
yλ
χ
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According to CTICM document:
The determined value for the coefficient corresponding to non-dimensional slenderness for the strong section, y-y
axis, yλ is: 8796.0=yχ as it can be observed from the conclusion extracted from CTCIM 2006-4:
b) over the strong axis of the section, z-z:
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:
34.0=α
zχ coefficient corresponding to non-dimensional slenderness zλ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=zzz
zλ
χ (6.49)
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zλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
zcr
yz
NfA
,
*=λ
Where: A is the cross section area; A=8446mm2; fy is the yielding strength of the material; fy=275N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
Outside the frame, the calculation can be made with more than the safety of taking in account a buckling length equal to the grater length of the two beam sections, 6m. A more accurate calculation is to perform a modal analysis of the column buckling outside the frame. The first eigen mode of instability corresponds to an amplification factor equal to critical 15.9=crα . The normal critical force can be directly calculated:
( ) Nmm
mmmmNl
IENfy
zzcr 153.1142396
²4890101318/210000²
²² 442
, =×××
=××
=ππ
42588.1153.1142396
/2758446 22
,
=×
=×
=mmNmm
NfA
zcr
yzλ
[ ] ( )[ ] 72497.142588.12.042588.134.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ
137096.042588.172497.172497.1
112222
≤=−+
=−Φ+Φ
=zzz
zλ
χ
According to CTICM document:
The determined value for the coefficient corresponding to non-dimensional slenderness for the strong section, z-z
axis, zλ is: 3711.0=zχ as it can be observed from the conclusion extracted from CTCIM 2006-4:
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13.17.2.7Lateral torsional buckling verification According to Advance Design calculations:
a) for the 3m part of the column:
The elastic moment for lateral-torsional buckling calculation, Mcr:
-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
z
t
z
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3) -where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
ψ is the fraction of the bending moment from the column extremities: 66667.097.25498.169
==kNmkNmψ
17932.1²66667.0252.066667.0423.0325.0
1²252.0423.0325.0
11 =
×+×+=
++=
ψψC
■ Flexion inertia moment around the Y axis: Iy=23130.00x104mm4 ■ Flexion inertia moment around the Z axis: Iz=1318.00x104 mm4 ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa. ■ Torsional moment of inertia: It=51.08x104 mm4 ■ Working inertial moment: Iw=490000x106mm6 ■ Length of the column part: L=3000mm ■ Shear modulus of rigidity: G=80800MPa
( )( )
Nmm
mmNmmmmN
mmmmNmmmm
mm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
2.806585210
33396.22534.303523217932.1101318/210000²
1008.51/808003000101318
1049
3000101318/210000²17932.1
²²
²²
442
4422
44
610
2
442
1
=
=××=×××
×××+
××
×
××××
×=××××
+×××
×=
π
ππ
π
Calculation of the non-dimensional slenderness factor, LTλ :
cr
yyLT
MfW
=λ
Plastic modulus,33101307 mmWy ×=
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66754.02.806585210
/275101307 233
=××
=×
=Nmm
mmNmmM
fW
cr
yyLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×= The cross section buckling curve will be chosen according to Table 6.4:
The imperfection factor α will be chosen according to Table 6.1:
34.0=α
( )[ ] 80228.0²66754.02.066754.034.015.0 =+−×+×=LTφ
180173.0²66754.0²80228.080228.0
1
²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
According to CTICM document:
The determined value for the coefficient corresponding to non-dimensional slenderness, LTλ is: 7877.0=LTχ as it can be observed from the conclusion extracted from CTCIM 2006-4:
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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
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3) -where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
ψ is the fraction of the bending moment from the column extremities: 0=ψ
77.11 =C ■ Flexion inertia moment around the Y axis: Iy=23130.00x104mm4 ■ Flexion inertia moment around the Z axis: Iz=1318.00x104 mm4 ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa. ■ Torsional moment of inertia: It=51.08x104 mm4 ■ Working inertial moment: Iw=490000x106mm6 ■ Length of the column part: L=6000mm ■ Shear modulus of rigidity: G=80800MPa
Nmm
mmNmmmmN
mmmmNmmmm
mm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
406423987
604.302085.75880877.1101318/210000²
1008.51/80800²6000101318
1049
²6000101318/210000²77.1
²²
²²
442
4422
44
610
2
442
1
=
=××=×××
×××+
××
×
××××
×=××××
+×××
×=
π
ππ
π
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Calculation of the non-dimensional slenderness factor, LTλ :
cr
yyLT
MfW
=λ
Plastic modulus,33101307 mmWy ×=
94040.0406423987
/275101307 233
=××
=×
=Nmm
mmNmmM
fW
cr
yyLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×= The cross section buckling curve will be chosen according to Table 6.4:
The imperfection factor α will be chosen according to Table 6.1:
34.0=α
( )[ ] 06804.1²94040.02.094040.034.015.0 =+−×+×=LTφ
163518.0²94040.0²06804.106804.1
1
²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
ADVANCE DESIGN VALIDATION GUIDE
71
According to CTICM document:
The determined value for the coefficient corresponding to non-dimensional slenderness, LTλ is: 694.0=LTχ as it can be observed from the conclusion extracted from CTCIM 2006-4:
13.17.2.8Bending and axial compression verification According to Advance Design calculations:
1
1
,
,,
1
,
,,
1
≤∆+
×+×
∆+×+
×M
Rkz
EdzEdzyz
M
RdyLT
RdyEdyyy
M
Rky
Ed
MMM
kMMM
kNN
γγχ
γχ
(6.61)
1
1
,
,,
1
,
,,
1
≤∆+
×+×
∆+×+
×M
Rkz
EdzEdzzz
M
RdyLT
RdyEdyzy
M
Rkz
Ed
MMM
kMMM
kNN
γγχ
γχ
(6.62)
The formulae can be simplified because:
There is no bending on the small inertia axis: 0, =EdzM
The section is considered to be a Class1: 0, =∆ RdyM and 0, =∆ RdzM
Therefore the formulae are:
( )
( )62.600.1
61.600.1
1
,
,
1
1
,
,
1
≤×
×+×
≤×
×+×
M
RkyLT
Edyzy
M
Rkz
Ed
M
RkyLT
Edyyy
M
Rky
Ed
MM
kNN
MM
kNN
γχ
γχ
γχ
γχ
ADVANCE DESIGN VALIDATION GUIDE
72
13.17.2.9Internal factor, yyk , calculation:
The internal factor yyk corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:
yy
ycr
Ed
ymLTmyyy C
NNCCk 1
1,
×−
××=µ
Auxiliary terms:
ycr
Edy
ycr
Ed
y
NN
NN
,
,
1
1
×−
−=
χµ
Where:
87968.0=yχ (previously calculated)
( ) Nmm
mmmmNl
IEN
fy
yycr 773.5918472
²900041023130/210000²
²² 42
, =×××
=××
=ππ
99741.0
773.591847212500087968.01
773.59184721250001
=×−
−= N
NN
yµ
According to CTICM document:
ADVANCE DESIGN VALIDATION GUIDE
73
The myC will be calculated according to Table A.1:
Calculation of the 0λ term:
LTC λλ ×= 10 -where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6) ψ is the fraction of the bending moment from the column extremities: 0=ψ
77.11 =C
ADVANCE DESIGN VALIDATION GUIDE
74
66754.02.806585210
/275101307 233
=××
=×
=Nmm
mmNmmM
fW
cr
yyLTλ
88811.066754.077.1110 =×=×=×= LTLT CC λλλ
Calculation of the 4
,,1 1120.0
−×
−××
TFcr
Ed
zcr
Ed
NN
NNC term:
Where:
-for a symmetrical section for the both axis, TcrTFcr NN ,, =
××+××=
²,
2
0,
Tcr
wtTcr L
IEIGIAN π
The mass moment of inertia 0I
20 gzy zAIII ⋅++=
Flexion inertia moment around the Y axis: Iy=23130.00x104mm4
Flexion inertia moment around the Z axis: Iz=1318.00x104 mm4
Cross section area: 28446mmA =
Distance between the section neutral axis and the section geometrical center: 0=gz
44444420 10244481013181023130 mmmmmmIIzAIII zygzy ×=×+×=+=⋅++=
-for simplification, it will be considered the same buckling length, TcrL , , for all the column parts:
mL Tcr 6, =
Torsional moment of inertia: It=51.08x104 mm4
Working inertial moment: Iw=490000x106mm6
Longitudinal elastic modulus: E = 210000 MPa
Shear modulus of rigidity: G=80800MPa
( )N
mmmmmmNmmmmN
mmmmN Tcr
788.2400423²6000
1049/2100001008.51/808001024448
8446 61022442
44
2
,
=
=
×××+×××
×=
π
NNN TcrTFcr 788.2400423,, ==
NN zcr 153.1142396, = (previously calculated)
25505.0
788.24004231250001
153.1142396125000177.120.01120.0 44
,,1
=
=
−×
−××=
−×
−××
NN
NN
NN
NNC
TFcr
Ed
zcr
Ed
ADVANCE DESIGN VALIDATION GUIDE
75
Therefore:
( )
≥
−×
−
×=
=
×+
××−+=
⇒=
−×
−××>=
1
11
11
25505.01120.088811.0
,,
2
0,
0,0,
4
,,10
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
a
aCCC
NN
NNC
ε
ε
λ
The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.
The coefficient must be calculated considering the column over the entire height.
( )LTy
LTymymymy a
aCCC
×+
××−+=
ε
ε
11 0,0,
yelEd
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 33, 101156 mmW yel ×=
90119.14101156
8446125000
1094.25433
26
,
, =×
××
=×=mm
mmNNmm
WA
NM
yelEd
Edyyξ
99779.01023130
1008.5111 44
44
=×⋅
−=−=mm
mmIyItaLT
The 0mC coefficient is defined according to the Table A.2:
The bending moment is null at the end of the column, therefore: 0=ψ
( )ycr
Edmy N
NC,
0, 33.036.021.079.0 ×−×+×+= ψψ
Where:
NN ycr 773.5918472, = (previously calculated)
( ) 78749.0773.5918472
12500033.0036.079.00, =×−×+=N
NCmy
( ) ( ) 95619.099779.090119.141
99779.090119.1478749.0178749.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
ADVANCE DESIGN VALIDATION GUIDE
76
According to CTICM document:
The mLTC coefficient takes into account the laterally restrained parts of the column. The mLTC coefficient must be calculated individually for each of the column parts.
1
11,,
2 ≥
−×
−
×=
Tcr
Ed
zcr
Ed
LTmymLT
NN
NN
aCC
a) for the 3m part of the column:
( )LTy
LTymymymmy a
aCCC
ξ
ξ
+−+=
11 0,0,3.
90119.14=yξ (previously calculated)
99779.0=LTa (previously calculated)
The 0mC coefficient is defined according to the Table A.2:
ψ is the fraction of the bending moment from the column part extremities: 66667.097.25498.169
==kNmkNmψ
ADVANCE DESIGN VALIDATION GUIDE
77
( )ycr
Edmy N
NC,
0, 33.036.021.079.0 ×−×+×+= ψψ
Where:
NN ycr 773.5918472, = (previously calculated)
( ) 93256.0773.5918472
12500033.066667.036.066667.021.079.00, =×−×+×+=N
NCmy
( ) ( ) 98609.099779.090119.141
99779.090119.1493256.0193256.01
1 0,0,3, =×+
××−+=
×+
×−+=
LTy
LTymymymmy a
aCCC
ξ
ξ
NN zcr 153.1142396, = (previously calculated)
NN Tcr 788.2400423, = (previously calculated)
99779.0=LTa (previously calculated)
05596.1
1
05596.1
788.24004231250001
153.11423961250001
99779.0²98609.0
11
3,
3,
,,
23,3,
=⇒
≥
=
−×
−
×=
=
−×
−
×=
mmLT
mmLT
Tcr
Ed
zcr
Ed
LTmmymmLT
C
CN
NN
N
NN
NN
aCC
b) for the 6m part of the column:
( )LTy
LTymymymmy a
aCCC
ξ
ξ
+−+=
11 0,0,6.
yelEd
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 33, 101156 mmW yel ×=
9353.9101156
8446125000
1098.16933
26
,
, =×
××
=×=mm
mmNNmm
WA
NM
yelEd
Edyyξ
99779.01023130
1008.5111 44
44
=×⋅
−=−=mm
mmIyItaLT
The 0mC coefficient is defined according to the Table A.2:
ADVANCE DESIGN VALIDATION GUIDE
78
ψ is the fraction of the bending moment from the column part extremities: 0=ψ
( )ycr
Edmy N
NC,
0, 33.036.021.079.0 ×−×+×+= ψψ
Where:
NN ycr 773.5918472, = (previously calculated)
( ) 78749.0773.5918472
12500033.0036.0021.079.00, =×−×+×+=N
NCmy
( ) ( ) 94873.099779.09353.91
99779.09353.978749.0178749.01
1 0,0,6, =×+
××−+=
×+
×−+=
LTy
LTymymymmy a
aCCC
ξ
ξ
NN zcr 153.1142396, = (previously calculated)
NN Tcr 788.2400423, = (previously calculated)
1
1
97746.0
788.24004231250001
153.11423961250001
99779.0²94873.0
11
6,
6,
,,
26,6,
=⇒
≥
=
−×
−
×=
=
−×
−
×=
mmLT
mmLT
Tcr
Ed
zcr
Ed
LTmmymmLT
C
CN
NN
N
NN
NN
aCC
In conclusion:
=
==
105596.1
6,
3,
mmLT
mmLTmLT C
CC
ADVANCE DESIGN VALIDATION GUIDE
79
The yyC coefficient is defined according to the Table A.1 on the “Auxiliary terms:” part:
ypl
yelLTplmy
ymy
yyyy W
WbnC
wC
wwC
,
,2max
2max
2 6.16.12)1(1 ≥
−×
××−××−×−+= λλ
Where:
■ 005.05.0,,
,20
,
,
,,
,20 =×
××××=×
××××=
RdyplLT
EdyLT
Rdpl
Edz
RdyplLT
EdyLTLT M
Ma
MM
MM
abχ
λχ
λ
■ 5.1,
, ≤=yel
yply W
Ww
■ Elastic modulus after the Y axis, 33, 101156 mmW yel ×=
■ Plastic modulus after the Y axis, 33, 101307 mmW ypl ×=
■ 5.113062.1101156101307
33
33
,
, ≤=××
==mmmm
WW
wyel
yply
■
1M
Rk
Edpl N
Nn
γ
=
■ NmmNmmfANN
M
yRdcRk 2322650
1/2758446 22
0, =
×=
×==
γ
■ 05382.0
12322650125000
1
=== NN
NNn
M
Rk
Edpl
γ
ADVANCE DESIGN VALIDATION GUIDE
80
■ ( )zy λλλ ;maxmax =
■ 62645.0=yλ (previously calculated)
■ 42504.1=zλ (previously calculated)
■ ( ) ( ) 42504.142504.1;62645.0max;maxmax === zy λλλ
■ 95619.0=myC (previously calculated)
98511.0
005382.0²42504.1²95619.013062.1
6.142504.1²95619.013062.1
6.12)113062.1(1
=
=
−×
××−××−×−+=yyC
88447.010130710115698511.0 33
33
,
, =××
=≥=mmmm
WW
Cypl
yelyy
In conclusion:
=×−
××=×−
××=
=×−
××=×−
××=
=98902.0
98511.01
773.59184721250001
99741.0195619.01
1
04409.198511.0
1
773.59184721250001
99741.005569.195619.01
1
,
6,6,
,
3,3`,
NNC
NNCCk
NNC
NNCCk
k
yy
ycr
Ed
ymmLTmymyy
yy
ycr
Ed
ymmLTmymyy
yy µ
µ
According to CTICM 2006-4:
ADVANCE DESIGN VALIDATION GUIDE
81
13.17.2.10Internal factor, zyk , calculation:
The internal factor zyk corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:
z
y
zy
ycr
Ed
zmLTmyzy w
wC
NNCCk ×××
−××= 6.01
1,
µ
Auxiliary terms:
zcr
Edz
zcr
Ed
z
NN
NN
,
,
1
1
×−
−=
χµ
Where:
37096.0=zχ (previously calculated)
( ) Nmm
mmmmNl
IENfy
zzcr 153.1142396
²4890101318/210000²
²² 442
, =×××
=××
=ππ
92826.0
153.114239612500037096.01
1533.11423961250001
=×−
−=
NN
NN
zµ
According to CTICM document:
ADVANCE DESIGN VALIDATION GUIDE
82
The myC will be calculated according to Table A.1:
( )
≥
−×
−
×=
=
×+
××−+=
⇒=
−×
−××>=
1
11
11
208.01120.088811.0
,,
2
0,
0,0,
4
,,10
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
a
aCCC
NN
NNC
ε
ε
λ
(previously calculated)
90119.14101156
8446125000
1094.25433
26
,
, =×
××
=×=mm
mmNNmm
WA
NM
yelEd
Edyyξ (previously calculated)
99779.01023130
1008.5111 44
44
=×⋅
−=−=mm
mmIyItaLT (previously calculated)
The 0mC coefficient is defined according to the Table A.2:
ADVANCE DESIGN VALIDATION GUIDE
83
The bending moment is null at the end of the column, therefore: 0=ψ
( ) 78749.033.036.021.079.0,
0, =×−×+×+=ycr
Edmy N
NC ψψ
(previously calculated)
( ) 95619.01
1 0,0, =+
−+=LTy
LTymymymy a
aCCC
ξ
ξ
(previously calculated)
=
==
−×
−
×=1
05596.1
116,
3,
,,
2
mmLT
mmLT
Tcr
Ed
zcr
Ed
LTmymLT C
C
NN
NN
aCC
(previously calculated)
The zyC coefficient is defined according to the Table A.1 on the “Auxiliary terms:” part:
ypl
yel
z
yLTpl
y
myyzy W
Www
dnw
CwC
,
,5
2max
2
6.0142)1(1 ××≥
−×
××−×−+=
λ
ADVANCE DESIGN VALIDATION GUIDE
84
Where:
001.0
21.0
2,,
,4
0
,,
,
,,
,4
0 =×××
×+
××=×
×××
×+
××=RdyplLTmy
Edy
zLT
Rdzplmz
Edz
RdyplLTmy
Edy
zLTLT MC
Ma
MCM
MCM
adχλ
λχλ
λ
5.113062.1101156101307
33
33
,
, ≤=××
==mmmm
WW
wyel
yply (previously calculated)
5.1,
, ≤=zel
zplz W
Ww
Elastic modulus after the Z axis, 33, 1040.146 mmW zel ×=
Plastic modulus after the Z axis, 33, 10229 mmW zpl ×=
5.15.1
564.11040.146
1022933
33
,
,
=⇒
≤
=×
×==
z
z
zel
zplz w
wmm
mmWW
w
1M
Rk
Edpl N
Nn
γ
=
05382.0
12322650125000
1
=== NN
NNn
M
Rk
Edpl
γ
(previously calculated)
( ) ( ) 42504.142504.1;62645.0max;maxmax === zy λλλ (previously calculated)
90887.0005382.013062.1
42504.198609.0142)113062.1(1 5
22
=
−×
××−×−+=zyC
46073.0101307101156
5.113062.16.06.0 33
33
,
, =××
××=××mmmm
WW
ww
ypl
yel
z
y
46073.06.090887.0142)1(1,
,5
2max
2
=××≥=
−×
××−×−+=
ypl
yel
z
yLTpl
y
myyzy W
Www
dnw
CwC
λ
In conclusion:
56307.0
5.113062.16.0
90887.01
773.59184721250001
923830.005569.198609.06.01
1,
=
=×××−
××=×××−
××=
NNw
wC
NNCCk
z
y
zy
ycr
Ed
zmLTmyzy
µ
ADVANCE DESIGN VALIDATION GUIDE
85
=×××−
××=
=×××−
××=
=×××−
××=
=×××−
××=
=
50752.05.1
13062.16.098511.0
1
773.59184721250001
99741.0195619.0
6.01
1
52306.05.1
13062.16.090887.0
1
773.59184721250001
99741.005569.195619.0
6.01
1
,
6,6,
,
3,3`,
NN
ww
CNNCCk
NN
ww
CNNCCk
k
z
y
zy
zcr
Ed
ymmLTmymzy
z
y
zy
zcr
Ed
ymmLTmymzy
zyµ
µ
According to CTICM 2006-4:
The bending and axial compression verifications are:
-for the 3m column part:
160789.046281.014508.01
35942500080173.0
25497000052306.0
1232265037096.0
125000
198501.092383.006119.01
35942500080173.0
25497000004409.1
1232265087968.0
125000
1
,
,
1
1
,
,
1
≤=+=
=×
×+×
=
=×
×+×
≤=+=
=×
×+×
=
=×
×+×
NmmNmm
NN
MM
kNN
NmmNmm
NN
MM
kNN
M
RkyLT
Edyzy
M
Rkz
Ed
M
RkyLT
Edyyy
M
Rky
Ed
γχ
γχ
γχ
γχ
ADVANCE DESIGN VALIDATION GUIDE
86
-for the 6m column part:
152073.037565.014508.01
35942500063518.0
16898000050752.0
1232265037096.0
125000
179322.073204.006118.01
35942500063518.0
16898000098902.0
1232265087968.0
125000
1
,
,
1
1
,
,
1
≤=+=
=×
×+×
=
=×
×+×
≤=+=
=×
×+×
=
=×
×+×
NmmNmm
NN
MM
kNN
NmmNmm
NN
MM
kNN
M
RkyLT
Edyzy
M
Rkz
Ed
M
RkyLT
Edyyy
M
Rky
Ed
γχ
γχ
γχ
γχ
According to CTICM 2006-4:
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
87
Ratio of the design normal force to design compresion resistance
Column subjected to axial and shear force to the top Work ratio - Fx
Ratio of the design share force to design share resistance
Column subjected to axial and shear force to the top Work ratio - Fz
ADVANCE DESIGN VALIDATION GUIDE
88
Ratio of the design share force to design share resistance
Column subjected to axial and shear force to the top Work ratio - oblique
yχ coefficient corresponding to non-dimensional slenderness yλ
Column subjected to axial and shear force to the top
yχ
ADVANCE DESIGN VALIDATION GUIDE
89
zχ coefficient corresponding to non-dimensional slenderness zλ
Column subjected to axial and shear force to the top
zχ
Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
ADVANCE DESIGN VALIDATION GUIDE
90
Internal factor, yzk
Column subjected to axial and shear force to the top
yzk
13.17.2.11Reference results
Result name Result description Reference value Work ratio - Fx Ratio of the design normal force to design compression resistance
5.38
Work ratio - Fz Ratio of the design share force to design share resistance
4.18 Work ratio - Oblique Ratio of the design moment resistance to design bending resistance
one the principal axis
70.94
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
0.88
zχ zχ coefficient corresponding to non-dimensional slenderness zλ
0.37
yyk Internal factor, yyk for the 3m segment
1.04
yyk Internal factor, yyk for the 6m segment
0.99
zyk Internal factor, zyk for the 3m segment
0.52
zyk Internal factor, zyk for the 6m segment
0.51
ADVANCE DESIGN VALIDATION GUIDE
91
13.17.3Calculated results Result name Result description Value Error Work ratio - Fx Ratio of the design normal force to design compression
resistance 5.38178 % 0.0331 %
Work ratio - Fz Ratio of the design share force to design share resistance 4.17973 % -0.0065 % Work ratio - Oblique
Ratio of the design moment resistance to design bending resistance one the principal axis
70.9383 % -0.0024 %
Xy Coefficient corresponding to non-dimensional slenderness 0.879684 adim
-0.0359 %
Xz Coefficient corresponding to non-dimensional slenderness 0.370957 adim
0.2586 %
Kyy Internal factor,kyy for the 3m segment 1.03159 adim
-0.8087 %
Kyy Internal factor,kyy for the 6m segment 0.983324 adim
-0.6743 %
Kzy Internal factor,kzy for the 3m segment 0.537037 adim
3.2763 %
Kzy Internal factor,kzy for the 6m segment 0.511305 adim
0.2559 %
ADVANCE DESIGN VALIDATION GUIDE
92
13.18 EC3 Test 20: Verifying the buckling resistance of a RC3020100 column
Test ID: 5700
Test status: Passed
13.18.1Description The test verifies the buckling of a RC3020100 column made of S355 steel.
The verifications are made according to Eurocode3 French Annex.
13.18.2Background Verification of buckling under compression efforts for a rectangular hollow, RC3020100 column made of S235 steel. The column is fixed at its base and free on the top. The column is subjected to a compression force (200 000 N) applied at its top. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.18.2.1Model 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=11819x104mm4 ■ Flexion inertia moment around the Z axis: Iz=6278x104 mm4
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Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (x = 5.00).
■ Inner: None. ■ Buckling lengths Lfy and Lfz are both imposed (10m)
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 5.0: FZ = N = -200 000 N, ■ Internal: None.
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13.18.2.2Buckling in the strong inertia of the profile (along Y-Y) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.
%100100,
≤×Rdb
Ed
NN
(6.46)
The design buckling resistance of the compressed element is calculated using the next formula:
1,
M
yRdb
fAN
γχ ××
= (6.47)
Where:
Coefficient corresponding to non-dimensional slenderness after the Y-Y axis
χ coefficient corresponding to non-dimensional slenderness λ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=λ
χ (6.49)
λ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
cr
y
NfA*
=λ
Where: A is the cross section area; A=5380mm2; fy is the yielding strength of the material; fy=235N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
( )N
mmmmMPa
LIEN
fy
zcr 943.2449625
100001011819210000²
²²
2
44
=×××
=××
=ππ
954.0943.2449625
/2359490 22
=×
=×
=N
mmNmmN
fA
cr
yλ
[ ]²)2.0(15.0 λλαφ +−+=
It will be used the following buckling curve:
The imperfection factor α corresponding to the appropriate buckling curve will be 0.21:
[ ] ( )[ ] 034.1954.02.0954.021.015.0²)2.0(15.0 2 =+−×+×=+−×+×= λλαφ
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Therefore:
1698.0954.0034.1034.1
112222
≤=−+
=−+
=λφφ
χ
1Mγ is a safety coefficient, 11 =Mγ
NmmNmmN Rdb 7.15566441
/2359490698.0 22
, =××
=
NNEd 200000=
%848.121007.1556644
200000100,
=×=×N
NNN
Rdb
Ed
13.18.2.3Buckling in the weak inertia of the profile (along Z-Z) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.
%100100,
≤×Rdb
Ed
NN
(6.46)
The design buckling resistance of the compressed element is calculated using the next formula:
1,
M
yRdb
fAN
γχ ××
= (6.47)
Where:
Coefficient corresponding to non-dimensional slenderness after the Z-Z axis
χ coefficient corresponding to non-dimensional slenderness λ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=λ
χ (6.49)
λ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
cr
y
NfA*
=λ
Where: A is the cross section area; A=5380mm2; fy is the yielding strength of the material; fy=235N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
( )N
mmmmMPa
LIEN
fy
zcr 905.1301188
10000106278210000²
²²
2
44
=×××
=××
=ππ
309.1905.1301188
/2359490 22
=×
=×
=N
mmNmmN
fA
cr
yλ
[ ]²)2.0(15.0 λλαφ +−+=
It will be used the following buckling curve:
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The imperfection factor α corresponding to the appropriate buckling curve will be 0.21:
[ ] ( )[ ] 473.1309.12.0309.121.015.0²)2.0(15.0 2 =+−×+×=+−×+×= λλαφ
Therefore:
1465.0309.1473.1473.1
112222
≤=−+
=−+
=λφφ
χ
1Mγ is a safety coefficient, 11 =Mγ
NmmNmmN Rdb 75.10370191
/2359490465.0 22
, =××
=
NNEd 200000=
%286.1910075.1037019
200000100,
=×=×N
NNN
Rdb
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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13.18.2.4Reference results
Result name Result description Reference value
yχ coefficient corresponding to non-dimensional slenderness after Y-Y axis
0.698
zχ coefficient corresponding to non-dimensional slenderness after Z-Z axis
0.465
Work ratio (y-y) Ratio of the design normal force to design buckling resistance (strong inertia) [%]
12.85%
Work ratio (z-z) Ratio of the design normal force to design buckling resistance (weak inertia) [%]
19.29%
13.18.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness
after Y-Y axis 0.697433 adim
0.0000 %
Xz coefficient corresponding to non-dimensional slenderness after Z-Z axis
0.465226 adim
0.0000 %
SNy Ratio of the design normal force to design buckling resistance in the strong inertia of the profile
0.128586 adim
0.0000 %
SNz Ratio of the design normal force to design buckling resistance in the weak inertia of the profile
0.192767 adim
0.0000 %
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13.19 EC3 Test 13: Verifying the resistance of a rectangular hollow section column subjected to bending and shear efforts
Test ID: 5707
Test status: Passed
13.19.1Description Verifies the resistance of a rectangular hollow section column (made of S235 steel) subjected to bending and shear efforts.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.19.2Background Verifies the adequacy of a rectangular hollow section column made of S235 steel to resist shear and bending efforts. Verification of the shear resistance at ultimate limit state, as well as the design resistance for bending, is realised. The name of the cross-section is RC3020100 and can be found in the Advance Design OTUA library. The column is fixed at its base and it is subjected to a punctual horizontal load applied to the middle height (200 000 N). The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.19.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: 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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Units
Metric System
Geometry
Below are described the column cross section characteristics:
■ Height: h = 300 mm, ■ Width: b = 200 mm, ■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm, ■ Column height: L = 5000 mm, ■ Section area: A = 9490 mm2 , ■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,
■ Partial factor for resistance of cross sections: 0.10 =Mγ .
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (z = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (z = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at z = 2.5: V= Fx = 200 000 N,
■ Internal: None.
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13.19.2.2Reference results for calculating the design plastic shear resistance of the cross-section The design plastic shear resistance of the cross-section is determined with formula (6.18) from EN 1993-1-1:2001. Before using it, the shear area (Av) has to be determined.
Shear area of the cross section
For a rectangular hollow section of uniform thickness the shear area is determined according to chapter 6.2.6 (3) from EN 1993-1-1. As the load is parallel to depth, the shear area is:
22
56942003003009490 mm
mmmmmmmm
hbhAAv =
+×
=+×
=
Design plastic shear resistance of the cross section
EN 1993-1-1: 2011 provides the following formula to calculate the design plastic shear resistance of the cross-section:
N
MPammf
AV
M
yv
Rdpl 6.7725460.1
32355694
32
0, =
×=
×=
γ
Work ratio
The verification of the design plastic shear resistance is done with relationship (6.17) from EN 1993-1-1. The corresponding work ratio is:
Work ratio = %89.251006.772546
200000100,
=×=×RdplV
V
13.19.2.3Reference results for calculating the design resistance for bending Before calculating the design resistance for bending, the cross section class has to be determined.
Cross section class
The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.
In this case, the column is subjected to bending efforts, therefore the stresses distribution is like in the picture below:
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Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class for compressed parts. The picture below shows an extract from this table.
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Taking into account that the top wing part is subjected to compression stresses, its class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to compression).
1510
10215220022=
×−×−=
×−×−=
mmmmmmmm
ttrb
tc
0.1235==
yfε
Therefore:
333315 =≤= εtc
This means that the top wing is Class 1. Because the bottom wing is tensioned, it will be classified as Class 1.
The left/right web is subjected to bending stresses. Their class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to bending):
2510
10215230022=
×−×−=
×−×−=
mmmmmmmm
ttrh
tc
0.1235==
yfε
Therefore:
727225 =≤= εtc
This means that the left/right web is Class 1.
Because a cross-section is classified according to the least favorable classification of its compression elements (chapter 5.5.2(6) from EN 1993-1-1: 2001), this means that the cross-section is Class 1.
Design resistance for bending
The design resistance for bending, for Class 1 cross-section, is determined with formula (6.13) from EN 1993-1-1:2001.
NmmMPammfWM
M
yyplRdc 224660000
0.1235956000 3
0
,, =
×=
×=
γ
Work ratio
The verification of the design resistance for bending is done with relationship (6.12) from EN 1993-1-1. The corresponding work ratio is:
Work ratio = %56.222100224660000
25000200000
1002100,,
=××
=××
=×Nmm
mmN
M
LV
MM
RdcRdc
Ed
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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13.19.2.4Reference 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 %
13.19.3Calculated results Result name Result description Value Error Work ratio - Fz Work ratio of the design plastic shear resistance 25.8793 % -0.0413 % Work ratio - Oblique
Work ratio of the design resistance for bending 222.559 % -0.0004 %
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13.20 EC3 test 10: Verifying the classification and the bending resistance of a welded built-up beam
Test ID: 5692
Test status: Passed
13.20.1Description Verifies the classification and the bending resistance of a welded built-up beam made of S355 steel. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.20.2Background Classification and bending resistance verification of a welded built-up beam made of S355 steel. The beam is simply supported and it is loaded by a uniformly distributed load (15 000 N/ml). The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.20.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = -15 000 N/ml,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 630 mm, ■ Flange width: b = 500 mm, ■ Flange thickness: tf = 18 mm, ■ Web thickness: tw = 8 mm, ■ Beam length: L = 5000 mm, ■ Section area: A = 22752 mm2,
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■ Partial factor for resistance of cross sections: 0.10 =Mγ .
Materials properties
S355 steel material is used. The following characteristics are used:
■ Yield strength fy = 355 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation along Y, Z axis and restrained in rotation along
X axis. ■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Uniformly distributed load: q = Fz = -15 000 N/ml,
■ Internal: None.
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13.20.2.2Cross-section classification Before calculating the design resistance for bending, the cross-section class has to be determined.
