Ad validation guide 2013 en parte1 by Mahabarata.gtc

Advance Design Validation Guide

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 â&#x20AC;&#x153;operational versionâ&#x20AC;?, 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 Officer

ADVANCE DESIGN VALIDATION GUIDE

CONTENTS 1

FINITE ELEMENTS ANALYSIS ..........................................................................................................21 1.1

System of two bars with three hinges (01-0002SSLLB_FEM)......................................................................................22

1.2

Slender beam with variable section (fixed-free) (01-0004SDLLB_FEM) ......................................................................25

1.3

Thin lozenge-shaped plate fixed on one side (alpha = 15 째) (01-0008SDLSB_FEM)...................................................28

1.4

Thin circular ring fixed in two points (01-0006SDLLB_FEM) ........................................................................................31

1.5

Thin lozenge-shaped plate fixed on one side (alpha = 45 째) (01-0010SDLSB_FEM)...................................................35

1.6

Vibration mode of a thin piping elbow in plane (case 2) (01-0012SDLLB_FEM)..........................................................38

1.7

Thin circular ring hanged on an elastic element (01-0014SDLLB_FEM)......................................................................41

1.8

Tied (sub-tensioned) beam (01-0005SSLLB_FEM) .....................................................................................................45

1.9

Circular plate under uniform load (01-0003SSLSB_FEM)............................................................................................50

1.10

Double fixed beam (01-0016SDLLB_FEM) ..................................................................................................................53

1.11

Short beam on simple supports (eccentric) (01-0018SDLLB_FEM).............................................................................57

1.12

Thin lozenge-shaped plate fixed on one side (alpha = 30 째) (01-0009SDLSB_FEM)...................................................61

1.13

Vibration mode of a thin piping elbow in plane (case 3) (01-0013SDLLB_FEM)..........................................................64

1.14

Short beam on simple supports (on the neutral axis) (01-0017SDLLB_FEM)..............................................................67

1.15

Thin lozenge-shaped plate fixed on one side (alpha = 0 째) (01-0007SDLSB_FEM).....................................................71

1.16

Vibration mode of a thin piping elbow in plane (case 1) (01-0011SDLLB_FEM)..........................................................74

1.17

Double fixed beam with a spring at mid span (01-0015SSLLB_FEM)..........................................................................77

1.18

Rectangular thin plate simply supported on its perimeter (01-0020SDLSB_FEM) .......................................................80

1.19

Slender beam on two fixed supports (01-0024SSLLB_FEM) .......................................................................................84

1.20

Annular thin plate fixed on a hub (repetitive circular structure) (01-0022SDLSB_FEM) ...............................................89

1.21

Bimetallic: Fixed beams connected to a stiff element (01-0026SSLLB_FEM)..............................................................91

1.22

Cantilever beam in Eulerian buckling (01-0021SFLLB_FEM) ......................................................................................94

1.23

Slender beam on three supports (01-0025SSLLB_FEM) .............................................................................................96

1.24

Fixed thin arc in out of plane bending (01-0028SSLLB_FEM) ...................................................................................100

1.25

Portal frame with lateral connections (01-0030SSLLB_FEM) ....................................................................................103

1.26

Double hinged thin arc in planar bending (01-0029SSLLB_FEM) ..............................................................................106

1.27

Thin square plate fixed on one side (01-0019SDLSB_FEM)......................................................................................109

1.28

Bending effects of a symmetrical portal frame (01-0023SDLLB_FEM) ......................................................................113

1.29

Fixed thin arc in planar bending (01-0027SSLLB_FEM) ............................................................................................116

1.30

Beam on elastic soil, hinged ends (01-0034SSLLB_FEM).........................................................................................119

1.31

Square plate under planar stresses (01-0039SSLSB_FEM) ......................................................................................123

1.32

Beam on elastic soil, free ends (01-0032SSLLB_FEM) .............................................................................................126

1.33

EDF Pylon (01-0033SFLLA_FEM) .............................................................................................................................129

1.34

Thin cylinder under a uniform radial pressure (01-0038SSLSB_FEM).......................................................................133

ADVANCE DESIGN VALIDATION GUIDE

1.35

Caisson beam in torsion (01-0037SSLSB_FEM) ....................................................................................................... 135

1.36

Beam on two supports considering the shear force (01-0041SSLLB_FEM) .............................................................. 138

1.37

Thin cylinder under a hydrostatic pressure (01-0043SSLSB_FEM)........................................................................... 141

1.38

Thin cylinder under a uniform axial load (01-0042SSLSB_FEM) ............................................................................... 144

1.39

Truss with hinged bars under a punctual load (01-0031SSLLB_FEM) ...................................................................... 147

1.40

Simply supported square plate (01-0036SSLSB_FEM) ............................................................................................. 150

1.41

Stiffen membrane (01-0040SSLSB_FEM) ................................................................................................................. 153

1.42

Torus with uniform internal pressure (01-0045SSLSB_FEM) .................................................................................... 156

1.43

Spherical dome under a uniform external pressure (01-0050SSLSB_FEM).............................................................. 159

1.44

Pinch cylindrical shell (01-0048SSLSB_FEM) ........................................................................................................... 162

1.45

Simply supported rectangular plate under a uniform load (01-0052SSLSB_FEM) .................................................... 164

1.46

Spherical shell under internal pressure (01-0046SSLSB_FEM) ................................................................................ 166

1.47

Simply supported square plate under a uniform load (01-0051SSLSB_FEM) ........................................................... 169

1.48

Simply supported rectangular plate loaded with punctual force and moments (01-0054SSLSB_FEM) ..................... 171

1.49

Triangulated system with hinged bars (01-0056SSLLB_FEM) .................................................................................. 173

1.50

Shear plate perpendicular to the medium surface (01-0055SSLSB_FEM) ................................................................ 176

1.51

Thin cylinder under its self weight (01-0044SSLSB_MEF) ........................................................................................ 178

1.52

Spherical shell with holes (01-0049SSLSB_FEM) ..................................................................................................... 180

1.53

Simply supported rectangular plate under a uniform load (01-0053SSLSB_FEM) .................................................... 183

1.54

A plate (0.01333 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0058SSLSB_FEM)................ 185

1.55

A plate (0.05 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0060SSLSB_FEM)...................... 187

1.56

A plate (0.02 m thick), fixed on its perimeter, loaded with a punctual force (01-0064SSLSB_FEM) .......................... 189

1.57

A plate (0.01 m thick), fixed on its perimeter, loaded with a punctual force (01-0062SSLSB_FEM) .......................... 191

1.58

A plate (0.02 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0059SSLSB_FEM)...................... 194

1.59

A plate (0.01333 m thick), fixed on its perimeter, loaded with a punctual force (01-0063SSLSB_FEM) .................... 196

1.60

A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (01-0066SSLSB_FEM) ............................ 199

1.61

Vibration mode of a thin piping elbow in space (case 2) (01-0068SDLLB_FEM)....................................................... 201

1.62

Vibration mode of a thin piping elbow in space (case 1) (01-0067SDLLB_FEM)....................................................... 204

1.63

A plate (0.01 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0057SSLSB_FEM)...................... 207

1.64

A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0061SSLSB_FEM)........................ 209

1.65

A plate (0.05 m thick), fixed on its perimeter, loaded with a punctual force (01-0065SSLSB_FEM) .......................... 211

1.66

Reactions on supports and bending moments on a 2D portal frame (Rafters) (01-0077SSLPB_FEM) ..................... 213

1.67

Slender beam of variable rectangular section (fixed-fixed) (01-0086SDLLB_FEM) ................................................... 215

1.68

Short beam on two hinged supports (01-0084SSLLB_FEM) ..................................................................................... 218

1.69

Double fixed beam in Eulerian buckling with a thermal load (01-0091HFLLB_FEM) ................................................. 220

1.70

Reactions on supports and bending moments on a 2D portal frame (Columns) (01-0078SSLPB_FEM) .................. 222

1.71

Plane portal frame with hinged supports (01-0089SSLLB_FEM)............................................................................... 224

1.72

A 3D bar structure with elastic support (01-0094SSLLB_FEM) ................................................................................. 226

ADVANCE DESIGN VALIDATION GUIDE

1.73

Fixed/free slender beam with eccentric mass or inertia (01-0096SDLLB_FEM) ........................................................233

1.74

Fixed/free slender beam with centered mass (01-0095SDLLB_FEM)........................................................................237

1.75

Vibration mode of a thin piping elbow in space (case 3) (01-0069SDLLB_FEM) .......................................................242

1.76

Slender beam of variable rectangular section with fixed-free ends (Ă&#x;=5) (01-0085SDLLB_FEM) .............................245

1.77

Cantilever beam in Eulerian buckling with thermal load (01-0092HFLLB_FEM) ........................................................250

1.78

Simple supported beam in free vibration (01-0098SDLLB_FEM)...............................................................................252

1.79

Membrane with hot point (01-0099HSLSB_FEM) ......................................................................................................255

1.80

Beam on 3 supports with T/C (k = -10000 N/m) (01-0102SSNLB_FEM) ...................................................................258

1.81

Beam on 3 supports with T/C (k = 0) (01-0100SSNLB_FEM) ....................................................................................261

1.82

Non linear system of truss beams (01-0104SSNLB_FEM) ........................................................................................264

1.83

Linear system of truss beams (01-0103SSLLB_FEM) ...............................................................................................267

1.84

Linear element in combined bending/tension - without compressed reinforcements - Partially tensioned section (02-0158SSLLB_B91) ................................................................................................................................................270

1.85

Design of a Steel Structure according to CM66 (03-0206SSLLG_CM66) ..................................................................275

1.86

Linear element in simple bending - without compressed reinforcement (02-0162SSLLB_B91) .................................284

1.87

Double cross with hinged ends (01-0097SDLLB_FEM) .............................................................................................288

1.88

Beam on 3 supports with T/C (k -> infinite) (01-0101SSNLB_FEM)...........................................................................291

1.89

Study of a mast subjected to an earthquake (02-0112SMLLB_P92)..........................................................................294

1.90

Design of a concrete floor with an opening (03-0208SSLLG_BAEL91) .....................................................................300

1.91

Design of a 2D portal frame (03-0207SSLLG_CM66) ................................................................................................308

1.92

Cantilever rectangular plate (01-0001SSLSB_FEM) ..................................................................................................315

1.93

Calculating torsors using different mesh sizes for a concrete wall subjected to a horizontal force (TTAD #13175) ...318

1.94

Verifying torsors on a single story coupled walls subjected to horizontal forces ........................................................318

1.95

Verifying diagrams for Mf Torsors on divided walls (TTAD #11557)...........................................................................318

1.96

Verifying the level mass center (TTAD #11573, TTAD #12315).................................................................................318

1.97

Generating results for Torsors NZ/Group (TTAD #11633) .........................................................................................318

1.98

Verifying Sxx results on beams (TTAD #11599).........................................................................................................319

1.99

Verifying forces results on concrete linear elements (TTAD #11647).........................................................................319

1.100 Verifying diagrams after changing the view from standard (top, left,...) to user view (TTAD #11854) ........................319 1.101 Verifying stresses in beam with "extend into wall" property (TTAD #11680) ..............................................................319 1.102 Verifying constraints for triangular mesh on planar elements (TTAD #11447) ...........................................................319 1.103 Verifying the displacement results on linear elements for vertical seism (TTAD #11756) ..........................................320 1.104 Verifying forces for triangular meshing on planar element (TTAD #11723)................................................................320 1.105 Generating planar efforts before and after selecting a saved view (TTAD #11849) ...................................................320 1.106 Verifying tension/compression supports on nonlinear analysis (TTAD #11518).........................................................320 1.107 Verifying tension/compression supports on nonlinear analysis (TTAD #11518).........................................................321 1.108 Verifying the display of the forces results on planar supports (TTAD #11728)...........................................................321 1.109 Verifying results on punctual supports (TTAD #11489) ..............................................................................................321 1.110 Generating a report with torsors per level (TTAD #11421).........................................................................................321 9

ADVANCE DESIGN VALIDATION GUIDE

1.111 Verifying nonlinear analysis results for frames with semi-rigid joints and rigid joints (TTAD #11495) ........................ 322 1.112 Verifying forces on a linear elastic support which is defined in a user workplane (TTAD #11929) ............................ 322 1.113 Verifying the internal forces results for a simple supported steel beam ..................................................................... 322 1.114 Verifying the main axes results on a planar element (TTAD #11725) ........................................................................ 322

CAD, RENDERING AND VISUALIZATION ...................................................................................... 323 2.1

Verifying annotation on selection (TTAD #12700)...................................................................................................... 324

2.2

Verifying rotation for steel beam with joint (TTAD #12592)........................................................................................ 324

2.3

Verifying hide/show elements command (TTAD #11753) .......................................................................................... 324

2.4

System stability during section cut results verification (TTAD #11752)...................................................................... 324

2.5

Verifying the grid text position (TTAD #11704) .......................................................................................................... 324

2.6

Verifying the grid text position (TTAD #11657) .......................................................................................................... 324

2.7

Generating combinations (TTAD #11721) ................................................................................................................. 325

2.8

Verifying the coordinates system symbol (TTAD #11611) ......................................................................................... 325

2.9

Verifying descriptive actors after creating analysis (TTAD #11589)........................................................................... 325

2.10

Creating a circle (TTAD #11525) ............................................................................................................................... 325

2.11

Creating a camera (TTAD #11526)............................................................................................................................ 325

2.12

Verifying the local axes of a section cut (TTAD #11681) ........................................................................................... 326

2.13

Verifying the snap points behavior during modeling (TTAD #11458) ......................................................................... 326

2.14

Verifying the representation of elements with HEA cross section (TTAD #11328)..................................................... 326

2.15

Verifying the descriptive model display after post processing results in analysis mode (TTAD #11475) ................... 326

2.16

Verifying holes in horizontal planar elements after changing the level height (TTAD #11490) .................................. 326

2.17

Verifying the display of elements with compound cross sections (TTAD #11486) ..................................................... 327

2.18

Modeling using the tracking snap mode (TTAD #10979) ........................................................................................... 327

2.19

Turning on/off the "ghost" rendering mode (TTAD #11999) ....................................................................................... 327

2.20

Moving a linear element along with the support (TTAD #12110) ............................................................................... 327

2.21

Verifying the "ghost" display after changing the display colors (TTAD #12064)......................................................... 327

2.22

Verifying the "ghost display on selection" function for saved views (TTAD #12054).................................................. 327

2.23

Verifying the steel connections modeling (TTAD #11698) ......................................................................................... 328

2.24

Verifying the fixed load scale function (TTAD #12183). ............................................................................................. 328

2.25

Verifying the dividing of planar elements which contain openings (TTAD #12229).................................................... 328

2.26

Verifying the program behavior when trying to create lintel (TTAD #12062).............................................................. 328

2.27

Verifying the program behavior when launching the analysis on a model with overlapped loads (TTAD #11837) .... 328

2.28

Verifying the display of punctual loads after changing the load case number (TTAD #11958) .................................. 329

2.29

Verifying the display of a beam with haunches (TTAD #12299)................................................................................ 329

2.30

Creating base plate connections for non-vertical columns (TTAD #12170) ............................................................... 329

2.31

Verifying drawing of joints in y-z plan (TTAD #12453) ............................................................................................... 329

ADVANCE DESIGN VALIDATION GUIDE

CLIMATIC GENERATOR ..................................................................................................................331 3.1

EC1: generating snow loads on a 3 slopes 3D portal frame with parapets (NF EN 1991-1-3/NA) (TTAD #11111) ...332

3.2

EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.3 - Wind - Example C) ....332

3.3

EC1: generating wind loads on a 3D portal frame with one slope roof (NF EN 1991-1-4/NA) (VT : 3.2 - Wind - Example B) ....................................................................................................................................332

3.4

EC1: generating wind loads on a triangular based lattice structure with compound profiles and automatic calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12276)............................................................................................332

3.5

EC1: generating snow loads on a 2 slopes 3D portal frame (NF EN 1991-1-3/NA) (VT : 3.4 - Snow - Example A)...333

3.6

EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.1 - Wind - Example A).....333

3.7

EC1: wind loads on a triangular based lattice structure with compound profiles and user defined "n" (NF EN 1991-1-4/NA) (TTAD #12276) .......................................................................................................................333

3.8

EC1: Verifying the geometry of wind loads on an irregular shed. (TTAD #12233) .....................................................333

3.9

EC1: Verifying the wind loads generated on a building with protruding roof (TTAD #12071, #12278) .......................333

3.10

EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11531) ..............................................................334

3.11

EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11569) ....334

3.12

Generating the description of climatic loads report according to EC1 Romanian standards (TTAD #11688).............334

3.13

EC1: generating snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapet height (TTAD #11943)...........................................................................................................................................................334

3.14

EC1: generating wind loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11687) .....335

3.15

EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11699) .............................................................335

3.16

EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11570) ....335

3.17

NV2009: generating wind loads and snow loads on a simple structure with planar support (TTAD #11380).............335

3.18

EC1: verifying the snow loads generated on a monopitch frame (TTAD #11302) ......................................................336

3.19

EC1: generating wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls (TTAD #11937) ...............336

3.20

EC1: generating snow loads on two side by side roofs with different heights, according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)...................................................................................................................336

3.21

EC1: generating wind loads on a 55m high structure according to German standards (DIN EN 1991-1-4/NA) (DEV2012 #3.12)........................................................................................................................................................336

3.22

EC1: generating wind loads on double slope 3D portal frame according to Czech standards (CSN EN 1991-1-4) (DEV2012 #3.18)........................................................................................................................................................337

3.23

EC1: generating snow loads on duopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)........................................................................................................................................................337

3.24

EC1: generating snow loads on two close roofs with different heights according to Czech standards (CSN EN 1991-1-3) (DEV2012 #3.18) .......................................................................................................................337

3.25

EC1: generating snow loads on monopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)........................................................................................................................................................337

3.26

EC1: snow on a 3D portal frame with horizontal roof and parapet with height reduction (TTAD #11191) ..................338

3.27

EC1: generating snow loads on a 2 slopes 3D portal frame with gutter (TTAD #11113)............................................338

3.28

EC1: generating snow loads on a 3D portal frame with horizontal roof and gutter (TTAD #11113) ...........................338

3.29

EC1: generating snow loads on a 3D portal frame with a roof which has a small span (< 5m) and a parapet (TTAD #11735)...........................................................................................................................................................338

3.30

EC1: generating wind loads on double slope 3D portal frame with a fully opened face (DEV2012 #1.6)...................339

3.31

EC1: generating wind loads on an isolated roof with two slopes (TTAD #11695) ......................................................339

ADVANCE DESIGN VALIDATION GUIDE

3.32

EC1: generating wind loads on duopitch multispan roofs with pitch < 5 degrees (TTAD #11852) ............................. 339

3.33

EC1: wind load generation on a simple 3D structure with horizontal roof .................................................................. 339

3.34

EC1 NF: generating wind loads on a 3D portal frame with 2 slopes roof (TTAD #11932) ......................................... 340

3.35

EC1: wind load generation on simple 3D portal frame with 4 slopes roof (TTAD #11604)......................................... 340

3.36

EC1: wind load generation on a signboard ................................................................................................................ 340

3.37

EC1: wind load generation on a simple 3D portal frame with 2 slopes roof (TTAD #11602)...................................... 340

3.38

EC1: wind load generation on a building with multispan roofs ................................................................................... 341

3.39

EC1: wind load generation on a high building with horizontal roof ............................................................................. 341

3.40

EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) ......................................................... 341

3.41

EC1: Generating snow loads on a single slope with lateral parapets (TTAD #12606) ............................................... 341

3.42

EC1: generating wind loads on a square based lattice structure with compound profiles and automatic calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12744) ........................................................................................... 342

3.43

EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) ......................................................... 342

3.44

EC1: Generating snow loads on two side by side buildings with gutters (TTAD #12806) .......................................... 342

3.45

EC1: Generating wind loads on a square based structure according to UK standards (BS EN 1991-1-4:2005) (TTAD #12608) .......................................................................................................................................................... 342

3.46

EC1: Generating snow loads on a 4 slopes with gutters building. (TTAD #12719) .................................................... 342

3.47

EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12808) ....................................................... 343

3.48

EC1: Generating snow loads on a 4 slopes with gutters building (TTAD #12716) ..................................................... 343

3.49

EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12835) ....................................................... 343

3.50

EC1: Generating snow loads on a 2 slope building with gutters and parapets. (TTAD #12878) ................................ 343

3.51

EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12841) ....................................................... 343

3.52

NV2009: Verifying wind and snow reports for a protruding roof (TTAD #11318) ....................................................... 344

3.53

NV2009: Generating wind loads on a 2 slopes 3D portal frame at 15m height (TTAD #12604) ................................ 344

COMBINATIONS............................................................................................................................... 345 4.1

Verifying combinations for CZ localization (TTAD #12542)........................................................................................ 346

4.2

Generating combinations (TTAD #11673) ................................................................................................................. 346

4.3

Defining concomitance rules for two case families (TTAD #11355) ........................................................................... 346

4.4

Generating load combinations with unfavorable and favorable/unfavorable predominant action (TTAD #11357) ..... 346

4.5

Generating combinations for NEWEC8.cbn (TTAD #11431) ..................................................................................... 346

4.6

Generating load combinations after changing the load case number (TTAD #11359)............................................... 347

4.7

Generating the concomitance matrix after adding a new dead load case (TTAD #11361) ........................................ 347

4.8

Generating a set of combinations with seismic group of loads (TTAD #11889) ......................................................... 347

4.9

Generating the concomitance matrix after switching back the effect for live load (TTAD #11806) ............................ 347

4.10

Performing the combinations concomitance standard test no. 5 (DEV2012 #1.7) ..................................................... 348

4.11

Performing the combinations concomitance standard test no.9 (DEV2012 #1.7) ...................................................... 348

4.12

Performing the combinations concomitance standard test no.4 (DEV2012 #1.7) ..................................................... 349

4.13

Performing the combinations concomitance standard test no.6 (DEV2012 #1.7) ...................................................... 349

4.14

Performing the combinations concomitance standard test no.8 (DEV2012 #1.7) ...................................................... 350

ADVANCE DESIGN VALIDATION GUIDE

4.15

Performing the combinations concomitance standard test no.10 (DEV2012 #1.7) ....................................................350

4.16

Performing the combinations concomitance standard test no.7 (DEV2012 #1.7) ......................................................351

4.17

Generating a set of combinations with Q group of loads (TTAD #11960) ..................................................................351

4.18

Performing the combinations concomitance standard test no.1 (DEV2010#1.7) .......................................................351

4.19

Generating a set of combinations with different Q "Base" types (TTAD #11806) .......................................................352

4.20

Performing the combinations concomitance standard test no.2 (DEV2012 #1.7) ......................................................352

4.21

Performing the combinations concomitance standard test no.3 (DEV2012 #1.7) ......................................................353

CONCRETE DESIGN ........................................................................................................................355 5.1

EC2: Verifying the minimum reinforcement area for a simply supported beam ..........................................................356

5.2

EC2: Verifying the longitudinal reinforcement area of a beam under a linear load .....................................................356

5.3

EC2: Verifying the longitudinal reinforcement area for a beam subjected to point loads............................................356

5.4

Modifying the "Design experts" properties for concrete linear elements (TTAD #12498) ...........................................356

5.5

EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - horizontal level behavior law.....356

5.6

EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - bilinear stress-strain diagram....357

5.7

Verifying the longitudinal reinforcement for a horizontal concrete bar with rectangular cross section ........................357

5.8

Verifying the reinforced concrete results on a structure with 375 load cases combinations (TTAD #11683) .............357

5.9

Verifying the longitudinal reinforcement bars for a filled circular column (TTAD #11678) ..........................................357

5.10

Verifying the reinforced concrete results on a fixed beam (TTAD #11836) ................................................................358

5.11

Verifying the longitudinal reinforcement for a fixed linear element (TTAD #11700)....................................................358

5.12

Verifying the longitudinal reinforcement for linear elements (TTAD #11636) .............................................................358

5.13

EC2 : calculation of a square column in traction (TTAD #11892) ...............................................................................358

5.14

Verifying Aty and Atz for a fixed concrete beam (TTAD #11812) ...............................................................................359

5.15

Verifying concrete results for planar elements (TTAD #11583) ..................................................................................359

5.16

Verifying concrete results for linear elements (TTAD #11556) ...................................................................................359

5.17

Verifying the reinforcement of concrete columns (TTAD #11635) ..............................................................................359

5.18

Verifying the minimum transverse reinforcement area results for an articulated beam (TTAD #11342).....................360

5.19

Verifying the minimum transverse reinforcement area results for articulated beams (TTAD #11342)........................360

5.20

EC2: column design with â&#x20AC;&#x153;Nominal Stiffness methodâ&#x20AC;? square section (TTAD #11625) ..............................................360

5.21

EC2: Verifying the transverse reinforcement area for a beam subjected to linear loads ............................................360

5.22

EC2 Test 1: Verifying a rectangular cross section beam made from concrete C25/30 to resist simple bending - Bilinear stress-strain diagram...................................................................................................................................361

5.23

EC2 Test 5: Verifying a T concrete section, without compressed reinforcement - Bilinear stress-strain diagram ......362

5.24

EC2 Test 9: Verifying a rectangular concrete beam with compressed reinforcement - Inclined stress-strain diagram ..................................................................................................................................367

5.25

EC2 Test 10: Verifying a T concrete section, without compressed reinforcement - Inclined stress-strain diagram ....377

5.26

EC2 Test 12: Verifying a rectangular concrete beam subjected to uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD3) ...................................................................383

5.27

EC2 Test 15: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram .....389

5.28

EC2 Test 16: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram .....395

ADVANCE DESIGN VALIDATION GUIDE

5.29

EC2 Test 11: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ....................................................................................... 401

5.30

EC2 Test 19: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ...................................................... 401

5.31

EC2 Test 20: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ...................................................... 402

5.32

EC2 Test 14: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement- Bilinear stress-strain diagram (Class XD1) ........................................................................................ 409

5.33

EC2 Test 18: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ....................................................................................... 415

5.34

EC2 Test 24: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 420

5.35

EC2 Test 25: Verifying the shear resistance for a rectangular concrete beam with inclined transversal reinforcement Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 420

5.36

EC2 Test 23: Verifying the shear resistance for a rectangular concrete - Bilinear stress-strain diagram (Class XC1)................................................................................................................................................................ 420

5.37

EC2 Test 13: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) ....................................................................................... 420

5.38

EC2 Test 17: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Inclined stress-strain diagram (Class XD1) ....................................................................................... 421

5.39

EC2 Test28: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class X0)................................................................................................................................ 427

5.40

EC2 Test32: Verifying a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature- Bilinear stress-strain diagram (Class XC1) .................................................................. 431

5.41

EC2 Test29: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Inclined stress-strain diagram (Class XC1) ............................................................................................................................. 441

5.42

EC2 Test33: Verifying a square concrete column subjected to compression by nominal rigidity method- Bilinear stress-strain diagram (Class XC1) ............................................................................................................................. 445

5.43

EC2 Test 27: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 452

5.44

EC2 Test31: Verifying a square concrete column subjected to compression and rotation moment to the top Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 457

5.45

EC2 Test35: Verifying a rectangular concrete column subjected to compression to top Based on nominal rigidity method - Bilinear stress-strain diagram (Class XC1) ........................................................ 471

5.46

EC2 Test36: Verifying a rectangular concrete column using the method based on nominal curvature Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 481

5.47

EC2 Test 37: Verifying a square concrete column using the simplified method – Professional rules Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 482

5.48

EC2 Test 26: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 485

5.49

EC2 Test30: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 489

5.50

EC2 Test34: Verifying a rectangular concrete column subjected to compression on the top – Method based on nominal curvature - Bilinear stress-strain diagram (Class XC1)................................................................................. 489

5.51

EC2 Test 38: Verifying a rectangular concrete column using the simplified method – Professional rules Bilinear stress-strain diagram (Class XC1) ................................................................................................................ 498

5.52

EC2 Test 40: Verifying a square concrete column subjected to a small compression force and significant rotation moment to the top - Bilinear stress-strain diagram (Class XC1) ................................................................................ 501

ADVANCE DESIGN VALIDATION GUIDE

5.53

EC2 Test 41: Verifying a square concrete column subjected to a significant compression force and small rotation moment to the top - Bilinear stress-strain diagram (Class XC1).................................................................................508

5.54

EC2 Test 44: Verifying a rectangular concrete beam subjected to eccentric loading - Bilinear stress-strain diagram (Class X0) ..................................................................................................................................................................520

5.55

EC2 Test 45: Verifying a rectangular concrete beam supporting a balcony - Bilinear stress-strain diagram (Class XC1) ................................................................................................................................................................521

5.56

EC2 Test 47: Verifying a rectangular concrete beam subjected to a tension distributed load - Bilinear stress-strain diagram (Class XD2) ..................................................................................................................................................528

5.57

EC2: Verifying the longitudinal reinforcement area of a beam under a linear load Inclined stress strain behavior law..............................................................................................................................528

5.58

EC2 Test 3: Verifying a rectangular concrete beam subjected to uniformly distributed load, with compressed reinforcement- Bilinear stress-strain diagram .............................................................................................................529

5.59

EC2 Test 2: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram ............................................................................................................538

5.60

EC2 Test 6: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram .......539

5.61

EC2 Test 7: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram .......543

5.62

EC2 Test 8: Verifying a rectangular concrete beam without compressed reinforcement – Inclined stress-strain diagram ......................................................................................................................................................................547

5.63

EC2 Test 4: Verifying a rectangular concrete beam subjected to Pivot A efforts – Inclined stress-strain diagram .....554

5.64

EC2 Test 39: Verifying a circular concrete column using the simplified method – Professional rules Bilinear stress-strain diagram (Class XC1).................................................................................................................555

5.65

EC2 Test 43: Verifying a square concrete column subjected to a small rotation moment and significant compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1)...................................559

5.66

EC2 Test 46 I: Verifying a square concrete beam subjected to a normal force of traction - Inclined stress-strain diagram (Class X0).....................................................................................................................................................568

5.67

EC2 Test 42: Verifying a square concrete column subjected to a significant rotation moment and small compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1)...................................571

5.68

EC2 Test 46 II: Verifying a square concrete beam subjected to a normal force of traction - Bilinear stress-strain diagram (Class X0).....................................................................................................................................................580

GENERAL APPLICATION ................................................................................................................583 6.1

Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238)................................................584

6.2

Verifying the precision of linear and planar concrete covers (TTAD #12525).............................................................584

6.3

Defining the reinforced concrete design assumptions (TTAD #12354) ......................................................................584

6.4

Verifying the synthetic table by type of connection (TTAD #11422) ...........................................................................584

6.5

Importing a cross section from the Advance Steel profiles library (TTAD #11487) ....................................................584

6.6

Creating and updating model views and post-processing views (TTAD #11552).......................................................585

6.7

Creating system trees using the copy/paste commands (DEV2012 #1.5)..................................................................585

6.8

Creating system trees using the copy/paste commands (DEV2012 #1.5)..................................................................585

6.9

Creating a new Advance Design file using the "New" command from the "Standard" toolbar (TTAD #12102) ..........585

6.10

Verifying mesh, CAD and climatic forces - LPM meeting ...........................................................................................585

6.11

Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724) ........................................................586

6.12

Changing the default material (TTAD #11870) ...........................................................................................................586

6.13

Verifying the objects rename function (TTAD #12162)...............................................................................................586

ADVANCE DESIGN VALIDATION GUIDE

6.14

Launching the verification of a model containing steel connections (TTAD #12100) ................................................. 586

6.15

Verifying the appearance of the local x orientation legend (TTAD #11737) ............................................................... 586

6.16

Verifying geometry properties of elements with compound cross sections (TTAD #11601) ...................................... 587

6.17

Verifying material properties for C25/30 (TTAD #11617) ........................................................................................... 587

6.18

Verifying element creation using commas for coordinates (TTAD #11141) ............................................................... 587

IMPORT / EXPORT ........................................................................................................................... 589 7.1

Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952) ....................................................... 590

7.2

Exporting an Advance Design model to DO4 format (DEV2012 #1.10) ..................................................................... 590

7.3

Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) ....................................................................... 590

7.4

Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172) ................................. 590

7.5

Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3) ....................................................................... 590

7.6

Exporting linear elements to IFC format (TTAD #10561) ........................................................................................... 591

7.7

Importing IFC files containing continuous foundations (TTAD #12410) ..................................................................... 591

7.8

Importing GTC files containing "PH.RDC" system (TTAD #12055)............................................................................ 591

7.9

Exporting a meshed model to GTC (TTAD #12550) .................................................................................................. 591

7.10

Verifying the load case properties from models imported as GTC files (TTAD #12306) ............................................ 591

7.11

Verifying the releases option of the planar elements edges after the model was exported and imported via GTC format (TTAD #12137) ............................................................................................................................................... 592

7.12

System stability when importing AE files with invalid geometry (TTAD #12232)........................................................ 592

7.13

Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9)........................... 592

7.14

Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197)........... 592

7.15

Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197)........... 592

CONNECTION DESIGN .................................................................................................................... 593 8.1

Deleting a welded tube connection - 1 gusset bar (TTAD #12630)............................................................................ 594

8.2

Creating connections groups (TTAD #11797)............................................................................................................ 594

MESH................................................................................................................................................. 595 9.1

Verifying the mesh for a model with generalized buckling (TTAD #11519)................................................................ 596

9.2

Verifying mesh points (TTAD #11748) ....................................................................................................................... 596

9.3

Creating triangular mesh for planar elements (TTAD #11727) .................................................................................. 596

REPORTS GENERATOR.................................................................................................................. 597 10.1

Verifying the modal analysis report (TTAD #12718) .................................................................................................. 598

10.2

Verifying the shape sheet strings display (TTAD #12622) ......................................................................................... 598

10.3

Verifying the shape sheet for a steel beam (TTAD #12455) ...................................................................................... 598

10.4

Verifying the global envelope of linear elements stresses (on the start point of super element) (TTAD #12230) ...... 598

10.5

Verifying the Max row on the user table report (TTAD #12512) ................................................................................. 598

10.6

Verifying the steel shape sheet display (TTAD #12657) ............................................................................................ 599

10.7

Verifying the EC2 calculation assumptions report (TTAD #11838) ............................................................................ 599

ADVANCE DESIGN VALIDATION GUIDE

10.8

Verifying the shape sheet report (TTAD #12353) .......................................................................................................599

10.9

Verifying the model geometry report (TTAD #12201).................................................................................................599

10.10 Verifying the global envelope of linear elements forces result (on end points and middle of super element) (TTAD #12230)...........................................................................................................................................................599 10.11 Verifying the global envelope of linear elements forces result (on start and end of super element) (TTAD #12230) .600 10.12 Verifying the global envelope of linear elements stresses (on the end point of super element) (TTAD #12230, TTAD #12261) ..................................................................................................................................600 10.13 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element) (TTAD #12230) .....600 10.14 Verifying the global envelope of linear elements displacements (on all quarters of super element) (TTAD #12230) .600 10.15 Verifying the global envelope of linear elements forces result (on all quarters of super element) (TTAD #12230) ....601 10.16 Verifying the global envelope of linear elements displacements (on the start point of super element) (TTAD #12230)...........................................................................................................................................................601 10.17 Verifying the global envelope of linear elements displacements (on the end point of super element) (TTAD #12230, TTAD #12261)............................................................................................................................................................601 10.18 Verifying the global envelope of linear elements forces result (on the end point of super element) (TTAD #12230, #12261).............................................................................................................................................601 10.19 Verifying the global envelope of linear elements displacements (on start and end of super element) (TTAD #12230)...........................................................................................................................................................602 10.20 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element) (TTAD #12230).........602 10.21 Verifying the global envelope of linear elements forces result (on the start point of super element) (TTAD #12230)...........................................................................................................................................................602 10.22 Verifying the global envelope of linear elements displacements (on end points and middle of super element) (TTAD #12230)...........................................................................................................................................................602 10.23 Verifying the global envelope of linear elements stresses (on each 1/4 of mesh element) (TTAD #12230)...............603 10.24 Verifying the global envelope of linear elements stresses (on start and end of super element) (TTAD #12230) .......603 10.25 Verifying the global envelope of linear elements stresses (on end points and middle of super element) (TTAD #12230)...........................................................................................................................................................603 10.26 Verifying the Min/Max values from the user reports (TTAD# 12231)..........................................................................603 10.27 Verifying the global envelope of linear elements stresses (on all quarters of super element) (TTAD #12230)...........604 10.28 Creating the rules table (TTAD #11802).....................................................................................................................604 10.29 Creating the steel materials description report (TTAD #11954) .................................................................................604 10.30 System stability when the column releases interfere with support restraints (TTAD #10557) ....................................604 10.31 Generating the critical magnification factors report (TTAD #11379)...........................................................................605 10.32 Modal analysis: eigen modes results for a structure with one level ............................................................................605 10.33 Generating a report with modal analysis results (TTAD #10849) ...............................................................................605

SEISMIC ANALYSIS .........................................................................................................................607 11.1

Verifying the spectrum results for EC8 seism (TTAD #11478) ...................................................................................608

11.2

Verifying the spectrum results for EC8 seism (TTAD #12472) ...................................................................................608

11.3

Verifying the combinations description report (TTAD #11632) ...................................................................................608

11.4

EC8 : Verifying the displacements results of a linear element according to Czech seismic standards (CSN EN 1998-1) (DEV2012 #3.18)........................................................................................................................................................608

11.5

Verifying signed concomitant linear elements envelopes on Fx report (TTAD #11517) .............................................608

ADVANCE DESIGN VALIDATION GUIDE

11.6

EC8 French Annex: verifying torsors on walls ........................................................................................................... 609

11.7

EC8 French Annex: verifying torsors on grouped walls from a multi-storey concrete structure ................................. 609

11.8

Verifying the damping correction influence over the efforts in supports (TTAD #13011). .......................................... 609

11.9

EC8 French Annex: verifying seismic results when a design spectrum is used (TTAD #13778) ............................... 609

11.10 EC8: verifying the sum of actions on supports and nodes restraints (TTAD #12706) ................................................ 609 11.11 Seismic norm PS92: verifying efforts and torsors on planar elements (TTAD #12974) ............................................. 609 11.12 EC8 Fr Annex: Generating forces results per modes on linear and planar elements (TTAD #13797) ....................... 610

STEEL DESIGN................................................................................................................................. 611 12.1

EC3 Test 2: Class section classification and share verification of an IPE300 beam subjected to linear uniform loading .......................................................................................................................................................... 612

12.2

EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 column ................ 619

12.3

EC3 Test 3: Class section classification and share and bending moment verification of an IPE300 column............. 626

12.4

EC3 Test 5: Class section classification and combined axial force with bending moment verification of an IPE300 column...................................................................................................................................................... 633

12.5

EC3 test 4: Class section classification and bending moment verification of an IPE300 column............................... 639

12.6

EC3 Test 1: Class section classification and compression verification of an IPE300 column .................................... 645

12.7

Generating the shape sheet by system (TTAD #11471) ............................................................................................ 651

12.8

Verifying the calculation results for steel cables (TTAD #11623) ............................................................................... 651

12.9

Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545) .......................................................... 651

12.10 Verifying the shape sheet results for a column (TTAD #11550)................................................................................. 651 12.11 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522) ................................................ 651 12.12 Verifying the cross section optimization according to EC3 (TTAD #11516) ............................................................... 652 12.13 Verifying shape sheet on S275 beam (TTAD #11731)............................................................................................... 652 12.14 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770) ........................................ 652 12.15 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533) ............................................ 652 12.16 Changing the steel design template for a linear element (TTAD #12491).................................................................. 652 12.17 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389) .......................................................................................................................................................... 653 12.18 EC3: Verifying the buckling length results (TTAD #11550) ........................................................................................ 653 12.19 EC3 fire verification: Verifying the work ratios after performing an optimization for steel profiles (TTAD #11975)..... 653 12.20 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method ........................................... 653 12.21 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method .............................................. 653 12.22 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method................................................. 654 12.23 Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873) .......................................................................................................................................................... 654 12.24 Verifying the buckling coefficient Xy on a class 2 section .......................................................................................... 654

TIMBER DESIGN .............................................................................................................................. 655 13.1

Modifying the "Design experts" properties for timber linear elements (TTAD #12259) .............................................. 656

13.2

Verifying the timber elements shape sheet (TTAD #12337) ...................................................................................... 656

ADVANCE DESIGN VALIDATION GUIDE

13.3

Verifying the units display in the timber shape sheet (TTAD #12445) ........................................................................656

13.4

EC5: Verifying the fire resistance of a timber purlin subjected to simple bending ......................................................656

13.5

EC5: Verifying a C24 timber beam subjected to shear force......................................................................................657

13.6

EC5: Verifying a timber column subjected to compression forces..............................................................................661

13.7

EC5: Verifying a timber beam subjected to combined bending and axial tension ......................................................665

13.8

EC5: Verifying a timber column subjected to tensile forces........................................................................................671

13.9

EC5: Shear verification for a simply supported timber beam......................................................................................674

13.10 EC5: Verifying a timber purlin subjected to oblique bending ......................................................................................675 13.11 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression ...679 13.12 EC5: Verifying a timber beam subjected to simple bending .......................................................................................680 13.13 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression ...................................................684 13.14 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes ..............................................689

XML TEMPLATE FILES ....................................................................................................................693 14.1

Loading a template with the properties of a planar element (DEV2012 #1.4) ............................................................694

14.2

Loading a template with the properties of a linear element (DEV2012 #1.4)..............................................................694

14.3

Saving the properties of a planar element as a template (DEV2012 #1.4).................................................................694

14.4

Saving the properties of a linear element as a template (DEV2012 #1.4) ..................................................................694

1 Finite Elements Analysis

ADVANCE DESIGN VALIDATION GUIDE

1.1

System of two bars with three hinges (01-0002SSLLB_FEM) Test ID: 2434 Test status: Passed

1.1.1 Description On a system of two bars (AC and BC) with three hinges, a punctual load in applied in point C. The vertical displacement in point C and the tensile stress on the bars are verified.

1.1.2 Background

1.1.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 09/89; Analysis type: linear static; Element type: linear. System of two bars with three hinges

Scale =1/33

0002SSLLB_FEM

4.5 00 m

0m 0 4.5

C Y Z

Units I. S. Geometry ■ ■ ■

Bars angle relative to horizontal: θ = 30°, Bars length: l = 4.5 m, -4 2 Bar section: A = 3 x 10 m .

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa.

Boundary conditions ■ ■

Outer: Hinged in A and B, Inner: Hinge on C

30°

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: Punctual load in C: F = -21 x 103 N. Internal: None.

1.1.2.2 Displacement of the model in C Reference solution uc = -3 x 10-3 m Finite elements modeling ■ ■ ■

Linear element: beam, imposed mesh, 21 nodes, 20 linear elements.

Displacement shape System of two bars with three hinges Displacement in C

Scale =1/33

0002SSLLB_FEM

1.1.2.3 Bars stresses Reference solutions σAC bar = 70 MPa σBC bar = 70 MPa Finite elements modeling ■ ■ ■

Linear element: beam, imposed mesh, 21 nodes, 20 linear elements.

ADVANCE DESIGN VALIDATION GUIDE

1.1.2.4 Shape of the stress diagram System of two bars with three hinges Bars stresses

Scale =1/34 0002SSLLB_FEM

1.1.2.5 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [cm]

-0.30

CM2

Sxx

Tensile stress on AC bar [MPa]

CM2

Sxx

Tensile stress on BC bar [MPa]

1.1.3 Calculated results Result name DZ

Result description Vertical displacement in point C [cm]

Value -0.299954 cm

Error 0.02%

Sxx

Tensile stress on AC bar [MPa]

69.9998 MPa

0.00%

Sxx

Tensile stress on BC bar [MPa]

69.9998 MPa

0.00%

ADVANCE DESIGN VALIDATION GUIDE

1.2

Slender beam with variable section (fixed-free) (01-0004SDLLB_FEM) Test ID: 2436 Test status: Passed

1.2.1 Description Verifies the first eigen mode frequencies for a slender beam with variable section, subjected to its own weight.

1.2.2 Background

1.2.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 09/89; Analysis type: modal analysis; Element type: linear. Slender beam with variable section (fixed-free)

Scale =1/4

01-0004SDLLB_FEM

Units I. S. Geometry ■ ■

Beam length: l = 1 m, Initial section (in A): ► Height: h1 = 0.04 m, ► Width: b1 = 0.04 m, ► Section: A1 = 1.6 x 10-3 m2, ► Flexure moment of inertia relative to z-axis: Iz1 = 2.1333 x 10-7 m4,

ADVANCE DESIGN VALIDATION GUIDE

■

Final section (in B): ► Height: h2 = 0.01 m, ► Width: b2 = 0.01 m, ► Section: A2 = 10-4 m2, ► Flexure moment of inertia relative to z-axis: Iz2 = 8.3333 x 10-10 m4.

Materials properties ■ ■

Longitudinal elastic modulus: E = 2 x 1011 Pa, Density: 7800 kg/m3.

Boundary conditions ■ ■

Outer: Fixed in A, Inner: None.

Loading ■ ■

External: None, Internal: None.

1.2.2.2 Eigen mode frequencies Reference solutions Precise calculation by numerical integration of the differential equation of beams bending (Euler-Bernoulli theories): 2 ∂2v ∂2v ∂ 2 (EIz 2 ) = -ρA ∂x ∂x2 ∂x

The result is: fi = λ1 23.289

where Iz and A vary with the abscissa.

1 h2 λ 2 2π i l λ2 73.9

E 12ρ λ3 165.23

λ4 299.7

Finite elements modeling ■ ■ ■

Linear element: variable beam, imposed mesh, 31 nodes, 30 linear elements.

λ5 478.1

ADVANCE DESIGN VALIDATION GUIDE

Eigen mode shapes

1.2.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

54.18

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

171.94

CM2

Eigen mode

Eigen mode 3 frequency [Hz]

384.4

CM2

Eigen mode

Eigen mode 4 frequency [Hz]

697.24

CM2

Eigen mode

Eigen mode 5 frequency [Hz]

1112.28

1.2.3 Calculated results Result name

Result description

Value

Error

Eigen mode 1 frequency [Hz]

54.01 Hz

-0.31%

Eigen mode 2 frequency [Hz]

170.58 Hz

-0.79%

Eigen mode 3 frequency [Hz]

378.87 Hz

-1.44%

Eigen mode 4 frequency [Hz]

681.31 Hz

-2.28%

Eigen mode 5 frequency [Hz]

1075.7 Hz

-3.29%

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1.3

Thin lozenge-shaped plate fixed on one side (alpha = 15 °) (01-0008SDLSB_FEM) Test ID: 2440 Test status: Passed

1.3.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozenge-shaped plate fixed on one side, subjected to its own weight only.

1.3.2 Background

1.3.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar. Thin lozenge-shaped plate fixed on one side

Scale =1/10 01-0008SDLSB_FEM

Units I. S. Geometry

■ ■

Thickness: t = 0.01 m, Side: a = 1 m,

■ ■

α = 15° Points coordinates: ► A(0;0;0) ► B(a;0;0) ► C ( 0.259a ; 0.966a ; 0 ) ► D ( 1.259a ; 0.966a ; 0 )

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Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■ ■

Outer: AB side fixed, Inner: None.

Loading ■ ■

External: None, Internal: None.

1.3.2.2 Eigen modes frequencies function by α angle Reference solution M. V. Barton formula for a lozenge of side "a" leads to the frequencies: fj =

1 ⋅ λi2 2π ⋅ a 2

Et 2 12ρ(1 − ν 2 )

where i = 1,2, or λi2 = g(α).

α = 15° λ1 3.601 λ22 8.872 M. V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method. 2

Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.

Eigen mode shapes

1.3.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

8.999

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

22.1714

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1.3.3 Calculated results

Result name

Result description

Value

Error

Eigen mode 1 frequency [Hz]

8.95 Hz

-0.54%

Eigen mode 2 frequency [Hz]

21.69 Hz

-2.17%

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1.4

Thin circular ring fixed in two points (01-0006SDLLB_FEM) Test ID: 2438 Test status: Passed

1.4.1 Description Verifies the first eigen modes frequencies for a thin circular ring fixed in two points, subjected to its own weight only.

1.4.2 Background

1.4.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 12/89; Analysis type: modal analysis, plane problem; Element type: linear. Thin circular ring fixed in two points

Scale =1/2 01-0006SDLLB_FEM

Units I. S. Geometry ■ ■ ■

Average radius of curvature: OA = OB = R = 0.1 m, Angular spacing between points A and B: 120° ; Rectangular straight section: ► Thickness: h = 0.005 m, ► Width: b = 0.010 m, ► Section: A = 5 x 10-5 m2, ► Flexure moment of inertia relative to the vertical axis: I = 1.042 x 10-10 m4,

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■

Point coordinates: ► O (0 ;0), ►

A (-0.05 3 ; -0.05),

►

B (0.05 3 ; -0.05).

Materials properties ■

Longitudinal elastic modulus: E = 7.2 x 1010 Pa

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 2700 kg/m3.

Boundary conditions ■ ■

Outer: Fixed at A and B, Inner: None.

Loading ■ ■

External: None, Internal: None.

1.4.2.2 Eigen mode frequencies Reference solutions The deformation of the fixed ring is calculated from the deformations of the free-free thin ring ■

Symmetrical mode: ►

u’i = i cos(iθ)

►

v’i = sin (iθ) 1-i2 θ’i = R sin (iθ)

►

■

Antisymmetrical mode: ►

u’i = i sin(iθ)

►

v’i = -cos (iθ) 1-i2 θ’i = R cos (iθ)

►

From Green’s method results: fj =

1 h λj ⋅ 2π R 2

E 12ρ

with a support angle of 120°.

i Symmetrical mode Antisymmetrical mode

1 4.8497 1.9832

Finite elements modeling ■ ■ ■

Linear element: beam, without meshing, 32 nodes, 32 linear elements.

2 14.7614 9.3204

3 23.6157 11.8490

4 21.5545

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Eigen mode shapes

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1.4.2.3 Theoretic results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency - 1 antisymmetric 1 [Hz]

235.3

CM2

Eigen mode

Eigen mode 2 frequency - 2 symmetric 1 [Hz]

575.3

CM2

Eigen mode

Eigen mode 3 frequency - 3 antisymmetric 2 [Hz]

1105.7

CM2

Eigen mode

Eigen mode 4 frequency - 4 antisymmetric 3 [Hz]

1405.6

CM2

Eigen mode

Eigen mode 5 frequency - 5 symmetric 2 [Hz]

1751.1

CM2

Eigen mode

Eigen mode 6 frequency - 6 antisymmetric 4 [Hz]

2557

CM2

Eigen mode

Eigen mode 7 frequency - 7 symmetric 3 [Hz]

2801.5

1.4.3 Calculated results

Result name

Result description Eigen mode 1 frequency - 1 antisymmetric 1 [Hz]

Value 236.32 Hz

Error 0.43%

Eigen mode 2 frequency - 2 symmetric 1 [Hz]

578.52 Hz

0.56%

Eigen mode 3 frequency - 3 antisymmetric 2 [Hz]

1112.54 Hz

0.62%

Eigen mode 4 frequency - 4 antisymmetric 3 [Hz]

1414.22 Hz

0.61%

Eigen mode 5 frequency - 5 symmetric 2 [Hz]

1760 Hz

0.51%

Eigen mode 6 frequency - 6 antisymmetric 4 [Hz]

2569.97 Hz

0.51%

Eigen mode 7 frequency - 7 symmetric 3 [Hz]

2777.43 Hz

-0.86%

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1.5

Thin lozenge-shaped plate fixed on one side (alpha = 45 °) (01-0010SDLSB_FEM) Test ID: 2442 Test status: Passed

1.5.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozenge-shaped plate fixed on one side, subjected to its own weight only.

1.5.2 Background

1.5.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar. Thin lozenge-shaped plate fixed on one side

Scale =1/10 01-0010SDLSB_FEM

Units I. S. Geometry ■ ■

Thickness: t = 0.01 m, Side: a = 1 m,

■

α = 45°

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■

Points coordinates: ► A(0;0;0) ► B(a;0;0) ►

2 2 a; a;0) 2 2

►

2+ 2 2 a; a;0) 2 2

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■ ■

Outer: AB side fixed, Inner: None.

Loading ■ ■

External: None, Internal: None.

1.5.2.2 Eigen mode frequencies relative to the α angle Reference solution M. V. Barton formula for a lozenge of side "a" leads to the frequencies: fj =

1 ⋅ λi2 2π ⋅ a 2

Et 2 12ρ(1 − ν 2 )

where i = 1,2, or λi2 = g(α).

α = 45° λ12 4.4502 λ22 10.56 M. V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.

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Eigen mode shapes

1.5.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

11.1212

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

26.3897

1.5.3 Calculated results

Result name

Result description Eigen mode 1 frequency [Hz]

Value 11.28 Hz

Error 1.43%

Eigen mode 2 frequency [Hz]

28.08 Hz

6.41%

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1.6

Vibration mode of a thin piping elbow in plane (case 2) (01-0012SDLLB_FEM) Test ID: 2444 Test status: Passed

1.6.1 Description Verifies the vibration modes of a thin piping elbow (1 m radius) extended by two straight elements of length L, subjected to its self weight only.

1.6.2 Background

1.6.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (plane problem); Element type: linear. Vibration mode of a thin piping elbow Case 2

Scale = 1/11 01-0012SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■ ■ 38

Average radius of curvature: OA = R = 1 m, L = 0.6 m, Straight circular hollow section: Outer diameter de = 0.020 m, Inner diameter di = 0.016 m, -4 2 Section: A = 1.131 x 10 m , Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, -9 4 Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10 m , -9 4 Polar inertia: Ip = 9.274 x 10 m .

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■

Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■

Outer: Fixed at points C and D At A: translation restraint along y and z, ► At B: translation restraint along x and z, Inner: None. ► ►

■

Loading ■ ■

External: None, Internal: None.

1.6.2.2 Eigen mode frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■ fj =

in plane bending:

λi

2π ⋅ R

EI z ρA

where i = 1,2,

Finite elements modeling ■ ■ ■

Linear element: beam, 23 nodes, 22 linear elements.

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Eigen mode shapes

1.6.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode frequency in plane 1 [Hz]

CM2

Eigen mode

Eigen mode frequency in plane 2 [Hz]

180

1.6.3 Calculated results

Result name

Result description Eigen mode frequency in plane 1 [Hz] Eigen mode frequency in plane 2 [Hz]

Value 94.62 Hz 184.68 Hz

Error 0.66% 100%

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1.7

Thin circular ring hanged on an elastic element (01-0014SDLLB_FEM) Test ID: 2446 Test status: Passed

1.7.1 Description Verifies the first eigen modes frequencies of a circular ring hanged on an elastic element, subjected to its self weight only.

1.7.2 Background

1.7.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 13/89; Analysis type: modal analysis, plane problem; Element type: linear. Thin circular ring hang from an elastic element

Scale = 1/1 01-0014SDLLB_FEM

Units I. S.

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Geometry ■ ■ ■

Average radius of curvature: OB = R = 0.1 m, Length of elastic element: AB = 0.0275 m ; Straight rectangular section: ► Ring Thickness: h = 0.005 m, Width: b = 0.010 m, Section: A = 5 x 10-5 m2, Flexure moment of relative to the vertical axis: I = 1.042 x 10-10 m4, ► Elastic element Thickness: h = 0.003 m,

Width: b = 0.010 m, Section: A = 3 x 10-5 m2, Flexure moment of inertia relative to the vertical axis: I = 2.25 x 10-11 m4, ■

Points coordinates: ► O ( 0 ; 0 ), ► A ( 0 ; -0.0725 ), ► B ( 0 ; -0.1 ).

Materials properties ■

Longitudinal elastic modulus: E = 7.2 x 1010 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 2700 kg/m3.

Boundary conditions ■ ■

Outer: Fixed in A, Inner: None.

Loading ■ ■

External: None, Internal: None.

1.7.2.2 Eigen mode frequencies Reference solutions The reference solution was established from experimental results of a mass manufactured aluminum ring. Finite elements modeling ■ ■ ■

Linear element: beam, 43 nodes, 43 linear elements.

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Eigen mode shapes

1.7.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 Asymmetrical frequency [Hz]

28.80

CM2

Eigen mode

Eigen mode 2 Symmetrical frequency [Hz]

189.30

CM2

Eigen mode

Eigen mode 3 Asymmetrical frequency [Hz]

268.80

CM2

Eigen mode

Eigen mode 4 Asymmetrical frequency [Hz]

641.00

CM2

Eigen mode

Eigen mode 5 Symmetrical frequency [Hz]

682.00

CM2

Eigen mode

Eigen mode 6 Asymmetrical frequency [Hz]

1063.00

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1.7.3 Calculated results

Result name

Result description

Value

Error

Eigen mode 1 Asymmetrical frequency [Hz]

28.81 Hz

0.03%

Eigen mode 2 Symmetrical frequency [Hz]

189.69 Hz

0.21%

Eigen mode 3 Asymmetrical frequency [Hz]

269.38 Hz

0.22%

Eigen mode 4 Asymmetrical frequency [Hz]

642.15 Hz

0.18%

Eigen mode 5 Symmetrical frequency [Hz]

683.9 Hz

0.28%

Eigen mode 6 Asymmetrical frequency [Hz]

1065.73 Hz

0.26%

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1.8

Tied (sub-tensioned) beam (01-0005SSLLB_FEM) Test ID: 2437 Test status: Passed

1.8.1 Description Verifies the tension force on a beam reinforced by a system of hinged bars, subjected to a uniform linear load.