The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.
In this case, the beam is subjected to a uniformly distributed load; therefore the stresses distribution is like in the picture below:
Table 5.2 - sheet 2, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class of the compressed flange (top flange). The picture below shows an extract from this table.
The top flange class can be determined by considering the cross-section geometrical properties and the conditions described in Table 5.2 - sheet 2 (above extract):
67.1318
2/)8500(=
−=
mmmmmm
tc
8136.0235==
yfε
Therefore:
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39.111467.13 =>= εtc
This means that the top flange is Class 4. Because the bottom flange is tensioned, it will be classified as Class 1.
Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the web class. The picture below shows an extract from this table. The web part is subjected to bending stresses.
The web class can be determined by considering the cross-section geometrical properties and the conditions described in Table 5.2 - sheet 1 (above extract):
25.748
218630=
×−=
mmmmmm
tc
8136.0235==
yfε
Therefore:
89.10012425.74 =≤= εtc
This means that the beam web is Class 3.
A cross-section is classified according to the least favorable classification of its compression elements (chapter 5.5.2(6) from EN 1993-1-1: 2001).
According to the calculation above, the beam section have Class 4 for top flange, Class 3 for web and Class 1 for bottom flange; therefore the class section for the entire beam section will be considered Class 4.
13.20.2.3Reference results for calculating the design resistance for bending The design resistance for bending for Class 4 cross-section is determined with the formula (6.15) from EN 1993-1-1:2001.
Before verifying this formula, it is necessary to determine the effective section modulus of the cross-section.
The effective section modulus of the cross section takes into account the reduction factor, ρ, which is applying only to parts in compression (top flange in this case).
The following parameters have to be determined in order to calculate the reduction factor: the buckling factor, the stress ratio and the plate modified slenderness.
The buckling factor (kσ) and the stress ratio(Ψ) - for flanges
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Table 4.2 from EN 1993-1-5 offers detailed information about determining the buckling factor and the stress ratio for flange. The below picture presents an extract from this table.
Taking into account that the stress distribution on the top flange is linear, the stress ratio becomes:
43.00.11
2 =→== σσσψ k
The plate modified slenderness (λp)
The formula used to determine the plate modified slenderness for top flange is:
( )( ) 902.043.08136.04.2818/2/8500
4.28/
=××
−==
mmmmmmk
tcp
σελ
The reduction factor (ρ)
The reduction factor for top flange is determined with relationship (4.3) from EN 1993-1-5. Because λp > 0.748, the reduction factor has the following formula:
0.1188.0
2 ≤−
=p
p
λλ
ρ
The effective width of the flange part can now be calculated:
( ) mmmmmmcb feff 89.2152
85008776.0, =−
×=×= ρ
Effective section modulus
The effective section modulus is determined considering the following cross-section:
■ Top flange width: beff,t = beff,f + tw + beff,f = 439.78 mm; ■ Top flange thickness: tf = 18 mm; ■ Web and bottom flange have the same dimensions as the original section.
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3,sup, 91.5204392 mmW yeff =
3,inf, 4.5736064 mmW yeff =
( ) 3,inf,,sup,min,, 91.5204392,min mmWWW yeffyeffyeff ==
Design resistance for bending
For Class 4 cross-section, EN 1993-1-1: 2001 provides (6.15) formula in order to calculate the design resistance for bending:
NmmMPammfWM
M
yeffRdc 1847559483
0.135591.5204392 3
0
min,, =
×=
×=
γ
Work ratio
Work ratio = ( ) %54.2100
18475594838/5000/151008/100
2
,
2
,
=××
=××
=×NmmmmmmN
MLq
MM
RdcRdc
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
13.20.2.4Reference results
Result name Result description Reference value Work ratio - Oblique Design resistance for bending work ratio [%] 2.54 %
13.20.3Calculated results Result name Result description Value Error Work ratio - Oblique
Design resistance for bending work ratio 2.53713 % -0.1130 %
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13.21 EC3 test 11: Cross section classification and compression resistance verification of a rectangular hollow section column
Test ID: 5705
Test status: Passed
13.21.1Description Verifies the cross section classification and the compression resistance of a rectangular hollow section column.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.21.2Background Classification and verification under compression efforts of a hot rolled rectangular hollow section column made of S235 steel. The name of the cross-section is RC3020100 and can be found in the Advance Design OTUA library. The column is fixed at its base and free on the top. It is subjected to a compression force (100 000 N) applied at its top. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.21.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = -100 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
Geometry
Below are described the column cross section characteristics:
■ Height: h = 300 mm, ■ Width: b = 200 mm,
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■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm, ■ Column length: L = 5000 mm, ■ Section area: A = 9490 mm2 ,
■ Partial factor for resistance of cross sections: .
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ 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: N = Fz = -100 000 N,
■ Internal: None.
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13.21.2.2Reference results for calculating the cross section class The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.
In this case, the column is subjected to a punctual compression load, therefore the stresses distribution is like in the picture below:
Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class for compressed parts. The picture below shows an extract from this table. The entire cross-section is subjected to compression stresses.
The cross-section class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1, and it is calculated for the most defavourable compressed part:
2510
10215230022=
×−×−=
×−×−=
mmmmmmmm
ttrh
tc
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0.1235==
yfε
Therefore:
333325 =≤= εtc
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.
13.21.2.3Reference results for calculating the compression resistance of the cross-section The compression resistance for Class 1 cross-section is determined with formula (6.10) from EN 1993-1-1:2001.
Compression resistance of the cross section
For Class 1 cross-section, EN 1993-1-1: 2011 provides the following formula in order to calculate the compression resistance of the cross-section:
NMPammfAN
M
yRdc 2230150
0.12359490 2
0, =
×=
×=
γ
Work ratio
Work ratio = %48.41002230150100000100
,
=×=×N
NN
N
Rdc
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 force Work ratio - Fx
13.21.2.4Reference results
Result name Result description Reference value Work ratio - Fx Compression resistance work ratio [%] 4.48 %
13.21.3Calculated results Result name Result description Value Error Work ratio - Fx Compression resistance work ratio 4.484 % 0.0893 %
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13.22 EC3 Test 25: Verifying an user defined I section class 4 column fixed on the bottom and with a displacement restraint at 2.81m from the bottom
Test ID: 5712
Test status: Passed
13.22.1Description The test verifies a user defined cross section column.
The cross section has an “I symmetric” shape with: 880mm height; 220mm width; 5mm center thickness; 15mm flange thickness; 0mm fillet radius and 0mm rounding radius.
The column is subjected to a -328kN axial compression force; 1274 kNm bending moment after the Y axis and 127.4 kNm bending moment after the Z axis. All the efforts are applied on the top of the column. The column height is 5.62m and has a restraint of displacement at 2.81m from the bottom over the weak axis.
The calculations are made according to Eurocode 3 French Annex.
13.22.2Background An I880*5+220*15 shaped column subjected to compression and bending, made from S275 steel. The column has a 880x5mm web and 220x15mm flanges. The column is hinged at its base and at his top the end is translation is permitted only on vertical direction and the rotation is blocked for the long axis of the column. The column is subjected to an axial compression load -328000 N, a 127400Nm bending moment after the X axis and a1274000Nm bending moment after the Y axis.
The column has lateral restraints against torsional buckling placed in at 2.81m from the column end (in the middle).
This test was evaluated by the French control office SOCOTEC.
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13.22.2.1Model 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: 210850mmA =
■ Overall breadth: mmb 220=
■ Flange thickness: mmt f 15=
■ Root radius: mmr 0=
■ Web thickness: mmtw 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 mmWy ×=
■ Elastic modulus after the Z axis, 33, 1008.242 mmW zel ×=
■ Plastic modulus after the Z axis, 33, 1031.368 mmW zpl ×=
■ Flexion inertia moment around the Y axis: Iy=149058.04x104mm4 ■ Flexion inertia moment around the Z axis: Iz=2662.89x104 mm4 ■ Torsional moment of inertia: It=51.46x104 mm4 ■ Working inertial moment: Iw=4979437.37x106mm6
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength fy = 275 MPa, ■ Longitudinal elastic modulus: E = 210000 MPa. ■ Shear modulus of rigidity: G=80800MPa
ADVANCE DESIGN VALIDATION GUIDE
120
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at the end point (z = 5.62) restrained in translation along Y and Z axis and restrained rotation
along X axis. ■ Inner:
Lateral buckling restraint in the middle of the column (z=2.81).
Loading
The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =--328000N; Mx=127400Nm and My=1274000Nm
ADVANCE DESIGN VALIDATION GUIDE
121
13.22.2.2Cross section Class According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
178.012750109.0
328.0212 −>−=−×
⋅=−⋅
⋅=y
Ed
fANψ
5.064.0005.085.0275
328.01211
21
>=
××+⋅=
××+⋅=
dtfN
y
Edα
ADVANCE DESIGN VALIDATION GUIDE
122
924.0275235235
===yf
ε
06.94)78.0(33.067.0
924.04233.067.0
42170924
1705
152880=
−×+×
=×+
×>=⇒
=
=×−
=ψ
ε
ε tc
mmmmmm
tc
therefore the beam web is considered to be Class 4
- for beam flange:
316.8924.0961.7924
61.715
5.107=×≤=⇒
=
==tc
tc
ε therefore the haunch is considered to be Class1.
In conclusion, the section is considered to be Class 4.
ADVANCE DESIGN VALIDATION GUIDE
123
13.22.2.3Effective cross-sections of Class4 cross-sections - the section is composed from Class 4 web and Class 1 flanges, therefore will start the web calculation:
- in order to simplify the calculations the web will be considered compressed only
1705
152880=
×−=
mmmmmm
tc
:
41 =⇒= σψ k
According to EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1
σελ
kt
bw
p ××=
4.28
wb is the width of the web; mmbw 850=
t is the web thickness; t=5mm
9244.0275235235
===yf
ε
261.349244.04.28
5850
=××
= mmmm
pλ
According to EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4
-the web is considered to be an internal compression element, therefore:
( )286.0
261.34055.0261.33055.0
043
673.0261.322 =
×−=
+×−=⇒
≥=+
>=
p
pp
λ
ψλρ
ψ
λ
=×==×=
=×=
⇒
×=
×=
×=
mmmmbmmmmbmmmmb
bbbb
bb
e
e
eff
effe
effe
weff
55.1211.2435.055.1211.2435.0
1.243850286.0
5.0
5.0
2
1
2
1
ρ
221, 5.121555.121555.1215 mmmmmmmmmmbtbtA ewewwebeff =×+×=×+×=
2, 330022015 mmmmmmbtA ffflangeeff =×=×=
222,, 2.7815330025.12152 mmmmmmAAA flangeeffwebeffeff =×+=×+=
ADVANCE DESIGN VALIDATION GUIDE
124
13.22.2.4Effective elastic section modulus of Class4 cross-sections - In order to simplify the calculation the section will be considered in pure bending
mmmmbb tc 4252
8501sup
inf ===⇒−==σσψ
9.231 =⇒−= σψ k
According to EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1
1705
152880=
×−=
mmmmmm
tc
σελ
kt
bw
p ××=
4.28
wb is the width of the web; mmbw 850=
t is the web thickness; t=5mm
9244.0275235235
===yf
ε
325.19.239244.04.28
5850
=××
= mmmm
pλ
According to EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4
( )692.0
325.12055.0325.13055.0
023
673.0325.122 =
×−=
+×−=⇒
≥=+
>=
p
pp
λ
ψλρ
ψ
λ
( )
=×==×=
=−−
×=
⇒
×=
×=−
×=×=
mmmmbmmmmb
mmmmb
bbbb
bbb
e
e
eff
effe
effe
wceff
46.1761.2946.064.1171.2944.0
1.29411
850692.0
6.0
4.01
2
1
2
1
ψρρ
ADVANCE DESIGN VALIDATION GUIDE
125
-the weight center coordinate is:
( ) ( ) ( ) ( )
mm
yG
53.155.10195095.158330
15220546.601564.1171522027.124546.6015.4321522018.366564.1175.43215220
−=−
=
=×+×+×+×
××−××−××+××=
-the inertial moment along the strong axis is:
4
23
23
23
23
14482344295.6993800726.748854356
97.4161522012
2201574.107546.60112
546.601
71.381564.11712
564.11703.4481522012
22015
mm
I y
=+=
=××+×
+××+×
+
+××+×
+××+×
=
423
23
23
23
63.2662749001522012
152200546.60112
46.6015
0564.11712
64.117501522012
15220
mm
I z
=××+×
+××+×
+
+××+×
+××+×
=
34
max, 533.3179229
53.4551448234429 mm
mmmm
zI
W yyel ===
34
max, 10.242068
11063.26627490 mm
mmmm
yIW z
zel ===
ADVANCE DESIGN VALIDATION GUIDE
126
13.22.2.5Buckling verification a) over the strong axis of the section, y-y:
- the imperfection factor α will be selected according to Tables 6.1 and 6.2:
34.0=α
ADVANCE DESIGN VALIDATION GUIDE
127
Coefficient corresponding to non-dimensional slenderness after Y-Y axis:
yχ coefficient corresponding to non-dimensional slenderness yλ will be determined from the relevant buckling
curve according to:
1122
≤−Φ+Φ
=yyy
yλ
χ (6.49)
yλ the non-dimensional slenderness corresponding to Class 4 cross-sections:
ycr
yeffy N
fA
,
*=λ
Where: A is the effective cross section area; 22.7815 mmAeff = ; fy is the yielding strength of the material;
fy=275N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
( ) kNNmm
mmmmNl
IEN
fy
yycr 37.9503544.95035371
²56201448234429/210000²
²² 42
, ==××
=××
=ππ
15.044.95035371
/2752.7815 22
,
=×
=×
=N
mmNmmN
fA
ycr
yeffyλ
[ ] ( )[ ] 503.015.02.015.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ
1
1
017.115.0503.0503.0
112222 =⇒
≥
=−+
=−Φ+Φ
=y
y
yyy
yχ
χ
λχ
ADVANCE DESIGN VALIDATION GUIDE
128
b) over the weak axis of the section, z-z:
- the imperfection factor α will be selected according to Tables 6.1 and 6.2:
49.0=α
Coefficient corresponding to non-dimensional slenderness after Z-Z axis:
zχ coefficient corresponding to non-dimensional slenderness zλ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=zzz
zλ
χ (6.49)
zλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
zcr
yeffz
NfA
,
*=λ
Where: A is the effective cross section area; 22.7815 mmAeff = ; fy is the yielding strength of the material;
fy=275N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
ml fz 81.2= because of the torsional buckling restraint from the middle of the column
ADVANCE DESIGN VALIDATION GUIDE
129
( ) kNNmm
mmmmNl
IENfz
zzcr 35.698962.6989347
²281063.26627490/210000²
²² 42
, ==××
=××
=ππ
555.062.6989347
/2752.7815 22
,
=×
=×
=N
mmNmmN
fA
zcr
yeffzλ
[ ] ( )[ ] 741.0555.02.0555.049.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ
812.0
1
812.0555.0741.0741.0
112222 =⇒
≤
=−+
=−Φ+Φ
=z
z
zzz
zχ
χλ
χ
13.22.2.6Lateral-torsional buckling verification a) for the top part of the column:
The elastic moment for lateral-torsional buckling calculation, Mcr:
- the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
z
t
z
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3)
- where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
ψ is the fraction of the bending moment from the column extremities: 50.01274637
==kNmkNmψ
ADVANCE DESIGN VALIDATION GUIDE
130
31.15.01274637
1 =⇒== CkNmkNmψ
According to EN 1993-1-1-AN France; Chapter 3.2; Table 1
Flexion inertia moment around the Y axis: 41448234429mmI y =
Flexion inertia moment around the Z axis: 463.26627490 mmI 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
2fz
w
thII
−×=
h cross section height; h=880mm
ft flange thickness; mmt f 15=
( ) 61124
10808.494
15880mm 326627490.6 mmmmmmIw ×=−×
=
According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column part: L=2810mm
Shear modulus of rigidity: G=80800N/mm2
( )( )
kNmNmm
mmNmmmmNmmmmNmm
mmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
43.40224022433856
32.439626.698934731.163.26627490/210000
514614/80800281063.26627490
10808.49
281063.26627490/21000031.1
²²
²²
422
422
4
611
2
422
1
==
=××=××
××+
××
×××
×=××××
+×××
×=
π
ππ
π
The elastic modulus : 34
max, 533.3179229
53.4551448234429 mm
mmmm
zI
W yyel ===
466.04022433856
/275533.3179229 23, =
×==
NmmmmNmm
MfW
cr
yyeffLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=
The cross section buckling curve will be chosen according to Table 6.4:
24220880
>==mmmm
bh
ADVANCE DESIGN VALIDATION GUIDE
131
The imperfection factor α will be chose according to Table 6.3:
76.0=α
( )[ ] ( )[ ] 710.0²466.02.0466.076.015.0²2.015.0 =+−×+=+−×+= LTLTLTLT λλαφ
1803.0²466.0²710.0710.0
1
²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
b) for the bottom part of the column:
The elastic moment for lateral-torsional buckling calculation, Mcr:
- the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
z
t
z
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3)
- where:
C1 is a coefficient that depends on several parameters, such as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
ADVANCE DESIGN VALIDATION GUIDE
132
ψ is the fraction of the bending moment from the column extremities: 0637
0==
kNmψ
77.10 1 =⇒= Cψ
According to EN 1993-1-1-AN France; Chapter 3.2; Table 1
Flexion inertia moment around the Y axis: 41448234429mmI y =
Flexion inertia moment around the Z axis: 463.26627490 mmI 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
2fz
w
thII
−×=
h cross section height; h=880mm
ft flange thickness; mmt f 15=
( ) 61124
10808.494
15880mm 326627490.6 mmmmmmIw ×=−×
=
According to EN1993-1-1-AN France; Chapter 2 (…4)
Length of the column part: L=2810mm
Shear modulus of rigidity: G=80800N/mm2
ADVANCE DESIGN VALIDATION GUIDE
133
( )( )
kNmNmm
mmNmmmmNmmmmNmm
mmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
89.54345434891269
32.439626.698934777.163.26627490/210000
514614/80800281063.26627490
10808.49
281063.26627490/21000077.1
²²
²²
422
422
4
611
2
422
1
==
=××=××
××+
××
×××
×=××××
+×××
×=
π
ππ
π
The elastic modulus : 34
max, 533.3179229
53.4551448234429 mm
mmmm
zI
W yyel ===
401.05434891269
/275533.3179229 23, =
×==
NmmmmNmm
MfW
cr
yyeffLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with the formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=
The cross section buckling curve will be chose according to Table 6.4:
24220880
>==mmmm
bh
The imperfection factor α will be chose according to Table 6.3:
76.0=α
( )[ ] ( )[ ] 657.0²401.02.0401.076.015.0²2.015.0 =+−×+=+−×+= LTLTLTLT λλαφ
1849.0²401.0²657.0657.0
1
²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
13.22.2.7Internal 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:
ADVANCE DESIGN VALIDATION GUIDE
134
a) for the top part of the column:
ycr
Ed
ymLTmyyy
NNCCk
,
1−××=
µ
ycr
Edy
ycr
Ed
y
NN
NN
,
,
1
1
×−
−
=χ
µ
1=yχ (previously calculated)
kNNEd 328=
kNNl
IEN
fy
yycr 37.9503544.95035371
²²
, ==××
=π
(previously calculated)
1
44.9503537132800011
44.950353713280001
1
1
,
, =×−
−=
×−
−
=
NN
NN
NN
NN
ycr
Edy
ycr
Ed
y
χµ
ADVANCE DESIGN VALIDATION GUIDE
135
The myC will be calculated according to Table A.1:
Calculation of the 0λ term:
0
,0
cr
yyeff
MfW ×
=λ
According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2
34
max, 533.3179229
53.4551448234429 mm
mmmm
zI
W yyel ===
The calculation the 0crM will be calculated using 11 =C and 02 =C , therefore:
( )( )
kNmNmm
mmNmmmmNmmmmNmm
mmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
56.30703070560199
32.439626.6989347163.26627490/210000
514614/80800281063.26627490
10808.49
281063.26627490/2100001
²²
²²
422
422
4
611
2
422
10
==
=××=××
××+
××
×××
×=××××
+×××
×=
π
ππ
π
534.03070560199
/275533.3179229 23
0
,0 =
×=
×=
NmmmmNmm
MfW
cr
yyeffλ
ADVANCE DESIGN VALIDATION GUIDE
136
Calculation of the 4
,,1 1120.0
−×
−××
TFcr
Ed
zcr
Ed
NN
NNC term:
Where:
- for a symmetrical section for the both axis, TcrTFcr NN ,, =
××+××=
²,
2
0,
Tcr
wtTcr L
IEIGIAN π
The mass moment of inertia 0I
20 gzy zAIII ⋅++=
Flexion inertia moment around the Y axis: 467.1490580416 mmI y =
Flexion inertia moment around the Z axis: 417.26628854 mmI z =
Cross section area: 210850mmA =
Distance between the section neutral axis and the section geometrical center: 0=gz
44420 151720927017.2662885467.1490580416 mmmmmmIIzAIII zygzy =+=+=⋅++=
- the buckling length, TcrL , ,
mL Tcr 81.2, =
Torsional moment of inertia: 475.514614 mmIt =
Working inertial moment: 61110808.49 mmIw ×= (previously calculated)
Longitudinal elastic modulus: E = 210000 MPa
Shear modulus of rigidity: G=80800MPa
( )N
mmmmmmNmmmmN
mmmmN Tcr
24.9646886²2810
10808.49/21000075.514614/808001517209270
10850 6112242
4
2
,
=
=
×××+××=
π
NNEd 328000=
NNN TcrTFcr 24.9646886,, ==
kNNl
IENfz
zzcr 35.698962.6989347
²²
, ==××
=π
(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
24.96468863280001
62.6989347328000131.120.01120.0 44
,,1
=
=
−×
−××=
−×
−××
NN
NN
NN
NNC
TFcr
Ed
zcr
Ed
ADVANCE DESIGN VALIDATION GUIDE
137
Therefore: For the top part of the column:
( )
≥
−×
−
×=
=
×+
××−+=
⇒
=
−⋅
−⋅⋅>=
=
−×
−××
=
1
11
11
224.01120.0534.0
224.01120.0
534.0
,,
2
0,
0,0,
4
,,10
4
,,1
0
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
a
aCCC
NN
NNC
NN
NNC
ε
ε
λ
λ
The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.
The coefficient must be calculated considering the column over the entire height.
( )LTy
LTymymymy a
aCCC
×+
××−+=
ε
ε
11 0,0,
yeff
eff
Ed
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 3,, 533.3179229 mmWW yeffyel ==
55.9533.3179229
2.7815328000
1012743
26
,
, =××
=×=mm
mmNNmm
WA
NM
yeff
eff
Ed
Edyyξ
19996.01448234429
75.51461411 4
4
≈=−=−=mm
mmIIa
y
tLT
The 0mC coefficient is defined according to the Table A.2:
The bending moment is null at the end of the column, therefore: 0=ψ
( )ycr
Edmy N
NC,
0, 33.036.021.079.0 ×−×+×+= ψψ
ADVANCE DESIGN VALIDATION GUIDE
138
Where:
kNNl
IEN
fy
yycr 37.9503544.95035371
²²
, ==××
=π
(previously calculated)
( ) 79.044.95035371
32800033.0036.079.00, =×−×+=N
NCmy
( ) ( ) 949.0155.91
155.979.0179.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
Equivalent uniform moment factor, mLTC , calculation
- mLTC must be calculated separately for each column part, separated by the lateral buckling restraint
1
11,,
2 ≥
−×
−
×=
Tcr
Ed
zcr
Ed
LTmymLT
NN
NN
aCC
-the myC term used for mLTC calculation, must be recalculated for the corresponding column part (in this case the top column part)
-this being the case, the myC will be calculated using 5.0=ψ :
( )LTy
LTymymymy a
aCCC
ξ
ξ
+−+=
11 0,0,
( ) ( ) 895.044.95035371
32800033.05.036.05.021.079.033.036.021.079.0,
0, =×−×+×+=×−×+×+=N
NNNC
ycr
Edmy ψψ
55.9533.3179229
2.7815328000
1012743
26
,
, =××
=×=mm
mmNNmm
WA
NM
yeff
eff
Ed
Edyyξ (previously calculated)
( ) ( ) 974.0155.91
155.9895.01895.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
1
1
989.0
644.9328.01
9897.6328.01
1974.0
11
2
,,
2
=⇒
≤
=
−×
−
×=
−×
−
×=
mLT
mLT
Tcr
Ed
zcr
Ed
LTmymLT
C
C
NN
NN
aCC
Therefore the yyk term corresponding to the top part of the column will be:
952.0
44.950353713280001
11949.01
,
=−
××=−
××=
NN
NN
CCk
ycr
Ed
ymLTmyyy
µ
ADVANCE DESIGN VALIDATION GUIDE
139
b) for the bottom part of the column:
ycr
Ed
ymLTmyyy
NNCCk
,
1−××=
µ
534.03070560199
/275533.3179229 23
0
,0 =
×=
×=
NmmmmNmm
MfW
cr
yyeffλ
For the bottom part of the column:
260.0
24.96468863280001
62.6989347328000177.120.01120.0 44
,,1
=
=
−×
−××=
−×
−××
NN
NN
NN
NNC
TFcr
Ed
zcr
Ed
ADVANCE DESIGN VALIDATION GUIDE
140
Therefore:
For the bottom part of the column:
( )
≥
−×
−
×=
=
×+
××−+=
⇒
=
−⋅
−⋅⋅>=
=
−×
−××
=
1
11
11
260.01120.0534.0
260.01120.0
534.0
,,
2
0,
0,0,
4
,,10
4
,,1
0
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
a
aCCC
NN
NNC
NN
NNC
ε
ε
λ
λ
The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.
The coefficient must be calculated considering the column over the entire height.
( )LTy
LTymymymy a
aCCC
×+
××−+=
ε
ε
11 0,0,
yeff
eff
Ed
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 3,, 533.3179229 mmWW yeffyel ==
55.9533.3179229
2.7815328000
1012743
26
,
, =××
=×=mm
mmNNmm
WA
NM
yeff
eff
Ed
Edyyξ
19996.01448234429
75.51461411 4
4
≈=−=−=mm
mmIIa
y
tLT
The 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 ===kNmM
M
Ed
Edψ
ADVANCE DESIGN VALIDATION GUIDE
141
( )ycr
Edmy N
NC,
0, 33.036.021.079.0 ×−×+×+= ψψ
Where:
kNNl
IEN
fy
yycr 37.9503544.95035371
²²
, ==××
=π
(previously calculated)
( ) 79.044.95035371
32800033.0036.079.00, =×−×+=N
NCmy
( ) ( ) 949.0155.91
155.979.0179.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
Equivalent uniform moment factor, mLTC , calculation
- mLTC must be calculated separately for each column part, separated by the lateral buckling restraint
1
11,,
2 ≥
−×
−
×=
Tcr
Ed
zcr
Ed
LTmymLT
NN
NN
aCC
-the myC term used for mLTC calculation, must be recalculated for the corresponding column part (in this case the top column part)
-this being the case, the myC will be calculated using 5.0=ψ :
( )LTy
LTymymymy a
aCCC
ξ
ξ
+−+=
11 0,0,
( ) ( ) 895.044.95035371
32800033.05.036.05.021.079.033.036.021.079.0,
0, =×−×+×+=×−×+×+=N
NNNC
ycr
Edmy ψψ
77.4533.3179229
2.7815328000
106373
26
,
, =××
=×=mm
mmN
NmmWA
NM
yeff
eff
Ed
Edyyξ
( ) ( ) 967.0177.41
177.4895.01895.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
1
1
974.0
24.96468863280001
62.69893473280001
1967.0
11
2
,,
2
=⇒
≤
=
−×
−
×=
−×
−
×=
mLT
mLT
Tcr
Ed
zcr
Ed
LTmymLT
C
C
NN
NN
NN
NN
aCC
952.0
44.950353713280001
11949.01
,
=−
××=−
××=
NN
NN
CCk
ycr
Ed
ymLTmyyy
µ
ADVANCE DESIGN VALIDATION GUIDE
142
Note:
The software does not give the results of the lower section because it is not the most solicited segment.
13.22.2.8Internal factor, yzk , calculation
a) for the top part of the column:
zcr
Ed
ymzyz
NN
Ck
,
1−×=
µ
The
mzC term must be calculated for the hole column length
=== 0
12740
,sup
,inf
kNmMM
Ed
Edψ
( ) ( ) 784.062.6989347
32800033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
823.0
62.69893473280001
1784.01
,
=−
×=−
×=
NN
NN
Ck
zcr
Ed
ymzyz
µ
b) for the bottom part of the column:
823.0
62.69893473280001
1784.01
,
=−
×=−
×=
NN
NN
Ck
zcr
Ed
ymzyz
µ
Note:
The software does not give the results of the lower section because it is not the most solicited segment.
13.22.2.9Internal factor, zyk , calculation
a) for the top part of the column:
ycr
Ed
zmLTmyzy
NN
CCk
,
1−××=
µ
991.0
62.6989347328000812.01
62.69893473280001
1
1
,
, =×−
−=
×−
−=
NN
NN
NN
NN
zcr
Edz
zcr
Ed
z
χµ
944.0
44.950353713280001
991.01949.01
,
=−
××=−
××=
NN
NN
CCk
ycr
Ed
zmLTmyzy
µ
ADVANCE DESIGN VALIDATION GUIDE
143
b) for the bottom part of the column:
944.0
44.950353713280001
991.01949.01
,
=−
××=−
××=
NN
NN
CCk
ycr
Ed
zmLTmyzy
µ
Note:
The software does not give the results of the lower section because it is not the most solicited segment.
13.22.2.10Internal factor, zzk , calculation
a) for the top part of the column:
815.0
62.69893473280001
991.0784.01
,
=−
×=−
×=
NN
NN
Ck
zcr
Ed
zmzzz
µ
b) for the bottom part of the column:
815.0
62.69893473280001
991.0784.01
,
=−
×=−
×=
NN
NN
Ck
zcr
Ed
zmzzz
µ
Note:
The software does not give the results of the lower section because it is not the most solicited segment.
13.22.2.11Bending and axial compression verification
∆+×+
×
∆+×+
×
∆+×+
×
∆+×+
×
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rkz
RdzEdzzz
M
RkyLT
RdyEdyzy
M
Rkz
Ed
M
Rkz
RdzEdzyz
M
RkyLT
RdyEdyyy
M
Rky
Ed
MMM
kMMM
kNN
MMM
kMMM
kNN
γγχ
γχ
γγχ
γχ
iyRk AfN ×=
ADVANCE DESIGN VALIDATION GUIDE
144
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
1/27510.242068
104.127823.0
1/275533.3179229803.0
101274952.0
1/2752.78151
328000
23
6
23
6
22
23
6
23
6
22
=++=×
××+
+×
×
××+
××
=++=×
××+
+×
×
××+
××
mmNmmNmm
mmNmmNmm
mmNmmN
mmNmmNmm
mmNmmNmm
mmNmmN
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
=++=×
××+
+×
×
××+
××
=++=×
××+
+×
×
××+
××
mmNmmNmm
mmNmmNmm
mmNmmN
mmNmmNmm
mmNmmNmm
mmNmmN
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.
ADVANCE DESIGN VALIDATION GUIDE
145
yχ coefficient corresponding to non-dimensional slenderness yλ
Column subjected to axial and shear force to the top
yχ
zχ coefficient corresponding to non-dimensional slenderness zλ
Column subjected to axial and shear force to the top
zχ
ADVANCE DESIGN VALIDATION GUIDE
146
Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yzk
Column subjected to axial and shear force to the top
yzk
ADVANCE DESIGN VALIDATION GUIDE
147
Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zzk
Column subjected to axial and shear force to the top
zzk
ADVANCE DESIGN VALIDATION GUIDE
148
Bending and axial compression verification term depending of the compression effort over the Y axis: SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
SNy
Bending and axial compression verification term depending of the Y bending moment over the Y axis: SMyy
Bending and axial compression verification term depending of the Y bending moment over the Y axis
SMyy
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149
Bending and axial compression verification term depending of the Z bending moment over the Y axis: SMyz
Bending and axial compression verification term depending of the Z bending moment over the Y axis
SMyz
Bending and axial compression verification term depending of the compression effort over the Z axis: SNz
Bending and axial compression verification term depending of the compression effort over the Z axis
SNz
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150
Bending and axial compression verification term depending of the Y bending moment over the Z axis: SMzy
Bending and axial compression verification term depending of the Y bending moment over the Z axis
SMzy
Bending and axial compression verification term depending of the Z bending moment over the Z axis: SMzz
Bending and axial compression verification term depending of the Z bending moment over the Z axis
SMzz
ADVANCE DESIGN VALIDATION GUIDE
151
13.22.2.12Reference results
Result name Result description Reference value
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
1
zχ zχ coefficient corresponding to non-dimensional slenderness zλ
0.81
yyk Internal factor, yyk
0.95
yzk Internal factor, yzk
0.82
zyk Internal factor, zyk
0.94
zyk Internal factor, zyk
0.82
SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
0.15
SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis
1.72
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
1.58
SNz
Bending and axial compression verification term depending of the compression effort over the z axis
0.19
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
1.71
SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
1.56
13.22.3Calculated results Result name Result description Value Error Xy Coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 % Xz Coefficient corresponding to non-dimensional slenderness 0.811841
adim 0.2273 %
Kyy Internal factor, kyy 0.950358 adim
0.0377 %
Kyz Internal factor, kyz 0.823048 adim
0.3717 %
Kzy Internal factor, kzy 0.941635 adim
0.1739 %
Kzz Internal factor, kzz 0.815493 adim
-0.5496 %
SNy Bending and axial compression verification term depending of the compression effort over the Y axis
0.152455 adim
1.6367 %
SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis
1.72357 adim
0.2076 %
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
1.57508 adim
-0.3114 %
SNz Bending and axial compression verification term depending of the compression effort over the z axis
0.187789 adim
-1.1637 %
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
1.70775 adim
-0.1316 %
SMzy Bending and axial compression verification term depending of the Z bending moment over the Z axis
1.70775 adim
-0.1316 %
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152
13.23 EC3 Test 19: Verifying the buckling resistance for a IPE300 column
Test ID: 5699
Test status: Passed
13.23.1Description The test verifies the buckling resistance for a IPE300 column made of S235 steel.
The verifications are made according to Eurocode3 French Annex.
13.23.2Background Classification and verification under compression efforts for an IPE 300 column made of S235 steel. The column is fixed at its base and free on the top. The column is subjected to a compression force (200 000 N) applied at its top. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.23.2.1Model 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=603.80x104mm4 ■ Flexion inertia moment around the Z axis: Iz=8356x104 mm4
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
ADVANCE DESIGN VALIDATION GUIDE
153
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.
13.23.2.2Buckling in the strong inertia of the profile (along Y-Y) The calculations are made in order to obtain the buckling resistance work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.
%100100,
≤×Rdb
Ed
NN
(6.46)
The design buckling resistance of the compressed element is calculated using the next formula:
1,
M
yRdb
fAN
γχ ××
= (6.47)
Where:
Coefficient corresponding to non-dimensional slenderness for Y-Y axis
χ coefficient corresponding to non-dimensional slenderness λ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=λ
χ (6.49)
λ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
cr
y
NfA*
=λ
Where: A is the cross section area; A=5380mm2; fy is the yielding strength of the material; fy=235N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
( )N
mmmmMPa
LIEN
fy
zcr 70.1731878
10000108356210000²
²²
2
44
=×××
=××
=ππ
854.070.1731878
/2355380 22
=×
=×
=N
mmNmmN
fA
cr
yλ
[ ]²)2.0(15.0 λλαφ +−+=
ADVANCE DESIGN VALIDATION GUIDE
154
The following buckling curve will be used:
The imperfection factor α corresponding to the appropriate buckling curve will be 0.21:
[ ] ( )[ ] 933.0854.02.0854.021.015.0²)2.0(15.0 2 =+−×+×=+−×+×= λλαφ
Therefore:
1764.0854.0933.0933.0
112222
≤=−+
=−+
=λφφ
χ
1Mγ is a safety coefficient, 11 =Mγ
NmmNmmN Rdb 089.9660511
/2355380764.0 22
, =××
=
NNEd 200000=
%703.20100089.966051
200000100,
=×=×N
NNN
Rdb
Ed
ADVANCE DESIGN VALIDATION GUIDE
155
13.23.2.3Buckling in the weak inertia of the profile (along Z-Z) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.
%100100,
≤×Rdb
Ed
NN
(6.46)
The design buckling resistance of the compressed element is calculated using the next formula:
1,
M
yRdb
fAN
γχ ××
= (6.47)
Where:
Coefficient corresponding to non-dimensional slenderness for Z-Z axis
χ coefficient corresponding to non-dimensional slenderness λ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=λ
χ (6.49)
λ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
cr
y
NfA*
=λ
Where: A is the cross section area; A=5380mm2; fy is the yielding strength of the material; fy=235N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
( )N
mmmmMPa
LIEN
fy
zcr 610.125144
100001080.603210000²
²²
2
44
=×××
=××
=ππ
178.3610.125144
/2355380 22
=×
=×
=N
mmNmmN
fA
cr
yλ
[ ]²)2.0(15.0 λλαφ +−+=
It will be used the following buckling curve:
ADVANCE DESIGN VALIDATION GUIDE
156
The imperfection factor α corresponding to the appropriate buckling curve will be 0.34:
[ ] ( )[ ] 056.6178.32.0178.334.015.0²)2.0(15.0 2 =+−×+×=+−×+×= λλαφ
Therefore:
1089.0178.3056.6056.6
112222
≤=−+
=−+
=λφφ
χ
1Mγ is a safety coefficient, 11 =Mγ
NmmNmmN Rdb 78.1127711
/2355380089.0 22
, =××
=
NNEd 200000=
%349.17710078.112771
200000100,
=×=×N
NNN
Rdb
Ed
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
157
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)
ADVANCE DESIGN VALIDATION GUIDE
158
13.23.2.4Reference results
Result name Result description Reference value
yχ coefficient corresponding to non-dimensional slenderness after Y-Y axis
0.764
zχ coefficient corresponding to non-dimensional slenderness after Z-Z axis
0.089
Work ratio (y-y) Ratio of the design normal force to design buckling resistance (strong inertia) [%]
0.21 %
Work ratio (z-z) Ratio of the design normal force to design buckling resistance (weak inertia) [%]
1.77 %
13.23.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness
after Y-Y axis 0.763129 adim
-0.1140 %
Xz coefficient corresponding to non-dimensional slenderness after Z-Z axis
0.0891543 adim
0.1734 %
SNy Ratio of the design normal force to design buckling resistance in the strong inertia of the profile
0.207253 adim
-1.3081 %
SNz Ratio of the design normal force to design buckling resistance in the weak inertia of the profile
1.77401 adim
0.2266 %
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13.24 EC3 Test 12: Verifying the design plastic shear resistance of a rectangular hollow section beam
Test ID: 5706
Test status: Passed
13.24.1Description Verifies the design plastic shear resistance of a rectangular hollow section beam made of S275 steel.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.24.2Background Verifies the adequacy of a rectangular hollow section beam made of S275 steel to resist shear. Verification of the shear resistance at ultimate limit state is realised. The name of the cross-section is RC3020100 and can be found in the Advance Design OTUA library. The beam is simply supported and it is subjected to an uniformly distributed load (50 000 N/ml) applied at its top. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.24.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = -50 000 N/ml,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in millimeters (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 ,
■ Partial factor for resistance of cross sections: 0.10 =Mγ .
ADVANCE DESIGN VALIDATION GUIDE
160
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength fy = 275 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation along Y, Z axis and restrained in rotation along
X axis. ■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ External: Uniformly distributed load: q = Fz = -50 000 N/ml,
■ Internal: None.
13.24.2.2Reference results for calculating the design plastic shear resistance of the cross-section The design plastic shear resistance of the cross-section is determined with formula (6.18) from EN 1993-1-1:2001. Before using it, the shear area (Av) has to be determined.
Shear area of the cross section
For a rectangular hollow section of uniform thickness the shear area is determined according to chapter 6.2.6 (3) from EN 1993-1-1. As the load is parallel to depth, the shear area is:
22
56942003003009490 mm
mmmmmmmm
hbhAAv =
+×
=+×
=
Design plastic shear resistance of the cross section
EN 1993-1-1: 2011 provides the following formula to calculate the design plastic shear resistance of the cross-section:
N
MPammf
AV
M
yv
Rdpl 9040440.1
32755694
32
0, =
×=
×=
γ
Work ratio
The verification of the design plastic shear resistance is done with relationship (6.17) from EN 1993-1-1. The corresponding work ratio is:
Work ratio = %83.13100904044125000100
9040442
5000/50
1002100,,
=×=×
×
=×
×
=×NN
N
mmmmN
V
Lq
VV
RdplRdpl
Ed
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
161
Finite elements results
Work ratio of the design shear resistance
Beam subjected to uniformly distributed load Work ratio - Fz
13.24.2.3Reference results
Result name Result description Reference value Work ratio - Fz Work ratio of the design shear resistance [%] 13.83 %
13.24.3Calculated results Result name Result description Value Error Work ratio - Fz Shear resistance work ratio 13.8219 % 0.1587 %
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162
13.25 EC3 Test 27: Verifying an user defined I section class 3 beam simply supported with a displacement restraint
Test ID: 5717
Test status: Passed
13.25.1Description The test verifies a user defined cross section beam. The beam is hinged at one end and the translations over the Y and Z axis and rotation after the X axis are blocked.
The cross section has an “I symmetric” shape with: 530mm height; 190mm width; 12mm center thickness; 19mm flange thickness; 0mm fillet radius and 0mm rounding radius.
The beam is subjected to 10 kN/m linear force applied vertically, 5 kN/m linear force applied horizontally and 3700 kN punctual force applied on the end of the beam.
The calculations are made according to Eurocode 3 French Annex.
13.25.2Background An I53*1.2+22*1.9 beam column subjected to axial compression, uniform distributed vertical force and uniform distributed horizontal force, made from S235 steel. The beam has a 53x12mm web and 220x19mm flanges. The beam is simply supported. The beam is subjected to an axial compression load 3700000 N, 10000 N/m uniform distributed load over the Z axis and 5000 N/m horizontal uniform distributed force after the Y axis.
This test was evaluated by the French control office SOCOTEC.
13.25.2.1Model 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).
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163
Units
Metric System
Geometrical properties
■ Column length: L=5000mm
■ Cross section area: 214264mmA =
■ Overall breadth: mmb 220=
■ Flange thickness: mmt f 19=
■ Root radius: mmr 0=
■ Web thickness: mmtw 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 mmWy =
■ Elastic modulus after the Z axis, 3, 41.307177 mmW zel =
■ Plastic modulus after the Z axis, 3, 00.477512 mmW zpl =
■ Flexion inertia moment around the Y axis: 467.665089874 mmI y =
■ Flexion inertia moment around the Z axis: 467.33789514 mmI z =
■ Torsional moment of inertia: 473.1269555 mmIt =
■ Working inertial moment: 667.6662201162989 mmI w =
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 210000 MPa. ■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis,
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Support at the end point (x = 5) restrained in translation along Y and Z axis and restrained rotation along X axis.
■ Inner: Lateral buckling restraint in the middle of the column (x=2.50).
Loading
The beam is subjected to the following loadings: ■ External: Point load from X=f.00m and z=.00m: Fx =-3700000N; ■ External: vertical uniform distributed linear load from X=0.00 to X=5.00: Fz=-10000N/m ■ External: horizontal uniform distributed linear load from X=0.00 to X=5.00: Fy=-5000N/m
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13.25.2.2Cross section Class According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
194.24694.246
sup
inf ===MpaMpa
σσψ
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1235235235
===yf
ε
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42424138381
4112
192530=×≤=<=×⇒
=
=×−
=εε
ε tc
mmmmmm
tc
therefore the beam web is
considered to be Class 3
-for beam flange:
91957.4
1
47.5192
12220
=×≤=⇒
=
=
−
=tc
tc
ε
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 3.
13.25.2.3Buckling verification a) over the strong axis of the section, y-y:
-the imperfection factor α will be selected according to the Table 6.1 and 6.2:
34.0=α
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Coefficient corresponding to non-dimensional slenderness after Y-Y axis:
yχ coefficient corresponding to non-dimensional slenderness yλ will be determined from the relevant buckling
curve according to:
1122
≤−Φ+Φ
=yyy
yλ
χ (6.49)
yλ the non-dimensional slenderness corresponding to Class 3 cross-sections:
ycr
yy N
fA
,
*=λ
Cross section area: 214264mmA =
Flexion inertia moment around the Y axis: 467.665089874 mmI y =
( ) kNNmm
mmmmNl
IEN
fy
yycr 06.5513921.55139061
²500067665089874./210000²
²² 42
, ==××
=××
=ππ
247.021.55139061
/23514264 22
,
=×
=×
=N
mmNmmN
fA
ycr
yyλ
[ ] ( )[ ] 538.0247.02.0247.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ
984.0
1
984.0247.0538.0538.0
112222 =⇒
≤
=−+
=−Φ+Φ
=y
y
yyy
yχ
χ
λχ
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:
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169
49.0=α
Coefficient corresponding to non-dimensional slenderness after Z-Z axis:
zχ coefficient corresponding to non-dimensional slenderness zλ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=zzz
zλ
χ (6.49)
zλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
zcr
yz
NfA
,
*=λ
ml fz 50.2=
Flexion inertia moment around the Z axis: 467.33789514 mmI z =
Cross section area: 214264mmA =
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( ) kNNmm
mmmmNl
IEN
fz
zzcr 235.1120519.11205235
²250067.33789514/210000²
²² 42
, ==××
=××
=ππ
547.019.11205235
/23514264 22
,
=×
=×
=N
mmNmmN
fA
zcr
yzλ
[ ] ( )[ ] 735.0547.02.0547.049.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ
816.0
1
819.0547.0735.0735.0
112222 =⇒
≤
=−+
=−Φ+Φ
=z
z
zzz
zχ
χλ
χ
13.25.2.4Lateral torsional buckling verification The elastic moment for lateral-torsional buckling calculation, Mcr:
-it must be studied separately for each beam segment
-however, the two sections are symmetrical, the same result will be obtained
µ is the isotactic moment report (for simply supported bar) due to Q load ant the maxim moment value
( ) 25.01025.318
²2500/100008
²3 =
×××
=××
=Nm
mmmNMLqµ
-the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
z
t
z
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3) -where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
According to EN 1993-1-1-AN France; Chapter 3; …(6)
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0=ψ therefore:
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31.11 =C
Flexion inertia moment around the Y axis: 467.665089874 mmI y =
Flexion inertia moment around the Z axis: 467.33789514 mmI z =
Longitudinal elastic modulus: 2/210000 mmNE =
Torsional moment of inertia: 473.1269555 mmIt =
Working inertial moment: 667.6662201162989 mmI w =
Shear modulus of rigidity: 2/80800 mmNG =
Buckling length of the beam mmL 2500=
Elastic modulus after the Y axis, 3, 11.2509773 mmW yel =
( )( )
Nmm
mmNmmmmN
mmmmNmmmmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
4001163141
58.27219.1120523531.167.33789514/210000
73.1269555/80800250067.337895141001163.22
250067.33789514/21000031.1
²²
²²
422
422
4
611
2
422
1
=
=××=××
××+
××
×××
×=××××
+×××
×=
π
ππ
π
384.04001163141
/23511.2509773 23, =
×==
NmmmmNmm
MfW
cr
yyeffLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined below:
Note:
36
1081.74001163141
1025.31 −×=×
=Nmm
MM
cr
Ed
0156.000781.0
0156.0125.05302203.03.0
04.01081.7
20.0384.0
20,
20,0,
20,3
=≤=⇒
=⇒=×=×=
≤⇒
≤×=
>=
−
LT
cr
Ed
LTLT
LT
cr
Ed
cr
Ed
LT
MM
hb
MM
MM
λ
λλ
λλ
According to EN 1993-1-1-AN France; AN.3; Chapter 6.3.2.2(4)
For slendernesses 2
0,LT
cr
Ed
MM λ≤ (see 6.3.2.3) lateral torsional buckling effects may be ignored and only cross sectional checks
apply.
-therefore:
1=LTχ
According to EN 1993-1-1-AN (2005); Chapter 6.3.2.2(4)
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173
13.25.2.5Internal factor, yyk , calculation
The internal factor yyk corresponding to a Class 4 section will be calculated according to Annex A, Table a.1, and will be
calculated separately for the two column parts separate by the middle torsional lateral restraint:
ycr
Ed
ymLTmyyy
NNCCk
,
1−××=
µ
ycr
Edy
ycr
Ed
y
NN
NN
,
,
1
1
×−
−
=χ
µ
984.0=yχ (previously calculated)
kNN Ed 3700=
( ) Nmm
mmmmNl
IEN
fy
yycr 21.55139061
²500067665089874./210000²
²² 42
, =××
=××
=ππ
(previously
calculated)
999.0
21.551309613700000984.01
21.5513096137000001
1
1
,
, =×−
−=
×−
−
=
NN
NN
NN
NN
ycr
Edy
ycr
Ed
y
χµ
- Cmy coefficient takes into account the behavior in the plane of bending (buckling in the plan and distribution of the bending moment).
- Must be calculated considering the beam along its length.
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174
The myC will be calculated according to Table A.1:
Calculation of the 0λ term:
0
,0
cr
yyeff
MfW ×
=λ
-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2
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175
34
max, 11.2509773
26567.665089874 mm
mmmm
zI
W yyel ===
The calculation the 0crM will be calculated using 11 =C and 02 =C , therefore:
( )( )
Nmm
mmNmmmmN
mmmmNmmmmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
3054323008
58.27219.11205235167.33789514/210000
73.1269555/80800250067.337895141001163.22
250067.33789514/2100001
²²
²²
422
422
4
611
2
422
10,
=
=××=××
××+
××
×××
×=××××
+×××
×=
π
ππ
π
439.03054323008
/23511.2509773 23
0,
,0 =
×==
NmmmmNmm
MfW
cr
yyeffλ
Calculation of the 4
,,1 1120.0
−×
−××
TFcr
Ed
zcr
Ed
NN
NNC term:
Where:
-for a symmetrical section for the both axis, TcrTFcr NN ,, =
××+××=
²1
,
2
20
,Tcr
wtTcr L
IEIGi
N π
0491.020
20
2220 =+++= zyiii zy
Torsional moment of inertia: 473.1269555 mmI t =
Working inertial moment: 667.6662201162989 mmI w =
- the buckling length, TcrL , ,
mL Tcr 50.2, =
( )N
mmmmmmNmmmmN
mmmmN Tcr
13
62242
4
2
,
10696.1
²250067.6662201162989/21000073.1269555/80800
0491.01
×=
=
××+××=
π
NN Ed 3700000=
NNN TcrTFcr13
,, 10696.1 ×==
( ) Nmm
mmmmNl
IENfz
zzcr 19.11205235
²250067.33789514/210000²
²² 42
, =××
=××
=ππ
(previously calculated)
C1=1
181.0
10696.137000001
19.1120523537000001120.01120.0 4
134
,,1
=
=
×−×
−××=
−×
−××
NN
NN
NN
NN
CTFcr
Ed
zcr
Ed
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Therefore:
( )
≥
−×
−
×=
=
×+
××−+=
⇒
=
−⋅
−⋅⋅>=
=
−×
−××
=
1
11
11
181.01120.0439.0
181.01120.0
439.0
,,
2
0,
0,0,
4
,,10
4
,,1
0
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
a
aCCC
NN
NNC
NN
NNC
ε
ε
λ
λ
The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.
The coefficient must be calculated considering the column over the entire height.
( )LTy
LTymymymy a
aCCC
×+
××−+=
ε
ε
11 0,0,
yeff
eff
Ed
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 3,, 11.2509773 mmWW yeffyel ==
048.011.2509773
142643700000
1025.313
26
,
, =××
=×=mm
mmNNmm
WA
NM
yeff
eff
Ed
Edyyξ
998.067.665089874
73.126955511 4
4
=−=−=mm
mmIIa
y
tLT
The 0mC coefficient is defined according to the Table A.2:
ycr
Edmy N
NC,
0, 03.01 ×+=
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Where:
( ) Nmm
mmmmNl
IEN
fy
yycr 21.55139061
²500067665089874./210000²
²² 42
, =××
=××
=ππ
(previously calculated)
NN Ed 3700000=
002.121.55139061
370000003.010, =×+=N
NCmy
( ) ( ) 002.1100481
1048.0002.11002.11
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
Equivalent uniform moment factor, mLTC , calculation
- It must be calculated for each of the two sections.
−×
−
×=
Tcr
Ed
zcr
Ed
LTmymLT
NN
NN
aCC
,,
2
11
It must again calculate the coefficient Cmy, but only for the left section.
mz 0001396.0−=δ
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( )999.0
21.5513906137000001
1025.3125000001396.067665089874./2100001
11
62
422
,,2
2
0,
=×
−
×××××
+=
=×
−
×
×××+=
NN
NmmmmmmmmN
NN
MLIE
Cycr
Ed
Edy
ymy
π
δπ
( ) ( ) 999.0998.0219.01
998.0219.0999.01999.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ε
ε
377.1
10696.137000001
19.1120523537000001
998.0²999.0
11 13,,
2 =
×−×
−
×=
−×
−
×=
NN
NN
NN
NN
aCC
Tcr
Ed
zcr
Ed
LTmymLT
Therefore the yyk term corresponding to the top part of the column will be:
476.1
21.5513096137000001
1377.1002.11
,
=−
××=−
××=
NN
NN
CCk
ycr
Ed
ymLTmyyy
µ
13.25.2.6Internal factor, yzk , calculation
zcr
Ed
ymzyz
NN
Ck
,
1−×=
µ
Cmz coefficient must be calculated considering the beam along its length.
010.119.11205235
370000003.0103.01,
0, =×+=×+==N
NNN
CCzcr
Edmzmz
506.1
19.1120523537000001
1776.01
,
=−
×=−
×=
NN
NN
Ck
zcr
Ed
ymzyz
µ
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13.25.2.7Internal factor, zyk , calculation
ycr
Ed
zmLTmyzy
NN
CCk
,
1−××=
µ
917.0
19.112052353700000816.01
19.1120523537000001
1
1
,
, =×−
−=
×−
−=
NN
NN
NN
NN
zcr
Edz
zcr
Ed
z
χµ
816.0=zχ (previously calculated)
355.1
21.5513096137000001
917.0377.1002.11
,
=−
××=−
=
NN
NN
CCk
ycr
Ed
zmLTmyzy
µ
13.25.2.8Internal factor, zzk , calculation
383.1
19.1120523537000001
917.0776.01
,
=−
×=−
×=
NN
NNCk
zcr
Ed
zmzzz
µ
13.25.2.9Bending and axial compression verification
∆+×+
×
∆+×+
×
∆+×+
×
∆+×+
×
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rkz
RdzEdzzz
M
RkyLT
RdyEdyzy
M
Rkz
Ed
M
Rkz
RdzEdzyz
M
RkyLT
RdyEdyyy
M
Rky
Ed
MMM
kMMM
kNN
MMM
kMMM
kNN
γγχ
γχ
γγχ
γχ
iyRk AfN ×=
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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
=++=×
××+
+×
×
××+
××
=++=×
××+
+×
×
××+
××
mmNmmNmm
mmNmmNmm
mmNmmN
mmNmmNmm
mmNmmNmm
mmNmmN
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
yχ coefficient corresponding to non-dimensional slenderness yλ
Column subjected to axial and shear force to the top
yχ
zχ coefficient corresponding to non-dimensional slenderness zλ
Column subjected to axial and shear force to the top
zχ
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Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yzk
Column subjected to axial and shear force to the top
yzk
Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
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Internal factor, zzk
Column subjected to axial and shear force to the top
zzk
Bending and axial compression verification term depending of the compression effort over the Y axis: SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
SNy
Bending and axial compression verification term depending of the Y bending moment over the Y axis: SMyy
Bending and axial compression verification term depending of the Y bending moment over the Y axis
SMyy
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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 over the Y axis
SMyz
Bending and axial compression verification term depending of the compression effort over the Z axis: SNz
Bending and axial compression verification term depending of the compression effort over the Z axis
SNz
Bending and axial compression verification term depending of the Y bending moment over the Z axis: SMzy
Bending and axial compression verification term depending of the Y bending moment over the Z axis
SMzy
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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 moment over the Z axis
SMzz
13.25.2.10Reference results
Result name Result description Reference value
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
0.98
zχ zχ coefficient corresponding to non-dimensional slenderness zλ
0.82
yyk Internal factor, yyk
1.47
yzk Internal factor, yzk
1.51
zyk Internal factor, zyk
1.35
zzk Internal factor, zzk
1.38
SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
1.12
SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis
0.08
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
0.33
SNz
Bending and axial compression verification term depending of the compression effort over the z axis
1.35
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
0.07
SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
0.30
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13.25.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness 0.983441
adim 0.3511 %
Xz coefficient corresponding to non-dimensional slenderness 0.816369 adim
-0.4428 %
Kyy Internal factor, kyy 1.47276 adim
0.1878 %
Kyz Internal factor, kyz 1.50599 adim
-0.2656 %
Kzy Internal factor, kzy 1.35211 adim
0.1563 %
Kzz Internal factor, kzz 1.38261 adim
0.1891 %
SNy Bending and axial compression verification term depending of the compression effort over the Y
1.12239 adim
0.2134 %
SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis
0.078033 adim
-2.4587 %
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
0.325974 adim
-1.2200 %
SNz Bending and axial compression verification term depending of the compression effort over the z axis
1.35209 adim
0.1548 %
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
0.0716405 adim
2.3436 %
SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
0.29927 adim
-0.2433 %
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13.26 EC3 Test 26: Verifying an user defined I section class 3 column fixed on the bottom
Test ID: 5714
Test status: Passed
13.26.1Description The test verifies a user defined cross section column.
The cross section has an “I symmetric” shape with: 408mm height; 190mm width; 9.4mm center thickness; 14.6mm flange thickness; 0mm fillet radius and 0mm rounding radius.
The column is subjected to 1000kN axial compression force and a 200kNm bending moment after the Y axis. All the efforts are applied on the top of the column.
The calculations are made according to Eurocode 3 French Annex.
13.26.2Background An I40.8*0.94+19*1.46 shaped column subjected to compression and bending, made from S275 steel. The column has a 40.8x9.4mm web and 190x14.6mm flanges. The column is fixed at it’s base The column is subjected to an axial compression load -1000000 N, a 200000Nm bending moment after the Y axis and a 5000N lateral force after the Y axis.
This test was evaluated by the French control office SOCOTEC.
13.26.2.1Model 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
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187
■ 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=2000mm
■ Cross section area: 272.9108 mmA =
■ Overall breadth: mmb 190=
■ Flange thickness: mmt f 6.14=
■ Root radius: mmr 0=
■ Web thickness: mmtw 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 mmWy =
■ Elastic modulus after the Z axis, 3, 65.175962 mmW zel =
■ Plastic modulus after the Z axis, 69.271897, =zplW
■ Flexion inertia moment around the Y axis: Iy=257332751mm4 ■ Flexion inertia moment around the Z axis: Iz=16716452.10 mm4 ■ Torsional moment of inertia: It=492581.13 mm4 ■ Working inertial moment: Iw=645759981974.33mm6
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength fy = 275 MPa, ■ Longitudinal elastic modulus: E = 210000 MPa. ■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation 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
13.26.2.2Cross section Class According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
- for beam web:
The web dimensions are 378.8x9.4mm.
120.012750091.0
000.1212 −>−=−×
⋅=−⋅
⋅=y
Ed
fANψ
5.001.10094.03788.0275
11211
21
>=
××+⋅=
××+⋅=
dtfN
y
Edα
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189
924.0275235235
===yf
ε
25.64)20.0(33.067.0
924.04233.067.0
42170924.0
30.404.9
6.142408=
−×+×
=×+
×>=⇒
=
=×−
=ψ
ε
ε tc
mmmmmm
tc
therefore the beam web is considered to be Class 3.
- for beam flange:
316.8924.0918.6924.0
18.66.1430.90
=×≤=⇒
=
==tc
tc
ε therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 3.
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13.26.2.3Buckling verification a) over the strong axis of the section, y-y:
- the imperfection factor α will be selected according to Tables 6.1 and 6.2:
34.0=α
Coefficient corresponding to non-dimensional slenderness after Y-Y axis:
yχ coefficient corresponding to non-dimensional slenderness yλ will be determined from the relevant buckling
curve according to:
1122
≤−Φ+Φ
=yyy
yλ
χ (6.49)
yλ the non-dimensional slenderness corresponding to Class 4 cross-sections:
ycr
yy N
fA
,
*=λ
Cross section area: 272.9108 mmA =
Flexion inertia moment around the Y axis: Iy=257332751mm4
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191
( ) kNNmm
mmmmNl
IEN
fy
yycr 05.1333387.133338053
²2000257332751/210000²
²² 42
, ==××
=××
=ππ
137.07.133338053
/27572.9108 22
,
=×
=×
=N
mmNmmN
fA
ycr
yyλ
[ ] ( )[ ] 499.0137.02.0137.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ
1
1
022.1137.0499.0499.0
112222 =⇒
≤
=−+
=−Φ+Φ
=y
y
yyy
yχ
χ
λχ
b) over the weak axis of the section, z-z:
- the imperfection factor α will be selected according to Tables 6.1 and 6.2:
49.0=α
Coefficient corresponding to non-dimensional slenderness after Z-Z axis:
zχ coefficient corresponding to non-dimensional slenderness zλ will be determined from the relevant buckling curve according to:
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192
1122
≤−Φ+Φ
=zzz
zλ
χ (6.49)
zλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
zcr
yz
NfA
,
*=λ
ml fz 00.2=
Flexion inertia moment around the Z axis: Iz=16716452.10 mm4
Cross section area: 272.9108 mmA =
( ) kNNmm
mmmmNl
IENfz
zzcr 70.866138.8661700
²200010.16716452/210000²
²² 42
, ==××
=××
=ππ
538.038.8661700
/27572.9108 22
,
=×
=×
=N
mmNmmN
fA
zcr
yzλ
[ ] ( )[ ] 728.0538.02.0538.049.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ
821.0
1
821.0538.0728.0728.0
112222 =⇒
≤
=−+
=−Φ+Φ
=z
z
zzz
zχ
χλ
χ
13.26.2.4Lateral torsional buckling verification The elastic moment for lateral-torsional buckling calculation, Mcr:
- the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
z
t
z
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3)
-where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
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193
- in order to simplify the calculation, we will consider 11 =C
Flexion inertia moment around the Y axis: Iy=257332751mm4
Flexion inertia moment around the Z axis: Iz=16716452.10 mm4
Longitudinal elastic modulus: E = 210000 N/mm2.
Torsional moment of inertia: It=492581.13 mm4
Warping inertial moment:
IW is the warping inertia (deformation inertia moment):
( )4
2fz
w
thII
−×=
h cross section height; h=408mm
ft flange thickness; mmt f 6.14=
( ) 61124
1046774.64
6.144080mm16716452.1 mmmmmmIw ×=−×
=
According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column: L=2000mm
Shear modulus of rigidity: G=80800N/mm2
( )( )
kNmNmm
mmNmmmmN
mmmmNmmmmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
089.18021802088994
052.208384.8661700110.16716452/210000
13.492581/80800200010.16716452
1046774.6
200010.16716452/2100001
²²
²²
422
422
4
611
2
422
1
==
=××=××
××+
××
×××
×=××××
+×××
×=
π
ππ
π
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194
The elastic modulus : 34
max, 172.1261434
204257332751 mm
mmmm
zI
W yyel ===
439.01802088994
/275172.1261434 23, =
×==
NmmmmNmm
MfW
cr
yyeffLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=
The cross section buckling curve will be chose according to Table 6.4:
2147.2190408
>==mmmm
bh
The imperfection factor α will be chose according to Table 6.3:
( )[ ] ( )[ ] 687.0²439.02.0439.076.015.0²2.015.0 =+−×+×=+−×+×= LTLTLTLT λλαφ
1813.0²439.0²687.0687.0
1
²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.
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13.26.2.5Internal factor, yyk , calculation
The internal factor yyk corresponding to a Class 4 section will be calculated according to Annex A, Table a.1, and will be
calculated separately for the two column parts separate by the middle torsional lateral restraint:
ycr
Ed
ymLTmyyy
NNCCk
,
1−××=
µ
ycr
Edy
ycr
Ed
y
NN
NN
,
,
1
1
×−
−
=χ
µ
022.1=yχ (previously calculated)
kNNEd 1000=
kNNl
IEN
fy
yycr 05.1333387.133338053
²²
, ==××
=π
(previously calculated)
1
7.133338053100000011
7.13333805310000001
1
1
,
, =×−
−=
×−
−
=
NNN
N
NN
NN
ycr
Edy
ycr
Ed
y
χµ
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196
The myC will be calculated according to Table A.1:
Calculation of the 0λ term:
0
,0
cr
yyeff
MfW ×
=λ
According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2
34
max, 172.1261434
204257332751 mm
mmmm
zI
W yyel ===
The calculation the 0crM will be calculated using 11 =C and 02 =C , therefore:
( )( )
kNmNmm
mmNmmmmN
mmmmNmmmmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
089.18021802088994
052.208384.8661700110.16716452/210000
13.492581/80800200010.16716452
1046774.6
200010.16716452/2100001
²²
²²
422
422
4
611
2
422
1
==
=××=××
××+
××
×××
×=××××
+×××
×=
π
ππ
π
439.01802088994
/275172.1261434 23,
0 =×
==Nmm
mmNmmM
fW
cr
yyeffλ
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197
Calculation of the 4
,,1 1120.0
−×
−××
TFcr
Ed
zcr
Ed
NN
NNC term:
Where:
- for a symmetrical section for the both axis, TcrTFcr NN ,, =
××+××=
²1
,
2
20
,Tcr
wtTcr L
IEIGi
N π
The mass moment of inertia 0I
0301.020
20
2220 =+++= zyiii zy
Torsional moment of inertia: It=492581.13 mm4
Working inertial moment: Iw=645759981974.33mm6
- the buckling length, TcrL , ,
mL Tcr 00.2, =
( )N
mmmmmmNmmmmN
mmmmN Tcr
13
62242
4
2
,
10244.1
²200033.746457599819/21000013.492581/80800
0301.01
×=
=
××+××=
π
NNEd 1000000=
NNN TcrTFcr13
,, 10244.1 ×==
( ) Nmm
mmmmNl
IENfz
zzcr 38.8661700
²200010.16716452/210000²
²² 42
, =××
=××
=ππ
(previously calculated)
C1=1 for the top part of the column
For the top part of the column:
172.0
10244.110000001
38.866170010000001120.01120.0 4
134
,,1
=
=
×−×
−××=
−×
−××
NN
NN
NN
NNC
TFcr
Ed
zcr
Ed
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198
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
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
a
aCCC
NN
NNC
NN
NNC
ε
ε
λ
λ
The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.