1.8.2 Background

1.8.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 13/89; Analysis type: static, thermoelastic (plane problem); Element type: linear. Tied (sub-tensioned) beam

Scale =1/37 01-0005SSLLB_FEM

Units I. S. Geometry ■

Length: AD = FB = a = 2 m, ► DF = CE = b = 4 m, ► CD = EF = c = 0.6 m, ► AC = EB = d = 2.088 m, ► Total length: L = 8 m, ►

ADVANCE DESIGN VALIDATION GUIDE

■

■ ■ ■

AD, DF, FB Beams: ► Section: A = 0.01516 m2, ► Shear area: Ar = A / 2.5, ► Inertia moment: I = 2.174 x 10-4 m4, CE Bar: ► Section: A1 = 4.5 x 10-3 m2, AC, EB bar: ► Section: A2 = 4.5 x 10-3 m2, CD, EF bars: ► Section: A3 = 3.48 x 10-3 m2.

Materials properties ■ ■ ■

Isotropic linear elastic material, Longitudinal elastic modulus: E = 2.1 x 1011 Pa, Shearing module: G = 0.4x E.

Boundary conditions ■ ■

Outer: Hinged in A, support connection in B (blocked vertical translation), Inner: Hinged at bar ends: AC, CD, EF, EB.

Loading ■

External: Uniform linear load p = -50000 N/ml,

■

Internal: Shortening of the CE tie of δ = 6.52 x 10-3 m (dilatation coefficient: αCE = 1 x 10-5 /°C and temperature variation ΔT = -163°C).

1.8.2.2 Compression force in CE bar Reference solution The solution is established by considering the deformation effects due to the shear force and normal force: 4 a μ=1-3 xL A k = A = 2.5 r t=

I A

γ = (L/c)2 x (1+ (A/A1) x (b/L) + 2 x (A/A2) x (d/a)2 x (d/L) + 2 x (A/A3) (c/a)2 x (c/L) τ = k x [(2Et2) / (GaL)] ρ=μ+γ+τ μ0 = 1 – (a/L)2 x (2 – a/L) τ0 = 6k x (E/G) x (t/L)2 x (1 + b/L) ρ0 = μ0 + τ0 NCE = - (1/12) x (pL2/c) x (ρ0 /ρ) + (EI/(Lc2)) x (δ/ρ) = 584584 N Finite elements modeling Linear element: without meshing, ■ ■ ■ ■

AD, DF, FB: S beam (considering the shear force deformations), AC, CD, EF, EB: bar, CE: beam, 6 nodes.

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Force diagrams Tied (sub-tensioned) beam

Scale =1/31 Compression force in CE bar

1.8.2.3 Bending moment at point H Reference solution MH = - (1/8) x pL2 x [1- (2/3) x (ρ0/ρ)] – (EI/(Lc)) x (δ/p) = 49249.5 N Finite elements modeling Linear element: without meshing, ■ ■ ■ ■

AD, DF, FB: S beam (considering the shear force deformations), AC, CD, EF, EB: bar, CE: beam, 6 nodes.

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Shape of the bending moment diagram Tied (sub-tensioned) beam

Scale =1/31

Mz bending moment

1.8.2.4 Vertical displacement at point D Reference solution The reference displacement vD provided by AFNOR is determined by averaging the results of several software with implemented finite elements method. vD = -0.5428 x 10-3 m Finite elements modeling ■

Linear element: without meshing, AD, DF, FB: S beam (considering the shear force deformations), ► AC, CD, EF, EB: bar, ► CE: beam, 6 nodes. ►

■

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Deformed shape Tied (sub-tensioned) beam

Scale =1/31 Deformed

1.8.2.5 Theoretical results Solver

Result name

Result description

Reference value

CM2

Tension force on CE bar [N]

584584

1.8.3 Calculated results

Result name Fx

Result description Tension force on CE bar [N]

Value 584580 N

Error 0.00%

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1.9

Circular plate under uniform load (01-0003SSLSB_FEM) Test ID: 2435 Test status: Passed

1.9.1 Description On a circular plate of 5 mm thickness and 2 m diameter, an uniform load, perpendicular on the plan of the plate, is applied. The vertical displacement on the plate center is verified.

1.9.2 Background

1.9.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 03/89; Analysis type: linear static; Element type: planar. Circular plate under uniform load 01-0003SSLSB_FEM

Units I. S. Geometry ■ ■

Circular plate radius: r = 1m, Circular plate thickness: h = 0.005 m.

Materials properties

■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Scale =1/10

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Boundary conditions ■

Outer: Plate fixed on the side (in all points of its perimeter), For the modeling, we consider only a quarter of the plate and we impose symmetry conditions on some nodes (see the following model; yz plane symmetry condition):translation restrained nodes along x and rotation restrained nodes along y and z: translation restrained nodes along x and rotation restrained nodes along y and z:

■

Inner: None.

Loading ■ ■

External: Uniform loads perpendicular on the plate: pZ = -1000 Pa, Internal: None.

1.9.2.2 Vertical displacement of the model at the center of the plate Reference solution Circular plates form: pr4 -1000 x 14 u = 64D = 64 x 2404 = - 6.50 x 10-3 m with the plate radius coefficient: D =

2.1 x 1011 x 0.0053 Eh3 2 = 12(1-0.32) 12(1-ν )

D = 2404 Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 70 nodes, 58 planar elements. Circular plate under uniform load Meshing

Scale =1.5

01-0003SSLSB_FEM

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Deformed shape Circular plate under uniform load Deformed

Scale =1.5

01-0003SSLSB_FEM

1.9.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement on the plate center [mm]

-6.50

1.9.3 Calculated results

Result name DZ

Result description Vertical displacement on the plate center [mm]

Value -6.47032 mm

Error 0.46%

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1.10 Double fixed beam (01-0016SDLLB_FEM) Test ID: 2448 Test status: Passed

1.10.1 Description Verifies the eigen modes frequencies and the vertical displacement on the middle of a beam consisting of eight elements of length "l", having identical characteristics. A punctual load of -50000 N is applied.

1.10.2 Background

1.10.2.1 Model description ■ ■ ■

Reference: internal GRAITEC test (beams theory); Analysis type: static linear, modal analysis; Element type: linear.

Units I. S. Geometry ■ ■ ■

Length: l = 16 m, Axial section: S=0.06 m2 4 Inertia I = 0.0001 m

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7850 kg/m3

Boundary conditions ■ ■

Outer: Fixed at both ends x = 0 and x = 8 m, Inner: None.

Loading ■ ■

External: Punctual load P = -50000 N at x = 4m, Internal: None.

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1.10.2.2 Displacement of the model in the linear elastic range Reference solution The reference vertical displacement v5, is calculated at the middle of the beam at x = 2 m.

v5 =

Pl 3 50000 × 16 3 = = 0.05079 m 192EI 192 × 2.1E11× 0.0001

Finite elements modeling ■ ■ ■

Linear element: beam, imposed mesh, 9 nodes, 8 elements.

Deformed shape Double fixed beam Deformed

1.10.2.3 Eigen mode frequencies of the model in the linear elastic range Reference solution Knowing that the first four eigen mode frequencies of a double fixed beam are given by the following formula:

fn =

χ 2.π .L2 2 n

⎧ χ 12 ⎪ 2 E.I ⎪χ 2 where for the first 4 eigen modes frequencies ⎨ 2 ρ.S ⎪χ 3 ⎪ 2 ⎩χ 4

Finite elements modeling ■ ■ ■

Linear element: beam, imposed mesh, 9 nodes, 8 elements.

= 22.37 → f1 = 2.937 Hz = 61.67 → f 2 = 8.095 Hz = 120.9 → f 3 = 15.871 Hz = 199.8 → f 4 = 26.228 Hz

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Modal deformations Double fixed beam Mode 1

Double fixed beam Mode 2

Double fixed beam Mode 3

Double fixed beam Mode 4

ADVANCE DESIGN VALIDATION GUIDE

1.10.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement on the middle of the beam [m]

-0.05079

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

2.937

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

8.095

CM2

Eigen mode

Eigen mode 3 frequency [Hz]

15.870

CM2

Eigen mode

Eigen mode 4 frequency [Hz]

26.228

1.10.3 Calculated results

Result name Dz

Result description

Value

Error

Vertical displacement on the middle of the beam [m]

-0.0507937 m

-0.01%

Eigen mode 1 frequency [Hz]

2.94 Hz

0.10%

Eigen mode 2 frequency [Hz]

8.09 Hz

-0.06%

Eigen mode 3 frequency [Hz]

15.79 Hz

-0.50%

Eigen mode 4 frequency [Hz]

25.76 Hz

-1.78%

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1.11 Short beam on simple supports (eccentric) (01-0018SDLLB_FEM) Test ID: 2450 Test status: Passed

1.11.1 Description Verifies the first eigen mode frequencies of a short beam on simple supports (the supports are eccentric relative to the neutral axis).

1.11.2 Background

1.11.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; Analysis type: modal analysis, (plane problem); Element type: linear. Short beam on simple supports (eccentric)

Scale = 1/5 01-0018SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■

Height: h = 0.2m, Length: l = 1 m, Width: b = 0.1 m, -2 4 Section: A = 2 x 10 m , -5 4 Flexure moment of inertia relative to z-axis: Iz = 6.667 x 10 m .

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■

Outer: Hinged at A (null horizontal and vertical displacements), ► Simple support at B. Inner: None. ►

■

Loading ■ ■

External: None. Internal: None.

1.11.2.2 Eigen modes frequencies Reference solution The problem has no analytical solution, the solution is determined by averaging several software: Timoshenko model with shear force deformation effects and rotation inertia. The bending modes and the traction-compression are coupled. Finite elements modeling ■ ■ ■

Linear element: S beam, imposed mesh, 10 nodes, 9 linear elements.

Eigen modes shape Short beam on simple supports (eccentric) Mode 1

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Short beam on simple supports (eccentric) Mode 2

Short beam on simple supports (eccentric) Mode 3

Short beam on simple supports (eccentric) Mode 4

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Short beam on simple supports (eccentric) Mode 5

1.11.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

392.8

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

902.2

CM2

Eigen mode

Eigen mode 3 frequency [Hz]

1591.9

CM2

Eigen mode

Eigen mode 4 frequency [Hz]

2629.2

CM2

Eigen mode

Eigen mode 5 frequency [Hz]

3126.2

1.11.3 Calculated results

Result name

Result description

Value

Error

Eigen mode 1 frequency [Hz]

393.7 Hz

0.23%

Eigen mode 2 frequency [Hz]

945.35 Hz

4.78%

Eigen mode 3 frequency [Hz]

1595.94 Hz

0.25%

Eigen mode 4 frequency [Hz]

2526.22 Hz

-3.92%

Eigen mode 5 frequency [Hz]

3118.91 Hz

-0.23%

ADVANCE DESIGN VALIDATION GUIDE

1.12 Thin lozenge-shaped plate fixed on one side (alpha = 30 °) (01-0009SDLSB_FEM) Test ID: 2441 Test status: Passed

1.12.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozenge-shaped plate fixed on one side, subjected to its own weight only.

1.12.2 Background

1.12.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar. Thin lozenge-shaped plate fixed on one side

Scale =1/10 01-0009SDLSB_FEM

Units I. S. Geometry ■ ■

Thickness: t = 0.01 m, Side: a = 1 m,

■ ■

α = 30° Points coordinates: ► A(0;0;0) ► B(a;0;0)

ADVANCE DESIGN VALIDATION GUIDE

►

3 C ( 0.5a ; 2 a ; 0 )

►

3 D ( 1.5a ; 2 a ; 0 )

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■ ■

Outer: AB side fixed, Inner: None.

Loading ■ ■

External: None, Internal: None.

1.12.2.2 Eigen mode frequencies relative to the α angle Reference solution M. V. Barton formula for a lozenge of side "a" leads to the frequencies: fj =

1 ⋅ λi2 2 2π ⋅ a

Et 2 12ρ(1 − ν 2 )

where i = 1,2, or λi2 = g(α).

α = 30° λ12 3.961 λ22 10.19 M. V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.

Eigen mode shapes

ADVANCE DESIGN VALIDATION GUIDE

1.12.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

9.8987

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

25.4651

1.12.3 Calculated results

Result name

Result description

Value

Error

Eigen mode 1 frequency [Hz]

9.82 Hz

-0.80%

Eigen mode 2 frequency [Hz]

23.44 Hz

-7.95%

ADVANCE DESIGN VALIDATION GUIDE

1.13 Vibration mode of a thin piping elbow in plane (case 3) (01-0013SDLLB_FEM) Test ID: 2445 Test status: Passed

1.13.1 Description Verifies the vibration modes of a thin piping elbow (1 m radius) extended by two straight elements of length L, subjected to its self weight only.

1.13.2 Background

1.13.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (plane problem); Element type: linear. Vibration mode of a thin piping elbow Case 3

Scale = 1/12 01-0013SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■

Average radius of curvature: OA = R = 1 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, -4 2 Section: A = 1.131 x 10 m , -9 4 Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10 m , -9 4 Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10 m , -9 4 Polar inertia: Ip = 9.274 x 10 m .

ADVANCE DESIGN VALIDATION GUIDE

■

Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■

Outer: Fixed at points C and Ds, At A: translation restraint along y and z, ► At B: translation restraint along x and z, Inner: None. ► ►

■

Loading ■ ■

External: None, Internal: None.

1.13.2.2 Eigen mode frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■ fj =

in plane bending:

λi

2π ⋅ R

EI z ρA

where i = 1,2,

Finite elements modeling ■ ■ ■

Linear element: beam, 41 nodes, 40 linear elements.

ADVANCE DESIGN VALIDATION GUIDE

Eigen mode shapes

1.13.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode frequency in plane 1 [Hz]

25.300

CM2

Eigen mode

Eigen mode frequency in plane 2 [Hz]

27.000

1.13.3 Calculated results

Result name

Result description

Value

Error

Eigen mode frequency in plane 1 [Hz]

24.96 Hz

-1.34%

Eigen mode frequency in plane 2 [Hz]

26.71 Hz

100%

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1.14 Short beam on simple supports (on the neutral axis) (01-0017SDLLB_FEM) Test ID: 2449 Test status: Passed

1.14.1 Description Verifies the first eigen mode frequencies of a short beam on simple supports (the supports are located on the neutral axis), subjected to its own weight only.

1.14.2 Background ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; Analysis type: modal analysis (plane problem); Element type: linear. Short beam on simple supports on the neutral axis

Scale = 1/6 01-0017SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■

Height: h = 0.2 m, Length: l = 1 m, Width: b = 0.1 m, -2 4 Section: A = 2 x 10 m , Flexure moment of inertia relative to z-axis: Iz = 6.667 x 10-5 m4.

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Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■

■

Outer: ► Hinged at A (null horizontal and vertical displacements), ► Simple support in B. Inner: None.

Loading ■ ■

External: None. Internal: None.

1.14.2.1 Eigen modes frequencies Reference solution The bending beams equation gives, when superimposing, the effects of simple bending, shear force deformations and rotation inertia, Timoshenko formula. The reference eigen modes frequencies are determined by a numerical simulation of this equation, independent of any software. The eigen frequencies in tension-compression are given by: fi =

λi ⋅ 2πl

where λi =

(2i − 1) 2

Finite elements modeling ■ ■ ■

Linear element: S beam, imposed mesh, 10 nodes, 9 linear elements.

ADVANCE DESIGN VALIDATION GUIDE

Eigen mode shapes

1.14.2.2 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

431.555

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

1265.924

CM2

Eigen mode

Eigen mode 3 frequency [Hz]

1498.295

CM2

Eigen mode

Eigen mode 4 frequency [Hz]

2870.661

CM2

Eigen mode

Eigen mode 5 frequency [Hz]

3797.773

CM2

Eigen mode

Eigen mode 6 frequency [Hz]

4377.837

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1.14.3 Calculated results

Result name

Result description

Value

Error

Eigen mode 1 frequency [Hz]

437.12 Hz

1.29%

Eigen mode 2 frequency [Hz]

1264.32 Hz

-0.13%

Eigen mode 3 frequency [Hz]

1537.16 Hz

2.59%

Eigen mode 4 frequency [Hz]

2911.46 Hz

1.42%

Eigen mode 5 frequency [Hz]

3754.54 Hz

-1.14%

Eigen mode 6 frequency [Hz]

4281.23 Hz

-2.21%

ADVANCE DESIGN VALIDATION GUIDE

1.15 Thin lozenge-shaped plate fixed on one side (alpha = 0 °) (01-0007SDLSB_FEM) Test ID: 2439 Test status: Passed

1.15.1 Description Verifies the eigen modes frequencies for a 10 mm thick lozenge-shaped plate fixed on one side, subjected to its own weight only.

1.15.2 Background

1.15.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLS 02/89; Analysis type: modal analysis; Element type: planar. Thin lozenge-shaped plate fixed on one side

Scale =1/10 01-0007SDLSB_FEM

Units I. S. Geometry ■ ■

Thickness: t = 0.01 m, Side: a = 1 m,

■ ■

α = 0° Points coordinates: ► A(0;0;0) ► B(a;0;0) ► C(0;a;0) ► D(a;a;0)

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■ ■

Outer: AB side fixed, Inner: None.

Loading ■ ■

External: None, Internal: None.

1.15.2.2 Eigen mode frequencies relative to the α angle Reference solution M. V. Barton formula for a side "a" lozenge, leads to the frequencies: fj =

1 ⋅ λi2 2π ⋅ a 2

Et 2 12ρ(1 − ν 2 )

where i = 1,2, and λi2 = g(α).

α=0 λ12 3.492 λ22 8.525 M.V. Barton noted the sensitivity of the result relative to the mode and the α angle. He acknowledged that the λi values were determined with a limited development of an insufficient order, which led to consider a reference value that is based on an experimental result, verified by an average of seven software that use the finite elements calculation method. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 61 nodes, 900 surface quadrangles.

Eigen mode shapes

1.15.2.3 Theoretical results

Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

8.7266

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

21.3042

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1.15.3 Calculated results

Result name

Result description Eigen mode 1 frequency [Hz] Eigen mode 2 frequency [Hz]

Value 8.67 Hz 21.21 Hz

Error -0.65% -0.44%

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1.16 Vibration mode of a thin piping elbow in plane (case 1) (01-0011SDLLB_FEM) Test ID: 2443 Test status: Passed

1.16.1 Description Verifies the vibration modes of a thin piping elbow (1 m radius) with fixed ends and subjected to its self weight only.

1.16.2 Background

1.16.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (plane problem); Element type: linear. Vibration mode of a thin piping elbow in plane Case 1

Scale = 1/7 01-0011SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■

Average radius of curvature: OA = R = 1 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, Section: A = 1.131 x 10-4 m2, -9 4 Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10 m , Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4,

ADVANCE DESIGN VALIDATION GUIDE

■ ■

Polar inertia: Ip = 9.274 x 10-9 m4. Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0)

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■ ■

Outer: Fixed at points A and B , Inner: None.

Loading ■ ■

External: None, Internal: None.

1.16.2.2 Eigen mode frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■ fj =

in plane bending:

λi

2π ⋅ R

EI z ρA

where i = 1,2,

Finite elements modeling ■ ■ ■

Linear element: beam, 11 nodes, 10 linear elements.

Eigen mode shapes

1.16.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode frequency in plane 1 [Hz]

119

CM2

Eigen mode

Eigen mode frequency in plane 2 [Hz]

227

ADVANCE DESIGN VALIDATION GUIDE

1.16.3 Calculated results

Result name

Result description

Value

Error

Eigen mode frequency in plane 1 [Hz]

120.09 Hz

0.92%

Eigen mode frequency in plane 2 [Hz]

227.1 Hz

100%

ADVANCE DESIGN VALIDATION GUIDE

1.17 Double fixed beam with a spring at mid span (01-0015SSLLB_FEM) Test ID: 2447 Test status: Passed

1.17.1 Description Verifies the vertical displacement on the middle of a beam consisting of four elements of length "l", having identical characteristics. A punctual load of -10000 N is applied.

1.17.2 Background

1.17.2.1 Model description ■ ■ ■

Reference: internal GRAITEC test; Analysis type: linear static; Element type: linear.

Units I. S. Geometry ■ ■ ■

l=1m 2 S = 0.01 m 4 I = 0.0001 m

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■ ■

Outer: Fixed at ends x = 0 and x = 4 m,

■ ■

Elastic support with k = EI/l rigidity Inner: None.

Loading ■ ■

External: Punctual load P = -10000 N at x = 2m, Internal: None.

ADVANCE DESIGN VALIDATION GUIDE

1.17.2.2 Displacement of the model in the linear elastic range Reference solution The reference vertical displacement v3, is calculated at the middle of the beam at x = 2 m. Rigidity matrix of a plane beam:

Given the symmetry / X and load of the structure, it is unnecessary to consider the degrees of freedom associated with normal work (u2, u3, u4). The same symmetry allows the deduction of: ■

v2 = v4

■

β2 = -β4

■

β3 = 0

⎡ 12 ⎢ l3 ⎢ 6 ⎢ 2 ⎢ l ⎢ 12 ⎢− l 3 ⎢ 6 ⎢ 2 ⎢ l ⎢ ⎢ EI ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣

6 l2 4 l 6 − 2 l 2 l

12 l3 6 − 2 l 24 l3

−

0 −

12 l3 6 l2

6 l2 2 l 0 8 l 6 − 2 l 2 l

12 l3 6 − 2 l ⎛ 24 1 ⎞ ⎜ 3 + ⎟ l⎠ ⎝l −

0 −

12 l3 6 l2

6 l2 2 l 0 8 l 6 − 2 l 2 l

12 l3 6 − 2 l 24 l3

−

0 −

12 l3 6 l2

6 l2 2 l 0 8 l 6 − 2 l 2 l

12 l3 6 − 2 l 12 l3 6 − 2 l

−

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎧ v1 ⎫ ⎧ R1 ⎫ ⎥⎪ β ⎪ ⎪ M ⎪ ⎥⎪ 1 ⎪ ⎪ 1 ⎪ ⎥ ⎪v 2 ⎪ ⎪ 0 ⎪ (1) ⎪ ⎥⎪ ⎪ ⎪ ⎥ ⎪β 2 ⎪ ⎪ 0 ⎪ (2) ⎥ ⎪⎪v 3 ⎪⎪ ⎪⎪− P ⎪⎪ (3 ) ⎥⎨ ⎬ = ⎨ ⎬ ⎥ ⎪ β3 ⎪ ⎪ 0 ⎪ (4 ) ⎥ ⎪v ⎪ ⎪ 0 ⎪ (5 ) 4 ⎪ 6 ⎥⎪ ⎪ ⎪ ⎥ ⎪β 4 ⎪ ⎪ 0 ⎪ (6 ) 2 l ⎥⎪ ⎪ ⎪ ⎪ 2 ⎥ ⎪v 5 ⎪ ⎪ R5 ⎪ l ⎥ ⎪⎩ β5 ⎪⎭ ⎪⎩ M5 ⎪⎭ 6⎥ − 2⎥ l ⎥ 4 ⎥ l ⎥⎦

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[ ]

The elementary rigidity matrix of the spring in its local axis system, k 5 = the global axis system by means of the rotation matrix (90° rotation):

⎡0 0 ⎢ ⎢0 1 ⎢ [K 5 ] = EI ⎢0 0 l ⎢0 0 ⎢0 − 1 ⎢ ⎣⎢0 0 →

0 0 0 0 0

⎡ 1 − 1⎤ (U 3 ) , must be expressed in ⎢ ⎥ ⎣− 1 1 ⎦ (U 6 )

0⎤ (u 3 ) ⎥ − 1 0⎥ (v 3 ) 0 0⎥ (β 3 ) ⎥ 0 0⎥ (u 6 ) 1 0⎥ (v 6 ) ⎥ 0 0⎦⎥ (β 6 ) 0

6 2 8 3 v3 + β3 + β4 = 0 ⇒ β4 = − v3 2 l l 4l l

→ −

→

0 0

EI l

12 6 24 v 3 − 2 β 3 + 3 v 4 = 0 ⇒ 2v 4 = v 3 3 l l l

6 2 8 6 2 v + β 2 + β 3 − 2 v 4 + β 4 = 0 ⇒ v 4 = v 2 (usually unnecessary) 2 2 l l l l l

(3) → Finite elements modeling ■ ■ ■

Linear element: beam, imposed mesh, 6 nodes, 4 linear elements + 1 spring,

Deformed shape Double fixed beam with a spring at mid span Deformed

Note: the displacement is expressed here in μm

1.17.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement on the middle of the beam [mm]

-0.11905

1.17.3 Calculated results

Result name Dz

Result description Vertical displacement on the middle of the beam [mm]

Value -0.119048 mm

Error 0.00%

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1.18 Rectangular thin plate simply supported on its perimeter (01-0020SDLSB_FEM) Test ID: 2452 Test status: Passed

1.18.1 Description Verifies the first eigen mode frequencies of a thin rectangular plate simply supported on its perimeter.

1.18.2 Background

1.18.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLS 03/89; Analysis type: modal analysis; Element type: planar. Rectangular thin plate simply supported on its perimeter

Scale = 1/8 01-0020SDLSB_FEM

Units I. S. Geometry ■ ■ ■ ■

Length: a = 1.5 m, Width: b = 1 m, Thickness: t = 0.01 m, Points coordinates in m: ► A (0 ;0 ;0) ► B (0 ;1.5 ;0) ► C (1 ;1.5 ;0) ► D (1 ;0 ;0)

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■

Outer: Simple support on all sides, ► For the modeling: hinged at A, B and D. Inner: None. ►

■

Loading ■ ■

External: None. Internal: None.

1.18.2.2 Eigen modes frequencies Reference solution M. V. Barton formula for a rectangular plate with supports on all four sides, leads to: fij =

j π i [ ( )2 + ( )2 ] 2 a b

Et 2 12ρ(1 − ν 2 )

where: i = number of half-length of wave along y ( dimension a) j = number of half-length of wave along x ( dimension b) Finite elements modeling ■ ■ ■

Planar element: shell, 496 nodes, 450 planar elements.

ADVANCE DESIGN VALIDATION GUIDE

Eigen mode shapes

1.18.2.3 Theoretical results

Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 1; j = 1. [Hz]

35.63

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 2; j = 1. [Hz]

68.51

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 1; j = 2. [Hz]

109.62

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 3; j = 1. [Hz]

123.32

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 2; j = 2. [Hz]

142.51

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 3; j = 2. [Hz]

197.32

ADVANCE DESIGN VALIDATION GUIDE

1.18.3 Calculated results

Result name

Result description Eigen mode "i" - "j" frequency, for i = 1; j = 1 (Mode 1) [Hz]

Value 35.58 Hz

Error -0.14%

Eigen mode "i" - "j" frequency, for i = 2; j = 1 (Mode 2) [Hz]

68.29 Hz

-0.32%

Eigen mode "i" - "j" frequency, for i = 1; j = 2 (Mode 3) [Hz]

109.98 Hz

0.33%

Eigen mode "i" - "j" frequency, for i = 3; j = 1 (Mode 4) [Hz]

123.02 Hz

-0.24%

Eigen mode "i" - "j" frequency, for i = 2; j = 2 (Mode 5) [Hz]

141.98 Hz

-0.37%

Eigen mode "i" - "j" frequency, for i = 3; j = 2 (Mode 6) [Hz]

195.55 Hz

-0.90%

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1.19 Slender beam on two fixed supports (01-0024SSLLB_FEM) Test ID: 2456 Test status: Passed

1.19.1 Description A straight slender beam with fixed ends is loaded with a uniform load, several punctual loads and a torque. The shear force, bending moment, vertical displacement and horizontal reaction are verified.

1.19.2 Background

1.19.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 01/89; Analysis type: linear static; Element type: linear. Slender beam on two fixed supports

Scale = 1/4 01-0024SSLLB_FEM

Units I. S. Geometry ■ ■

Length: L = 1 m, Beam inertia: I = 1.7 x 10-8 m4.

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa.

Boundary conditions ■ ■

Outer: Fixed at A and B, Inner: None.

Loading ■

■

External: ► Uniformly distributed load from A to B: py = p = -24000 N/m, ► Punctual load at D: Fx = F1 = 30000 N, ► Torque at D: Cz = C = -3000 Nm, ► Punctual load at E: Fx = F2 = 10000 N, ► Punctual load at E: Fy = F = -20000 N. Internal: None.

1.19.2.2 Shear force at G Reference solution Analytical solution: ■

Shear force at G: VG

VG = 0.216F – 1.26

C L

Finite elements modeling ■ ■ ■

Linear element: beam, 5 nodes, 4 linear elements.

ADVANCE DESIGN VALIDATION GUIDE

Results shape Slender beam on two fixed supports

Scale = 1/5 Shear force

1.19.2.3 Bending moment in G Reference solution Analytical solution: ■

Bending moment at G: MG pL2 MG = 24 - 0.045LF – 0.3C

Finite elements modeling ■ ■ ■

Linear element: beam, 5 nodes, 4 linear elements.

ADVANCE DESIGN VALIDATION GUIDE

Results shape Slender beam on two fixed supports

Scale = 1/5 Bending moment

1.19.2.4 Vertical displacement at G Reference solution Analytical solution: ■

Vertical displacement at G: vG pl4 0.003375FL3 0.015CL2 vG = 384EI + + EI EI

Finite elements modeling ■ ■ ■

Linear element: beam, 5 nodes, 4 linear elements.

ADVANCE DESIGN VALIDATION GUIDE

Results shape Slender beam on two fixed supports

Scale = 1/4 Deformed

1.19.2.5 Horizontal reaction at A Reference solution Analytical solution: ■ Horizontal reaction at A: HA HA = -0.7F1 –0.3F2 Finite elements modeling ■ ■ ■

Linear element: beam, 5 nodes, 4 linear elements.

1.19.2.6 Theoretical results Solver

Result name

Result description

Reference value

CM2

Shear force in point G. [N]

-540

CM2

Bending moment in point G. [Nm]

-2800

CM2

Vertical displacement in point G. [cm]

-4.90

CM2

Horizontal reaction in point A. [N]

24000

1.19.3 Calculated results Result name Fz

Result description Shear force in point G [N]

Value -540 N

Error 0.00%

Bending moment in point G [Nm]

-2800 N*m

0.00%

Vertical displacement in point G [cm]

-4.90485 cm

-0.10%

Horizontal reaction in point A [N]

24000 N

0.00%

ADVANCE DESIGN VALIDATION GUIDE

1.20 Annular thin plate fixed on a hub (repetitive circular structure) (01-0022SDLSB_FEM) Test ID: 2454 Test status: Passed

1.20.1 Description Verifies the eigen mode frequencies of a thin annular plate fixed on a hub.

1.20.2 Background

1.20.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLS 04/89; Analysis type: modal analysis; Element type: planar element. Annular thin plate fixed on a hub (repetitive circular structure)

Scale = 1/3

01-0022SDLSB_FEM

Units I. S. Geometry ■ ■ ■

Inner radius: Ri = 0.1 m, Outer radius: Re = 0.2 m, Thickness: t = 0.001 m.

Material properties ■

Longitudinal elastic modulus: E = 2.1 x 10

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Pa,

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: Fixed on a hub at any point r = Ri. Inner: None.

Loading ■ ■

External: None. Internal: None.

1.20.2.2 Eigen modes frequencies Reference solution The solution of determining the frequency based on Bessel functions leads to the following formula: fij =

1 λ2 2πRe2 ij

Et2 12ρ(1-ν2)

where: i = the number of nodal diameters j = the number of nodal circles and λij2 such as: j \ i 0 1

0 13.0 85.1

1 13.3 86.7

2 14.7 91.7

3 18.5 100

Finite elements modeling ■ ■ ■

Planar element: plate, 360 nodes, 288 planar elements.

1.20.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 0; j = 0. [Hz]

79.26

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 0; j = 1. [Hz]

518.85

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 1; j = 0. [Hz]

81.09

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 1; j = 1. [Hz]

528.61

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 2; j = 0. [Hz]

89.63

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 2; j = 1. [Hz]

559.09

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 3; j = 0. [Hz]

112.79

CM2

Eigen mode

Eigen mode “i" - “j” frequency, for i = 3; j = 1. [Hz]

609.70

1.20.3 Calculated results Result name

Result description Eigen mode "i" - "j" frequency, for i = 0; j = 0 (Mode 1) [Hz]

Value 79.05 Hz

Error -0.26%

Eigen mode "i" - "j" frequency, for i = 0; j = 1 (Mode 18) [Hz]

521.84 Hz

0.58%

Eigen mode "i" - "j" frequency, for i = 1; j = 0 (Mode 2) [Hz]

80.52 Hz

-0.70%

Eigen mode "i" - "j" frequency, for i = 1; j = 1 (Mode 20) [Hz]

529.49 Hz

0.17%

Eigen mode "i" - "j" frequency, for i = 2; j = 0 (Mode 4) [Hz]

88.43 Hz

-1.34%

Eigen mode “i" - “j” frequency, for i = 2; j = 1 (Mode 22) [Hz]

552.43 Hz

-1.19%

Eigen mode "i" - "j" frequency, for i = 3; j = 0 (Mode 7) [Hz]

110.27 Hz

-2.23%

Eigen mode "i" - "j" frequency, for i = 3; j = 1 (Mode 25) [Hz]

593.83 Hz

-2.60%

ADVANCE DESIGN VALIDATION GUIDE

1.21 Bimetallic: Fixed beams connected to a stiff element (01-0026SSLLB_FEM) Test ID: 2458 Test status: Passed

1.21.1 Description Two beams fixed at one end and rigidly connected to an undeformable beam is loaded with a punctual load. The deflection, vertical reaction and bending moment are verified in several points.

1.21.2 Background

1.21.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 05/89; Analysis type: linear static; Element type: linear. Fixed beams connected to a stiff element 01-0026SSLLB_FEM

Scale = 1/10

Units I. S. Geometry ■

■ ■

Lengths: ► L = 2 m, ► l = 0.2 m, -8 4 Beams inertia moment: I = (4/3) x 10 m , The beam sections are squared, of side: 2 x 10-2 m.

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2 x 1011 Pa.

Boundary conditions ■ ■

Outer: Fixed in A and C, Inner: The tangents to the deflection of beams AB and CD at B and D remain horizontal; practically, we restraint translations along x and z at nodes B and D.

Loading ■ ■

External: In D: punctual load F = Fy = -1000N. Internal: None.

1.21.2.2 Deflection at B and D Reference solution The theory of slender beams bending (Euler-Bernouilli formula) leads to a deflection at B and D: The resolution of the hyperstatic system of the slender beam leads to: FL3 vB = vD = 24EI Finite elements modeling ■ ■ ■

Linear element: beam, 4 nodes, 3 linear elements.

Results shape Fixed beams connected to a stiff element Deformed

Scale = 1/10

ADVANCE DESIGN VALIDATION GUIDE

1.21.2.3 Vertical reaction at A and C Reference solution Analytical solution. Finite elements modeling ■ ■ ■

Linear element: beam, 4 nodes, 3 linear elements.

1.21.2.4 Bending moment at A and C Reference solution Analytical solution. Finite elements modeling ■ ■ ■

Linear element: beam, 4 nodes, 3 linear elements

1.21.2.5 Theoretical results Solver

Result name

Result description

Reference value

CM2

Deflection in point B [m]

0.125

CM2

Deflection in point D [m]

0.125

CM2

Vertical reaction in point A [N]

-500

CM2

Vertical reaction in point C [N]

-500

CM2

Bending moment in point A [Nm]

500

CM2

Bending moment in point C [Nm]

500

1.21.3 Calculated results

Result name D

Result description

Value

Error

Deflection in point B [m]

0.125376 m

0.30%

Deflection in point D [m]

0.125376 m

0.30%

Vertical reaction in point A [N]

-500 N

0.00%

Vertical reaction in point C [N]

-500 N

0.00%

Bending moment in point A [Nm]

500.083 N*m

0.02%

Bending moment in point C [Nm]

500.083 N*m

0.02%

ADVANCE DESIGN VALIDATION GUIDE

1.22 Cantilever beam in Eulerian buckling (01-0021SFLLB_FEM) Test ID: 2453 Test status: Passed

1.22.1 Description Verifies the critical load result on node 5 of a cantilever beam in Eulerian buckling. A punctual load of -100000 is applied.

1.22.2 Background

1.22.2.1 Model description ■ ■ ■

Reference: internal GRAITEC test (Euler theory); Analysis type: Eulerian buckling; Element type: linear.

Units I. S. Geometry ■ ■ ■

L = 10 m 2 S=0.01 m 4 I = 0.0002 m

Materials properties ■

Longitudinal elastic modulus: E = 2.0 x 10

■

Poisson's ratio: ν = 0.1.

N/m2,

Boundary conditions ■ ■

Outer: Fixed at end x = 0, Inner: None.

Loading ■ ■

External: Punctual load P = -100000 N at x = L, Internal: None.

ADVANCE DESIGN VALIDATION GUIDE

1.22.2.2 Critical load on node 5 Reference solution The reference critical load established by Euler is:

Pcritique =

98696 π 2EI = 98696 N ⇒ λ = = 0.98696 2 100000 4L

Finite elements modeling ■ ■ ■

Planar element: beam, imposed mesh, 5 nodes, 4 elements.

Deformed shape

1.22.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Critical load on node 5. [N]

-98696

1.22.3 Calculated results

Result name Fx

Result description Critical load on node 5 (mode 1) [N]

Value -100000 N

Error -1.32%

ADVANCE DESIGN VALIDATION GUIDE

1.23 Slender beam on three supports (01-0025SSLLB_FEM) Test ID: 2457 Test status: Passed

1.23.1 Description A straight slender beam on three supports is loaded with two punctual loads. The bending moment, vertical displacement and reaction on the center are verified.

1.23.2 Background

1.23.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 03/89; Analysis type: static (plane problem); Element type: linear. Slender beam on three supports

Scale = 1/49 01-0025SSLLB_FEM

Units I. S. Geometry ■ ■

Length: L = 3 m, Beam inertia: I = 6.3 x 10-4 m4.

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Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 Pa. Boundary conditions ■

Outer: Hinged at A, ► Elastic support at B (Ky = 2.1 x 106 N/m), ► Simple support at C. Inner: None. ►

■

Loading ■ ■

External: 2 punctual loads F = Fy = -42000N. Internal: None.

1.23.2.2 Bending moment at B Reference solution The resolution of the hyperstatic system of the slender beam leads to: k= ■

6EI L3Ky Bending moment at B: MB

MB = ±

L ( −6 + 2k )F (8 + k ) 2

Finite elements modeling ■ ■ ■

Linear element: beam, 5 nodes, 4 linear elements.

Results shape Slender beam on three supports

Scale = 1/49 Bending moment

ADVANCE DESIGN VALIDATION GUIDE

1.23.2.3 Reaction in B Reference solution ■

Compression force in the spring: VB -11F VB = 8 + k Finite elements modeling ■ ■ ■

Linear element: beam, 5 nodes, 4 linear elements.

1.23.2.4 Vertical displacement at B Reference solution ■

Deflection at the spring location: vB 11F vB = Ky(8 + k)

Finite elements modeling ■ ■ ■

Linear element: beam, 5 nodes, 4 linear elements.

Results shape Slender beam on three supports Deformed

1.23.2.5 Theoretical results

Solver

Result name

Result description

Reference value

CM2

Bending moment in point B. [Nm]

-63000

CM2

Vertical displacement in point B. [cm]

-1.00

CM2

Reaction in point B. [N]

-21000

ADVANCE DESIGN VALIDATION GUIDE

1.23.3 Calculated results

Result name My

Result description Bending moment in point B [Nm]

Value -63000 N*m

Error 0.00%

Vertical displacement in point B [cm]

-1 cm

0.00%

Reaction in point B [N]

-21000 N

0.00%

ADVANCE DESIGN VALIDATION GUIDE

1.24 Fixed thin arc in out of plane bending (01-0028SSLLB_FEM) Test ID: 2460 Test status: Passed

1.24.1 Description An arc of a circle fixed at one end is loaded with a punctual force at its free end, perpendicular to the plane. The out of plane displacement, torsion moment and bending moment are verified.

1.24.2 Background

1.24.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 07/89; Analysis type: static linear; Element type: linear. Fixed thin arc in out of plane bending 01-0028SSLLB_FEM

Units I. S.

100

Scale = 1/6

ADVANCE DESIGN VALIDATION GUIDE

Geometry ■ ■

Medium radius: R = 1 m , Circular hollow section: ► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.

Materials properties ■

Longitudinal elastic modulus: E = 2 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■ ■

Outer: Fixed at A. Inner: None.

Loading ■ ■

External: Punctual force in B perpendicular on the plane: Fz = F = 100 N. Internal: None.

1.24.2.2 Displacements at B Reference solution Displacement out of plane at point B: FR3 π EIx 3π uB = EI [ 4 + K ( 4 - 2)] x T where KT is the torsional rigidity for a circular section (torsion constant is 2Ix). KT = 2GIx =

EIx 1+ν

FR3 π 3π ⇒ uB = EI [ 4 + (1 + ν) ( 4 - 2)] x

Finite elements modeling ■ ■ ■

Linear element: beam, 46 nodes, 45 linear elements.

1.24.2.3 Moments at θ = 15° Reference solution ■

Torsion moment: Mx’ = Mt = FR(1 - sinθ)

■

Bending moment: Mz’ = Mf = -FRcosθ

Finite elements modeling ■ ■ ■

Linear element: beam, 46 nodes, 45 linear elements.

101

ADVANCE DESIGN VALIDATION GUIDE

1.24.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Displacement out of plane in point B [m]

0.13462

CM2

Torsion moment in θ = 15° [Nm]

74.1180

CM2

Bending moment in θ = 15° [Nm]

-96.5925

1.24.3 Calculated results

Result name D

102

Result description

Value

Error

Displacement out of plane in point B [m]

0.135156 m

0.40%

Torsion moment in Theta = 15° [Nm]

74.103 N*m

-0.02%

Bending moment in Theta = 15° [Nm]

-96.5925 N*m

0.00%

ADVANCE DESIGN VALIDATION GUIDE

1.25 Portal frame with lateral connections (01-0030SSLLB_FEM) Test ID: 2462 Test status: Passed

1.25.1 Description Verifies the rotation about z-axis and the bending moment on a portal frame with lateral connections.

1.25.2 Background

1.25.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 10/89; Analysis type: static linear; Element type: linear. Portal frame with lateral connections 01-0030SSLLB_FEM

Scale = 1/21

Units I. S. Geometry Beam AB

Length lAB = 4 m

lAC = 1 m

lAD = 1 m

lAE = 2 m

Moment of inertia 64 IAB = 3 x 10-8 m4 1 IAC = 12 x 10-8 m4 1 IAD = 12 x 10-8 m4 4 -8 4 IAE = 3 x 10 m

103

ADVANCE DESIGN VALIDATION GUIDE

■ ■

G is in the middle of DA. The beams have square sections: ► AAB = 16 x 10-4 m ► AAD = 1 x 10-4 m ► AAC = 1 x 10-4 m ► AAE = 4 x 10-4 m

Materials properties Longitudinal elastic modulus: E = 2 x 1011 Pa, Boundary conditions ■

Outer: Fixed at B, D and E, Hinge at C, Inner: None.

► ►

■

Loading ■

External: Punctual force at G: Fy = F = - 105 N, ► Distributed load on beam AD: p = - 103 N/m. Internal: None. ►

■

1.25.2.2 Displacements at A Reference solution Rotation at A about z-axis: EIAn We say: kAn = l An

where n = B, C, D or E

3 K = kAB + kAD + kAE + 4 kAC kAn rAn = K FlAD plAB2 C1 = 8 - 12 C1 θ = 4K Finite elements modeling ■ ■ ■

104

Linear element: beam, 6 nodes, 5 linear elements.

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Displacements shape Portal frame with lateral connections Deformed

1.25.2.3 Moments in A Reference solution ■

plAB2 MAB = 12 + rAB x C1

■

FlAD MAD = - 8 + rAD x C1

■ ■

MAE = rAE x C1 MAC = rAC x C1

Finite elements modeling ■ ■ ■

Linear element: beam, 6 nodes, 5 linear elements

1.25.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Rotation θ about z-axis in point A [rad]

-0.227118

CM2

Bending moment in point A (MAB) [Nm]

11023.72

CM2

Bending moment in point A (MAC) [Nm]

113.559

CM2

Bending moment in point A (MAD) [Nm]

12348.588

CM2

Bending moment in point A (MAE) [Nm]

1211.2994

1.25.3 Calculated results Result name RY

Result description Rotation Theta about z-axis in point A [rad]

Value -0.227401 Rad

Error -0.12%

Bending moment in point A (Moment AB) [Nm]

11021 N*m

-0.02%

Bending moment in point A (Moment AC) [Nm]

113.704 N*m

0.13%

Bending moment in point A (Moment AD) [Nm]

12347.5 N*m

-0.01%

Bending moment in point A (Moment AE) [Nm]

1212.77 N*m

0.12%

105

ADVANCE DESIGN VALIDATION GUIDE

1.26 Double hinged thin arc in planar bending (01-0029SSLLB_FEM) Test ID: 2461 Test status: Passed

1.26.1 Description Verifies the rotation about Z-axis, the vertical displacement and the horizontal displacement on several points of a double hinged thin arc in planar bending.

1.26.2 Background

1.26.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 08/89; Analysis type: static linear (plane problem); Element type: linear. Double hinged thin arc in planar bending 01-0029SSLLB_FEM

Units I. S. Geometry ■ ■

106

Medium radius: R = 1 m , Circular hollow section: ► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.

Scale = 1/8

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

Outer: Hinge at A, At B: allowed rotation along z, vertical displacement restrained along y. Inner: None.

► ►

■

Loading ■ ■

External: Punctual load at C: Fy = F = - 100 N. Internal: None.

1.26.2.2 Displacements at A, B and C Reference solution ■

Rotation about z-axis

π FR2 θA = - θB = ( 2 - 1) 2EI ■

Displacement;

π FR 3π FR3 Vertical at C: vC = 8 EA + ( 4 - 2) 2EI FR FR3 Horizontal at B: uB = 2EA - 2EI Finite elements modeling ■ ■ ■

Linear element: beam, 37 nodes, 36 linear elements.

Displacements shape Fixed thin arc in planar bending

Scale = 1/11 Deformed

107

ADVANCE DESIGN VALIDATION GUIDE

1.26.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Rotation about Z-axis in point A [rad]

0.030774

CM2

Rotation about Z-axis in point B [rad]

-0.030774

CM2

Vertical displacement in point C [cm]

-1.9206

CM2

Horizontal displacement in point B [cm]

5.3912

1.26.3 Calculated results

Result name RY

108

Result description

Value

Error

Rotation about Z-axis in point A [rad]

0.0307785 Rad

0.01%

Rotation about Z-axis in point B [rad]

-0.0307785 Rad

-0.01%

Vertical displacement in point C [cm]

-1.92019 cm

0.02%

Horizontal displacement in point B [cm]

5.386 cm

-0.10%

ADVANCE DESIGN VALIDATION GUIDE

1.27 Thin square plate fixed on one side (01-0019SDLSB_FEM) Test ID: 2451 Test status: Passed

1.27.1 Description Verifies the first eigen modes frequencies of a thin square plate fixed on one side.

1.27.2 Background

1.27.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLS 01/89; Analysis type: modal analysis; Element type: planar. Thin square plate fixed on one side

Scale = 1/6 01-0019SDLSB_FEM

Units I. S. Geometry ■ ■ ■

Side: a = 1 m, Thickness: t = 1 m, Points coordinates in m: ► A (0 ;0 ;0) ► B (1 ;0 ;0) ► C (1 ;1 ;0) ► D (0 ;1 ;0)

109

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■ ■

Outer: Edge AD fixed. Inner: None.

Loading ■ ■

External: None. Internal: None.

1.27.2.2 Eigen modes frequencies Reference solution M. V. Barton formula for a square plate with side "a", leads to: fj =

1 λi2 2 2π ⋅ a i λi

Et 2 12ρ(1 − ν 2 ) 1 3.492

Finite elements modeling ■ ■ ■

110

Planar element: shell, 959 nodes, 900 planar elements.

where i = 1,2, . . . 2 8.525

3 21.43

4 27.33

5 31.11

6 54.44

ADVANCE DESIGN VALIDATION GUIDE

Eigen mode shapes Thin square plate fixed on one side Mode 1

Thin square plate fixed on one side Mode 2

Thin square plate fixed on one side Mode 3

111

ADVANCE DESIGN VALIDATION GUIDE

1.27.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 frequency [Hz]

8.7266

CM2

Eigen mode

Eigen mode 2 frequency [Hz]

21.3042

CM2

Eigen mode

Eigen mode 3 frequency [Hz]

53.5542

CM2

Eigen mode

Eigen mode 4 frequency [Hz]

68.2984

CM2

Eigen mode

Eigen mode 5 frequency [Hz]

77.7448

CM2

Eigen mode

Eigen mode 6 frequency [Hz]

136.0471

1.27.3 Calculated results

Result name

112

Result description

Value

Error

Eigen mode 1 frequency [Hz]

8.67 Hz

-0.65%

Eigen mode 2 frequency [Hz]

21.22 Hz

-0.40%

Eigen mode 3 frequency [Hz]

53.13 Hz

-0.79%

Eigen mode 4 frequency [Hz]

67.74 Hz

-0.82%

Eigen mode 5 frequency [Hz]

77.15 Hz

-0.77%

Eigen mode 6 frequency [Hz]

134.65 Hz

-1.03%

ADVANCE DESIGN VALIDATION GUIDE

1.28 Bending effects of a symmetrical portal frame (01-0023SDLLB_FEM) Test ID: 2455 Test status: Passed

1.28.1 Description Verifies the first eigen mode frequencies of a symmetrical portal frame with fixed supports.

1.28.2 Background

1.28.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 01/89; Analysis type: modal analysis; Element type: linear. Bending effects of a symmetrical portal frame

Scale = 1/5 01-0023SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■

Straight rectangular sections for beams and columns: Thickness: h = 0.0048 m, Width: b = 0.029 m, -4 2 Section: A = 1.392 x 10 m , Flexure moment of inertia relative to z-axis: Iz = 2.673 x 10-10 m4,

113

ADVANCE DESIGN VALIDATION GUIDE

■

Points coordinates in m: A -0.30 0

x y

B 0.30 0

C -0.30 0.36

D 0.30 0.36

E -0.30 0.81

F 0.30 0.81

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■ ■

Outer: Fixed at A and B, Inner: None.

Loading ■ ■

External: None. Internal: None.

1.28.2.2 Eigen modes frequencies Reference solution Dynamic radius method (slender beams theory). Finite elements modeling ■ ■ ■

Linear element: beam, 60 nodes, 60 linear elements.

Deformed shape Bending effects of a symmetrical portal frame Mode 13

114

Scale = 1/7

ADVANCE DESIGN VALIDATION GUIDE

1.28.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode 1 antisymmetric frequency [Hz]

8.8

CM2

Eigen mode

Eigen mode 2 antisymmetric frequency [Hz]

29.4

CM2

Eigen mode

Eigen mode 3 symmetric frequency [Hz]

43.8

CM2

Eigen mode

Eigen mode 4 symmetric frequency [Hz]

56.3

CM2

Eigen mode

Eigen mode 5 antisymmetric frequency [Hz]

96.2

CM2

Eigen mode

Eigen mode 6 symmetric frequency [Hz]

102.6

CM2

Eigen mode

Eigen mode 7 antisymmetric frequency [Hz]

147.1

CM2

Eigen mode

Eigen mode 8 symmetric frequency [Hz]

174.8

CM2

Eigen mode

Eigen mode 9 antisymmetric frequency [Hz]

178.8

CM2

Eigen mode

Eigen mode 10 antisymmetric frequency [Hz]

206

CM2

Eigen mode

Eigen mode 11 symmetric frequency [Hz]

266.4

CM2

Eigen mode

Eigen mode 12 antisymmetric frequency [Hz]

320

CM2

Eigen mode

Eigen mode 13 symmetric frequency [Hz]

335

1.28.3 Calculated results

Result name

Result description Eigen mode 1 antisymmetric frequency [Hz]

Value 8.78 Hz

Error -0.23%

Eigen mode 2 antisymmetric frequency [Hz]

29.43 Hz

0.10%

Eigen mode 3 symmetric frequency [Hz]

43.85 Hz

0.11%

Eigen mode 4 symmetric frequency [Hz]

56.3 Hz

0.00%

Eigen mode 5 antisymmetric frequency [Hz]

96.05 Hz

-0.16%

Eigen mode 6 symmetric frequency [Hz]

102.7 Hz

0.10%

Eigen mode 7 antisymmetric frequency [Hz]

147.08 Hz

-0.01%

Eigen mode 8 symmetric frequency [Hz]

174.96 Hz

0.09%

Eigen mode 9 antisymmetric frequency [Hz]

178.92 Hz

0.07%

Eigen mode 10 antisymmetric frequency [Hz]

206.23 Hz

0.11%

Eigen mode 11 symmetric frequency [Hz]

266.62 Hz

0.08%

Eigen mode 12 antisymmetric frequency [Hz]

319.95 Hz

-0.02%

Eigen mode 13 symmetric frequency [Hz]

334.96 Hz

-0.01%

115

ADVANCE DESIGN VALIDATION GUIDE

1.29 Fixed thin arc in planar bending (01-0027SSLLB_FEM) Test ID: 2459 Test status: Passed

1.29.1 Description Arc of a circle fixed at one end, subjected to two punctual loads and a torque at its free end. The horizontal displacement, vertical displacement and rotation about Z-axis are verified.

1.29.2 Background

1.29.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 06/89; Analysis type: static linear (plane problem); Element type: linear. Fixed thin arc in planar bending 01-0027SSLLB_FEM

Units I. S. Geometry ■ ■

116

Medium radius: R = 3 m , Circular hollow section: ► de = 0.02 m, ► di = 0.016 m, ► A = 1.131 x 10-4 m2, ► Ix = 4.637 x 10-9 m4.

Scale = 1/24

ADVANCE DESIGN VALIDATION GUIDE

Materials properties Longitudinal elastic modulus: E = 2 x 1011 Pa. Boundary conditions ■ ■

Outer: Fixed in A. Inner: None.

Loading ■

External: At B: punctual load F1 = Fx = 10 N, punctual load F2 = Fy = 5 N, ► bending moment about Oz, Mz = 8 Nm. Internal: None. ► ►

■

1.29.2.2 Displacements at B Reference solution At point B: ■

R2 displacement parallel to Ox: u = 4EI [F1πR + 2F2R + 4Mz]

■

R2 displacement parallel to Oy: v = 4EI [2F1πR + (3π - 8)F2R + 2(π - 2)Mz]

■

R rotation around Oz: θ = 4EI [4F1R + 2(π - 2)F2R + 2πMz]

Finite elements modeling ■ ■ ■

Linear element: beam, 31 nodes, 30 linear elements.

117

ADVANCE DESIGN VALIDATION GUIDE

Results shape Fixed thin arc in planar bending

Scale = 1/19 Deformed

1.29.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Horizontal displacement in point B [m]

0.3791

CM2

Vertical displacement in point B [m]

0.2417

CM2

Rotation about Z-axis in point B [rad]

-0.1654

1.29.3 Calculated results

Result name DX

118

Result description

Value

Error

Horizontal displacement in point B [m]

0.378914 m

-0.05%

Vertical displacement in point B [m]

0.241738 m

0.02%

Rotation about Z-axis in point B [rad]

-0.165362 Rad

0.02%

ADVANCE DESIGN VALIDATION GUIDE

1.30 Beam on elastic soil, hinged ends (01-0034SSLLB_FEM) Test ID: 2466 Test status: Passed

1.30.1 Description A beam under a punctual load, a distributed load and two torques lays on a soil of constant linear stiffness. The rotation around z-axis, the vertical reaction, the vertical displacement and the bending moment are verified in several points.

1.30.2 Background

1.30.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 16/89; Analysis type: static linear (plane problem); Element type: linear. Beam on elastic soil, hinged ends 01-0034SSLLB_FEM

Scale = 1/27

Units I. S. Geometry ■ ■

L = (π 10 )/2, I = 10-4 m4.

119

ADVANCE DESIGN VALIDATION GUIDE

Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 Pa. Boundary conditions ■

Outer: Free A and B ends, ► Soil with a constant linear stiffness ky = K = 840000 N/m2. Inner: None. ►

■

Loading ■

External: Punctual force at D: Fy = F = - 10000 N, ► Uniformly distributed force from A to B: fy = p = - 5000 N/m, ► Torque at A: Cz = -C = -15000 Nm, ► Torque at B: Cz = C = 15000 Nm. Internal: None. ►

■

1.30.2.2 Displacement and support reaction at A Reference solution β=

K/(4EI)

ϕ = βL/2 λ = ch(2ϕ) + cos(2ϕ) ■

Vertical support reaction: VA = -p(sh(2ϕ) + sin(2ϕ)) - 2βFch(ϕ)cos(ϕ) + 2β2C(sh(2ϕ) - sin(2ϕ)) x

■

Rotation about z-axis: θA = p(sh(2ϕ) – sin(2ϕ)) + 2βFsh(ϕ)sin(ϕ) - 2β2C(sh(2ϕ) + sin(2ϕ)) x

Finite elements modeling ■ ■ ■

120

1 2βλ

Linear element: beam, 50 nodes, 49 linear elements.