The coefficient must be calculated considering the column over the entire height.
( )LTy
LTymymymy a
aCCC
×+
××−+=
ε
ε
11 0,0,
yeff
eff
Ed
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 3,, 06.1261435 mmWW yeffyel ==
444.106.1261435
72.91081000000
102003
26
,
, =××
=×=mm
mmN
NmmWA
NM
yeff
eff
Ed
Edyyξ
1998.0257332751
492581.1311 4
4
≈=−=−=mmmm
IIa
y
tLT
The 0mC coefficient is defined according to the Table A.2:
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199
The bending moment has the same value on both ends of the column: 1=ψ
( )ycr
Edmy N
NC,
0, 33.036.021.079.0 ×−×+×+= ψψ
Where:
( ) Nmm
mmmmNl
IEN
fy
yycr 7.133338053
²2000257332751/210000²
²² 42
, =××
=××
=ππ
(previously calculated)
NNEd 1000000=
( ) 002.17.133338053
100000033.0136.0121.079.00, =×−×+×+=N
NCmy
( ) ( ) 001.11444.11
1444.1002.11002.11
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
Equivalent uniform moment factor, mLTC , calculation
065.1
1
065.1
10244.110000001
38.866170010000001
1001.1
11 13
2
,,
2
=⇒
≥
=
×−×
−
×=
−×
−
×=
mLT
mLT
Tcr
Ed
zcr
Ed
LTmymLT
C
C
NN
NN
NN
NN
aCC
Therefore the yyk term corresponding to the top part of the column will be:
074.1
7.13333805310000001
1065.1001.11
,
=−
××=−
××=
NN
NN
CCk
ycr
Ed
ymLTmyyy
µ
13.26.2.6Internal factor, yzk , calculation
zcr
Ed
ymzyz
NN
Ck
,
1−×=
µ
mzC term must be calculated for the whole column length
=== 1
200200
,sup
,inf
kNmkNm
MM
Ed
Edψ
( ) kNNmm
mmmmNl
IENfz
zzcr 70.866138.8661700
²200010.16716452/210000²
²² 42
, ==××
=××
=ππ
( ) ( ) 776.038.8661700
100000033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
878.0
38.866170010000001
1776.01
,
=−
×=−
×=
NN
NNCk
zcr
Ed
ymzyz
µ
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13.26.2.7Internal factor, zyk , calculation
ycr
Ed
zmLTmyzy
NN
CCk
,
1−××=
µ
799.0
38.86617001000000821.01
38.866170010000001
1
1
,
, =×−
−=
×−
−
=
NN
NN
NN
NN
zcr
Edz
zcr
Ed
z
χµ
821.0=zχ (previously calculated)
050.1
7.13333805310000001
977.0065.1001.11
,
=−
××=−
××=
NN
NN
CCk
ycr
Ed
zmLTmyzy
µ
13.26.2.8Internal factor, zzk , calculation
857.0
38.866170010000001
977.0776.01
,
=−
×=−
×=
NN
NN
Ck
zcr
Ed
zmzzz
µ
13.26.2.9Bending and axial compression verification
∆+×+
×
∆+×+
×
∆+×+
×
∆+×+
×
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rkz
RdzEdzzz
M
RkyLT
RdyEdyzy
M
Rkz
Ed
M
Rkz
RdzEdzyz
M
RkyLT
RdyEdyyy
M
Rky
Ed
MMM
kMMM
kNN
MMM
kMMM
kNN
γγχ
γχ
γγχ
γχ
iyRk AfN ×=
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41.118.074.049.0
1/2755.175962
1010857.0
1/27506.1261435813.0
10200050.1
1/27572.9108821.0
1000000
34.118.076.040.0
1/2755.175962
1010878.0
1/27506.1261435813.0
10200074.1
1/27572.91081
1000000
23
6
23
6
222
23
6
23
6
22
=++=×
××+
+×
×
××+
××
=++=×
××+
+×
×
××+
××
mmNmmNmm
mmNmmNmm
mmNmmmmN
mmNmmNmm
mmNmmNmm
mmNmmN
Finite elements modeling
■ Linear element: S beam, ■ 5 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
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yχ coefficient corresponding to non-dimensional slenderness yλ
Column subjected to axial and shear force to the top
yχ
zχ coefficient corresponding to non-dimensional slenderness zλ
Column subjected to axial and shear force to the top
zχ
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Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yzk
Column subjected to axial and shear force to the top
yzk
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Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zzk
Column subjected to axial and shear force to the top
zzk
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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 effort over the Y axis
SNy
Bending and axial compression verification term depending of the Y bending moment over the Y axis: SMyy
Bending and axial compression verification term depending of the Y bending moment over the Y axis
SMyy
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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 over the Y axis
SMyz
Bending and axial compression verification term depending of the compression effort over the Z axis: SNz
Bending and axial compression verification term depending of the compression effort over the Z axis
SNz
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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 moment over the Z axis
SMzy
Bending and axial compression verification term depending of the Z bending moment over the Z axis: SMzz
Bending and axial compression verification term depending of the Z bending moment over the Z axis
SMzz
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13.26.2.10Reference results
Result name Result description Reference value
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
1
zχ zχ coefficient corresponding to non-dimensional slenderness zλ
0.821
yyk Internal factor, yyk
1.074
yzk Internal factor, yzk
0.878
zyk Internal factor, zyk
1.050
zyk Internal factor, zyk
0.857
SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
0.40
SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis
0.76
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
0.18
SNz
Bending and axial compression verification term depending of the compression effort over the z axis
0.49
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
0.74
SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
0.18
13.26.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 % Xz coefficient corresponding to non-dimensional slenderness 0.821634
adim 0.0772 %
Kyy Internal factor kyy 1.11767 adim
4.0661 %
Kyz Internal factor kyz 0.877605 adim
-0.0450 %
Kzy Internal factor kzy 1.09224 adim
4.0229 %
Kzz Internal factor kzz 0.857639 adim
0.0746 %
#SNy Bending and axial compression verification term depending of the compression effort over the Y axis
0.399218 adim
-0.1955 %
SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis
0.783306 adim
3.0666 %
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
0.181362 adim
0.7567 %
SNz Bending and axial compression verification term depending of the compression effort over the z axis
0.485883 adim
-0.8402 %
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
0.765485 adim
3.4439 %
SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
0.177236 adim
-1.5356 %
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13.27 EC3 test 9: Verifying the classification and the compression resistance of a welded built-up column
Test ID: 5674
Test status: Passed
13.27.1Description Verifies the cross-section classification and the compression resistance of a welded built-up column made of S355 steel. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.27.2Background Classification and verification of a welded built-up column made of S355 steel. The column is fixed at its base and free on the top. It is loaded by a compression force (100 000 N), applied at its top. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.27.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = -100 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
Units
Metric System
Geometry
Below are described the column cross section characteristics:
■ Height: h = 630 mm, ■ Flange width: b = 500 mm, ■ Flange thickness: tf = 18 mm,
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■ Web thickness: tw = 8 mm, ■ Column length: L = 5000 mm, ■ Section area: A = 22752 mm2 ,
■ Partial factor for resistance of cross sections: 0.10 =Mγ .
Materials properties
S355 steel material is used. The following characteristics are used:
■ Yield strength fy = 355 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (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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13.27.2.2Cross-section classification Before calculating the compression resistance, the cross-section class has to be determined.
The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.
In this case, the column is subjected to a punctual compression load, therefore the stresses distribution is like in the picture below:
Table 5.2 - sheet 2, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the flanges class. The picture below shows an extract from this table.
The top flange class can be determined by considering the cross-section geometrical properties and the conditions described in Table 5.2 - sheet 2:
67.1318
2/)8500(=
−=
mmmmmm
tc
81.0235==
yfε
Therefore:
34.111467.13 =>= εtc
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This means that the top column flange is Class 4. Having the same dimensions, the bottom column flange is also Class 4.
Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the web class. The picture below shows an extract from this table.
The web class can be determined by considering the cross-section geometrical properties and the conditions described in Table 5.2 - sheet 1:
25.748
218630=
×−=
mmmmmm
tc
81.0235==
yfε
Therefore:
02.344225.74 =>= εtc
This means that the column web is Class 4.
A cross-section is classified according to the least favorable classification of its compression elements (chapter 5.5.2(6) from EN 1993-1-1: 2001).
According to the calculation above, the column section have Class 4 web and Class 4 flanges; therefore the class section for the entire column section will be considered Class 4.
13.27.2.3Reference results for calculating the compression resistance of the cross-section The compression resistance for Class 4 cross-section is determined with the formula (6.11) from EN 1993-1-1:2001.
In order to verify the compression resistance for Class 4 cross-section, it is necessary to determine the effective area of the cross-section.
The effective area of the cross section takes into account the reduction factor, ρ, which is applying to both parts in compression (flanges and web).
The following parameters have to be determined, for each part in compression, in order to calculate the reduction factor: the buckling factor, the stress ratio and the plate modified slenderness.
The buckling factor (kσ) and the stress ratio(Ψ) - for flanges
Table 4.2 from EN 1993-1-5 offers detailed information about determining the buckling factor and the stress ratio for flanges. The below picture presents an extract from this table.
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Taking into account that the stress distribution on flanges is linear, the stress ratio becomes:
43.00.11
2 =→== σσσψ k
The buckling factor (kσ) and the stress ratio(Ψ) - for web
Table 4.1 from EN 1993-1-5 offers detailed information about determining the buckling factor and the stress ratio for web. The below picture presents an extract from this table.
Taking into account that the stress distribution on web is linear, the stress ratio becomes:
0.40.11
2 =→== σσσψ k
The plate modified slenderness (λp) – for flanges
The formula used to determine the plate modified slenderness for flanges is:
( )( ) 906.043.081.04.2818/2/8500
4.28/
=××
−==
mmmmmmk
tcp
σελ
The plate modified slenderness (λp) – for web
The formula used to determine the plate modified slenderness for web is:
( ) 614.10.481.04.288/182630
4.28/
=××
×−==
mmmmmmk
tbp
σελ
The reduction factor (ρ) – for flanges
The reduction factor for flanges is determined with relationship (4.3) from EN 1993-1-5. Because λp > 0.748, the reduction factor has the following formula:
0.1188.0
2 ≤−
=p
p
λλ
ρ
The effective width of the flange part can now be calculated:
( ) mmmmmmcb feff 25.2152
8500875.0, =−
×=×= ρ
The reduction factor (ρ) – for web
The reduction factor for web is determined with relationship (4.2) from EN 1993-1-5. Because λp > 0.673, the reduction factor has the following formula:
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214
( )0.1
3055.02 ≤
+×−=
p
p
λψλ
ρ
The effective width of the web can now be calculated:
( ) mmmmmmbb weff 8.317182630535.0, =×−×=×= ρ
Effective area
The effective area is determined considering the following:
weffwwfefffefffeff bttbbtA ,,, )(2 ×+++××=
Compression resistance of the cross section
For Class 4 cross-section, EN 1993-1-1: 2001 provides (6.11) formula in order to calculate the compression resistance of the cross-section:
NMPammfAN
M
yeffRdc 6506582
0.13554.18328 2
0, =
×=
×=
γ
Work ratio
Work ratio = %54.11006506582100000100
,
=×=×N
NN
NRdc
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 compression axial force Work ratio - Fx
13.27.2.4Reference results
Result name Result description Reference value Work ratio - Fx Compression resistance work ratio [%] 1.53 %
13.27.3Calculated results Result name Result description Value Error Work ratio - Fx Compression resistance work ratio 1.5322 % 0.1438 %
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13.28 EC3 Test 21: Verifying the buckling resistance of a CHS219.1x6.3H column
Test ID: 5701
Test status: Passed
13.28.1Description The test verifies the buckling resistance of a CHS219.1x6.3H made of S355.
The tests are made according to Eurocode 3 French Annex.
13.28.2Background Buckling verification under compression efforts for an CHS219.1x6.3H column made of S355 steel. The column is fixed at its base and free on the top. The column is subjected to a compression force (100 000 N) applied at its top. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.28.2.1Model 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=2366x104mm4 ■ Flexion inertia moment around the Z axis: Iz=2366x104 mm4
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Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 355 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (z = 3.00).
■ Inner: None. ■ Buckling lengths Lfy and Lfz are both imposed with 6m value
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 3.0: FZ = N = -100 000 N, ■ Internal: None.
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13.28.2.2Buckling in the strong inertia of the profile (along Y-Y) The calculations are made in order to obtain the work ratio of the analyzed element. The work ratio of the element is calculated using the percentage of the design buckling resistance of the compressed element (Nb,Rd) from the compression force applied to the element (NEd). The design buckling resistance of the compressed member, Nb,Rd, is calculated according to Eurocode 3 1993-1-1-2005, Chapter 6.3.1.1.
%100100,
≤×Rdb
Ed
NN
(6.46)
Cross-class classification is made according to Table 5.2
778.343.6
1.219==
mmmm
td
814.0355235235
===yf
ε
381.46814.05080778.34 2 =×=×≤= εtd
therefore the section is considered to be Class 2
It will be used the following buckling curve corresponding to Table 6.2:
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219
The imperfection factor α corresponding to the appropriate buckling curve will be 0.21:
The design buckling resistance of the compressed element is calculated using the next formula:
1,
M
yRdb
fAN
γχ ××
= (6.47)
Where:
Coefficient corresponding to non-dimensional slenderness after the Y-Y axis
χ coefficient corresponding to non-dimensional slenderness λ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=yyy
yλ
χ (6.49)
λ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
1
*λ
λ×
==iL
NfA cr
cr
yy (6.50)
41.76355
2100001 === ππλ
yfE
mmmm
mmAI
i y 283.754210
238600002
4
===
043.141.76283.75
6000
1
=×
=×
=mm
mmiLcr
y λλ
[ ] [ ] 132.1²043.1)2.0043.1(21.015.0)2.0(15.0 2 =+−×+×=+−+×= yyy λλαφ
1636.0²043.1²132.1132.1
1≤=
−+=yχ
A is the cross section area; A=4210mm2; fy is the yielding strength of the material; fy=355N/mm2 and 1Mγ is a safety
coefficient, 11 =Mγ
The design buckling resistance of the compression member will be:
NmmNmmfAN
M
yyRdb 8.950533
1/3554210636.0 22
1, =
××=
××=
γχ
NNEd 100000=
%520.101008.950533
100000100,
=×=×N
NNN
Rdb
Ed
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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
LTχ
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221
Ratio of the design normal force to design buckling resistance in the strong inertia of the profile
Column subjected to axial force Adimensional - SNy
13.28.2.3Reference results
Result name Result description Reference value
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
0.636
ySN Ratio of the design normal force to design buckling resistance in the strong inertia of the profile
0.1052
13.28.3Calculated results Result name Result description Value Error Xy coefficient corresponding to non-dimensional slenderness 0.635463
adim 0.0000 %
SNy Ratio of the design normal force to design buckling resistance in the strong inertia of the profile
0.105293 adim
0.0000 %
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13.29 EC3 Test 24: Verifying an user defined I section class 4 column fixed on the bottom and without any other restraint
Test ID: 5709
Test status: Passed
13.29.1Description The test verifies an user defined cross section column.
The cross section has an “I symmetric” shape with: 880mm height; 220mm width; 5mm center thickness; 15mm flange thickness; 0mm fillet radius and 0mm rounding radius.
The column is subjected to 328 kN axial compression force; 1274 kNm bending moment after the Y axis and 127.4 kNm bending moment after the Z axis. All the efforts are applied on the top of the column. The column height is 5.62m and has no restraints over its length.
The calculations are made according to Eurocode 3 French Annex.
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223
13.29.2Background An I880*5+220*15 shaped column subjected to compression and bending, made from S275 steel. The column has a 880x5mm web and 220x15mm flanges. The column is hinged at its base and, at his top end, translation is permitted only on vertical direction and the rotation is blocked for the long axis of the column. The column is subjected to an axial compression load 328000 N, 127400Nm bending moment after the X axis and 1274000Nm bending moment after the Y axis.
This test was evaluated by the French control office SOCOTEC.
13.29.2.1Model 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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224
Geometrical properties
■ Column length: L=5620mm
■ Cross section area: 210850mmA =
■ Overall breadth: mmb 220=
■ Flange thickness: mmt f 15=
■ Root radius: mmr 0=
■ Web thickness: mmtw 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 mmWy ×=
■ Elastic modulus after the Z axis, 33, 1008.242 mmW zel ×=
■ Plastic modulus after the Z axis, 33, 1031.368 mmW zpl ×=
■ Flexion inertia moment around the Y axis: Iy=149058.04x104mm4 ■ Flexion inertia moment around the Z axis: Iz=2662.89x104 mm4 ■ Torsional moment of inertia: It=51.46x104 mm4 ■ Working inertial moment: Iw=4979437.37x106mm6
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength fy = 275 MPa, ■ Longitudinal elastic modulus: E = 210000 MPa. ■ Shear modulus of rigidity: G=80800MPa
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at the end point (z = 5.62) restrained in translation along Y and Z axis and restrained rotation
along X axis.
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225
Loading
The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =--328000N; Mx=127400Nm and My=1274000Nm
13.29.2.2Cross section Class According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
-for beam web:
The web dimensions are 850x5mm.
178.012750109.0
328.0212 −>−=−×
⋅=−⋅
⋅=y
Ed
fANψ
5.064.0005.0850.0275
328.01211
21
>=
××+⋅=
××+⋅=
dtfN
y
Edα
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226
924.0275235235
===yf
ε
057.94)78.0(33.067.0
924.04233.067.0
421701
1705
152880=
−×+×
=×+
×>=⇒
=
=×−
=ψ
ε
ε tc
mmmmmm
tc
therefore the beam web is considered to be Class 4
-for beam flange:
9961.71
61.715
5.107=×≤=⇒
=
==ε
ε tc
tc
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 4
13.29.2.3Effective cross-sections of Class4 cross-sections -the section is composed from Class 4 web and Class 1 flanges, therefore will start the web calculation:
-in order to simplify the calculations the web will be considered compressed only
1705
152880=
×−=
mmmmmm
tc
:
41 =⇒= σψ k
-according EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1
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227
σελ
kt
bw
p ××=
4.28
wb is the width of the web; mmbw 850=
t is the web thickness; t=5mm
9244.0275235235
===yf
ε
261.349244.04.28
5850
=××
= mmmm
pλ
-according EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4
-the web is considered to be an internal compression element, therefore:
( )286.0
261.34055.0261.33055.0
043
673.0261.322 =
×−=
+×−=⇒
≥=+
>=
p
pp
λ
ψλρ
ψ
λ
=×==×=
=×=
⇒
×=
×=
×=
mmmmbmmmmbmmmmb
bbbb
bb
e
e
eff
effe
effe
weff
55.1211.2435.055.1211.2435.0
1.243850286.0
5.0
5.0
2
1
2
1
ρ
221, 5.121555.121555.1215 mmmmmmmmmmbtbtA ewewwebeff =×+×=×+×=
2, 330022015 mmmmmmbtA ffflangeeff =×=×=
222,, 2.7815330025.12152 mmmmmmAAA flangeeffwebeffeff =×+=×+=
13.29.2.4Effective elastic section modulus of Class4 cross-sections -In order to simplify the calculation the section will be considered in pure bending
mmmmbb tc 4252
8501sup
inf ===⇒−==σσψ
9.231 =⇒−= σψ k
-according EC3 Part 1,5 – EN 1993-1-5-2004; Table 4.1
1705
152880=
×−=
mmmmmm
tc
σελ
kt
bw
p ××=
4.28
wb is the width of the web; mmbw 850=
t is the web thickness; t=5mm
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228
9244.0275235235
===yf
ε
325.19.239244.04.28
5850
=××
= mmmm
pλ
-according EC3 Part 1,5 – EN 1993-1-5-2004; Chapter4.4
( )692.0
325.12055.0325.13055.0
023
673.0325.122 =
×−=
+×−=⇒
≥=+
>=
p
pp
λ
ψλρ
ψ
λ
( )
=×==×=
=−−
×=
⇒
×=
×=−
×=×=
mmmmbmmmmb
mmmmb
bbbb
bbb
e
e
eff
effe
effe
wceff
46.1761.2946.064.1171.2944.0
1.29411
850692.0
6.0
4.01
2
1
2
1
ψρρ
-the weight center coordinate is:
( ) ( ) ( ) ( )
mm
yG
53.155.10195095.158330
15220546.601564.1171522027.124546.6015.4321522018.366564.1175.43215220
−=−
=
=×+×+×+×
××−××−××+××=
-the inertial moment along the strong axis is:
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229
4
23
23
23
23
14482344295.6993800726.748854356
97.4161522012
2201574.107546.60112
546.601
71.381564.11712
564.11703.4481522012
22015
mm
I y
=+=
=××+×
+××+×
+
+××+×
+××+×
=
423
23
23
23
63.2662749001522012
152200546.60112
46.6015
0564.11712
64.117501522012
15220
mm
I z
=××+×
+××+×
+
+××+×
+××+×
=
34
max, 533.3179229
53.4551448234429 mm
mmmm
zI
W yyel ===
34
max, 10.242068
11063.26627490 mm
mmmm
yIW z
zel ===
13.29.2.5Buckling 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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230
34.0=α
Coefficient corresponding to non-dimensional slenderness after Y-Y axis:
yχ coefficient corresponding to non-dimensional slenderness yλ will be determined from the relevant buckling
curve according to:
1122
≤−Φ+Φ
=yyy
yλ
χ (6.49)
yλ the non-dimensional slenderness corresponding to Class 4 cross-sections:
ycr
yeffy N
fA
,
*=λ
Where: A is the effective cross section area; 22.7815 mmAeff = ; fy is the yielding strength of the material;
fy=275N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
( ) kNNmm
mmmmNl
IEN
fy
yycr 37.9503544.95035371
²56201448234429/210000²
²² 42
, ==××
=××
=ππ
15.044.95035371
/2752.7815 22
,
=×
=×
=N
mmNmmN
fA
ycr
yeffyλ
[ ] ( )[ ] 503.015.02.015.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ
1
1
017.115.0503.0503.0
112222 =⇒
≥
=−+
=−Φ+Φ
=y
y
yyy
yχ
χ
λχ
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:
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231
49.0=α
Coefficient corresponding to non-dimensional slenderness after Z-Z axis:
zχ coefficient corresponding to non-dimensional slenderness zλ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=zzz
zλ
χ (6.49)
zλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
zcr
yeffz
NfA
,
*=λ
Where: A is the effective cross section area; 22.7815 mmAeff = ; fy is the yielding strength of the material;
fy=275N/mm2 and Ncr is the elastic critical force for the relevant buckling mode based on the gross cross sectional properties:
( ) kNNmm
mmmmNl
IENfz
zzcr 34.1747905.1747336
²562063.26627490/210000²
²² 42
, ==××
=××
=ππ
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232
109.1905.1747336
/2752.7815 22
,
=×
=×
=N
mmNmmN
fA
zcr
yeffzλ
[ ] ( )[ ] 338.1109.12.0109.149.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ
479.0
1
479.0109.1338.1338.1
112222 =⇒
≤
=−+
=−Φ+Φ
=z
z
zzz
zχ
χλ
χ
13.29.2.6Lateral torsional buckling verification -the transversal load is applied to the shear center, therefore C2zg=0 and the moment formula will be:
z
t
z
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3) -where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
ψ is the fraction of the bending moment from the column extremities: 01274
0==
kNmkNmψ
77.101274
01 =⇒== C
kNmkNmψ
According to EN 1993-1-1-AN France; Chapter 3.2; Table 1
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233
Flexion inertia moment around the Y axis: 41448234429mmI y =
Flexion inertia moment around the Z axis: 463.26627490 mmI 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
2fz
w
thII
−×=
h cross section height; h=880mm
ft flange thickness; mmt f 15=
( ) 61124
10808.494
15880mm 326627490.6 mmmmmmIw ×=−×
=
-according to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column part: L=5620mm
Shear modulus of rigidity: G=80800N/mm2
( )( )
kNmNmm
mmNmmmmNmmmmNmm
mmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
163.14201420163158
185.459905.174733677.163.26627490/210000
514614/80800562063.26627490
10808.49
562063.26627490/21000077.1
²²
²²
422
422
4
611
2
422
1
==
=××=××
××+
××
×××
×=××××
+×××
×=
π
ππ
π
The elastic modulus : 34
max, 533.3179229
53.4551448234429 mm
mmmm
zI
W yyel ===
785.01420163158
/275533.3179229 23, =
×==
NmmmmNmm
MfW
cr
yyeffLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=
ADVANCE DESIGN VALIDATION GUIDE
234
The cross section buckling curve will be chose according to Table 6.4:
24220880
>==mmmm
bh
The imperfection factor α will be chose according to Table 6.3:
76.0=α
( )[ ] ( )[ ] 030.1²785.02.0785.076.015.0²2.015.0 =+−×+=+−×+= LTLTLTLT λλαφ
1589.0²785.0²030.1030.1
1
²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
13.29.2.7Internal factor, yyk , calculation
The internal factor yyk corresponding to a Class 1 section will be calculated according to Annex A, Table a.1:
ADVANCE DESIGN VALIDATION GUIDE
235
ycr
Ed
ymLTmyyy
NNCCk
,
1−××=
µ
ycr
Edy
ycr
Ed
y
NN
NN
,
,
1
1
×−
−
=χ
µ
1=yχ (previously calculated)
kNNEd 328=
kNNl
IEN
fy
yycr 37.9503544.95035371
²²
, ==××
=π
(previously calculated)
1
44.9503537132800011
44.950353713280001
1
1
,
, =×−
−=
×−
−
=
NN
NN
NN
NN
ycr
Edy
ycr
Ed
y
χµ
The myC will be calculated according to Table A.1:
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236
Calculation of the 0λ term:
0
,0
cr
yyeff
MfW ×
=λ
-according to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2 33
, 1066.3387 mmW yel ×=
kNmNmmM cr 163.14201420163158 == (previously calculated)
810.01420163158
/2753387660 23
0
,0 =
×=
×=
NmmmmNmm
MfW
cr
yyeffλ
Calculation of the 4
,,1 1120.0
−×
−××
TFcr
Ed
zcr
Ed
NN
NNC term:
Where:
-for a symmetrical section for the both axis, TcrTFcr NN ,, =
××+××=
²,
2
0,
Tcr
wtTcr L
IEIGIAN π
The mass moment of inertia 0I
20 gzy zAIII ⋅++=
Flexion inertia moment around the Y axis: 467.1490580416 mmI y =
Flexion inertia moment around the Z axis: 417.26628854 mmI z =
Cross section area: 210850mmA =
Distance between the section neutral axis and the section geometrical center: 0=gz
44420 151720927017.2662885467.1490580416 mmmmmmIIzAIII zygzy =+=+=⋅++=
- the buckling length, TcrL , ,
mL Tcr 62.5, =
Torsional moment of inertia: 475.514614 mmIt =
Working inertial moment: 61110808.49 mmIw ×= (previously calculated)
Longitudinal elastic modulus: E = 210000 MPa
Shear modulus of rigidity: G=80800MPa
( )N
mmmmmmNmmmmN
mmmmN Tcr
14.2634739²5620
10808.49/21000075.514614/808001517209270
10850 6112242
4
2
,
=
=
×××+××=
π
NNEd 328000=
NNN TcrTFcr 14.2634739,, ==
ADVANCE DESIGN VALIDATION GUIDE
237
Nl
IENfz
zzcr 905.1747336
²²
, =××
=π
(previously calculated)
244.0
14.26347393280001
905.1747336328000177.120.01120.0 44
,,1
=
=
−×
−××=
−×
−××
NN
NN
NN
NNC
TFcr
Ed
zcr
Ed
Therefore:
( )
≥
−×
−
×=
=
×+
××−+=
⇒
=
−⋅
−⋅⋅>=
=
−×
−××
=
1
11
11
244.01120.0810.0
244.01120.0
810.0
,,
2
0,
0,0,
4
,,10
4
,,1
0
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
a
aCCC
NN
NNC
NN
NNC
ε
ε
λ
λ
The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.
The coefficient must be calculated considering the column over the entire height.
( )LTy
LTymymymy a
aCCC
×+
××−+=
ε
ε
11 0,0,
yeff
eff
Ed
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 3,, 533.3179229 mmWW yeffyel ==
55.9533.3179229
2.7815328000
1012743
26
,
, =××
=×=mm
mmNNmm
WA
NM
yeff
eff
Ed
Edyyξ
19996.01448234429
75.51461411 4
4
≈=−=−=mm
mmIIa
y
tLT
ADVANCE DESIGN VALIDATION GUIDE
238
The 0mC coefficient is defined according to the Table A.2:
The bending moment is null at the end of the column, therefore: 0=ψ
( )ycr
Edmy N
NC,
0, 33.036.021.079.0 ×−×+×+= ψψ
Where:
kNNl
IEN
fy
yycr 37.9503544.95035371
²²
, ==××
=π
(previously calculated)
( ) 79.044.95035371
32800033.0036.079.00, =×−×+=N
NCmy
( ) ( ) 949.0155.91
155.979.0179.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
1
11,,
2 ≥
−×
−
×=
Tcr
Ed
zcr
Ed
LTmymLT
NN
NN
aCC
068.1
14.26347393280001
905.17473363280001
1949.0 2 =
−×
−
×=
NN
NN
CmLT
0161.1
44.950353713280001
1068.1949.01
,
=−
××=−
××=
NN
NNCCk
ycr
Ed
ymLTmyyy
µ
ADVANCE DESIGN VALIDATION GUIDE
239
13.29.2.8Internal factor, yzk , calculation
zcr
Ed
ymzyz
NN
Ck
,
1−×=
µ
( ) ( ) 7677.0905.1747336
32800033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
945.0
905.17473363280001
17677.01
,
=−
×=−
×=
NN
NNCk
zcr
Ed
ymzyz
µ
13.29.2.9Internal factor, zyk , calculation
ycr
Ed
zmLTmyzy
NN
CCk
,
1−××=
µ
893.0
905.1747336328000479.01
905.17473363280001
1
1
,
, =×−
−=
×−
−=
NN
NN
NN
NN
zcr
Edz
zcr
Ed
z
χµ
908.0
44.950353713280001
893.0068.1949.01
,
=−
××=−
××=
NN
NNCCk
ycr
Ed
zmLTmyzy
µ
13.29.2.10Internal factor, zzk , calculation
844.0
905.17473363280001
893.07677.01
,
=−
×=−
×=
NN
NNCk
zcr
Ed
zmzzz
µ
13.29.2.11Bending and axial compression verification
∆+×+
×
∆+×+
×
∆+×+
×
∆+×+
×
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rkz
RdzEdzzz
M
RkyLT
RdyEdyzy
M
Rkz
Ed
M
Rkz
RdzEdzyz
M
RkyLT
RdyEdyyy
M
Rky
Ed
MMM
kMMM
kNN
MMM
kMMM
kNN
γγχ
γχ
γγχ
γχ
iyRk AfN ×=
ADVANCE DESIGN VALIDATION GUIDE
240
14.461.124.232.0
1/27510.242068
104.127844.0
1/275533.3179229589.0
101274908.0
1/2752.7815479.0
328000
47.4808.151.215.0
1/27510.242068
104.127945.0
1/275533.3179229589.0
1012740161.1
1/2752.78151
328000
23
6
23
6
22
23
6
23
6
22
=++=×
××+
+×
×
××+
××
=++=×
××+
+×
×
××+
××
mmNmmNmm
mmNmmNmm
mmNmmN
mmNmmNmm
mmNmmNmm
mmNmmN
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
241
yχ coefficient corresponding to non-dimensional slenderness yλ
Column subjected to axial and shear force to the top
yχ
zχ coefficient corresponding to non-dimensional slenderness zλ
Column subjected to axial and shear force to the top
zχ
ADVANCE DESIGN VALIDATION GUIDE
242
Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yzk
Column subjected to axial and shear force to the top
yzk
ADVANCE DESIGN VALIDATION GUIDE
243
Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zzk
Column subjected to axial and shear force to the top
zzk
ADVANCE DESIGN VALIDATION GUIDE
244
Bending and axial compression verification term depending of the compression effort over the Y axis: SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
SNy
Bending and axial compression verification term depending of the Y bending moment over the Y axis: SMyy
Bending and axial compression verification term depending of the Y bending moment over the Y axis
SMyy
ADVANCE DESIGN VALIDATION GUIDE
245
Bending and axial compression verification term depending of the Z bending moment over the Y axis: SMyz
Bending and axial compression verification term depending of the Z bending moment over the Y axis
SMyz
Bending and axial compression verification term depending of the compression effort over the Z axis: SNz
Bending and axial compression verification term depending of the compression effort over the Z axis
SNz
ADVANCE DESIGN VALIDATION GUIDE
246
Bending and axial compression verification term depending of the Y bending moment over the Z axis: SMzy
Bending and axial compression verification term depending of the Y bending moment over the Z axis
SMzy
Bending and axial compression verification term depending of the Z bending moment over the Z axis: SMzz
Bending and axial compression verification term depending of the Z bending moment over the Z axis
SMzz
ADVANCE DESIGN VALIDATION GUIDE
247
13.29.2.12Reference results
Result name Result description Reference value
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
1
zχ zχ coefficient corresponding to non-dimensional slenderness zλ
0.48
yyk Internal factor, yyk
1.011
yzk Internal factor, yzk
0.95
zyk Internal factor, zyk
0.902
zyk Internal factor, zyk
0.84
SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
0.15
SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis
2.50
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
1.81
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
2.23
SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
1.61
13.29.3Calculated results Result name Result description Value Error Xy Coefficient corresponding to non-dimensional slenderness 1 adim 0.0000 % Xz Coefficient corresponding to non-dimensional slenderness 0.479165
adim -0.1740 %
Kyy Internal factor kyy 1.01066 adim
-0.0336 %
Kyz Internal factor kyz 0.9451 adim -0.5158 % Kzy Internal factor kzy 0.902094
adim 0.0104 %
Kzz Internal factor kzz 0.843573 adim
0.4254 %
#SNy Bending and axial compression verification term depending of the compression effort over the Y axis
0.152455 adim
1.6367 %
#SMyy Bending and axial compression verification term depending of the Y bending moment over the Y axis
2.49874 adim
-0.0504 %
#SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
1.80865 adim
-0.0746 %
#SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
2.23031 adim
0.0139 %
#SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
1.61436 adim
0.2708 %
ADVANCE DESIGN VALIDATION GUIDE
248
13.30 EC3 Test 29: Verifying an user defined I section class 1, column hinged on base and restrained on top for the X, Y translation and Z rotation
Test ID: 5729
Test status: Passed
13.30.1Description The test verifies a user defined cross section column.
The column is an “I symmetric” shape with: 260mm height; 150mm width; 7.1mm web thickness; 10.7mm flange thickness; 0mm fillet radius and 0mm rounding radius. The column is made of S275 steel.
The column is subjected to 328 kN axial compression force and 50kNm bending moment after the Y axis and 10kNm bending moment after the Z axis. All the efforts are applied to the top of the column.
The calculations are made according to Eurocode 3 French Annex.
13.30.2Background An I260*7.1+150*10.7 shaped column subjected to compression and bending, made from S275 steel. The column has a 260x7.1mm web and 150x10.7mm flanges. The column is fixed for all translations and free for all rotations, at its base, and on the top end, the translations over the X and Y axes and the rotation over the Z axis are not permitted. In the middle of the column there is a restraint over the Y axis, therefore the bucking length for the XY plane is equal to half of the column length. The column is subjected to an axial compression load -328000 N, a 10000Nm bending moment after the X axis and a 50000Nm bending moment after the Y axis
This test was evaluated by the French control office SOCOTEC.
13.30.2.1Model 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).
ADVANCE DESIGN VALIDATION GUIDE
249
Units
Metric System
Geometrical properties
■ Column length: L=5620mm
■ Cross section area: 206.4904 mmA =
■ Overall breadth: mmb 150=
■ Flange thickness: mmt f 70.10=
■ Root radius: mmr 0=
■ Web thickness: mmtw 10.7=
■ Depth of the web: mmhw 260=
■ Elastic modulus after the Y axis, 3, 63.445717 mmW yel =
■ Plastic modulus after the Y axis, 318.501177 mmWy =
■ Elastic modulus after the Z axis, 3, 89.80344 mmW zel =
■ Plastic modulus after the Z axis, 3, 96.123381 mmW zpl =
■ Flexion inertia moment around the Y axis: 464.57943291 mmI y =
■ Flexion inertia moment around the Z axis: 446.6025866 mmIz =
■ Torsional moment of inertia: 497.149294 mmIt =
■ Working inertial moment: 688.19351706542 mmIw =
Materials properties
S275 steel material is used. The following characteristics are used:
■ Yield strength fy = 275 MPa, ■ Longitudinal elastic modulus: E = 210000 MPa. ■ Shear modulus of rigidity: G=80800MPa
ADVANCE DESIGN VALIDATION GUIDE
250
Boundary conditions
The boundary conditions are described below:
■ Outer: ► Support at start point (x = 0) restrained in translation along X, Y and Z axis, ► Restraint of translation over the Y axis at half (z=2.81) ► Support at start point (z = 5.62) restrained in translation along X and Y axis, and restrained in rotation
along Z axis,
Loading
The column is subjected to the following loadings: ■ External: Point load From X=0.00m and z=5.62m: FZ =-328000N; Mx=10000Nm and My=50000Nm
ADVANCE DESIGN VALIDATION GUIDE
251
13.30.2.2Cross section Class According to Advance Design calculations:
Cross-class classification is made according to Table 5.2
- for beam web:
The web dimensions are 238.6x7.1mm.
178.012750109.0
328.0212 −>−=−×
⋅=−⋅
⋅=y
Ed
fANψ
5.085.00071.02386.0275
328.01211
21
>=
××+⋅=
××+⋅=
dtfN
y
Edα
ADVANCE DESIGN VALIDATION GUIDE
252
924.0275235235
===yf
ε
41.36185.013
924.0396113
3966.33924.0
61.331.7
7.102260=
−××
=−×
×≤=⇒
=
=×−
=α
ε
ε tc
mmmmmm
tc
therefore the beam
web is considered to be Class 1
-for beam flange:
316.8924.0968.6
924.0
68.67.10
21.7150
=×≤=⇒
=
=
−
=tc
tc
ε
therefore the haunch is considered to be Class1
In conclusion, the section is considered to be Class 1.
ADVANCE DESIGN VALIDATION GUIDE
253
13.30.2.3Buckling verification a) over the strong axis of the section, y-y:
- the imperfection factor α will be selected according to Tables 6.1 and 6.2:
34.0=α
Coefficient corresponding to non-dimensional slenderness after Y-Y axis:
yχ coefficient corresponding to non-dimensional slenderness yλ will be determined from the relevant buckling
curve according to:
1122
≤−Φ+Φ
=yyy
yλ
χ (6.49)
yλ the non-dimensional slenderness corresponding to Class 1 cross-sections:
ycr
yy N
fA
,
*=λ
Cross section area: 206.4904 mmA =
Flexion inertia moment around the Y axis: 464.57943291 mmI y =
ADVANCE DESIGN VALIDATION GUIDE
254
( ) kNNmm
mmmmNl
IEN
fy
yycr 33.380295.3802327
²562064.57943291/210000²
²² 42
, ==××
=××
=ππ
5956.095.3802327
/27506.4904 22
,
=×
=×
=N
mmNmmN
fA
ycr
yyλ
[ ] ( )[ ] 7446.05956.02.05956.034.015.0²)2.0(15.0 2 =+−×+×=+−×+×= yyy λλαφ
839.0
1
839.05956.07446.07446.0
112222 =⇒
≤
=−+
=−Φ+Φ
=y
y
yyy
yχ
χ
λχ
b) over the weak axis of the section, z-z:
- the imperfection factor α will be selected according to Tables 6.1 and 6.2:
49.0=α
Coefficient corresponding to non-dimensional slenderness after Z-Z axis:
ADVANCE DESIGN VALIDATION GUIDE
255
zχ coefficient corresponding to non-dimensional slenderness zλ will be determined from the relevant buckling curve according to:
1122
≤−Φ+Φ
=zzz
zλ
χ (6.49)
zλ the non-dimensional slenderness corresponding to Class 1, 2 and 3 cross-sections:
zcr
yz
NfA
,
*=λ
Flexion inertia moment around the Z axis: 446.6025866 mmI z =
Cross section area: 206.4904 mmA =
( ) kNNmm
mmmmNl
IENfz
zzcr 71.3158151.1581706
²281046.6025866/210000²
²² 42
, ==××
=××
=ππ
923.051.1581706
/27506.4904 22
,
=×
=×
=N
mmNmmN
fA
zcr
yzλ
[ ] ( )[ ] 103.1923.02.0923.049.015.0²)2.0(15.0 2 =+−×+×=+−×+×= zzz λλαφ
586.01
586.0923.0103.1103.1
112222 =⇒
≤
=−+
=−Φ+Φ
=z
z
zzz
zχ
χλ
χ
ADVANCE DESIGN VALIDATION GUIDE
256
13.30.2.4Lateral 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
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3)
- where:
C1 is a coefficient that depends of several parameters as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
31.15.02525
1,
, =⇒=== CkNkN
MM
topy
botomyψ
ADVANCE DESIGN VALIDATION GUIDE
257
Flexion inertia moment around the Y axis: 464.57943291 mmI y =
Flexion inertia moment around the Z axis: 446.6025866 mmI z =
Torsional moment of inertia: 497.149294 mmIt =
Working inertial moment: 688.19351706542 mmI w =
Yield strength fy = 275 MPa,
Longitudinal elastic modulus: E = 210000 MPa.
Shear modulus of rigidity: G=80800MPa
Warping inertial moment (recalculated):
IW is the warping inertia (deformation inertia moment):
( )4
2fz
w
thII
−×=
h cross section height; mmh 260=
ft flange thickness; mmt f 7.10=
( ) 624
093627638294
7.1026046.6025866 mmmmmmmmI w =−×
=
According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column: L=2810mm
( )( )
kNmNmm
mmNmmmmN
mmmmNmmmmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
36.31574.315363380
20.15251.158170631.146.6025866/210000
97.149294/80800281046.6025866
09362763829
281046.6025866/21000031.1
²²
²²
422
422
4
6
2
422
1
==
=××=××
××+×
×××
×=××××
+×××
×=
π
ππ
π
661.074.315363380
/27518.501177 23, =
×==
NmmmmNmm
MfW
cr
yyplLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=
ADVANCE DESIGN VALIDATION GUIDE
258
The cross section buckling curve will be chose according to Table 6.4:
2733.1150260
≤==mmmm
bh
The imperfection factor α will be chose according to Table 6.3:
49.0=LTα
( )[ ] ( )[ ] 831.0²661.02.0661.049.015.0²2.015.0 =+−×+×=+−×+×= LTLTLTLT λλαφ
1749.0²661.0²831.0831.0
1²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
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
wzcr IE
IGLII
LIECM
××××
+×××
×=²²
²²
1 ππ
According to EN 1993-1-1-AN France; Chapter 2; …(3)
- where:
C1 is a coefficient that depends of several parameters, such as: section properties; support conditions; moment diagram allure
²252.0423.0325.01
1 ψψ ++=C
According to EN 1993-1-1-AN France; Chapter 3; …(6)
ADVANCE DESIGN VALIDATION GUIDE
259
ψ is the fraction of the bending moment from the column extremities: 0637
0==
kNmψ
77.10 1 =⇒= Cψ According to EN 1993-1-1-AN France; Chapter 3.2; Table 1
Flexion inertia moment around the Y axis: 464.57943291 mmI y =
Flexion inertia moment around the Z axis: 446.6025866 mmI z =
Torsional moment of inertia: 497.149294 mmIt =
Working inertial moment: 688.19351706542 mmI w =
Yield strength fy = 275 MPa,
Longitudinal elastic modulus: E = 210000 MPa.
Shear modulus of rigidity: G=80800MPa
Warping inertial moment (recalculated):
IW is the warping inertia (deformation inertia moment):
( )4
2fz
w
thII
−×=
h cross section height; mmh 260=
ft flange thickness; mmt f 7.10=
( ) 624
093627638294
7.1026046.6025866 mmmmmmmmI w =−×
=
According to EN1993-1-1-AN France; Chapter 2 (…4) Length of the column: L=2810mm
ADVANCE DESIGN VALIDATION GUIDE
260
( )( )
kNmNmm
mmNmmmmN
mmmmNmmmmmm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
10.4266.426102243
20.15251.158170677.146.6025866/210000
97.149294/80800281046.6025866
09362763829
281046.6025866/21000077.1
²²
²²
422
422
4
6
2
422
1
==
=××=××
××+×
×××
×=××××
+×××
×=
π
ππ
π
569.06.426102243
/27518.501177 23, =
×==
NmmmmNmm
MfW
cr
yyplLTλ
Calculation of the LTχ for appropriate non-dimensional slenderness LTλ will be determined with formula:
1²²
1≤
−+=
LTLTLT
LTλφφ
χ (6.56)
( )[ ]²2.015.0 LTLTLTLT λλαφ +−×+×=
The cross section buckling curve will be chose according to Table 6.4:
2733.1150260
≤==mmmm
bh
The imperfection factor α will be chose according to Table 6.3:
49.0=LTα
( )[ ] ( )[ ] 752.0²569.02.0569.049.015.0²2.015.0 =+−×+×=+−×+×= LTLTLTLT λλαφ
1804.0²569.0²752.0752.0
1²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ
ADVANCE DESIGN VALIDATION GUIDE
261
13.30.2.5Internal 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
ymLTmyyy C
NNCCk 1̀
1,
×−
××=µ
ycr
Edy
ycr
Ed
y
NN
NN
,
,
1
1
×−
−
=χ
µ
839.0=yχ (previously calculated)
kNNEd 328=
kNNl
IEN
fy
yycr 33.380295.3802327
²²
, ==××
=π
(previously calculated)
985.0
95.3802327328000839.01
95.38023273280001
1
1
,
, =×−
−=
×−
−=
NN
NN
NN
NN
ycr
Edy
ycr
Ed
y
χµ
ADVANCE DESIGN VALIDATION GUIDE
262
The myC will be calculated according to Table A.1:
Calculation of the 0λ term:
0
,0
cr
yypl
MfW ×
=λ
According to Eurocode 3 EN 1993-1-1-2005; Chapter 6.3.2.2 318.501177 mmWy =
The calculation the 0crM will be calculated using 11 =C and 02 =C , therefore:
( )( )
kNmNmm
mmNmmmmN
mmmmNmmmm
mm
mmmmmmN
IEIGL
II
LIECM
z
t
z
wzcr
73.2408.240735730
20.15251.1581706146.6025866/210000
97.149294/80800281046.6025866
88.19351706542
281046.6025866/2100001
²²
²²
422
422
4
6
2
422
1
==
=××=××
××+×
×××
×=××××
+×××
×=
π
ππ
π
757.08.240735730
/27518.501177 23,
0 =×
==Nmm
mmNmmM
fW
cr
yyplλ
ADVANCE DESIGN VALIDATION GUIDE
263
Calculation of the 4
,,1 1120.0
−×
−××
TFcr
Ed
zcr
Ed
NN
NNC term:
Where:
- for a symmetrical section for the both axis, TcrTFcr NN ,, =
××+××=
²1
,
2
0,
Tcr
wtTcr L
IEIGI
N π
The mass moment of inertia 0I
44420 1.6396915846.602586664.57943291 mmmmmmIIzAIII zygzy =+=+=×++=
Torsional moment of inertia: 497.149294 mmIt =
Working inertial moment: 688.19351706542 mmIw =
- the buckling length, TcrL , ,
mL Tcr 81.2, =
( )kNN
mmmmmmNmmmmN
mmmmN Tcr
63.280668.2806625²2810
88.19351706542/21000097.149294/808001.63969158
06.4904 62242
4
2
,
==
=
××+××=
π
NNEd 328000=
NNN TcrTFcr 68.2806625,, ==
kNNl
IENfz
zzcr 71.3158151.1581706
²²
, ==××
=π
(previously calculated)
C1=1 for the top part of the column
For the top part of the column:
183.0
68.28066253280001
51.15817063280001120.01120.0 44
,,1
=
=
−×
−××=
−×
−××
NN
NN
NN
NNC
TFcr
Ed
zcr
Ed
ADVANCE DESIGN VALIDATION GUIDE
264
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
Tcr
Ed
zcr
Ed
LTmymLT
mzmz
LTy
LTymymymy
TFcr
Ed
zcr
Ed
TFcr
Ed
zcr
Ed
NN
NN
aCC
CC
aa
CCC
NN
NNC
NN
NNC
ε
ε
λ
λ
The myC coefficient takes into account the column behavior in the buckling plane: the buckling and bending moment distribution.
The coefficient must be calculated considering the column over the entire height.
( )LTy
LTymymymy a
aCCC
×+
××−+=
ε
ε
11 0,0,
yeff
eff
Ed
Edyy W
AN
M
,
, ×=ξ
Elastic modulus after the Y axis, 3,, 63.445717 mmWW yeffyel ==
677.163.445717
06.4904328000
10503
26
,
, =××
=×=mm
mmN
NmmWA
NM
yeff
eff
Ed
Edyyξ
1997.064.57943291
97.14929411 4
4
≈=−=−=mm
mmIIa
y
tLT
The 0mC coefficient is defined according to the Table A.2:
ADVANCE DESIGN VALIDATION GUIDE
265
The bending moment in null at one end of the column, therefore: 0=ψ
( )ycr
Ed
ycr
Edmy N
NNNC
,,0, 33.036.079.033.036.021.079.0 ××−=×−×+×+= ψψ
Where:
kNNl
IEN
fy
yycr 33.380295.3802327
²²
, ==××
=π
(previously calculated)
NNEd 328000=
780.095.3802327
32800033.036.079.00, =××−=N
NCmy
( ) ( ) 904.01677.11
1677.1780.01780.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
Equivalent uniform moment factor, mLTC , calculation
Equivalent uniform moment factor, mLTC , calculation
- mLTC must be calculated separately for each column part, separated by the lateral buckling restraint
1
11,,
2 ≥
−×
−
×=
Tcr
Ed
zcr
Ed
LTmymLT
NN
NN
aCC
- the myC term used for mLTC calculation, must be recalculated for the corresponding column part (in this case the top column
part)
- this being the case, myC will be calculated using 5.0=ψ :
( )LTy
LTymymymy a
aCCC
ξ
ξ
+−+=
11 0,0,
( ) ( ) 900.095.3802327
23800033.05.036.05.021.079.033.036.021.079.0,
0, =×−×+×+=×−×+×+=N
NNNC
ycr
Edmy ψψ
677.163.445717
06.4904328000
10503
26
,
, =××
=×=mm
mmN
NmmWA
NM
yeff
eff
Ed
Edyyξ (previously calculated)
( ) ( ) 956.01677.11
1677.1900.01900.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
ADVANCE DESIGN VALIDATION GUIDE
266
089.1
1
089.1
68.28066253280001
51.15817063280001
997.0956.0
11
2
,,
2
=⇒
≥
=
−×
−
×=
=
−×
−
×=
mLT
mLT
Tcr
Ed
zcr
Ed
LTmymLT
C
CN
NN
N
NN
NN
aCC
The yyC coefficient is defined according to the Table A.1, Auxiliary terms:
ypl
yelLTplmy
ymy
yyyy W
WbnC
wC
wwC
,
,2
max2
max2 6.16.12)1(1 ≥
−×
××−××−×−+=
−−
λλ
Rdzpl
Edz
RdyplLT
EdyLTLT M
MM
Mab
,,
,
,,
,2
05.0 ××
×××=−
χλ
- LTb must be calculated separately for each of the two column parts, depending on 0
−
λ and LTχ :
757.0661.031.110 =×=×=−−
LTC λλ (for the top part of the column)
ADVANCE DESIGN VALIDATION GUIDE
267
041.0/27596.123381
10000000/27518.501177749.0
50000000757.0997.05.0
5.05.0
23232
,
,
,
,2
0
,,
,
,,
,2
0
=
=×
×××
×××=
××
×××××=×
××××=
−−
mmNmmNmm
mmNmmNmm
fWM
fWM
aMM
MM
abyzpl
Edz
yyplLT
EdyLT
Rdzpl
Edz
RdyplLT
EdyLTLT χ
λχ
λ
5.1124.163.44571718.501177
3
3
,
, ≤===mmmm
WW
wyel
yply
243.0
1/27506.4904
32800022
1
=×
==mmNmm
NNNn
M
Rk
Edpl
γ
( ) 923.0923.0;5956.0max;maxmax ==
=
−−
zy λλλ
993.0041.0243.0²923.0²904.0124.16.1923.0²904.0
124.16.12)1124.1(1 =
−×
××−××−×−+=yyC
993.0889.018.50117763.445717
993.0
,
,
3
3
,
, =⇒
≥
==
=
yy
ypl
yelyy
ypl
yel
yy
C
WW
C
mmmm
WW
C
kNNl
IEN
fy
yycr 33.380295.3802327
²²
, ==××
=π
(previously calculated)
Therefore, the yyk term corresponding to the top part of the column will be:
069.1993.01
95.38023273280001
985.0089.1904.01̀
1,
=×−
××=×−
××=
NNC
NNCCk
yy
ycr
Ed
ymLTmyyy
µ
ADVANCE DESIGN VALIDATION GUIDE
268
b) for the bottom part of the column:
yy
ycr
Ed
ymLTmyyy C
NNCCk 1̀
1,
×−
××=µ
- the terms: mLTC ; LTb and yyC must be recalculated:
1
11,,
2 ≥
−×
−
×=
Tcr
Ed
zcr
Ed
LTmymLT
NN
NN
aCC
- the myC term must be calculated corresponding to the bottom part of the column (with )0=ψ :
( ) ( ) 780.095.3802327
23800033.036.079.033.036.021.079.0,
0, =×−×+=×−×+×+=N
NNNC
ycr
Edmy ψψ
839.063.445717
06.4904328000
10253
26
,
, =××
=×=mm
mmN
NmmWA
NM
yeff
eff
Ed
Edyyξ
( ) ( ) 885.01839.01
1839.0780.01780.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ
1
1
933.0
68.28066253280001
51.15817063280001
997.0885.0
11
2
,,
2
=⇒
≥
=
−×
−
×=
=
−×
−
×=
mLT
mLT
Tcr
Ed
zcr
Ed
LTmymLT
C
CN
NN
N
NN
NN
aCC
ADVANCE DESIGN VALIDATION GUIDE
269
ypl
yelLTplmy
ymy
yyyy W
WbnC
wC
wwC
,
,2
max2
max2 6.16.12)1(1 ≥
−×
××−××−×−+=
−−
λλ
Rdzpl
Edz
RdyplLT
EdyLTLT M
MM
Mab
,,
,
,,
,2
05.0 ××
×××=−
χλ
-the LTb must be calculated separately for each of the two column parts, depending of 0
−
λ and LTχ :
757.08.240735730
/27518.501177 23,
0 =×
==Nmm
mmNmmM
fW
cr
yyplλ (for the bottom part of the column)
1804.0²569.0²752.0752.0
1²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ (for the bottom part of the column)
0095.0/27596.123381
5000000/27518.501177804.0
25000000757.0997.05.0
5.05.0
23232
,
,
,
,2
0
,,
,
,,
,2
0
=
=×
×××
×××=
××
×××××=×
××××=
−−
mmNmmNmm
mmNmmNmm
fWM
fWM
aMM
MM
abyzpl
Edz
yyplLT
EdyLT
Rdzpl
Edz
RdyplLT
EdyLTLT χ
λχ
λ
5.1124.163.44571718.501177
3
3
,
, ≤===mmmm
WW
wyel
yply
243.0
1/27506.4904
32800022
1
=×
==mmNmm
NNNn
M
Rk
Edpl
γ
( ) 923.0923.0;5956.0max;maxmax ==
=
−−
zy λλλ
997.00095.0243.0²923.0²904.0124.16.1923.0²904.0
124.16.12)1124.1(1 =
−×
××−××−×−+=yyC
997.0889.018.50117763.445717
997.0
,
,
3
3
,
, =⇒
≥
==
=
yy
ypl
yelyy
ypl
yel
yy
C
WW
C
mmmm
WW
C
kNNl
IEN
fy
yycr 33.380295.3802327
²²
, ==××
=π
(previously calculated)
ADVANCE DESIGN VALIDATION GUIDE
270
Therefore the yyk term corresponding to the bottom part of the column will be:
977.0997.01
95.38023273280001
985.01904.01̀
1,
=×−
××=×−
××=
NNC
NNCCk
yy
ycr
Ed
ymLTmyyy
µ
Note: The software does not give the results of the lower section because it is not the most requested segment.
13.30.2.6Internal factor, yzk , calculation
y
z
yz
zcr
Ed
ymzyz w
wC
NNCk ×××
−×= 6.01
1,
µ
-the mzC ter will be considered for the entire column length (with )0=ψ :
( ) ( ) 765.051.1581706
32800033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
985.0
95.3802327328000839.01
95.38023273280001
1
1
,
, =×−
−=
×−
−
=
NN
NN
NN
NN
ycr
Edy
ycr
Ed
y
χµ
(previously calculated)
Nl
IENfz
zzcr 51.1581706
²²
, =××
=π
(previously calculated)
−×
×
×−×−+=−
LTplz
mzzyz cn
wCwC 5
2
max2
142)1(1 λ
Rdyplltmy
Edy
z
LTLT MCM
ac,,
,4
2
0
510
×××
+××=
−
−
χλ
λ
997.064.57943291
97.14929411 4
4
=−=−=mm
mmIIa
y
tLT (previously calculated)
757.08.240735730
/27518.501177 23,
0 =×
==Nmm
mmNmmM
fW
cr
yyplλ (previously calculated)
923.051.1581706
/27506.4904 22
,
=×
=×
=N
mmNmmN
fA
zcr
yzλ (previously calculated)
NmM Edy 50000, =
5.15.1
5.1536.189.8034496.123381
3
3
,
,
=⇒
≤
≤===z
z
zel
zplz w
wmmmm
WW
w
ADVANCE DESIGN VALIDATION GUIDE
271
5.1124.163.44571718.501177
3
3
,
, ≤===mmmm
WW
wyel
yply
- the myC term will be considered separately for each column part:
a) for the top part of the column:
( ) ( ) 956.01677.11
1677.1900.01900.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ (previously calculated)
1749.0²661.0²831.0831.0
1²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ (previously calculated)
NmmmmNmmfWM yyRdypl 5.137823724/27518.501177 23,, =×=×=
506.05.137823724749.0956.0
50000000923.05
²757.0997.010
510
4
,,
,4
2
0
=××
×+
××=
=××
×+
××=−
−
NmmNmm
MCM
acRdyplltmy
Edy
z
LTLT χλ
λ
Nl
IENfz
zzcr 51.1581706
²²
, =××
=π
(previously calculated
( ) ( ) 765.051.1581706
32800033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
( ) 923.0923.0;5956.0max;maxmax ==
=
−−
zy λλλ
243.0
1
==
M
Rk
Edpl N
Nn
γ
(previously calculated)
- Therefore:
( ) 878.0506.0243.05.1
923.0765.014215.11
142)1(1
5
22
5
2
max2
=
−×
××−×−+=
=
−×
×
×−×−+=
−
LTplz
mzzyz cn
wCwC λ
750.0124.1
5.16.0878.01
51.15817063280001
985.0765.06.01
1,
=×××−
×=×××−
×=
NNw
wC
NNCk
y
z
yz
zcr
Ed
ymzyz
µ
ADVANCE DESIGN VALIDATION GUIDE
272
b) for the bottom part of the column:
( ) ( ) 885.01839.01
1839.0780.01780.01
1 0,0, =×+
××−+=
×+
××−+=
LTy
LTymymymy a
aCCC
ξ
ξ (previously calculated)
1804.0²569.0²752.0752.0
1²²
1≤=
−+=
−+=
LTLTLT
LTλφφ
χ (previously calculated)
NmmmmNmmfWM yyRdypl 5.137823724/27518.501177 23,, =×=×=
254.05.137823724804.0885.0
25000000923.05
²757.0997.010
510
4
,,
,4
2
0
=××
×+
××=
=××
×+
××=−
−
NmmNmm
MCM
acRdyplltmy
Edy
z
LTLT χλ
λ
Nl
IENfz
zzcr 51.1581706
²²
, =××
=π
(previously calculated
( ) ( ) 765.051.1581706
32800033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
( ) 923.0923.0;5956.0max;maxmax ==
=
−−
zy λλλ
243.0
1
==
M
Rk
Edpl N
Nn
γ
(previously calculated)
-Therefore:
( ) 0043.1254.0243.05.1
923.0765.014215.11
142)1(1
5
22
5
2
max2
=
−×
××−×−+=
=
−×
×
×−×−+=
−
LTplz
mzzyz cn
wCwC λ
656.0124.1
5.16.00043.11
51.15817063280001
985.0765.06.01
1,
=×××−
×=×××−
×=
NNw
wC
NNCk
y
z
yz
zcr
Ed
ymzyz
µ
Note: The software does not give the results of the lower section because it is not the most requested segment.
ADVANCE DESIGN VALIDATION GUIDE
273
13.30.2.7Internal factor, zyk , calculation
a) for the top part of the column:
y
z
zy
ycr
Ed
zmLTmyzy w
wC
NNCCk ×××
−××= 6.01
1,
µ
902.0
51.1581706328000586.01
51.15817063280001
1
1
,
, =×−
−=
×−
−=
NN
NN
NN
NN
zcr
Edz
zcr
Ed
z
χµ
586.0=zχ (previously calculated)
ypl
yel
z
yLTpl
y
myyzy W
Www
dnw
CwC
,
,5
2
max2
6.0142)1(1 ××≥
−×
×
×−×−+=
−
λ
NmmmmNmmfWM yyRdypl 5.137823724/27518.501177 23,, =×=×=
NmmmmNmmfWM zzRdzpl 33930039/27596.123381 23,, =×=×=
-in order to calculate the LTd term, the terms myC and mzC must be recalculated for each column part;
-the term mzC must be recalculated for the top column part only, using 5.0=ψ :
( )
( ) 908.051.1581706
32800033.05.036.05.021.079.0
33.05.036.05.021.079.0,
0,
=×−×+×+=
=×−×+×+==
NN
NNCC
zcr
Edmzmz
301.033930039908.0
100000005.137823724749.0956.0
50000000923.01.0
757.0997.02
1.02
4
,,
,
,,
,4
0
=
=×
×××
×+
××=
=×
×××
×+
××=−
−
NmmNmm
NmmNmm
MCM
MCM
adRdzplmz
Edz
RdyplLTmy
Edy
z
LTLT χλ
λ
ADVANCE DESIGN VALIDATION GUIDE
274
859.0
6.0
616.018.50117763.445717
124.15.16.06.0
859.0301.0243.0124.1
847.1904.0142)1124.1(1
142)1(1
,
,
3
3
,
,
5
22
5
2
max2
=⇒
××≥
=××=××
=
−×
××−×−+=
=
−×
×
×−×−+=
−
zy
ypl
yel
z
yzy
ypl
yel
z
y
LTply
myyzy
C
WW
ww
C
mmmm
WW
ww
dnw
CwC
λ
- for the calculation of the zyC term, myC will be used for the entire column and LTd will be used for the top column part:
588.05.1
124.16.0859.01
95.38023273280001
902.0089.1904.06.01
1,
=×××−
××=×××−
××=
NNw
wC
NNCCk
z
y
zy
ycr
Ed
zmLTmyzy
µ
b) for the bottom part of the column:
y
z
zy
ycr
Ed
zmLTmyzy w
wC
NNCCk ×××
−××= 6.01
1,
µ
902.0
51.1581706328000586.01
51.15817063280001
1
1
,
, =×−
−=
×−
−=
NN
NN
NN
NN
zcr
Edz
zcr
Ed
z
χµ
586.0=zχ (previously calculated)
ypl
yel
z
yLTpl
y
myyzy W
Www
dnw
CwC
,
,5
2
max2
6.0142)1(1 ××≥
−×
×
×−×−+=
−
λ
NmmmmNmmfWM yyRdypl 5.137823724/27518.501177 23,, =×=×=
NmmmmNmmfWM zzRdzpl 33930039/27596.123381 23,, =×=×=
- in order to calculate the LTd term, the terms myC and mzC must be recalculated for each column part;
- the term mzC must be recalculated for the top column part only, using 0=ψ :
( ) ( ) 765.051.1581706
32800033.036.079.033.036.079.0,
0, =×−×+=×−×+==N
NNNCC
zcr
Edmzmz
ADVANCE DESIGN VALIDATION GUIDE
275
090.033930039765.0
50000005.137823724804.0885.0
25000000923.01.0
757.0997.02
1.02
4
,,
,
,,
,4
0
=
=×
×××
×+
××=
=×
×××
×+
××=−
−
NmmNmm
NmmNmm
MCM
MCM
adRdzplmz
Edz
RdyplLTmy
Edy
z
LTLT χλ
λ
( ) 923.0923.0;5956.0max;maxmax ==
=
−−
zy λλλ
892.0
6.0
462.018.50117763.445717
5.1124.16.06.0
892.0090.0243.0124.1
923.0885.0142)1124.1(1
142)1(1
,
,
3
3
,
,
5
22
5
2
max2
=⇒
××≥
=××=××
=
−×
××−×−+=
=
−×
×
×−×−+=
−
zy
ypl
yel
z
yzy
ypl
yel
z
y
LTply
myyzy
C
WW
ww
C
mmmm
WW
ww
dnw
CwC
λ
- for the calculation of the
zyC term, myC will be used for the entire column and LTd will be used for the top column part:
566.05.1
124.16.0892.01
95.38023273280001
902.0089.1904.06.01
1,
=×××−
××=×××−
××=
NNw
wC
NNCCk
z
y
zy
ycr
Ed
zmLTmyzy
µ
Note: The software does not give the results of the lower section because it is not the most requested segment.