1 (K/β)λ

ADVANCE DESIGN VALIDATION GUIDE

Deformed shape Beam on elastic soil, hinged ends

Scale = 1/20 Deformed

1.30.2.3 Displacement and bending moment at D Reference solution ■

Vertical displacement: vD = 2p(λ - 2ch(ϕ)cos(ϕ)) + βF(sh(2ϕ) – sin(2ϕ)) - 8β2Csh(ϕ)sin(ϕ) x

■

1 2Kλ

Bending moment: MD = 4psh(ϕ)sin(ϕ) + βF(sh(2ϕ) + sin(2ϕ)) - 8β2Cch(ϕ)cos(ϕ) x

1 4β2λ

Finite elements modeling ■ ■ ■

Linear element: beam, 50 nodes, 49 linear elements.

121

ADVANCE DESIGN VALIDATION GUIDE

Bending moment diagram Beam on elastic soil, hinged ends

Scale = 1/20 Bending moment

1.30.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Rotation around z-axis in point A [rad]

0.003045

CM2

Vertical reaction in point A [N]

-11674

CM2

Vertical displacement in point D [cm]

-0.423326

CM2

Bending moment in point D [Nm]

-33840

1.30.3 Calculated results

Result name RY

Value

Error

Rotation around z-axis in point A [rad]

0.00304333 Rad

-0.05%

Vertical reaction in point A [N]

-11709 N

-0.30%

Vertical displacement in point D [cm]

-0.423297 cm

0.01%

Bending moment in point D [Nm]

-33835.9 N*m

0.01%

122

Result description

ADVANCE DESIGN VALIDATION GUIDE

1.31 Square plate under planar stresses (01-0039SSLSB_FEM) Test ID: 2470 Test status: Passed

1.31.1 Description Verifies the vertical displacement and the stresses on a square plate of 2 x 2 m, fixed on 3 sides with a uniform surface load on its surface.

1.31.2 Background

1.31.2.1 Model description ■ ■ ■

Reference: Internal GRAITEC test; Analysis type: static linear; Element type: planar (membrane). Square plate under planar stresses

Scale = 1/19 Modeling

ξ ,η ∈ [− 1;1] Units I. S. Geometry ■ ■

Thickness: e = 1 m, 4 square elements of side h = 1 m.

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

123

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Boundary conditions ■ ■

Outer: Fixed on 3 sides, Inner: None.

Loading ■ ■

External: Uniform load p = -1. 108 N/ml on the upper surface, Internal: None.

1.31.2.2 Displacement of the model in the linear elastic range Reference solution The reference displacements are calculated on nodes 7 and 9. v9 =

-6ph(3 + ν)(1 - ν2) = -0.1809 x 10-3 m, E(8(3 - ν)2 - (3 + ν)2)

v7 =

4(3 - ν) v9 = -0.592 x 10-3 m, 3+ν

For element 1.4: (For the stresses calculated above, the abscissa point (x = 0; y = 0) corresponds to node 8.) ξ = -1 ; σyy = 0 ξ = 0 ; σyy = -47.44 MPa ξ = 1 ; σyy = -94.88 MPa

E (v9 - v7) σyy = (1 + ξ) for 1 - ν2 2h

σxx = νσyy

ξ = -1 ; σxx = 0 ξ = 0 ; σxx = -14.23 MPa ξ = 1 ; σxx = -28.46 MPa

for

E (v9 + v7) + η(v9 - v7) σxy = (1 + ξ) 4h 1+ν

for

Finite elements modeling ■ ■ ■

124

Planar element: membrane, imposed mesh, 9 nodes, 4 surface quadrangles.

η = -1 ; ξ = 0 ; σxy = -47.82 MPa η = 0 ; ξ = 0 ; σxy = -31.21 MPa η= 1 ; ξ = 0 ; σxy = -14.61 MPa

ADVANCE DESIGN VALIDATION GUIDE

Deformed shape

1.31.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement on node 7 [mm]

-0.592

CM2

Vertical displacement on node 5 [mm]

-0.1809

CM2

sxx_mid

σxx stresses on Element 1.4 in x = 0 m [MPa]

CM2

sxx_mid

σxx stresses on Element 1.4 in x = 0.5 m [MPa]

-14.23

CM2

sxx_mid

σxx stresses on Element 1.4 in x = 1 m [MPa]

-28.46

CM2

syy_mid

σyy stresses on Element 1.4 in x = 0 m [MPa]

CM2

syy_mid

σyy stresses on Element 1.4 in x = 0.5 m [MPa]

-47.44

CM2

syy_mid

σyy stresses on Element 1.4 in x = 1 m [MPa]

-94.88

CM2

sxy_mid

σxy stresses on Element 1.4 in y = 0 m [MPa]

-14.66

CM2

sxy_mid

σxy stresses on Element 1.4 in y = 1 m [MPa]

-47.82

1.31.3 Calculated results Result name DZ

Result description Vertical displacement on node 7 [mm]

Value -0.59203 mm -0.180898 mm

Error -0.01%

Vertical displacement on node 5 [mm]

0.00%

sxx_mid

Sigma xx stresses on Element 1.4 in x = 0 m [MPa]

7.45058e-015 MPa

0.00%

sxx_mid

Sigma xx stresses on Element 1.4 in x = 0.5 m [MPa]

-14.2315 MPa

-0.01%

sxx_mid

Sigma xx stresses on Element 1.4 in x = 1 m [MPa]

-28.463 MPa

-0.01%

syy_mid

Sigma yy stresses on Element 1.4 in x = 0 m [MPa]

1.49012e-014 MPa

0.00%

syy_mid

Sigma yy stresses on Element 1.4 in x = 0.5 m [MPa]

-47.4383 MPa

0.00%

syy_mid

Sigma yy stresses on Element 1.4 in x = 1 m [MPa]

-94.8767 MPa

0.00%

sxy_mid

Sigma xy stresses on Element 1.4 in y = 0 m [MPa]

-14.611 MPa

0.33%

sxy_mid

Sigma xy stresses on Element 1.4 in y = 1 m [MPa]

-47.8178 MPa

0.00%

125

ADVANCE DESIGN VALIDATION GUIDE

1.32 Beam on elastic soil, free ends (01-0032SSLLB_FEM) Test ID: 2464 Test status: Passed

1.32.1 Description A beam under 3 punctual loads lays on a soil of constant linear stiffness. The bending moment, vertical displacement and rotation about z-axis on several points of the beam are verified.

1.32.2 Background

1.32.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 15/89; Analysis type: static linear (plane problem); Element type: linear. Beam on elastic soil, free ends 01-0032SSLLB_FEM

Units I. S. Geometry ■ ■

126

L = (π 10 )/2, I = 10-4 m4.

Scale = 1/21

ADVANCE DESIGN VALIDATION GUIDE

Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 Pa. Boundary conditions ■

Outer: Free A and B extremities, ► Constant linear stiffness of soil ky = K = 840000 N/m2. Inner: None. ►

■

Loading ■ ■

External: Punctual load at A, C and B: Fy = F = - 10000 N. Internal: None.

1.32.2.2 Bending moment and displacement at C Reference solution β=

K/(4EI)

ϕ = βL/2 λ = sh (2ϕ) + sin (2ϕ) ■

Bending moment: MC = (F/(4β))(ch(2ϕ) - cos (2ϕ) – 8sh(ϕ)sin(ϕ))/λ

■

Vertical displacement: vC = - (Fβ/(2K))( ch(2ϕ) + cos (2ϕ) + 8ch(ϕ)cos(ϕ) + 2)/λ

Finite elements modeling ■ ■ ■

Linear element: beam, 72 nodes, 71 linear elements.

Bending moment diagram Beam on elastic soil, free ends

Scale = 1/20 Bending moment

127

ADVANCE DESIGN VALIDATION GUIDE

1.32.2.3 Displacements at A Reference solution ■

Vertical displacement: vA = (2Fβ/K)( ch(ϕ)cos(ϕ) + ch(2ϕ) + cos(2ϕ))/λ

■

Rotation about z-axis θA = (-2Fβ2/K)( sh(ϕ)cos(ϕ) - sin(ϕ)ch(ϕ) + sh(2ϕ) - sin(2ϕ))/λ

Finite elements modeling ■ ■ ■

Linear element: beam, 72 nodes, 71 linear elements

1.32.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Bending moment in point C [Nm]

5759

CM2

Vertical displacement in point C [m]

-0.006844

CM2

Vertical displacement in point A [m]

-0.007854

CM2

Rotation θ about z-axis in point A [rad]

-0.000706

1.32.3 Calculated results

Result name My Dz

128

Result description

Value

Error

Bending moment in point C [Nm]

5779.54 N*m

0.36%

Vertical displacement in point C [m]

-0.00684369 m

0.00%

Vertical displacement in point A [m]

-0.00786073 m

-0.09%

Rotation Theta about z-axis in point A [rad]

-0.000707427 Rad

-0.20%

ADVANCE DESIGN VALIDATION GUIDE

1.33 EDF Pylon (01-0033SFLLA_FEM) Test ID: 2465 Test status: Passed

1.33.1 Description Verifies the displacement at the top of an EDF Pylon and the dominating buckling results. Three punctual loads corresponding to wind loads are applied on the main arms, on the upper arm and on the lower horizontal frames of the pylon.

1.33.2 Background

1.33.2.1 Model description ■ ■ ■

Reference: Internal GRAITEC test; Analysis type: static linear, Eulerian buckling; Element type: linear

Units I. S. Geometry

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

Outer: Hinged support, ► For the modeling, a fixed restraint and 4 beams were added at the pylon supports level. Inner: None. ►

■

129

ADVANCE DESIGN VALIDATION GUIDE

Loading ■

External: Punctual loads corresponding to a wind load. FX = 165550 N, FY = - 1240 N, FZ = - 58720 N on the main arms, FX = 50250 N, FY = - 1080 N, FZ = - 12780 N on the upper arm, ► FX = 11760 N, FY = 0 N, FZ = 0 N on the lower horizontal frames Internal: None. ► ►

■

1.33.2.2 Displacement of the model in the linear elastic range Reference solution

130

ADVANCE DESIGN VALIDATION GUIDE

Software Max deflection (m) λ dominating mode

ANSYS 5.3 0.714 2.77

NE/NASTRAN 7.0 0.714 2.77

Finite elements modeling ■ ■ ■

Linear element: beam, imposed mesh, 402 nodes, 1034 elements.

Deformed shape

131

ADVANCE DESIGN VALIDATION GUIDE

Buckling modal deformation (dominating mode)

1.33.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Displacement at the top of the pylon [m]

0.714

Dominating buckling - critical 位, mode 4 [Hz]

2.77

CM2

1.33.3 Calculated results

Result name D

132

Result description

Value

Error

Displacement at the top of the pylon [m]

0.71254 m

-0.20%

Dominating buckling - critical Lambda - mode 4 [Hz]

2.83 Hz

2.17%

ADVANCE DESIGN VALIDATION GUIDE

1.34 Thin cylinder under a uniform radial pressure (01-0038SSLSB_FEM) Test ID: 2469 Test status: Passed

1.34.1 Description Verifies the stress, the radial deformation and the longitudinal deformation of a cylinder loaded with a uniform internal pressure.

1.34.2 Background

1.34.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 06/89; Analysis type: static elastic; Element type: planar. Thin cylinder under a uniform radial pressure 01-0038SSLSB_FEM

Scale = 1/18

Units I. S. Geometry ■ ■ ■

Length: L = 4 m, Radius: R = 1 m, Thickness: h = 0.02 m.

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

133

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■

Outer: Free conditions For the modeling, only ¼ of the cylinder is considered and the symmetry conditions are applied. On the other side, we restrained the displacements at a few nodes in order to make the model stable. Inner: None.

► ►

■

Loading ■ ■

External: Uniform internal pressure: p = 10000 Pa, Internal: None.

1.34.2.2 Stresses in all points Reference solution Stresses in the planar elements coordinate system (x axis is parallel with the length of the cylinder): ■ ■

σxx = 0 pR σyy = h

Finite elements modeling ■ ■ ■

Planar element: shell, 209 nodes, 180 planar elements.

1.34.2.3 Cylinder deformation in all points ■

Radial deformation: pR2 δR = Eh

■

Longitudinal deformation: δL =

-pRνL Eh

1.34.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

syy_mid

σyy stress in all points [Pa]

500000.000000

CM2

δL radial deformation of the cylinder in all points [µm]

2.380000

CM2

δL longitudinal deformation of the cylinder in all points [µm]

-2.860000

1.34.3 Calculated results

Result name syy_mid

134

Result description

Value

Error

Sigma yy stress in all points [Pa]

499521 Pa

-0.10%

Delta R radial deformation of the cylinder in all points [µm]

2.39213 µm

0.51%

Delta L longitudinal deformation of the cylinder in all points [µm]

-2.85445 µm

0.19%

ADVANCE DESIGN VALIDATION GUIDE

1.35 Caisson beam in torsion (01-0037SSLSB_FEM) Test ID: 2468 Test status: Passed

1.35.1 Description A torsion moment is applied on the free end of a caisson beam fixed on one end. For both ends, the displacement, the rotation about Z-axis and the stress are verified.

1.35.2 Background

1.35.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 05/89; Analysis type: static linear; Element type: planar. Caisson beam in torsion

Scale = 1/4 01-0037SSLSB_FEM

Units I. S. Geometry ■ ■ ■

Length; L = 1m, Square section of side: b = 0.1 m, Thickness = 0.005 m.

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

135

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: Beam fixed at end x = 0; Inner: None.

Loading ■ ■

External: Torsion moment M = 10N.m applied to the free end (for modeling, 4 forces of 50 N). Internal: None.

1.35.2.2 Displacement and stress at two points Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. Points coordinates: ■ A (0,0.05,0.5) ■ B (-0.05,0,0.8) Note: point O is the origin of the coordinate system (x,y,z). Finite elements modeling ■ ■ ■

Planar element: shell, 90 nodes, 88 planar elements.

Deformed shape Caisson beam in torsion

Scale = 1/4 Deformed

136

ADVANCE DESIGN VALIDATION GUIDE

1.35.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Displacement in point A [m]

-0.617 x 10-6

CM2

Displacement in point B [m]

-0.987 x 10-6

CM2

Rotation about Z-axis in point A [rad]

0.123 x 10-4

CM2

Rotation about Z-axis in point B [rad]

0.197 x 10-4

CM2

sxy_mid

σxy stress in point A [MPa]

-0.11

CM2

sxy_mid

σxy stress in point B [MPa]

-0.11

1.35.3 Calculated results

Result name D

Result description

Value

Error

Displacement in point A [µm]

0.615909 µm

-0.18%

Displacement in point B [µm]

0.986806 µm

-0.02%

Rotation about Z-axis in point A [rad]

-1.23211e-005 Rad

-0.17%

Rotation about Z-axis in point B [rad]

-1.97172e-005 Rad

-0.09%

sxy_mid

Sigma xy stress in point A [MPa]

-0.100037 MPa

-0.04%

sxy_mid

Sigma xy stress in point B [MPa]

-0.100212 MPa

-0.21%

137

ADVANCE DESIGN VALIDATION GUIDE

1.36 Beam on two supports considering the shear force (01-0041SSLLB_FEM) Test ID: 2472 Test status: Passed

1.36.1 Description Verifies the vertical displacement on a 300 cm long beam, consisting of an I shaped profile of a total height of 20.04 cm, a 0.96 cm thick web and 20.04 cm wide / 1.46 cm thick flanges.

1.36.2 Background

1.36.2.1 Model description ■ ■ ■

Reference: Internal GRAITEC test; Analysis type: static linear (plane problem); Element type: linear.

Units I. S. Geometry l= h= b= tw = tf = Sx= Iz = Sy =

300 20.04 20.04 1.46 0.96 74.95 5462 16.43

cm cm cm cm cm cm2 cm4 cm2

Materials properties

138

■ ■

Longitudinal elastic modulus: E = 2285938 daN/cm2, 2 Transverse elastic modulus G = 879207 daN/cm

■

Poisson's ratio: ν = 0.3.

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■

Outer: Simple support on node 11, For the modeling, put an hinge at node 1 (instead of a simple support). Inner: None.

► ►

■

Loading ■ ■

External: Vertical punctual load P = -20246 daN at node 6, Internal: None.

1.36.2.2 Vertical displacement of the model in the linear elastic range Reference solution The reference displacement is calculated in the middle of the beam, at node 6. shear 6flexion 78 6 78 3 Pl Pl − 20246 x3003 − 20246 x300 v6 = + = + = −0.912 − 0.105 = −1.017 cm 48 EI z 4GS y 48 x 2285938 x5462 4 x 2285938 x16.43 2(1 + 0.3)

Finite elements modeling ■ ■ ■

Planar element: S beam, imposed mesh, 11 nodes, 10 linear elements.

Deformed shape

139

ADVANCE DESIGN VALIDATION GUIDE

1.36.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement at node 6 [cm]

-1.017

1.36.3 Calculated results

Result name DZ

140

Result description Vertical displacement at node 6 [cm]

Value -1.01722 cm

Error -0.02%

ADVANCE DESIGN VALIDATION GUIDE

1.37 Thin cylinder under a hydrostatic pressure (01-0043SSLSB_FEM) Test ID: 2474 Test status: Passed

1.37.1 Description Verifies the stress, the longitudinal deformation and the radial deformation of a thin cylinder under a hydrostatic pressure.

1.37.2 Background

1.37.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 08/89; Analysis type: static, linear elastic; Element type: planar. Thin cylinder under a hydrostatic pressure 01-0043SSLSB_FEM

Scale = 1/25

Units I. S. Geometry ■ ■ ■

Thickness: h = 0.02 m, Length: L = 4 m, Radius: R = 1 m.

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

141

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■

Outer: For the modeling, we consider only a quarter of the cylinder, so we impose the symmetry conditions on the nodes that are parallel with the cylinder’s axis. Inner: None.

■

Loading ■

z External: Radial internal pressure varies linearly with the "p" height, p = p0 L ,

■

Internal: None.

1.37.2.2 Stresses Reference solution x axis of the local coordinate system of planar elements is parallel to the cylinders axis. σxx = 0 σyy =

p0Rz Lh

Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 209 nodes, 180 surface quadrangles.

1.37.2.3 Cylinder deformation Reference solution ■

δL longitudinal deformation of the cylinder: δL =

■

-p0Rνz2 2ELh

δL radial deformation of the cylinder: p0R2z δR = ELh

Finite elements modeling ■ ■ ■

142

Planar element: shell, imposed mesh, 209 nodes, 180 surface quadrangles.

ADVANCE DESIGN VALIDATION GUIDE

Deformation shape Thin cylinder under a hydrostatic pressure Deformed

1.37.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

syy_mid

ﾏペy stress in z = L/2 [Pa]

500000.000000

CM2

ﾎｴL longitudinal deformation of the cylinder at the inferior extremity [mm]

-0.002860

CM2

ﾎｴL radial deformation of the cylinder in z = L/2 [mm]

0.002380

1.37.3 Calculated results

Result name syy_mid

Result description

Value

Error

Sigma yy stress in z = L/2 [Pa]

504489 Pa

0.90%

Delta L longitudinal deformation of the cylinder at the inferior extremity [mm]

-0.00285442 mm

0.20%

Delta L radial deformation of the cylinder in z = L/2 [mm]

0.00238372 mm

0.16%

143

ADVANCE DESIGN VALIDATION GUIDE

1.38 Thin cylinder under a uniform axial load (01-0042SSLSB_FEM) Test ID: 2473 Test status: Passed

1.38.1 Description Verifies the stress, the longitudinal deformation and the radial deformation of a cylinder under a uniform axial load.

1.38.2 Background

1.38.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 07/89; Analysis type: static elastic; Element type: planar. Thin cylinder under a uniform axial load 01-0042SSLSB_FEM

Units I. S. Geometry ■ ■ ■

Thickness: h = 0.02 m, Length: L = 4 m, Radius: R = 1 m.

Materials properties

144

■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Scale = 1/19

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■

Outer: Null axial displacement at the left end: vz = 0, For the modeling, only a ¼ of the cylinder is considered. Inner: None.

► ►

■

Loading ■ ■

External: Uniform axial load q = 10000 N/m Inner: None.

1.38.2.2 Stress in all points Reference solution x axis of the local coordinate system of planar elements is parallel to the cylinders axis. q σxx = h σyy = 0 Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 697 nodes, 640 surface quadrangles.

1.38.2.3 Cylinder deformation at the free end Reference solution ■

δL longitudinal deformation of the cylinder: qL δL = Eh

■

δR radial deformation of the cylinder: δR =

-qνR Eh

Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 697 nodes, 640 surface quadrangles.

145

ADVANCE DESIGN VALIDATION GUIDE

Deformation shape Thin cylinder under a uniform axial load

Scale = 1/22

Deformation shape

1.38.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

sxx_mid

ﾏベx stress at all points [Pa]

5 x 10

CM2

syy_mid

ﾏペy stress at all points [Pa]

CM2

ﾎｴL longitudinal deformation at the free end [m]

9.52 x 10-6

CM2

ﾎｴR radial deformation at the free end [m]

-7.14 x 10-7

1.38.3 Calculated results

Result name sxx_mid

146

Result description

Value

Error

Sigma xx stress at all points [Pa]

500000 Pa

0.00%

syy_mid

Sigma yy stress at all points [Pa]

1.05305e-009 Pa

0.00%

Delta L longitudinal deformation at the free end [mm]

-0.00952381 mm

-0.04%

Delta R radial deformation at the free end [mm]

0.000710887 mm

-0.44%

ADVANCE DESIGN VALIDATION GUIDE

1.39 Truss with hinged bars under a punctual load (01-0031SSLLB_FEM) Test ID: 2463 Test status: Passed

1.39.1 Description Verifies the horizontal and the vertical displacement in several points of a truss with hinged bars, subjected to a punctual load.

1.39.2 Background ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 11/89; Analysis type: static linear (plane problem); Element type: linear.

1.39.2.1 Model description Truss with hinged bars under a punctual load

Scale = 1/10 01-0031SSLLB_FEM

Units I. S. Geometry Elements AC CB CD BD

Length (m) 0.5 2 0.5 2 2.5 2

Area (m2) 2 x 10-4 2 x 10-4 1 x 10-4 1 x 10-4

147

ADVANCE DESIGN VALIDATION GUIDE

Materials properties Longitudinal elastic modulus: E = 1.962 x 1011 Pa. Boundary conditions ■ ■

Outer: Hinge at A and B, Inner: None.

Loading ■ ■

External: Punctual force at D: Fy = F = - 9.81 x 103 N. Internal: None.

1.39.2.2 Displacements at C and D Reference solution Displacement method. Finite elements modeling ■ ■ ■

Linear element: beam, 4 nodes, 4 linear elements.

Displacements shape Truss with hinged bars under a punctual load

Scale = 1/9 Deformed

148

ADVANCE DESIGN VALIDATION GUIDE

1.39.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Horizontal displacement in point C [mm]

0.26517

CM2

Horizontal displacement in point D [mm]

3.47902

CM2

Vertical displacement in point C [mm]

0.08839

CM2

Vertical displacement in point D [mm]

-5.60084

1.39.3 Calculated results

Result name DX

Result description Horizontal displacement in point C [mm]

Value 0.264693 mm

Error -0.18%

Horizontal displacement in point D [mm]

3.47531 mm

-0.11%

Vertical displacement in point C [mm]

0.0881705 mm

-0.25%

Vertical displacement in point D [mm]

-5.595 mm

0.10%

149

ADVANCE DESIGN VALIDATION GUIDE

1.40 Simply supported square plate (01-0036SSLSB_FEM) Test ID: 2467 Test status: Passed

1.40.1 Description Verifies the vertical displacement in the center of a simply supported square plate.

1.40.2 Background

1.40.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 02/89; Analysis type: static linear; Element type: planar. Simply supported square plate 01-0036SSLSB_FEM

Units I. S. Geometry ■ ■

Side = 1 m, Thickness h = 0.01m.

Materials properties

150

■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7950 kg/m3.

Scale = 1/9

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■

Outer: Simple support on the plate perimeter, For the modeling, we add a fixed support at B. Inner: None.

► ►

■

Loading ■ ■

External: Self weight (gravity = 9.81 m/s2). Internal: None.

1.40.2.2 Vertical displacement at O Reference solution According to Love- Kirchhoff hypothesis, the displacement w at a point (x,y): w(x,y) = Σ wmnsinmπxsinnπy where wmn =

192ρg(1 - ν2) mn(m2 + n2)π6Eh2

Finite elements modeling ■ ■ ■

Planar element: shell, 441 nodes, 400 planar elements.

Deformed shape Simply supported square plate

Scale = 1/6 Deformed

151

ADVANCE DESIGN VALIDATION GUIDE

1.40.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point O [μm]

-0.158

1.40.3 Calculated results

Result name Dz

152

Result description Vertical displacement in point O [µm]

Value -0.164901 µm

Error -4.37%

ADVANCE DESIGN VALIDATION GUIDE

1.41 Stiffen membrane (01-0040SSLSB_FEM) Test ID: 2471 Test status: Passed

1.41.1 Description Verifies the horizontal displacement and the stress on a plate (8 x 12 cm) fixed in the middle on 3 supports with a punctual load at its free node.

1.41.2 Background

1.41.2.1 Model description ■ ■ ■

Reference: Klaus-Jürgen Bathe - Finite Element Procedures in Engineering Analysis, Example 5.13; Analysis type: static linear; Element type: planar (membrane).

ξ ,η ∈ [− 1;1] Units I. S. Geometry ■ ■ ■

Thickness: e = 0.1 cm, Length: l = 8 cm, Width: B = 12 cm.

Materials properties ■

Longitudinal elastic modulus: E = 30 x 106 N/cm2,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■ ■

Outer: Fixed on 3 sides, Inner: None.

153

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Loading ■ ■

External: Uniform load Fx = F = 6000 N at A, Internal: None.

1.41.2.2 Results of the model in the linear elastic range Reference solution Point B is the origin of the coordinate system used for the results positions.

uA =

6000 F F = = = 9.3410 − 4 cm 6 6 K eabE ⎛ ⎞ ES 2.67410 + 3.7510 2 1 ⎜⎜ 2 + 2 ⎟+ 2 b (1 + ν ) ⎟⎠ 2a 3 ⎝ a 1 −ν

(

)

η = 1; σ xx1 = 0 ⎧ E uA ⎪ (1 − η ) for ⎨ η = 0; σ xx1 = 1924 N/cm 2 = 19.24 MPa σ xx1 = 2 1 − ν 2a ⎪η = −1; σ = 3849 N/cm 2 = 38.49 MPa xx1 ⎩ ⎧ η = 1; σ yy1 = 0 ⎪ σ yy1 = νσ xx1 for ⎨ η = 0; σ yy1 = 577 N/cm 2 = 5.77 MPa ⎪η = −1; σ = 1155 N/cm 2 = 11.55 MPa yy1 ⎩ ⎧ ξ = −1; σ xy1 = 0 E uA ⎪ (1 + ξ ) for ⎨ ξ = 0; σ xy1 = −898 N/cm2 = −8.98 MPa σ xy1 = − 1 + ν 8b ⎪ξ = 1; σ = −1796 N/cm 2 = −17.96 MPa xy1 ⎩ Finite elements modeling ■ ■ ■

Planar element: membrane, imposed mesh, 6 nodes, 2 quadrangle planar elements and 1 bar.

Deformed shape

154

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1.41.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Horizontal displacement Element 1 in A [cm]

9.340000

CM2

sxx_mid

σxx stress Element 1 in y = 0 cm [MPa]

38.490000

CM2

sxx_mid

σxx stress Element 1 in y = 6 cm [MPa]

CM2

syy_mid

σyy stress Element 1 in y = 0 cm [MPa]

11.550000

CM2

syy_mid

σyy stress Element 1 in y = 6 cm [MPa]

CM2

sxy_mid

σxy stress Element 1 in x = 0 cm [MPa]

CM2

sxy_mid

σxy stress Element 1 in x = 4 cm [MPa]

-8.980000

CM2

sxy_mid

σxy stress Element 1 in x = 8 cm [MPa]

-17.960000

1.41.3 Calculated results

Result name DX

Result description

Value

Error

Horizontal displacement Element 1 in A [µm]

9.33999 µm

0.00%

sxx_mid

Sigma xx stress Element 1 in y = 0 cm [MPa]

38.489 MPa

0.00%

sxx_mid

Sigma xx stress Element 1 in y = 6 cm [MPa]

3.63798e-015 MPa

0.00%

syy_mid

Sigma yy stress Element 1 in y = 0 cm [MPa]

11.5467 MPa

-0.03%

syy_mid

Sigma yy stress Element 1 in y = 6 cm [MPa]

-9.09495e-016 MPa

0.00%

sxy_mid

Sigma xy stress Element 1 in x = 0 cm [MPa]

-2.96059e-015 MPa

0.00%

sxy_mid

Sigma xy stress Element 1 in x = 4 cm [MPa]

-8.98076 MPa

-0.01%

sxy_mid

Sigma xy stress Element 1 in x = 8 cm [MPa]

-17.9615 MPa

-0.01%

155

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1.42 Torus with uniform internal pressure (01-0045SSLSB_FEM) Test ID: 2476 Test status: Passed

1.42.1 Description Verifies the stress and the radial deformation of a torus with uniform internal pressure.

1.42.2 Background

1.42.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 10/89; Analysis type: static, linear elastic; Element type: planar.

Torus with uniform internal pressure 01-0045SSLSB_FEM

Units I. S. Geometry ■ ■ ■

156

Thickness: h = 0.02 m, Transverse section radius: b = 1 m, Average radius of curvature: a = 2 m.

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■ ■

1 Outer: For the modeling, only /8 of the cylinder is considered, so the symmetry conditions are imposed to end nodes. Inner: None.

Loading ■ ■

External: Uniform internal pressure p = 10000 Pa Internal: None.

1.42.2.2 Stresses Reference solution (See stresses description on the first scheme of the overview) If a – b ≤ r ≤ a + b pb r + a σ11 = 2h r pb σ22 = 2h Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 361 nodes, 324 surface quadrangles.

1.42.2.3 Cylinder deformation Reference solution ■

δR radial deformation of the torus: pb δR = 2Eh (r - ν(r + a))

Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 361 nodes, 324 surface quadrangles.

1.42.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

syy_mid

σ11 stresses for r = a - b [Pa]

7.5 x 10

CM2

syy_mid

σ11 stresses for r = a + b [Pa]

4.17 x 10

CM2

sxx_mid

σ22 stress for all r [Pa]

2.50 x 10

CM2

δL radial deformations of the torus for r = a - b [m]

1.19 x 10-7

CM2

δL radial deformations of the torus for r = a + b [m]

1.79 x 10-6

157

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1.42.3 Calculated results

Result name syy_mid syy_mid

158

Result description

Value

Error

Sigma 11 stresses for r = a - b [Pa]

742770 Pa

-0.96%

Sigma 11 stresses for r = a + b [Pa]

415404 Pa

-0.38%

sxx_mid

Sigma 22 stress for all r [Pa]

250331 Pa

0.13%

Delta L radial deformations of the torus for r = a - b [mm]

-0.000117352 mm

1.38%

Delta L radial deformations of the torus for r = a + b [mm]

0.00180274 mm

0.71%

ADVANCE DESIGN VALIDATION GUIDE

1.43 Spherical dome under a uniform external pressure (01-0050SSLSB_FEM) Test ID: 2480 Test status: Passed

1.43.1 Description A spherical dome of radius (a) is subjected to a uniform external pressure. The horizontal displacement and the external meridian stresses are verified.

1.43.2 Background

1.43.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 22/89; Analysis type: static, linear elastic; Element type: planar.

Spherical dome under a uniform external pressure 01-0050SSLSB_FEM

Units I. S.

159

ADVANCE DESIGN VALIDATION GUIDE

Geometry ■ ■

Radius: a = 2.54 m, Thickness: h = 0.0127 m,

■

Angle: θ = 75°.

Materials properties ■

Longitudinal elastic modulus: E = 6.897 x 1010 Pa,

■

Poisson's ratio: ν = 0.2.

Boundary conditions ■ ■

Outer: Fixed on the dome perimeter, Inner: None.

Loading ■ ■

External: Uniform pressure p = 0.6897 x 106 Pa, Internal: None.

1.43.2.2 Horizontal displacement and exterior meridian stress Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution. Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 401 nodes, 400 planar elements.

Deformed shape

160

ADVANCE DESIGN VALIDATION GUIDE

1.43.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Horizontal displacements in ψ = 15°

1.73 x 10-3

CM2

Horizontal displacements in ψ = 45°

-1.02 x 10-3

CM2

syy_mid

σyy external meridian stresses in ψ = 15°

-74

CM2

sxx_mid

σXX external meridian stresses in ψ = 45°

-68

1.43.3 Calculated results

Result name DX

Result description

Value

Error

Horizontal displacements in Psi = 15° [mm]

1.73064 mm

0.04%

Horizontal displacements in Psi = 45° [mm]

-1.01367 mm

0.62%

syy_mid

Sigma yy external meridian stresses in Psi = 15° [MPa]

-72.2609 MPa

2.35%

sxx_mid

Sigma XX external meridian stresses in Psi = 45° [MPa]

-68.9909 MPa

-1.46%

161

ADVANCE DESIGN VALIDATION GUIDE

1.44 Pinch cylindrical shell (01-0048SSLSB_FEM) Test ID: 2478 Test status: Passed

1.44.1 Description A cylinder of length L is pinched by 2 diametrically opposite forces (F). The vertical displacement is verified.

1.44.2 Background ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 20/89; Analysis type: static, linear elastic; Element type: planar.

1.44.2.1 Model description A cylinder of length L is pinched by 2 diametrically opposite forces (F).

Pinch cylindrical shell 01-0048SSLSB_FEM

Units I. S.

162

ADVANCE DESIGN VALIDATION GUIDE

Geometry ■ ■ ■

Length: L = 10.35 m (total length), Radius: R = 4.953 m, Thickness: h = 0.094 m.

Materials properties ■

Longitudinal elastic modulus: E = 10.5 x 106 Pa,

■

Poisson's ratio: ν = 0.3125.

Boundary conditions ■ ■

Outer: For the modeling, we consider only half of the cylinder, so we impose symmetry conditions (nodes in the horizontal xz plane are restrained in translation along y and in rotation along x and z), Inner: None.

Loading ■ ■

External: 2 punctual loads F = 100 N, Internal: None.

1.44.2.2 Vertical displacement at point A Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution. Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 777 nodes, 720 surface quadrangles.

1.44.2.3 Theoretical result Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point A [m]

-113.9 x 10-3

1.44.3 Calculated results

Result name DZ

Result description Vertical displacement in point A [mm]

Value -113.3 mm

Error 0.53%

163

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1.45 Simply supported rectangular plate under a uniform load (01-0052SSLSB_FEM) Test ID: 2482 Test status: Passed

1.45.1 Description A rectangular plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.

1.45.2 Background

1.45.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 24/89; Analysis type: static, linear elastic; Element type: planar. Simply supported rectangular plate under a uniform load 01-0052SSLSB_FEM

Units I. S. Geometry ■ ■ ■

Width: a = 1 m, Length: b = 2 m, Thickness: h = 0.01 m,

Materials properties

164

■

Longitudinal elastic modulus: E = 1.0 x 107 Pa,

■

Poisson's ratio: ν = 0.3.

Scale = 1/11

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: Simple support on the plate perimeter (null displacement along z-axis), Inner: None.

Loading ■ ■

External: Normal pressure of plate p = pZ = -1.0 Pa, Internal: None.

1.45.2.2 Vertical displacement and bending moment at the center of the plate Reference solution Love-Kirchhoff thin plates theory. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 435 nodes, 392 surface quadrangles.

1.45.2.3 Theoretical background Solver

Result name

Result description

Reference value

CM2

Vertical displacement at plate center [m]

-1.1060 x 10-2

CM2

Mxx

MX bending moment at plate center [Nm]

-0.1017

CM2

Myy

MY bending moment at plate center [Nm]

-0.0464

1.45.3 Calculated results

Result name DZ

Result description

Value

Error

Vertical displacement at plate center [cm]

-1.10238 cm

0.33%

Mxx

Mx bending moment at plate center [Nm]

-0.101737 N*m

-0.04%

Myy

My bending moment at plate center [Nm]

-0.0462457 N*m

0.33%

165

ADVANCE DESIGN VALIDATION GUIDE

1.46 Spherical shell under internal pressure (01-0046SSLSB_FEM) Test ID: 2477 Test status: Passed

1.46.1 Description A spherical shell is subjected to a uniform internal pressure. The stress and the radial deformation are verified.

1.46.2 Background

1.46.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 14/89; Analysis type: static, linear elastic; Element type: planar.

Spherical shell under internal pressure 01-0046SSLSB_FEM

Units I. S.

166

ADVANCE DESIGN VALIDATION GUIDE

Geometry ■ ■

Thickness: h = 0.02 m, Radius: R2 = 1 m,

■

θ = 90° (hemisphere).

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

Outer: Simple support (null displacement along vertical displacement) on the shell perimeter. For modeling, we consider only half of the hemisphere, so we impose symmetry conditions (DOF restrains placed in the vertical plane xy in translation along z and in rotation along x and y). In addition, the node at the top of the shell is restrained in translation along x to assure the stability of the structure during calculation).

■

Inner: None.

Loading ■ ■

External: Uniform internal pressure p = 10000 Pa Internal: None.

1.46.2.2 Stresses Reference solution (See stresses description on the first scheme of the overview) If 0° ≤ θ ≤ 90° σ11 = σ22 =

pR22 2h

Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 343 nodes, 324 planar elements.

1.46.2.3 Cylinder deformation Reference solution ■

δR radial deformation of the calotte: δR =

pR22 (1 - ν) sin θ 2Eh

Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 343 nodes, 324 planar elements.

167

ADVANCE DESIGN VALIDATION GUIDE

Deformed shape Spherical shell under internal pressure

Scale = 1/11 Deformed

1.46.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

sxx_mid

σ11 stress for all θ [Pa]

2.50 x 10

CM2

syy_mid

σ22 stress for all θ [Pa]

2.50 x 10

CM2

δR radial deformations for θ = 90° [m]

8.33 x 10-7

1.46.3 Calculated results

Result name sxx_mid

168

Result description

Value

Error

Sigma 11 stress for all Theta [Pa]

250202 Pa

0.08%

syy_mid

Sigma 22 stress for all Theta [Pa]

249907 Pa

-0.04%

Delta R radial deformations for Theta = 90° [mm]

0.000832794 mm

-0.02%

ADVANCE DESIGN VALIDATION GUIDE

1.47 Simply supported square plate under a uniform load (01-0051SSLSB_FEM) Test ID: 2481 Test status: Passed

1.47.1 Description A square plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.

1.47.2 Background

1.47.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 24/89; Analysis type: static, linear elastic; Element type: planar. Simply supported square plate under a uniform load 01-0051SSLSB_FEM

Scale = 1/9

Units I. S. Geometry ■ ■

Side: a =b = 1 m, Thickness: h = 0.01 m,

Materials properties ■

Longitudinal elastic modulus: E = 1.0 x 107 Pa,

■

Poisson's ratio: ν = 0.3.

169

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: Simple support on the plate perimeter (null displacement along z-axis), Inner: None

Loading ■ ■

External: Normal pressure of plate p pZ = -1.0 Pa, Internal: None.

1.47.2.2 Vertical displacement and bending moment at the center of the plate Reference solution Love-Kirchhoff thin plates theory. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 361 nodes, 324 planar elements.

1.47.2.3 Theoretical result Solver

Result name

Result description

Reference value

CM2

Vertical displacement at plate center [m]

-4.43 x 10-3

CM2

Mxx

MX bending moment at plate center [Nm]

0.0479

CM2

Myy

MY bending moment at plate center [Nm]

0.0479

1.47.3 Calculated results

Result name DZ

170

Result description

Value

Error

Vertical displacement at plate center [m]

-0.00435847 m

1.61%

Mxx

Mx bending moment at plate center [Nm]

0.0471381 N*m

-1.59%

Myy

My bending moment at plate center [Nm]

0.0471381 N*m

-1.59%

ADVANCE DESIGN VALIDATION GUIDE

1.48 Simply supported rectangular plate loaded with punctual force and moments (010054SSLSB_FEM) Test ID: 2484 Test status: Passed

1.48.1 Description A rectangular plate simply supported is subjected to a punctual force and moments. The vertical displacement is verified.

1.48.2 Background

1.48.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 26/89; Analysis type: static linear; Element type: planar. Simply supported rectangular plate loaded with punctual force and moments 01-0054SSLSB_FEM

Units I. S. Geometry ■ ■ ■

Width: DA = CB = 20 m, Length: AB = DC = 5 m, Thickness: h = 1 m,

Materials properties ■

Longitudinal elastic modulus: E =1000 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■ ■

Outer: Punctual support at A, B and D (null displacement along z-axis), Inner: None. 171

ADVANCE DESIGN VALIDATION GUIDE

Loading ■

■

External: ► In A: MX = 20 Nm, MY = -10 Nm, ► In B: MX = 20 Nm, MY = 10 Nm, ► In C: FZ = -2 N, MX = -20 Nm, MY = 10 Nm, ► In D: MX = -20 Nm, MY = -10 Nm, Internal: None.

1.48.2.2 Vertical displacement at C Reference solution Love-Kirchhoff thin plates theory. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 867 nodes, 800 surface quadrangles.

Deformed shape Simply supported rectangular plate loaded with punctual force and moments Deformed

1.48.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-12.480

1.48.3 Calculated results Result name Dz

172

Result description Vertical displacement in point C [m]

Value -12.6677 m

Error -1.50%

ADVANCE DESIGN VALIDATION GUIDE

1.49 Triangulated system with hinged bars (01-0056SSLLB_FEM) Test ID: 2486 Test status: Passed

1.49.1 Description A truss with hinged bars is placed on three punctual supports (subjected to imposed displacements) and is loaded with two punctual forces. A thermal load is applied to all the bars. The traction force and the vertical displacement are verified.

1.49.2 Background

1.49.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 12/89; Analysis type: static (plane problem); Element type: linear.

Units I. S. Geometry ■ ■ ■

θ = 30°, Section A1 = 1.41 x 10-3 m2, -3 2 Section A2 = 2.82 x 10 m .

Materials properties ■

Longitudinal elastic modulus: E =2.1 x 1011 Pa,

■

Coefficient of linear expansion: α = 10-5 °C-1.

Boundary conditions ■

Outer: Hinge at A (uA = vA = 0), Roller supports at B and C ( uB = v’C = 0), Inner: None.

► ►

■

173

ADVANCE DESIGN VALIDATION GUIDE

Loading ■

■

External: ► Support displacement: vA = -0.02 m ; vB = -0.03 m ; v’C = -0.015 m , ► Punctual loads: FE = -150 KN ; FF = -100 KN, ► Expansion effect on all bars for a temperature variation of 150° in relation with the assembly temperature (specified geometry), Internal: None.

1.49.2.2 Tension force in BD bar Reference solution Determining the hyperstatic unknown with the section cut method. Finite elements modeling ■ ■ ■

Linear element: S beam, automatic mesh, 11 nodes, 17 S beams + 1 rigid S beam for the modeling of the simple support at C.

1.49.2.3 Vertical displacement at D Reference solution vD displacement was determined by several software with implemented finite elements method. Finite elements modeling ■ ■ ■

Linear element: S beam, automatic mesh, 11 nodes, 17 S beams + 1 rigid S beam for the modeling of simple support at C.

Deformed shape Triangulated system with hinged bars 01-0056SSLLB_FEM

174

ADVANCE DESIGN VALIDATION GUIDE

1.49.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

FX traction force on BD bar [N]

43633

CM2

Vertical displacement on point D [m]

-0.01618

1.49.3 Calculated results

Result name Fx DZ

Result description

Value

Error

Fx traction force on BD bar [N]

42870.9 N

-1.75%

Vertical displacement on point D [m]

-0.0162358 m

-0.34%

175

ADVANCE DESIGN VALIDATION GUIDE

1.50 Shear plate perpendicular to the medium surface (01-0055SSLSB_FEM) Test ID: 2485 Test status: Passed

1.50.1 Description Verifies the vertical displacement of a rectangular shear plate fixed at one end, loaded with two forces.

1.50.2 Background

1.50.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 27/89; Analysis type: static; Element type: planar. Shear plate

Scale = 1/50 01-0055SSLSB_FEM

Units I. S. Geometry ■ ■ ■

Length: L = 12 m, Width: l = 1 m, Thickness: h = 0.05 m,

Materials properties ■

Longitudinal elastic modulus: E = 1.0 x 107 Pa,

■

Poisson's ratio: ν = 0.25.

Boundary conditions ■ ■

176

Outer: Fixed AD edge, Inner: None.

ADVANCE DESIGN VALIDATION GUIDE

Loading ■

■

External: ► At B: Fz = -1.0 N, ► At C: FZ = 1.0 N, Internal: None.

1.50.2.2 Vertical displacement at C Reference solution Analytical solution. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 497 nodes, 420 surface quadrangles.

Deformed shape Shear plate

Scale = 1/35 Deformed

1.50.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

35.37 x 10-3

1.50.3 Calculated results Result name Dz

Result description Vertical displacement in point C [m]

Value 35.6655 mm

Error 0.84%

177

ADVANCE DESIGN VALIDATION GUIDE

1.51 Thin cylinder under its self weight (01-0044SSLSB_MEF) Test ID: 2475 Test status: Passed

1.51.1 Description Verifies the stress, the longitudinal deformation and the radial deformation of a thin cylinder subjected to its self weight only.

1.51.2 Background

1.51.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLS 09/89; ■ Analysis type: static, linear elastic; ■ Element type: planar. A cylinder of R radius and L length subject of self weight only. Thin cylinder under its self weight 01-0044SSLSB_FEM

Units I. S. Geometry ■ ■ ■

Thickness: h = 0.02 m, Length: L = 4 m, Radius: R = 1 m.

Materials properties

178

■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: γ = 7.85 x 104 N/m3.

Scale = 1/24

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■

Outer: Null axial displacement at z = 0, For the modeling, we consider only a quarter of the cylinder, so we impose the symmetry conditions on the nodes that are parallel with the cylinder’s axis. Inner: None.

► ►

■

Loading ■ ■

External: Cylinder self weight, Internal: None.

1.51.2.2 Stresses Reference solution x axis of the local coordinate system of planar elements is parallel to the cylinders axis. σxx = γz σyy = 0 Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 697 nodes, 640 surface quadrangles.

1.51.2.3 Cylinder deformation Reference solution ■

δL longitudinal deformation of the cylinder: γz2 δL = 2E

■

δR radial deformation of the cylinder: δR =

-γνRz E

1.51.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

sxx_mid

σxx stress for z = L [Pa]

-314000.000000

CM2

δL longitudinal deformation for z = L [mm]

0.002990

CM2

δR radial deformation for z = L[mm]

-0.000440

* To obtain this result, you must generate a calculation note “Planar elements stresses by load case in neutral fiber" with results on center.

1.51.3 Calculated results Result name sxx_mid

Result description

Value

Error

Sigma xx stress for z = L [Pa]

-309143 Pa

1.55%

Delta L longitudinal deformation for z = L [mm]

0.00298922 mm

-0.03%

Delta R radial deformation for z = L [mm]

-0.000443587 mm

-0.82%

179

ADVANCE DESIGN VALIDATION GUIDE

1.52 Spherical shell with holes (01-0049SSLSB_FEM) Test ID: 2479 Test status: Passed

1.52.1 Description A spherical shell with holes is subjected to 4 forces, opposite 2 by 2. The horizontal displacement is verified.

1.52.2 Background ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 21/89; Analysis type: static, linear elastic; Element type: planar.

1.52.2.1 Model description

Spherical shell with holes 01-0049SSLSB_FEM

Units I. S. Geometry

180

■ ■

Radius: R = 10 m Thickness: h = 0.04 m,

■

Opening angle of the hole: ϕ0 = 18°.

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 6.285 x 107 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

■

Outer: For modeling, we consider only a quarter of the shell, so we impose symmetry conditions (nodes in the vertical yz plane are restrained in translation along x and in rotation along y and z. Nodes on the vertical xy plane are restrained in translation along z and in rotation along x and y), Inner: None.

Loading ■ ■

External: Punctual loads F = 1 N, according to the diagram, Internal: None.

1.52.2.2 Horizontal displacement at point A Reference solution The reference solution is determined by averaging the results of several calculation software with implemented finite elements method. 2% uncertainty about the reference solution. Finite elements modeling ■ ■ ■

Planar element: shell, imposed mesh, 99 nodes, 80 surface quadrangles.

Deformed shape Spherical shell with holes

Scale = 1/79 Deformed

181

ADVANCE DESIGN VALIDATION GUIDE

1.52.2.3 Theoretical background Solver

Result name

Result description

Reference value

CM2

Horizontal displacement at point A(R,0,0) [mm]

94.0

1.52.3 Calculated results

Result name DX

182

Result description Horizontal displacement at point A(R,0,0) [mm]

Value 92.6205 mm

Error -1.47%

ADVANCE DESIGN VALIDATION GUIDE

1.53 Simply supported rectangular plate under a uniform load (01-0053SSLSB_FEM) Test ID: 2483 Test status: Passed

1.53.1 Description A rectangular plate simply supported is subjected to a uniform load. The vertical displacement and the bending moments at the plate center are verified.

1.53.2 Background

1.53.2.1 Model description ■ ■ ■

Scale = 1/25

Units I. S. Geometry ■ ■ ■

Width: a = 1 m, Length: b = 5 m, Thickness: h = 0.01 m,

Materials properties ■

Longitudinal elastic modulus: E = 1.0 x 107 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■ ■

Outer: Simple support on the plate perimeter (null displacement along z-axis), Inner: None.

183

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: Normal pressure of plate p = pZ = -1.0 Pa, Internal: None.

1.53.2.2 Vertical displacement and bending moment at the center of the plate Reference solution Love-Kirchhoff thin plates theory. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 793 nodes, 720 surface quadrangles.

1.53.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement at plate center [m]

1.416 x 10-2

CM2

Mxx

MX bending moment at plate center [Nm]

0.1246

CM2

Myy

MY bending moment at plate center [Nm]

0.0375

1.53.3 Calculated results

Result name DZ

184

Result description

Value

Error

Vertical displacement at plate center [cm]

-1.40141 cm

1.03%

Mxx

Mx bending moment at plate center [Nm]

-0.124082 N*m

0.42%

Myy

My bending moment at plate center [Nm]

-0.0375624 N*m

-0.17%

ADVANCE DESIGN VALIDATION GUIDE

1.54 A plate (0.01333 m thick), fixed on its perimeter, loaded with a uniform pressure (010058SSLSB_FEM) Test ID: 2488 Test status: Passed

1.54.1 Description Verifies the vertical displacement for a square plate (0.01333 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.

1.54.2 Background

1.54.2.1 Model description ■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; ■ Analysis type: static; ■ Element type: planar. Square plate of side "a", for the modeling, only a quarter of the plate is considered.

Units I. S. Geometry ■ ■ ■

Side: a = 1 m, Thickness: h = 0.01333 m, a Slenderness: λ = h = 75.

Materials properties ■ ■

Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

Outer: Fixed sides: AB and BD, For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),

■

Inner: None.

185

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: 1 MPa uniform pressure, Internal: None.

1.54.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.

1.54.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-2.8053 x 10-2

1.54.3 Calculated results Result name Dz

186

Result description Vertical displacement in point C [cm]

Value -2.79502 cm

Error 0.37%

ADVANCE DESIGN VALIDATION GUIDE

1.55 A plate (0.05 m thick), fixed on its perimeter, loaded with a uniform pressure (010060SSLSB_FEM) Test ID: 2490 Test status: Passed

1.55.1 Description Verifies the vertical displacement for a square plate (0.05 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.

1.55.2 Background

1.55.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.

Units I. S. Geometry ■ ■ ■

Side: a = 1 m, Thickness: h = 0.05 m, a Slenderness: λ = h = 20.

Materials properties ■ ■

Reinforcement, 11 Longitudinal elastic modulus: E = 2.1 x 10 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

Outer: Fixed edges: AB and BD, For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x and restrained rotation around y and z) and CD side (restrained displacement along y and restrained rotation around x and z),

■

Inner: None.

187

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: 1 MPa uniform pressure, Internal: None.

1.55.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.

1.55.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-0.55474 x 10-3

1.55.3 Calculated results

Result name Dz

188

Result description Vertical displacement in point C [cm]

Value -0.0549874 cm

Error 0.88%

ADVANCE DESIGN VALIDATION GUIDE

1.56 A plate (0.02 m thick), fixed on its perimeter, loaded with a punctual force (010064SSLSB_FEM) Test ID: 2494 Test status: Passed

1.56.1 Description Verifies the vertical displacement for a square plate (0.02 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.

1.56.2 Background

1.56.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar. 0.02 m thick plate fixed on its perimeter

Scale = 1/5

01-0064SSLSB_FEM

Units I. S. Geometry ■ ■

Side: a = 1 m, Thickness: h = 0.02 m,

■

Slenderness: λ = 50.

Materials properties ■ ■

Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

189

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: Fixed edges, Inner: None.

Loading ■ ■

External: punctual force applied in the center of the plate: FZ = -106 N, Internal: None.

1.56.2.2 Vertical displacement at point C (the center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.

1.56.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-0.037454

1.56.3 Calculated results

Result name DZ

190

Result description Vertical displacement in point C [m]

Value -0.0369818 m

Error 1.26%

ADVANCE DESIGN VALIDATION GUIDE

1.57 A plate (0.01 m thick), fixed on its perimeter, loaded with a punctual force (010062SSLSB_FEM) Test ID: 2492 Test status: Passed

1.57.1 Description Verifies the vertical displacement for a square plate (0.01 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.

1.57.2 Background ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.

1.57.2.1 Model description Square plate of side "a". 0.01 m thick plate fixed on its perimeter

Scale = 1/5

01-0062SSLSB_FEM

Units I. S. Geometry ■ ■ ■

Side: a = 1 m, Thickness: h = 0.01 m, a Slenderness: λ = h = 100.

191

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■ ■

Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■ ■

Outer: Fixed edges, Inner: None.

Loading ■ ■

6 External: Punctual force applied on the center of the plate: FZ = -10 N, Internal: None.

1.57.2.2 Vertical displacement at point C (center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

192

Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.

ADVANCE DESIGN VALIDATION GUIDE

1.57.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-0.29579

1.57.3 Calculated results

Result name DZ

Result description Vertical displacement in point C [m]

Value -0.292146 m

Error 1.23%

193

ADVANCE DESIGN VALIDATION GUIDE

1.58 A plate (0.02 m thick), fixed on its perimeter, loaded with a uniform pressure (010059SSLSB_FEM) Test ID: 2489 Test status: Passed

1.58.1 Description Verifies the vertical displacement for a square plate (0.02 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.

1.58.2 Background

1.58.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.

Units I. S. Geometry ■ ■ ■

Side: a = 1 m, Thickness: h = 0.02 m, a Slenderness: λ = h = 50.

Materials properties ■ ■

Reinforcement, 11 Longitudinal elastic modulus: E = 2.1 x 10 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

■

194

Inner: None.

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: 1 MPa uniform pressure, Internal: None.

1.58.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.

1.58.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-0.83480 x 10-2

1.58.3 Calculated results

Result name Dz

Result description Vertical displacement in point C [cm]

Value -0.82559 cm

Error 1.10%

195

ADVANCE DESIGN VALIDATION GUIDE

1.59 A plate (0.01333 m thick), fixed on its perimeter, loaded with a punctual force (010063SSLSB_FEM) Test ID: 2493 Test status: Passed

1.59.1 Description Verifies the vertical displacement for a square plate (0.01333 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.

1.59.2 Background

1.59.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar. 0.01333 m thick plate fixed on its perimeter 01-0063SSLSB_FEM

Units I. S. Geometry ■ ■ ■

196

Side: a = 1 m, Thickness: h = 0.01333 m, a Slenderness: λ = h = 75.

Scale = 1/5

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■ ■

Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■ ■

Outer: Fixed sides, Inner: None.

Loading ■ ■

6 External: Punctual force applied on the center of the plate: FZ = -10 N, Internal: None.

1.59.2.2 Vertical displacement at point C (the center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.

197

ADVANCE DESIGN VALIDATION GUIDE

1.59.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-0.12525

1.59.3 Calculated results

Result name DZ

198

Result description Vertical displacement in point C [m]

Value -0.124583 m

Error 0.53%

ADVANCE DESIGN VALIDATION GUIDE

1.60 A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (01-0066SSLSB_FEM) Test ID: 2496 Test status: Passed

1.60.1 Description Verifies the vertical displacement for a square plate (0.1 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.

1.60.2 Background

1.60.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar. 0.1 m thick plate fixed on its perimeter

Scale = 1/5

01-0066SSLSB_FEM

Units I. S. Geometry ■ ■

Side: a = 1 m, Thickness: h = 0.1 m,

■

Slenderness: λ = 10.

Materials properties ■ ■

Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

199

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: Fixed edges, Inner: None.

Loading ■ ■

External: punctual force applied in the center of the plate: FZ = -106 N, Internal: None.