13.30.2.8Internal factor, zzk , calculation
a) for the top part of the column:
zz
zcr
Ed
zmzzz C
NNCk 1
1,
×−
×=µ
zpl
zelLTplmz
zmz
zzzz W
WenC
wC
wwC
,
,max
2max
2 6.16.12)1(1 ≥
−×
××−××−×−+= λλ
NmmmmNmmfWM yyRdypl 5.137823724/27518.501177 23,, =×=×=
- in calculating the LTe , the myC term must be used accordingly with the corresponding column part
787.0/27518.501177749.0956.0
50000000923.01.0
757.0997.07.1
1.07.1
234
,,
,4
0
=×××
×+
××=
=××
×+
××=−
−
mmNmmNmm
MCM
aeRdyplltmy
Edy
z
LTLT χλ
λ
ADVANCE DESIGN VALIDATION GUIDE
276
- for the calculation of the
zzC term, mzC will be used for the entire column and LTe will be used for the top column part:
013.1651.0
96.12338189.80344
013.1243.0787.0847.1765.05.16.1847.1765.0
5.16.12)15.1(1
6.16.12)1(1
,
,
3
3
,
,
222
2max
2max
2
=⇒
≥
==
=×
−
××−××−×−+=
=×
−
××−××−×−+=
zz
zpl
zelzz
zpl
zel
plLTmzz
mzz
zzz
C
WW
C
mmmm
WW
neCw
Cw
wC λλ
860.0013.11
51.15817063280001
902.0765.01
1,
=×−
×=×−
×=
NNC
NNCk
zz
zcr
Ed
zmzzz
µ
b) for the bottom part of the column:
zz
zcr
Ed
zmzzz C
NNCk 1
1,
×−
×=µ
zpl
zelLTplmz
zmz
zzzz W
WenC
wC
wwC
,
,max
2max
2 6.16.12)1(1 ≥
−×
××−××−×−+= λλ
NmmmmNmmfWM yyRdypl 5.137823724/27518.501177 23,, =×=×=
- in calculating the LTe , the myC term must be used according to the corresponding column part
396.0/27518.501177804.0885.0
25000000923.01.0
757.0997.07.1
1.07.1
234
,,
,4
0
=×××
×+
××=
=××
×+
××=−
−
mmNmmNmm
MCM
aeRdyplltmy
Edy
z
LTLT χλ
λ
- for the calculation of the
zzC term, mzC will be used for the entire column and LTe will be used for the top column part:
( ) 923.0923.0;5956.0max;maxmax ==
=
−−
zy λλλ
060.1651.0
96.12338189.80344
060.1243.0396.0923.0765.05.16.1923.0765.0
5.16.12)15.1(1
6.16.12)1(1
,
,
3
3
,
,
222
2max
2max
2
=⇒
≥
==
=×
−
××−××−×−+=
=×
−
××−××−×−+=
zz
zpl
zelzz
zpl
zel
plLTmzz
mzz
zzz
C
WW
C
mmmm
WW
neCw
Cw
wC λλ
ADVANCE DESIGN VALIDATION GUIDE
277
821.0060.11
51.15817063280001
902.0765.01
1,
=×−
×=×−
×=
NNC
NNCk
zz
zcr
Ed
zmzzz
µ
Note: The software does not give the results of the lower section because it is not the most requested segment.
13.30.2.9Bending and axial compression verification
∆+×+
×
∆+×+
×
∆+×+
×
∆+×+
×
1
,
,,
1
,
,,
1
1
,
,,
1
,
,,
1
M
Rkz
RdzEdzzz
M
RkyLT
RdyEdyzy
M
Rkz
Ed
M
Rkz
RdzEdzyz
M
RkyLT
RdyEdyyy
M
Rky
Ed
MMM
kMMM
kNN
MMM
kMMM
kNN
γγχ
γχ
γγχ
γχ
iyRk AfN ×=
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
=++=×
××+
+×
×
××+
××
=++=×
××+
+×
×
××+
××
mmNmmNmm
mmNmmNmm
mmNmmN
mmNmmNmm
mmNmmNmm
mmNmmN
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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
=++=×
××+
+×
×
××+
××
=++=×
××+
+×
×
××+
××
mmNmmNmm
mmNmmNmm
mmNmmN
mmNmmNmm
mmNmmNmm
mmNmmN
Finite elements modeling
■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
279
yχ coefficient corresponding to non-dimensional slenderness yλ
Column subjected to axial and shear force to the top
yχ
zχ coefficient corresponding to non-dimensional slenderness zλ
Column subjected to axial and shear force to the top
zχ
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Internal factor, yyk
Column subjected to axial and shear force to the top
yyk
Internal factor, yzk
Column subjected to axial and shear force to the top
yzk
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281
Internal factor, zyk
Column subjected to axial and shear force to the top
zyk
Internal factor, zzk
Column subjected to axial and shear force to the top
zzk
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282
Bending and axial compression verification term depending of the compression effort over the Y axis: SNy
Bending and axial compression verification term depending of the compression effort over the Y axis
SNy
Bending and axial compression verification term depending of the Y bending moment over the Y axis: SMyy
Bending and axial compression verification term depending of the Y bending moment over the Y axis
SMyy
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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 over the Y axis
SMyz
Bending and axial compression verification term depending of the compression effort over the Z axis: SNz
Bending and axial compression verification term depending of the compression effort over the Z axis
SNz
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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 moment over the Z axis
SMzy
Bending and axial compression verification term depending of the Z bending moment over the Z axis: SMzz
Bending and axial compression verification term depending of the Z bending moment over the Z axis
SMzz
ADVANCE DESIGN VALIDATION GUIDE
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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; support conditions; 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
LTχ
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13.30.2.10Reference results a) for the top part of the column:
Result name Result description Reference value
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
0.839
zχ zχ coefficient corresponding to non-dimensional slenderness zλ
0.586
yyk Internal factor, yyk
1.069
yzk Internal factor, yzk
0.750
zyk Internal factor, zyk
0.588
zzk Internal factor, zzk
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 bending moment over the Y axis
0.52
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
0.21
SNz
Bending and axial compression verification term depending of the compression effort over the z axis
0.41
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
0.27
SMzz Bending and axial compression verification term depending of the Z bending moment 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
LTχ 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
yχ yχ coefficient corresponding to non-dimensional slenderness yλ
0.839
zχ zχ coefficient corresponding to non-dimensional slenderness zλ
0.586
yyk Internal factor, yyk
0.977
yzk Internal factor, yzk
0.656
zyk Internal factor, zyk
0.566
zzk Internal factor, zzk
0.821
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 bending moment over the Y axis
0.22
SMyz Bending and axial compression verification term depending of the Z bending moment over the Y axis
0.97
SNz
Bending and axial compression verification term depending of the compression effort over the z axis
0.42
SMzy Bending and axial compression verification term depending of the Y bending moment over the Z axis
0.13
SMzz Bending and axial compression verification term depending of the Z bending moment over the Z axis
0.12 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 426.10
LTχ 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.
13.30.3Calculated results Result name Result description Value Error Xy Coefficient corresponding to non-dimensional slenderness
by Y axis 0.839285 adim
0.0340 %
Xz Coefficient corresponding to non-dimensional slenderness by the Z axis
0.585533 adim
-0.0797 %
Kyy Internal coefficient kyy 1.07027 adim
0.1188 %
Kyy Internal coefficient kyy (bottom) 0.954475 adim
-2.3055 %
Kyz Internal coefficient kyz 0.750217 adim
0.0289 %
Kyz Internal coefficient kyz (bottom) 0.656481 adim
0.0733 %
Kzy Internal coefficient kzy 0.593445 adim
0.9260 %
Kzy Internal coefficient kzy (bottom) 0.508819 adim
-0.0356 %
Kzz Internal coefficient kzz 0.860237 adim
0.0276 %
Kzz Internal coefficient kzz (bottom) 0.821717 adim
0.0873 %
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13.31 EC3 Test 31: Verifying IPE450 column fixed on base subjected to axial compression and bending moment, both applied on top
Test ID: 5731
Test status: Passed
13.31.1Description The test verifies an IPE450 column made of S275 steel.
The column is subjected to a -1000kN compression effort and a 200kNm bending moment by the Y axis.
The calculations are made according to Eurocode 3 French Annex.
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13.32 EC3 Test 15: Verifying a rectangular hollow section column subjected to bending and axial efforts
Test ID: 5735
Test status: Passed
13.32.1Description Verifies a rectangular hollow section column made of S235 steel subjected to bending and axial efforts.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.32.2Background Verifies the adequacy of a rectangular hollow section column made of S235 steel to resist bending and axial efforts. The name of the cross-section is RC3020100 and it can be found in the Advance Design OTUA library. The column is fixed at its base and it is subjected to a uniformly distributed load over its height and a punctual axial load applied on the top. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.32.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = - 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).
ADVANCE DESIGN VALIDATION GUIDE
291
Units
Metric System
Geometry
Below are described the column cross section characteristics:
■ Height: h = 300 mm, ■ Width: b = 200 mm, ■ Thickness: t = 10 mm, ■ Outer radius: r = 15 mm, ■ Column height: L = 5000 mm, ■ Section area:A = 9490 mm2, ■ Plastic section modulus about y-y axis: Wpl,y = 956000 mm3,
■ Partial factor for resistance of cross sections: 0.10 =Mγ .
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (z = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (z = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at z = 5.0: Fz = - 500 000 N, Uniformly distributed load: q = Fx = 5 000 N/ml
■ Internal: None.
ADVANCE DESIGN VALIDATION GUIDE
292
13.32.2.2Reference results for calculating the column subjected to bending and axial force In order to verify the steel column subjected to bending and axial force, the design resistance for uniform compression (Nc,Rd) and also the design plastic moment resistance (Mpl,Rd) have to be compared with the design values of the corresponding efforts.
The design resistance for uniform compression is verified considering the relationship (6.9) from chapter 6.2.4 (EN 1993-1-1), while the design plastic moment resistance is verified considering the criterion (6.12) from chapter 6.2.5 (EN 1993-1-1).
Before starting the above verifications, the cross-section class has to be determined.
Cross section class
The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.
In this case, the stresses distribution is like in the picture below:
Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class for compressed parts. The picture below shows an extract from this table.
ADVANCE DESIGN VALIDATION GUIDE
293
Taking into account that the top wing part is subjected to compression stresses, its class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to compression).
1510
10215220022=
×−×−=
×−×−=
mmmmmmmm
ttrb
tc
0.1235==
yfε
Therefore:
333315 =≤= εtc
This means that the top wing is Class 1. Because the bottom wing is tensioned, it will be classified as Class 1.
The left/right web is subjected to bending stresses. Their class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to bending and compression). It is also necessary to determine which portion of the web is compressed (α). α is determined considering the stresses distribution on the web.
5.0832.063.26
1321
>=→=−
αα
αMPa
MPa
2510
10215230022=
×−×−=
×−×−=
mmmmmmmm
ttrh
tc
0.1235==
yfε
Therefore:
34.40113
39625 =−
≤=α
εtc
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 EN 1993-1-1:2001.
NMPammfAN
M
yRdc 2230150
0.12359490 2
0, =
×=
×=
γ
The verification of the design resistance for uniform compression is done with relationship (6.9) from EN 1993-1-1. The corresponding work ratio is:
Work ratio = %42.221002230150500000100100
,,
=×=×=×N
NN
FNN
Rdc
z
Rdc
Ed
Verifying the design plastic moment resistance
The design plastic moment resistance, for Class 1 cross-section, is determined with formula (6.13) from EN 1993-1-1:2001.
NmmMPammfWMM
M
yyplRdplRdc 224660000
0.1235956000 3
0
,,, =
×=
×==
γ
The verification of the design resistance for bending is done with relationship (6.12) from EN 1993-1-1. The corresponding work ratio is:
Work ratio = %82.27100224660000
250005000/5
1002100,,
=×××
=×××
=×Nmm
mmmmmmN
M
LLq
MM
RdcRdc
Ed
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
ADVANCE DESIGN VALIDATION GUIDE
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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
ADVANCE DESIGN VALIDATION GUIDE
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13.32.2.3Reference 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 %
13.32.3Calculated results Result name Result description Value Error Work ratio - Fx Compression resistance work ratio 22.42 % 0.0000 % Work ratio - Oblique
Work ratio of the design resistance for bending 27.8198 % -0.0007 %
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13.33 EC3 Test 33: Verifying UPN300 simple supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and punctual vertical force by Z axis
Test ID: 5733
Test status: Passed
13.33.1Description The test verifies an upn300 beam made of S235 steel.
The beam is subjected to 20 kN compression force, 50 kN PUNCTUAL vertical load applied to the middle of the beam and 5kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
13.34 EC3 Test 35: Verifying C310x30.8 class 4 cantilever, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load applied on the free end
Test ID: 5737
Test status: Passed
13.34.1Description The test verifies a C310x30.8 beam made of S355 steel.
The beam is subjected to 3.00 kN compression force, 1.80 kN punctual vertical load applied on the free end of the beam and 1.2.kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
13.35 EC3 Test 37: Verifying RHS350x150x8.5H class 3 column, loaded with centric compression, punctual lateral load and bending moment, all applied to the top of the column
Test ID: 5739
Test status: Passed
13.35.1Description The test verifies a RHS350x150x8.5H column made of S275 steel.
The column is subjected to 680 kN compression force, 5 kN horizontal load applied on Y axis direction and 200 kNm bending moment after the Y axis. All loads are applied on the top of the column.
The calculations are made according to Eurocode 3 French Annex.
13.36 EC3 Test 30: Verifying IPE300 beam, simply supported, loaded with centric compression and uniform linear efforts by Y and Z axis
Test ID: 5730
Test status: Passed
13.36.1Description The test verifies an IPE 300 beam made of
The beam is subjected to a 20kN compression effort, a -10kN/m uniform linear effort applied vertically and a -5kN/m linear uniform load applied horizontal.
The calculations are made according to Eurocode 3 French Annex.
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13.37 EC3 Test 34: Verifying C310x30.8 class 3beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle
Test ID: 5734
Test status: Passed
13.37.1Description The test verifies an C310x30.8 beam made of S235 steel.
The beam is subjected to a 12 kN compression force, 8 kN PUNCTUAL vertical load applied to the middle of the beam and 1.5 kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
13.38 EC3 Test 36: Verifying RHS300x150x9H class 1 simply supported beam, loaded with centric compression, uniform linear horizontal efforts and a vertical punctual load in the middle
Test ID: 5738
Test status: Passed
13.38.1Description The test verifies an RHS300x150x9H beam made of S275 steel.
The beam is subjected to 12 kN axial compression force, 7 kN punctual vertical load applied to the middle of the beam and 3 kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
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13.39 EC3 Test 14: Verifying the bending resistance of a rectangular hollow section column made of S235 steel
Test ID: 5728
Test status: Passed
13.39.1Description Verifies the design resistance for bending of a rectangular hollow section column made of S235 steel.
The verification is made according to Eurocode 3 (EN 1993-1-1) French annex.
13.39.2Background Verifies the adequacy of a rectangular hollow section column made of S235 steel to resist bending efforts. Verification of the design resistance for bending at ultimate limit state is realised. The name of the cross-section is RC3020100 and can be found in the Advance Design OTUA library. The column is fixed at its base and it is subjected to a punctual horizontal load applied to the middle height (50 000 N). The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.39.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fx = 50 000 N,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
ADVANCE DESIGN VALIDATION GUIDE
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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 =Mγ .
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (z = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (z = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at z = 2.5: V= Fx = 50 000 N,
■ Internal: None.
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13.39.2.2Reference results for calculating the design resistance for bending Before calculating the design resistance for bending, the cross section class has to be determined.
Cross section class
The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.
In this case, the column is subjected to bending efforts, therefore the stresses distribution is like in the picture below:
Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class for compressed parts. The picture below shows an extract from this table.
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Taking into account that the top wing part is subjected to compression stresses, its class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to compression).
1510
10215220022=
×−×−=
×−×−=
mmmmmmmm
ttrb
tc
0.1235==
yfε
Therefore:
333315 =≤= εtc
This means that the top wing is Class 1. Because the bottom wing is tensioned, it will be classified as Class 1.
The left/right web is subjected to bending stresses. Their class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to bending):
2510
10215230022=
×−×−=
×−×−=
mmmmmmmm
ttrh
tc
0.1235==
yfε
Therefore:
727225 =≤= εtc
This means that the left/right web is Class 1.
Because a cross-section is classified according to the least favorable classification of its compression elements (chapter 5.5.2(6) from EN 1993-1-1: 2001), this means that the cross-section is Class 1.
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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.
NmmMPammfWM
M
yyplRdc 224660000
0.1235956000 3
0
,, =
×=
×=
γ
Work ratio
The verification of the design resistance for bending is done with relationship (6.12) from EN 1993-1-1. The corresponding work ratio is:
Work ratio = %64.55100224660000
2500050000
1002100,,
=××
=××
=×Nmm
mmN
M
LV
MM
RdcRdc
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
13.39.2.3Reference results
Result name Result description Reference value Work ratio - Oblique Work ratio of the design resistance for bending [%] 55.64 %
13.39.3Calculated 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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13.40 EC3 Test 32: Verifying IPE600 simple supported beam, loaded with centric compression and uniform linear efforts by Y and Z axis
Test ID: 5732
Test status: Passed
13.40.1Description The test verifies an IPE600 beam made of S275 steel.
The beam is subjected to a -3700kN compression force, a -10kN/m linear uniform vertical load and a -5kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
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13.41 EC3 Test 16: Verifying a simply supported rectangular hollow section beam subjected to biaxial bending
Test ID: 5736
Test status: Passed
13.41.1Description Verifies a simply supported rectangular hollow section beam made of S235 steel subjected to biaxial bending.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.41.2Background Verifies the adequacy of a rectangular hollow section beam made of S235 steel to resist bi-axial bending efforts. The name of the cross-section is RC3020100 and it can be found in the Advance Design OTUA library. The beam is simply supported and it is subjected to uniformly distributed loads over its length. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.41.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear; ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = - 10 000 N/ml, Fy = 10 000 N/ml,
■ 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,
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■ 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 =Mγ .
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (z = 0) restrained in translation along X, Y and Z axis, Support at end point (z = 5.00) restrained in translation along Y and Z axis, and restrained in rotation
about the X axis. ■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ External: Uniformly distributed load: q1 = Fz = -10 000 N/ml Uniformly distributed load: q2 = Fy = 10 000 N/ml
■ Internal: None.
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13.41.2.2Reference results for calculating the beam subjected to bi-axial bending In order to verify the steel beam subjected to bi-axial bending, the criterion (6.41) from chapter 6.2.9.1 (EN 1993-1-1) has to be used.
Before verifying this criterion, the cross-section class has to be determined.
Cross section class
The following results are determined according to Eurocode 3: Design of steel structures - Part 1-1: General rules and rules for buildings (EN 1993-1-1: 2001), Chapter 5.5.2.
In this case, the stresses distribution is like in the picture below:
Table 5.2 - sheet 1, from Chapter 5.5.2 (EN 1993-1-1: 2001), establish the rules to determine the class for compressed parts. The picture below shows an extract from this table.
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Taking into account that the entire cross-section is subjected to bending stresses, its class can be determined by considering the geometrical properties and the conditions described in Table 5.2 - sheet 1 (the above highlighted extract – part subject to bending).
2510
10215230022=
×−×−=
×−×−=
mmmmmmmm
ttrb
tc
0.1235==
yfε
Therefore:
727225 =≤= εtc
This means that the left/right web is Class 1. As the dimensions for top/bottom wing are smaller than the left/right web, they will be also classified as Class 1.
Because a cross-section is classified according to the least favorable classification of its compression elements (chapter 5.5.2(6) from EN 1993-1-1: 2001), this means that the cross-section is Class 1.
Determining the design plastic moment resistance
Before verifying for bi-axial bending a rectangular structural hollow section of uniform thickness, the design plastic moment resistance reduced due to the axial force (MN,Rd) needs to be determined. Its determination has to be made about 2 axes (according to the bending efforts) and it will be done with formulae (6.39) and (6.40) from EN 1993-1-1. Other terms involved in calculation have to be determined: aw, af, n.
■ Ratio of design normal force to design plastic resistance to normal forces, n:
→=Rdpl
Ed
NNn
,
as the beam is not subjected to axial efforts n = 0.
■ Determination of aw for hollow section:
5.05.02=→≤
××−= ww a
AtbAa
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■ Determination of af for hollow section:
3678.05.02=→≤
××−= ff a
AthAa
■ Determination of aw for hollow section:
5.05.02=→≤
××−= ww a
AtbAa
■ Determination of design plastic moment resistance (about y-y axis) reduced due to the axial force, MN,y,Rd:
( )( ) Rdypl
w
RdyplRdyN M
anM
M ,,,,
,, 5.011
≤×−
−×=
In order to fulfill the above relationship MN,y,Rd must be equal to Mpl,y,Rd.
NmmMPammfWMM
M
yyplRdyplRdyN
43
0
,,,,, 1022466
0.1235956000
×=×
=×
==γ
■ Determination of design plastic moment resistance (about z-z axis) reduced due to the axial force, MN,z,Rd:
( )( ) Rdzpl
f
RdzplRdzN M
anM
M ,,,,
,, 5.011
≤×−
−×=
In order to fulfill the above relationship MN,z,Rd must be equal to Mpl,z,Rd.
NmmMPammfWMM
M
yzplRdzplRdzN
43
0
,,,,, 105.16943
0.1235721000
×=×
=×
==γ
Verifying for bi-axial bending
Criterion (6.41) from EN 1993-1-1 has to be fulfilled:
0.1,,
,
,,
, ≤
+
βα
RdzN
Edz
RdyN
Edy
MM
MM
■ Determination of constants α and β for rectangular hollow section:
66.1613.1166.1
2 ==→≤×−
== βαβαn
■ Determination of design bending moments (My,Ed and Mz,Ed) at the middle of the beam:
NmmmmmmNLqM Edy
4222
, 10312585000/10
81
×=×
=×
=
NmmmmmmNLqM Edz
4222
, 10312585000/10
82
×=×
=×
=
■ Verifying criterion (6.41) from EN 1993-1-1:
0.1098.00604.00378.0105.16943
1031251022466
10312566.1
4
466.1
4
4
≤=+=
×
×+
×
×Nmm
NmmNmm
Nmm
Work ratio = %8.9100
0.1098.0
=×
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Verifying for simple bending about y-y axis
For class 1 cross-section, design resistance for simple bending about y-y axis is verified using the criterion (6.12) from EN 1993-1-1:
0.1139.01022466
1031254
4
,,
, ≤=×
×=
NmmNmm
MM
Rdypl
Edy
Work ratio = %9.13100
0.1139.0
=×
Verifying for simple bending about z-z axis
For class 1 cross-section, design resistance for simple bending about z-z axis is verified using the criterion (6.12) from EN 1993-1-1:
0.11844.0105.16943
1031254
4
,,
, ≤=×
×=
NmmNmm
MM
Rdzpl
Edz
Work ratio = %44.18100
0.11844.0
=×
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.
Finite elements results
Work ratio of the design resistance for bending
Beam subjected to bending efforts Work ratio – Oblique
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13.41.2.3Reference results
Result name Result description Reference value Work ratio - Oblique Work ratio of the design resistance for biaxial bending 18.44 %
13.41.3Calculated results Result name Result description Value Error Work ratio - Oblique
Work ratio of the design resistance for biaxial bending 18.4437 % 0.2375 %
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13.42 EC3 Test 18: Verifying a simply supported circular hollow section element subjected to torsional efforts
Test ID: 5743
Test status: Passed
13.42.1Description Verifies a simply supported circular hollow section element made of S235 steel subjected to torsional efforts.
The verification is made according to Eurocode3 (EN 1993-1-1) French Annex.
13.43 EC3 Test 39: Verifying CHS323.9x6.3H class 2 beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle
Test ID: 5741
Test status: Passed
13.43.1Description The test verifies a CHS323.9x6.3H beam made of S275 steel.
The beam is subjected to 20 kN axial compression force, 50 kN punctual vertical load applied to the middle of the beam and 4 kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
13.44 EC3 Test 38: Verifying RHS350x150x5H class 4 column, loaded with centric compression, punctual horizontal force by Y and a bending moment, all applied to the top
Test ID: 5740
Test status: Passed
13.44.1Description The test verifies a RHS350x150x5H column made of S355 steel.
The column is subjected to 680 kN compression force, 5 kN punctual horizontal load and 200 kNm bending moment, all applied to the top.
The calculations are made according to Eurocode 3 French Annex.
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13.45 EC3 Test 45: Comparing the shear resistance of a welded built-up beam made from different steel materials
Test ID: 5745
Test status: Passed
13.45.1Description The shear resistance of a welded built-up beam made of S275 steel is compared with the shear resistance of the same built-up beam made of a user-defined steel material.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.45.2Background Verifies the shear resistance of a welded built-up beam made of 500 MPa yield strength user-defined steel. The beam is simply supported and it is subjected to a uniformly distributed load (20 000 N/ml) applied over its length. The dead load will be neglected.
Also verifies the shear resistance of the same welded built-up beam made of S275 steel. The loading and support conditions are the same.
This test was evaluated by the French control office SOCOTEC.
13.45.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear; ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = - 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 =Mγ .
Materials properties
500 MPa yield strength user-defined material and S275 steel are used. The following characteristics are used:
■ Yield strength fy = 500 MPa, ■ Yield strength (for S275 steel) fy = 275 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (z = 0) restrained in translation along X, Y and Z axis, Support at end point (z = 5.00) restrained in translation along X, Y and Z axis.
► Inner: None.
Loading
The beam is subjected to the following loadings:
■ External: Uniformly distributed load: q = Fz = -20 000 N/ml
■ Internal: None.
13.45.2.2Reference results for calculating the design plastic shear resistance of the cross section In order to verify the steel beam subjected to shear, the criterion (6.18) from chapter 6.2.6 (EN 1993-1-1) has to be used:
0.1,
≤Rdpl
Ed
VV
■ VEd represents the design value of the shear force:
NmmmlNLqVEd 500002
5000/200002
=×
=×
=
■ Vpl,Rd represents the design plastic shear resistance. The design plastic shear resistance of the cross-section is determined with formula (6.18) from EN 1993-1-1:2001. Before using it, the shear area (Av) has to be determined.
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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 (η) may be conservatively taken equal 1.0.
For a welded I sections, the shear area is determined according to chapter 6.2.6 (3) from EN 1993-1-1. As the load is parallel to web, the shear area is:
( )∑ =××=××= 206.1978)1.76.278(0.1 mmmmmmthA wwv η
Shear area of the cross section made of S275 steel
According to chapter 5.1 from EN 1993-1-5, because the steel grade used for beam is up to S460, the factor for shear area (η) is 1.2.
As the load is parallel to web, the shear area becomes:
( )∑ =××=××= 267.2373)1.76.278(2.1 mmmmmmthA wwv η
Design plastic shear resistance of the cross section made of 500 MPa yield strength user-defined material
EN 1993-1-1 provides the following formula to calculate the design plastic shear resistance of the cross-section:
N
MPammf
AV
M
yv
Rdpl 7.5710160.1
350006.1978
32
0, =
×=
×=
γ
The verification of the design plastic shear resistance of the cross section is done with criterion (6.18) from EN 1993-1-1:
0.10876.07.571016
50000
,
≤==N
NVV
Rdpl
Ed
The corresponding work ratio is:
Work ratio = %76.81007.571016
50000100,
=×=×N
NVV
Rdpl
Ed
Design plastic shear resistance of the cross section made of S275 steel
EN 1993-1-1 provides the following formula to calculate the design plastic shear resistance of the cross-section:
N
MPammf
AV
M
yv
Rdpl 7.3768700.1
327567.2373
32
0, =
×=
×=
γ
The verification of the design plastic shear resistance of the cross section is done with criterion (6.18) from EN 1993-1-1:
0.1133.07.376870
50000
,
≤==N
NVV
Rdpl
Ed
The corresponding work ratio is:
Work ratio = %27.131007.376870
50000100,
=×=×N
NVV
Rdpl
Ed
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
13.45.2.3Reference 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 %
13.45.3Calculated results Result name Result description Value Error Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 275
MPa) 13.2671 % -0.0219 %
Work ratio - Fz Work ratio of the design plastic shear resistance (fy = 500 MPa)
8.75631 % -0.0421 %
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13.46 EC3 Test 43: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange
Test ID: 5750
Test status: Passed
13.46.1Description Determines the lateral torsional buckling parameters for a I-shaped laminated beam made of S235 steel, considering the load applied on the lower flange.
The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.
13.47 EC3 Test 41: Determining lateral torsional buckling parameters for a I-shaped laminated beam considering the load applied on the lower flange
Test ID: 5753
Test status: Passed
13.47.1Description Determines the lateral torsional buckling parameters for a I-shaped laminated beam made of S235 steel, considering the load applied on the lower flange. The loadings applied on the beam are: a uniformly distributed load and 2 punctual bending moments, acting opposite to each other, applied at beam extremities.
The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.
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13.48 EC3 test 8: Verifying the classification and the resistance of a column subjected to bending and axial load
Test ID: 5632
Test status: Passed
13.48.1Description Verifies the classification and the resistance for an IPE 600 column made of S235 steel subjected to bending and axial force. The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
13.48.2Background Classification and verification of an IPE 600 column, made of S235 steel, subjected to bending and axial force. The column is fixed at its base and free on the top. The column is loaded by a compression force (1 000 000 N), applied at its top, and a uniformly distributed load (50 000 N/ml). The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.48.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load case (Q1) and load combination are used:
■ Exploitation loadings (category A), Q1: Fz = -1 000 000 N, Fx = 50 000 N/ml,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q1
■ 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 ypl = ,
■ Partial factor for resistance of cross sections: 0.10 =Mγ .
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Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, Free at end point (x = 5.00).
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 5.0: N = FZ = -1 000 000 N, Uniformly distributed load: q = Fx = 50 000 N/ml
■ Internal: None.
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13.48.2.2Reference results for calculating the column subjected to bending and axial force In order to verify the steel column subjected to bending and axial force, the design resistance for uniform compression (Nc,Rd) and also the design plastic moment resistance reduced due to the axial force (MN,Rd) have to be compared with the design values of the corresponding efforts.
The design resistance for uniform compression is verified considering the relationship (6.9) from chapter 6.2.4 (EN 1993-1-1), while for bi-axial bending, the criterion (6.41) from chapter 6.2.9.1 (EN 1993-1-1) has to be satisfied.
Before starting the above verifications, the cross-section class has to be determined.
Cross section class
Considering that the column is subjected to combined bending and axial compression, and also that its axial effort is bigger than 835 kN, the following classification is made according to the CTICM journal no. 4 – 2005 (extracted of journal):
So, according to this table, the column cross-section is Class 2.
Verifying the design resistance for uniform compression
Expression (6.10) from EN 1993-1-1 is used to determine the design compression resistance, Nc,Rd:
NMPammfAN
M
yRdc 3666000
0.123515600 2
0, =
×=
×=
γ
In order to verify the design resistance for uniform compression, the criterion (6.9) from chapter 6.2.4 (EN 1993-1-1) has to be satisfied:
%100%3.270.1273.036660001000000
,,
≤→≤===NN
NN
NN
RdcRdc
Ed
Verifying the column subjected to bending and axial force
According to paragraph 6.2.9.1 (4) from EN 1993-1-1, allowance will not be made for the effect of the axial force on the plastic resistance moment about the y-y axis if relationship (6.33) is fulfilled.
NNNNN RdplEd 916500366600025.0100000025.0 , =×>→×≤
In this case, because the above verification is not fulfilled, the axial force has an impact on the plastic resistance moment about the y-y axis.
In order to verify the column subjected to bending and axial force, the criterion (6.41) from EN 1993-1-1 has to be used. Supplementary terms need to be determined: design resistance for bending (Mpl,Rd), ratio of design normal force to design plastic resistance to normal force of the gross cross-section (n), ratio of web area to gross area (a), design plastic moment resistance reduced due to the axial force (MN,Rd).
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■ Design plastic moment resistance:
NmmMPammfWM
M
yyplRdypl 825320000
0.12353512000 3
0
,,, =
×=
×=
γ
■ Ratio of design normal force to design plastic resistance to normal force of the gross cross-section:
273.036660001000000
,
===NN
NNn
Rdpl
■ Ratio of web area to gross area:
464.015600
1922021560022
2
=××−
=××−
=mm
mmmmmmA
tbAa f
■ Design plastic moment resistance reduced due to the axial force is determined according to expression (6.36) from EN 1993-1-1:
anMM RdyplRdyN ×−
−×=
5.011
,,,, but RdyplRdyN MM ,,,, ≤
NmmNmmNmmM RdyN 825320000781259948464.05.01
273.01825320000,, ≤=×−
−×=
■ The column subjected to bending and axial force is verified with criterion (6.41) from EN 1993-1-1:
0.1,,
,
,,
, ≤
+
βα
RdzN
Edz
RdyN
Edy
MM
MM
Because the column doesn’t have bending moment about z axis, the second term from criterion (6.41) is
neglected. The verification becomes: 0.12/
,,
2
,,
, ≤
×=
RdyNRdyN
Edy
MLq
MM
( ) %100%9.790.1799.0781259948
2/5000/502/ 2
,,
2
<→≤=
×=
×Nmm
mmmmNM
Lq
RdyN
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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13.48.2.3Reference 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 %
13.48.3Calculated results
Result name Result description Value Error Work ratio - Fx Work ratio of the design resistance for uniform
compression 27.2777 % -0.0084 %
Work ratio - Oblique
Work ratio of the design resistance for oblique bending 79.9691 % -0.0386 %
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13.49 EC3 Test 17: Verifying a simply supported rectangular hollow section beam subjected to torsional efforts
Test ID: 5742
Test status: Passed
13.49.1Description Verifies a simply supported rectangular hollow section beam made of S235 steel subjected to torsional efforts.
The verification is made according to Eurocode 3 (EN 1993-1-1) French Annex.
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13.50 EC3 Test 44: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange
Test ID: 5749
Test status: Passed
13.50.1Description Determines the lateral torsional buckling parameters for a I-shaped welded built-up beam made of S235 steel, considering the load applied on the upper flange. The loadings applied on the beam are: a uniformly distributed load and 2 punctual bending moments, acting opposite to each other, applied at beam extremities.
The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.
13.50.2Background Determines the elastic critical moment (Mcr) and factors (C1, C2, χLT) involved in the torsional buckling verification for a simply supported steel beam. The beam is made of S235 steel and it is subjected to a uniformly distributed load (50 000 N/ml) applied over its length and concentrated bending moments applied at its extremities (loads are applied to the upper fibre). The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.50.2.1Model description ■ Reference: Guide d’evaluation Advance Design, EN 1993-1-1: 2001; ■ Analysis type: static linear; ■ Element type: linear. The following load case and load combination are used:
■ Exploitation loadings (category A), Q: Fz = - 50 000 N/ml, My,1 = 142 x 106 Nmm, My,2 = - 113.6 x 106 Nmm,
■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
■ Cross section dimensions are in milimeters (mm).
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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 =Mγ .
Materials properties
S235 steel material is used. The following characteristics are used:
■ Yield strength fy = 235 MPa, ■ Longitudinal elastic modulus: E = 2.1 x 105 MPa; ■ Shear modulus of rigidity: G=80800MPa.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (x = 0) restrained in translation along X, Y and Z axis, Support at end point (x = 5.00) restrained in translation along Y and Z axis and restrained rotation
along X axis. ■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ External: Uniformly distributed load over its length: q = Fz = -50 000 N/ml Bending moment at x=0: My,1 = 142 x 106 Nmm Bending moment at x=5: My,2 = - 113.6 x 106 Nmm,
■ Internal: None.
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13.50.2.2Reference results for calculating the elastic critical moment of the cross section In order to determine the elastic critical moment of the cross section (Mcr), factors C1 and C2 have to be calculated. They are determined considering the method provided at chapter 3.5 from French Annex of EN 1993-1-1. C1 and C2 coefficients are depending on the bending moment diagram along the member segment between lateral restraints.
The simply supported beam has the following bending moment diagram (the values are in “Newton x meter”):
For a beam subjected to uniformly distributed load and concentrated bending moments applied at its extremities, the moments distribution is defined considering two parameters:
■ Ratio between the moments at extremities:
8.0142000113600
=−−
=NmNmψ
■ Ratio between the moment given by uniformly distributed load and the biggest bending moment from extremity:
( ) 1.11014285000/50
8 6
22
=×××
=××
=Nmm
mmmmNMLqµ
Its value is positive as both loadings are deforming the beam about the same fibre (chapter 3.4 from French Annex of EN 1993-1-1).
In order to determine C1 and C2 parameters, factors β, γ, a, b, c, A, B, d1, e1, r1, ξ, m, C10, d2, e2, r2 need to be calculated considering the analytical relationships provided in chapter 3.5 from French Annex of EN 1993-1-1:
■ 2.411.148.014 =−×+=−×+= µψβ
■ 84.81.182.48 22 =×−=×−= µβγ
■ ( ) 738.19126223.06960364.01413364.015.0 2 =+−++×= µβµγβa
■ ( ) 4765.14281556.19240091.01603341.015.0 2 =+−++×= µβµγβb
■ 0602.09352904.05940757.00900633.01801266.0 2 =−+−−= µβµγβc
■ 5625.22 =−×= cbaA
■ 2143.42
2 =+×=baB
■ ( ) 036.2152.01 =+×+= ψµd
■ 3.01 =e
■ As d1 > e1, the factor r1 is equal to 1.0
■ 0.14773.08
15.0 ≤=−
−=µψξ
■ ( ) ( ) 0.1002.21411 ≥=−×××+−×−= ξξµψξm
■ 5363.02
42
110 =×
×−−×=
AABBrC
■ 065.2675.0425.02 =×++= ψµd
■ 37.035.065.02 =×−= ψe
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■ As d2 > e2, the factor r2 is equal to 1.0 Having the above factors, C1 and C2 coefficients become:
■ 074.1101 =×= CmC
■ 235.0398.0 1022 =×××= CrC µ
The load being applied at the top fibre it tends to accentuate the lateral torsional buckling, so it will reduce the value of elastic critical moment. In this case, the distance from the shear centre to the point of load application (zg) will be positive:
■ mmzg 130+=
The French Annex of EN 1993-1-1 provides the analytical relationship used to determine the value of the elastic critical moment:
■ ( ) NmmzCzCIEIGL
II
LIECM gg
z
t
z
wzcr
62
222
2
2
2
1 1071772.91 ×=
×−×+××××
+×××
×=π
π
13.50.2.3Reference results for calculating the reduction factor for lateral torsional buckling The calculation of the reduction factor for lateral torsional buckling (χLT) is done using the formula (6.56) from chapter 6.3.2.2 (EN 1993-1-1).
Before determining the reduction factor for lateral torsional buckling (χLT), the following terms should be determined:
LTλ , imperfection factor αLT, φLT.
■ Non dimensional slenderness for lateral torsional buckling, LTλ :
133.11071772.91
/23518.5011776
23, =
××
=×
=Nmm
mmNmmM
fW
cr
yyplLTλ
■ In order to determine the imperfection factor αLT, the buckling curve must be chosen. According to table 6.4
from EN 1993-1-1, for welded I-sections which have the ratio h / b ≤ 2, the recommended lateral torsional buckling curve is “c”. In this case, table 6.3 from EN 1993-1-1 recommends the value for imperfection factor αLT:
49.0=LTα
■ The value used to determine the reduction factor χLT, φLT, becomes:
( )[ ] ( )[ ] 370.1133.12.0133.149.015.02.015.0 22 =+−×+×=+−×+×= LTLTLTLT λλαφ
■ The reduction factor for lateral torsional buckling is calculated using the formula (6.56) from EN 1993-1-1:
0.1467.0133.137.137.1
112222
≤=−+
=−+
=LTLTLT
LTλφφ
χ
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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13.50.2.4Reference 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
13.50.3Calculated results Result name Result description Value Error C1 C1 parameter 1.07375
adim -0.0233 %
C2 C2 parameter 0.234812 adim
-0.0800 %
Mcr Elastic critical moment 91.72 kN*m 0.0000 % XLT Reduction factor for lateral torsional buckling 0.466895
adim -0.0225 %
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13.51 EC3 Test 40: Verifying CHS508x8H class 3, simply supported beam, loaded with centric compression, uniform linear horizontal efforts by Y and a vertical punctual load in the middle
Test ID: 5744
Test status: Passed
13.51.1Description The test verifies a CHS508x8H beam made of S235 steel.
The beam is subjected to 20 kN axial compression force, 30 kN punctual vertical load applied to the middle of the beam and 7 kN/m linear uniform horizontal load.
The calculations are made according to Eurocode 3 French Annex.
13.52 EC3 Test 42: Determining lateral torsional buckling parameters for a I-shaped welded built-up beam considering the load applied on the upper flange
Test ID: 5752
Test status: Passed
13.52.1Description Determines the lateral torsional buckling parameters for a I-shaped welded built-up beam made of S235 steel, considering the load applied on the upper flange. The loadings applied on the beam are: a uniformly distributed load and 2 punctual negative bending moments applied at beam extremities.
The determination is made considering the provisions from Eurocode 3 (EN 1993-1-1) French Annex.
13.53 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533)
Test ID: 4549
Test status: Passed
13.53.1Description Verifies the shape sheet for a steel beam with circular cross-section when the lateral torsional buckling is computed and when it is not.
13.54 Changing the steel design template for a linear element (TTAD #12491)
Test ID: 4540
Test status: Passed
13.54.1Description Selects a different design template for steel linear elements.
13.55 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389)
Test ID: 4529
Test status: Passed
13.55.1Description Verifies the program behavior when the "Shape sheet" command is used for elements which were excluded from the specialized calculation (chords).
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13.56 EC3: Verifying the buckling length results (TTAD #11550)
Test ID: 4481
Test status: Passed
13.56.1Description Performs the steel calculation and verifies the buckling length results according to Eurocodes 3 - French standards. The shape sheet report is generated.
The model consists of a vertical linear element (IPE300 cross section, S275 material) with a rigid fixed support at the base. A punctual live load of 200.00 kN is applied.
13.57 EC3 fire verification: Verifying the work ratios after performing an optimization for steel profiles (TTAD #11975)
Test ID: 4484
Test status: Passed
13.57.1Description Runs the Steel elements verification and generates the "Envelopes and optimizing profiles" report in order to verify the work ratios. The verification is performed using the EC3 norm with Romanian annex.
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13.58 EC3 test 4: Class section classification and bending moment verification of an IPE300 column
Test ID: 5412
Test status: Passed
13.58.1Description Classification and verification of an IPE 300 column made of S235 steel.
The column is connected to the ground by a fixed connection and is free on the top part.
In the middle, the column is subjected to a 50 kN force applied on the web direction, defined as a live load.
The dead load will be neglected.
13.58.2Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 50kN force applied on the web direction, defined as a live load. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.58.2.1Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Exploitation loadings (category A): Q = 50kN, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q
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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.
13.58.2.2Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In this case, the column is subjected to a lateral load, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below.
The Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.
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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:
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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 class section for the entire column section will be considered Class 1.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)
13.58.2.3Reference results in calculating the bending moment resistance
1,
≤Rdc
Ed
MM
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.5(1)
0,
*
M
yplRdc
fwM
γ= for Class1 cross sections
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.5(2)
Where:
340.628 cmwpl =
fy nominal yielding strength for S235 fy=235MPa
0Mγ partial safety coefficient 10 =Mγ
Therefore:
MNmfw
MM
yplRdVy 147674.0
1235*10*40.628* 6
0,, ===
−
γ
MNmkNmkNmM Ed 125.012550*5.2 ===
%84148.0125.0
,,
==RdVy
Ed
MM
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
13.58.2.4Reference results
Result name Result description Reference value Combined oblique bending
Work ratio - Oblique 85 %
13.58.3Calculated results Result name Result description Value Error Work ratio - Oblique
Work ratio - Oblique 84.6459 % -0.4166 %
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13.59 EC3 Test 2: Class section classification and shear verification of an IPE300 beam subjected to linear uniform loading
Test ID: 5410
Test status: Passed
13.59.1Description Classification and verification of an IPE 300 beam made of S235 steel.
The beam is subjected to a 50 kN/m linear uniform load applied gravitationally.
The force is considered to be a live load and the dead load is neglected.
13.59.2Background Classification and verification of sections for an IPE 300 beam made from S235 steel. The beam is subjected to a 50 kN/m linear uniform load applied gravitationally. The force is considered to be a live load and the dead load is neglected.
This test was evaluated by the French control office SOCOTEC.
13.59.2.1Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Exploitation loadings (category A): Q = -50kN/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.
13.59.2.2Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In this case the 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 Chapter 5.5.2
The section geometrical properties are described in the picture below:
According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:
276.57.1045.56
==mmmm
tc
92.0=ε
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Therefore:
28.892.0*9*9276.57.1045.56
==≤== εmmmm
tc
this means that the column flanges are Class 1.
A cross-section is classified by quoting the heist (least favorable) class of its compression elements.
According to the calculation above, the beam section have a Class 1 web and Class 1 flanges; therefore the class section for the entire beam section will be considered Class 1.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)
13.59.2.3Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd shall be determined as follows:
0,
3*
M
yv
Rdpl
fA
Vγ
=
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(2)
Where:
Av: section shear area for rolled profiles fwfv trttbAA *)*2(**2 ++−=
A: cross-section area A=53.81cm2
b: overall breadth b=150mm
h: overall depth h=300mm
hw: depth of the web hw=248.6mm
r: root radius r=15mm
tf: flange thickness tf=10.7mm
tw: web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3)
fy: nominal yielding strength for S275 fy=275MPa
0Mγ : partial safety coefficient 10 =Mγ
Therefore:
kNMN
fA
VM
yv
Rdpl 7.4074077.01
3275*10*68.25
3* 4
0, ====
γ
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For more:
Verification of the shear buckling resistance for webs without stiffeners:
εη*72≤
w
w
th
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(7)
20.111*20.1*20.1
0
1 ===M
M
γγη
92.0275235235
===yf
ε
91.9392.020.1*72*7201.35
1.76.248
==≤==εη
w
w
th
There is no need for shear buckling resistance verification
According to: EC3 Part 1,5 EN 1993-1-5-2004 Chapter 5.1(2)
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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13.59.2.4Reference results
Result name Result description Reference value Fz Shear force 125 kN Work ratio Work ratio - Fz 31 %
13.59.3Calculated results Result name Result description Value Error Fz Fz -125 kN 0.0000 % Work ratio - Fz Work ratio Fz 30.6579 % -1.1035 %
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13.60 EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 beam
Test ID: 5424
Test status: Passed
13.60.1Description Classification and verification on combined bending of an IPE 300 beam made of S235 steel.
The beam is connected to its ends by a connection with all translation blocked and on the other end by a connection with translation blocked on the Y and Z axis and rotation blocked along X axis.
The beam is subjected to a -10 kN/m uniform linear force applied along the beam gravitational along the Z local axis, and a 10kN/m uniform linear force applied along the beam on the Y axis.
Both forces are considered as live loads.
The dead load will be neglected.
13.60.2Background Classification and verification on combined bending of sections for an IPE 300 beam made from S235 steel. The beam is connected to its ends by a connection with all translation blocked and on the other end by a connection with translation blocked on the Y and Z axis and rotation blocked along X axis. The beam is subjected to a -10kN/m uniform linear force applied along the beam gravitational along the Z local axis, and a 10kN/m uniform linear force applied along the beam on the Y axis. Both forces are considered live loads. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.60.2.1Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Exploitation loadings (category A): Q1 = -10kN/m, Q2 = 10kN/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 (x = 5.00) restrained in translation and rotation along Y, Z axis and rotation
blocked along X axis. ■ Inner: None.
13.60.2.2Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In this case, the beam is subjected to linear uniform equal loads, one vertical and one horizontal, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below:
Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.
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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:
According to Table 5.2 and the beam section geometrical properties, the next conclusions can be found:
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Therefore:
This means that the beam left top flanges are Class 1.
Overall the beam top flange cross-section class is Class 1.
In the same way will be determined that the beam bottom flange cross-section class is also Class 1
A cross-section is classified by quoting the heist (least favorable) class of its compression elements.
According to the calculation above, the column section have a Class 1 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 1.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)
13.60.2.3Reference results for calculating the combined biaxial bending
1,
,
,
, ≤
+
βα
EdNz
Edz
RdNy
EdY
MM
MM
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(5)
In which α and β are constants, which may conservatively be taken as unity, otherwise as follows:
For I and H sections:
2=α
)1;max(n=β
00,,
===RdplRdpl
Ed
NNNn therefore 1=β
Bending around Y:
For cross-sections without bolts holes, the following approximations may be used for standard rolled I or H sections and for welded I or H sections with equal flanges:
anM
M RdyplRdNy *5.01
)1(*,,, −
−= but RdplyRdNy MM ,, ≤
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(4)
00,,
===RdplRdpl
Ed
NNNn
5.0403.010*81.53
)0107.0*15.0*210*81.53()**2(4
4
≤=−
=−
= −
−
AtbA
a f
8.0)403.0*5.01(,,,,
,,RdyplRdypl
RdyN
MMM =
−=
RdyplRdyNRdyplRdyN MMMM ,,,,,,,,*8.0 >⇒= but RdplyRdNy MM ,, ≤
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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 sections and for welded I or H sections with equal flanges:
anM
M RdzplRdNz *5.01
)1(*,,, −
−= but RdplzRdNz MM ,, ≤
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(4)
RdzplRdzNRdzplRdzN MMMM ,,,,,,,,*8.0 >⇒= but RdplzRdNz MM ,, ≤ therefore it will be considered:
MNmfw
MMM
yzplRdzplRdzN 030.0
1235*10*20.125* 6
0
,,,,, ====
−
γ In conclusion:
11086.1029375.003125.0
148.003125.0 12
,
,
,
, >=
+
=
+
βα
EdNz
Edz
RdNy
EdY
MM
MM
The coresponding work ratio is:
WR = 1.1086 x 100 = 110.86 %
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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13.60.2.4Reference results
Result name Result description Reference value Combined oblique bending
Combined oblique bending [%] 110.86 %
13.60.3Calculated results Result name Result description Value Error Work ratio - Oblique
Work ratio-Oblique 110.691 % -0.2784 %
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13.61 EC3 Test 1: Class section classification and compression verification of an IPE300 column
Test ID: 5383
Test status: Passed
13.61.1Description 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.
13.61.2Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. On top, the column is subjected to a 100kN force applied gravitationally, defined as a live load. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.61.2.1Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Exploitation loadings (category A): Q = -100kN, ■ 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.
13.61.2.2Reference results for calculating the cross section class In this case, the column is subjected only to compression, therefore the distribution of stresses along the section is like in the picture below:
To determine the web class, we use Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
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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=ε 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:
276.57.1045.56
==mmmm
tc
1=ε
Therefore:
9*9276.57.1045.56
=≤== εmmmm
tc
this means that the column flanges are Class 1.
A cross-section is classified by quoting the heist (least favorable) class of its compression elements.
According to the calculation above, the column section have a Class 2 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 2.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)
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13.61.2.3Reference results in calculating the compressive resistance Nc,Rd The design resistance of the cross-section force Nc,Rd shall be determined as follows:
For Class 1, 2 or 3 cross-section 0
,
*
M
yRdc
fAN
γ=
Where:
A section area A=53.81cm2
Fy nominal yielding strength for S235 fy=235MPa
0Mγ partial safety coefficient 10 =Mγ
Therefore:
kNMNfA
NM
yRdc 54.1264264535.1
1235*10*81.53* 4
0, ====
−
γ
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.4(2)
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
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Finite elements results
Compressive resistance work ratio
Column subjected to compressive load Work ratio [%]
13.61.2.4Reference results
Result name Result description Reference value Work ratio Compressive resistance work ratio [%] 8 %
13.61.3Calculated results Result name Result description Value Error Work ratio - Fx Work ratio Fx 7.90805 % -1.1494 %
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13.62 EC3 Test 5: Class section classification and combined axial force with bending moment verification of an IPE300 column
Test ID: 5421
Test status: Passed
13.62.1Description Classification and verification of an IPE 300 column made of S235 steel.
The column is connected to the ground by a fixed connection and is free on the top part.
The column is subjected to a 500 kN compressive force applied on top and a 5 kN/m uniform linear load applied on all the length of the column, on the web direction, both defined as live loads.
The dead load will be neglected.
13.62.2Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. The column is subjected to a 500kN compressive force applied on top and a 5kN/m uniform linear load applied for all the length of the column, on the web direction, both defined as live loads. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.62.2.1Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Exploitation loadings (category A): 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.
13.62.2.2Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In this case the column is subjected to compression and lateral load, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below.
Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.
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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:
121.01235005381.0
500.0212 −>−=−×
⋅=−⋅
⋅=y
Ed
fANψ
5.010.10071.02486.0235
5.01211
21
>=
××+⋅=
××+⋅=
dtfN
y
Edα
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Therefore:
92.69)21.0(*33.067.0
1*4233.067.0
42=
−+=
+ ψε
Therefore:
92.6933.067.0
42014.35 =+
≤=ψ
εtc
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 class section for the entire column section will be considered Class 3.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6)
Cross sections for class 3, the maximum longitudinal stress should check:
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.4(3)
In this case:
In absence of shear force, for Class 3 cross-sections the maximum longitudinal stress shall satisfy the criterion:
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.2(1)
This means:
The corresponding work ratio is: WR = 0.8728 x 100 = 87.28 %
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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 [%]
13.62.2.3Reference results
Result name Result description Reference value Bending and axial compression
Work ratio – oblique [%] 87.28 %
13.62.3Calculated results Result name Result description Value Error Work ratio - Oblique
Work ratio- Oblique 87.2799 % 0.3217 %
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13.63 EC3 Test 3: Class section classification, shear and bending moment verification of an IPE300 column
Test ID: 5411
Test status: Passed
13.63.1Description 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.
13.63.2Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 200kN force applied on the web direction, defined as a live load. The dead load will be neglected.
This test was evaluated by the French control office SOCOTEC.
13.63.2.1Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Exploitation loadings (category A): Q = -200kN, ■ 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.
13.63.2.2Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2
In case the column is subjected to a lateral load, the stresses distribution on the most stressed point (the column base) is like in the picture below:
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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:
According to the 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 web is Class 1.
To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.
The section geometrical properties are described in the picture below:
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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 flanges are Class 1.
A cross-section is classified by quoting the heist (least favorable) class of its compression elements.
According to the calculation above, the column section have a Class 1 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 1.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6).
13.63.2.3Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd , is determined as follows:
0,
3*
M
yv
Rdpl
fA
Vγ
=
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(2)
Where:
Av section shear area for rolled profiles fwfv trttbAA *)*2(**2 ++−=
A cross-section area A=53.81cm2
b overall breadth b=150mm
h overall depth h=300mm
hw depth of the web hw=248.6mm
r root radius r=15mm
tf flange thickness tf=10.7mm
tw web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3)
fy nominal yielding strength for S235 fy=235MPa
0Mγ partial safety coefficient 10 =Mγ
Therefore:
kNMN
fA
VM
yv
Rdpl 42.3483484.01
3235*10*68.25
3* 4
0, ====
γ
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13.63.2.4Reference results in calculating the bending moment resistance
%50%4.5742.348
200,
>==Rdpl
Ed
VV
The shear force is greater than half of the plastic shear resistance. Its effect on the moment resistance must be taken into account.
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(1)(2)
Where:
0223.01348.0
200.0*212 22
,
=
−=
−=
Rdpl
Ed
VVρ
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(3)
fwfv trttbAA *)*2(**2 ++−=
Av section shear area for rolled profiles
A cross-section area A=53.81cm2
b overall breadth b=150mm
h overall depth h=300mm
hw depth of the web hw=248.6mm
r root radius r=15mm
tf flange thickness tf=10.7mm
tw web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=
According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.6(3)
fy nominal yielding strength for S235 fy=235MPa
0Mγ partial safety coefficient 10 =Mγ
Therefore:
MNmf
tAw
MM
yw
vpl
RdVy 146.01
235*0071.0*4
)²10*68.25(*0223.010*40.628*4
²* 46
0,, =
−
=
−
=
−−
γ
ρ
MNmkNmkNmM Ed 5.0500200*5.2 ===
%342146.0500.0
,,
==RdVy
Ed
MM
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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
13.63.2.5Reference results
Result name Result description Reference value Shear z direction work ratio
Work ratio - Fz 57 %
Combined oblique bending
Work ratio - Oblique 341 %
13.63.3Calculated results Result name Result description Value Error Work ratio - Fz Work ratio Fz 57.4021 % 0.7054 % Work ratio - Oblique
Work ratio - Oblique 341.348 % 0.1021 %
14 Timber design
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14.1 EC5: Verifying a timber purlin subjected to oblique bending
Test ID: 4878
Test status: Passed
14.1.1 Description Verifies a rectangular timber purlin made from solid timber C24 to resist oblique bending. The verification is made following the rules from Eurocode 5 French annex.
14.1.2 Background Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to oblique bending. The verification of the bending stresses at ultimate limit state is performed.
14.1.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test E.3; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent purlins (span): d = 1.8 m. The following load cases and load combination are used:
■ Loadings from the structure: G = 550 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2; All loads will be projected on the purlin direction since the roof slope is 17°.
Simply supported purlin subjected to loadings
Units
Metric System
Geometry
Purlin cross section characteristics:
■ Height: h = 0.20 m, ■ Width: b = 0.10 m, ■ Length: L = 3.5 m, ■ Section area: A = 0.02 m2 ,
■ Elastic section modulus about the strong axis, y: 322
000666.06
20.01.06
mhbWy =⋅
=×
= ,
■ Elastic section modulus about the strong axis, z: 322
000333.06
20.01.06
mhbWz =⋅
=×
= .
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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: fm,k = 24 x 106 Pa, ■ Service class 2.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (z=0) restrained in translation along X, Y, Z; Support at end point (z = 3.5) restrained in translation along X, Y, Z and restrained in rotation along X.
■ Inner: None.
Loading
The purlin is subjected to the following projected loadings (at ultimate limit state):
■ External: Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x sin17° =
1101.22 N/m, Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x cos17° =
3601.92 N/m, ■ Internal: None.
14.1.2.2 Reference results in calculating the timber purlin subjected to oblique bending In order to verify the timber purlin subjected to oblique bending at ultimate limit state, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, have to be determined. After this, the reference solution (which includes the design bending stress about y axis, the design bending stress about z axis and the maximum work ratio for strength verification) is calculated.
Reference solution for ultimate limit state verification
Before calculating the reference solution (design bending stress about y axis, design bending stress about z axis and maximum work ratio for strength verification) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).
■ Modification factor for duration of load (short term) and moisture content: kmod = 0.9 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties: γM = 1.3
■ Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0
■ System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)
■ Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections): km = 0.7
■ Design bending stress about y axis (induced by uniformly distributed load, qz):
σm,y,d = Pam
mmN
WLq
WM
y
z
y
y 63
222
102814.8000666.08
5.392.3601
8×=
×
×=
××
=
■ Design bending stress about z axis (induced by uniformly distributed load, qy):
σm,z,d = Pam
mmN
WLq
WM
z
y
z
z 63
222
100638.5000333.08
5.322.1101
8×=
×
×=
×
×=
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■ Design bending strength:
fm,y,d = fm,z,d = Pakkkf hsysM
km66mod
, 10615.160.10.13.19.01024 ×=××××=×××
γ
■ Maximum work ratio for strength verification; it represents the maximum value between the work ratios obtained with formulae 6.17 and 6.18 from EN 1995-1-1 norm:
1max
,,
,,
,,
,,
,,
,,
,,
,,
≤
+
+
dzm
dzm
dzm
dzmm
dym
dymm
dym
dym
f
fk
fk
fσ
σ
σ
σ
Finite elements modeling
■ Linear element: S beam, ■ 5 nodes, ■ 1 linear element.
Stress SMy diagram
Simply supported purlin subjected to uniformly distributed load, qz Stress SMy [Pa]
Stress SMz diagram
Simply supported purlin subjected to uniformly distributed load, qz Stress SMz [Pa]
Maximum work ratio for strength verification
Strength of a simply supported purlin subjected to oblique bending Work ratio [%]
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14.1.2.3 Reference results
Result name Result description Reference value Smy Design bending stress about y axis [Pa] 8281441 Pa SMz Design bending stress about z axis [Pa] 5063793 Pa
Work ratio Maximum work ratio for strength verification [%] 71.2 %
14.1.3 Calculated results Result name Result description Value Error Stress SMy Design bending stress about y axis 8.47672e+006 Pa 0.0000 % Stress SMz Design bending stress about z axis 5.18319e+006 Pa 0.0000 % Work ratio Maximum work ratio for strength verification 71.153 % 0.0000 %
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14.2 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes
Test ID: 4896
Test status: Passed
14.2.1 Description Verifies the residual cross section of a column exposed to fire for 60 minutes. The column is made from glued laminated timber GL24 and it has only 3 faces exposed to fire. The verification is made according to chapter 4.2.2 (Reduced cross section method) from EN 1995-1-2 norm.
14.2.2 Background Verifies the adequacy of the cross sectional resistance for a rectangular cross section, which is made from glued laminated timber GL24, exposed to fire for 60 minutes on 3 faces. The verification is made according to chapter 4.2.2 from EN 1995-1-2 norm.
14.2.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test F.1; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Timber column with fixed base
Units
Metric System
Geometry
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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Materials properties
Glued laminated timber GL24 is used. The following characteristics are used in relation to this material:
■ Density: ρ = 380 kg/m3, ■ Design charring rate: βn = 0.7 x 10-3 m/min,
Boundary conditions
The boundary conditions are described below:
■ Outer: Fixed at base (Z = 0), Support at top (Z = 5.00) restrained in translation along X and Y,
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at Z = 5.00: Fz = N = - 100000 N, ■ Internal: None.
14.2.2.2 Reference results in calculating the cross sectional resistance of a timber column exposed to fire
Reference solution
The reference solution (residual cross section) is determined by reducing the initial cross section dimensions by the effective charring depth according to chapter 4.2.2 from EN 1995-1-2. Before calculating the effective charring depth we need to determine some parameters involved in calculations (dchar,n, k0, d0).
■ Depth of layer with assumed zero strength and stiffness: d0 = 7 x 10-3 m; ■ Coefficient depending of fire resistance time and also depending if the members are protected or not:
k0 = 1.0 (according to table 4.1 from EN 1995-1-2)
■ Notional design charring depth:
dchar,n = m.minminm.tn 0420601070β 3 =⋅⋅=⋅ − (according to relation 3.2 from EN 1995-1-2)
■ Effective charring depth: def = 00 dkd n,char ⋅+
■ Residual cross section:
Afi = )db()dh( efef ⋅−×− 2
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
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Residual cross section
Column with fixed base exposed to fire for 60 minutes Afi
14.2.2.3 Reference results
Result name Result description Reference value Afi Residual cross section [m2] 0.056202 m2
14.2.3 Calculated results Result name Result description Value Error Afi Residual cross section 0.056202 m² 0.0000 %
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14.3 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending
Test ID: 4901
Test status: Passed
14.3.1 Description Verifies the fire resistance of a rectangular cross section purlin made from solid timber C24 to resist simple bending. The purlin is exposed to fire on 3 faces for 30 minutes. The verification is made according to chapter 4.2.2 (Reduced cross section method) from EN 1995-1-2 norm.
14.3.2 Background Verifies the adequacy of the fire resistance for a rectangular cross section purlin made from solid timber C24 to resist simple bending. The purlin is exposed to fire on 3 faces for 30 minutes (the top of the purlin is not exposed to fire). Verification of the bending stresses corresponding to frequent combination of actions is realized.
Chapter 1.1.1.3 presents the results obtained with the theoretical background explained at chapter 1.1.1.2.
14.3.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test F.2; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
Loadings from the structure: G = 500 N/m2, Snow load: S = 700 N/m2, Frequent combination of actions: CFQ = 1.0 x G + 0.5 x S = 850 N/m2
Purlin with 3 supports
Units
Metric System
Geometry
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: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Density: ρ = 350 kg/m3, ■ Design charring rate (softwood): βn = 0.8 x 10-3 m/min, ■ Service class 1.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (X = 0) restrained in translation along X, Y and Z, Support at middle point (X = 3.30) restrained in translation along X, Y and Z, Support at end point (X = 6.60) restrained in translation along X, Y, Z and restrained in rotation along
X. ■ Inner: None.
Loading
The beam is subjected to the following loadings:
■ External: Uniformly distributed load: q = CFQ x d = 850 N/m2 x 1.5 m = 1275 N/m,
■ Internal: None.
14.3.2.2 Reference results in calculating the fire resistance of a timber purlin subjected to simple bending
In order to verify the fire resistance for a timber purlin subjected to simple bending it is necessary to determine the residual cross section. After this, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod,fi, γM,fi, kfi, km, have to be determined.
Residual cross section
The residual cross section is determined by reducing the initial cross section dimensions by the effective charring depth according to chapter 4.2.2 from EN 1995-1-2. Before calculating the effective charring depth we need to determine some parameters involved in calculations (dchar,n, k0, d0).
■ Depth of layer with assumed zero strength and stiffness: d0 = 7 x 10-3 m; ■ Coefficient depending of fire resistance time and also depending if the members are protected or not:
k0 = 1.0 (according to table 4.1 from EN 1995-1-2)
■ Notional design charring depth:
dchar,n = mmtn 024.0min30min
108.0 3 =⋅⋅=⋅ −β (according to relation 3.2 from EN 1995-1-2)
■ Effective charring depth:
def = 00, dkd nchar ⋅+
■ Residual cross section:
Afi = )2()( efefefef dbdhbh ⋅−×−=×
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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, γM,fi, kfi).
■ Modification factor in case of a verification done with residual section: kmod,fi = 1.0 (according to paragraph 5 from chapter 4.2.2 from EN 1995-1-2)
■ Partial safety factor for timber in fire situations: γM,fi = 1.0
■ Factor kfi, taken from table 2.1 (EN 1995-1-2): kfi = 1.25 (for solid timber)
■ Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections) – according to paragraph 2 from chapter 6.1.6 (EN 1995-1-1): km = 0.7
■ Design bending stress (taking into account the residual cross section):
σm,d = 2
6
efef
y
y
y
hbM
WM
×
×=
The picture below shows the bending moment diagram (kNm). My from the above formula represents the maximum bending moment achieved from frequent combination of actions.
■ Design bending strength (for fire situation):
fm,d,fi = Pak
fkfiM
fikmfi
66
,
mod,, 1030
0.10.1102425.1 ×=×××=××
γ
■ Work ratio according to formulae 6.17 from EN 1995-1-1 norm (considering that the axial effort, as well as the bending moment about z axis, are null):
0.1,
, ≤dm
dm
fσ
■ Work ratio according to formulae 6.18 from EN 1995-1-1 norm (considering that the axial effort, as well as the bending moment about z axis, are null):
0.1,
, ≤×dm
dmm f
kσ
■ Maximum work ratio for bending verification for fire situation:
100100;max,
,
,
,
,
, ×=×
×=
dm
dm
dm
dmm
dm
dm
ffk
fWR
σσσ
Finite elements modeling
■ Linear element: S beam, ■ 8 nodes, ■ 1 linear element.