1.60.2.2 Vertical displacement at point C (center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.

1.60.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-0.42995 x 10-3

1.60.3 Calculated results

Result name DZ

200

Result description Vertical displacement in point C [mm]

Value -0.412094 mm

Error 4.15%

ADVANCE DESIGN VALIDATION GUIDE

1.61 Vibration mode of a thin piping elbow in space (case 2) (01-0068SDLLB_FEM) Test ID: 2498 Test status: Passed

1.61.1 Description Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, extended with two straight elements (0.6 m long) and subjected to its self weight only.

1.61.2 Background

1.61.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (in space); Element type: linear. Vibration mode of a thin piping elbow

Scale = 1/11

01-0068SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■

Average radius of curvature: OA = R = 1 m, L = 0.6 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, Section: A = 1.131 x 10-4 m2, -9 4 Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10 m , -9 4 Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10 m ,

201

ADVANCE DESIGN VALIDATION GUIDE

■ ■

Polar inertia: Ip = 9.274 x 10-9 m4. Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■

Outer: Fixed at points C and D In A: translation restraint along y and z, ► In B: translation restraint along x and z, Inner: None. ► ►

■

Loading ■ ■

External: None, Internal: None.

1.61.2.2 Eigen modes frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■

transverse bending: fj =

μi2 2π R2

GIp ρA

where i = 1,2.

Finite elements modeling ■ ■ ■

Linear element: beam, 23 nodes, 22 linear elements.

Eigen mode shapes

202

ADVANCE DESIGN VALIDATION GUIDE

1.61.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode Transverse 1 frequency [Hz]

33.4

CM2

Eigen mode

Eigen mode Transverse 2 frequency [Hz]

100

1.61.3 Calculated results

Result name

Result description

Value

Error

Eigen mode Transverse 1 frequency [Hz]

33.19 Hz

-0.63%

Eigen mode Transverse 2 frequency [Hz]

94.62 Hz

-5.38%

203

ADVANCE DESIGN VALIDATION GUIDE

1.62 Vibration mode of a thin piping elbow in space (case 1) (01-0067SDLLB_FEM) Test ID: 2497 Test status: Passed

1.62.1 Description Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, fixed on its ends and subjected to its self weight only.

1.62.2 Background

1.62.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (space problem); Element type: linear. Vibration mode of a thin piping elbow

Scale = 1/7

01-0067SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■

204

Average radius of curvature: OA = R = 1 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, -4 2 Section: A = 1.131 x 10 m , Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9 m4, -9 4 Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10 m ,

ADVANCE DESIGN VALIDATION GUIDE

■ ■

Polar inertia: Ip = 9.274 x 10-9 m4. Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0)

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■ ■

Outer: Fixed at points A and B, Inner: None.

Loading ■ ■

External: None, Internal: None.

1.62.2.2 Eigen modes frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■ fj =

transverse bending: μi2 2π R2

GIp ρA

where i = 1,2.

Finite elements modeling ■ ■ ■

Linear element: beam, 11 nodes, 10 linear elements.

Eigen mode shapes

1.62.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode Transverse 1 frequency [Hz]

44.23

CM2

Eigen mode

Eigen mode Transverse 2 frequency [Hz]

125

205

ADVANCE DESIGN VALIDATION GUIDE

1.62.3 Calculated results

Result name

206

Result description

Value

Error

Eigen mode Transverse 1 frequency [Hz]

44.12 Hz

-0.25%

Eigen mode Transverse 2 frequency [Hz]

120.09 Hz

-3.93%

ADVANCE DESIGN VALIDATION GUIDE

1.63 A plate (0.01 m thick), fixed on its perimeter, loaded with a uniform pressure (010057SSLSB_FEM) Test ID: 2487 Test status: Passed

1.63.1 Description Verifies the vertical displacement for a square plate (0.01 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.

1.63.2 Background

1.63.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.

Units I. S. Geometry ■ ■ ■

Side: a = 1 m, Thickness: h = 0.01 m, a Slenderness: λ = h = 100.

Materials properties ■ ■

Reinforcement, 11 Longitudinal elastic modulus: E = 2.1 x 10 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

■

Inner: None.

207

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: 1 MPa uniform pressure, Internal: None.

1.63.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.

1.63.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-6.639 x 10-2

1.63.3 Calculated results

Result name Dz

208

Result description Vertical displacement in point C [cm]

Value -6.56563 cm

Error 1.11%

ADVANCE DESIGN VALIDATION GUIDE

1.64 A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (010061SSLSB_FEM) Test ID: 2491 Test status: Passed

1.64.1 Description Verifies the vertical displacement for a square plate (0.1 m thick), of side "a", fixed on its perimeter, loaded with a uniform pressure.

1.64.2 Background

1.64.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar.

Units I. S. Geometry ■ ■ ■

Side: a = 1 m, Thickness: h = 0.1 m, a Slenderness: λ = h = 10.

Materials properties ■ ■

Reinforcement, 11 Longitudinal elastic modulus: E = 2.1 x 10 Pa,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

■

Inner: None.

209

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: 1 MPa uniform pressure, Internal: None.

1.64.2.2 Vertical displacement at C Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 289 nodes, 256 surface quadrangles.

1.64.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-0.78661 x 10-4

1.64.3 Calculated results

Result name Dz

210

Result description Vertical displacement in point C [mm]

Value -0.0781846 mm

Error 0.61%

ADVANCE DESIGN VALIDATION GUIDE

1.65 A plate (0.05 m thick), fixed on its perimeter, loaded with a punctual force (010065SSLSB_FEM) Test ID: 2495 Test status: Passed

1.65.1 Description Verifies the vertical displacement for a square plate (0.05 m thick), of side "a", fixed on its perimeter, loaded with a punctual force in the center.

1.65.2 Background

1.65.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLV 09/89; Analysis type: static; Element type: planar. 0.05 m thick plate fixed on its perimeter

Scale = 1/5

01-0065SSLSB_FEM

Units I. S. Geometry ■ ■

Side: a = 1 m, Thickness: h = 0.05 m,

■

Slenderness: λ = 20.

Materials properties ■ ■

Reinforcement, Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

211

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: Fixed sides, Inner: None.

Loading ■ ■

External: Punctual force applied at the center of the plate: FZ = -106 N, Internal: None.

1.65.2.2 Vertical displacement at point C center of the plate) Reference solution This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained with Serendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these values at ± 5%. Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 961 nodes, 900 surface quadrangles.

1.65.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement in point C [m]

-0.2595 x 10-2

1.65.3 Calculated results

Result name DZ

212

Result description Vertical displacement in point C [m]

Value -0.00257232 m

Error 0.86%

ADVANCE DESIGN VALIDATION GUIDE

1.66 Reactions on supports and bending moments on a 2D portal frame (Rafters) (010077SSLPB_FEM) Test ID: 2500 Test status: Passed

1.66.1 Description Moments and actions on supports calculation on a 2D portal frame. The purpose of this test is to verify the results of Advance Design for the M. R. study of a 2D portal frame.

1.66.2 Background

1.66.2.1 Model description ■ ■ ■

Reference: Design and calculation of metal structures. Analysis type: static linear; Element type: linear.

213

ADVANCE DESIGN VALIDATION GUIDE

1.66.2.2 Moments and actions on supports M.R. calculation on a 2D portal frame. RDM results, for the linear load perpendicular on the rafters, are:

VA = VE =

qL 2

H A = HE =

MB = MD = −Hh

MC =

qL² 8h + 5f =H 32 h²(k + 3) + f (3h + f ) qL ² − H(h + f ) 8

1.66.2.3 Theoretical results Comparison between theoretical results and the results obtained by Advance Design for a linear load perpendicular on the chords Solver

Result name

Result description

Reference value

CM2

Vertical reaction V in A [DaN]

-1000

CM2

Vertical reaction V in E [DaN]

-1000

CM2

Horizontal reaction H in A [DaN]

-332.9

CM2

Horizontal reaction H in E [DaN]

-332.9

CM2

Moment in node B [DaNm]

2496.8

CM2

Moment in node D [DaNm]

-2496.8

CM2

Moment in node C [DaNm]

-1671

1.66.3 Calculated results

Result name Fz

214

Result description

Value

Error

Vertical reaction V on node A [daN]

-1000 daN

0.00%

Vertical reaction V on node E [daN]

-1000 daN

0.00%

Horizontal reaction H on node A [daN]

-332.665 daN

0.07%

Horizontal reaction H on node E [daN]

-332.665 daN

0.07%

Moment in node B [daNm]

2494.99 daN*m

-0.07%

Moment in node D [daNm]

-2494.99 daN*m

0.07%

Moment in node C [daNm]

-1673.35 daN*m

-0.14%

ADVANCE DESIGN VALIDATION GUIDE

1.67 Slender beam of variable rectangular section (fixed-fixed) (01-0086SDLLB_FEM) Test ID: 2504 Test status: Passed

1.67.1 Description Verifies the eigen modes (flexion) for a slender beam with variable rectangular section (fixed-fixed).

1.67.2 BackgroundOverview

1.67.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 10/89; Analysis type: modal analysis (plane problem); Element type: linear.

Units I. S. Geometry ■ ■ ■

■

Length: L = 0.6 m, Constant thickness: h = 0.01 m Initial section: ► b0 = 0.03 m ► A0 = 3 x 10-4 m² Section variation: ►

with (α = 1)

►

b = b0e-2αx

►

A = A0e-2αx

Materials properties ■

E = 2 x 1011 Pa

■

ν = 0.3

■

ρ = 7800 kg/m3

Boundary conditions ■

Outer: Fixed at end x = 0, Fixed at end x = 0.6 m. Inner: None.

► ►

■

Loading ■ ■

External: None, Internal: None.

215

ADVANCE DESIGN VALIDATION GUIDE

1.67.2.2 Reference results Calculation method used to obtain the reference solution ωi pulsation is given by the roots of the equation:

1 − cos(rl)ch(sl) +

s² − r ² sh(sl)sin(rl) = 0 2rs

with

λ4i =

(

)

ρA 0 ωi2 si ; r = α ² + λ2i ; s = λ2i − α ² ⎯⎯→ λ2i − α ² > 0 EIzo

Therefore, the ν translation components of φi(x) mode, are:

⎤ ⎡ cos(rl) − ch(sl) (s sin(rx ) − rsh( sx ))⎥ Φ i (x ) = e αx ⎢cos(rx ) − ch(sx ) + rsh(sl) − s sin(rl) ⎦ ⎣ Uncertainty about the reference: analytical solution: Reference values Eigen mode Frequency (Hz) Eigen mode φi(x)* order x=0 0.1 0.2 0.3 0.4 1 143.303 0 0.237 0.703 1 0.859 2 396.821 0 -0.504 -0.818 0 0.943 3 779.425 0 0.670 0.210 -0.831 0.257 4 1289.577 0 -0.670 0.486 0 -0.594 * φi(x) eigen modes* standardized to 1 at the point of maximum amplitude. Eigen modes

216

0.5 0.354 0.752 1 1

0.6 0 0 0 0

ADVANCE DESIGN VALIDATION GUIDE

1.67.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Eigen mode

Frequency of eigen mode 1 [Hz]

143.303

CM2

Eigen mode

Frequency of eigen mode 2 [Hz]

396.821

CM2

Eigen mode

Frequency of eigen mode 3 [Hz]

779.425

CM2

Eigen mode

Frequency of eigen mode 4 [Hz]

1289.577

1.67.3 Calculated results

Result name

Result description

Value

Error

Frequency of eigen mode 1 [Hz]

145.88 Hz

1.80%

Frequency of eigen mode 2 [Hz]

400.26 Hz

0.87%

Frequency of eigen mode 3 [Hz]

783.15 Hz

0.48%

Frequency of eigen mode 4 [Hz]

1293.42 Hz

0.30%

217

ADVANCE DESIGN VALIDATION GUIDE

1.68 Short beam on two hinged supports (01-0084SSLLB_FEM) Test ID: 2502 Test status: Passed

1.68.1 Description Verifies the deflection magnitude on a non-slender beam with two hinged supports.

1.68.2 Background

1.68.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 02/89 Analysis type: static linear (plane problem); Element type: linear.

Units I. S. Geometry ■ ■ ■ ■

Length: L = 1.44 m, Area: A = 31 x 10-4 m² Inertia: I = 2810 x 10-8 m4 Shearing coefficient: az = 2.42 = A/Ar

Materials properties ■

E = 2 x 1011 Pa

■

ν = 0.3

Boundary conditions ■ ■

218

Hinge at end x = 0, Hinge at end x = 1.44 m.

ADVANCE DESIGN VALIDATION GUIDE

Loading Uniformly distributed force of p = -1. X 105 N/m on beam AB.

1.68.2.2 Reference results Calculation method used to obtain the reference solution The deflection on the middle of a non-slender beam considering the shear force deformations given by the Timoshenko function:

5 pl 4 l2p + 384 EI 8 A r G

where G =

E 2(1 + Ď&#x2026;)

and A r = A

where "Ar" is the reduced area and "az" the shear coefficient calculated on the transverse section. Uncertainty about the reference: analytical solution: Reference values Point C

Magnitudes and units V, deflection (m)

Value -1.25926 x 10-3

1.68.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Deflection magnitude in point C [m]

-1.25926

1.68.3 Calculated results

Result name Dz

Result description Deflection magnitude in node C [m]

Value -0.00125926 m

Error 0.00%

219

ADVANCE DESIGN VALIDATION GUIDE

1.69 Double fixed beam in Eulerian buckling with a thermal load (01-0091HFLLB_FEM) Test ID: 2506 Test status: Passed

1.69.1 Description Verifies the normal force on the nodes of a double fixed beam in Eulerian buckling with a thermal load.

1.69.2 Background

1.69.2.1 Model description ■ ■ ■

Reference: Euler theory; Analysis type: Eulerian buckling; Element type: linear.

Units I. S. Geometry L = 10 m Cross Section IPE200

Sx m² Vx m3 0.002850 0.00000000

Sy m² V1y m3 0.001400 0.00002850

Sz m² V1z m3 0.001799 0.00019400

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,

■

Poisson's ratio: ν = 0.3.

■

Coefficient of thermal expansion: α = 0.00001

Boundary conditions ■ ■

220

Outer: Fixed at end x = 0, Inner: None.

Ix m4 V2y m3 0.0000000646 0.00002850

Iy m4 V2z m3 0.0000014200 0.00019400

Iz m4 0.0000194300

ADVANCE DESIGN VALIDATION GUIDE

Loading ■

External: Punctual load FZ = 1 N at = L/2 (load that initializes the deformed shape),

■

Internal: ΔT = 5°C corresponding to a compression force of:

N = ESαΔT = 2.1E11 × 0.00285 × 0.00001 × 5 = 29.925 kN

1.69.2.2 Displacement of the model in the linear elastic range Reference solution The reference critical load established by Euler is:

Pcritical =

π 2 EI ⎛L⎞ ⎜ ⎟ ⎝2⎠

= 117.724 kN ⇒ λ =

29.925 = 3.93 117.724

Observation: in this case, the thermal load has no effect over the critical coefficient Finite elements modeling ■ ■ ■

Linear element: beam, imposed mesh, 11 nodes, 10 elements.

Deformed shape of mode 1

1.69.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Normal Force on Node 6 - Case 101 [kN]

-29.925

CM2

Normal Force on Node 6 - Case 102 [kN]

-117.724

1.69.3 Calculated results Result name Fx Fx

Result description Normal Force Fx on Node 6 - Case 101 [kN] Normal Force Fx on Node 6 - Case 102 [kN]

Value -29.904 kN -118.081 kN

Error 0.07% -0.30%

221

ADVANCE DESIGN VALIDATION GUIDE

1.70 Reactions on supports and bending moments on a 2D portal frame (Columns) (010078SSLPB_FEM) Test ID: 2501 Test status: Passed

1.70.1 Description Moments and actions on supports calculation on a 2D portal frame. The purpose of this test is to verify the results of Advance Design for the M. R. study of a 2D portal frame.

1.70.2 Background

1.70.2.1 Model description ■ ■ ■

222

Reference: Design and calculation of metal structures. Analysis type: static linear; Element type: linear.

ADVANCE DESIGN VALIDATION GUIDE

1.70.2.2 Moments and reactions on supports M.R. calculation on a 2D portal frame. RDM results, for the linear load perpendicular on the column, are:

VA = −VE = −

MB =

qh² 2L

HE =

qh² − HEh 2

qh ² 5kh + 6 (2h + f ) 16 h ² (k + 3 ) + f (3h + f )

H A = HE − qh

qh² − HE (h + f ) 4

MD = −HE h

MC =

1.70.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical reaction V in A [DaN]

140.6

CM2

Vertical reaction V in E [DaN]

-140.6

CM2

Horizontal reaction H in A [DaN]

579.1

CM2

Horizontal reaction H in E [DaN]

170.9

CM2

Moment in B [DaNm]

-1530.8

CM2

Moment in D [DaNm]

-1281.7

CM2

Moment in C [DaNm]

302.7

1.70.3 Calculated results

Result name Fz

Result description

Value

Error

Vertical reaction V on node A [daN]

140.625 daN

0.02%

Vertical reaction V on node E [daN]

-140.625 daN

-0.02%

Horizontal reaction H on node A [daN]

579.169 daN

0.01%

Horizontal reaction H on node E [daN]

170.831 daN

-0.04%

Moment in node B [daNm]

-1531.27 daN*m

-0.03%

Moment in node D [daNm]

-1281.23 daN*m

0.04%

Moment in node C [daNm]

302.063 daN*m

-0.21%

223

ADVANCE DESIGN VALIDATION GUIDE

1.71 Plane portal frame with hinged supports (01-0089SSLLB_FEM) Test ID: 2505 Test status: Passed

1.71.1 Description Calculation of support reactions of a 2D portal frame with hinged supports.

1.71.2 Background

1.71.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLL 14/89; Analysis type: static linear; Element type: linear.

Units I. S. Geometry ■ ■ ■ ■ ■ ■

Length: L = 20 m, I1 = 5.0 x 10-4 m4 a=4m h=8m b = 10.77 m -4 4 I2 = 2.5 x 10 m

Materials properties ■ ■

224

Isotropic linear elastic material. E = 2.1 x 1011 Pa

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions Hinged base plates A and B (uA = vA = 0 ; uB = vB = 0). Loading

■ ■ ■ ■

p = -3 000 N/m F1 = -20 000 N F2 = -10 000 N M = -100 000 Nm

1.71.2.2 Calculation method used to obtain the reference solution ■ ■ ■ ■ ■ ■ ■ ■

K = (I2/b)(h/I1) p = a/h m=1+p B = 2(K + 1) + m C = 1 + 2m N = B + mC VA = 3pl/8 + F1/2 – M/l + F2h/l HA = pl²(3 + 5m)/(32Nh) + (F1l/(4h))(C/N) + F2(1-(B + C)/(2N)) + (3M/h)((1 + m)/(2N))

1.71.2.3 Reference values Point A A C

Magnitudes and units V, vertical reaction (N) H, horizontal reaction (N) vc (m)

Value 31 500.0 20 239.4 -0.03072

1.71.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical reaction V in point A [N]

-31500

CM2

Horizontal reaction H in point A [N]

-20239.4

CM2

vc displacement in point C [m]

-0.03072

1.71.3 Calculated results Result name Fz

Result description

Value

Error

Vertical reaction V in point A [N]

-31500 N

0.00%

Horizontal reaction H in point A [N]

-20239.3 N

0.00%

Displacement in point C [m]

-0.0307191 m

0.00%

225

ADVANCE DESIGN VALIDATION GUIDE

1.72 A 3D bar structure with elastic support (01-0094SSLLB_FEM) Test ID: 2508 Test status: Passed

1.72.1 Description A 3D bar structure with elastic support is subjected to a vertical load of -100 kN. The V2 magnitude on node 5, the normal force magnitude, the reaction magnitude on supports and the action magnitude are verified.

1.72.1.1 Model description ■ ■ ■

Reference: Internal GRAITEC; Analysis type: static linear; Element type: linear.

Units I. S. Geometry For all bars: ■ ■ ■

H=3m B=3m S = 0.02 m2 Element 1 (bar) 2 (bar) 3 (bar) 4 (bar) 5 (spring)

Node i 1 2 3 4 5

Node j 5 5 5 5 6

Materials properties ■ ■

226

Isotropic linear elastic materials Longitudinal elastic modulus: E = 2.1 E8 N/m2,

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: At node 5: K = 50000 kN/m ; Inner: None.

Loading ■ ■

External: Vertical load at node: P = -100 kN, Internal: None.

1.72.1.2 Theoretical results System solution

L= H2 + ■

B2 . Also, U1 = V1 = U5 = U6 = V6 = 0 2

Stiffness matrix of bar 1

227

ADVANCE DESIGN VALIDATION GUIDE

1 1 x x u ( x) = (1 − ).ui + .u j ⇔ u (ξ ) = (1 − ξ ).ui + (1 + ξ ).u j 2 2 L L 2x −1 where ξ = L in the local coordinate system:

⎧ 1⎫ ⎪− 2 ⎪⎧ 1 T [k1 ] = ∫ [B] [H ][B]dVe = ES ∫0 [B ] [B ]dx = ES ∫−1 ⎨ 1 ⎬⎨− ve ⎪ ⎪⎩ 2 ⎩ 2 ⎭ T

⎡1 ⎢ ES ⎢ 0 [k 1 ] = ⎢ L −1 ⎢ ⎣0

1⎫ 2 ⎬ dξ 2⎭ L

⎡1 1⎤ ⎢ − ⎥ 4 dξ = ES ⎡ 1 − 1⎤ (ui ) = ES ⎢ 0 1 ⎥ L ⎢⎣− 1 1 ⎥⎦ (u j ) L ⎢− 1 ⎥ ⎢ 4 ⎦ ⎣0

⎡ 1 2 ES ⎢ 4 = ⎢ L −∫1 ⎢− 1 ⎣ 4 1

where

0 − 1 0⎤ (ui ) 0 0 0⎥⎥ (vi ) 0 1 0⎥ (u j ) ⎥ 0 0 0⎦ (v j )

0 − 1 0⎤ (u1 ) ⎥ 0 0 0⎥ ( v 1 ) 0 1 0⎥ (u 5 ) ⎥ 0 0 0⎦ ( v 5 )

[ ]

The elementary matrix k e expressed in the global coordinate system XY is the following: (θ angle allowing the transition from the global base to the local base):

0 0 ⎤ ⎡ cos θ sin θ ⎢ ⎥ − θ θ sin cos 0 0 ⎥ T − [K e ] = [R e ] [k e ][R e ] avec [R e ] = ⎢ ⎢ 0 0 cos θ sin θ ⎥ ⎢ ⎥ − sin θ cos θ⎦ 0 ⎣ 0 ⎡ cos 2 θ − cos 2 θ − cos θ sin θ⎤ cos θ sin θ ⎢ ⎥ 2 − cos θ sin θ − sin 2 θ ⎥ sin θ ES ⎢ cos θ sin θ [K e ] = ⎢ L − cos 2 θ − cos θ sin θ cos 2 θ cos θ sin θ ⎥ ⎢ ⎥ 2 − sin θ cos θ sin θ sin 2 θ ⎦⎥ ⎣⎢− cos θ sin θ

Knowing that

cosθ =

B L 2

and sinθ =

H L

⎧ B2 2 θ = cos ⎪ 2L2 ⎪ ⎪⎪ H2 , then: ⎨sin 2 θ = 2 L ⎪ ⎪ HB ⎪sin θ cos θ = 2L2 ⎩⎪

⎡ B2 ⎢ ⎢ 2 ⎢ HB H ES ⎢ 2 for element 1 nodes 1 → 5, θ = arctan ( ) : [K1 ] = 3 ⎢ B2 D L ⎢− ⎢ 2 ⎢ HB ⎢− 2 ⎣ 228

HB 2 H2 −

HB 2

−H2

B2 2 HB − 2 2 B 2 HB 2 −

HB ⎤ ⎥ 2⎥ (U 1 ) −H2⎥ ⎥ (V1 ) HB ⎥ (U 5 ) ⎥ 2 ⎥ (V5 ) ⎥ H2 ⎥ ⎦

−

ADVANCE DESIGN VALIDATION GUIDE

■

Stiffness matrix of spring support 5

We say K ′ =

K 4

⎡1 ⎢0 ⎡ 1 − 1⎤ (ui ) ⎢ ′ = K in the local coordinate system : [k 5 ] = K ′⎢ ⎥ ⎢− 1 ⎣− 1 1 ⎦ (u j ) ⎢ ⎣0

⎡0 ⎢0 for element 5 nodes 5 → 6, θ = 90° : [K 5 ] = K ' ⎢ ⎢0 ⎢ ⎣0

0 − 1 0⎤ (ui ) 0 0 0⎥⎥ (vi ) 0 1 0⎥ (u j ) ⎥ 0 0 0⎦ (v j )

0 ⎤ (U 5 ) 1 0 − 1⎥⎥ (V5 ) 0 0 0 ⎥ (U 6 ) ⎥ 1 0 − 1⎦ (V6 ) 0 0

[ ]

System K {Q} = {F}

■

⎡ ES B 2 ⎢ 3 ⎢ L 2 ⎢ ES HB ⎢ L3 2 ⎢ 2 ⎢ − ES B 3 ⎢ L 2 ⎢ ⎢− ES HB ⎢ L3 2 ⎢ 0 ⎢ 0 ⎣⎢

ES HB L3 2 ES 2 H L3 ES HB − 3 L 2 ES 2 − 3 H L 0 0

ES B 2 L3 2 ES HB − 3 L 2 ES B 2 L3 2 ES HB L3 2 0 0 −

ES HB L3 2 ES − 3 H2 L ES HB L3 2 ES 2 H + K′ L3 0 − K′ −

0 0 0 0 0 0

⎤ 0 ⎥ ⎥ ⎧U ⎫ ⎧R ⎫ X1 ⎥ 1 0 ⎥ ⎪ V ⎪ ⎪R ⎪ 1 Y1 ⎪ ⎥ ⎪⎪ ⎪⎪ ⎪⎪ R U ⎪ 5 ⎪ ⎪ X5 ⎪⎪ 0 ⎥⎨ ⎬ = ⎨ ⎬ ⎥ V5 −P⎪ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − K ′⎥ ⎪U 6 ⎪ ⎪R X 6 ⎪ ⎥ ⎪V ⎪ ⎪R ⎪ ⎩ 6 ⎭ ⎩ Y6 ⎭ 0 ⎥ ⎥ K ′ ⎦⎥

If U1 = V1 = U5 = U6 = V6 = 0, then:

V5 =

−P

ES 2 H + K′ L3

−P

4 = −0.001885 m ES 2 K + H 4 L3

ES HB ES HB V5 = 1015 N R X5 = 3 V5 = −1015 N R X6 = 0 L3 2 L 2 ES K R Y 6 = − V5 = 23563 N = − 3 H 2 V5 = 1436 N 4 L

R X1 = − And

R Y1 Note: ■ ■

The values on supports specified by Advance Design correspond to the actions, RY6 calculated value must be multiplied by 4 in relation to the double symmetry,

■

value is similar to the one found by Advance Design by dividing this by

229

ADVANCE DESIGN VALIDATION GUIDE

Effort in bar 1:

⎡ B ⎢L 2 ⎧u1 ⎫ ⎢ H ⎪v ⎪ ⎢ − ⎪ 1⎪ ⎢ L ⎨ ⎬=⎢ ⎪u5 ⎪ ⎢ 0 ⎪⎩v5 ⎪⎭ ⎢ ⎢ 0 ⎢⎣

H L B L 2

0 0 B

L 2 H − L

⎧u1 ⎫ ⎡ cosθ ⎪ v ⎪ ⎢− sin θ ⎪ 1⎪ ⎢ ⎨ ⎬= ⎪u5 ⎪ ⎢ 0 ⎪⎩v5 ⎪⎭ ⎢⎣ 0

⎤ 0 ⎥ ⎥ ⎧U1 ⎫ 0 ⎥⎪V ⎪ ⎥ ⎪ 1 ⎪ and ES ⎡ 1 − 1⎤ ⎧u1 ⎫ = ⎧ N1 ⎫ = ⎧ 1759 ⎫ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ H ⎥ ⎨U 5 ⎬ L ⎢⎣− 1 1 ⎥⎦ ⎩u5 ⎭ ⎩ N 5 ⎭ ⎩− 1759⎭ ⎪ ⎪ L ⎥⎥ ⎪V ⎪ ⎩ 5⎭ B ⎥ L 2 ⎥⎦

sin θ cosθ

0 0

cosθ − sin θ

0 ⎤ ⎧U1 ⎫ 0 ⎥⎥ ⎪⎪ V1 ⎪⎪ ES ⎡ 1 − 1⎤ ⎧u1 ⎫ ⎧ N1 ⎫ ⎨ ⎬ and ⎨ ⎬=⎨ ⎬ sin θ ⎥ ⎪U 5 ⎪ L ⎢⎣− 1 1 ⎥⎦ ⎩u5 ⎭ ⎩ N 5 ⎭ ⎥ cosθ ⎦ ⎪⎩V5 ⎪⎭

Reference values

Point 5 All bars Supports 1, 3, 4 and 5 Supports 1, 3, 4 and 5

Magnitude V2 Normal force Fz action Action Fx=Fy

Units m N N N

Support 6

Fz action

Finite elements modeling ■ ■ ■

230

Linear element: beam, automatic mesh, 5 nodes, 4 linear elements.

Value -1.885 10-3 -1759 -1436

1015 / 2 = ±718 23563 x 4=94253

ADVANCE DESIGN VALIDATION GUIDE

Deformed shape

Normal forces diagram

231

ADVANCE DESIGN VALIDATION GUIDE

1.72.1.3 Reference values Solver

Result name

Result description

Reference value

CM2

V2 magnitude on node 5 [m]

-1.885 10-3

CM2

Normal force magnitude on bar 1 [N]

-1759

CM2

Normal force magnitude on bar 2 [N]

-1759

CM2

Normal force magnitude on bar 3 [N]

-1759

CM2

Normal force magnitude on bar 4 [N]

-1759

CM2

Fz reaction magnitude on support 1 [N]

1436

CM2

Fz reaction magnitude on support 3 [N]

1436

CM2

Fz reaction magnitude on support 4 [N]

1436

CM2

Fz reaction magnitude on support 5 [N]

1436

CM2

Action Fx magnitude on support 1 [N]

-718

CM2

Action Fx magnitude on support 3 [N]

718

CM2

Action Fx magnitude on support 4 [N]

718

CM2

Action Fx magnitude on support 5 [N]

-718

CM2

Action Fy magnitude on support 1 [N]

-718

CM2

Action Fy magnitude on support 3 [N]

-718

CM2

Action Fy magnitude on support 4 [N]

718

CM2

Action Fy magnitude on support 5 [N]

718

CM2

Fz reaction magnitude on support 6 [N]

23563 x 4=94253

1.72.2 Calculated results

Result name D

232

Result description

Value

Error

Displacement on node 5 [mm]

1.88508 mm

0.00%

Normal force magnitude on bar 1 [N]

-1759.4 N

-0.02%

Normal force magnitude on bar 2 [N]

-1759.4 N

-0.02%

Normal force magnitude on bar 3 [N]

-1759.4 N

-0.02%

Normal force magnitude on bar 4 [N]

-1759.4 N

-0.02%

Fz reaction magnitude on support 1 [N]

1436.55 N

0.04%

Fz reaction magnitude on support 2 [N]

1436.55 N

0.04%

Fz reaction magnitude on support 3 [N]

1436.55 N

0.04%

Fz reaction magnitude on support 4 [N]

1436.55 N

0.04%

Action Fx magnitude on support 1 [N]

-718.274 N

-0.04%

Action Fx magnitude on support 2 [N]

718.274 N

0.04%

Action Fx magnitude on support 3 [N]

718.274 N

0.04%

Action Fx magnitude on support 4 [N]

-718.274 N

-0.04%

Action Fy magnitude on support 1 [N]

-718.274 N

-0.04%

Action Fy magnitude on support 2 [N]

-718.274 N

-0.04%

Action Fy magnitude on support 3 [N]

718.274 N

0.04%

Action Fy magnitude on support 4 [N]

718.274 N

0.04%

Action Fy magnitude on support 6 [N]

94253.8 N

0.00%

ADVANCE DESIGN VALIDATION GUIDE

1.73 Fixed/free slender beam with eccentric mass or inertia (01-0096SDLLB_FEM) Test ID: 2510 Test status: Passed

1.73.1 Description Fixed/free slender beam with eccentric mass or inertia. Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending, transverse bending, punctual mass.

1.73.2 Background

1.73.2.1 Model description ■ ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 15/89; Analysis type: modal analysis; Element type: linear. Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending, transverse bending, punctual mass..

1.73.2.2 Problem data

Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■

Outer diameter: Inner diameter: Beam length: Distance BC: Area: Inertia: Polar inertia: Punctual mass:

de= 0.35 m, di = 0.32 m, l = 10 m, lBC = 1 m A =1.57865 x 10-2 m2 -4 4 Iy = Iz = 2.21899 x 10 m -4 4 Ip = 4.43798 x 10 m mc = 1000 kg

233

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elasticity modulus of AB element: E = 2.1 x 1011 Pa,

■

Density of the linear element AB: ρ = 7800 kg/m3

■ ■

Poisson's ratio ν=0.3(this coefficient was not specified in the AFNOR test , the value 0.3 seems to be the more appropriate to obtain the correct frequency value of modes No. 4 and 5 with NE/NASTRAN: Elastic modulus of BC element: E = 1021 Pa

■

Density of the linear element BC: ρ = 0 kg/m3

Boundary conditions Fixed at point A, x = 0, Loading None for the modal analysis

1.73.2.3 Reference frequencies Reference solutions The different eigen frequencies are determined using a finite elements model of Euler beam (slender beam). fz + t0 = flexion x,z + torsion fy + tr = flexion x,y + traction Mode 1 (fz + t0) 2 (fy + tr) 3 (fy + tr) 4 (fz + t0) 5 (fz + t0) 6 (fy + tr) 7 (fz + t0) 1 (fz + t0)

Units Hz Hz Hz Hz Hz Hz Hz Hz

Reference 1.636 1.642 13.460 13.590 28.900 31.960 61.610 63.930

Uncertainty about the reference solutions The uncertainty about the reference solutions: ± 1% Finite elements modeling ■ ■ ■ ■

Linear element AB: Beam Imposed mesh: 50 elements. Linear element BC: Beam Without meshing

Modal deformations

234

ADVANCE DESIGN VALIDATION GUIDE

1.73.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Frequency

Eigen mode 1 frequency (fz + t0) [Hz]

1.636

CM2

Frequency

Eigen mode 2 frequency (fy + tr) [Hz]

1.642

CM2

Frequency

Eigen mode 3 frequency (fy + tr) [Hz]

13.46

CM2

Frequency

Eigen mode 4 frequency (fz + t0) [Hz]

13.59

CM2

Frequency

Eigen mode 5 frequency (fz + t0) [Hz]

28.90

CM2

Frequency

Eigen mode 6 frequency (fy + tr) [Hz]

31.96

CM2

Frequency

Eigen mode 7 frequency (fz + t0) [Hz]

61.61

CM2

Frequency

Eigen mode 8 frequency (fy + tr) [Hz]

63.93

Note: fz + t0 = flexion x,z + torsion fy + tr = flexion x,y + traction Observation: because the mass matrix of Advance Design is condensed and not consistent, the torsion modes obtained are not taking into account the self rotation mass inertia of the beam.

235

ADVANCE DESIGN VALIDATION GUIDE

1.73.3 Calculated results

Result name

236

Result description

Value

Error

Eigen mode 1 frequency [Hz]

1.64 Hz

0.24%

Eigen mode 2 frequency [Hz]

1.64 Hz

-0.12%

Eigen mode 3 frequency [Hz]

13.45 Hz

-0.07%

Eigen mode 4 frequency [Hz]

13.65 Hz

0.44%

Eigen mode 5 frequency [Hz]

29.72 Hz

2.84%

Eigen mode 6 frequency [Hz]

31.96 Hz

0.00%

Eigen mode 7 frequency [Hz]

63.09 Hz

2.40%

Eigen mode 8 frequency [Hz]

63.93 Hz

100%

ADVANCE DESIGN VALIDATION GUIDE

1.74 Fixed/free slender beam with centered mass (01-0095SDLLB_FEM) Test ID: 2509 Test status: Passed

1.74.1 Description Fixed/free slender beam with centered mass. Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion, plane bending, transverse bending, punctual mass.

1.74.2 Background ■ ■ ■ ■

1.74.2.1 Test data

Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■

Outer diameter Inner diameter: Beam length: Area: Polar inertia: Inertia: Punctual mass: Beam self-weight:

de = 0.35 m, di = 0.32 m, l = 10 m, A = 1.57865 x 10-2 m2 -4 4 IP = 4.43798 x 10 m Iy = Iz = 2.21899 x 10-4m4 mc = 1000 kg M

237

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Density: ρ = 7800 kg/m3

■

Poisson's ratio: ν=0.3 (this coefficient was not specified in the AFNOR test , the value 0.3 seems to be the more appropriate to obtain the correct frequency value of mode No. 8 with NE/NASTRAN)

Boundary conditions ■ ■

Outer: Fixed at point A, x = 0, Inner: none

Loading None for the modal analysis

1.74.2.2 Reference results Reference frequency For the first mode, the Rayleigh method gives the approximation formula

f1 = 1 / 2Π x

3 EI z I (m c + 0.24M) 3

Mode 1 2 3 4 5 6 7 8 9 10 Comment:

Shape Flexion Flexion Flexion Flexion Flexion Flexion Traction Torsion Flexion Flexion

Units Hz Hz Hz Hz Hz Hz Hz Hz Hz Hz

Reference 1.65 1.65 16.07 16.07 50.02 50.02 76.47 80.47 103.2 103.2

The mass matrix associated with the beam torsion on two nodes, is expressed as:

ρ × l × IP ⎡ 1 1/ 2⎤ ×⎢ ⎥ 3 ⎣1/ 2 1 ⎦ And to the extent that Advance Design uses a condensed mass matrix, the value of the torsion mass inertia introduced in the model is set to:

ρ × l × Ip 3

Uncertainty about the reference frequencies ■

Analytical solution mode 1

■

Other modes: ± 1%

Finite elements modeling ■ ■

238

Linear element AB: Beam Beam meshing: 20 elements.

ADVANCE DESIGN VALIDATION GUIDE

Modal deformations

239

ADVANCE DESIGN VALIDATION GUIDE

Observation: the deformed shape of mode No. 8 that does not really correspond to a torsion deformation, is actually the display result of the translations and not of the rotations. This is confirmed by the rotation values of the corresponding mode.

Node 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

DX -3.336e-033 -5.030e-013 -1.005e-012 -1.505e-012 -2.002e-012 -2.495e-012 -2.983e-012 -3.464e-012 -3.939e-012 -4.406e-012 -4.863e-012 -5.310e-012 -5.746e-012 -6.169e-012 -6.580e-012 -6.976e-012 -7.357e-012 -7.723e-012 -8.072e-012 -8.403e-012 -8.717e-012

DY 6.479e-031 1.575e-008 6.185e-008 1.365e-007 2.381e-007 3.648e-007 5.149e-007 6.867e-007 8.785e-007 1.088e-006 1.315e-006 1.556e-006 1.811e-006 2.077e-006 2.353e-006 2.637e-006 2.928e-006 3.224e-006 3.524e-006 3.826e-006 4.130e-006

Eigen modes vector 8 DZ RX -6.316e-031 1.055e-022 -1.520e-008 1.472e-002 -5.966e-008 2.944e-002 -1.317e-007 4.416e-002 -2.296e-007 5.887e-002 -3.517e-007 7.359e-002 -4.963e-007 8.831e-002 -6.618e-007 1.030e-001 -8.464e-007 1.177e-001 -1.049e-006 1.325e-001 -1.267e-006 1.472e-001 -1.499e-006 1.619e-001 -1.744e-006 1.766e-001 -2.000e-006 1.913e-001 -2.265e-006 2.061e-001 -2.539e-006 2.208e-001 -2.819e-006 2.355e-001 -3.104e-006 2.502e-001 -3.393e-006 2.649e-001 -3.685e-006 2.797e-001 -3.977e-006 2.944e-001

With NE/NASTRAN, the results associated with mode No. 8, are:

240

RY 5.770e-028 6.022e-008 1.171e-007 1.705e-007 2.206e-007 2.673e-007 3.106e-007 3.506e-007 3.873e-007 4.207e-007 4.508e-007 4.777e-007 5.015e-007 5.221e-007 5.396e-007 5.541e-007 5.658e-007 5.746e-007 5.808e-007 5.844e-007 5.856e-007

RZ 5.980e-028 6.243e-008 1.214e-007 1.769e-007 2.289e-007 2.774e-007 3.225e-007 3.641e-007 4.023e-007 4.371e-007 4.684e-007 4.964e-007 5.210e-007 5.423e-007 5.605e-007 5.755e-007 5.874e-007 5.965e-007 6.028e-007 6.065e-007 6.077e-007

ADVANCE DESIGN VALIDATION GUIDE

1.74.2.3 Theoretical results Solver

Result description

Reference value

CM2

Result name

Eigen mode 1 frequency [Hz]

1.65

CM2

Eigen mode 2 frequency [Hz]

1.65

CM2

Eigen mode 3 frequency [Hz]

16.07

CM2

Eigen mode 4 frequency [Hz]

16.07

CM2

Eigen mode 5 frequency [Hz]

50.02

CM2

Eigen mode 6 frequency [Hz]

50.02

CM2

Eigen mode 7 frequency [Hz]

76.47

CM2

Eigen mode 9 frequency [Hz]

103.20

CM2

Eigen mode 10 frequency [Hz]

103.20

Comment: The difference between the reference frequency of torsion mode (mode No. 8) and the one found by Advance Design may be explained by the fact that Advance Design is using a lumped mass matrix (see the corresponding description sheet).

1.74.3 Calculated results

Result name

Result description

Value

Error

Eigen mode 1 frequency [Hz]

1.65 Hz

0.00%

Eigen mode 2 frequency [Hz]

1.65 Hz

0.00%

Eigen mode 3 frequency [Hz]

16.06 Hz

-0.06%

Eigen mode 4 frequency [Hz]

16.06 Hz

-0.06%

Eigen mode 5 frequency [Hz]

50 Hz

-0.04%

Eigen mode 6 frequency [Hz]

50 Hz

-0.04%

Eigen mode 7 frequency [Hz]

76.46 Hz

-0.01%

Eigen mode 9 frequency [Hz]

103.14 Hz

-0.06%

Eigen mode 10 frequency [Hz]

103.14 Hz

100%

241

ADVANCE DESIGN VALIDATION GUIDE

1.75 Vibration mode of a thin piping elbow in space (case 3) (01-0069SDLLB_FEM) Test ID: 2499 Test status: Passed

1.75.1 Description Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, extended with two straight elements (2 m long) and subjected to its self weight only.

1.75.1.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 14/89; Analysis type: modal analysis (space problem); Element type: linear. Vibration mode of a thin piping elbow

Scale = 1/12

01-0069SDLLB_FEM

Units I. S. Geometry ■ ■ ■ ■ ■ ■ ■ ■ ■

242

Average radius of curvature: OA = R = 1 m, L = 2 m, Straight circular hollow section: Outer diameter: de = 0.020 m, Inner diameter: di = 0.016 m, Section: A = 1.131 x 10-4 m2, -9 4 Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10 m , -9 4 Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10 m , -9 4 Polar inertia: Ip = 9.274 x 10 m .

ADVANCE DESIGN VALIDATION GUIDE

■

Points coordinates (in m): ► O(0;0;0) ► A(0;R;0) ► B(R;0;0) ► C ( -L ; R ; 0 ) ► D ( R ; -L ; 0 )

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Density: ρ = 7800 kg/m3.

Boundary conditions ■

Outer: Fixed at points C and D At A: translation restraint along y and z, ► At B: translation restraint along x and z, Inner: None. ► ►

■

Loading ■ ■

External: None, Internal: None.

1.75.1.2 Eigen modes frequencies Reference solution The Rayleigh method applied to a thin curved beam is used to determine parameters such as: ■

transverse bending: fj =

μi2 2π R2

GIp ρA

where i = 1,2 with i = 1,2:

Finite elements modeling ■ ■ ■

Linear element: beam, 41 nodes, 40 linear elements.

Eigen mode shapes

243

ADVANCE DESIGN VALIDATION GUIDE

1.75.1.3 Theoretical results Reference Solver

Result name

Result description

Reference value

CM2

Eigen mode

Eigen mode Transverse 1 frequency [Hz]

17.900

CM2

Eigen mode

Eigen mode Transverse 2 frequency [Hz]

24.800

1.75.2 Calculated results

Result name

244

Result description

Value

Error

Eigen mode Transverse 1 frequency [Hz]

17.65 Hz

-1.40%

Eigen mode Transverse 2 frequency [Hz]

24.43 Hz

-1.49%

ADVANCE DESIGN VALIDATION GUIDE

1.76 Slender beam of variable rectangular section with fixed-free ends (ß=5) (01-0085SDLLB_FEM) Test ID: 2503 Test status: Passed

1.76.1 Description Verifies the eigen modes (bending) for a slender beam with variable rectangular section (fixed-free).

1.76.2 Background

1.76.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SDLL 09/89; Analysis type: modal analysis (plane problem); Element type: linear.

Units I. S. Geometry ■ ■

■

Length: L = 1 m, Straight initial section: ► h0 = 0.04 m ► b0 = 0.05 m ► A0 = 2 x 10-3 m² Straight final section ► h1 = 0.01 m ► b1 = 0.01 m ► A1 = 10-4 m²

Materials properties ■

E = 2 x 1011 Pa

■

ρ = 7800 kg/m3

Boundary conditions ■

Outer: Fixed at end x = 0, ► Free at end x = 1 Inner: None. ►

■

Loading ■ ■

External: None, Internal: None.

245

ADVANCE DESIGN VALIDATION GUIDE

1.76.2.2 Reference results Calculation method used to obtain the reference solution Precise calculation by numerical integration of the differential equation of beams bending (Euler-Bernoulli theories):

δ2 ⎛ δ² ν ⎞ δ²ν ⎜ EIz ⎟ = −ρA 2 δ δt ² x ² δx ⎝ ⎠ where Iz and A vary with the abscissa. The result is:

hο ⎧ α= =4 1 h1 E ⎪⎪ h1 with ⎨ fi = λi(α, β ) 12ρ 2π l² ⎪ β = bο = 5 ⎪⎩ b1 β=5

λ1 24.308

λ2 75.56

λ3 167.21

λ4 301.9

Uncertainty about the reference: analytical solution: Reference values Eigen mode type Flexion

MODE 1

246

1 2 3 4 5

Frequency (Hz) 56.55 175.79 389.01 702.36 1117.63 Scale = 1/4

λ5 480.4

ADVANCE DESIGN VALIDATION GUIDE

MODE 2

MODE 3

Scale = 1/4

247

ADVANCE DESIGN VALIDATION GUIDE

MODE 4

Scale = 1/4

MODE 5

Scale = 1/4

1.76.2.3 Theoretical results Theoretical Frequency

248

Result name

Result description

Reference value

Eigen mode

Frequency of eigen mode 1 [Hz]

56.55

Eigen mode

Frequency of eigen mode 2 [Hz]

175.79

Eigen mode

Frequency of eigen mode 3 [Hz]

389.01

Eigen mode

Frequency of eigen mode 4 [Hz]

702.36

Eigen mode

Frequency of eigen mode 5 [Hz]

1117.63

ADVANCE DESIGN VALIDATION GUIDE

1.76.3 Calculated results

Result name

Result description Frequency of eigen mode 1 [Hz]

Value 58.49 Hz

Error 3.43%

Frequency of eigen mode 2 [Hz]

177.67 Hz

1.07%

Frequency of eigen mode 3 [Hz]

388.85 Hz

-0.04%

Frequency of eigen mode 4 [Hz]

697.38 Hz

-0.71%

Frequency of eigen mode 5 [Hz]

1106.31 Hz

-1.01%

249

ADVANCE DESIGN VALIDATION GUIDE

1.77 Cantilever beam in Eulerian buckling with thermal load (01-0092HFLLB_FEM) Test ID: 2507 Test status: Passed

1.77.1 Description Verifies the vertical displacement and the normal force on a cantilever beam in Eulerian buckling with thermal load.

1.77.2 Background

1.77.2.1 Model description ■ ■ ■

Reference: Euler theory; Analysis type: Eulerian buckling; Element type: linear.

Units I. S. Geometry ■ ■ ■

L = 10 m 2 S=0.01 m 4 I = 0.0002 m

Materials properties ■

Longitudinal elastic modulus: E = 2.0 x 1010 N/m2,

■

Poisson's ratio: ν = 0.1.

■

Coefficient of thermal expansion: α = 0.00001

Boundary conditions ■ ■

Outer: Fixed at end x = 0, Inner: None.

Loading ■ ■

External: Punctual load P = -100000 N at x = L, Internal: T = -50°C (Contraction equivalent to the compression force) ( ε0 =

250

N −100000 = = 0.0005 = αΔT = 0.00001 × −50 ) ES 2.10 10 × 0.01

ADVANCE DESIGN VALIDATION GUIDE

1.77.2.2 Displacement of the model in the linear elastic range Reference solution The reference critical load established by Euler is:

Pcritical =

π 2 EI 2

= 98696 N ⇒ λ =

98696 = 0.98696 100000

Observation: in this case, the thermal load has no effect over the critical coefficient Finite elements modeling ■ ■ ■

Linear element: beam, imposed mesh, 5 nodes, 4 elements.

Deformed shape

1.77.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement v5 on Node 5 - Case 101 [cm]

-1.0

CM2

Normal Force on Node A - Case 101 [N]

-100000

CM2

Normal Force on Node A - Case 102 [N]

-98696

1.77.3 Calculated results

Result name DZ

Result description

Value

Error

Vertical displacement on Node 5 [cm]

-1 cm

0.00%

Normal Force - Case 101 [N]

-100000 N

0.00%

Normal Force - Case 102 [N]

-98699.3 N

0.00%

251

ADVANCE DESIGN VALIDATION GUIDE

1.78 Simple supported beam in free vibration (01-0098SDLLB_FEM) Test ID: 2512 Test status: Passed

1.78.1 Description Simple supported beam in free vibration. Tested functions: Shear force, eigen frequencies.

1.78.2 Background ■ ■ ■

Reference: NAFEMS, FV5 Analysis type: modal analysis; Tested functions: Shear force, eigen frequencies.

1.78.2.1 Problem data

Units I. S. Geometry

Full square section: ■ ■ ■

252

Dimensions: Area: Inertia:

a x b = 2m x 2 m 2 A = 4m IP = 2.25 m4 4 Iy = Iz = 1.333 m

ADVANCE DESIGN VALIDATION GUIDE

Materials properties ■

Longitudinal elastic modulus: E = 2 x 1011 Pa,

■

Poisson's ratio: ν = 0.3.

■

Density: ρ = 8000 kg/m3

Boundary conditions ■

Outer: x = y = z = Rx = 0 at A ; y = z =0 at B ; Inner: None.

► ►

■

Loading None for the modal analysis

1.78.2.2 Reference frequencies Mode 1 2 3 4 5 6 7 8 9

Shape Flexion Flexion Torsion Traction Flexion Flexion Torsion Flexion Flexion

Units Hz Hz Hz Hz Hz Hz Hz Hz Hz

Reference 42.649 42.649 77.542 125.00 148.31 148.31 233.10 284.55 284.55

Comment: Due to the condensed (lumped) nature of the mass matrix of Advance Design, the frequencies values of 3 and 7 modes cannot be found by this software. The same modeling done with NE/NASTRAN gave respectively for mode 3 and 7: 77.2 and 224.1 Hz. Finite elements modeling ■ ■

Straight elements: linear element Imposed mesh: 5 meshes

253

ADVANCE DESIGN VALIDATION GUIDE

Modal deformations

1.78.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Frequency of eigen mode 1 [Hz]

42.649

CM2

Frequency of eigen mode 2 [Hz]

42.649

CM2

Frequency of eigen mode 3 [Hz]

77.542

CM2

Frequency of eigen mode 4 [Hz]

125.00

CM2

Frequency of eigen mode 5 [Hz]

148.31

CM2

Frequency of eigen mode 6 [Hz]

148.31

CM2

Frequency of eigen mode 7 [Hz]

233.10

Comment: The torsion modes No. 3 and 7 that are calculated with NASTRAN cannot be calculated with Advance Design CM2 solver and therefore the mode No. 3 of the Advance Design analysis corresponds to mode No. 4 of the reference. The same problem in the case of No. 7 - Advance Design, that corresponds to mode No. 8 of the reference.

1.78.3 Calculated results Result name

254

Result description

Value

Error

Frequency of eigen mode 1 [Hz]

43.11 Hz

1.08%

Frequency of eigen mode 2 [Hz]

43.11 Hz

1.08%

Frequency of eigen mode 3 [Hz]

124.49 Hz

-0.41%

Frequency of eigen mode 4 [Hz]

149.38 Hz

0.72%

Frequency of eigen mode 5 [Hz]

149.38 Hz

0.72%

Frequency of eigen mode 6 [Hz]

269.55 Hz

-5.27%

Frequency of eigen mode 7 [Hz]

269.55 Hz

-5.27%

ADVANCE DESIGN VALIDATION GUIDE

1.79 Membrane with hot point (01-0099HSLSB_FEM) Test ID: 2513 Test status: Passed

1.79.1 Description Membrane with hot point. Tested functions: Stresses.

1.79.2 Background ■ ■ ■

Reference: NAFEMS, Test T1 Analysis type: static, thermo-elastic; Tested functions: Stresses.

1.79.2.1 Problem data

Observation: the units system of the initial NAFEMS test, defined in mm, was transposed in m for practical reasons. However, this has no influence on the results values.

255

ADVANCE DESIGN VALIDATION GUIDE

Units I. S. Geometry / meshing A quarter of the structure is modeled by incorporating the terms of symmetries. Thickness: 1 m

Materials properties ■

Longitudinal elastic modulus: E = 1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3,

■

Elongation coefficient α = 0.00001.

Boundary conditions ■

Outer: For all nodes in y = 0, uy =0; For all nodes in x = 0, ux =0; Inner: None.

► ►

■

256

■

External: None,

■

Internal: Hot point, thermal load ΔT = 100°C;

ADVANCE DESIGN VALIDATION GUIDE

1.79.3 σyy stress at point A: Reference solution:

Reference value: σyy = 50 MPa in A Finite elements modeling ■ ■ ■

Planar elements: membranes, 28 planar elements, 39 nodes.

1.79.3.1 Theoretical results Solver

Result name

Result description

Reference value

CM2

syy_mid

σyy in A [MPa]

Note: This value (50.87) is obtained with a vertical cross section through point A. The value represents σyy at the left end of the diagram. With CM2, it is essential to display the results with the “Smooth results on planar elements” option deactivated.

1.79.4 Calculated results

Result name syy_mid

Result description Sigma yy in A [MPa]

Value 50.8666 MPa

Error 1.73%

257

ADVANCE DESIGN VALIDATION GUIDE

1.80 Beam on 3 supports with T/C (k = -10000 N/m) (01-0102SSNLB_FEM) Test ID: 2516 Test status: Passed

1.80.1 Description Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k = -10000 N/m).

1.80.2 Background

1.80.2.1 Model description ■ ■ ■

Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, T/C.

Units I. S. Geometry ■ ■

L = 10 m Section: IPE 200, Iz = 0.00001943 m4

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

Outer: Support at node 1 restrained along x and y (x = 0), ► Support at node 2 restrained along y (x = 10 m), ► T/C ky Rigidity = -10000 N/m (the – sign corresponds to an upwards restraint), Inner: None. ►

■

Loading ■ ■

258

External: Vertical punctual load P = -100 N at x = 5 m, Internal: None.

ADVANCE DESIGN VALIDATION GUIDE

1.80.2.2 References solutions Displacements

( ) = −0.000129 rad 32EI (3EI + 2k L ) PL (3EI + k L ) =− = 0.000106 rad 16EI (3EI + 2k L ) 3PL2 2EI z + k y L3

β1 =

β2

v3 = β3 =

− 3PL3 = 0.00058 m 16 3EI z + 2k y L3

(

)

) = 0.000034 rad + 2k L )

PL2 − 6EI z + k y L3

(

32EI z 3EI z

Mz Moments

M z1 = 0 Mz2 =

(

3k y PL4

16 3EI z + 2k y L3

M z ( x = 5m) =

) = −58.15 N.m

PL (M z 2 − M z1 ) + = −220.9 N.m 4 2

Finite elements modeling ■ ■ ■

Linear element: S beam, automatic mesh, 3 nodes, 2 linear elements + 1 T/C.

Deformed shape

259

ADVANCE DESIGN VALIDATION GUIDE

Moment diagram

1.80.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Rotation Ry in node 1 [rad]

-0.000129

CM2

Rotation Ry in node 2 [rad]

0.000106

CM2

Displacement - node 3 [m]

0.00058

CM2

Rotation Ry in node 3 [rad]

0.000034

CM2

M moment - node 1 [Nm]

CM2

M moment - node 2 [Nm]

-58.15

CM2

M moment - middle span 1 [Nm]

-220.9

1.80.3 Calculated results

Result name RY RY

260

Result description

Value

Error

Rotation Ry in node 1 [rad]

0.000129488 Rad

0.38%

Rotation Ry in node 2 [rad]

-0.000105646 Rad

0.33%

Displacement - node 3 [m]

0.000581169 m

0.20%

Rotation Ry in node 3 [rad]

-3.44295e-005 Rad

-1.26%

M moment - node 1 [Nm]

1.77636e-015 N*m

0.00%

M moment - node 2 [Nm]

58.1169 N*m

-0.06%

M moment - middle span 1 [Nm]

-220.942 N*m

-0.02%

ADVANCE DESIGN VALIDATION GUIDE

1.81 Beam on 3 supports with T/C (k = 0) (01-0100SSNLB_FEM) Test ID: 2514 Test status: Passed

1.81.1 Description Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k = 0).

1.81.2 Background

1.81.2.1 Model description ■ ■ ■

Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, T/C.