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Residual cross-section area
Simply supported beam subjected to bending (fire situation) Residual area [m2]
Design bending stress taking into account the residual cross-section
Simply supported beam subjected to bending (fire situation) Design bending stress for residual cross-section [Pa]
Maximum work ratio for bending verification (fire situation)
Simply supported beam subjected to bending (fire situation) Work ratio [%]
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14.3.2.3 Reference results
Result name Result description Reference value Afi Residual area [m2] 0.002197 m2
Stress Design bending stress for residual cross-section [Pa] 27568524 Pa Work ratio Maximum work ratio for fire verification [%] 91.9 %
14.3.3 Calculated results Result name Result description Value Error Afi Residual area 0.002197 m² 0.0000 % Stress Design bending stress for residual cross-section 2.75626e+007 Pa 0.0000 % Work ratio Maximum work ratio for fire verification 91.8752 % 0.0000 %
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14.4 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression
Test ID: 4877
Test status: Passed
14.4.1 Description Verifies the lateral torsional stability for a rectangular timber beam subjected to combined bending and axial compression. The verification is made following the rules from Eurocode 5 French annex.
14.4.2 Background Verifies the lateral torsional stability of a rectangular cross section made from solid timber C24 subjected to simple bending (about the strong axis) and axial compression.
14.4.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test E.2; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent rafters (span): d = 0.5 m. The following load cases and load combination are used:
■ Loadings from the structure: G = 450 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m2; All loads will be projected on the rafter direction, since its slope is 50% (26.6°).
Simply supported rafter subjected to projected loadings
Units
Metric System
Geometry
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: 322
000333.06
20.005.06
mhbWy =⋅
=×
= .
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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: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Longitudinal elastic modulus: E = 1.1 x 1010 Pa, ■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 1010 Pa, ■ Service class 2.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (z = 0) restrained in translation along Y, Z and restrained in rotation along X. Support at end point (z = 5.00) restrained in translation along X, Y, Z.
■ Inner: None.
Loading
The rafter is subjected to the following projected loadings (at ultimate limit state):
■ External: Uniformly distributed load: q = Cmax x d x cos26.6° = 1957.5 N/m2 x 0.5 m x cos26.6° = 875.15 N/m, Compressive load component: N = Cmax x d x sin26.6° x L = 1957.5 N/m2 x 0.5m x sin26.6° x 5.00m =
= 2191.22 N ■ Internal: None.
14.4.2.2 Reference results in calculating the timber beam subjected to combined stresses In order to verify the lateral torsional stability of a timber beam subjected to combined stresses at ultimate limit state, the formula (6.35) from EN 1995-1-1 norm is used. Before applying this formula we need to determine some parameters involved in calculations like: slenderness ratios, relative slenderness ratios, instability factors. After this, we calculate the reference solution which includes: the design compressive stress, the design bending stress and the work ratio based on formula (6.35) from EN 1995-1-1.
Slenderness ratios
In professional practice the slenderness ratio is limited to 120. The slenderness ratios corresponding to bending about y and z axes are determined as follows:
■ Slenderness ratio corresponding to bending about the z axis:
4.34605.051121212 =
×=
×==
mm
blm
bl gc
zλ
It is necessary to reduce the buckling length about the z axis, because λz exceeded the value 120. A restraint is placed in each tierce of the rafter, so that the slenderness ratio corresponding to bending about the z axis become:
1205.11505.0667.11121212 <=
×=
×==
mm
blm
bl gc
zλ
■ Slenderness ratio corresponding to bending about the y axis:
6.862.051121212 =
×=
×==
mm
hlm
hl gc
yλ
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Relative slenderness ratios
The relative slenderness ratios are:
■ Relative slenderness ratio corresponding to bending about the z axis:
958.11074.0
10215.115
,10
6
050
,0,, =
××
==Pa
PaEf kcz
zrel ππλλ
■ Relative slenderness ratio corresponding to bending about the y axis:
468.11074.0
10216.86
,10
6
050
,0,, =
××
==Pa
PaEf kcy
yrel ππλ
λ
■ Maximum relative slenderness ratio:
( ) 958.1,max ,,max, == yrelzrelrel λλλ
So there is a risk of buckling because λrel,max ≥ 0.3.
Relative slenderness for bending
The effective length of the beam and the critical bending stress must be determined before calculating the relative slenderness for bending.
■ Effective length of the beam; its calculation is made according to table 6.1 from EN 1995-1-1 and it is based on the loading type and support conditions. The effective length is increased by “2⋅h” because the load is applied at the compressed fiber of the beam:
mmmhlleff 9.42.020.59.029.0 =⋅+⋅=⋅+⋅=
■ Critical bending stress (determined according to formula 6.32 from EN 1995-1-1):
σm,y,crit = Pamm
Pamlh
Ebef
61022
05.02
10724.149.42.0
1074.005.078.078.0×=
⋅⋅⋅⋅
=⋅
⋅
■ Relative slenderness for bending (determined according to formula 6.30 from EN 1995-1-1):
λrel,m = 277.110724.14
10246
6
,
, =⋅
⋅=
PaPaf
critm
km
σ
Instability factors
In order to determine the instability factors we need to determine the βc factor. It is a factor for solid timber members within the straightness limits defined in Section 10 from EN 1995-1-1:
βc = 0.2 (according to relation 6.29 from EN 1995-1-1)
The instability factors are: kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z
2] (according to relation 6.28 from EN 1995-1-1)
ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y2] (according to relation 6.27 from EN 1995-1-1)
2,
2,1
zrelzz
zckk
kλ−+
= (according to relation 6.26 from EN 1995-1-1)
2,
2,1
yrelyy
yckk
kλ−+
= (according to relation 6.25 from EN 1995-1-1)
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Reference solution for ultimate limit state verification
Before calculating the reference solution (the design compressive stress, the design bending stress and the work ratio based on formula (6.35) from EN 1995-1-1) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km, kcrit).
■ Modification factor for duration of load (short term) and moisture content: kmod = 0.9 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties: γM = 1.3
■ Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0
■ System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)
■ Factor which takes into account the reduced bending strength due to lateral buckling: Kcrit = 1.56 – 0.75λrel,m (because 0.75 < λrel,m < 1.4)
■ Design compressive stress (induced by the compressive component, N):
σc,d = PamN
AN 219122
101022.2191
23 =×
= −
■ Design compressive strength:
fc,0,d = PakfM
kc66mod
,0, 10538.143.19.01021 ×=××=×
γ
■ Design bending stress (induced by uniformly distributed load, q):
σm,d = Pam
mmN
WLq
WM
yy
y 63
222
10213.8000333.08
00.515.875
8×=
×
×=
××
=
■ Design bending strength:
fm,y,d = Pakkkf hsysM
km66mod
, 10615.160.10.13.19.01024 ×=××××=×××
γ
■ Work ratio according to formula 6.35 from EN 1995-1-1 norm:
1,0,,
,
2
,,
, ≤⋅
+
⋅ dczc
dc
dymcrit
dm
fkfkσσ
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 [%]
14.4.2.3 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 %
14.4.3 Calculated results Result name Result description Value Error Stress SFx Design compressive stress 219124 Pa 0.0000 % Stress SMy Design bending stress 8.20519e+006 Pa 0.0000 % Kcrit kcrit factor 0.592598 adim 0.0000 % Work ratio Work ratio according to formula 6.35 0.759538 adim 0.0000 %
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14.5 EC5: Shear verification for a simply supported timber beam
Test ID: 4822
Test status: Passed
14.5.1 Description Verifies a rectangular cross section beam made from solid timber C24 to shear efforts. The verification of the shear stresses at ultimate limit state is performed.
14.6 Modifying the "Design experts" properties for timber linear elements (TTAD #12259)
Test ID: 4509
Test status: Passed
14.6.1 Description Defines the "Design experts" properties for a timber linear element, in a model created with a previous version of the program.
14.7 Verifying the timber elements shape sheet (TTAD #12337)
Test ID: 4538
Test status: Passed
14.7.1 Description Verifies the timber elements shape sheet.
14.8 Verifying the units display in the timber shape sheet (TTAD #12445)
Test ID: 4539
Test status: Passed
14.8.1 Description Verifies the Afi units display in the timber shape sheet.
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14.9 EC5: Verifying a timber beam subjected to simple bending
Test ID: 4682
Test status: Passed
14.9.1 Description Verifies a rectangular cross section beam made from solid timber C24 to resist simple bending. Verifies the bending stresses at ultimate limit state, as well as the deflections at serviceability limit state.
14.9.2 Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending. Verification of the bending stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit state are performed.
14.9.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test C; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 0.5 kN/m2, ■ Exploitation loadings (category A): Q = 1.5 kN/m2, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2 ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q
Simply supported beam
Units
Metric System
Geometry
Below are described the beam cross section characteristics:
■ Height: h = 0.20 m, ■ Width: b = 0.075 m, ■ Length: L = 4.50 m, ■ Distance between adjacent beams (span): d = 0.5 m, ■ Section area: A = 15.0 x 10-3 m2 ,
■ Elastic section modulus about the strong axis y: 322
0005.06
20.0075.06
mhbWy =⋅
=×
=
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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: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,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.
14.9.2.2 Reference results in calculating the timber beam subjected to uniformly distributed loads In order to verify the timber beam bending stresses at ultimate limit state, the formulae (6.11) and (6.12) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, must be calculated. After this, the reference solution, which includes the design bending stress about the principal y axis, the design bending strength and the corresponding work ratios, is calculated.
A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the effective values with the limiting values for deflections specified in EN 1995-1-1 norm.
Reference solution for ultimate limit state verification
Before calculating the reference solution (design bending stress, design bending strength and work ratios) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).
■ Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties: γM = 1.3
■ Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0
■ System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)
■ Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections): km = 0.7
■ Design bending stress (induced by the applied forces):
σm,d = PahbLq
WM
y
y 62
2
2
2
104039.72.0075.085.44625.16
86
×=××××
=××××
=
■ Design bending strength:
fm,d = Pakkkf hsysM
km66mod
, 10769.140.10.13.18.01024 ×=××××=×××
γ
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■ Work ratio according to formulae 6.11 from EN 1995-1-1 norm:
0.1,
, ≤dm
dm
fσ
■ Work ratio according to formulae 6.12 from EN 1995-1-1 norm:
0.1,
, ≤×dm
dmm f
kσ
Reference solution for serviceability limit state verification
The following limiting values for instantaneous deflection (for a base variable action), final deflection and net deflection are considered:
300)( LQwinst ≤
125Lwfin ≤
200,Lw finnet ≤
For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are the followings:
■ Instantaneous deflection (for a base variable action):
8.600)(00749.0)( LQwmQw instinst =⇒=
■ Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):
45.45000999.0 Lwmdw instCQinst =⇒==
■ In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the
deformation factor (kdef) has to be chosen:
6.0=defk (calculated value for service class 1, according to table 3.2 from EN 1995-1-1)
95.157800285.000475.06.06.0 Lwmmdw creepQPcreep =⇒=×=×=
■ Final deflection:
47.35001284.000285.000999.0 Lwmmmwww fincreepinstfin =⇒=+=+=
■ Net deflection:
47.35001284.0001284.0 ,,
Lwmmmwww finnetcfinfinnet =⇒=+=+=
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
14.9.2.3 Reference results
Result name Result description Reference value σm,d Design bending stress [Pa] 7403906.25 Pa Strength work ratio Work ratio (6.11) [%] 50 % winst (Q) Deflection for a base variable action [m] 0.00749 m
dCQ Deflection for a characteristic combination of actions [m] 0.00999 m
winst Instantaneous deflection [m] 0.00999 m kdef Deformation coefficient 0.6 dQP Deflection for a quasi-permanent combination of actions [m] 0.00475 m wfin Final deflection [m] 0.01284 m wnet,fin Net deflection [m] 0.01284 m
14.9.3 Calculated results Result name Result description Value Error Stress Design bending stress 7.40391e+006
Pa 0.0000 %
Work ratio Work ratio (6.11) 50.1306 % 0.0000 % D Deflection for a base variable action 0.00749224 m 0.0000 % D Deflection for a characteristic combination of actions 0.00998966 m 0.0000 % Winst Instantaneous deflection 0.00998966 m 0.0000 % Kdef Deformation coefficient 0.6 adim 0.0000 % D Deflection for a quasi-permanent combination of actions 0.00474509 m 0.0000 % Wfin Final deflection 0.0128367 m 0.0000 % Wnet,fin Net deflection 0.0128367 m 0.0000 %
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14.10 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression
Test ID: 4879
Test status: Passed
14.10.1Description Verifies the stability of a rectangular timber purlin made from solid timber C24 subjected to biaxial bending and axial compression. The verification is made following the rules from Eurocode 5 French annex.
14.10.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to biaxial bending and axial compression. The verification is made according to formulae (6.23) and (6.24) from EN 1995-1-1 norm.
14.10.2.1Model description ■ Reference: Guide de validation Eurocode 5, test E.4; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent purlins (span): d = 1.8 m. The following load cases and load combination are used:
■ The ultimate limit state (ULS) combination is: CULS = 1.35 x G + 1.5 x S + 0.9 x W; ■ Loadings from the structure: G = 550 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ Axial compression force due to wind effect on the supporting elements: W = 15000 N; ■ Uniformly distributed load corresponding to the ultimate limit state combination:
Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2.
All loads will be projected on the purlin direction since its slope is 30% (17°).
Simply supported purlin subjected to loadings
Units
Metric System
Geometry
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
20.01.06
mhbWy =⋅
=×
= ,
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■ Elastic section modulus about the strong axis, z: 322
000333.06
20.01.06
mhbWz =⋅
=×
= .
Materials properties
Rectangular solid timber C24 is used. The followings characteristics are used in relation to this material:
■ Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Longitudinal elastic modulus: E = 1.1 x 1010 Pa, ■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 1010 Pa, ■ Service class 2.
Boundary conditions
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.
14.10.2.2Reference results in calculating the timber purlin subjected to combined stresses In order to verify the stability of a timber purlin subjected to biaxial bending and axial compression at ultimate limit state, the formulae (6.23) and (6.24) from EN 1995-1-1 norm are used. Before applying these formulae we need to determine some parameters involved in calculations like: slenderness ratios, relative slenderness ratios, instability factors. After this, we’ll calculate the maximum work ratio for stability verification, which represents in fact the reference solution.
Slenderness ratios
The slenderness ratios corresponding to bending about y and z axes are determined as follows:
■ Slenderness ratio corresponding to bending about the z axis:
24.1211.0
5.31121212 =×
=×
==m
mb
lmbl gc
zλ
■ Slenderness ratio corresponding to bending about the y axis:
62.602.0
5.31121212 =×
=×
==m
mh
lmhl gc
yλ
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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
102124.12110
6
05,0
,0,, =
××
==Pa
PaEf kcz
zrel ππλλ
■ Relative slenderness ratio corresponding to bending about the y axis:
028.11074.0
102162.60
,10
6
050
,0,, =
××
==Pa
PaEf kcy
yrel ππλ
λ
■ Maximum relative slenderness ratio:
( ) 056.2,max ,,max, == yrelzrelrel λλλ
So there is a risk of buckling because λrel,max ≥ 0.3.
Instability factors
In order to determine the instability factors we need to determine the βc factor. It is a factor for solid timber members within the straightness limits defined in Section 10 from EN 1995-1-1:
βc = 0.2 (according to relation 6.29 from EN 1995-1-1)
The instability factors are: kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z
2] (according to relation 6.28 from EN 1995-1-1)
ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y2] (according to relation 6.27 from EN 1995-1-1)
2,
2,1
zrelzz
zckk
kλ−+
= (according to relation 6.26 from EN 1995-1-1)
2,
2,1
yrelyy
yckk
kλ−+
= (according to relation 6.25 from EN 1995-1-1)
Reference solution for ultimate limit state verification
Before calculating the reference solution (the maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1) it is necessary to determine the design compressive stress, the design compressive strength, the design bending stress, the design bending strength and some parameters involved in calculations (kmod, γM, kh, ksys, km).
■ Design compressive stress (induced by the axial compressive load from the corresponding ULS combination, N):
σc,0,d = Pam
NAN 675000
02.013500
2 ==
■ Design bending stress about the y axis (induced by uniformly distributed load, qz):
σm,y,d = Pam
mmN
WLq
WM
y
z
y
y 63
222
102814.8000666.08
50.392.3601
8×=
×
×=
××
=
■ Design bending stress about the z axis (induced by uniformly distributed load, qy):
σm,z,d = Pam
mmN
WLq
WM
z
y
z
z 63
222
100638.5000333.08
50.322.1101
8×=
×
×=
×
×=
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■ Modification factor for duration of load (instantaneous action) and moisture content (service class 2):
kmod = 1.1 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties: γM = 1.3
■ Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0
■ System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)
■ Design compressive strength:
fc,0,d = PakfM
kc66mod
,0, 10769.173.11.11021 ×=××=×
γ
■ Design bending strength:
fm,y,d = fm,z,d = Pakkkf hsysM
km66mod
, 10308.200.10.13.11.11024 ×=××××=×××
γ
■ Maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1:
1max
,,
,,
,,
,,
,0,,
,0,
,,
,,
,,
,,
,0,,
,0,
≤
+⋅+⋅
⋅++⋅
dzm
dzm
dym
dymm
dczc
dc
dzm
dzmm
dym
dym
dcyc
dc
ffk
fk
fk
ffk
σσσ
σσσ
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
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Instability factor, kcy
Simply supported purlin subjected to biaxial bending and axial compression Kc,y
Instability factor, kcz
Simply supported purlin subjected to biaxial bending and axial compression Kc,z
Maximum work ratio for stability verification
Simply supported purlin subjected to biaxial bending and axial compression Work ratio [%]
14.10.2.3Reference 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 %
14.10.3Calculated results Result name Result description Value Error Kc,y Instability factor, kc,y 0.665025 adim 0.0000 % Kc,z Instability factor, kc,z 0.212166 adim 0.0000 % Work ratio Maximum work ratio for stability verification 70.6586 % -0.6208 %
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14.11 EC5: Verifying a timber column subjected to tensile forces
Test ID: 4693
Test status: Passed
14.11.1Description Verifies the tensile resistance of a rectangular cross section column (fixed at base) made from solid timber C24.
14.11.2Background Verifies the adequacy of the tension resistance for a rectangular cross section made from solid timber C24. The verification is made according to formula (6.1) from EN 1995-1-1 norm.
14.11.2.1Model description ■ Reference: Guide de validation Eurocode 5, test A; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Column with fixed base
Units
Metric System
Geometry
Cross section characteristics:
■ Height: h = 0.122 m, ■ Width: b = 0.036m, ■ Section area: A = 43.92 x 10-4 m2 ■ I = 5.4475 x 10-6 m4.
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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: ft,0,k = 14 x 106 Pa, ■ Service class 2.
Boundary conditions
The boundary conditions:
■ Outer: Fixed at base (z = 0),
Free at top (z = 5),
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at z = 5: Fz = N = 10000 N, ■ Internal: None.
14.11.2.2Reference results in calculating the timber column subjected to tension force
Reference solution
The reference solution is determined by formula (6.1) from EN 1995-1-1. Before applying this formula we need to determine some parameters involved in calculations (kmod, γM, kh). After this, the design tensile stress, the design tensile strength and the corresponding work ratio are calculated.
■ Modification factor for duration of load and moisture content: kmod = 0.9 ■ Partial factor for material properties: γM = 1.3 ■ Depth factor (“h” represents the width, because the element is tensioned):
kh = min
3.1
150 2.0
h
■ Design tensile stress (induced by the ultimate limit state force, N):
σt,0,d = AN
■ Design tensile strength:
ft,0,d = hM
kt kkf ××γ
mod,0,
■ Work ratio:
SFx = 0.1,0,
,0, ≤dt
dt
fσ
(according to relation 6.1 from EN 1995-1-1)
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
14.11.2.3Reference results
Result name Result description Reference value σt,0,d Design tensile stress [Pa] 2276867.03 Pa SFx Work ratio [%] 18 %
14.11.3Calculated results Result name Result description Value Error Stress SFx Design tensile stress 2.27687e+006 Pa 0.0000 % Work ratio Work ratio 18.0704 % 0.3911 %
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14.12 EC5: Verifying a timber column subjected to compression forces
Test ID: 4823
Test status: Passed
14.12.1Description Verifies the compressive resistance of a rectangular cross section column (hinged at base) made from solid timber C18.
14.12.2Background Verifies the adequacy of the compressive resistance for a rectangular cross section made from solid timber C18. The verification is made according to formula (6.35) from EN 1995-1-1 norm.
14.12.2.1Model description ■ Reference: Guide de validation Eurocode 5, test B; ■ Analysis type: static linear (plane problem); ■ Element type: linear.
Simply supported column
Units
Metric System
Geometry
Column cross section characteristics:
■ Height: h = 0.15 m, ■ Width: b = 0.10 m, ■ Section area: A = 15.0 x 10-3 m2
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,
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■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.6 x 1010 Pa, ■ Characteristic compressive strength along the grain: fc,0,k = 18 x 106 Pa, ■ Service class 1.
Boundary conditions
The boundary conditions:
■ Outer: Support at base (z=0) restrained in translation along X, Y and Z, Support at top (z = 3.2) restrained in translation along X, Y and restrained in rotation along Z.
■ Inner: None.
Loading
The column is subjected to the following loadings:
■ External: Point load at z = 3.2: Fz = N = -20000 N, ■ Internal: None.
14.12.2.2Reference results in calculating the timber column subjected to compression force The formula (6.35) from EN 1995-1-1 is used in order to verify a timber column subjected to compression force. Before applying this formula we need to determine some parameters involved in calculations, such as: slenderness ratios, relative slenderness ratios, instability factors. After this, the reference solution is calculated. This includes: the design compressive stress, the design compressive strength and the corresponding work ratio.
Slenderness ratios
The most important slenderness is calculated relative to the z axis, as it will be the axis of rotation if the column buckles.
■ Slenderness ratio corresponding to bending about the z axis:
85.1101.0
2.31121212 =×
=×
==m
mb
lmbl gc
zλ
■ Slenderness ratio corresponding to bending about the y axis (informative):
9.7315.0
2.31121212 =×
=×
==mm
hlm
hl gc
yλ
Relative slenderness ratios
The relative slenderness ratios are:
■ Relative slenderness ratio corresponding to bending about the z axis:
933.1106.01018
1.0122.31
,
12
,10
6
050
,0,
050
,0,, =
××
×××
=×
××==
PaPa
mm
Ef
blm
Ef kcgkcz
zrel πππλλ
■ Relative slenderness ratio corresponding to bending about the y axis (informative):
288.1106.01018
15.0122.31
,
12
,10
6
050
,0,
050
,0,, =
××
×××
=×
××==
PaPa
mm
Ef
hlm
Ef kcgkcy
yrel πππλ
λ
So there is a risk of buckling, because λrel,max ≥ 0.3.
Instability factors
In order to determine the instability factors we need to determine the βc factor. It is a factor for solid timber members within the straightness limits defined in Section 10 from EN 1995-1-1:
βc = 0.2 (according to relation 6.29 from EN 1995-1-1)
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The instability factors are: ■ kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z
2] (according to relation 6.28 from EN 1995-1-1) ■ ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y
2] (informative)
■ 2
,2,1
zrelzz
zckk
kλ−+
= (according to relation 6.26 from EN 1995-1-1)
■ 2
,2,1
yrelyy
yckk
kλ−+
= (informative)
Reference solution
Before calculating the reference solution (design compressive stress, design compressive strength and work ratio) we need to determine some parameters involved in calculations (kmod, γM).
■ Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties: γM = 1.3 ■ Design compressive stress (induced by the applied forces):
σc,0,d = AN
■ Design compressive strength:
fc,0,d = M
kckfγ
mod,0,
■ Work ratio:
Work ratio = 0.1,0,,
,0, ≤× dczc
dc
fkσ
(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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Work ratio diagram
Simply supported column subjected to compression force Work ratio
14.12.2.3Reference results
Result name Result description Reference value kc,z Instability factor 0.2400246 kc,y Instability factor (informative) 0.488869 σc,0,d Design compressive stress [Pa] 1333333 Pa Work ratio Work ratio [%] 50 %
14.12.3Calculated results Result name Result description Value Error Kc,z Instability factor 0.240107 adim 0.0000 % Kc,y Instability factor 0.488612 adim 0.0000 % Stress SFx Design compressive stress 1.33333e+006 Pa 0.0000 % Work ratio Work ratio 50.1319 % 0.2638 %
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14.13 EC5: Verifying a timber beam subjected to combined bending and axial tension
Test ID: 4872
Test status: Passed
14.13.1Description Verifies a rectangular cross section rafter made from solid timber C24 to resist combined bending and axial tension. The verification of the cross-section subjected to combined stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit state are performed.
14.13.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending and axial tension. The verification of the deflections at serviceability limit state is also performed.
14.13.2.1Model description ■ Reference: Guide de validation Eurocode 5, test E; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent rafters (span): d = 0.5 m. The following load cases and load combination are used:
■ Loadings from the structure: G = 450 N/m2; ■ Snow load (structure is located at an altitude > 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m2; ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x S; ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.2 x S. All loads will be projected on the rafter direction since its slope is 50% (26.6°).
Simply supported rafter subjected to projected loadings
Units
Metric System
Geometry
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: 322
000333.06
20.005.06
mhbWy =⋅
=×
= .
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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: ft,0,k = 14 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Service class 2.
Boundary conditions
The boundary conditions are described below:
■ Outer: Support at start point (z=0) restrained in translation along Y, Z and restrained in rotation along X. Support at end point (z = 5.00) restrained in translation along X, Y, Z.
■ Inner: None.
Loading
The rafter is subjected to the following projected loadings (at ultimate limit state):
■ External: Uniformly distributed load: q = Cmax x d x cos26.6° = 1957.5 N/m2 x 0.5 m x cos26.6° = 875.15 N/m, Tensile load component: N = Cmax x d x sin26.6° x L = 1957.5 N/m2 x 0.5m x sin26.6° x 5.00m =
= 2191.22 N ■ Internal: None.
14.13.2.2Reference results in calculating the timber beam subjected to combined stresses In order to verify the timber beam subjected to combined stresses at ultimate limit state, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, must be determined. After this the reference solution, which consists of the design tensile stress, the design tensile strength and the corresponding work ratio and also the work ratios of the combined stresses, is calculated.
A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the effective values with the limiting values for deflections specified in EN 1995-1-1 norm.
Reference solution for ultimate limit state verification
Before calculating the reference solution (the design tensile stress, the design tensile strength and the corresponding work ratio, and also the work ratios of the combined stresses) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).
■ Modification factor for duration of load (short term) and moisture content: kmod = 0.9 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties: γM = 1.3
■ Depth factor (“h” represents the width in millimeters because the element is tensioned):
kh = min 25.13.125.1
min3.150
150min
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
■ Design tensile stress (induced by the ultimate limit state force, N):
σt,0,d = PamN
AN 219122
101022.2191
23 =×
= −
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■ Design tensile strength:
ft,0,d = Pakkf hM
kt66mod
,0, 10115.1225.13.19.01014 ×=×××=××
γ
■ Work ratio:
SFx = 0.1,0,
,0, ≤dt
dt
fσ
(according to relation 6.1 from EN 1995-1-1)
■ Design bending stress (induced by the applied forces):
σm,y,d = Pamm
mmN
hbLq
WM
y
y 622
22
2
2
102045.82.005.08
00.515.8756
86
×=××
××=
××××
=
■ Design bending strength:
fm,y,d = Pakkkf hsysM
km66mod
, 10615.160.10.13.19.01024 ×=××××=×××
γ
■ Work ratio according to formulae 6.17 from EN 1995-1-1 norm:
1,,
,,
,,
,,
,0,
,0, ≤++dzm
dzmm
dym
dym
dt
dt
fk
ffσσσ
■ Work ratio according to formulae 6.18 from EN 1995-1-1 norm:
1,,
,,
,,
,,
,0,
,0, ≤++dzm
dzm
dym
dymm
dt
dt
ffk
fσσσ
Reference solution for serviceability limit state verification
The following limiting values for instantaneous deflection (for a base variable action), final deflection and net deflection are considered:
300)( LQwinst ≤
125Lwfin ≤
200,Lw finnet ≤
For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are:
■ Instantaneous deflection (for a base variable action):
05.547)(00914.0)( LQwmQw instinst =⇒=
■ Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):
7.36401371.0 Lwmdw instCQinst =⇒==
■ In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the
deformation factor (kdef) has to be chosen:
8.0=defk (value determined for service class 2, according to table 3.2 from EN 1995-1-1)
6.97600512.00064.08.08.0 Lwmmdw creepQPcreep =⇒=×=×=
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■ Final deflection:
5.26501883.000512.001371.0 Lwmmmwww fincreepinstfin =⇒=+=+=
■ Net deflection:
5.26501883.0001883.0 ,,
Lwmmmwww finnetcfinfinnet =⇒=+=+=
Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
SFx work ratio diagram
Simply supported beam subjected to tensile forces Work ratio SFx
Strength work ratio diagram
Simply supported beam subjected to combined stresses Strength work ratio
Instantaneous deflection winst(Q)
Simply supported beam subjected to snow loads Instantaneous deflection winst(Q) [m]
Instantaneous deflection winst(CQ)
Simply supported beam subjected to a characteristic load combination of actions Instantaneous deflection winst(CQ) [m]
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Final deflection wfin
Simply supported beam subjected to characteristic load combination of actions Instantaneous deflection winst(CQ) [m]
Net deflection wnet,fin
Simply supported beam subjected to characteristic load combination of actions Instantaneous deflection winst(CQ) [m]
14.13.2.3Reference results
Result name Result description Reference value SFx SFx work ratio [%] 1.808 % Strength work ratio Work ratio (6.17) [%] 51.19 % winst (Q) Deflection for a base variable action [m] 0.00914 m
dCQ Deflection for a characteristic combination of actions [m] 0.01371 m
winst Instantaneous deflection [m] 0.01371 m kdef Deformation coefficient 0.8 dQP Deflection for a quasi-permanent combination of actions [m] 0.0064 m wfin Final deflection [m] 0.01883 m wnet,fin Net deflection [m] 0.01883 m
14.13.3Calculated results Result name Result description Value Error Work ratio SFx SFx work ratio 1.81483 % 0.3778 % Work ratio Strength work ratio 51.1941 % 0.0000 % D w_inst(Q) 0.00914038 m 0.0000 % D deflection for a characteristic combination 0.0137106 m 0.0000 % Winst instantaneous deflection 0.0137105 m 0.0000 % Kdef deformation coefficient 0.8 adim 0.0000 % D deformation for a quasi-permanent combination 0.00639828 m 0.0000 % Wfin final deflection 0.0188291 m 0.0000 % Wnet,fin net final deflection 0.0188291 m 0.0000 %
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14.14 EC5: Verifying a C24 timber beam subjected to shear force
Test ID: 5036
Test status: Passed
14.14.1Description Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification of the shear stresses at ultimate limit state is performed.
14.14.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification of the shear stresses at ultimate limit state is performed.
14.14.2.1Model description ■ Reference: Guide de validation Eurocode 5, test D; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:
■ Loadings from the structure: G = 0.5 kN/m2, ■ Exploitation loadings (category A): Q = 1.5 kN/m2, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2
Simply supported beam
Units
Metric System
Geometry
Beam cross section characteristics:
■ Height: h = 0.225 m, ■ Width: b = 0.075 m, ■ Length: L = 5.00 m, ■ Distance between adjacent beams (span): d = 0.5 m, ■ Section area: A = 16.875 x 10-3 m2 ,
Materials properties
Rectangular solid timber C24 is used. The following characteristics are used in relation to this material:
■ Characteristic shear strength: fv,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.
14.14.2.2Reference results in calculating the timber beam subjected to uniformly distributed loads In order to verify the timber beam shear stresses at ultimate limit state, the formula (6.13) from EN 1995-1-1 norm is used. Before using it, some parameters involved in calculations, like kmod, kcr, γM, kf, beff, heff, have to be determined. After this the reference solution, which includes the design shear stress about the principal y axis, the design shear strength and the corresponding work ratios, is calculated.
Reference solution for ultimate limit state verification
Before calculating the reference solution (design shear stress, design shear strength and work ratio) it is necessary to determine some parameters involved in calculations (kmod, γM, kcr, kf, beff, heff).
■ Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)
■ Partial factor for material properties: γM = 1.3
■ Cracking factor, kcr : kcr = 0.67 (for solid timber)
■ Factor depending on the shape of the cross section, kf: kf = 3/2 (for a rectangular cross section)
■ Effective width, beff: beff = kcr x b = 0.67 x 0.075m = 0.05025m
■ Effective height, heff: heff = h = 0.225m
■ Design shear stress (induced by the applied forces):
τd = Pamm
N
hbFk
effeff
dvf 6, 10485075.0225.005025.0
25.365623
×=×
×=
×
×
■ Design shear strength:
fv,d = PaPakfM
kv66mod
, 10538.13.18.0105.2 ×=××=×
γ
■ Work ratio according to formulae 6.13 from EN 1995-1-1 norm:
0.1,
≤dv
d
fτ
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Finite elements modeling
■ Linear element: S beam, ■ 6 nodes, ■ 1 linear element.
Shear force, Fz, diagram
Simply supported beam subjected to bending Shear force diagram [N]
Design shear stress diagram
Simply supported beam subjected to bending Design shear stress [Pa]
Shear strength work ratio diagram
Simply supported beam subjected to bending Work ratio S_d [%]
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14.14.2.3Reference results
Result name Result description Reference value Fz Shear force [kN] 3.65625 kN Stress S_d Design shear stress [Pa] 485074.63 Pa
Work ratio S_d Shear work ratio (6.13) [%] 32 %
14.14.3Calculated results Result name Result description Value Error Fz Shear force -3.65625 kN 0.0000 % Stress S_d Design shear stress 485075 Pa 0.0001 % Working ratio S_d Shear strength work ratio 31.5299 % -1.4691 %
031516-0409-0622