Units I. S. Geometry ■ ■

L = 10 m 4 Section: IPE 200, Iz = 0.00001943 m

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

Outer: Support at node 1 restrained along x and y (x = 0), Support at node 2 restrained along y (x = 10 m), ► T/C stiffness ky = 0, Inner: None. ► ►

■

Loading ■ ■

External: Vertical punctual load P = -100 N at x = 5 m, Internal: None.

261

ADVANCE DESIGN VALIDATION GUIDE

1.81.2.2 References solutions ky being null, the non linear model behaves the same way as the structure without support 3. Displacements

( ) = −0.000153 rad 32EI (3EI + 2k L ) PL (3EI + k L ) =− = 0.000153 rad 16EI (3EI + 2k L ) 3PL2 2EI z + k y L3

β1 =

β2

v3 = β3 =

− 3PL3 = 0.00153 m 16 3EI z + 2k y L3

(

)

) = 0.000153 rad + 2k L )

PL2 − 6EI z + k y L3

(

32EI z 3EI z

Mz Moments

M z1 = 0 Mz2 =

(

3k y PL4

16 3EI z + 2k y L3

M z ( x = 5m) =

)=0

PL (M z 2 − M z1 ) + = −250 N.m 4 2

Finite elements modeling ■ ■ ■

Linear element: S beam, automatic mesh, 3 nodes, 2 linear elements + 1 T/C.

Deformed shape

262

ADVANCE DESIGN VALIDATION GUIDE

Moment diagrams

1.81.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Rotation Ry in node 1 [rad]

0.000153

CM2

Rotation Ry in node 2 [rad]

-0.000153

CM2

Displacement V in node 3 [m]

0.00153

CM2

Rotation Ry in node 3 [rad]

0.000153

CM2

Moment M in node 1 [Nm]

CM2

Moment M - middle span 1 [Nm]

-250

1.81.3 Calculated results

Result name RY

Result description

Value

Error

Rotation Ry in node 1 [rad]

0.000153175 Rad

0.11%

Rotation Ry in node 2 [rad]

-0.000153175 Rad

-0.11%

Displacement V in node 3 [m]

0.00153175 m

0.11%

Rotation Ry in node 3 [rad]

-0.000153175 Rad

-0.11%

Moment M in node 1 [Nm]

5.05771e-014 N*m

0.00%

Moment M - middle span 1 [Nm]

-250 N*m

0.00%

263

ADVANCE DESIGN VALIDATION GUIDE

1.82 Non linear system of truss beams (01-0104SSNLB_FEM) Test ID: 2518 Test status: Passed

1.82.1 Description Verifies the displacement and the normal force for a bar system containing 4 elements of the same length and 2 diagonals.

1.82.2 Background

1.82.2.1 Model description ■ ■ ■

Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, bar, tie.

Units I. S. Geometry ■ ■

L=5m Section S = 0.005 m2

Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 N/m2. Boundary conditions ■

Outer: Support at node 1 restrained along x and y, ► Support at node 2 restrained along x and y, Inner: None. ►

■

264

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: Horizontal punctual load P = 50000 N at node 3, Internal: None.

1.82.2.2 References solutions In non linear analysis without large displacement, the introduction of ties for the diagonal bars removes bar 5 (test No. 0103SSLLB_FEM allows finding an compression force in this bar at the linear calculation). Displacements

u3 = u 4 = v3 = −

5PL = 0.001195 m 11ES

PL = −0.000238 m ES

v4 = 0 N normal forces

N12 = 0 N 23 = −P = −50000 N N 43 = 0

N14 = 0 N13 = 2P = 70711 N N 42 = 0

Finite elements modeling ■ ■ ■

Linear element: bar, without meshing, 4 nodes, 6 linear elements.

Deformed shape

265

ADVANCE DESIGN VALIDATION GUIDE

Normal forces

1.82.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2 CM2

u3 displacement on Node 3 [m]

0.001195

v3 displacement on Node 3 [m]

-0.000238

CM2 CM2

u4 displacement on Node 4 [m]

0.001195

v4 displacement on Node 4 [m]

CM2

N12 normal force on Element 1 [N]

CM2

N23 normal force on Element 2 [N]

-50000

CM2

N34 normal force on Element 3 [N]

CM2

N14 normal force on Element 4 [N]

CM2

N13 normal effort on Element 5 [N]

70711

CM2

N24 normal force on Element 6 [N]

1.82.3 Calculated results Result name DX DZ

266

Result description u3 displacement on Node 3 [m] v3 displacement on Node 3 [m]

Value 0.00119035 m -0.000238095 m

Error -0.39% -0.04%

u4 displacement on Node 4 [m]

0.00119035 m

-0.39%

v4 displacement on Node 4 [m]

9.94686e-316 m

0.00%

N12 normal force on Element 1 [N]

0.00%

N23 normal force on Element 2 [N]

-50000 N

0.00%

N34 normal force on Element 3 [N]

0.00%

N14 normal force on Element 4 [N]

2.08884e-307 N

0.00%

N13 normal effort on Element 5 [N]

70710.7 N

0.00%

N24 normal force on Element 6 [N]

0.00%

ADVANCE DESIGN VALIDATION GUIDE

1.83 Linear system of truss beams (01-0103SSLLB_FEM) Test ID: 2517 Test status: Passed

1.83.1 Description Verifies the displacement and the normal force for a bar system containing 4 elements of the same length and 2 diagonals.

1.83.2 Background

1.83.2.1 Model description ■ ■ ■

Reference: internal GRAITEC test; Analysis type: static linear; Element type: linear, bar.

Units I. S. Geometry ■ ■

L=5m Section S = 0.005 m2

Materials properties Longitudinal elastic modulus: E = 2.1 x 1011 N/m2. Boundary conditions ■

Outer: Support at node 1 restrained along x and y, ► Support at node 2 restrained along x and y, Inner: None. ►

■

267

ADVANCE DESIGN VALIDATION GUIDE

Loading ■ ■

External: Horizontal punctual load P = 50000 N at node 3, Internal: None.

1.83.2.2 References solutions Displacements

30PL = 0.000649 m 11ES − 6PL v3 = = −0.000129 m 11ES 25PL u4 = = 0.000541 m 11ES 5PL v4 = = 0.000108 m 11ES u3 =

N normal forces

N12 = 0 N 23 = − N 43 =

N14 = 6 P = −27272 N 11

5 P = 22727 N 11

N13 =

5 P = 22727 N 11

6 2 P = 38569 N 11

N 42 = −

5 2 P = −32141 N 11

Finite elements modeling ■ ■ ■

Linear element: bar, without meshing, 4 nodes, 6 linear elements.

Deformed shape

268

ADVANCE DESIGN VALIDATION GUIDE

Normal forces

1.83.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2 CM2

u3 displacement on Node 3 [m]

0.000649

v3 displacement on Node 3 [m]

-0.000129

CM2 CM2

u4 displacement on Node 4 [m]

0.000541

v4 displacement on Node 4 [m]

0.000108

CM2

N12 normal force on Element 1 [N]

CM2

N23 normal force on Element 2 [N]

-27272

CM2

N43 normal force on Element 3 [N]

22727

CM2

N14 normal force on Element 4 [N]

22727

CM2

N13 normal effort on Element 5 [N]

38569

CM2

N42 normal force on Element 6 [N]

-32141

1.83.3 Calculated results Result name DX DZ

Result description u3 displacement on Node 3 [m] v3 displacement on Node 3 [m]

Value 0.000649287 m

Error 0.04%

-0.000129871 m

-0.68%

u4 displacement on Node 4 [m]

0.000541063 m

0.01%

v4 displacement on Node 4 [m]

0.000108224 m

0.21%

N12 normal force on Element 1 [N]

0.00%

N23 normal force on Element 2 [N]

-27272.9 N

0.00%

N43 normal force on Element 3 [N]

22727.1 N

0.00%

N14 normal force on Element 4 [N]

22727.1 N

0.00%

N13 normal effort on Element 5 [N]

38569.8 N

0.00%

N42 normal force on Element 6 [N]

-32140.9 N

0.00%

269

ADVANCE DESIGN VALIDATION GUIDE

1.84 Linear element in combined bending/tension - without compressed reinforcements - Partially tensioned section (02-0158SSLLB_B91) Test ID: 2520 Test status: Passed

1.84.1 Description Verifies the reinforcement results for a concrete beam with 8 isostatic spans subjects to uniform loads and compression normal forces.

1.84.1.1 Model description ■ ■ ■

Reference: J. Perchat (CHEC) reinforced concrete course Analysis type: static linear; Element type: planar.

Units ■ ■ ■ ■

Forces: kN Moment: kN.m Stresses: MPa Reinforcement density: cm²

Geometry ■ ■

Beam dimensions: 0.2 x 0.5 ht Length: l = 48 m in 8 spans of 6m,

Materials properties ■

Longitudinal elastic modulus: E = 20000 MPa,

■

Poisson's ratio: ν = 0.

Boundary conditions ■

Outer: Hinged at end x = 0, Vertical support at the same level with all other supports Inner: Hinged at each beam end (isostatic)

► ►

■

Loading ■

External: Case 1 (DL):uniform linear load g= -5kN/m (on all spans except 8) Fx = 10 kN at x = 42m: Ng = -10 kN for spans from 6 to 7

►

Fx = 140 kN at x = 32m: Ng = -150 kN for span 5 Fx = -50 kN at x = 24m: Ng = -100 kN for span 4 Fx = 50 kN at x = 18m: Ng = -50 kN for span 3 Fx = 50 kN at x = 12m: Ng = -100 kN for span 2 Fx = -70 kN at x = 6m: Ng = -30 kN for span 1 ►

Case 10 (DL):uniform linear load g = -5 kN/m (span 8) Fx = 10 kN at x = 48m: Ng = -10 kN Fx = -10 kN at x = 42m

270

ADVANCE DESIGN VALIDATION GUIDE

Case 2 to 8 (LL):uniform linear load q = -9 kN/m (on spans 1, 3 to 7) uniform linear load q = -15 kN/m (on span 2)

►

Fx = 30 kN at x = 6m (case 2 span 1) Fx = -50 kN at x = 6m (case 3 span 2) Fx = 50 kN at x = 12m (case 3 span 2) Fx = -40 kN at x = 12m (case 4 span 3) Fx = 40 kN at x = 18m (case 4 span 3) Fx = -100 kN at x = 18m (case 5 span 4) Fx = 100 kN at x = 24m (case 5 span 4) Fx = -150 kN at x = 24m (case 6 span 5) Fx = 150 kN at x = 30m (case 6 span 5) Fx = -8 kN at x = 30m (case 7 span 6) Fx = 8 kN at x = 36m (case 7 span 6) Fx = -8 kN at x = 36m (case 8 span 7 Fx = 8 kN at x = 42m (case 8 span 7) Case 9 (ACC):uniform linear load a = -25 kN/m (on 8th span) Fx = 8 kN at x = 36m (case 9 span 8)

►

Fx = -8 kN at x = 42m (case 9 span 8) Comb BAELUS: 1.35xDL+1.5xLL with duration of more than 24h (comb 101, 104 to 107) Comb BAEULI: 1.35xDL+1.5xLL with duration between 1h and 24h (comb 102) Comb BAELUC: 1.35xDL + 1.5xLL with duration of less than 1h (comb 103) Comb BAELS: 1xDL + 1*LL (comb 108 to 114) Comb BAELA: 1xDL + 1xACC with duration of less than 1h (comb 115) ■

Internal: None.

Reinforced concrete calculation hypothesis: All concrete covers are set to 5 cm BAEL 91 calculation (according to 99 revised version) Span 1

Concrete B20

Reinforcement HA fe500

Application D>24h

Concrete No

B35

Adx fe235

1h<D<24h

B50

HA fe 400

D<1h

Yes

4 5

B25 B25

HA fe500 HA fe500

D>24h D>24h

Yes No

6 7 8

B30 B40 B45

Adx fe235 HA fe500 HA fe500

D>24h D>24h D<1h

Yes Yes Yes

Cracking Non prejudicial Non prejudicial Non prejudicial Prejudicial Very prejudicial Prejudicial 160 MPa Non prejudicial

271

ADVANCE DESIGN VALIDATION GUIDE

1.84.1.2 Reinforcement calculation Reference solution fc28 ft28 fe teta gamb gams h

Span 1 20 1.8 500 1 1.5 1.15 1.6

Span 2 35 2.7 235 0.9 1.5 1.15 1

Span 3 50 3.6 400 0.85 1.5 1.15 1.6

Span 4 25 2.1 500 1 1.5 1.15 1.6

Span 5 25 2.1 500 1 1.5 1.15 1.6

Span 6 30 2.4 235 1 1.5 1.15 1

Span 7 40 3 500 1 1.5 1.15 1.6

Span 8 45 3.3 500 0.85 1.15 1 1.6

fbu fed sigpreju sigtpreju

11.33 434.78 250.00 200.00

22.04 204.35 156.67 125.33

33.33 347.83 264.00 211.20

14.17 434.78 250.00 200.00

17.00 204.35 156.67 125.33

22.67 434.78 160.00 160.00

39.13 500.00 252.76 202.21

g q pu pser G Q l Mu Nu Mser Nser Vu

5.00 9.00 20.25 14.00 -30.00 -30.00 6.00 91.13 -85.50 63.00 -60.00 60.75

5.00 9.00 20.25 14.00 -10.00 -8.00 6.00 91.13 -25.50 63.00 -18.00 60.75

5.00 25.00 30.00

Mu/A ubu

74.03 0.161

5.00 5.00 5.00 5.00 5.00 15.00 9.00 9.00 9.00 9.00 29.25 20.25 20.25 20.25 20.25 20.00 14.00 14.00 14.00 14.00 -100.00 -50.00 -100.00 -150.00 -10.00 -50.00 -40.00 -100.00 -100.00 -8.00 6.00 6.00 6.00 6.00 6.00 131.63 91.13 91.13 91.13 91.13 -210.00 -127.50 -285.00 -352.50 -25.50 90.00 63.00 63.00 63.00 63.00 -150.00 -90.00 -200.00 -250.00 -18.00 87.75 60.75 60.75 60.75 60.75 Main reinforcement calculation according to ULS 89.63 65.63 34.13 20.63 86.03 0.100 0.049 0.059 0.036 0.125

a z Au Mser/A a Mrb A B C D alpha1 z Aserp Mser/A a Mrb A B C

272

-10.00 -8.00 6.00 135.00 -18.00

90.00

86.03 0.094

131.40 0.083

0.221 0.133 0.062 0.077 0.046 0.167 0.123 0.410 0.426 0.439 0.436 0.442 0.420 0.428 6.12 20.57 7.97 8.35 9.18 11.27 5.21 Main reinforcement calculation with prejudicial cracking according to SLS 51.000 60.000 45.000 23.000 13.000 59.400 59.400

0.108 0.430 6.46

0.4186 87.53 1.0000 -3.0000 -0.4533 0.4533

0.4737 0.4737 0.6328 0.6923 121.16 121.16 182.01 258.82 1.0000 1.0000 1.0000 1.0000 -3.0000 -3.0000 -3.0000 -3.0000 -0.2044 -0.1156 -0.8426 -0.8250 0.2044 0.1156 0.8426 0.8250 0.238 0.432 0.428 0.414 0.385 0.386 10.22 10.99 10.75 Main reinforcement calculation with very prejudicial cracking according to SLS 51.00 60.00 45.00 23.00 13.00 59.40 59.40

0.47 96.93 1.0000 -3.0000 -0.5667

0.6678 220.78 1.0000 -3.0000 -0.8511 0.8511

0.72 231.67 1.0000 -3.0000 -1.0638

0.6303 302.44 1.0000 -3.0000 -0.3788 0.3788

0.68 319.66 1.0000 -3.0000 -0.4735

0.53 132.43 1.0000 -3.0000 -0.2556

0.53 132.43 1.0000 -3.0000 -0.1444

0.68 192.27 1.0000 -3.0000 -1.0532

0.69 258.82 1.0000 -3.0000 -0.8250

0.000 0.6157 267.55 1.0000 -3.0000 0.0000 0.0000

0.00 0.67 283.60 1.0000 -3.0000 0.0000

ADVANCE DESIGN VALIDATION GUIDE

Span 1 0.5667

Span 2 1.0638

tu k At/st

0.68 0.57 1.87

0.98 0.40 7.08

Aflex e0 Aminfsimp Aminfcomp At Atmin

6.12 -0.95 0.75 0.83 1.87 1.60

20.57 -1.67 2.38 2.54 7.08 3.40

D alpha1 z Asertp

Span 3 0.4735

Span 4 0.2556

Span 5 0.1444 0.203 0.420 14.049 Transverse reinforcement calculation 0.68 0.68 0.68 0.00 -0.14 -0.41 4.31 3.90 4.77 Recapitulation 7.97 10.22 14.05 -1.43 -3.17 -3.97 1.86 0.87 0.87 2.01 0.90 0.90 4.31 3.90 4.77 2.00 1.60 1.60

Span 6 1.0532

Span 7 0.8250

Span 8 0.0000

0.68 0.00 7.34

0.68 0.00 3.45

1.00 0.00 4.44

11.27 -0.29 2.11 2.80 7.34 3.40

10.75 -0.29 1.24 1.65 3.45 1.60

6.46 -0.13 1.37 0.30 4.44 1.60

Finite elements modeling ■ ■ ■

Linear elements: beams with imposed mesh 29 nodes, 28 linear elements.

1.84.1.3 Theoretical results Solver CM2

Result name Az

Result description

Reference value 2

Inf. main reinf. T1 [cm ] 2

6.12

CM2

Amin

Min. main reinf. T1 [cm ]

0.75

CM2

Atz

Trans. reinf. T1 [cm2]

1.87 2

CM2

Inf. main reinf. T2 [cm ]

20.57

CM2

Amin

Min. main reinf. T2 [cm2]

2.38

CM2

Atz

Trans. reinf. T2 [cm ]

CM2

Inf. main reinf. T3 [cm2]

7.08 2

7.97

CM2

Amin

Min. main reinf. T3 [cm ]

1.86

CM2

Atz

Trans. reinf. T3 [cm2]

4.31

CM2

Inf. main reinf. T4 [cm ] 2

10.22

CM2

Amin

Min. main reinf. T4 [cm ]

0.87

CM2

Atz

Trans. reinf. T4 [cm2]

3.90 2

CM2

Inf. main reinf. T5 [cm ]

14.05

CM2

Amin

Min. main reinf. T5 [cm2]

4.20

CM2

Atz

Trans. reinf. T5 [cm ]

4.77

CM2

Inf. main reinf. T6 [cm2]

11.27

CM2

Amin

Min. main reinf. T6 [cm2]

2.11

CM2

Atz

Trans. reinf. T6 [cm2]

7.34 2

CM2

Inf. main reinf. T7 [cm ]

10.75

CM2

Amin

Min. main. reinf. T7 [cm2]

1.24

CM2

Atz

Trans. reinf. T7 [cm ]

3.45 2

CM2

Inf. main reinf. T8 [cm ]

6.46

CM2

Amin

Min. main reinf. T8 [cm2]

1.37

CM2

Atz

Trans. reinf. T8 [cm ]

4.44

The "Mu limit" method must be applied in order to achieve the same results.

273

ADVANCE DESIGN VALIDATION GUIDE

1.84.2 Calculated results

Result name Az

274

Result description

Value

Error

Inf. main reinf. T1 [cm²]

-6.11718 cm²

0.05%

Amin

Min. main reinf. T1 [cm²]

0.7452 cm²

-0.64%

Atz

Trans. reinf. T1 [cm²]

1.8699 cm²

-0.01%

Inf. main reinf. T2 [cm²]

-20.5688 cm²

0.01%

Amin

Min. main reinf. T2 [cm²]

2.3783 cm²

-0.07%

Atz

Trans. reinf. T2 [cm²]

7.07943 cm²

-0.01%

Inf. main reinf. T3 [cm²]

-7.96552 cm²

0.06%

Amin

Min. main reinf. T3 [cm²]

1.863 cm²

0.16%

Atz

Trans. reinf. T3 [cm²]

4.3125 cm²

0.06%

Inf. main reinf. T4 [cm²]

-10.2301 cm²

-0.10%

Amin

Min. main reinf. T4 [cm²]

0.8694 cm²

-0.07%

Atz

Trans. reinf. T4 [cm²]

3.9008 cm²

0.02%

Inf. main reinf. T5 [cm²]

-14.0512 cm²

-0.01%

Amin

Min. main reinf. T5 [cm²]

4.2 cm²

0.00%

Atz

Trans. reinf. T5 [cm²]

4.7702 cm²

0.00%

Inf. main reinf. T6 [cm²]

-11.2742 cm²

-0.04%

Amin

Min. main reinf. T6 [cm²]

2.11404 cm²

0.19%

Atz

Trans. reinf. T6 [cm²]

7.34043 cm²

0.01%

Inf. main reinf. T7 [cm²]

-10.7634 cm²

-0.12%

Amin

Min. main. reinf. T7 [cm²]

1.242 cm²

0.16%

Atz

Trans. reinf. T7 [cm²]

3.45 cm²

0.00%

Inf. main reinf. T8 [cm²]

-6.47718 cm²

-0.27%

Amin

Min. main reinf. T8 [cm²]

1.3662 cm²

-0.28%

Atz

Trans. reinf. T8 [cm²]

4.44444 cm²

0.10%

ADVANCE DESIGN VALIDATION GUIDE

1.85 Design of a Steel Structure according to CM66 (03-0206SSLLG_CM66) Test ID: 2522 Test status: Passed

1.85.1 Description Verifies the steel calculation results (maximum displacement, normal force, bending moment, deflections, buckling lengths, lateral-torsional buckling and cross section optimization) for a simple metallic framework with a concrete floor, according to CM66.

1.85.2 Background

1.85.2.1 Model description ■ ■

■

Calculation model: Simple metallic framework with a concrete floor. Load case: ► Permanent loads: 150 kg/m² for the floor and 25kg/m² for the roof. ► Overloads: 250 kg/m² on the floor. ► Wind loads on region II for a normal location ► Snow loads on region 2B at an altitude of 750m. CM66 Combinations

Model preview

275

ADVANCE DESIGN VALIDATION GUIDE

Structure’s load case Code CMP CMS CMV CMV CMV CMV CMV CMV CMV CMV CMN

No. 1 2 3 4 5 6 7 8 9 10 11

Type Static Static Static Static Static Static Static Static Static Static Static

Title SW + Dead loads Overloads for usage Wind overloads along +X in overpressure Wind overloads along +X in depression Wind overloads along -X in overpressure Wind overloads along -X in depression Wind overloads along +Z in overpressure Wind overloads along +Z in depression Wind overloads along -Z in overpressure Wind overloads along -Z in depression Normal snow overloads

1.85.2.2 Effel Structure results Displacement Envelope (“CMCD" load combinations)

Env.

Case

No.

Max(D) Min(D) Max(DX) Min(DX) Max(DY) Min(DY) Max(DZ) Min(DZ)

213 188 204 204 213 213 201 203

148 1.1 72.1 313 148 61.5 371 370

Envelope of linear element forces Max. D DX location (cm) (cm) CENTER 12.115 0.037 START 0.000 0.000 START 3.138 3.099 END 2.872 -1.872 CENTER 12.115 0.037 END 9.986 -0.118 CENTER 4.149 -0.006 CENTER 4.124 -0.006

DY (cm) 12.035 0.000 0.434 -0.129 12.035 -9.985 -0.188 -0.240

DZ (cm) -1.393 0.000 0.244 -2.174 -1.393 0.046 4.145 -4.118

Envelope of forces on linear elements (“CMCFN” load combinations)

Env.

Case

No.

Max (Fx) Min (Fx) Max(Fy) Min(Fy) Max(Fz) Min(Fz) Max(Mx) Min(Mx) Max(My) Min(My) Max (Mz) Min (Mz)

120 138 120 120 177 187 120 120 177 179 120 120

4.1 98 57 60 371 370 111 21 371 370 57 59.2

Envelope of linear element forces Fx Fy Fz MaxSite (T) (T) (T) START 19.423 -4.108 -1.384 START -41.618 -0.962 -0.192 END -13.473 16.349 -0.016 START -15.994 -16.112 -0.006 START -3.486 -0.118 2.655 START -3.666 -0.147 -2.658 END 3.933 4.840 0.278 END -22.324 13.785 -0.191 CENTER -3.099 -0.118 -0.323 CENTER -3.283 -0.155 0.321 END -13.473 16.349 -0.016 END -19.455 -8.969 -0.702

Mx (T*m) -0.003 0.000 -0.003 -3E-004 0.000 0.000 0.028 -0.028 0.000 0.000 -0.003 -0.003

My (T*m) 1.505 0.000 0.002 6E-006 0.000 0.000 -4E-005 -0.004 4.403 -4.373 0.002 -0.001

Mz (T*m) 7.551 0.000 55.744 53.096 0.000 0.000 11.531 42.562 -0.500 -0.660 55.744 -57.105

Envelope of linear element stresses (“CMCFN” load combinations)

Env.

Case

No.

Max(sxxMax) Min(sxxMax) Max(sxyMax) Min(sxyMax) Max(sxzMax) Min(sxzMax) Max(sFxx) Min(sFxx) Max(sMxxMax) Min(sMxxMax)

120 120 120 120 185 179 120 120 120 1

59.2 292 57 60 371 370 293 292 59.2 1.1

276

Envelope of linear element stresses sxxMax sxyMax MaxSite (MPa) (MPa) END 273.860 -14.696 START -150.743 0.000 START 262.954 37.139 END 241.643 -36.595 START -2.949 -0.183 START -3.104 -0.255 END 161.095 9E-005 START -150.743 0.000 END 273.860 -14.696 START -4.511 3.155

sxzMax (MPa) -1.024 0.000 -0.030 -0.011 3.876 -3.882 -0.002 0.000 -1.024 -0.646

sFxx (MPa) -16.453 -150.743 -15.609 -18.536 -2.949 -3.104 161.095 -150.743 -16.453 -4.511

sMxxMax (MPa) 290.312 0.000 278.562 260.179 0.000 0.000 0.000 0.000 290.312 0.000

ADVANCE DESIGN VALIDATION GUIDE

1.85.2.3 CM66 Effel Expertise results Hypotheses For columns ■

Deflections:

1/150 Envelopes deflections calculation.

■

Buckling XY plane:

Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on fixed nodes

■

Lateral-torsional buckling: Ldi automatic calculation: hinged restraint Lds automatic calculation: hinged restraint

For rafters ■

Deflections:

1/200 Envelopes deflections calculation.

■

Buckling:

XY plane: Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on fixed nodes

■

Lateral-torsional buckling: Ldi automatic calculation: no restraint Lds automatic calculation: hinged restraint

For columns ■

Deflections:

1/150 Envelopes deflections calculation.

■

Buckling:

XY plane: Automatic calculation of the structure on displaceable nodes XZ plane: Automatic calculation of the structure on displaceable nodes

■

Lateral-torsional buckling: Ldi automatic calculation: hinged restraint Lds automatic calculation: hinged restraint

Optimization parameters ■ ■ ■

Work ratio optimization between 90 and 100% All the sections from the library are available. Labels optimization.

The results of the optimization given below correspond to an iteration of the finite elements calculation.

277

ADVANCE DESIGN VALIDATION GUIDE

Deflection verification Ratio

Max values on the element ■ ■ ■

278

Columns: L / 168 Rafter: L / 96 Column: L / 924

ADVANCE DESIGN VALIDATION GUIDE

CM Stress diagrams Work ratio

Stresses

Max values on the element ■ Columns: 375.16 MPa ■ Rafter: 339.79 MPa ■ Column: 180.98 MPa

279

ADVANCE DESIGN VALIDATION GUIDE

Buckling lengths Lfy

Lfz

280

ADVANCE DESIGN VALIDATION GUIDE

Lateral-torsional buckling lengths Ldi

Lds

281

ADVANCE DESIGN VALIDATION GUIDE

Optimization

1.85.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Maximum displacement (CMCD) [cm]

12.115

CM2

Envelope normal force (CMCFN) Min (Fx) [T]

-41.618

CM2

Envelope normal force (CMCFN) Max (Fx) [T]

19.423

CM2

Envelope bending moment (CMCFN) Min (Mz) [Tm]

-57.105

CM2

Envelope bending moment (CMCFN) Max (Mz) [Tm]

55.744

Warning, the Mz bending moment of Effel Structure corresponds to the My bending moment of Advance Design.

282

Solver

Result name

Result description

Reference value

CM2

Deflection

CM deflections on Columns [adm]

L / 168 (89%)

CM2

Deflection

CM deflections on Rafters [adm]

L / 96 (208%)

CM2

Deflection

CM deflections on Columns [adm]

L / 924 (16%)

CM2

Stress

CM stresses on Columns [MPa]

374.67

CM2

Stress

CM stresses on Rafters [MPa]

339.74

CM2

Stress

CM stresses on Columns [MPa]

180.98

CM2

Lfy

Buckling lengths on Columns Lfy [m]

8.02

CM2

Lfz

Buckling lengths on Columns Lfz [m]

24.07

CM2

Lfy

Buckling lengths on Rafters Lfy [m]

1.72

CM2

Lfz

Buckling lengths on Rafters Lfz [m]

20.25

CM2C

Lfy

Buckling lengths on Columns Lfy [m]

4.20

CM2

Lfz

Buckling lengths on Columns Lfz [m]

5.67

ADVANCE DESIGN VALIDATION GUIDE

Warning, the local axes in Effel Structure are opposite to those in Advance Design. Solver

Result name

Result description

Reference value

CM2

Ldi

Lateral-torsional buckling lengths on Columns Ldi [m]

8.5

CM2

Lds

Lateral-torsional buckling lengths on Columns Lds [m]

8.5

CM2

Ldi

Lateral-torsional buckling lengths on Rafters Ldi [m]

8.61

CM2

Lds

Lateral-torsional buckling lengths on Rafters Lds [m]

1.72

CM2

Ldi

Lateral-torsional buckling lengths on Columns Ldi [m]

CM2

Lds

Lateral-torsional buckling lengths on Columns Lds [m]

Solver

Result name

Result description

CM2

Work ratio

IPE500 columns - section optimization [adm]

1.59

IPE600

CM2

Work ratio

IPE400 rafters - section optimization [adm]

1.45

IPE500

CM2

Work ratio

IPE400 columns - section optimization [adm]

0.77

IPE360

Rate (%)

Final section

1.85.3 Calculated results

Result name D

Result description

Value

Error

Maximum displacement (CMCD) [cm]

12.143 cm

0.23%

Envelope normal force (CMCFN) Min (Fx) [T]

-41.6548 T

-0.09%

Envelope normal force (CMCFN) Max (Fx) [T]

19.4828 T

0.31%

Envelope bending moment (CMCFN) Min (Mz)[Tm]

-57.1283 T*m

-0.04%

Envelope bending moment (CMCFN) Max (Mz) [Tm]

55.7785 T*m

0.06%

Deflection

CM deflections on Columns [adm]

167.736 Adim.

-0.16%

Deflection

CM deflections on Rafters [adm]

96.1185 Adim.

0.12%

Deflection

CM deflections on Columns [adm]

924.996 Adim.

0.11%

Stress

CM stresses on Columns [MPa]

374.568 MPa

-0.03%

Stress

CM stresses on Rafters [MPa]

352.227 MPa

3.68%

Stress

CM stresses on Columns [MPa]

180.7 MPa

-0.15%

Lfy

Buckling lengths on Columns Lfy [m]

7.96718 m

-0.66%

Lfz

Buckling lengths on Columns Lfz [m]

24.0693 m

0.00%

Lfy

Buckling lengths on Rafters Lfy [m]

6.63414 m

0.06%

Lfz

Buckling lengths on Rafters Lfz [m]

20.2452 m

-0.02%

Lfy

Buckling lengths on Columns Lfy [m]

4.19567 m

-0.10%

Lfz

Buckling lengths on Columns Lfz [m]

5.67211 m

0.04%

Ldi

Lateral-torsional buckling lengths on Columns Ldi [m]

8.5 m

0.00%

Lds

Lateral-torsional buckling lengths on Columns Lds [m]

8.5 m

0.00%

Ldi

Lateral-torsional buckling lengths on Rafters Ldi [m]

8.61187 m

0.02%

Lds

Lateral-torsional buckling lengths on Rafters Lds [m]

8.61187 m

0.02%

Ldi

Lateral-torsional buckling lengths on Columns Ldi [m]

0.00%

Lds

Lateral-torsional buckling lengths on Columns Lds [m]

0.00%

Work ratio

IPE500 columns - section optimization [adm]

1.59391 Adim.

0.25%

Work ratio

IPE400 rafters - section optimization [adm]

1.49884 Adim.

3.37%

Work ratio

IPE400 columns - section optimization [adm]

0.768938 Adim.

-0.14%

283

ADVANCE DESIGN VALIDATION GUIDE

1.86 Linear element in simple bending - without compressed reinforcement (02-0162SSLLB_B91) Test ID: 2521 Test status: Passed

1.86.1 Description Verifies the reinforcement results for a concrete beam with 8 isostatic spans subjected to uniform loads.

1.86.1.1 Model description ■ ■ ■

Reference: J. Perchat (CHEC) reinforced concrete course Analysis type: static linear; Element type: planar.

Units ■ ■ ■ ■

Forces: kN Moment: kN.m Stresses: MPa 2 Reinforcement density: cm

Geometry ■ ■

Beam dimensions: 0.2 x 0.5 ht Length: l = 42 m in 7 spans of 6m,

Materials properties ■

Longitudinal elastic modulus: E = 20000 MPa,

■

Poisson's ratio: ν = 0.

Boundary conditions ■

Outer: Hinged at end x = 0, Vertical support at the same level with all other supports Inner: Hinge z at each beam end (isostatic)

► ►

■

Loading ■

External: ► Case 1 (DL):uniform linear load g = -5 kN/m (on all spans except 8) ►

Case 2 to 8 (LL):uniform linear load q = -9 kN/m (on spans 1, 3 to 7) uniform linear load q = -15 kN/m (on span 2)

►

Case 9 (ACC): uniform linear load a = -25 kN/m (on 8th span)

►

Case 10 (DL):uniform linear load g = -5 kN/m (on 8th span) Comb BAELUS: 1.35xDL+1.5xLL with duration of more than 24h (comb 101, 104 to 107) Comb BAEULI: 1.35xDL+1.5xLL with duration between 1h and 24h (comb 102) Comb BAELUC: 1.35xDL + 1.5xLL with duration of less than 1h (comb 103) Comb BAELS: 1xDL + 1*LL (comb 108 to 114) Comb BAELUA: 1xDL + 1xACC (comb 115)

■

284

Internal: None.

ADVANCE DESIGN VALIDATION GUIDE

Reinforced concrete calculation hypothesis: â&#x2013; â&#x2013;

All concrete covers are set to 5 cm BAEL 91 calculation with the revised version 99 Span 1

Concrete B20

Reinforcement HA fe500

Application D>24h

Concrete No

B35

Adx fe235

1h<D<24h

B50

HA fe 400

D<1h

Yes

4 5

B25 B60

HA fe500 HA fe500

D>24h D>24h

Yes No

6 7 8

B30 B40 B45

Adx fe235 HA fe500 HA fe500

D>24h D>24h D<1h

Yes Yes Yes

Cracking Non prejudicial Non prejudicial Non prejudicial Prejudicial Very prejudicial Prejudicial 160 MPa Non prejudicial

1.86.1.2 Reinforcement calculation Reference solution fc28 ft28 fe teta gamb gams h

Span 1 20 1.8 500 1 1.5 1.15 1.6

Span 2 35 2.7 235 0.9 1.5 1.15 1

Span 3 50 3.6 400 0.85 1.5 1.15 1.6

Span 4 25 2.1 500 1 1.5 1.15 1.6

Span 5 60 4.2 500 1 1.5 1.15 1.6

Span 6 30 2.4 235 1 1.5 1.15 1

Span 7 40 3 500 1 1.5 1.15 1.6

Span 8 45 3.3 500 0.85 1.15 1 1.6

fbu fed sigpreju sigtpreju

11.33 434.78 250.00 200.00

22.04 204.35 156.67 125.33

33.33 347.83 264.00 211.20

14.17 434.78 250.00 200.00

34.00 434.78 285.15 228.12

17.00 204.35 156.67 125.33

22.67 434.78 160.00 160.00

39.13 500.00 252.76 202.21

g q pu pser l Mu Mser Vu

5.00 9.00 20.25 14.00 6.00 91.13 63.00 60.75

ubu a z Au A B C D alpha1 z Aserp A B C D alpha1 z Asertp

5.00 5.00 5.00 5.00 5.00 5.00 15.00 9.00 9.00 9.00 9.00 9.00 29.25 20.25 20.25 20.25 20.25 20.25 20.00 14.00 14.00 14.00 14.00 14.00 6.00 6.00 6.00 6.00 6.00 6.00 131.63 91.13 91.13 91.13 91.13 91.13 90.00 63.00 63.00 63.00 63.00 63.00 87.75 60.75 60.75 60.75 60.75 60.75 Longitudinal reinforcement calculation according to ELU 0.199 0.147 0.068 0.159 0.066 0.132 0.099 0.279 0.200 0.087 0.217 0.086 0.178 0.131 0.400 0.414 0.434 0.411 0.435 0.418 0.426 5.24 15.56 6.03 5.10 4.82 10.67 4.91 Main reinforcement calculation with prejudicial cracking according to SLS 1.000 1.000 1.000 1.000 1.000 1.000 1.000 -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 -0.56000 -0.89362 -0.87500 0.56000 0.89362 0.87500 0.367 0.442 0.438 0.395 0.384 0.384 6.38 10.48 10.25 Main reinforcement calculation with very prejudicial cracking according to SLS 1.00 1.00 1.00 1.00 1.00 1.00 1.00 -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 -0.70 -1.60 -0.66 -0.70 -0.61371 -1.12 -0.88 0.70 1.60 0.66 0.70 0.61371 1.12 0.88 0.381 0.393 7.030

5.00 25.00 30.00 6.00 135.00 90.00 0.085 0.111 0.430 6.28 1.000 -3.000

1.00 -3.00 0.00 0.00

285

ADVANCE DESIGN VALIDATION GUIDE

Span 1

Span 2

tu k At/st

0.68 1.00 0.69

0.98 1.00 1.79

Aflex Aminflex At Atmin

5.24 0.75 0.69 1.60

15.56 2.38 1.79 3.40

Span 3 Span 4 Span 5 Transversal reinforcement calculation 0.68 0.68 0.68 0.00 0.00 0.00 4.31 3.45 3.45 Recapitulation 6.03 6.38 7.03 1.86 0.87 1.74 4.31 3.45 3.45 2.00 1.60 1.60

Span 6

Span 7

Span 8

0.68 0.00 7.34

0.68 0.00 3.45

1.00 0.00 4.44

10.67 2.11 7.34 3.40

10.25 1.24 3.45 1.60

6.28 1.37 4.44 1.60

Finite elements modeling ■ ■ ■

Linear elements: beams with imposed mesh 29 nodes, 28 linear elements.

1.86.1.3 Theoretical results Reference Solver

Result name

Result description

Reference value

CM2

Inf. main reinf. T1 [cm2]

5.24

CM2

Amin

Min. main reinf. T1 [cm2]

0.75

CM2

Atz

Trans. reinf. T1 [cm ]

0.69

CM2

Inf. main reinf. T2 [cm2]

15.56

CM2

Amin

Min. main reinf. T2 [cm2]

2.38

CM2

Atz

Trans. reinf. T2 [cm2]

1.79 2

CM2

Inf. main reinf. T3 [cm ]

6.03

CM2

Amin

Min. main reinf. T3 [cm2]

1.86

CM2

Atz

Trans. reinf. T3 [cm ]

CM2

Inf. main reinf. T4 [cm2]

CM2

Amin

4.31 2

Min. main reinf. T4 [cm ] 2

CM2

Atz

Trans. reinf. T4 [cm ]

CM2

Inf. main reinf. T5 [cm2]

0.87 3.45

7.03

CM2

Amin

Min. main reinf. T5 [cm ]

1.74

CM2

Atz

Trans. reinf. T5 [cm2]

3.45 2

CM2

Inf. main reinf. T6 [cm ]

10.67

CM2

Amin

Min. main reinf. T6 [cm2]

2.11

CM2

Atz

Trans. reinf. T6 [cm ]

7.34

CM2

Inf. main reinf. T7 [cm2]

10.25

CM2

Amin

Min. main. reinf. T7 [cm2]

1.24

CM2

Atz

Trans. reinf. T7 [cm2]

3.45

CM2

Inf. main reinf. T8 [cm2] 2

6.28

CM2

Amin

Min. main reinf. T8 [cm ]

1.37

CM2

Atz

Trans. reinf. T8 [cm2]

4.44

The "Mu limit" method must be applied to attain the same results.

286

6.38

ADVANCE DESIGN VALIDATION GUIDE

1.86.2 Calculated results

Result name Az

Result description

Value

Error

Inf. main reinf. T1 [cm²]

-5.24348 cm²

-0.07%

Amin

Min. main reinf. T1 [cm²]

0.7452 cm²

-0.64%

Atz

Trans. reinf. T1 [cm²]

0.69 cm²

0.00%

Inf. main reinf. T2 [cm²]

-15.5613 cm²

-0.01%

Amin

Min. main reinf. T2 [cm²]

2.3783 cm²

-0.07%

Atz

Trans. reinf. T2 [cm²]

1.79433 cm²

0.24%

Inf. main reinf. T3 [cm²]

-6.03286 cm²

-0.05%

Amin

Min. main reinf. T3 [cm²]

1.863 cm²

0.16%

Atz

Trans. reinf. T3 [cm²]

4.3125 cm²

0.06%

Inf. main reinf. T4 [cm²]

-6.38336 cm²

-0.05%

Amin

Min. main reinf. T4 [cm²]

0.8694 cm²

-0.07%

Atz

Trans. reinf. T4 [cm²]

3.45 cm²

0.00%

Inf. main reinf. T5 [cm²]

-7.03527 cm²

-0.07%

Amin

Min. main reinf. T5 [cm²]

1.7388 cm²

-0.07%

Atz

Trans. reinf. T5 [cm²]

3.45 cm²

0.00%

Inf. main reinf. T6 [cm²]

-10.6698 cm²

0.00%

Amin

Min. main reinf. T6 [cm²]

2.11404 cm²

0.19%

Atz

Trans. reinf. T6 [cm²]

7.34043 cm²

0.01%

Inf. main reinf. T7 [cm²]

-10.2733 cm²

-0.23%

Amin

Min. main. reinf. T7 [cm²]

1.242 cm²

0.16%

Atz

Trans. reinf. T7 [cm²]

3.45 cm²

0.00%

Inf. main reinf. T8 [cm²]

-6.29338 cm²

-0.21%

Amin

Min. main reinf. T8 [cm²]

1.3662 cm²

-0.28%

Atz

Trans. reinf. T8 [cm²]

4.44444 cm²

0.10%

287

ADVANCE DESIGN VALIDATION GUIDE

1.87 Double cross with hinged ends (01-0097SDLLB_FEM) Test ID: 2511 Test status: Passed

1.87.1 Description Double cross with hinged ends. Tested functions: Eigen frequencies, crossed beams, in plane bending.

1.87.2 Background ■ ■ ■

Reference: NAFEMS, FV2 test Analysis type: modal analysis; Tested functions: Eigen frequencies, Crossed beams, In plane bending.

1.87.2.1 Problem data

Units I. S. Geometry

Full square section: ■ ■ ■ ■

Arm length: Dimensions: Area: Inertia:

L=5m a x b = 0.125 x 0.125 -2 2 A = 1.563 10 m -5 4 IP = 3.433 x 10 m Iy = Iz = 2.035 x 10-5m4

Materials properties

288

■

Longitudinal elastic modulus: E = 2 x 1011 Pa,

■

Density: ρ = 8000 kg/m3

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: A, B, C, D, E, F, G, H points restraint along x and y; Inner: None.

Loading None for the modal analysis

1.87.2.2 Reference frequencies Mode 1 2,3 4 to 8 9 10,11 12 to 16

Units Hz Hz Hz Hz Hz Hz

Reference 11.336 17.709 17.709 45.345 57.390 57.390

Finite elements modeling ■ ■

Linear elements type: Beam Imposed mesh: 4 Elements / Arms

Modal deformations

289

ADVANCE DESIGN VALIDATION GUIDE

1.87.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Frequency

Frequency of Eigen Mode 1 [Hz]

11.336

CM2

Frequency

Frequency of Eigen Mode 2 [Hz]

17.709

CM2

Frequency

Frequency of Eigen Mode 3 [Hz]

17.709

CM2

Frequency

Frequency of Eigen Mode 4 [Hz]

17.709

CM2

Frequency

Frequency of Eigen Mode 5 [Hz]

17.709

CM2

Frequency

Frequency of Eigen Mode 6 [Hz]

17.709

CM2

Frequency

Frequency of Eigen Mode 7 [Hz]

17.709

CM2

Frequency

Frequency of Eigen Mode 8 [Hz]

17.709

CM2

Frequency

Frequency of Eigen Mode 9 [Hz]

45.345

CM2

Frequency

Frequency of Eigen Mode 10 [Hz]

57.390

CM2

Frequency

Frequency of Eigen Mode 11 [Hz]

57.390

CM2

Frequency

Frequency of Eigen Mode 12 [Hz]

57.390

CM2

Frequency

Frequency of Eigen Mode 13 [Hz]

57.390

CM2

Frequency

Frequency of Eigen Mode 14 [Hz]

57.390

CM2

Frequency

Frequency of Eigen Mode 15 [Hz]

57.390

CM2

Frequency

Frequency of Eigen Mode 16 [Hz]

57.390

1.87.3 Calculated results Result name

290

Result description

Value

Error

Frequency of Eigen Mode 1 [Hz]

11.33 Hz

-0.05%

Frequency of Eigen Mode 2 [Hz]

17.66 Hz

-0.28%

Frequency of Eigen Mode 3 [Hz]

17.66 Hz

-0.28%

Frequency of Eigen Mode 4 [Hz]

17.69 Hz

-0.11%

Frequency of Eigen Mode 5 [Hz]

17.69 Hz

-0.11%

Frequency of Eigen Mode 6 [Hz]

17.69 Hz

-0.11%

Frequency of Eigen Mode 7 [Hz]

17.69 Hz

-0.11%

Frequency of Eigen Mode 8 [Hz]

17.69 Hz

-0.11%

Frequency of Eigen Mode 9 [Hz]

45.02 Hz

-0.72%

Frequency of Eigen Mode 10 [Hz]

56.06 Hz

-2.32%

Frequency of Eigen Mode 11 [Hz]

56.06 Hz

-2.32%

Frequency of Eigen Mode 12 [Hz]

56.34 Hz

-1.83%

Frequency of Eigen Mode 13 [Hz]

56.34 Hz

-1.83%

Frequency of Eigen Mode 14 [Hz]

56.34 Hz

-1.83%

Frequency of Eigen Mode 15 [Hz]

56.34 Hz

-1.83%

Frequency of Eigen Mode 16 [Hz]

56.34 Hz

-1.83%

ADVANCE DESIGN VALIDATION GUIDE

1.88 Beam on 3 supports with T/C (k -> infinite) (01-0101SSNLB_FEM) Test ID: 2515 Test status: Passed

1.88.1 Description Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length and identical characteristics with 3 T/C supports (k -> infinite).

1.88.2 Background

1.88.2.1 Model description ■ ■ ■

Reference: internal GRAITEC test; Analysis type: static non linear; Element type: linear, T/C.

Units I. S. Geometry ■ ■

L = 10 m 4 Section: IPE 200, Iz = 0.00001943 m

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 N/m2,

■

Poisson's ratio: ν = 0.3.

Boundary conditions ■

Outer: ► ►

Support at node 1 restrained along x and y (x = 0), Support at node 2 restrained along y (x = 10 m),

T/C stiffness ky → ∞ (1.1030N/m), Inner: None.

►

■

Loading ■ ■

External: Vertical punctual load P = -100 N at x = 5 m, Internal: None.

291

ADVANCE DESIGN VALIDATION GUIDE

1.88.2.2 References solutions ky being infinite, the non linear model behaves the same way as a beam on 3 supports. Displacements

( ) = −0.000115 rad 32EI (3EI + 2k L ) PL (3EI + k L ) =− = 0.000077 rad 16EI (3EI + 2k L ) 3PL2 2EI z + k y L3

β1 =

β2

v3 = β3 =

− 3PL3 =0 16 3EI z + 2k y L3

(

)

) = −0.000038 rad + 2k L )

PL2 − 6EI z + k y L3

(

32EI z 3EI z

Mz Moments

M z1 = 0 Mz2 =

(

3k y PL4

16 3EI z + 2k y L3

M z ( x = 5m) =

) = −93.75 N.m

PL (M z 2 − M z1 ) + = −203.13 N.m 4 2

Finite elements modeling ■ ■ ■

Linear element: S beam, automatic mesh, 3 nodes, 2 linear elements + 1 T/C.

Deformed shape

292

ADVANCE DESIGN VALIDATION GUIDE

Moment diagram

1.88.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Rotation Ry - node 1 [rad]

0.000115

CM2

Rotation Ry - node 2 [rad]

-0.000077

CM2

Displacement - node 3 [m]

CM2

Rotation Ry - node 3 [rad]

0.000038

CM2

Moment M - node 1 [Nm]

CM2

Moment M - node 2 [Nm]

93.75

CM2

Moment M - middle span 1 [Nm]

-203.13

1.88.3 Calculated results

Result name RY

Result description

Value

Error

Rotation Ry in node 1 [rad]

0.000115005 Rad

0.00%

Rotation Ry in node 2 [rad]

-7.65875e-005 Rad

0.54%

Displacement - node 3 [m]

9.36486e-030 m

0.00%

Rotation Ry in node 3 [rad]

3.81695e-005 Rad

0.45%

Moment M - node 1 [Nm]

1.3145e-013 N*m

0.00%

Moment M - node 2 [Nm]

93.6486 N*m

-0.11%

Moment M - middle span 1 [Nm]

-203.176 N*m

-0.02%

293

ADVANCE DESIGN VALIDATION GUIDE

1.89 Study of a mast subjected to an earthquake (02-0112SMLLB_P92) Test ID: 2519 Test status: Passed

1.89.1 Description A structure consisting of 2 beams and 2 punctual masses, subjected to a lateral earthquake along X. The frequency modes, the eigen vectors, the participation factors, the displacement at the top of the mast and the forces at the top of the mast are verified.

1.89.1.1 Model description ■ ■ ■

Reference: internal GRAITEC test; Analysis type: modal and spectral analyses; Element type: linear, mass.

1.89.1.2 Material strength model

Units I. S. Geometry ■ ■ ■

Length: Outer radius: Inner radius:

L = 35 m, Rext = 3.00 m Rint = 2.80 m

■ ■ ■

Axial section: Polar inertia: Bending inertias:

S= 3.644 m 4 Ip = 30.68 m Ix =15.34 m4 Iy = 15.34 m4

Masses ■ ■

M1 =203873.6 kg M2 =101936.8 kg

Materials properties

294

■

Longitudinal elastic modulus:

E = 1.962 x 1010 N/m2,

■

Poisson's ratio:

ν = 0.1,

■

Density:

ρ = 25 kN/m3

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■

Outer: Fixed in X = 0, Y = 0 m,

Loading ■

External: Seismic excitation on X direction

Finite elements modeling Linear element: beam, automatic mesh,

1.89.1.3 Seismic hypothesis in conformity with PS92 regulation ■ ■ ■ ■ ■

Zone: Nice Sophia Antipolis (Zone II). Site: S1 (Medium soil, 10m thickness). Construction type class: B Behavior coefficient: 3 Material damping: 4% (Reinforced concrete).

1.89.1.4 Modal analysis Eigen periods reference solution Substract the value of structure’s specific horizontal periods by solving the following equation:

(

)

det K − Mω 2 = 0 48 EI 7 L3

≡

⎛ M ≡ ⎜⎜ ⎝ 0

Eigen modes 1 2

⎛ 16 ⎜⎜ ⎝ − 5 0 ⎞ ⎟ M 2 ⎟⎠

− 5 2

Units Hz Hz

⎞ ⎟⎟ ⎠

Reference 2.085 10.742

295

ADVANCE DESIGN VALIDATION GUIDE

Modal vectors For ω1:

⎛ 48EI ⎛ 16 − 5 ⎞ ⎛ M ω 2 ⎜ ⎜ ⎟−⎜ 1 1 ⎜ 7L3 ⎜⎝ − 5 2 ⎟⎠ ⎜ 0 ⎝ ⎝

⎛U ⎞ ⎛ 1 ⎞ 0 ⎞⎟ ⎞⎟⎛ U1 ⎞ ⎜ ⎟⎟ = 0 ⇒ χ 1 ⎜⎜ 1 ⎟⎟ = ⎜⎜ ⎟⎟ 2 ⎟ ⎟⎜ M 2 ω 1 ⎠ ⎠⎝ U 2 ⎠ ⎝ U 2 ⎠ ⎝ 3.055 ⎠

For ω2:

1 ⎞ ⎛U ⎞ ⎛ ⎟⎟ χ 2 ⎜⎜ 1 ⎟⎟ = ⎜⎜ ⎝ U 2 ⎠ ⎝ − 0.655 ⎠ Normalizing relative to the mass

⎛ 9.305 × 10 −4 ⎞ ⎛ 2.01× 10 −3 ⎞ ⎟ ; χ2 ⎜ ⎟ χ 1 ⎜⎜ −3 ⎟ ⎜ − 1.316 × 10 −3 ⎟ ⎠ ⎝ 2.842 × 10 ⎠ ⎝ Modal deformations

296

ADVANCE DESIGN VALIDATION GUIDE

1.89.1.5 Spectral study Design spectrum

Nominal acceleration:

f1 = 2.085Hz ⇒ a n = 5.5411m s 2

f 2 = 10.742Hz ⇒ a n = 6.25 m s 2 Observation: the gap between pulses is greater than 10%, so the modal responses can be regarded as independent. Reference participation factors

γ i = χ iMΔ earthquake direction vector Eigen modes 1 2

Reference 479.427 275.609

297

ADVANCE DESIGN VALIDATION GUIDE

Pseudo-acceleration

Γi = a i × ξ × γ i × χ i in (m/s2) ⎛ 5% ⎞ ⎟⎟ ξ = ⎜⎜ ⎝ η ⎠

0 .4

: Damping correction factor.

η: Structure damping.

⎛ 2.7026 ⎞ ⎟⎟ Γ1 = ⎜⎜ ⎝ 8.2556 ⎠

⎛ 3.7852 ⎞ ⎟⎟ Γ1 = ⎜⎜ ⎝ - 2.4783 ⎠

Reference modal displacement

⎛ 1.576E − 02 ⎞ ⎟⎟ ψ 1 = ⎜⎜ ⎝ 4.814E − 02 ⎠

⎛ 8.318E − 04 ⎞ ⎟⎟ ψ 2 = ⎜⎜ ⎝ - 5.446E − 04 ⎠

Equivalent static forces

⎛ 5.510E + 05 ⎞ ⎟⎟ F1 = ⎜⎜ ⎝ 8.415E + 05 ⎠

⎛ 7.717E + 05 ⎞ ⎟⎟ F2 = ⎜⎜ ⎝ - 2.526E + 05 ⎠

Displacement at the top of the mast

U1 =

((4.81E − 02)

+ (− 5.446E − 04 )

Units m

)

Reference 4.814 E-02

Shear force at the top of the mast

((8.415E + 05)

T1 =

+ (− 2.526E + 05 )

)

3: Being the behavior coefficient of forces Units N

Reference 2.929 E+05

Moment at the base

Units N.m

Reference 1.578 E+07

1.89.1.6 Theoretical results Reference

298

Solver

Result name

Result description

Reference value

CM2

Frequency

Frequency Mode 1 [Hz]

2.085

CM2

Frequency

Frequency Mode 2 [Hz]

10.742

CM2

Displacement at the top of the mast [cm]

4.814

CM2

Forces at the top of the mast [N]

2.929E+05

ADVANCE DESIGN VALIDATION GUIDE

1.89.2 Calculated results

Result name

Result description Frequency Mode 1 [Hz]

Value 2.08 Hz

Error -0.24%

Frequency Mode 2 [Hz]

10.74 Hz

-0.02%

Displacement at the top of the mast [cm]

4.81159 cm

-0.05%

Forces at the top of the mast [N]

292677 N

-0.08%

299

ADVANCE DESIGN VALIDATION GUIDE

1.90 Design of a concrete floor with an opening (03-0208SSLLG_BAEL91) Test ID: 2524 Test status: Passed

1.90.1 Description Verifies the displacements, bending moments and reinforcement results for a 2D concrete slab with supports and punctual loads.

1.90.2 Background

1.90.2.1 Model description ■

Calculation model: 2D concrete slab. Slab thickness: 20 cm ► Slab length: 20m ► Slab width: 10m ► The supports (punctual and linear) are considered as hinged. ► Supports positioning (see scheme below) ► 1,50m*2,50m opening => see positioning on the following scheme Materials: ► Concrete B25 ► Young module: E= 36000 MPa Load case: ► Permanent loads: 100 kg/m2 ► Permanent loads: 200 kg/ml around the opening ► Punctual loads of 2T in permanent loads (see the following definition) ► Usage overloads: 250 kg/m2 Mesh density: 0.5 m ►

■

300

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Slab geometry

Support positions

301

ADVANCE DESIGN VALIDATION GUIDE

Positions of punctual loads

Global loading overview

302

ADVANCE DESIGN VALIDATION GUIDE

Load Combinations Code BAGMAX BAQ BAELS BAELU

Numbers 1 2 101 102

Type Static Static Comb_Lin Comb_Lin

Title Permanent loads + self weight Usage overloads Gmax+Q 1.35Gmax+1.5Q

1.90.2.2 Effel Structure Results SLS max displacements (load combination 101)

Mx bending moment for ULS load combination

303

ADVANCE DESIGN VALIDATION GUIDE

My bending moment for ULS load combination

Mxy bending moment for ULS load combination

1.90.2.3 Effel RC Expert Results Main hypothesis ■ ■ ■ ■ ■

304

Top and bottom concrete covers: 3 cm Slightly dangerous cracking Concrete B25 => Fc28= 25 MPa Reinforcement calculation according to Wood method. Calculation starting from non averaged forces.

ADVANCE DESIGN VALIDATION GUIDE

Axi reinforcements

Ayi reinforcements

305

ADVANCE DESIGN VALIDATION GUIDE

Axs reinforcements

Ays reinforcements

306

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1.90.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Max displacement for SLS (load combination 101) [cm]

0.176

CM2

Myy

Mx and My bending moments for ULS (load combination 102) Max(Mx) [kN.m]

25.20

CM2

Myy

Mx and My bending moments for ULS (load combination 102) Min(Mx) [kN.m]

-15.71

CM2

Mxx

Mx and My bending moments for ULS (load combination 102) Max(My) [kN.m]

31.17

CM2

Mxx

Mx and My bending moments for ULS (load combination 102) Min(My) [kN.m]

-18.79

CM2

Mxy

Mx and My bending moments for ULS (load combination 102) Max (Mxy) [kN.m]

10.26

CM2

Mxy

Mx and My bending moments for ULS (load combination 102) Min (Mxy) [kN.m]

-10.14

CM2

Axi

Theoretic reinforcements Axi [cm2]

3.84

CM2

Axs

Theoretic reinforcements Axs [cm2]

3.55

CM2

Ayi

Theoretic reinforcements Ayi [cm2]

3.75

CM2

Ays

Theoretic reinforcements Ays [cm2]

4.53

These values are obtained from the maximum values from the mesh.

1.90.3 Calculated results

Result name D

Result description

Value

Error

Max displacement for SLS (load combination 101) [cm]

0.174641 cm

-0.77%

Myy

Mx and My bending moments for ULS (load combination 102) Max(Mx) [kNm]

25.2594 kN*m

0.24%

Myy

Mx and My bending moments for ULS (load combination 102) Min(Mx) [kNm]

-15.6835 kN*m

0.17%

Mxx

Mx and My bending moments for ULS (load combination 102) Max(My) [kNm]

31.2449 kN*m

0.24%

Mxx

Mx and My bending moments for ULS (load combination 102) Min(My) [kNm]

-18.7726 kN*m

0.09%

Mxy

Mx and My bending moments for ULS (load combination 102) Max (Mxy) [kNm]

10.1558 kN*m

-1.02%

Mxy

Mx and My bending moments for ULS (load combination 102) Min (Mxy) [kNm]

-10.2508 kN*m

-1.09%

Axi

Theoretic reinforcements Axi [cm2]

3.83063 cm²

-0.24%

Axs

Theoretic reinforcementsAxs [cm2]

3.629 cm²

2.23%

Ayi

Theoretic reinforcements Ayi [cm2]

3.72879 cm²

-0.57%

Ays

Theoretic reinforcements Ays [cm2]

4.61909 cm²

1.97%

307

ADVANCE DESIGN VALIDATION GUIDE

1.91 Design of a 2D portal frame (03-0207SSLLG_CM66) Test ID: 2523 Test status: Passed

1.91.1 Description Verifies the steel calculation results (displacement at ridge, normal forces, bending moments, deflections, stresses, buckling lengths, lateral torsional buckling lengths and cross section optimization) for a 2D metallic portal frame, according to CM66.

1.91.2 Background

1.91.2.1 Model description ■

■

Calculation model: 2D metallic portal frame. ► Column section: IPE500 ► Rafter section: IPE400 ► Base plates: hinged. ► Portal frame width: 20m ► Columns height: 6m ► Portal frame height at the ridge: 7.5m Load case: ► Permanent loads: 150 kg/m on the roof + elements self weight. ► Usage overloads: 800 kg/ml on the roof Mesh density: 1m

Model preview

Combinations Code CMP CMS CMCFN CMCFN CMCFN CMCD

308

Numbers 1 2 101 102 103 104

Type Static Static Comb_Lin Comb_Lin Comb_Lin Comb_Lin

Title Permanent load + self weight Usage overloads 1.333P 1.333P+1.5S P+1.5S P+S

ADVANCE DESIGN VALIDATION GUIDE

1.91.2.2 Effel Structure Results Ridge displacements (combination 104)

Diagram of normal force envelope

309

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Envelope of bending moments diagram

1.91.2.3 Effel Expert CM results Main hypotheses For columns ■

Deflections:

1/150 Envelopes deflections calculation.

■

Buckling:

XY plane: Automatic calculation of the structure on fixed nodes XZ plane: Automatic calculation of the structure on fixed nodes Ka-Kb Method

■

Lateral-torsional buckling:

Ldi automatic calculation: no restraints Lds imposed value: 2 m

For the rafters ■

Deflections:

1/200 Envelopes deflections calculation.

■

Buckling:

XY plane: Automatic calculation of the structure on fixed nodes XZ plane: Automatic calculation of the structure on fixed nodes Ka-Kb Method

■

Lateral-torsional buckling:

Ldi automatic calculation: No restraints Lds imposed value: 1.5m

Optimization criteria ■ ■

310

Work ratio optimization between 90 and 100% Labels optimization (on Advance Design templates)

ADVANCE DESIGN VALIDATION GUIDE

Deflection verification Ratio

CM Stress diagrams Work ratio

Stresses

311

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Buckling lengths Lfy

Lfz

Lateral-torsional buckling lengths Ldi

312

ADVANCE DESIGN VALIDATION GUIDE

Lds

Optimization

313

ADVANCE DESIGN VALIDATION GUIDE

1.91.2.4 Theoretical results Solver

Result name

Result description

Reference value

CM2

Displacement at the ridge [cm]

9.36

CM2

Envelope normal forces on Columns (min) [T]

-15.77

CM2

Envelope normal forces on Rafters (max) [T]

-1.02

CM2

Envelope bending moments on Columns (min) [T.m]

-42.41

CM2

Envelope bending moments on Rafters (max) [T.m]

42.41

CM2

Deflection

CM deflections on Columns [%]

L / 438 (34%)

CM2

Deflection

CM deflections on Rafters [%]

L / 111 (180%)

CM2

Stress

CM stresses on Columns [MPa]

230.34

CM2

Stress

CM stresses on Rafters [MPa]

458.38

CM2

Lfy

Buckling lengths on Columns - Lfy [m]

5.84

CM2

Lfz

Buckling lengths on Columns - Lfz [m]

CM2

Lfy

Buckling lengths on Rafters - Lfy [m]

7.08

CM2

Lfz Buckling lengths on Rafters - Lfz [m] Warning, the local axes in Effel Structure have different orientation in Advance Design. Result description

10.11

Solver

Result name

Reference value

CM2

Ldi

Lateral-torsional buckling lengths on Columns Ldi [m]

CM2

Lds

Lateral-torsional buckling lengths on Columns Lds [m]

CM2

Ldi

Lateral-torsional buckling lengths on Rafters Ldi [m]

10.11

CM2

Lds

Lateral-torsional buckling lengths on Rafters Lds [m]

1.5

Solver

Result name

Result description

Final section

CM2

Work ratio

IPE500 columns - section optimization

IPE500

CM2

Work ratio

IPE400 rafters - section optimization

195

IPE550

Rate (%)

1.91.3 Calculated results Result name D

314

Result description Displacement at the ridge [cm]

Value 9.36473 cm

Error 0.05%

Envelope normal forces on Columns (min) [T]

-15.7798 T

-0.06%

Envelope normal forces on Rafters (max) [T]

-1.0161 T

0.38%

Envelope bending moments on Columns (min) [Tm]

42.4226 T*m

0.03%

Envelope bending moments on Rafters (max) [Tm]

42.4226 T*m

0.03%

Deflection

CM deflections on Columns [adm]

438.077 Adim.

0.02%

Deflection

CM deflections on Rafters [adm]

111.325 Adim.

0.29%

Stress

CM stresses on Columns [adm]

230.331 MPa

0.00%

Stress

CM stresses on Rafters [adm]

458.367 MPa

0.00%

Lfy

Buckling lengths on Columns - Lfy [m]

0.00%

Lfz

Buckling lengths on Columns - Lfz [m]

5.84401 m

0.07%

Lfy

Buckling lengths on Rafters - Lfy [m]

10.1119 m

0.02%

Lfz

Buckling lengths on Rafters - Lfz [m]

7.07904 m

-0.01%

Ldi

Lateral-torsional buckling lengths on Columns Ldi [m]

0.00%

Lds

Lateral-torsional buckling lengths on Columns Lds [m]

0.00%

Ldi

Lateral-torsional buckling lengths on Rafters Ldi [m]

10.1119 m

0.02%

Lds

Lateral-torsional buckling lengths on Rafters Lds [m]

1.5 m

0.00%

Work ratio

IPE500 columns - section optimization [adm]

0.980131 Adim.

0.01%

Work ratio

IPE400 rafters - section optimization [adm]

1.9505 Adim.

0.03%

ADVANCE DESIGN VALIDATION GUIDE

1.92 Cantilever rectangular plate (01-0001SSLSB_FEM) Test ID: 2433 Test status: Passed

1.92.1 Description Verifies the vertical displacement on the free extremity of a cantilever rectangular plate fixed on one side. The plate is 1 m long, subjected to a uniform planar load.

1.92.2 Background

1.92.2.1 Model description ■ ■ ■

Reference: Structure Calculation Software Validation Guide, test SSLS 01/89. Analysis type: linear static. Element type: planar. Cantilever rectangular plate

Scale =1/4

01-0001SSLSB_FEM

Units S.I. Geometry ■ ■ ■

Thickness: e = 0.005 m, Length: l = 1 m, Width: b = 0.1 m.

Materials properties ■

Longitudinal elastic modulus: E = 2.1 x 1011 Pa,

■

Poisson's ratio: ν = 0.3. 315

ADVANCE DESIGN VALIDATION GUIDE

Boundary conditions ■ ■

Outer: Fixed at end x = 0, Inner: None.

Loadings ■ ■

External: Uniform load p = -1700 Pa on the upper surface, Internal: None.

1.92.2.2 Displacement of the model in the linear elastic range Reference solution The reference displacement is calculated for the unsupported end located at x = 1m. bl4p u = 8EI = z

0.1 x 14 x 1700 0.1 x 0.0053 11 8 x 2.1 x 10 x 12

= -9.71 cm

Finite elements modeling ■ ■ ■

Planar element: plate, imposed mesh, 1100 nodes, 990 surface quadrangles.

Deformed shape Deformed cantilever rectangular plate 01-0001SSLSB_FEM

316

Scale =1/4

ADVANCE DESIGN VALIDATION GUIDE

1.92.2.3 Theoretical results Solver

Result name

Result description

Reference value

CM2

Vertical displacement on the free extremity [cm]

-9.71

1.92.3 Calculated results

Result name DZ

Result description Vertical displacement on the free extremity [cm]

Value -9.58696 cm

Error 1.27%

317

ADVANCE DESIGN VALIDATION GUIDE

1.93 Calculating torsors using different mesh sizes for a concrete wall subjected to a horizontal force (TTAD #13175) Test ID: 5088 Test status: Passed

1.93.1 Description Calculates torsors using different mesh sizes for a concrete wall subjected to a horizontal force.

1.94 Verifying torsors on a single story coupled walls subjected to horizontal forces Test ID: 4804 Test status: Passed

1.94.1 Description Verifies torsors on a single story coupled walls subjected to horizontal forces

1.95 Verifying diagrams for Mf Torsors on divided walls (TTAD #11557) Test ID: 3610 Test status: Passed

1.95.1 Description Performs the finite elements calculation and verifies the results diagrams for Mf torsors on a high wall divided in 6 walls (by height). The loads applied on the model: self weight, two live load cases and seism loads according to Eurocodes 8.

1.96 Verifying the level mass center (TTAD #11573, TTAD #12315) Test ID: 3609 Test status: Passed

1.96.1 Description Performs the finite elements calculation on a model with two planar concrete elements with a linear support. Verifies the level mass center and generates the "Excited total masses" and "Level modal mass and rigidity centers" reports. The model consists of two planar concrete elements with a linear fixed support. The loads applied on the model: self weight, a planar live load of -1 kN and seism loads according to French standards of Eurocodes 8.

1.97 Generating results for Torsors NZ/Group (TTAD #11633) Test ID: 3594 Test status: Passed

1.97.1 Description Performs the finite elements calculation on a complex concrete structure with four levels. Generates results for Torsors NZ/Group. Verifies the legend results. The structure has 88 linear elements, 30 planar elements, 48 windwalls, etc.

318

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1.98 Verifying Sxx results on beams (TTAD #11599) Test ID: 3595 Test status: Passed

1.98.1 Description Performs the finite elements calculation on a complex model with concrete, steel and timber elements. Verifies the Sxx results on beams. Generates the maximum stresses report. The structure has 40 timber linear elements, 24 concrete linear elements, 143 steel elements. The loads applied on the structure: dead loads, live loads, snow loads, wind loads and temperature loads (according to Eurocodes).

1.99 Verifying forces results on concrete linear elements (TTAD #11647) Test ID: 3551 Test status: Passed

1.99.1 Description Verifies forces results on concrete beams consisting of a linear element and on beams consisting of two linear elements. Generates the linear elements forces by load case report.

1.100 Verifying diagrams after changing the view from standard (top, left,...) to user view (TTAD #11854) Test ID: 3539 Test status: Passed

1.100.1Description Verifies the results diagrams display after changing the view from standard (top, left,...) to user view.

1.101 Verifying stresses in beam with "extend into wall" property (TTAD #11680) Test ID: 3491 Test status: Passed

1.101.1Description Verifies the results on two concrete beams which have the "Extend into the wall" option enabled. One of the beams is connected to 2 walls on both sides and one with a wall and a pole. Generates the linear elements forces by elements report.

1.102 Verifying constraints for triangular mesh on planar elements (TTAD #11447) Test ID: 3460 Test status: Passed

1.102.1Description Performs the finite elements calculation, verifies the stresses for triangular mesh on a planar element and generates a report for planar elements stresses in neutral fiber. The planar element is 20 cm thick, C20/25 material with a linear rigid support. A linear load of 30.00 kN is applied on FX direction.

319

ADVANCE DESIGN VALIDATION GUIDE

1.103 Verifying the displacement results on linear elements for vertical seism (TTAD #11756) Test ID: 3442 Test status: Passed

1.103.1Description Verifies the displacements results on an inclined steel bar for vertical seism according to Eurocodes 8 localization and generates the corresponding report. The steel bar has a rigid support and IPE100 cross section and is subjected to self weight and seism load on Z direction (vertical).

1.104 Verifying forces for triangular meshing on planar element (TTAD #11723) Test ID: 3463 Test status: Passed

1.104.1Description Performs the finite elements calculation, verifies the forces for triangular meshing on a planar element and generates a report for planar elements forces by load case. The planar element is a square shell (5 m) with a thickness of 20 cm, C20/25 material with a linear rigid support. A linear load of -10.00 kN is applied on FZ direction.

1.105 Generating planar efforts before and after selecting a saved view (TTAD #11849) Test ID: 3454 Test status: Passed

1.105.1Description Generates efforts for all planar elements before and after selecting the third saved view.

1.106 Verifying tension/compression supports on nonlinear analysis (TTAD #11518) Test ID: 4198 Test status: Passed

1.106.1Description Verifies the behavior of supports with several rigidities fields defined. Performs the finite elements calculation and generates the "Displacements of linear elements by element" report. The model consists of a vertical linear element (concrete B20, R20*30 cross section) with a rigid punctual support at the base and a T/C punctual support at the top. A value of 15000.00 kN/m is defined for the KTX and KTZ stiffeners of the T/C support. Two loads of 500.00 kN are applied.

320

ADVANCE DESIGN VALIDATION GUIDE

1.107 Verifying tension/compression supports on nonlinear analysis (TTAD #11518) Test ID: 4197 Test status: Passed

1.107.1Description Verifies the supports behavior when the rigidity has a high value. Performs the finite elements calculation and generates the "Displacements of linear elements by element" report. The model consists of a vertical linear element (concrete B20, R20*30 cross section) with a rigid punctual support at the base and a T/C punctual support at the top. A large value of the KTX stiffener of the T/C support is defined. Two loads of 500.00 kN are applied.

1.108 Verifying the display of the forces results on planar supports (TTAD #11728) Test ID: 4375 Test status: Passed

1.108.1Description Performs the finite elements calculation and verifies the display of the forces results on a planar support. The model consists of a concrete vertical element with a planar support.

1.109 Verifying results on punctual supports (TTAD #11489) Test ID: 3693 Test status: Passed

1.109.1Description Performs the finite elements calculation and generates the punctual supports report, containing the following tables: "Displacements of point supports by load case", "Displacements of point supports by element", "Point support actions by load case", "Point support actions by element" and "Sum of actions on supports and nodes restraints". The structure consists of concrete, steel and timber linear elements with punctual supports.

1.110 Generating a report with torsors per level (TTAD #11421) Test ID: 3774 Test status: Passed

1.110.1Description Generates a report with the torsors per level results.

321

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1.111 Verifying nonlinear analysis results for frames with semi-rigid joints and rigid joints (TTAD #11495) Test ID: 3795 Test status: Passed

1.111.1Description Verifies the nonlinear analysis results for two frames with one level. One of the frames has semi-rigid joints and the other has rigid joints.

1.112 Verifying forces on a linear elastic support which is defined in a user workplane (TTAD #11929) Test ID: 4553 Test status: Passed

1.112.1Description Verifies forces on a linear elastic support, which is defined in a user workplane, and generates a report with forces for linear support in global and local workplane.

1.113 Verifying the internal forces results for a simple supported steel beam Test ID: 4533 Test status: Passed

1.113.1Description Performs the finite elements calculation for a horizontal element (S235 material and IPE180 cross section) with two hinge rigid supports at each end. One of the supports has translation restraints on X, Y and Z, the other support has restraints on Y and Z. Verifies the internal forces My, Fz. Validated according to: Example: 3.1 - Simple beam bending without the stability loss Publication: Steel structures members - Examples according to Eurocodes By: F. Wald a kol.

1.114 Verifying the main axes results on a planar element (TTAD #11725) Test ID: 4310 Test status: Passed

1.114.1Description Verifies the main axes results on a planar element. Performs the finite elements calculation for a concrete wall (20 cm thick) with a linear support. Displays the forces results on the planar element main axes.

322

2 CAD, Rendering and Visualization

ADVANCE DESIGN VALIDATION GUIDE

2.1

Verifying annotation on selection (TTAD #12700) Test ID: 4575 Test status: Passed

2.1.1 Description Verifying annotation on selection (TTAD #12700)

2.2

Verifying rotation for steel beam with joint (TTAD #12592) Test ID: 4560 Test status: Passed

2.2.1 Description Verifying rotation for steel beam with joint at one end (TTAD #12592)

2.3

Verifying hide/show elements command (TTAD #11753) Test ID: 3443 Test status: Passed

2.3.1 Description Verifies the hide/show elements command for the whole structure using the right-click option.

2.4

System stability during section cut results verification (TTAD #11752) Test ID: 3457 Test status: Passed

2.4.1 Description Performs the finite elements calculation and verifies the section cut results on a concrete planar element with an opening.

2.5

Verifying the grid text position (TTAD #11704) Test ID: 3464 Test status: Passed

2.5.1 Description Verifies the grid text position from different views.

2.6

Verifying the grid text position (TTAD #11657) Test ID: 3465 Test status: Passed

2.6.1 Description Verifies the grid text position from different views.

324

ADVANCE DESIGN VALIDATION GUIDE

2.7

Generating combinations (TTAD #11721) Test ID: 3468 Test status: Passed

2.7.1 Description Generates combinations for three types of loads: live loads, dead loads and snow (with an altitude > 1000 m and base effect); generates the combinations description report.

2.8

Verifying the coordinates system symbol (TTAD #11611) Test ID: 3550 Test status: Passed

2.8.1 Description Verifies the coordinates system symbol display from different views.

2.9

Verifying descriptive actors after creating analysis (TTAD #11589) Test ID: 3579 Test status: Passed

2.9.1 Description Generates the finite elements calculation on a complex concrete structure (C35/45 material). Verifies the descriptive actors after creating the analysis. The structure consists of 42 linear elements, 303 planar elements, 202 supports, etc. 370 planar loads are applied: live loads, dead loads and temperature.

2.10 Creating a circle (TTAD #11525) Test ID: 3607 Test status: Passed

2.10.1 Description Creates a circle.

2.11 Creating a camera (TTAD #11526) Test ID: 3608 Test status: Passed

2.11.1 Description Verifies the camera creation and visibility.

325

ADVANCE DESIGN VALIDATION GUIDE

2.12 Verifying the local axes of a section cut (TTAD #11681) Test ID: 3637 Test status: Passed

2.12.1 Description Changes the local axes of a section cut in the descriptive model and verifies if the local axes are kept in analysis model.

2.13 Verifying the snap points behavior during modeling (TTAD #11458) Test ID: 3644 Test status: Passed

2.13.1 Description Verifies the snap points behavior when the "Allowed deformation" function is enabled (stretch points) and when it is disabled (grip points).

2.14 Verifying the representation of elements with HEA cross section (TTAD #11328) Test ID: 3701 Test status: Passed

2.14.1 Description Verifies the representation of elements with HEA340 cross section.

2.15 Verifying the descriptive model display after post processing results in analysis mode (TTAD #11475) Test ID: 3733 Test status: Passed

2.15.1 Description Performs the finite elements calculation and displays the forces results on linear elements. Returns to the model mode to verify the descriptive model display.

2.16 Verifying holes in horizontal planar elements after changing the level height (TTAD #11490) Test ID: 3740 Test status: Passed

2.16.1 Description Verifies holes in horizontal planar elements after changing the level height.

326

ADVANCE DESIGN VALIDATION GUIDE

2.17 Verifying the display of elements with compound cross sections (TTAD #11486) Test ID: 3742 Test status: Passed

2.17.1 Description Creates an element with compound cross section (CS1 IPE400 IPE240) and verifies the cross section display.

2.18 Modeling using the tracking snap mode (TTAD #10979) Test ID: 3745 Test status: Passed

2.18.1 Description Enables the "tracking" snap mode to model structure elements.

2.19 Turning on/off the "ghost" rendering mode (TTAD #11999) Test ID: 4304 Test status: Passed

2.19.1 Description Verifies the on/off function for the "ghost" rendering mode when the workplane display is disabled.

2.20 Moving a linear element along with the support (TTAD #12110) Test ID: 4302 Test status: Passed

2.20.1 Description Moves a linear element along with the element support, after selecting both elements.

2.21 Verifying the "ghost" display after changing the display colors (TTAD #12064) Test ID: 4349 Test status: Passed

2.21.1 Description Verifies the "ghost" display on selected elements after changing the element display color.

2.22 Verifying the "ghost display on selection" function for saved views (TTAD #12054) Test ID: 4347 Test status: Passed

2.22.1 Description Verifies the display of saved views which contain elements with the "ghost on selection" function enabled.

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2.23 Verifying the steel connections modeling (TTAD #11698) Test ID: 4440 Test status: Passed

2.23.1 Description Verifies the modeling of steel connections.

2.24 Verifying the fixed load scale function (TTAD #12183). Test ID: 4429 Test status: Passed

2.24.1 Description Verifies the "fixed load scale" function.

2.25 Verifying the dividing of planar elements which contain openings (TTAD #12229) Test ID: 4483 Test status: Passed

2.25.1 Description Verifies the dividing of planar elements which contain openings.

2.26 Verifying the program behavior when trying to create lintel (TTAD #12062) Test ID: 4507 Test status: Passed

2.26.1 Description Verifies the program behavior when trying to create lintel on a planar element with an inappropriate opening.

2.27 Verifying the program behavior when launching the analysis on a model with overlapped loads (TTAD #11837) Test ID: 4511 Test status: Passed

2.27.1 Description Verifies the program behavior when launching the analysis on a model that had overlapped loads.

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2.28 Verifying the display of punctual loads after changing the load case number (TTAD #11958) Test ID: 4508 Test status: Passed

2.28.1 Description Creates a punctual load and verifies the display of the load after placing it in another load case using the load case number from the properties window.

2.29 Verifying the display of a beam with haunches (TTAD #12299) Test ID: 4513 Test status: Passed

2.29.1 Description Verifies the display of a beam with haunches, in the "Linear contour" rendering mode.

2.30 Creating base plate connections for non-vertical columns (TTAD #12170) Test ID: 4534 Test status: Passed

2.30.1 Description Creates a base plate connection on a non-vertical column.

2.31 Verifying drawing of joints in y-z plan (TTAD #12453) Test ID: 4551 Test status: Passed

2.31.1 Description Verifying drawing of joints in y-z plan (TTAD #12453)

329

3 Climatic Generator

ADVANCE DESIGN VALIDATION GUIDE

3.1

EC1: generating snow loads on a 3 slopes 3D portal frame with parapets (NF EN 1991-1-3/NA) (TTAD #11111) Test ID: 4546 Test status: Passed

3.1.1 Description Generates snow loads on a 3 slopes 3D portal frame with parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA)

3.2

EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.3 - Wind - Example C) Test ID: 4523 Test status: Passed

3.2.1 Description Generates wind loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 19911-4/NA).

3.3

EC1: generating wind loads on a 3D portal frame with one slope roof (NF EN 1991-1-4/NA) (VT : 3.2 - Wind - Example B) Test ID: 4521 Test status: Passed

3.3.1 Description Generates wind loads on a 3D portal frame with one slope roof, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA).

3.4

EC1: generating wind loads on a triangular based lattice structure with compound profiles and automatic calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12276) Test ID: 4525 Test status: Passed

3.4.1 Description Generates the wind loads on a triangular based lattice structure with compound profiles, using automatic calculation of "n" - eigen mode frequency (NF EN 1991-1-4/NA). The wind loads are generated according to Eurocodes 1 French standards.

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3.5

EC1: generating snow loads on a 2 slopes 3D portal frame (NF EN 1991-1-3/NA) (VT : 3.4 Snow - Example A) Test ID: 4518 Test status: Passed

3.5.1 Description Generates snow loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 19911-3/NA)

3.6

EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.1 - Wind - Example A) Test ID: 4520 Test status: Passed

3.6.1 Description Generates wind loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 19911-4/NA).

3.7

EC1: wind loads on a triangular based lattice structure with compound profiles and user defined "n" (NF EN 1991-1-4/NA) (TTAD #12276) Test ID: 4526 Test status: Passed

3.7.1 Description Generates wind loads on a triangular based lattice structure with compound profiles using user defined "n" - eigen mode frequency - (NF EN 1991-1-4/NA) (TTAD #12276).

3.8

EC1: Verifying the geometry of wind loads on an irregular shed. (TTAD #12233) Test ID: 4478 Test status: Passed

3.8.1 Description Verifies the geometry of wind loads on an irregular shed. The wind loads are generated according to Eurocodes 1 French standard.

3.9

EC1: Verifying the wind loads generated on a building with protruding roof (TTAD #12071, #12278) Test ID: 4510 Test status: Passed

3.9.1 Description Generates wind loads on the windwalls of a concrete structure with protruding roof, according to the Eurocodes 1 French standard. Verifies the wind loads from both directions and generates the "Description of climatic loads" report. The structure has concrete columns and beams (R2*3 cross section and B20 material) and rigid supports.

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3.10 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11531) Test ID: 4090 Test status: Passed

3.10.1 Description Generates wind loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 French standards Martinique wind speed. The structure consists of concrete (C20/25) beams and columns with rigid fixed supports, with rectangular cross section (R20*30).

3.11 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11569) Test ID: 4087 Test status: Passed

3.11.1 Description Generates snow loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanian standards. The structure consists of concrete (C20/25) beams and columns with rigid fixed supports, with rectangular cross section (R20*30).

3.12 Generating the description of climatic loads report according to EC1 Romanian standards (TTAD #11688) Test ID: 4104 Test status: Passed

3.12.1 Description Generates the "Description of climatic loads" report according to EC1 Romanian standards. The model consists of a steel portal frame with rigid fixed supports. Haunches are defined at both ends of the beams. Dead loads and SR EN 1991-1-4/NB wind loads are generated.

3.13 EC1: generating snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapet height (TTAD #11943) Test ID: 3706 Test status: Passed

3.13.1 Description Generates snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapet height, according to Eurocodes 1 French standard. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

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3.14 EC1: generating wind loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11687) Test ID: 4055 Test status: Passed

3.14.1 Description Generates wind loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanian standards. The structure consists of concrete (C20/25) beams and columns with rigid fixed supports.

3.15 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11699) Test ID: 4085 Test status: Passed

3.15.1 Description Generates wind loads on a 2 slopes 3D portal frame according to Eurocodes 1 French standards, using the "Case 1" formula for calculating the turbulence factor. The structure consists of steel elements with hinge rigid supports.

3.16 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11570) Test ID: 4086 Test status: Passed

3.16.1 Description Generates snow loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanian standards. The structure consists of concrete (C20/25) beams and columns with rigid fixed supports, with rectangular cross section (R20*30).

3.17 NV2009: generating wind loads and snow loads on a simple structure with planar support (TTAD #11380) Test ID: 4091 Test status: Passed

3.17.1 Description Generates wind loads and snow loads on the windwalls of a concrete structure with a planar support, according to NV2009 French standards. The structure has concrete columns and beams (R20*30 cross section and C20/25 material).

335

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3.18 EC1: verifying the snow loads generated on a monopitch frame (TTAD #11302) Test ID: 3713 Test status: Passed

3.18.1 Description Generates wind loads on the windwalls of a monopitch frame, according to the Eurocodes 1 French standard. The structure has concrete beams and columns (R20*30 cross section and B20 material) with rigid fixed supports.

3.19 EC1: generating wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls (TTAD #11937) Test ID: 3705 Test status: Passed

3.19.1 Description Generates wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls, according to the Eurocodes 1 French standards. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with rigid fixed supports.

3.20 EC1: generating snow loads on two side by side roofs with different heights, according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13) Test ID: 3615 Test status: Passed

3.20.1 Description Generates snow loads on two side by side roofs with different heights, according to Eurocodes 1 German standards. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

3.21 EC1: generating wind loads on a 55m high structure according to German standards (DIN EN 1991-1-4/NA) (DEV2012 #3.12) Test ID: 3618 Test status: Passed

3.21.1 Description Generates wind loads on the windwalls of a 55m high structure, according to Eurocodes 1 German standards. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

336

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3.22 EC1: generating wind loads on double slope 3D portal frame according to Czech standards (CSN EN 1991-1-4) (DEV2012 #3.18) Test ID: 3621 Test status: Passed

3.22.1 Description Generates wind loads on the windwalls of a double slope 3D portal frame, according to the Eurocodes 1 Czech standard (CSN EN 1991-1-4). The structure has concrete columns and beams (R20*30 cross section and C20/25 material).

3.23 EC1: generating snow loads on duopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13) Test ID: 3613 Test status: Passed

3.23.1 Description Generates snow loads on the windwalls of duopitch multispan roofs structure, according to Eurocodes 1 German standards. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

3.24 EC1: generating snow loads on two close roofs with different heights according to Czech standards (CSN EN 1991-1-3) (DEV2012 #3.18) Test ID: 3623 Test status: Passed

3.24.1 Description Generates snow loads on two close roofs with different heights, according to Eurocodes 1 Czech standards. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

3.25 EC1: generating snow loads on monopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13) Test ID: 3614 Test status: Passed

3.25.1 Description Generates snow loads on the windwalls of a monopitch multispan roofs structure, according to Eurocodes 1 German standards. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

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3.26 EC1: snow on a 3D portal frame with horizontal roof and parapet with height reduction (TTAD #11191) Test ID: 3605 Test status: Passed

3.26.1 Description Generates snow loads on the windwalls of a 3D portal frame with horizontal roof and 2 parapets, according to Eurocodes 1. The height of one parapet is reduced. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

3.27 EC1: generating snow loads on a 2 slopes 3D portal frame with gutter (TTAD #11113) Test ID: 3603 Test status: Passed

3.27.1 Description Generates snow loads on the windwalls of a 2 slopes 3D portal frame with gutter, according to Eurocodes 1. The structure consists of concrete (C20/25) beams and columns with rigid fixed supports.

3.28 EC1: generating snow loads on a 3D portal frame with horizontal roof and gutter (TTAD #11113) Test ID: 3604 Test status: Passed

3.28.1 Description Generates snow loads on the windwalls of a 3D portal frame with horizontal roof and gutter, according to Eurocodes 1. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

3.29 EC1: generating snow loads on a 3D portal frame with a roof which has a small span (< 5m) and a parapet (TTAD #11735) Test ID: 3606 Test status: Passed

3.29.1 Description Generates snow loads on the windwalls of a 3D portal frame with a roof which has a small span (< 5m) and a parapet, according to Eurocodes 1. The structure has concrete beams and columns (R20*30 cross section and C20/25 material) with fixed rigid supports.

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3.30 EC1: generating wind loads on double slope 3D portal frame with a fully opened face (DEV2012 #1.6) Test ID: 3535 Test status: Passed

3.30.1 Description Generates wind loads on the windwalls of a concrete structure, according to the Eurocodes 1 French standard. The structure has concrete columns and beams (R20*30 cross section and C20/25 material), a double slope roof and a fully opened face.

3.31 EC1: generating wind loads on an isolated roof with two slopes (TTAD #11695) Test ID: 3529 Test status: Passed

3.31.1 Description Generates wind loads on a concrete structure, according to the Eurocodes 1 French standard. The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and an isolated roof with two slopes.

3.32 EC1: generating wind loads on duopitch multispan roofs with pitch < 5 degrees (TTAD #11852) Test ID: 3530 Test status: Passed

3.32.1 Description Generates wind loads on the windwalls of a steel structure, according to the Eurocodes 1 French standard. The structure has steel columns and beams (I cross section and S275 material), rigid hinge supports and multispan roofs with pitch < 5 degrees.

3.33 EC1: wind load generation on a simple 3D structure with horizontal roof Test ID: 3099 Test status: Passed

3.33.1 Description Generates wind loads on the windwalls of a concrete structure, according to the Eurocodes 1 standard. The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and horizontal roof.

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3.34 EC1 NF: generating wind loads on a 3D portal frame with 2 slopes roof (TTAD #11932) Test ID: 3004 Test status: Passed

3.34.1 Description Generates the wind loads on a concrete structure according to the French Eurocodes 1 standard. The structure has a roof with two slopes, concrete columns and beams (R20*30 cross section and C20/25 material). The columns have rigid supports.

3.35 EC1: wind load generation on simple 3D portal frame with 4 slopes roof (TTAD #11604) Test ID: 3104 Test status: Passed

3.35.1 Description Generates wind loads on the windwalls of a concrete structure with 4 slopes roof, according to the Eurocodes 1 standard. The structure has 6 concrete columns (R20*30 cross section and C20/25 material) with rigid supports and C20/25 concrete walls.

3.36 EC1: wind load generation on a signboard Test ID: 3107 Test status: Passed

3.36.1 Description Generates wind loads on the windwall of a concrete signboard, according to the Eurocodes 1 standard. The signboard has concrete elements (R20*30 cross section and C20/25 material) and rigid supports.

3.37 EC1: wind load generation on a simple 3D portal frame with 2 slopes roof (TTAD #11602) Test ID: 3103 Test status: Passed

3.37.1 Description Generates wind loads on the windwalls of a concrete structure with 2 slopes roof, according to the Eurocodes 1 standard. The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and rigid supports.

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3.38 EC1: wind load generation on a building with multispan roofs Test ID: 3106 Test status: Passed

3.38.1 Description Generates wind loads on the windwalls of a concrete structure with multispan roofs, according to the Eurocodes 1 standard. The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and rigid supports.

3.39 EC1: wind load generation on a high building with horizontal roof Test ID: 3101 Test status: Passed

3.39.1 Description Generates wind loads on the windwalls of a concrete structure, according to the Eurocodes 1 standard. The structure is 63m high, has 4 columns and 4 beams (R20*30 cross section and C20/25 material), rigid supports and horizontal roof.

3.40 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) Test ID: 4569 Test status: Passed

3.40.1 Description Generates snow loads on a 4 slopes shed with gutters on each slope and middle parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

3.41 EC1: Generating snow loads on a single slope with lateral parapets (TTAD #12606) Test ID: 4570 Test status: Passed

3.41.1 Description Generates snow loads on a single slope with lateral parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

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3.42 EC1: generating wind loads on a square based lattice structure with compound profiles and automatic calculation of "n" (NF EN 1991-1-4/NA) (TTAD #12744) Test ID: 4580 Test status: Passed

3.42.1 Description Generates the wind loads on a square based lattice structure with compound profiles, using automatic calculation of "n" - eigen mode frequency. The wind loads are generated according to Eurocodes 1 - French standards (NF EN 1991-1-4/NA).

3.43 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) Test ID: 4568 Test status: Passed

3.43.1 Description Generates snow loads on a 4 slopes shed with gutters on each slope and lateral parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

3.44 EC1: Generating snow loads on two side by side buildings with gutters (TTAD #12806) Test ID: 4848 Test status: Passed

3.44.1 Description Generates snow loads on two side by side buildings with gutters, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

3.45 EC1: Generating wind loads on a square based structure according to UK standards (BS EN 1991-1-4:2005) (TTAD #12608) Test ID: 4845 Test status: Passed

3.45.1 Description Generates the wind loads on a square based structure. The wind loads are generated according to Eurocodes 1 - UK standards (BS EN 1991-1-4:2005).

3.46 EC1: Generating snow loads on a 4 slopes with gutters building. (TTAD #12719) Test ID: 4847 Test status: Passed

3.46.1 Description Generates snow loads on a model from CTCIM which contains 4 slopes with gutters, according to the Eurocodes 1 French standard (NF EN 1991-1-3/NA).

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3.47 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12808) Test ID: 4849 Test status: Passed

3.47.1 Description Generates snow loads on 2 closed building with gutters, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

3.48 EC1: Generating snow loads on a 4 slopes with gutters building (TTAD #12716) Test ID: 4846 Test status: Passed

3.48.1 Description Generates snow loads on a model from CTCIM which contains 4 slopes with gutters and lateral parapets, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

3.49 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12835) Test ID: 4850 Test status: Passed

3.49.1 Description Generates snow loads on 2 closed building with gutters. The lower building is longer. The wind loads are generated according to Eurocodes 1 - French standards (NF EN 1991-1-3/NA).

3.50 EC1: Generating snow loads on a 2 slope building with gutters and parapets. (TTAD #12878) Test ID: 4852 Test status: Passed

3.50.1 Description Generates snow loads on a 2 slope building with gutters and lateral parapets on all sides, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

3.51 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12841) Test ID: 4851 Test status: Passed

3.51.1 Description Generates snow loads on 2 closed building with gutters. The lower building is longer and has a 4 slope shed and the higher building has a 2 slope roof. The snow loads are generated according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

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3.52 NV2009: Verifying wind and snow reports for a protruding roof (TTAD #11318) Test ID: 4536 Test status: Passed

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

3.53 NV2009: Generating wind loads on a 2 slopes 3D portal frame at 15m height (TTAD #12604) Test ID: 4567 Test status: Passed

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

344

4 Combinations

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4.1

Verifying combinations for CZ localization (TTAD #12542) Test ID: 4550 Test status: Passed

4.1.1 Description Verifies simplified combinations for CZ localization.

4.2

Generating combinations (TTAD #11673) Test ID: 3471 Test status: Passed

4.2.1 Description Generates concomitance between three types of loads applied on a structure (live loads, dead loads and seismic loads - EN 1998-1), using the quadratic combination function. Generates the combinations description report and the point support actions by element report.

4.3

Defining concomitance rules for two case families (TTAD #11355) Test ID: 3749 Test status: Passed

4.3.1 Description Generates live loads and dead loads on a steel structure. Defines the concomitance rules between the two load case families and generates the concomitance matrix.

4.4

Generating load combinations with unfavorable and favorable/unfavorable predominant action (TTAD #11357) Test ID: 3751 Test status: Passed

4.4.1 Description Generates load combinations with unfavorable and favorable/unfavorable predominant action. Predominant action is a case family with 2 static load cases.

4.5

Generating combinations for NEWEC8.cbn (TTAD #11431) Test ID: 3746 Test status: Passed

4.5.1 Description Generates combinations for NEWEC8.cbn.

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4.6

Generating load combinations after changing the load case number (TTAD #11359) Test ID: 3756 Test status: Passed

4.6.1 Description Generates load combinations with concomitance matrix after changing the load case number. A report with the combinations description is generated.

4.7

Generating the concomitance matrix after adding a new dead load case (TTAD #11361) Test ID: 3766 Test status: Passed

4.7.1 Description Creates a new dead load case, after two case families were created, and generates the concomitance matrix. A report with the combinations description is generated.

4.8

Generating a set of combinations with seismic group of loads (TTAD #11889) Test ID: 4350 Test status: Passed

4.8.1 Description Generates a set of combinations with seismic group of loads.

4.9

Generating the concomitance matrix after switching back the effect for live load (TTAD #11806) Test ID: 4219 Test status: Passed

4.9.1 Description Generates the concomitance matrix and the combinations description reports after switching back the effect for live load.

347

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4.10 Performing the combinations concomitance standard test no. 5 (DEV2012 #1.7) Test ID: 4394 Test status: Passed

4.10.1 Description Generates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base or Acco) 3 - E (Base) - only one seismic load ! 4 - Q (Base or Acco)

Set value "0" (exclusive) between seismic and all Q loads (seism only combination).

Generates combinations. Generates the combinations report.

4.11 Performing the combinations concomitance standard test no.9 (DEV2012 #1.7) Test ID: 4408 Test status: Passed

4.11.1 Description Generates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base or Acco) 3 - E Group (Base) 3 - EX 4 - EY 5 - EZ

Un-group 3 - E Group to independent loads : 3 - EX 4 - EY 5 - EZ Generates combinations. Generates the combinations report.

348

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4.12 Performing the combinations concomitance standard test no.4 (DEV2012 #1.7) Test ID: 4391 Test status: Passed

4.12.1 Description Generates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base) 3 - G (Favorable or Unfavorable) 4 - Q (Base)

Generates combinations. Generates the combinations report.

4.13 Performing the combinations concomitance standard test no.6 (DEV2012 #1.7) Test ID: 4397 Test status: Passed

4.13.1 Description Generates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base or Acco) 3 - E (Base) - only one seismic load

Generates combinations. Generates the combinations report.

349

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4.14 Performing the combinations concomitance standard test no.8 (DEV2012 #1.7) Test ID: 4407 Test status: Passed

4.14.1 Description Generates loads:

1 - G (Favorable or Unfavorable)

2 - Q Group(Base or Acco) 2-Q 3-Q 4-Q

5 - Q (Base or Acco) 6 - Snow (Base or Acco)

Generates combinations. Generates the combinations report.

Un-group 2-Q Group to independent 2-Q, 3-Q, 4-Q loads. Generates combinations. Generates the combinations report.

4.15 Performing the combinations concomitance standard test no.10 (DEV2012 #1.7) Test ID: 4409 Test status: Passed

4.15.1 Description Generates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base or Acco) 3 - E Group (Base) 3 - EX 4 - EY 5 - EZ

Defines the seismic group type as "Quadratic". Generates the corresponding combinations and the combinations report.

Resets the combination set. Defines the seismic group type as "Newmark". Generates the corresponding combinations and the combinations report.

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4.16 Performing the combinations concomitance standard test no.7 (DEV2012 #1.7) Test ID: 4405 Test status: Passed

4.16.1 Description Creates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base or Acco) 3 - G (Favorable or Unfavorable) 4 - Q (Acco) 5 - Acc (Base)

Generates combinations. Generates the combinations report.

4.17 Generating a set of combinations with Q group of loads (TTAD #11960) Test ID: 4353 Test status: Passed

4.17.1 Description Generates a set of combinations with Q group of loads.

4.18 Performing the combinations concomitance standard test no.1 (DEV2010#1.7) Test ID: 4382 Test status: Passed

4.18.1 Description Generates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base or Acco) 3 - Q (Acco)

Generates combinations. Generates the combinations report.

351

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4.19 Generating a set of combinations with different Q "Base" types (TTAD #11806) Test ID: 4357 Test status: Passed

4.19.1 Description Generates a set of combinations with different Q "Base" types 1-G 2 - Q - Base 3-G 4 - Q -Base or acco Generates the first set of combinations and the first combinations report. 1-G 2 - Q - Base 3-G 4 - Q -Base Generates the second set of combinations and the second combinations report.

4.20 Performing the combinations concomitance standard test no.2 (DEV2012 #1.7) Test ID: 4384 Test status: Passed

4.20.1 Description Generates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base or Acco) 3 - G (Favorable or Unfavorable) 3 - Q (Base)

Generates combinations. Generates the combinations report.

352

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4.21 Performing the combinations concomitance standard test no.3 (DEV2012 #1.7) Test ID: 4386 Test status: Passed

4.21.1 Description Generates loads:

1 - G (Favorable or Unfavorable) 2 - Q (Base or Acco) 3 - G (Favorable or Unfavorable) 4 - Q (Base or Acco)

Generates combinations. Generates the combinations report.

353

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5.1

EC2: Verifying the minimum reinforcement area for a simply supported beam Test ID: 4517 Test status: Passed

5.1.1 Description Verifies the minimum reinforcement area for a simply supported concrete beam subjected to self weight. The verification is made with Eurocodes 2 - French annex.

5.2

EC2: Verifying the longitudinal reinforcement area of a beam under a linear load Test ID: 4519 Test status: Passed

5.2.1 Description Verifies the longitudinal reinforcement area of a beam under a linear load (horizontal level behavior law). Verification is done with Eurocodes 2 norm French Annex.

5.3

EC2: Verifying the longitudinal reinforcement area for a beam subjected to point loads Test ID: 4527 Test status: Passed

5.3.1 Description Verifies the longitudinal reinforcement area for a beam subjected to point loads (applied at the middle of the beam). The verification is performed according to EC2 norm with French Annex.

5.4

Modifying the "Design experts" properties for concrete linear elements (TTAD #12498) Test ID: 4542 Test status: Passed

5.4.1 Description Defines the "Design experts" properties for a concrete (EC2) linear element in analysis model and verifies properties from descriptive model.

5.5

EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - horizontal level behavior law Test ID: 4557 Test status: Passed

5.5.1 Description Verifies the longitudinal reinforcement area of a beam under self-weight and linear loads - horizontal level behavior law. The verification is made according to EC2 norm with French Annex.

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5.6

EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - bilinear stress-strain diagram Test ID: 4541 Test status: Passed

5.6.1 Description Verifies the longitudinal reinforcement area of a simply supported beam under a linear load - bilinear stress-strain diagram. Verification is done according to Eurocodes 2 norm with French Annex.

5.7

Verifying the longitudinal reinforcement for a horizontal concrete bar with rectangular cross section Test ID: 4179 Test status: Passed

5.7.1 Description Performs the finite elements calculation and the reinforced concrete calculation according to the Eurocodes 2 French DAN. Verifies the longitudinal reinforcement and generates the corresponding report: "Longitudinal reinforcement linear elements". The model consists of a concrete linear element with rectangular cross section (R18*60) with rigid hinge supports at both ends and two linear vertical loads: -15.40 kN and -9.00 kN.

5.8

Verifying the reinforced concrete results on a structure with 375 load cases combinations (TTAD #11683) Test ID: 3475 Test status: Passed

5.8.1 Description Verifies the reinforced concrete results for a model with more than 100 load cases combinations. On a concrete structure there are applied: dead loads, self weight, live loads, wind loads (according to NV2009) and accidental loads. A number of 375 combinations are obtained. Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforcement areas planar elements report.

5.9

Verifying the longitudinal reinforcement bars for a filled circular column (TTAD #11678) Test ID: 3543 Test status: Passed

5.9.1 Description Verifies the longitudinal reinforcement for a vertical concrete bar. Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforcement report. The bar has a circular cross section with a radius of 40.00 cm and a rigid hinge support. A vertical punctual load of 5000.00 kN is applied.

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5.10 Verifying the reinforced concrete results on a fixed beam (TTAD #11836) Test ID: 3542 Test status: Passed

5.10.1 Description Verifies the concrete results on a fixed horizontal beam. Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforced concrete calculation results report. The beam has a R25*60 cross section, C25/30 material and has a rigid hinge support at one end and a rigid support with translation restraints on Y and Z at the other end. A linear dead load (-28.75 kN) and a live load (-50.00 kN) are applied.

5.11 Verifying the longitudinal reinforcement for a fixed linear element (TTAD #11700) Test ID: 3547 Test status: Passed

5.11.1 Description Verifies the longitudinal reinforcement for a horizontal concrete bar. Performs the finite elements calculation and the reinforced concrete calculation. Verifies the longitudinal reinforcement and generates the reinforcement report. The bar has a rectangular cross section R40*80, has a rigid hinge support at one end and a rigid support with translation restraints on Y and Z. Four loads are applied: a linear dead load of -50.00 kN on FZ, a punctual dead load of -30.00 kN on FZ, a linear live load of -60.00 kN on FZ and a punctual live load of -25.00 kN on FZ.

5.12 Verifying the longitudinal reinforcement for linear elements (TTAD #11636) Test ID: 3545 Test status: Passed

5.12.1 Description Verifies the longitudinal reinforcement for a vertical concrete bar. Performs the finite elements calculation and the reinforced concrete calculation. Verifies the longitudinal reinforcement and generates the reinforcement report. The bar has a square cross section of 30.00 cm, a rigid fixed support at the base and a support with translation restraints on X and Y. A vertical punctual load of -1260.00 kN is applied.

5.13 EC2 : calculation of a square column in traction (TTAD #11892) Test ID: 3509 Test status: Passed

5.13.1 Description The test is performed on a single column in tension, according to Eurocodes 2. The column has a section of 20 cm square and a rigid support. A permanent load (traction of 100 kN) and a live load (40 kN) are applied. Performs the finite elements calculation and the reinforced concrete calculation. Generates the longitudinal reinforcement report.

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5.14 Verifying Aty and Atz for a fixed concrete beam (TTAD #11812) Test ID: 3528 Test status: Passed

5.14.1 Description Performs the finite elements calculation and the reinforced concrete calculation of a model with a horizontal concrete beam. The beam has a R20*50 cross section and two hinge rigid supports. Verifies Aty and Atz for the fixed concrete beam.

5.15 Verifying concrete results for planar elements (TTAD #11583) Test ID: 3548 Test status: Passed

5.15.1 Description Verifies the reinforcement results on planar elements. Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforced concrete analysis report: data and results. The model consists of two planar elements (C20/25 material) with rigid fixed linear supports. On each element, a punctual load of 50.00 kN on FX is applied.

5.16 Verifying concrete results for linear elements (TTAD #11556) Test ID: 3549 Test status: Passed

5.16.1 Description Verifies the reinforcement results for a horizontal concrete bar. Performs the finite elements calculation and the reinforced concrete calculation. Verifies the reinforcement and generates the reinforced concrete analysis report: data and results. The bar has a rectangular cross section R20*50, a rigid hinge support at one end, a rigid support with translation restraints on X, Y and Z and rotation restraint on X.

5.17 Verifying the reinforcement of concrete columns (TTAD #11635) Test ID: 3564 Test status: Passed

5.17.1 Description Verifies the reinforcement of a concrete column.

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5.18 Verifying the minimum transverse reinforcement area results for an articulated beam (TTAD #11342) Test ID: 3638 Test status: Passed

5.18.1 Description Verifies the minimum transverse reinforcement area for an articulated horizontal beam. Performs the finite elements calculation and the reinforced concrete calculation and generates the "Transverse reinforcement linear elements" report. The beam has a rectangular cross section (R20*50), B25 material and two hinge rigid supports at both ends.

5.19 Verifying the minimum transverse reinforcement area results for articulated beams (TTAD #11342) Test ID: 3639 Test status: Passed

5.19.1 Description Verifies the minimum transverse reinforcement area for two articulated horizontal beams. Performs the finite elements calculation and the reinforced concrete calculation and generates the "Transverse reinforcement linear elements" report. Each beam has rectangular cross section (R30*70), B25 material and two hinge rigid supports at both ends. On each beam there are applied: - Dead loads: a linear load of -25.00 kN and two punctual loads of -55.00 kN and -65.00 kN - Live loads: a linear load of -20.00 kN and two punctual loads of -40.00 kN and -35.00 kN.

5.20 EC2: column design with â&#x20AC;&#x153;Nominal Stiffness methodâ&#x20AC;? square section (TTAD #11625) Test ID: 3001 Test status: Passed

5.20.1 Description Verifies and generates the corresponding report for the longitudinal reinforcement bars of a column. The column is designed with "Nominal stiffness method", with a square cross section (C40).

5.21 EC2: Verifying the transverse reinforcement area for a beam subjected to linear loads Test ID: 4555 Test status: Passed

5.21.1 Description Verifies the transverse reinforcement area for a beam subjected to linear loads. The verification is made according to EC2 norm with French Annex.

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ADVANCE DESIGN VALIDATION GUIDE

5.22 EC2 Test 1: Verifying a rectangular cross section beam made from concrete C25/30 to resist simple bending - Bilinear stress-strain diagram Test ID: 4969 Test status: Passed

5.22.1 Description Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses will be made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

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5.23 EC2 Test 5: Verifying a T concrete section, without compressed reinforcement - Bilinear stress-strain diagram Test ID: 4978 Test status: Passed

5.23.1 Description Verifies a T concrete section, without compressed reinforcement - Bilinear stress-strain diagram. The purpose of this test is to verify the My resulted stresses for the USL load combination and the results of the theoretical reinforcement area Az.

The objective is to verify: - The stresses results - The theoretical reinforcement area results

5.23.2 Background This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.23.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ Loadings from the structure: G = 52.3 kN/m ■ Exploitation loadings (category A): Q = 13kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stress-strain diagram The objective is to verify: ■ ■

The stresses results The theoretical reinforcement area results

Simply supported beam

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Units Metric System Geometry Beam cross section characteristics:

■ ■ ■

Beam length: 7m Concrete cover: c=3.50 cm Effective height: d=h-(0.6*h+ebz)=0.595 m; d’=ebz=0.035m

Materials properties Rectangular solid concrete C16/20 and S400B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Exposure class XC1 Concrete density: 16kN/m3 Reinforcement steel ductility: Class A The calculation is made considering bilinear stress-strain diagram

■

Concrete C16/20:

f cd =

f ck

γc

16 = 10.67 MPa 1,5

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N ■

2/3 fctm = 0.30 * fck = 0.30 * 16 2 / 3 = 1.90MPa

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1 ■

Steel S400B :

f yd =

f yk

γs

400 = 347.8MPa 1,15 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

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 end point (x = 7) restrained in translation along Y and Z, and restrained rotation along X. Inner: None.

► ►

■

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Loading The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*352.3+1.5*13=90.105kN/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=52.3+13=65.3kN/ml Load calculations:

M Ed =

70.105 * 7² = 551.89kNm 8

M Ecq =

65.3 * 7² = 399.96kNm 8

5.23.2.2 Reference results in calculating the concrete beam moment At first, it will be determined the moment resistance of the concrete section only:

h ⎛ Mbtu = beff * hf * ⎜ d − f 2 ⎝

0,10 ⎞ ⎞ ⎛ ⎟ * 10,67 = 523kNm ⎟ * fcd = 0,9 * 0,10 * ⎜ 0,595 − 2 ⎠ ⎠ ⎝

Comparing Mbtu with MEd:

M btu = 523kNm < M Ed = 551.89kNm Therefore the concrete section is not entirely compressed; Therefore, calculations considering the T section are required.

5.23.2.3 Reference reinforcement calculation: Theoretical section 2: The moment corresponding to this section is:

hf ⎛ M Ed 2 = (beff − bw ) * h f * f cd * ⎜⎜ d − 2 ⎝ = 0.418MNm

⎞ 0.1 ⎞ ⎛ ⎟⎟ = (0.9 − 0.18) * 0.1*10.67 * ⎜ 0595 − ⎟= 2 ⎠ ⎝ ⎠

According to this value, the steel section is:

A2 =

364

M Ed 2 0.418 = = 22.08cm ² hf ⎞ 0 .1 ⎞ ⎛ ⎛ ⎜⎜ d − ⎟⎟ * f yd ⎜ 0.595 − ⎟ * 347.8 2 ⎝ ⎠ 2 ⎠ ⎝

ADVANCE DESIGN VALIDATION GUIDE

Theoretical section 1: The theoretical section 1 corresponds to a calculation for a rectangular shape beam section

M Ed 1 = M Ed − M Ed 2 = 0.552 − 0.418 = 133kNm

μ cu =

M Ed 1 0.133 = = 0.197 bw * d ² * Fcd 0.18 * 0.595 ² * 10.67

For a S400B reinforcement and for a XC1 exposure class, there will be a: μ cu < μ lu = 0.372 , therefore there will be no compressed reinforcement. There will be a calculation without compressed reinforcement:

[

]

[

]

α u = 1.25 * 1 − (1 − 2 * μcu ) = 1.25* 1 − (1 − 2 * 0.197) = 0.276 zc1 = d * (1 − 0.4 * α u ) = 0.595 * (1 − 0.40 * 0.276) = 0.529m A1 =

M Ed 1 0.133 = = 7.25cm² z c1 * f yd 0.529 * 347.8

Theoretical section 1: In conclusion the entire reinforcement steel area is A=A1+A2=4.25+22.08=29.33cm2 Finite elements modeling ■ ■ ■

Linear element: S beam, 8 nodes, 1 linear element.

ULS and SLS load combinations(kNm) Simply supported beam subjected to bending

ULS (reference value: 552kNm)

2 Theoretical reinforcement area(cm )

For Class B reinforcement steel ductility (reference value: A=29.33cm2)

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5.23.2.4 Reference results Result name

Result description

My,ULS

My corresponding to the 101 combination (ULS) [kNm]

Reference value 552 kNm

Az (Class B)

29.33 cm2

Theoretical reinforcement area [cm ]

5.23.3 Calculated results

Result name My Az

366

Result description

Value

Error

My USL

-540.63 kN*m

2.0408 %

-28.6439 cmÂ˛

2.3507 %

ADVANCE DESIGN VALIDATION GUIDE

5.24 EC2 Test 9: Verifying a rectangular concrete beam with compressed reinforcement – Inclined stress-strain diagram Test ID: 4982 Test status: Passed

5.24.1 Description Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. For these tests, the constitutive law for reinforcement steel, on the inclined stress-strain diagram is applied.

This test performs the verification of the theoretical reinforcement area for a rectangular concrete beam subjected to the defined loads. The test confirms the presence of the compressed reinforcement for this model.

5.24.2 Background Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses, the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage are performed. For these tests, the constitutive law for reinforcement steel, on the inclined stress-strain diagram is applied. This test performs the verification of the theoretical reinforcement area for a rectangular concrete beam subjected to the defined loads. The test confirms the presence of the compressed reinforcement for this model.

5.24.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Linear loadings : Loadings from the structure: G = 55 kN/m+ dead load, Live loads:

■

Point loads: Loadings from the structure: G = 35kN; Live loads:

■

Q=60kN/m

Q=25kN

ψ 2 = 0,8

■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.6 x Q ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering inclined stress-strain diagram The objective is to verify: ■ ■ ■

The stresses results The longitudinal reinforcement The minimum reinforcement percentage

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Simply supported beam

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

Height: h = 0.80 m, Width: b = 0.40 m, Length: L = 6.30 m, 2 Section area: A = 0.32 m , Concrete cover: c=4.50 cm Effective height: d=h-(0.6*h+ebz)=0.707 m; d’=ebz=0.045m

Materials properties Rectangular solid concrete C25/30 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■ ■

Exposure class XD1 3 Concrete density: 25kN/m Reinforcement steel ductility: Class A The calculation is performed considering inclined stress-strain diagram Cracking calculation required Concrete C25/30:

fcd =

fck 25 = = 16,67MPa γ c 1,5 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1

Ecm ■

⎛ f +8⎞ = 22000⎜ ck ⎟ ⎝ 10 ⎠

0.3

⎛ 25 + 8 ⎞ = 22000⎜ ⎟ ⎝ 10 ⎠

0.3

= 31476MPa

Steel S500 :

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

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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, Support at end point (x = 6.3) restrained in translation along Y and Z, and restrained rotation along X. Inner: None.

► ►

■

Loading The beam is subjected to the following load combinations: ■

Dead load: G’=0.4*0.8*2.5=8.00 kN/ml

■

Linear load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*(55+8)+1.5*60=175.05 kN/ml Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=55+8+60=123 kN/ml Quasi-permanent combination of actions: CQP = 1.0 x G + 0.6 x Q=55+8+0.8*60=111 kN/m

■

Point load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*35+1.5*25=84.75 kN Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=35+25=60 kN Quasi-permanent combination of actions: CQP = 1.0 x G + 0.6 x Q=35+0.8*25=55 kN

■

Load calculations:

M Ed =

175.05 * 6.30² 184.75 * 6.30 + = 1002kNm 8 4

M Ecq =

123 * 6.30² 60 * 6.30 + = 705kNm 8 4

M Eqp =

111* 6.30² 55 * 6.30 + = 637kNm 8 4

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ADVANCE DESIGN VALIDATION GUIDE

5.24.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.2

β ( f cm ) =

16.8 16.8 = = 2.925 25 + 8 f cm According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.4

at t0 = 28 days According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.5 The value of the φRH coefficient depends of the concrete quality:

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎝ ⎠ According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.3a

α1 = α 2 = 1

f CM ≤ 35MPa

⎛ 35 ⎞ If not: α1 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.7

and

⎛ 35 ⎞ ⎟⎟ α 2 = ⎜⎜ ⎝ f cm ⎠

0.2

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.8c In this case,

f cm = f ck + 8Mpa = 33Mpa

α1 = α 2 = 1 50 ⎛ ⎞ 1− ⎜ ⎟ 2 * Ac 2 * 400 * 800 100 ⎟ = 1.78 = = 266.67mm ⇒ ϕ RH = ⎜1 + h0 = 2 * (400 + 800) u ⎜ 0.1* 3 266.67 ⎟ ⎜ ⎟ ⎝ ⎠ According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.6

ϕ (∞, t 0 ) = ϕ RH * β ( f cm ) * β (t 0 ) = 1.78 * 2.92 * 0.488 = 2.54 The coefficient of equivalence is determined by the following formula:

αe =

Es Ecm 1 + ϕ (∞, t0 ) *

370

= M Eqp M Ecar

200000 = 20.94 31476 637 1 + 2..54 * 705

ADVANCE DESIGN VALIDATION GUIDE

5.24.2.3 Reference results in calculating the concrete beam reduced moment limit Due to exposure class XD3, we will verify the section having non-compressed steel reinforcement, determining the reduced moment limit, using the formula defined by Jeans Roux in his book “Practice of EC2”: This value can be determined by the next formula if

μ luc = K (α e ) *

f ck < 50MPa valid for a constitutive law to horizontal plateau:

f ck ( 4.62 − 1.66 * γ ) * f ck + (165.69 − 79.62 * γ )

(

K (α e ) = 10−4 * a + b * α e + c * α e

)

Where:

a = 75,3 * 25 − 189,8 = 1692 .7 b = −5,6 * 25 + 874,5 = 734 .5 c = 0,04 * 25 − 13 = −12 K (α e ) = 10 −4 * (1692 .7 + 734 .5 * 20 .94 − 12 * 20 .94 ² ) = 1.181 Then:

γ=

1002 = 1,422 705

μ luc = K (α e ) *

f ck = 0.271 (4.62 − 1.66 * γ ) * f ck + (165.69 − 79.62 * γ )

Calculation of reduced moment:

μ cu =

M Ed 1.002 = = 0.301 bw * d ² * f cd 0.40 * 0.707 ² * 16.67

μ cu = 0.301 > μ luc = 0.271

therefore the compressed reinforcement is present in the beam section

Reference reinforcement calculation at SLU: The calculation will be divided for theoretical sections: Calculation of the tension steel section (Section A1): The calculation of tensioned steel section must be conducted with the corresponding moment of μlu :

M Ed 1 = μlu * bw * d ² * f cd = 0.271 * 0.40 * 0.707² * 16.67 = 0.902 * 106 Nm ■

The α value:

[

]

[

]

α lu = 1.25 * 1 − (1 − 2 * μlu ) = 1.25 * 1 − (1 − 2 * 0.271) = 0.404 ■

Calculation of the lever arm zc:

zlu = d * (1 − 0.4 * α lu ) = 0.707 * (1 − 0.4 * 0.404) = 0.593m ■

Tensioned reinforcement elongation calculation:

ε su =

1 − αu

αu

* ε cu 2 =

1 − 0.404 * 3.5 = 5.17 0.404 ‰

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■

Tensioned reinforcement efforts calculation(S500A):

σ su = 432,71 + 952,38 * ε su ≤ 454MPa σ su = 432,71 + 952,38 * 0.00517 = 437.63MPa ≤ 454Mpa ■

Calculation of the reinforcement area:

A1 =

M Ed 1 0.902 *10 −6 Nm = = 34.76cm² z lu . f yd 0.593m * 437.63MPa

Compressed steel reinforcement reduction (Section As2): Reduction coefficient: ε sc =

3,5 3. 5 ( α lu * d − d' ) = (0.404 * 0.707 − 0.045 ) = 2.957 ‰ 1000 * α lu * d 1000 * 0.404 * 0.707

ε sc = 0.00295 > ε yd = 0.00217 ⇒ σ sc = 432.71 + 952.38 * 0.00295 = 435.52MPa Compressed reinforcement calculation:

As 2 =

M Ed − M Ed 1 1.002 − 0.902 = = 3.47 cm ² ( d − d ' )σ sc (0.707 − 0.045) * 435 .52

The steel reinforcement condition:

A2 = As 2 .

σ sc f yd

= 3.47 *

435.52 = 3.48cm² 437.64

Total area to be implemented: ■ ■

In the lower part: As1=A1+A2=38.24 cm2 (tensioned reinforcement) In the top part: As2=3.47 cm2 (compressed reinforcement)

Reference reinforcement calculation at SLS: The concrete beam design at SLS will be made considering a limitation of the characteristic compressive cylinder strength of concrete at 28 days, at 0.6*fck The assumptions are: ■

The SLS moment: MEcq = 705kNm

■

The equivalence coefficient: α e = 20.94

■

The stress on the concrete will be limited at 0.6*fck=15Mpa and the stress on steel at 0.8*fyk, or 400MPa

Calculation of the resistance moment MRd for detecting the presence of the compressed reinforcement:

x1 =

αe *σ c 20.94 * 15 * 0.707 = 0.311m *d = 20.94 * 15 + 400 αe *σ c + σ s

1 1 Fc = * bw * x1 *σ c = * 0.40 * 0.311×15 = 0.933 *10 6 N 2 2 zc = d −

x1 0.311 = 0.707 − = 0.603m 3 3

M rb = Fc * zc = 0.933 * 106 * 0.603 = 0.563 * 106 Nm Therefore the compressed reinforced established earlier was correct.

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Theoretical section 1 (tensioned reinforcement only)

M 1 = M rb = 0.563 * 106 Nm

α1 =

x1 0.311 = = 0.440 d 0.707

zc = d − A1 =

x1 0.311 = 0.707 − = 0.603m 3 3

M1 0.563 = = 23 .34 cm ² z c × σ s 0.603 × 400

Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)

M 2 = M ser ,cq − M rb = (0.705 − 0.591) * 10 6 = 0.142 * 10 6 Nm Compressed reinforcement stresses:

σ sc = α e * σ c *

α1 * d − d ' α1 * d

σ sc = 20.94 * 15 *

0.440 * 0.707 − 0.045 = 268.66MPa 0.440 * 0.707

Compressed reinforcement area:

A' =

M2 0.142 = = 7.98cm ² (d − d ' ) * σ sc (0.707 − 0.045 ) * 268 .66

Complementary tensioned reinforcement area:

As = A ' *

σ sc 268 .66 = 7.98 * = 5.36cm ² 400 σs

Section area: Tensioned reinforcement: 23.34+5.36=28.7 cm2 Compressed reinforcement: 7.98 cm2 Considering an envelope calculation of ULS and SLS, it will be obtained: Tensioned reinforcement ULS: A=38.24cm2 Compressed reinforcement SLS: A=7.98cm2 To optimize the reinforcement area, it is preferable a third iteration by recalculating with SLS as a baseline amount of tensioned reinforcement (after ULS: Au=38.24cm2) Reference reinforcement third calculation at SLS: For this third iteration, the calculation will begin considering the section of the tensile reinforcement found when 2 calculating for ULS: Au=38.24cm From this value, it will be calculated the stress obtained in the tensioned reinforcement:

σs =

AELS 28.70 *σ s = * 400 = 300.21MPa AELU 38.24

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ADVANCE DESIGN VALIDATION GUIDE

Calculating the moment resistance Mrb for detecting the presence of compressed steel reinforcement:

x1 =

α e *σ c 20.94 *15 * 0.707 = 0.361m *d = 20.94 *15 + 300.21 α e *σ c + σ s

1 1 Fc = * bw * x1 *σ c = * 0.40 * 0.361×15 = 1.083 *10 6 N 2 2 zc = d −

x1 0.361 = 0.707 − = 0.587 m 3 3

M rb = Fc * z c = 1.083 *10 6 * 0.587 = 0.636 *10 6 Nm Therefore the compressed reinforced established earlier was correct. Theoretical section 1 (tensioned reinforcement only)

M 1 = M rb = 0.636 *106 Nm

α1 =

x1 0.361 = = 0.511 d 0.707

zc = d − A1 =

x1 0.361 = 0.707 − = 0.587 m 3 3

M1 0.36 = = 36 .09 cm ² z c * σ s 0.578 * 400

Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)

M 2 = M ser ,cq − M rb = (0.705 − 0.636) *10 6 = 0.069 *10 6 Nm Compressed reinforcement stresses:

σ sc = α e * σ c *

α1 * d − d ' 0.511* 0.707 − 0.045 = 20.94 *15 * = 275MPa 0.511* 0.707 α1 * d

Compressed reinforcement area:

A' =

M2 0.069 = = 3.79cm ² ( d − d ' ) * σ sc (0.707 − 0.045 )* 275

Complementary tensioned reinforcement area:

As = A ' *

σ sc 275 = 3.79 * = 3.47 cm ² 300 .21 σs

Section area: Tensioned reinforcement: 36.09+3.47=39.56 cm2 Compressed reinforcement: 3.79 cm2

374

ADVANCE DESIGN VALIDATION GUIDE

Reference solution for minimal reinforcement area The minimum percentage for a rectangular beam in pure bending is:

f ct , eff ⎧ * bw * d ⎪0.26 * = Max ⎨ f yk ⎪ 0.0013 * b * d w ⎩

As , min

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2 Where:

f ct ,eff = f ctm = 2.56 MPa

from cracking conditions

Therefore:

As ,min

2.56 ⎧ * 0.40 * 0.707 = 3.77 * 10 −4 m² ⎪0.26 * = max ⎨ = 3.77cm² 500 ⎪⎩ 0.0013 * 0.40 * 0.707 = 3.68 * 10 −4 m²

Finite elements modeling ■ ■ ■

Linear element: S beam, 9 nodes, 1 linear element.

ULS and SLS load combinations(kNm) Simply supported beam subjected to bending ULS (reference value: 1002kNm)

SLS (reference value: 705kNm)

SLS –Quasi-permanent (reference value: 637kNm)

375

ADVANCE DESIGN VALIDATION GUIDE

Theoretical reinforcement area(cm2) For Class A reinforcement steel ductility (reference value: A=39.56cm2 and A’=3.79cm2)

Minimum reinforcement area(cm2) (reference value: 3.77cm2)

5.24.2.4 Reference results Result name

Result description

Reference value

My,ULS

My corresponding to the 101 combination (ULS) [kNm]

1001 kNm

My,SLS,cq

My corresponding to the 102 combination (SLS) [kNm]

705 kNm

My,SLS,qp

My corresponding to the 102 combination (SLS) [kNm]

637 cm2

Az (Class A)

Theoretical reinforcement area [cm ]

Amin

Minimum reinforcement area [cm2]

39.56 cm2 3.77 cm

5.24.3 Calculated results Result name My My

376

Result description

Value

Error

My USL

-985.226 kN*m

1.6662 %

My SLS cq

-692.949 kN*m

1.6694 %

My SLS qp

-626.623 kN*m

1.6760 %

-39.5901 cm²

0.0008 %

Amin

3.77193 cm²

0.0001 %

ADVANCE DESIGN VALIDATION GUIDE

5.25 EC2 Test 10: Verifying a T concrete section, without compressed reinforcement - Inclined stress-strain diagram Test ID: 4984 Test status: Passed

5.25.1 Description Simple Bending Design for Ultimate Limit State - The purpose of this test is to verify the My resulted stresses for the ULS load combination, the results of the theoretical reinforcement area, "Az" and the minimum reinforcement percentage, "Amin". This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.25.2 Background Verifies the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.25.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■ ■ ■ ■

Loadings from the structure: G = 500 kN/m Exploitation loadings (category A): Q = 300kN/m, The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q 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

■

ψ 2 = 0,3

■ Reinforcement steel ductility: Class B ■ The calculation is performed considering inclined stress-strain diagram The objective is to verify: ■ ■ ■

The stresses results The theoretical reinforcement area The reinforcement minimum percentage area

Simply supported beam

377

ADVANCE DESIGN VALIDATION GUIDE

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

■ ■ ■

Beam length: 6m Concrete cover: c=4.00 cm Effective height: d=h-(0.6*h+ebz)=0.900 m; d’=ebz=0.04m

Materials properties Rectangular solid concrete C30/37 and S500B reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Exposure class XC2 Reinforcement steel ductility: Class B The calculation is made considering the inclined stress-strain diagram Concrete C16/20:

f cd =

f ck 30 = = 20 MPa γc 1,5 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N

2/3 f ctm = 0.30 * f ck = 0.30 * 30 2 / 3 = 2.90 MPa

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1 ■

Steel S400B :

f yd =

f yk γs

500 = 434.78 MPa 1,15 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

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 end point (x = 7) restrained in translation along Y and Z, and restrained rotation along X. Inner: None.

► ►

■

378

ADVANCE DESIGN VALIDATION GUIDE

Loading The beam is subjected to the following load combinations: Load combinations: The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q=1.35*500+1.5*300=1125 kN/m Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=500+300=800 kN/m Quasi-permanent combination of actions CQP = 1.0 x G + 0.3 x Q=500+0.3*300=590 kN/m Load calculations:

MEd =

1125 * 6² = 5062.5 kNm 8

MEcq =

800 * 6² = 3600 kNm 8

MEqp =

590 * 6² = 2655 kNm 8

5.25.2.2 Reference results in calculating the concrete beam moment At first it will be determined the moment resistance of the concrete section only:

h ⎛ Mbtu = b eff * h f * ⎜⎜ d − f 2 ⎝

⎞ 0,20 ⎞ ⎛ ⎟⎟ * fcd = 1.40 * 0,20 * ⎜ 0,9 − ⎟ * 20 = 4.480 * 10 3 kNm 2 ⎠ ⎝ ⎠

Comparing Mbtu with MEd:

Mbtu = 4.480 * 10 3 kNm < MEd = 5.0625 * 10 3 kNm Therefore the concrete section is not entirely compressed; There are required calculations considering the T section. Reference reinforcement calculation:

For those calculations, the beam section will be divided in two theoretical section: Section 1: For the calculation of the concrete only Section 2: For the calculation of the compressed reinforcement Theoretical section 2:

The moment corresponding to this section is:

h ⎞ 0.20 ⎞ ⎛ ⎛ MEd2 = (b eff − b w ) * hf * fcd * ⎜ d − f ⎟ = (1.40 − 0.40) * 0.20 * 20 * ⎜ 0.90 − ⎟= 2 2 ⎠ ⎝ ⎠ ⎝ = 3.2 * 103 kNm Stress from the compressed steel reinforcement considering a steel grade S500B:

σ su = 432,71 + 727,27 * ε su ≤ 466 MPa

379

ADVANCE DESIGN VALIDATION GUIDE

In order to determine the ε su , the neutral axis position must be determined:

MEd1 = MEd − MEd2 = (5.063 − 3.200) * 103 = 1.863 * 103 kNm μ cu =

MEd1 1.863 = = 0.287 b w * d² * Fcd 0.40 * 0.9² * 20

[

]

[

]

αu = 1.25 * 1 − (1 − 2 * μ cu ) = 1.25 * 1 − (1 − 2 * 0.287) = 0.435 Depending of αu , the neutral axis position can be established:

ε su =

1 − αu 1 − 0.435 * ε cu2 = * 3.50 = 4.55 ‰ αu 0.435

Then we calculate the stress in the tensioned steel reinforcement:

σsu = 432,71 + 727,27 × 0.00455 = 436,02MPa ≤ 466MPa According to those above, the theoretical reinforcement can be calculated: A2 =

MEd2 h ⎞ ⎛ ⎜⎜ d − f ⎟⎟ * Fyd 2 ⎠ ⎝

3.200 = 91.74 cm² 0.20 ⎞ ⎛ ⎜ 0.9 − ⎟ * 436.02 2 ⎠ ⎝

Theoretical section 1:

The theoretical section 1 corresponds to a calculation for a rectangular shape beam section

MEd1 = MEd − MEd2 = (5.063 − 3.200) * 103 = 1.863 * 103 kNm μ cu =

MEd1 1.863 = = 0.287 < μ lu = 0.372 b w * d² * Fcd 0.40 * 0.9² * 20

For a S500B reinforcement and for a XC2 exposure class, there will be a: μcu = 0.287 < μlu = 0.372 , therefore there will be no compressed reinforcement. There will be a calculation without compressed reinforcement:

[

]

[

αu = 1.25 * 1 − (1 − 2 * μ cu ) = 1.25 * 1 −

(1 − 2 * 0.287) ] = 0.435

z c1 = d * (1 − 0.4 * αu ) = 0.9 * (1 − 0.40 * 0.435) = 0.743m A1 =

MEd1 1.863 = = 57.46cm² z c1 * f yd 0.743 * 436.02

Theoretical section 1:

In conclusion the entire reinforcement steel area is A=A1+A2=91.74+57.46=149.20cm2 Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is: fct,eff ⎧ * bw * d ⎪0.26 * A s,min = Max ⎨ f yk ⎪ 0.0013 * b * d w ⎩

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2 Where: fct,eff = fctm = 2.90MPa from cracking conditions

380

ADVANCE DESIGN VALIDATION GUIDE

Therefore: 2.90 ⎧ * 0.40 * 0.90 = 5.42 * 10 − 4 m² ⎪0.26 * A s,min = max ⎨ = 5.42 cm² 500 ⎪ 0.0013 * 0.40 * 0.90 = 4.68 * 10 − 4 m² ⎩ Finite elements modeling

■ ■ ■

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

ULS and SLS load combinations (kNm)

Simply supported beam subjected to bending

ULS (reference value: 5062.5kNm)

Theoretical reinforcement area(cm2)

For Class B reinforcement steel ductility (reference value: A=149.20cm2)

Minimum reinforcement area(cm2)

(reference value: 5.42cm2)

5.25.2.3 Reference results Result name

Result description

My,ULS

My corresponding to the 101 combination (ULS) [kNm]

Az (Class B) Amin

Reference value 2

Theoretical reinforcement area [cm ] 2

Minimum reinforcement area [cm ]

5062.5 kNm 149.20 cm2 5.42 cm2

381

ADVANCE DESIGN VALIDATION GUIDE

5.25.3 Calculated results

Result name My

382

Result description

Value

Error

My USL

-5062.5 kN*m

0.0000 %

-149.031 cm²

0.0001 %

Amin

-5.42219 cm²

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.26 EC2 Test 12: Verifying a rectangular concrete beam subjected to uniformly distributed load, without compressed reinforcement- Bilinear stress-strain diagram (Class XD3) Test ID: 4985 Test status: Passed

5.26.1 Description Simple Bending Design for Service State Limit Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. The verification of the bending stresses at service limit state is performed. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

5.26.2 Background Simple Bending Design for Service State Limit Verify the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During this test, the calculation of stresses, the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage are performed.

5.26.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ ■

Loadings from the structure: G = 25 kN/m (including dead load), Exploitation loadings (category A): Q = 15kN/m,

■ ■ ■

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

383

ADVANCE DESIGN VALIDATION GUIDE

Geometry

Beam cross section characteristics: ■ ■ ■ ■ ■ ■

Height: h = 0.65 m, Width: b = 0.28 m, Length: L = 6.40 m, Section area: A = 0.182 m2 , Concrete cover: c = 4.5 cm Effective height: d = h-(0.6*h+ebz) = 0.57m; d’ = ebz = 0.045 m

Materials properties

Rectangular solid concrete C30/37 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■

Exposure class XD3 Concrete density: 25 kN/m3 Stress-strain law for reinforcement: Bilinear stress-strain diagram The concrete age t0 = 28 days Humidity 50%

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 end point (x = 5.80) restrained in translation along Y, Z and restrained in rotation along X. Inner: None. ►

■

The beam is subjected to the following load combinations: ■

Load combinations: Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q = 25 + 15 = 40 kN/ml Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q = 25 + 0.3*15 = 29.5 kN/ml

■

384

Load calculations:

MEcq =

( 25 + 15) * 6.40² = 205 kNm 8

MEqp =

( 25 + 0.3 * 15) * 6.40² = 151 kNm 8

ADVANCE DESIGN VALIDATION GUIDE

5.26.2.2 Reference results in calculating the concrete final value of creep coefficient

ϕ (∞, t0 ) = ϕ RH β ( f cm )β (t0 ) Where:

β ( f cm ) = β (t 0 ) =

16.8 16.8 = = 2.725MPa f cm 30 + 8

1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t 0

t0 : concrete age t0=28days

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1*3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

α1 = α 2 = 1 if f ≤ 35 Mpa cm ⎛ 35 ⎞ ⎟ If not α1 = ⎜⎜ ⎟ ⎝ fcm ⎠

0.7

⎛ 35 ⎞ ⎟ and α 2 = ⎜⎜ ⎟ ⎝ fcm ⎠

0.2

f cm = f ck + 8Mpa = 38 Mpa > 35 Mpa

In this case

⎛ 35 ⎞ ⎟⎟ α1 = ⎜⎜ f cm ⎝ ⎠

0.7

⎛ 35 ⎞ ⎟⎟ α 2 = ⎜⎜ ⎝ f cm ⎠

0.2

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

therefore

0.7

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

= 0.944 0 .2

= 0.984

In this case: Humidity RH = 50 %

h0 =

ϕ RH

2 Ac 2 * 280 * 650 = = 195.70mm u 2 * (280 + 650)

50 100 =1+ * 0.984 = 1.78 ⇒ ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.78 * 2.73 * 0.488 = 2.37 0.1 * 3 195.70 1−

Calculating the equivalence coefficient:

The coefficient of equivalence is determined by the following formula:

αe =

Es Ecm 1 + ϕ (∞, t0 ) *

M Eqp M Ecar

385

ADVANCE DESIGN VALIDATION GUIDE

Where:

Ecm

⎛ f +8⎞ = 22 * ⎜ ck ⎟ ⎝ 10 ⎠

0 .3

⎛ 30 + 8 ⎞ = 22 * ⎜ ⎟ ⎝ 10 ⎠

0 .3

= 32837 MPa

Es = 200000MPa 1 + ϕ (∞, t 0 ) *

Es Ecm

αe =

M Eqp M Ecar

151 = 1 + 2.37 * = 2.75 205

1 + ϕ (∞, t0 ) *

M Eqp M Ecar

200000 = 16.76 32837 151 1 + 2.37 * 205

Material characteristics:

The maximum compression on the concrete is:

σ bc = 0,6 * fck = 0,6 * 30 = 18Mpa

For the maximum stress on the steel taut, we consider the constraint limit

σ s = 0,8 * f yk = 400 Mpa

Neutral axis position calculation:

The position of the neutral axis must be determined by calculating maximum stress on the concrete and reinforcement):

α1 =

α1

(position corresponding to the state of

16.76 *15 α e *σ c = = 0.430 α e *σ c + σ s 16.76 *18 + 400

Moment resistance calculation:

Knowing the α1 value, it can be determined the moment resistance of the concrete section, using the following formulas:

x1 = α1 * d = 0.430 * 0.570 = 0.243 m Mrb =

1 * b w * x1 * σ c 2

⎛ x ⎞ 0.243 ⎞ ⎛ * ⎜ d − 1 ⎟ = 0.5 * 0.28 * 0.243 * 18 * ⎜ 0.570 − ⎟ = 0.297 MNm ⎟ ⎜ 3 3 ⎠ ⎝ ⎝ ⎠

Where: Utile height : d = h – (0.06h + ebz) = 0.57 m The moment resistance Mrb = 297 KNm Because MEcq = 205kNm < Mrb = 297 kNm the supposition of having no compressed reinforcement is correct. Calculation of reinforcement area with max constraint on steel and concrete

The reinforcement area is calculated using the SLS load combination Neutral axis position: α1 = 0,430 ⎛ α ⎞ 0.430 ⎞ ⎛ Lever arm: z c = d * ⎜1 − 1 ⎟ = 0.57 * ⎜1 − ⎟ = 0.485 m ⎜ ⎟ 3 3 ⎠ ⎝ ⎝ ⎠

Reinforcement section: A s1,ser =

386

Mser zc * σ s

0.205 = 10.56 cm² 0.485 * 400

ADVANCE DESIGN VALIDATION GUIDE

Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is: fct,eff ⎧ * bw * d ⎪0.26 * A s,min = Max ⎨ fyk ⎪ 0.0013 * b * d w ⎩

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2 Where: fct,eff = fctm = 2.896MPa from cracking conditions

Therefore: 2.896 ⎧ * 0.28 * 0.57 = 2.40 * 10 − 4 m² ⎪0.26 = 2.40 cm² A s,min = max ⎨ 500 ⎪⎩ 0.0013 * 0.28 * 0.57 = 2.07 * 10 − 4 m² Finite elements modeling

■ ■ ■

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

ULS and SLS load combinations (kNm)

Simply supported beam subjected to bending

SLS (reference value: 205kNm)

Theoretical reinforcement area (cm2)

(reference value: 10.50cm2)

387

ADVANCE DESIGN VALIDATION GUIDE

Minimum reinforcement area (cm2)

(reference value: 2.40cm2)

5.26.2.3 Reference results Result name

Result description

Reference value

My,SLS

My corresponding to the 102 combination (SLS) [kNm] 2

10.50 cm2

Theoretical reinforcement area [cm ] 2

2.39 cm2

Minimum reinforcement area [cm ]

Amin

205 kNm

5.26.3 Calculated results

Result name My

388

Result description

Value

Error

My SLS

-200.533 kN*m

2.0833 %

-10.2949 cm²

0.5220 %

Amin

-2.38697 cm²

0.0001 %

ADVANCE DESIGN VALIDATION GUIDE

5.27 EC2 Test 15: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram Test ID: 4998 Test status: Passed

5.27.1 Description Simple Bending Design for Service Limit State The purpose of this test is to verify the My resulted stresses for the SLS load combination and the results of the theoretical reinforcement area Az and of the minimum reinforcement percentage.

5.27.2 Background This test performs the verification of the theoretical reinforcement area for a T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.27.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ Loadings from the structure: G = 20 kN/m (including the dead load) ■ Exploitation loadings (category A): Q = 10kN/m, ■ 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 ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering bilinear stress-strain diagram The objective is to verify: ■ ■ ■

The stresses results The theoretical reinforcement area The minimum reinforcement percentage

Simply supported beam

389

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics:

■ ■ ■ ■

Beam length: 8m Beam height: h=0.76m Concrete cover: c=4.50 cm Effective height: d=h-(0.6*h+ebz)=0.669 m; d’=ebz=0.045m

Materials properties

Rectangular solid concrete C20/25 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■ ■ ■

Exposure class XD1 Concrete density: 16kN/m3 Reinforcement steel ductility: Class A The calculation is performed considering bilinear stress-strain diagram The concrete age t0=28 days Humidity RH=50% Concrete: fck = 20MPa

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 end point (x = 7) restrained in translation along Y and Z, and restrained rotation along X. Inner: None.

► ►

■

The beam is subjected to the following load combinations: ■

Load combinations: Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=20+10=30kN/ml Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q=20+0.3*10=23kN/ml

390

ADVANCE DESIGN VALIDATION GUIDE

■

Load calculations:

M ser ,cq =

(20 + 10)* 8² = 240kNm

M ser ,qp =

(20 + 0.3 *10)* 8² = 184kNm

5.27.2.2 Reference results in calculating the concrete final value of the creep coefficient

ϕ (∞, t0 ) = ϕ RH β ( f cm )β (t0 ) Where:

β ( f cm ) = β (t 0 ) =

16.8 16.8 = = 3.17MPa f cm 20 + 8

1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t 0

t0 : concrete age t0=28days

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1*3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

α1 = α 2 = 1 if f ≤ 35 Mpa cm ⎛ 35 ⎞ If not α1 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.7

⎛ 35 ⎞ and α 2 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

In this case

0.2

f cm = f ck + 8 Mpa = 28 Mpa ≤ 35 Mpa

therefore

α1 = α 2 = 1 In this case: Humidity RH=50 %

h0 =

ϕ RH

2 Ac 2 * 318000 = = 162.24mm u 3920

50 100 = 1+ = 1.92 ⇒ ϕ (∞, t 0 ) = ϕ RH * β ( f cm ) * β (t 0 ) = 1.92 * 3.17 * 0.488 = 2.97 3 0.1* 162.24 1−

391

ADVANCE DESIGN VALIDATION GUIDE

Calculating the equivalence coefficient:

The coefficient of equivalence is determined by the following formula:

Es Ecm

αe =

1 + ϕ (∞, t0 ) *

M Eqp M Ecar

Where:

Ecm

⎛ f +8⎞ = 22 * ⎜ ck ⎟ ⎝ 10 ⎠

0 .3

⎛ 20 + 8 ⎞ = 22 * ⎜ ⎟ ⎝ 10 ⎠

0. 3

= 29962 MPa

Es = 200000MPa 1 + ϕ (∞, t0 ) *

M Eqp M Ecar

Es Ecm

αe =

1 + ϕ (∞, t0 ) *

184 = 1 + 2.97 * = 3.28 240 =

M Eqp M Ecar

200000 = 21.89 2962 184 1 + 2.97 * 240

Material characteristics:

The maximum compression on the concrete is:

σ bc = 0,6 * f ck = 0,6 * 20 = 12Mpa

For the maximum stress on the steel taut, we consider the constraint limit

σ s = 0,8 * f yk = 400 Mpa

Neutral axis position calculation; Calculation of Mtser:

d−

M tser

0,669 −

0,10

σs 3 * b * h 2 = 400 * 3 * 1,20 * 0,10 2 = 0,122MNm * = eff f 2 *αe d − hf 43.77 0,669 − 0,10

M ser = 240kNm > M tser = 122kNm the neutral axes is on the beam body Concrete compressive stresses

σm =

M ser beff * h f * (d −

⎛ σ ⎜σm + s αe ⎜ σ c = d *⎜ h ⎜ d− f ⎜ 2 ⎝

hf 2

= )

0,240 0,10 ) 1,20 * 0,10 * (0,669 − 2

⎞ 400 ⎞ ⎛ ⎟ ⎜ 3,23 + ⎟ ⎟ σs 21.89 ⎟ − 400 = 4,96 MPa ⎜ − = 0 , 669 * ⎟ α ⎜ 0,669 − 0,10 ⎟ 21.89 e ⎟ ⎜ ⎟ ⎟ 2 ⎝ ⎠ ⎠

σ c = 4.96MPa < σ c = 12MPa

392

= 3,23MPa

=> there is no compressed reinforcement

ADVANCE DESIGN VALIDATION GUIDE

The calculation of the tension reinforcement theoretical section As1

x1 =

α e *σ c 21.89 * 4,96 d= * 0,669 = 14.30cm 21.89 * 4,96 + 400 α e *σ c + σ s

N c1 = beff * x1 *

σc 2

= 1,20 * 0,1430 *

4,96 = 0,426MN 2

M1 = Nc1 * zc1 = 0,426 * 0,622 = 0,265MN.m and z c1 = 0,669 −

0,1430 = 0,622m 3

M1 0,265 = = 10,65cm² zc1 * σ s 0,622 * 400

As1 =

Calculation of the steel compressed reinforcement As2:

x 2 = x1 − hf = 0,1430 − 0,10 = 0,0430m σc * x 2 4,96 * 0,0430 = = 1,49MPa x1 0,1430

σc 2 =

Nc 2 = (b eff − b w ) * x 2

1,49 σc 2 = (1,20 − 0,3) * 0,0430 * = 0,029MN 2 2

M2 = Nc2 * zc2 = 0,029 * 0,555 = 0,016MNm with zc 2 = 0,669 − 0,10 −

A s2 =

M2 zc 2 * σ s

0,0430 = 0,555m 3

0,016 = 0,72cm² 0,555 * 400

Notions of serviceability moment M0 :

As0 = As1 − As2 = 10.65 − 0.72 = 9.93cm² M0 = M1 − M2 = 0,265 − 0,016 = 0,249MN.m Theoretical steel reinforcement section :

As =

A s0 * Mser 9,93 * 0,240 = = 9,58cm² M0 0,249

Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is: fct,eff ⎧ * bw * d ⎪0.26 * A s,min = Max ⎨ fyk ⎪ 0.0013 * b * d w ⎩

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2 Where: fct,eff = fctm = 2.21MPa from cracking conditions

Therefore:

2.21 ⎧ * 0.30 * 0.669 = 2.31 * 10− 4 m² ⎪0.26 * A s,min = max ⎨ = 2.61cm² 500 ⎪ 0.0013 * 0.30 * 0.669 = 2.61 * 10− 4 m² ⎩

393

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling

■ ■ ■

Linear element: S beam, 9 nodes, 1 linear element.

SLS load combinations(kNm)

Simply supported beam subjected to bending SLS (reference value: 240kNm)

Theoretical reinforcement area (cm2)

(reference value: As=9.58cm2)

Minimum reinforcement area (cm2)

(reference value: 2.57cm2)

5.27.2.3 Reference results Result name

Result description

Reference value

My,SLS

My corresponding to the 102 combination (SLS) [kNm]

240 kNm

9.58 cm2

Theoretical reinforcement area [cm ] 2

2.57 cm2

Minimum reinforcement area [cm ]

Amin

5.27.3 Calculated results Result name My

394

Result description

Value

Error

My SLS

-240 kN*m

0.0000 %

-9.64217 cm²

-0.0000 %

Amin

-2.574 cm²

-0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.28 EC2 Test 16: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram Test ID: 4999 Test status: Passed

5.28.1 Description Simple Bending Design for Service State Limit The purpose of this test is to verify the My resulted stresses for the SLS load combination and the results of the theoretical reinforcement area Az and of the minimum reinforcement percentage. This test performs verification for the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.28.2 Background This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.28.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ Loadings from the structure: G = 40 kN/m (including the dead load) ■ Exploitation loadings (category A): Q = 10kN/m, ■ 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 ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering bilinear stress-strain diagram The objective is to verify: ■ ■ ■

The stresses results The theoretical reinforcement area The minimum reinforcement percentage

Simply supported beam

395

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics:

■ ■ ■ ■

Beam length: 8m Beam height: h=0.67m Concrete cover: c=4.50 cm Effective height: d=h-(0.6*h+ebz)=0.585 m; d’=ebz=0.045m

Materials properties

Rectangular solid concrete C20/25 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■ ■

Exposure class XD1 Concrete: fck = 20MPa Reinforcement steel ductility: Class A The calculation is performed considering bilinear stress-strain diagram The concrete age t0=28 days Humidity RH=50%

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 end point (x = 7) restrained in translation along Y and Z, and restrained rotation along X. Inner: None. ►

■

The beam is subjected to the following load combinations: ■

Load combinations: Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=40+10=50kN/ml Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q=40+0.3*10=43kN/ml

■

Load calculations: M ser,cq = M ser,qp =

396

(40 + 10) * 8² 8

= 400 kNm

(40 + 0.3 * 10) * 8² 8

= 344 kNm

ADVANCE DESIGN VALIDATION GUIDE

5.28.2.2 Reference results in calculating the concrete final value of creep coefficient

ϕ (∞, t0 ) = ϕ RH β ( f cm )β (t0 ) Where:

16.8 16.8 = = 3.17MPa f cm 20 + 8

β ( f cm ) = β (t 0 ) =

1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t 0

t0 : concrete age t0=28days

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1*3 h0 ⎟ ⎜ ⎟ ⎝ ⎠ if

⎛ 35 ⎞ If not α1 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.7

⎛ 35 ⎞ and α 2 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.2

In this case,

f cm = f ck + 8 Mpa = 28 Mpa ≤ 35 Mpa , therefore:

α1 = α 2 = 1 In this case: Humidity RH=50 %

h0 =

2 Ac 2 *192600 = = 123mm 3140 u

ϕ RH

50 100 = 2 ⇒ ϕ (∞, t ) = ϕ * β ( f ) * β (t ) = 2 * 3.17 * 0.488 = 3.11 = 1+ 0 RH cm 0 0.1* 3 123 1−

Calculating the equivalence coefficient:

The coefficient of equivalence is determined by the following formula:

Es E cm

αe =

1 + (∞, t 0 ) *

MEqp MEcar

Where: ⎛ f +8⎞ Ecm = 22 * ⎜ ck ⎟ ⎝ 10 ⎠

0. 3

⎛ 20 + 8 ⎞ = 22 * ⎜ ⎟ ⎝ 10 ⎠

0. 3

= 29962MPa

Es = 200000 MPa 1 + (∞, t 0 ) *

MEqp MEcar

= 1 + 3.11 *

344 = 3.68 400

397

ADVANCE DESIGN VALIDATION GUIDE

αe =

Es Ecm 1 + ϕ (∞, t 0 ) *

= M Eqp M Ecar

200000 = 24.54 29962 344 1 + 3.11* 400

Material characteristics:

The maximum compression on the concrete is: For the maximum stress on the steel taut, we consider the constraint limit

σ s = 0,8 * f yk = 400 Mpa

Neutral axis position calculation; Calculation of Mtser:

d−

M tser

0,585 −

0,10

σs 3 *10.90 * 0,10 2 = 0,083MNm 3 * b * h 2 = 400 * = * eff f 2 *α e d − h f 2 * 24.54 0,585 − 0,10

M ser = 400kNm > M tser = 83kNm the neutral axes is on the beam body Concrete compressive stresses

σm =

M ser beff * h f * (d −

⎛ σ ⎜σm + s αe ⎜ σ c = d *⎜ h ⎜ d− f ⎜ 2 ⎝

hf 2

= )

0,400 0,10 0.90 * 0,10 * (0,585 − ) 2

= 8.31MPa

⎞ 400 ⎞ ⎛ ⎟ ⎟ ⎜ 8.31 + ⎟ σs 24.54 ⎟ − 400 = 10.61MPa ⎜ 0 , 585 * − = ⎟ α ⎜ 0,585 − 0,10 ⎟ 24.54 e ⎟ ⎟ ⎜ ⎟ 2 ⎠ ⎝ ⎠

σ c = 10.61MPa < σ c = 12MPa

=> there is no compressed reinforcement

The calculation of the tension reinforcement theoretical section As1

and

398

ADVANCE DESIGN VALIDATION GUIDE

Calculation of the steel compressed reinforcement As2:

x2 = x1 − h f = 0,2306 − 0,10 = 0,1306m

σ c2 =

σ c * x2 x1

1.61* 0,1306 = 6.01MPa 0,2306

N c 2 = (beff − bw ) * x2

σ c2 2

= (0.90 − 0,18) * 0,1306 *

6.01 = 0,282 MN 2

M 2 = Nc2 * zc 2 = 0,282* 0,441= 0,125MNm with

z c 2 = 0,585 − 0,10 −

As 2 =

0,1306 = 0,441m 3

M2 0,125 = = 7.06cm² zc 2 * σ s 0,441 * 400

Notions of serviceability moment M0 :

Theoretical steel reinforcement section :

As =

As 0 * M ser 20.46 * 0,400 = = 18.83cm² 0,435 M0

Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is:

As , min

f ct , eff ⎧ * bw * d ⎪0.26 * = Max ⎨ f yk ⎪ 0.0013 * b * d w ⎩ According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2 Where:

f ct ,eff = f ctm = 2.21MPa

from cracking conditions

Therefore:

2.21 ⎧ * 0.18 * 0.585 = 1.21*10 −4 m² ⎪0.26 * = 1.37cm² As ,min = max ⎨ 500 −4 ⎪⎩ 0.0013 * 0.18 * 0.585 = 1.37 *10 m² Finite elements modeling

■ ■ ■

Linear element: S beam, 9 nodes, 1 linear element.

399

ADVANCE DESIGN VALIDATION GUIDE

SLS load combinations(kNm)

Simply supported beam subjected to bending SLS (reference value: 400kNm)

Theoretical reinforcement area(cm2)

(reference value: As=18.83cm2)

Minimum reinforcement area(cm2)

(reference value: 1.37cm2)

5.28.2.3 Reference results Result name

Result description

Reference value

My,SLS

My corresponding to the 102 combination (SLS) [kNm]

400 kNm

Theoretical reinforcement area [cm2]

18.83 cm2

Amin

Minimum reinforcement area [cm2]

1.37 cm2

5.28.3 Calculated results

Result name My

400

Result description

Value

Error

My SLS

-400 kN*m

0.0000 %

-18.8948 cm²

0.0001 %

Amin

-1.36843 cm²

-0.0001 %

ADVANCE DESIGN VALIDATION GUIDE

5.29 EC2 Test 11: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement- Bilinear stress-strain diagram (Class XD1) Test ID: 4983 Test status: Passed

5.29.1 Description Simple Bending Design for Service State Limit Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

5.30 EC2 Test 19: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) Test ID: 5033 Test status: Passed

5.30.1 Description Simple Bending Design for Serviceability State Limit Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section; the maximum spacing of cracks and the crack openings are verified.

401

ADVANCE DESIGN VALIDATION GUIDE

5.31 EC2 Test 20: Verifying the crack openings for a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) Test ID: 5034 Test status: Passed

5.31.1 Description Simple Bending Design for Serviceability State Limit Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section; the maximum spacing of cracks and the crack openings are verified.

5.31.2 Background Simple Bending Design for Serviceability State Limit Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses will be made along with the calculation of compressing stresses in concrete section σc and compressing stresses in the steel reinforcement section σs; maximum spacing of cracks and the crack openings.

5.31.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ Loadings from the structure: G = 30 kN/m (including the dead load) ■ Exploitation loadings (category A): Q = 37.5 kN/m, ■ Structural class: S4 ■ Reinforcement steel ductility: Class B The objective is to verify: ■ ■ ■ ■ ■

The stresses results The compressing stresses in concrete section σc The compressing stresses in the steel reinforcement section σs. The maximum spacing of cracks The crack opening

Simply supported beam

402

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■ ■ ■

Height: h = 0.80 m, Width: b = 0.40 m, Length: L = 8.00 m, Section area: A = 0.32 m2 , Concrete cover: c=4.5cm Effective height: d=71cm;

Materials properties

Rectangular solid concrete C25/30 is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Exposure class XD1 Stress-strain law for reinforcement: Bilinear stress-strain diagram The concrete age t0=28 days Humidity RH=50%

■

Characteristic compressive cylinder strength of concrete at 28 days: fck = 25Mpa

■

Characteristic yield strength of reinforcement: f yk = 500Mpa

■

A st = 30.16cm² for 3 beds of 5HA16

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 end point (x = 8) restrained in translation along Y and Z Inner: None. ►

■

The beam is subjected to the following load combinations: ■

Load calculations: M0Ed = 774 kNm ► Mcar = 540 kNm ► Mfq = 390 kNm ► Mqp = 330 kNm ►

403

ADVANCE DESIGN VALIDATION GUIDE

5.31.2.2 Reference results in calculating the concrete final value of creep coefficient

ϕ (∞, t0 ) = ϕ RH β ( f cm )β (t0 ) Where:

β ( f cm ) = β (t0 ) =

16.8 16.8 = = 2.92MPa f cm 25 + 8

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

t0 : concrete age t0=28days

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 ⎜ * α1 ⎟ * α 2 = 1+ ⎜ 0.1*3 h0 ⎟ ⎜ ⎟ ⎝ ⎠ If

f cm ≤ 35 Mpa ,

α1 = α 2 = 1 If not,

⎛ 35 ⎞ ⎟⎟ α1 = ⎜⎜ ⎝ f cm ⎠

0.7

⎛ 35 ⎞ ⎟ α 2 = ⎜⎜ f cm ⎟⎠ ⎝ and

0.2

In this case,

f cm = f ck + 8 Mpa = 33 Mpa ≤ 35 Mpa , therefore α1 = α 2 = 1 In this case: Humidity RH=50 %

h0 =

ϕ RH

2 Ac 2 * 400 * 800 = = 266.67mm u 2 * (400 + 800)

50 100 =1+ = 1.78 ⇒ ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.78 * 2.92 * 0.488 = 2.55 0.1 * 3 266.67 1−

The coefficient of equivalence is determined by the following formula: Under quasi-permanent combinations:

αe =

Es Ecm 1 + ϕ (∞, t0 )

Where:

Ecm

⎛ f +8⎞ = 22 * ⎜ ck ⎟ ⎝ 10 ⎠

Es = 200000MPa

404

0. 3

⎛ 25 + 8 ⎞ = 22 * ⎜ ⎟ ⎝ 10 ⎠

0 .3

= 31476 MPa

ADVANCE DESIGN VALIDATION GUIDE

ϕ (∞, t0 ) = 2.55

αe =

Es 200000 = = 22.56 Ecm 31476 1 + ϕ (∞, t0 ) 1 + 2.55

Material characteristics:

The maximum compression on the concrete is:

σ bc = 0,6 f ck = 0,6 * 25 = 15Mpa

For the maximum stress on the steel taut, we consider the constraint limit

σ s = 0,8 * f yk = 400 Mpa

Neutral axis position calculation:

Neutral axis equation:

1 * b w * x1 ² − A st * α e * (d − x1 ) + A sc * α e * ( x1 − d' ) = 0 2

− α e * ( A sc + A st ) + α e2 * ( A sc + A st )² + 2 * b w * α e * (d'* A sc + d * A st )

x1 =

By simplifying the previous equation, by considering x1 = =

− α e * A st + α e2 * A st ² + 2 * b w * α e * d * A st ) b

Asc = 0 , it will be obtained:

− 22.56 * 30.16 + 22.56² * 30.16² + 2 * 40 * 22.56 * 71 * 30.16 = 35 cm 40

Calculating the second moment: ⎡ b * x3 ⎤ I = ⎢ w 1 + A st * α e * ( d − x1 )² + A sc * α e * ( x1 − d' )² ⎥ = ⎢⎣ 3 ⎥⎦ ⎡ 0.4 * 0.3503 ⎤ =⎢ + 30,16 * 10 − 4 * 22.56 * ( 0.71 − 0.350 )² ⎥ = 0.0145m 4 3 ⎣⎢ ⎦⎥

Stresses calculation:

σc =

M ser 0,330 * x1 = * 0,350 = 7.96 Mpa I 0,0145

σ st = α e * σ c *

d − x1 0,71 − 0,350 = 22.56 * 7.96 * = 184.7 Mpa ≤ σ s = 400Mpa x1 0,350

Maximum spacing of cracks:

Bottom reinforcement 3HA20+3HA16=15.46cm2

Ac , eff

⎧ ⎧ ⎪2 . 5 * ( h − d ) ⎪2.5 * (0.8 − 0.71) = 0.255 ⎪⎪ (h − x ) ⎪⎪ (0.8 − 0.350) = b * min ⎨ =0.40 * min ⎨ = 0.15 = 0.4 * 0.15 = 0.06m 2 3 3 ⎪ ⎪ h 0 .8 ⎪ ⎪ = 0. 4 ⎪⎩ ⎪ 2 2 ⎩ According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(2); Figure 7.1

405

ADVANCE DESIGN VALIDATION GUIDE

ρ p , eff =

φeq =

30.16 *10−4 As = = 0.0500 0.06 Ac , eff

n1 * φ12 + n2 * φ22 = 16mm n1 * φ1 + n2 * φ2 According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(3)

sr ,max = k3 * c +

0.425 * k1 * k 2 * φ

ρ p ,eff

Where: c=0.051m

⎛ 25 ⎞ k3 = 3.4 * ⎜ ⎟ ⎝ c ⎠

2/3

⎛ 25 ⎞ = 3.4 * ⎜ ⎟ ⎝ 51 ⎠

2/3

= 2.113

Therefore:

sr , max = k3 * c +

0.425 * k1 * k2 * φ

ρ p , eff

= 2.113 * 0.051 +

0.425 * 0.8 * 0.5 * (16 * 10−3 ) = 162mm 0.050

Calculation of average strain:

αe =

Es 200000 = = 6.41 Ecm 31187 σ s − kt *

ε sm − ε cm =

f ct ,eff

ρ p ,eff

* (1 + α e .ρ p ,eff )

184.71 − 0.4 *

2.56 * (1 + 6.41 * 0.050) σ 0.050 = 7.88 *10 −4 ≥ 0,6 * s = 5.54 *10 −4 Es 200000

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(2) Calculation of crack widths:

wk = sr , max * (ε sm − ε cm ) = 0.162 * (7.88 × 10−4 ) = 0.128mm According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(1) For an exposure class XD1, using the French national annex, we retained an opening crack of 0.20mm max. This criterion is satisfied. Finite elements modeling

■ ■ ■

406

Linear element: S beam, 11 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

SLS load combinations (kNm) and stresses (MPa)

Simply supported beam subjected to bending

ULS (reference value: MEd=774kNm)

SLS characteristic (reference value: Mser-cq=540kNm)

SLS quasi-permanent (reference value: Mser-qp=330kNm)

Compressing stresses in concrete section ﾏツ-qp (reference value: ﾏツ-qp =7.96MPa )

Compressing stresses in the steel reinforcement section ﾏピ-qp (reference value: ﾏピ-qp=185MPa)

Maximum cracking space Sr,max (reference value: Sr,max=162mm)

407

ADVANCE DESIGN VALIDATION GUIDE

Maximum crack opening Wk (reference value: Wk=0.128mm)

5.31.2.3 Reference results Result name

Result description

Reference value

MEd

My corresponding to the 102 combination (ULS) [kNm]

774 kNm

Mser-cq

My corresponding to the 104 combination (SLS) [kNm]

540 kNm

Mser-qp

My corresponding to the 108 combination (SLS) [kNm]

330 kNm

ﾏツ

Compressing stresses in concrete section ﾏツ [MPa]

7.96 MPa

ﾏピ

Compressing stresses in the steel reinforcement section ﾏピ [MPa]

Sr,max

Maximum cracking space Sr,max [mm]

162 mm

Maximum crack opening Wk

0.128 mm

185 Mpa

5.31.3 Calculated results

Result name My

408

Result description

Value

Error

My ULS

-774 kN*m

0.0000 %

My SLS cq

-540 kN*m

0.0000 %

My SLS qp

-330 kN*m

0.0000 %

Sc QP

7.76924 MPa

0.0000 %

Ss QP

-181.698 MPa

0.0001 %

Sr,max

16.1577 cm

-0.0003 %

-0.0125137 cm

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.32 EC2 Test 14: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement- Bilinear stress-strain diagram (Class XD1) Test ID: 4987 Test status: Passed

5.32.1 Description Simple Bending Design for Service State Limit - Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. The verification of the bending stresses at service limit state is performed.

5.32.2 Background Simple Bending Design for Service State Limit Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

5.32.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Linear loads: Loadings from the structure: G = 50 kN/m + dead load, Exploitation loadings (category A): Q = 60kN/m,

■

Punctual loads G=30kN Q=25kN

ψ 2 = 0,3 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

409

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■

Height: h = 0.80 m, Width: b = 0.40 m, Length: L = 6.30 m, Section area: A = 0.320 m2 , Concrete cover: c=4.5cm Effective height: d=h-(0.6*h+ebz)=0.707m; d’=ebz=0.045m

Materials properties

Rectangular solid concrete C25/30 and S500A reinforcement steel is used. The following characteristics are used in relation to this material: ■ ■ ■ ■ ■

Exposure class XD1 Concrete density: 25kN/m3 Stress-strain law for reinforcement: Bilinear stress-strain diagram The concrete age t0=28 days Humidity RH=50%

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 end point (x = 5.80) restrained in translation along Y, Z and restrained in rotation along X. Inner: None.

► ►

■

The beam is subjected to the following load combinations: ■

Dead load: 0.40*0.80*25 = 8kN/ml

■

Load combinations: Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q=8+50+60=118kN/ml Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q=8+50+0.3*60=76kN/ml

■

410

Load calculations:

M ser , cq =

(30 + 25) * 6.3 + (8 + 50 + 60) * 6.3² = 672.05kN .m

M ser , qp =

(30 + 0.3 * 25) * 6.3 + (8 + 50 + 0.3 * 60) × 6.3² = 436.11kN .m

ADVANCE DESIGN VALIDATION GUIDE

5.32.2.2 Reference results in calculating the concrete final value of creep coefficient

ϕ (∞, t0 ) = ϕ RH β ( f cm )β (t0 ) Where:

β ( fcm ) = β (t 0 ) =

16.8 16.8 = = 2.92MPa f cm 25 + 8

1 1 = = 0.488 0.20 0.1 + 280.20 0.1 + t 0

t0 : concrete age t0=28days

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1*3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

α1 = α 2 = 1 if f ≤ 35 Mpa cm ⎛ 35 ⎞ If not α1 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.7

⎛ 35 ⎞ and α 2 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.2

f cm = f ck + 8 Mpa = 33 Mpa ≤ 35 Mpa

In this case

therefore

α1 = α 2 = 1 In this case: Humidity RH=50 %

h0 =

2 Ac 2 * 400 * 800 = = 267mm u 2 * (400 + 800)

ϕ RH

50 100 = 1.78 ⇒ ϕ (∞, t ) = ϕ * β ( f ) * β (t ) = 1.78 * 2.92 * 0.488 = 2.54 =1+ 0 RH cm 0 3 0.1 * 267 1−

Calculating the equivalence coefficient:

The coefficient of equivalence is determined by the following formula:

Es Ecm

αe =

1 + ϕ (∞, t0 ) *

M Eqp M Ecar

Where:

Ecm

⎛ f +8⎞ = 22 * ⎜ ck ⎟ ⎝ 10 ⎠

0.3

⎛ 25 + 8 ⎞ = 22 * ⎜ ⎟ ⎝ 10 ⎠

0.3

= 31476 MPa

Es = 200000MPa

411

ADVANCE DESIGN VALIDATION GUIDE

1 + ϕ (∞, t0 ) * Es Ecm

αe =

1 + ϕ (∞, t0 ) *

M Eqp M Ecar

435 = 1 + 2.54 * = 2.65 672 =

M Eqp M Ecar

200000 = 16.82 31476 435 1 + 2.54 * 672

Material characteristics:

The maximum compression on the concrete is:

σ bc = 0,6 * f ck = 0,6 *15 = 15Mpa

For the maximum stress on the steel taut, we consider the constraint limit

σ s = 0,8 * f yk = 400 Mpa

Neutral axis position calculation:

The position of the neutral axis must be determined by calculating maximum stress on the concrete and reinforcement):

α1 =

(position corresponding to the state of

αe *σ c 16.82 *15 = = 0.387 α e * σ c + σ s 16.82 *18 + 400

Moment resistance calculation:

Knowing the α1 value, it can be determined the moment resistance of the concrete section, using the following formulas:

x1 = α1 * d = 0.387 * 0.707 = 0.273 m M rb =

1 x 0.273 ⎞ ⎛ * bw * x1 * σ c * (d − 1 ) = 0.5 * 0.4 * 0.273 *15 * ⎜ 0.707 − ⎟ = 0.505MNm 2 3 3 ⎠ ⎝

Where: Utile height : d = h – (0.06h + ebz) = 0.707m The moment resistance Mrb = 505KNm Because

M Ecq = 672kNm > M rb = 505kNm , the supposition of having no compressed reinforcement is

incorrect. The calculation of the tension reinforcement theoretical section A1

zc = d −

A1 =

412

0.273 x1 = 0.707 − = 0.616m 3 3

M rb 0.505 = = 20.51cm² zc * σ s 0.616 * 400

ADVANCE DESIGN VALIDATION GUIDE

Stress calculation for steel reinforcement σsc:

δ '=

d ' 0.045 = = 0.064 d 0.707

σ sc = α e * σ c *

0.387 − 0.067 α1 − δ ' = 16.82 *15 * = 210.78MPa 0.387 α1

Calculation of the steel compressed reinforcement A’:

A' =

M ser − M rb 0.672 − 0.505 = = 11.96cm² (d − d ' ) * σ sc (0.707 − 0.045) * 210.78

Calculation of the steel tensioned reinforcement A2:

A2 = A'*

σ sc 210.78 = 11.96 * = 6.30cm² 400 σs

Calculation of the steel reinforcement :

A = A1 + A2 = 20 .51 + 6.30 = 26 .81cm ² Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is:

As , min

f ct , eff ⎧ * bw * d ⎪0.26 * = Max ⎨ f yk ⎪ 0.0013 * b * d w ⎩

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2 Where:

f ct , eff = f ctm = 2.56 MPa

from cracking conditions

Therefore:

2.56 ⎧ * 0.40 * 0.707 = 3.76 * 10− 4 m² ⎪0.26 * = 3.76cm² As , min = max ⎨ 500 −4 ⎪⎩ 0.0013 * 0.40 * 0.707 = 3.68 *10 m² Finite elements modeling

■ ■ ■

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

413

ADVANCE DESIGN VALIDATION GUIDE

ULS and SLS load combinations (kNm)

Simply supported beam subjected to bending

SLS (reference value: 672kNm)

Theoretical reinforcement area (cm2) (reference value: As=26.81cm2; A’=11.96cm2)

Minimum reinforcement area (cm2)

(reference value: 3.76cm2)

5.32.2.3 Reference results Result name

Result description

Reference value

My,SLS

My corresponding to the 102 combination (SLS) [kNm]

672 kNm

Theoretical reinforcement area [cm2]

26.81 cm2

3.77 cm2

Minimum reinforcement area [cm ]

Amin

5.32.3 Calculated results Result name My

414

Result description My SLS

Value -658.615 kN*m

Error 1.9964 %

-26.5647 cm²

0.9099 %

Amin

-3.77193 cm²

-0.0001 %

ADVANCE DESIGN VALIDATION GUIDE

5.33 EC2 Test 18: Verifying a rectangular concrete beam subjected to a uniformly distributed load, with compressed reinforcement - Bilinear stress-strain diagram (Class XD1) Test ID: 5011 Test status: Passed

5.33.1 Description Simple Bending Design for Serviceability State Limit - Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section.

5.33.2 Background Simple Bending Design for Serviceability State Limit Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of compressing stresses in concrete section σc and compressing stresses in the steel reinforcement section σs.

5.33.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ Loadings from the structure: G = 37.5 kN/m (including the dead load), ■ Exploitation loadings (category A): Q = 37.5 kN/m, ■ Structural class: S4 ■ Reinforcement steel ductility: Class A ■ For the stress calculation the French annexes was used The objective is to verify: ■ ■ ■

The stresses results The compressing stresses in concrete section σc The compressing stresses in the steel reinforcement section σs.

Simply supported beam

415

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■

Height: h = 0.80 m, Width: b = 0.35 m, Length: L = 8.00 m, Section area: A = 0.28 m2 , Concrete cover: c=4 cm Effective height: d=72cm;

Materials properties

Rectangular solid concrete C25/30 is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Exposure class XD1 Stress-strain law for reinforcement: Bilinear stress-strain diagram The concrete age t0=28 days Humidity RH=50%

■

Characteristic compressive cylinder strength of concrete at 28 days: fck = 25Mpa

■

Characteristic yield strength of reinforcement: f yk = 500Mpa

■

A st = 37.70cm² for 3 HA20

■

A sc = 6.28cm² for 2 HA10

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 end point (x = 8) restrained in translation along Y and Z Inner: None.

► ►

■

The beam is subjected to the following load combinations:

416

■

Load combinations: ► Characteristic combination of actions: CCQ = 1.0*G+1.0*Q =75 kN/ml

■

Load calculations: ► M0Ed = 855 kNm ► Mcar = 600 kNm ► Mqp = 390 kNm

ADVANCE DESIGN VALIDATION GUIDE

5.33.2.2 Reference results in calculating the concrete final value of creep coefficient

ϕ (∞, t0 ) = ϕ RH β ( f cm )β (t0 ) Where:

β ( f cm ) = β (t0 ) =

16.8 16.8 = = 2.92MPa f cm 25 + 8

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

t0 : concrete age t0=28days

RH ⎛ ⎞ 1− ⎜ ⎟ 100 ⎜ = 1+ * α1 ⎟ * α 2 ⎜ 0.1*3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

ϕ RH

⎛ 35 ⎞ If not, α1 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.7

⎛ 35 ⎞ and α 2 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.2

In this case,

f cm = f ck + 8 Mpa = 33 Mpa ≤ 35 Mpa , therefore α1 = α 2 = 1 In this case, Humidity RH=50 %

h0 =

ϕ RH

2 Ac 2 * 350 * 800 = = 243.48mm u 2 * (350 + 800)

50 100 =1+ = 1.80 ⇒ ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.80 * 2.92 * 0.488 = 2.56 3 0.1 * 243.48 1−

The coefficient of equivalence is determined by the following formula:

Es Ecm

αe =

1 + ϕ (∞, t0 ) *

M Eqp M Ecar

Where:

⎛ f +8⎞ Ecm = 22 * ⎜ ck ⎟ ⎝ 10 ⎠

0.3

⎛ 25 + 8 ⎞ = 22 * ⎜ ⎟ ⎝ 10 ⎠

0.3

= 31476 MPa

Es = 200000MPa 1 + ϕ (∞, t0 ) *

M Eqp M Ecar

= 1 + 2.56 *

390 = 2.664 600

417

ADVANCE DESIGN VALIDATION GUIDE

αe =

Es Ecm 1 + ϕ (∞, t0 ) *

= M Eqp

200000 = 16.90 31476 2.664

M Ecar

Material characteristics:

The maximum compression on the concrete is:

σ bc = 0,6 fck = 0,6 * 25 = 15Mpa

For the maximum stress on the steel taut, we consider the constraint limit

σ s = 0,8 * f yk = 320 Mpa

Neutral axis position calculation:

Neutral axis equation:

x1 = =

1 * b w * x1 ² − A st * α e * (d − x1 ) + A sc * α e * ( x1 − d' ) = 0 2

− α e * ( A st + A sc ) + α e ² * ( A st + A sc )² + 2 * b w * α e * ( d * A st + d'* A sc ) b

− 16.90 * (37.70 + 6.28) + 16.90² * (37.70 + 6.28)² + 2 * 35 * 16.90 * ( 72 * 37.70 + 4 * 6.28) = 34,41cm 35

Calculating the second moment: ⎡ b * x3 ⎤ I = ⎢ w 1 + A st * α e * ( d − x1 )² + A sc * α e * ( x1 − d' )² ⎥ = ⎢⎣ 3 ⎥⎦ ⎡ 35 * 34,413 ⎤ =⎢ + 37,70 * 16.90( 72 − 34,45)² + 6,28 * 16.90 * (34.45 − 4)² ⎥ = 1472097 cm 4 = 0.01472m 4 3 ⎢⎣ ⎥⎦

Stresses calculation:

σc =

M ser 0,600 * x1 = * 0,3441 = 14 Mpa ≥ σ c = 12 Mpa I 0,01472

σ st = α e * σ c *

d − x1 0,72 − 0,3441 = 16.90 *14 * = 259Mpa ≤ σ s = 400Mpa x1 0,3441

Finite elements modeling

■ ■ ■

418

Linear element: S beam, 9 nodes, 1 linear element.

ADVANCE DESIGN VALIDATION GUIDE

SLS load combinations (kNm) and stresses (MPa)

Simply supported beam subjected to bending

SLS (reference value: Mscq=600kNm)

Compressing stresses in concrete section σc (reference value: σc =14MPa )

Compressing stresses in the steel reinforcement section σs (reference value: σs=260.15MPa)

5.33.2.3 Reference results Result name

Result description

Reference value

My,SLS

My corresponding to the 104 combination (SLS) [kNm]

600 kNm

σc

Compressing stresses in concrete section σc (MPa)

14 MPa

σs

Compressing stresses in the steel reinforcement section σs (MPa)

260.15 MPa

5.33.3 Calculated results

Result name My

Result description

Value

Error

My SLS

-600 kN*m

0.0000 %

Sc CQ

14.122 MPa

0.0142 %

Ss CQ

-260.116 MPa

0.0014 %

419

ADVANCE DESIGN VALIDATION GUIDE

5.34 EC2 Test 24: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement - Bilinear stress-strain diagram (Class XC1) Test ID: 5058 Test status: Passed

5.34.1 Description Verifies a rectangular cross section beam made from concrete C20/25 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) will be determined, along with the cross-sectional area of the shear reinforcement (Asw) calculation.

5.35 EC2 Test 25: Verifying the shear resistance for a rectangular concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class XC1) Test ID: 5065 Test status: Passed

5.35.1 Description Verifies the shear resistance for a rectangular concrete beam C20/25 with inclined transversal reinforcement Bilinear stress-strain diagram (Class XC1). For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) is calculated, along with the cross-sectional area of the shear reinforcement (Asw).

5.36 EC2 Test 23: Verifying the shear resistance for a rectangular concrete - Bilinear stress-strain diagram (Class XC1) Test ID: 5053 Test status: Passed

5.36.1 Description Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts (VRd,max) will be determined, along with the cross-sectional area of the shear reinforcement (Asw) calculation.

5.37 EC2 Test 13: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) Test ID: 4986 Test status: Passed

5.37.1 Description Simple Bending Design for Service State Limit - Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage. The verification of the bending stresses at service limit state is performed.

420

ADVANCE DESIGN VALIDATION GUIDE

5.38 EC2 Test 17: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Inclined stress-strain diagram (Class XD1) Test ID: 5000 Test status: Passed

5.38.1 Description Simple Bending Design for Serviceability State Limit Verifies the adequacy of a rectangular cross section made from concrete C20/25 to resist simple bending. During this test, the determination of stresses is made along with the determination of compressing stresses in concrete section and compressing stresses in the steel reinforcement section.

5.38.2 Background Simple Bending Design for Serviceability State Limit Verifies the adequacy of a rectangular cross section made from concrete C20/25 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of compressing stresses in concrete section σc and compressing stresses in the steel reinforcement section σs.

5.38.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■ Loadings from the structure: G = 9.375 kN/m (including the dead load), ■ Mfq = Mcar = Mqp = 75 kNm ■ Structural class: S4 ■ Characteristic combination of actions: CCQ = 1.0 x G ■ Reinforcement steel ductility: Class B The objective is to verify: ■ ■ ■

The stresses results The compressing stresses in concrete section σc The compressing stresses in the steel reinforcement section σs.

Simply supported beam

421

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■ ■ ■

Height: h = 0.50 m, Width: b = 0.20 m, Length: L = 8.00 m, Section area: A = 0.10 m2 , Concrete cover: c=4.5cm Effective height: d=44cm;

Materials properties

Rectangular solid concrete C20/25 is used. The following characteristics are used in relation to this material: ■ ■ ■ ■

Exposure class XD1 Stress-strain law for reinforcement: Inclined stress-strain diagram The concrete age t0=28 days Humidity RH=50%

■

Characteristic compressive cylinder strength of concrete at 28 days:

■

Characteristic yield strength of reinforcement:

Ast = 9,42cm²

■

f ck = 20Mpa

f yk = 400Mpa

for 3 HA20

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 end point (x = 8) restrained in translation along Y and Z Inner: None.

► ►

■

The beam is subjected to the following load combinations: ■

Characteristic combination of actions: CCQ = 1.0 x G =9.375kN/ml

■

Load calculations: Mfq = Mcar = Mqp = 75 kNm

5.38.2.2 Reference results in calculating the concrete final value of creep coefficient

ϕ (∞, t0 ) = ϕ RH β ( f cm )β (t0 ) Where:

β ( f cm ) = β (t0 ) =

16.8 16.8 = = 3.17MPa f cm 20 + 8

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

t0 : concrete age t0=28days

422

ADVANCE DESIGN VALIDATION GUIDE

ϕ RH

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1*3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

α1 = α 2 = 1 if f ≤ 35 Mpa cm ⎛ 35 ⎞ If not, α1 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.7

⎛ 35 ⎞ and α 2 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.2

f cm = f ck + 8Mpa = 28 Mpa ≤ 35 Mpa

In this case, Therefore

α1 = α 2 = 1

In this case: Humidity RH=50 %

h0 =

ϕ RH

2 Ac 2 * 200 * 500 = = 142.86mm u 2 * (200 + 500)

50 100 =1+ = 1.96 ⇒ ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.96 * 3.17 * 0.488 = 3.03 0.1 * 3 142.86 1−

The coefficient of equivalence is determined by the following formula:

Es Ecm

αe =

1 + ϕ (∞, t0 ) *

M Eqp M Ecar

Where:

Ecm

⎛ f +8⎞ = 22 * ⎜ ck ⎟ ⎝ 10 ⎠

0 .3

⎛ 20 + 8 ⎞ = 22 * ⎜ ⎟ ⎝ 10 ⎠

0. 3

= 29962 MPa

Es = 200000MPa 1 + ϕ (∞, t0 ) *

αe =

Es Ecm 1 + ϕ (∞, t0 ) *

M Eqp M Ecar

= 1 + 3.03 *1 = 4.03

= M Eqp

200000 = 26.90 29962 1 + 3.03 *1

M Ecar

423

ADVANCE DESIGN VALIDATION GUIDE

Material characteristics:

The maximum compression on the concrete is: σ bc = 0,6fck = 0,6 * 25 = 15Mpa For the maximum stress on the steel taut, we consider the constraint limit σ s = 0,8 * f yk = 320Mpa Checking inertia cracked or not:

Before computing the constraints, check whether the section is cracked or not. For this, we determine the cracking moment which corresponds to a tensile stress on the concrete equal to fctm :

M cr =

f ctm * I v

Where:

b * h3 0,20 * 0,503 = = 0,00208m 4 12 12

h = 0,25m 2

The average stress in concrete is: 2

f ctm = 0.30 * f ck3 = 0.30 * 20 3 = 2,21Mpa The critical moment of cracking is therefore:

M cr =

f ctm * I 2,21 * 0,00208 = = 0,018MNm v 0,25

The servility limit state moment is 0.075MNm therefore the cracking inertia is present. Neutral axis position calculation:

Neutral axis equation:

x1 =

1 * bw * x1 ² − Ast * α e * (d − x1 ) + Asc * α e * ( x1 − d ' ) = 0 2

− α e * ( Asc + Ast ) + α e2 * ( Asc + Ast )² + 2 * bw * α e * ( d '*Asc + d * Ast ) bw

By simplifying the previous equation, by considering

Asc = 0 , it will be obtained:

− α e * ( Ast ) + α e2 * Ast ² + 2 * bw * α e * d * Ast ) x1 = bw

x1 =

− 26.90 * 9,42 + 26.90² * 9,42² + 2 * 20 * 26,90 * (44 / 9,42) = 23,04cm 20

Calculating the second moment:

⎡ b * x3 ⎤ ⎡ 0,20 * 0,2303 ⎤ I = ⎢ w 1 + Ast * α e * (d − x1 )² ⎥ = ⎢ + 9,42 * 10 − 4 * 26.90 * (0,44 − 0,230)² ⎥ = 0,001929 m 4 3 ⎣ 3 ⎦ ⎣ ⎦

424

ADVANCE DESIGN VALIDATION GUIDE

Stresses calculation:

σc =

M ser 0,075 * x1 = * 0,230 = 8,94 Mpa ≤ σ c = 12 Mpa I 0,001929

σ st = α e * σ c *

d − x1 0,44 − 0,230 = 26.90 * 8,94 * = 219.57 Mpa ≤ σ s = 320Mpa x1 0,230

Finite elements modeling

■ ■ ■

Linear element: S beam, 9 nodes, 1 linear element.

SLS load combinations (kNm) and stresses (MPa)

Simply supported beam subjected to bending

SLS (reference value: Mscq=75kNm)

Compressing stresses in concrete section σc

(reference value: σc =8.94MPa )

Compressing stresses in the steel reinforcement section σs

(reference value: σs=218.57MPa)

5.38.2.3 Reference results Result name

Result description

Reference value

My,SLS

My corresponding to the 103 combination (SLS) [kNm]

75 kNm

σc

Compressing stresses in concrete section σc (MPa)

8.94 MPa

σs

Compressing stresses in the steel reinforcement section σs (MPa)

219.67 MPa

425

ADVANCE DESIGN VALIDATION GUIDE

5.38.3 Calculated results

Result name My

426

Result description

Value

Error

My SLS

-75 kN*m

-0.0000 %

Sc CQ

8.99322 MPa

0.0359 %

Ss CQ

-219.639 MPa

0.0006 %

ADVANCE DESIGN VALIDATION GUIDE

5.39 EC2 Test28: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class X0) Test ID: 5083 Test status: Passed

5.39.1 Description Verifies a T cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max. will be calculated, along with the cross-sectional area of the shear reinforcement, Asw, calculation. The test will not use the reduced shear force value.

5.39.2 Background Verifies a T cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the cross-sectional area of the shear reinforcement, Asw, calculation. The test will not use the reduced shear force value.

5.39.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Concrete: C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: The dead load will be represented by a linear load of 13.83kN/m

■

Exploitation loadings: The live load will be considered from one linear load of 26.60kN/m

■ ■ ■ ■

Structural class: S1 Reinforcement steel ductility: Class B Exposure class: X0 Concrete density: 25kN/m3

427

ADVANCE DESIGN VALIDATION GUIDE

■

The reinforcement will be displayed like in the picture below:

The objective is to: ■ ■ ■ ■

Verify the shear stresses results Verify the transverse reinforcement Verify the transverse reinforcement distribution by the Caqout method Identify the steel sewing

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■

Length: L = 10.00 m, Concrete cover: c=3.5cm Effective height: d=h-(0.06*h+ebz)=0.764m; d’=ehz=0.035m

■ ■

Stirrup slope: α= 90° Strut slope: θ=30˚

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 end point (x = 10.00) restrained in translation along Y, Z and rotation along X Inner: None.

► ►

■

428

ADVANCE DESIGN VALIDATION GUIDE

Loading:

For a beam under a uniformly distributed load Pu, the shear force is defined by the following equation:

V ( x) = Pu * x −

Pu * l 2

For the beam end, (x=0) the shear force will be:

VEd = −

Pu * l 58,57 *10 =− = −292,9kN 2 2

In the following calculations, the negative sign of the shear will be neglected, as this has no effect in the calculations.

Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to an unreduced shear, that will be used for further calculations.

5.39.2.2 Reference results in calculating the maximum design shear resistance

VRd , max =

v1 * f cd * zu * bw tgθ + cot θ According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 6.2.3.(3)

In this case:

θ = 30°

and

α = 90°

v1 strength reduction factor for concrete cracked in shear f ⎤ 25 ⎤ ⎡ ⎡ v1 = 0,6 * ⎢1 − ck ⎥ = 0.6 * ⎢1 − = 0.54 ⎣ 250 ⎦ ⎣ 250 ⎥⎦

zu = 0.9 * d = 0.9 * 0.764 = 0.688m VRd , max =

0.54 * 16 .67 * 0.688 * 0.22 = 0.590 MN tg 30 + cot 30

VEd = 0.293MN < VRd , max = 0.590 MN Calculation of transverse reinforcement:

The transverse reinforcement is calculated using the following formula:

Asw VEd . * tgθ 0.293 * tg 30° = = = 5.66cm² / ml 500 s zu * f ywd 0.688 * 1,15 Finite elements modeling

■ ■ ■

Linear element: S beam, 11 nodes, 1 linear element. 429

ADVANCE DESIGN VALIDATION GUIDE

Advance Design gives the following results for Atz (cm2/ml)

5.39.2.3 Reference results Result name

Result description

Reference value

Fz corresponding to the 101 combination (ULS) [kNm]

292.9 kN

At,z

Theoretical reinforcement area [cm2/ml]

5.66 cm2/ml

5.39.3 Calculated results

Result name Fz

Atz

430

Result description

Value

Error

-292.852 kN

-0.0002 %

Atz

5.65562 cmÂ˛

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.40 EC2 Test32: Verifying a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature- Bilinear stress-strain diagram (Class XC1) Test ID: 5102 Test status: Passed

5.40.1 Description Verifies a square concrete column subjected to compression and rotation moment to the top – Method based on nominal curvature- Bilinear stress-strain diagram (Class XC1). The column made of concrete C25/30. The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed. The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced.

5.40.2 Background Method based on nominal curvature. Verifies the adequacy of a square cross section made from concrete C25/30.

5.40.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: 15t axial force ► 1.5tm rotation moment applied to the column top ► The self-weight is neglected Exploitation loadings: ► 6.5t axial force ► 0.7tm rotation moment applied to the column top ►

■

► ► ► ► ► ► ► ►

ψ 2 = 0,3 The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q Concrete cover 5cm Transversal reinforcement spacing a=30cm Concrete C25/30 Steel reinforcement S500B The column is considered isolated and braced

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ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.40 m, Width: b = 0.40 m, Length: L = 6.00 m, Concrete cover: c = 5 cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free at the top. Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*15+1.5*6.5=30t=0.300MN MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm

■

e0 =

MEd 0.031 = = 0.10m NEd 0.300

5.40.2.2 Reference results in calculating the concrete column Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

l0 = 2 * l = 12m According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:

λ=

2 3 * l0 2 3 * 12 = = 104 a 0.40

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

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ADVANCE DESIGN VALIDATION GUIDE

Where:

ϕ (∞, t0 )

creep coefficient

M EQP

serviceability firs order moment under quasi-permanent load combination

M Ed

ULS first order moment (including the geometric imperfections)

First order eccentricity evaluation:

e1 = e0 + ei ei = 0.03m The first order moment provided by the quasi-permanent loads:

e1 = e0 + ei =

M Eqp 0 N Eqp 0

+ ei =

1.50 + 0.30 * 0.70 + 0.30 = 0.13m 15 + 0.30 * 0.65

N Eqp1 = 1.50 + 0.30 * 0.70 = 16.95t

M Eqp1 = N Eqp1 * e1 = 16.95 * 0.13 = 2.20tm = 0.022MNm The first order ULS moment is defined latter in this example:

The creep coefficient

ϕ (∞, t0 )

is defined as follows:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )

β ( f cm ) = β (t0 ) =

16.8 16.8 = = 2.92 f cm 25 + 8

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

(for t0= 28 days concrete age).

RH 100 =1+ 0.1 * 3 h0 1−

ϕ RH

h0 =

2 * Ac 2 * 400 * 400 = = 200 mm ⇒ ϕ RH u 2 * (400 + 400 )

50 100 = 1.85 =1+ 0.1 * 3 200 1−

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.85 * 2.92 * 0.488 = 2.64 The effective creep coefficient calculation:

ϕef = ϕ (∞, t0 ) *

M EQP M Ed

= 2.64 *

0.022 = 1.49 0.039 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

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ADVANCE DESIGN VALIDATION GUIDE

The necessity of buckling calculation (second order effect):

For an isolated column, the slenderness limit verification is done using the next formula:

λlim =

20 * A * B * C n According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

Where:

N Ed 0.300 = = 0.112 Ac * f cd 0.40² * 16.67

1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.49 = 0.77

B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70

λlim =

because the reinforcement ratio in not yet known

because the ratio of the first order moment is unknown

20 * 0.77 *1.1 * 0.7 = 35 .43 0.112

λ = 104 > λlim = 35.43 Therefore, the second order effects most be considered. The second order effects; The buckling calculation:

The stresses for the ULS load combination are: NEd= 1.35*15 + 1.50*6.50= 30 t = 0,300MN MEd= 1.35*1.50 + 1.50*0.7= 3.075 t = 0,031MNm Therefore, it must be determined: ■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:

e0 =

M Ed 0.031 = = 0.10 m N Ed 0.300

Additional eccentricity:

ei =

434

l0 12 = = 0.03m 400 400

ADVANCE DESIGN VALIDATION GUIDE

The first order eccentricity: stresses correction:

The forces correction, used for the combined flexural calculations:

N Ed = 0.300MN e1 = e0 + ei = 0.13m M Ed = e1 * N Ed = (0.10 + 0.03) * 0.300 = 0.039MNm M = N Ed * e0 ⎧⎪20mm ⎧⎪ 20mm ⎧ 20mm e0 = max ⎨ h = max ⎨ 400mm = max ⎨ = 20mm ⎩13.3mm ⎪⎩ 30 ⎪⎩ 30 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4) Reinforcement calculation in the first order situation:

To play the nominal rigidity method a starting section frame is needed. For this it will be used a concrete section considering only the first order effects. Advance Design iterates as many time as necessary. The needed frames will be determined assuming a compound bending with compressive stress. All the results above were obtained in the center of gravity for the concrete section alone. The stresses must be reduced in the centroid of the tensioned steel.

h 0.40 ⎞ ⎛ M ua = M G 0 + N * ( d − ) = 0.039 + 0.300 * ⎜ 0.35 − ⎟ = 0.084 MNm 2 ⎠ 2 ⎝ Verification if the section is partially compressed:

μ BC = 0,8 *

μ cu =

h h 0,40 0,40 * (1 − 0,4 * ) = 0,8 * * (1 − 0,4 * ) = 0,496 d d 0,35 0,35

M ua 0.084 = = 0.103 bw * d ² * f cd 0.40 * 0.35² * 16.67

μcu = 0.103 < 0.496 = μBC

therefore the section is partially compressed.

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ADVANCE DESIGN VALIDATION GUIDE

Calculations of steel reinforcement in pure bending:

Calculations of steel reinforcement in combined bending:

For the combined bending:

A = A'−

N 0.300 = 5.84 *10− 4 − = −1.06cm2 f yd 434.78

The minimum reinforcement percentage:

As , min =

0,10 * N Ed 0.10 * 0.300 = = 0.69cm ² ≥ 0.002 * Ac = 3.02cm 2 434 .78 f yd

The reinforcement will be 4HA10 representing a 3.14cm2 section The second order effects calculation:

The second order effect will be determined by applying the method of nominal curvature: Calculation of nominal curvature:

Considering a symmetrical reinforcement 4HA10 (3.14cm ²), the curvature can be determined by the following formula:

1 1 = K r * Kϕ * r r0 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.8.3 (1) With:

f yd

434.78 1 Es = = = 200000 = 0.0138m −1 r0 0,45 * d 0,45 * d 0.45 * 0.35

ε yd

: is a correction factor depending on axial load,=>

ω=

N Ed 0.300 = = 0.112 Ac * f cd 0.40² * 16.67

As * f yd Ac * f cd

3,14.10−4 * 434.78 = 0.0512 0.40² *16.67

nu = 1 + ω = 1 + 0.0512 = 1.0512 nbal = 0,4

436

Kr =

nu − n ≤1 nu − nbal

ADVANCE DESIGN VALIDATION GUIDE

Kr = Kϕ

1.0512 − 0.112 = 1.44 ≤ 1 ⇒ K r = 1 1.0512 − 0.40

: is a factor for taking account of creep=>

β = 0,35 +

K ϕ = 1 + β * ϕ ef ≥ 1

25 104 fck λ − = 0.35 + − = −0.218 200 150 200 150

K ϕ = 1 + β * ϕ ef = 1 − 0.218 * 1.49 = 0.675 ≥ 1 ⇒ K ϕ = 1 Therefore the curvature becomes:

1 1 = K r * Kϕ * = 0.0138 m −1 r r0 Calculation moment:

M Ed = M 0 Ed + M 2 Where:

M 0 Ed

: first order moment including the geometrical imperfections.

: second order moment

The second order moment must be calculated from the curvature:

M 2 = N Ed * e2 e2 =

1 l02 12 ² = 0.199 m * = 0.0138 * r c 10

M 2 = N Ed * e2 = 0.300 * 0.199 = 0.0597MNm M Ed = M 0 Ed + M 2 = 0.039 + 0.0597 = 0.099MNm The frame must be sized corresponding the demands of the second order effects:

N Ed = 0.300MN M Ed = 0.099 MNm

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ADVANCE DESIGN VALIDATION GUIDE

Reinforcement calculation corresponding to the second order:

The input parameters in the diagram are:

μ=

M Ed 0.099 = = 0.093 2 b * h * f cd 0.40 * 0.40² *16.67

N Ed 0.300 = = 0.112 b * h * f cd 0.40 * 0.40 * 16.67

ω = 0 . 15 ; =>

∑A

ω * b * h * f cd 0.15 * 0.40 ² * 16 .67 = = 9.20cm ² ; which means 4.60cm2 per side. f yd 434 .78

Buckling checking

The verification will be made considering the reinforcement found previously (9.20cm2) The curvature evaluation:

The frames have an influence on the Kr parameter only:

1 = 0.0138 m −1 r0 Kr =

nu − n ≤1 nu − nbal n = 0.112

ω=

As * f yd Ac * f cd

9,20 *10−4 * 434.78 = 0.15 0.40² *16.67

nu = 1 + ω = 1 + 0.15 = 1.15

nbal = 0,4 Kr =

438

1.15 − 0.112 = 1.384 ≤ 1 ⇒ K r = 1 1.15 − 0.40

ADVANCE DESIGN VALIDATION GUIDE

Kϕ = 1 : coefficient that takes account of the creep => K ϕ = 1 + β * ϕ ef ≥ 1 The curvature becomes:

1 1 = K r * Kϕ * = 0.0138 m −1 r r0 It was obtained the same curvature and therefore the same second order moment, which validates the section reinforcement found. Finite elements modeling

■ ■ ■

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

Theoretical reinforcement area (cm2)

(reference value: 9.20cm2)

Theoretical value (cm2)

(reference value: 9.28 cm2)

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ADVANCE DESIGN VALIDATION GUIDE

5.40.2.3 Reference results Result name

Result description

Reference value

Reinforcement area [cm2]

4.64 cm2

Theoretical reinforcement area [cm2]

9.28 cm2

5.40.3 Calculated results

Result name Az

440

Result description

Value

-4.64 cmÂ˛

Error

-0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.41 EC2 Test29: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Inclined stress-strain diagram (Class XC1) Test ID: 5092 Test status: Passed

5.41.1 Description Verifies a T cross section beam made from concrete C35/40 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max., will be determined, along with the cross-sectional area of the shear reinforcement, Asw, calculation. The test will not use the reduced shear force value.

The beam model was provided by Bouygues.

5.41.2 Background Verifies a T cross section beam made from concrete C35/40 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be determined, along with the cross-sectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value. The beam model was provided by Bouygues.

5.41.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Concrete: C35/40 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: The dead load will be represented by two linear loads of 22.63kN/m and 47.38kN/m

■

Exploitation loadings: The live load will be considered from one linear load of 80.00kN/m

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ADVANCE DESIGN VALIDATION GUIDE

■ ■ ■

Structural class: S3 Reinforcement steel ductility: Class B Exposure class: XC1

■

Characteristic compressive cylinder strength of concrete at 28 days:

■ ■ ■ ■ ■

Partial factor for concrete: Relative humidity: RH=50% Concrete age: t0=28days Design value of concrete compressive strength: fcd=23 MPa Secant modulus of elasticity of concrete: Ecm=34000 MPa

■ ■

Concrete density: Mean value of axial tensile strength of concrete: fctm=3.2 MPa

■

Final value of creep coefficient:

■ ■ ■

Characteristic yield strength of reinforcement: Steel ductility: Class B K coefficient: k=1.08

■ ■

Design yield strength of reinforcement: Design value of modulus of elasticity of reinforcing steel: Es=200000MPa

■

Steel density:

■

Characteristic strain of reinforcement or prestressing steel at maximum load:

■

Strain of reinforcement or prestressing steel at maximum load:

■

Slenderness ratio:

λ = 0.80

The objective is to verify: ■ ■ ■

The of shear stresses results The longitudinal reinforcement corresponding to a 5cm concrete cover The transverse reinforcement corresponding to a 2.7cm concrete cover

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■

Length: L = 10.00 m,

■ ■

Stirrup slope: α= 90° Strut slope: θ=29.74˚

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 end point (x = 10.00) restrained in translation along Y, Z and rotation along X Inner: None.

► ►

■

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ADVANCE DESIGN VALIDATION GUIDE

Loading:

Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to an unreduced shear, that will be used for further calculations.

5.41.2.2 Reference results in calculating the longitudinal reinforcement For a 5 cm concrete cover and dinit=1.125m, the reference value will be 108.3 cm2:

For a 2.7 cm concrete cover and dinit=1.148m, the reference value will be 105.7 cm2:

5.41.2.3 Reference results in calculating the transversal reinforcement

Asw VEd . * tgθ = s zu * f ywd Where:

zu = 0.9 * d = 0.9 *1.125 = 1.0125m cot θ = 1.75 ⇒ θ = 29.74°

α = 90° Asw VEd . * tgθ 1.502 * tg 29.74° = = = 19.50cm² / ml 500 s zu * f ywd 1.0125 * 1,15

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ADVANCE DESIGN VALIDATION GUIDE

The minimum reinforcement percentage calculation:

0,08 * f ck 0,08 * 35 Asw ≥ ρ w, min * bw * sin α = * bw * sin α = * 0.55 * sin 90° = 5.21cm² / ml f yk 500 s Finite elements modeling

■ Linear element: S beam, ■ 15 nodes, ■ 1 linear element. Advance Design gives the following results for Atz (cm2/ml)

5.41.2.4 Reference results Result name

Result description

Reference value

My corresponding to the 101 combination (ULS) [kNm]

5255.58 kNm

Fz corresponding to the 101 combination (ULS) [kNm]

1501.59 kN 2

Az(5cm cover)

Az longitudinal reinforcement corresp. to a 5cm cover [cm /ml]

108.30 cm2/ml

Az(2.7cm cover)

Az longitudinal reinforcement corresp. to a 2.7cm cover [cm2/ml]

105.69 cm2/ml

At,z,1

At,z transversal reinforcement for the beam end [cm /ml]

19.10 cm2/ml

At,z,2

At,z transversal reinforcement for the middle of the beam [cm2/ml]

5.21 cm2/ml

5.41.3 Calculated results

Result name My

444

Result description

Value

-5255.58 kN*m

Error

-0.0000 %

-1501.59 kN

-0.0003 %

Az 5cm

-105.694 cm²

2.4071 %

Az 2.7cm

-105.694 cm²

-0.0000 %

Atz

Atz end

19.0973 cm²

0.0000 %

Atz

Atz middle

5.20615 cm²

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.42 EC2 Test33: Verifying a square concrete column subjected to compression by nominal rigidity method- Bilinear stress-strain diagram (Class XC1) Test ID: 5109 Test status: Passed

5.42.1 Description Verifies a square concrete column subjected to compression by nominal rigidity method- Bilinear stress-strain diagram (Class XC1). The column is made of concrete C30/37. The verification of the axial force, applied on top, at ultimate limit state is performed. Nominal rigidity method - The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part, the translations along X and Y axis are also blocked. This test is based on the example from "Applications of Eurocode 2" (J. & JA Calgaro Cortade).

5.42.2 Background Nominal rigidity method. Verify the adequacy of a square cross section made from concrete C30/37.

5.42.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: 1260kN axial force ► The self-weight is neglected Concrete cover 5cm Concrete C30/37 Steel reinforcement S500B Concrete age 28 days Relative humidity 50% Buckling length: L0=0.7*4.47=3.32mm ►

■ ■ ■ ■ ■ ■

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ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.30 m, Width: b = 0.30 m, Length: L = 4.74 m, Concrete cover: c=5cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part, the translations along X and Y axis are also blocked. Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: NEd =1260kN

■

Proposed reinforcement: 2*6.28cm2=12.56cm2

5.42.2.2 Reference results in calculating the concrete column Load calculation:

NEd= 1.35*NG = 1700 KN Initial eccentricity:

e0 =

Mu 0 = = 0cm M u 1 .7

Additional eccentricity due to geometric imperfections:

L ⎞ 322 ⎞ ⎛ ⎛ ei = max ⎜ 2cm; 0 ⎟ = max ⎜ 2cm; ⎟ = max (2cm;0.805cm ) = 2cm 400 ⎠ 400 ⎠ ⎝ ⎝ According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.2(7) The first order eccentricity :

e1 = e0 + ei = 0.02 m

The necessity of buckling calculation (second order effect):

For an isolated column, the slenderness limit check is done using the next formula:

λlim =

20 * A * B * C n According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

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ADVANCE DESIGN VALIDATION GUIDE

Where:

N Ed 1 .7 = = 0.944 Ac * f cd 0.30² * 20

ϕ ef

1 1 + 0,2 * ϕ ef

A = 0.7

B = 1.1

if ω (reinforcement ratio) is not known, if it is,

In this case B

C = 0.70

is not known, if it is,

= 1+ 2*

B = 1 + 2 *ω = 1 + 2 *

As * f yd Ac * f cd

12.56 *10−4 * 434.78 = 1.27 0.3² * 20

if rm is not known, if it is,

C = 1,7 − rm

In this case:

20 * A * B * C 20 * 0.7 * 1.27 * 0.7 = = 12 .81 n 0.944

λlim =

λ=

L0 * 12 3.32 * 12 = = 38.34 h 0 .3

λ = 38.34 > λlim = 12.81 Therefore, the second order effects most be considered Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

ϕ (∞,t0 ) M EQP

M Ed

creep coefficient

serviceability firs order moment under quasi-permanent load combination ULS first order moment (including the geometric imperfections)

The ratio of the moment is in this case:

M 0 Eqp M 0 Ed

N eqp * e1 N ed *e1

N eqp N ed

1.26 = 0.74 1.7

ϕ (∞, t0 ) is the final value of the creep:

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ADVANCE DESIGN VALIDATION GUIDE

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) With:

t0 is the concrete age

RH ⎞ ⎛ 1− ⎟ ⎜ 100 * α ⎟ * α = ⎜1 + 1 2 ⎟ ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎠ ⎝

ϕ RH

α1 = α 2 = 1 if f cm

⎛ 35 ⎞ ⎟⎟ α1 = ⎜⎜ f ≤ 35 MPa ⎝ cm ⎠ if not

0.7

⎛ 35 ⎞ ⎟⎟ α 2 = ⎜⎜ f ⎝ cm ⎠ and

0.2

RH relative humidity: RH=50% h0= mean radius of the element in mm

h0 =

2 * Ac 2 * 300 * 300 = = 150 mm u 2 * (300 + 300 )

u=column section perimeter

f cm

ϕ RH

⎛ 35 ⎞ α1 = ⎜ ⎟ = 38 MPa ≥ 35 MPa ⇒ ⎝ 38 ⎠

0.7

⎛ 35 ⎞ = 0.944 α2 = ⎜ ⎟ ⎝ 38 ⎠ and

0.2

= 0.984

RH 50 ⎞ ⎛ ⎞ ⎛ 1− 1− ⎟ ⎜ ⎟ ⎜ 100 * 0.944 ⎟ * 0.984 = 1.86 100 * α ⎟ * α = ⎜1 + = ⎜1 + 1 2 ⎟ ⎜ 0.1 * 3 150 ⎟ ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎜ ⎠ ⎝ ⎠ ⎝

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.86 * 2.73 * 0.488 = 2.48 ϕef = ϕ (∞, t0 ) *

M 0 Eqp M 0 Ed

= 2.48 * 0.74 = 1.835

1 + ϕ ef = 1 + 1.835 = 2.835 Calculation of nominal rigidity:

The nominal rigidity of a post or frame member, it is estimated from the following formula:

EI = K c * Ecd * I c + K s * Es * I s According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.2 (1)

Kc =

k1 * k 2 1 + ϕe

1 + ϕ ef = 2.835

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ADVANCE DESIGN VALIDATION GUIDE

f ck = 20

k1 =

30 = 1.22 20

n*λ 170

k2 =

N Ed 1 .7 = = 0.944 Ac * f cd 0.3² * 20

λ = 38 .34

k2 =

n * λ 0.944 * 38.34 = = 0.213 ⇒ k2 = 0.20 170 170

Kc =

k1k 2 1.22 × 0.20 = = 0.086 1 + ϕe 2.835

E cd =

Ecm

E cm

γ CE

⎛ f +8⎞ = 22 * ⎜ ck ⎟ ⎝ 10 ⎠ E cd = Ic =

E cm

γ CE

0.3

⎛ 25 + 8 ⎞ = 22 * ⎜ ⎟ ⎝ 10 ⎠

0.3

= 32837 MPa

32837 = 27364 MPa 1 .2

b * h 3 0.3 * 0.33 = = 0.000675 m 4 12 12

Ks=1 Es = 200 Gpa 2

Is = 2*

Atheo ⎛ d − d ' ⎞ 12.56 *10−4 ⎛ 0.25 − 0.05 ⎞ −5 4 *⎜ = 2 * *⎜ ⎟ ⎟ = 1.256 *10 m 2 ⎝ 2 ⎠ 2 2 ⎝ ⎠

EI = 0.086 * 27364 * 0.000675 + 1 * 200000 * 1.256 * 10 −5 = 4.1MNm ²

NB =

β=

M Ed

π ² * EI

π² c0

Lf ² =

π² 8

π ² * 4.10 3.32²

= 3.67 MN

= 1.234

⎤ ⎡ ⎤ ⎡ ⎢ ⎢ β ⎥ 1.234 ⎥ ⎥ = 0.034 * ⎢1 + = 0.070 MNm = M 0 Ed * ⎢1 + NB 3.67 ⎥ ⎥ ⎢ ⎥ ⎢ −1 −1 ⎥⎦ ⎢⎣ N Ed 1 .7 ⎦ ⎣

The calculation made with flexural:

N Ed = 1.7MN and M Ed = 0.070MNm , Therefore, a 2*6.64cm2 reinforcement area is obtained.

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ADVANCE DESIGN VALIDATION GUIDE

The calculated reinforcement is very close to the initial assumption (2*6.28cm2) It is not necessary to continue the calculations; it retains a section of 2*6.64cm2 It sets up 4HA20 (2*6.28=12.57cm2) Finite elements modeling

■ ■ ■

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

Theoretical reinforcement area(cm2)

(reference value: 6.54cm2)

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ADVANCE DESIGN VALIDATION GUIDE

Theoretical value (cm2)

(reference value: 13.08 cm2)

5.42.2.3 Reference results Result name

Result description

Reference value 2

6.54 cm2

Reinforcement area [cm ] 2

13.08 cm2

Theoretical reinforcement area [cm ]

5.42.3 Calculated results

Result name Az

Result description

Value

-6.53808 cmÂ˛

Error

0.0001 %

451

ADVANCE DESIGN VALIDATION GUIDE

5.43 EC2 Test 27: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement - Bilinear stress-strain diagram (Class XC1) Test ID: 5076 Test status: Passed

5.43.1 Description Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram is generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max., will be calculated, along with the cross-sectional area of the shear reinforcement, Asw, and the theoretical reinforcement. For the calculation, the reduced shear force values will be used.

5.43.2 Background Description: Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram is generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max. will be calculated, along with the cross-sectional area of the shear reinforcement, Asw, and the theoretical reinforcement.

5.43.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Concrete C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

452

■

Loadings from the structure: The dead load will be represented by a linear load of 40kN/m

■

Exploitation loadings: The live load will be considered from one linear load of 25kN

■ ■ ■

Structural class: S1 Reinforcement steel ductility: Class B The reinforcement will be displayed like in the picture below:

ADVANCE DESIGN VALIDATION GUIDE

The objective is to verify: ■ ■

The shear stresses results The theoretical reinforcement value

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■ ■

Height: h = 0.70 m, Width: b = 0.35 m, Length: L = 5.75 m, Concrete cover: c=3.5cm Effective height: d=h-(0.06*h+ebz)=0.623m; d’=ehz=0.035m

■ ■

Stirrup slope: α= 90° Strut slope: θ=45˚

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 end point (x = 5.30) restrained in translation along Y, Z and rotation along X Inner: None. ►

■

Loading:

For a beam subjected to a point load, Pu, the shear stresses are defined by the formula below:

VEd =

Pu * l 2 According to EC2 the Pu point load is defined by the next formula: Pu = 1,35*Pg + 1,50*Pq=1.35*40kN+1.50*25kN=91.5kN

In this case :

VEd =

5.75 * 91.5 = 263 KN 2

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ADVANCE DESIGN VALIDATION GUIDE

5.43.2.2 Reference results in calculating the lever arm zc: The lever arm is calculated using the design formula for pure bending:

Pu = 91.5kN / ml M Ed =

μ cu =

91.5 * 5.75² = 378.15kNm 8

M Ed 0.378 = = 0.167 bw * d ² * f cd 0.35 * 0.623² * 16.67

(

)

(

)

αu = 1.25 * 1 − 1 − 2 * μcu = 1.25 * 1 − 1 − 2 * 0.167 = 0.230 zc = d * (1 − 0.4 * α u ) = 0.623 * (1 − 0.4 * 0.230) = 0.566m Calculation of reduced shear force

When the shear force curve has no discontinuities, the EC2 allows to consider, as a reduction, the shear force to a horizontal axis x = z. cot θ . In the case of a member subjected to a distributed load, the equation of the shear force is:

V ( x) = Pu * x −

Pu * l 2

Therefore:

x = z * cot θ = 0.566 * cot 45° = 0.566m VEd , red = 91.5 * 0.566 − 263 = −211kN Warning: Advance Design does not apply the reduction of shear force corresponding to the distributed loads (cutting edge at x = d).

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ADVANCE DESIGN VALIDATION GUIDE

Calculation of maximum design shear resistance:

VRd, max = α cw * ν1 * fcd * z u * b w *

(cotα + cotθ) 1 + cot 2θ According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 6.2.3.(3)

Where:

α cw = 1 coefficient taking account of the state of the stress in the compression chord and v1 = 0,6 * ⎡⎢1 − ⎣

f ck ⎤ 250 ⎥⎦

When the transverse frames are vertical, the above formula simplifies to:

VRd , max =

v1 * f cd * zu * bw tgθ + cot θ

In this case:

θ = 45°

and

α = 90°

v1 strength reduction factor for concrete cracked in shear 25 ⎤ f ⎤ ⎡ ⎡ v1 = 0,6 * ⎢1 − ck ⎥ = 0.6 * ⎢1 − = 0.54 ⎣ 250 ⎦ ⎣ 250 ⎥⎦

zu = 0.9 * d = 0.9 * 0.623 = 0.56m VRd , max =

0.54 *16.67 * 0.566 * 0.35 = 0.891MN = 891kN 2

VEd = 236 kN < VRd , max = 891kN Calculation of transversal reinforcement:

Given the vertical transversal reinforcement (α = 90°), the transverse reinforcement is calculated using the following formula:

Asw VEd . * tgθ 0,211 * tg 45 = = = 8,67cm² / ml 500 s zu * f ywd 0,566 * 1,15 Calculation of theoretical reinforcement value:

The French national annex indicates the formula:

Asw ≥ ρ w, min * bw * sin α s With:

ρ w, min =

0,08 * f ck 0,08 * 20 = = 7.15 × 10− 4 500 f yk

Asw ≥ ρ w, min * bw * sin α = 7.15 *10− 4 * 0.2 * sin 90° = 1.43cm² / ml s

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Finite elements modeling

■ Linear element: S beam, ■ 3 nodes, ■ 1 linear element. Advance Design gives the following results for At,min,z (cm2/ml)

5.43.2.3 Reference results Result name

Result description

Reference value

Fz corresponding to the 101 combination (ULS) [kNm]

263 kN

At,min,z

Theoretical reinforcement area [cm2/ml]

1.43 cm2/ml

5.43.3 Calculated results

Result name Fz

Atminz

456

Result description

Value

Error

-168.937 kN

-0.0003 %

Atmin,z

1.43108 cm²

0.0002 %

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5.44 EC2 Test31: Verifying a square concrete column subjected to compression and rotation moment to the top - Bilinear stress-strain diagram (Class XC1) Test ID: 5101 Test status: Passed

5.44.1 Description Nominal rigidity method. Verifies the adequacy of a rectangular cross section column made from concrete C25/30. The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced. Verifies the column to resist simple bending. The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed. The column is considered connected to the ground by a fixed connection and free to the top part.

5.44.2 Background Nominal rigidity method. Verifies the adequacy of a rectangular cross section made from concrete C25/30.

5.44.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

■

ψ 2 = 0,3

■ ■ ■ ■ ■ ■ ■

The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q Concrete cover 5cm Transversal reinforcement spacing a=30cm Concrete C25/30 Steel reinforcement S500B The column is considered isolated and braced

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Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.40 m, Width: b = 0.40 m, Length: L = 6.00 m, Concrete cover: c=5cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*15+150*6.5=30t=0.300MN MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm

■

e0 =

MEd 0.031 = = 0.10m NEd 0.300

5.44.2.2 Reference results in calculating the concrete column Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

l0 = 2 * l = 12m According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:

λ=

2 3 * l0 2 3 * 12 = = 104 a 0.40

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

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Where:

ϕ (∞,t0 )

creep coefficient

serviceability first order moment under quasi-permanent load combination ULS first order moment (including the geometric imperfections) First order eccentricity evaluation:

e1 = e0 + ei ei = 0.03m The first order moment provided by the quasi-permanent loads:

e1 = e0 + ei =

M Eqp 0 1.50 + 0.30 * 0.70 + ei = + 0.30 = 0.13m N Eqp 0 15 + 0.30 * 0.65

N Eqp1 = 1.50 + 0.30 * 0.70 = 16.95t M Eqp1 = N Eqp1 * e1 = 16.95 * 0.13 = 2.20tm = 0.022MNm The first order ULS moment is defined latter in this example:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )

β ( f cm ) = β (t0 ) =

ϕ RH

h0 =

16.8 16.8 = = 2.92 f cm 25 + 8

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

(for t0= 28 days concrete age).

RH 100 =1+ 0.1 * 3 h0 1−

2 * Ac 2 * 400 * 400 = = 200 mm ⇒ ϕ RH u 2 * (400 + 400 )

50 100 = 1.85 =1+ 3 0.1 * 200 1−

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.85 * 2.92 * 0.488 = 2.64 The effective creep coefficient calculation:

ϕef = ϕ (∞, t0 ) *

M EQP M Ed

= 2.64 *

0.022 = 1.49 0.039 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

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The second order effects; The buckling calculation:

For an isolated column, the slenderness limit check is done using the next formula:

λlim =

20 * A * B * C n According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

Where:

N Ed 0.300 = = 0.112 Ac * f cd 0.40² * 16.67

1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.49 = 0.77

B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70

λlim =

because the reinforcement ratio in not yet known

because the ratio of the first order moment is not known

20 * 0.77 *1.1 * 0.7 = 35 .43 0.112

λ = 104 > λlim = 35.43 Therefore, the second order effects most be considered. The second order effects; The buckling calculation:

The stresses for the ULS load combination are: NEd= 1.35*15 + 1.50*6.50= 30 t = 0,300MN MEd= 1.35*1.50 + 1.50*0.7= 3.075 t = 0,031MNm Therefore it must be determined: ■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:

e0 =

M Ed 0.031 = = 0.10 m N Ed 0.300

Additional eccentricity:

ei =

460

l0 12 = = 0.03m 400 400

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The first order eccentricity: stresses correction:

The forces correction, used for the combined flexural calculations:

N Ed = 0.300MN e1 = e0 + ei = 0.13m M Ed = e1 * N Ed = (0.10 + 0.03) * 0.300 = 0.039MNm M = N Ed * e0

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4) Reinforcement calculation in the first order situation:

h 0.40 ⎞ ⎛ M ua = M G 0 + N * (d − ) = 0.039 + 0.300 * ⎜ 0.35 − ⎟ = 0.084 MNm 2 2 ⎠ ⎝ Verification if the section is partially compressed:

μ BC = 0,8 *

μ cu =

h h 0,40 0,40 * (1 − 0,4 * ) = 0,8 * * (1 − 0,4 * ) = 0,496 d d 0,35 0,35

M ua 0.084 = = 0.103 bw * d ² * f cd 0.40 * 0.35² * 16.67

μcu = 0.103 < 0.496 = μBC

therefore the section is partially compressed.

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Calculations of steel reinforcement in pure bending:

μcu = 0.103

[

]

α u = 1,25 * 1 − (1 − 2 * 0,103) = 0,136 zc = d * (1 − 0,4 * α u ) = 0,35 * (1 − 0,4 * 0,136) = 0,331m

Calculations of steel reinforcement in combined bending:

For the combined bending:

A = A'−

N 0.300 = 5.84 *10− 4 − = −1.06cm2 f yd 434.78

The minimum reinforcement percentage:

As , min =

0,10 * N Ed 0.10 * 0.300 = = 0.69 cm ² ≥ 0.002 * Ac = 3.02 cm 2 f yd 434 .78

The reinforcement will be 4HA10 representing a 3.14cm2 section The second order effects calculation:

The second order effect will be determined by applying the method of nominal rigidity: Calculation of nominal rigidity:

The nominal rigidity of a post or frame member, it is estimated from the following formula:

EI = K c * Ecd * I c + K s * Es * I s According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.2 (1) With:

Ecd =

Ecm 1.2

f cm = f ck + 8Mpa = 33Mpa Ecm

⎛f ⎞ = 22000 * ⎜ cm ⎟ ⎝ 10 ⎠

Ecd = Ic =

0.3

⎛ 33 ⎞ = 22000 * ⎜ ⎟ ⎝ 10 ⎠

= 31476Mpa

Ecm 31476 = = 26230Mpa 1.2 1.2

b * h 3 0.40 4 = = 2,133 .10 −3 m 4 12 12

Es = 200000Mpa I s : Inertia

3,14.10−4 As ρ= = = 0.002 Ac 0.40 × 0.40 462

0.3

(cross section inertia)

ADVANCE DESIGN VALIDATION GUIDE

0.002 ≤ ρ =

f ck = 20

k1 = n=

As < 0.01 Ac

25 = 1.12 Mpa 20

0.300 N Ed = = 0.112 Ac * f cd 0.40² *16.67

k2 = n *

λ 170

= 0.112 *

104 = 0.069 ≤ 0.20 170

A ⎛h 3,14 *10−4 ⎛ 0.40 ⎞ ⎞ Is = 2 * s * ⎜ − c ⎟ = 2 * *⎜ − 0.05 ⎟ = 7,06 *10− 6 m4 2 ⎝2 2 2 ⎠ ⎝ ⎠

Ks = 1 and

Kc =

k1 * k 2 1.12 * 0.069 = = 0.031 1 + ϕ ef 1 + 1.49

Therefore:

EI = 0.031* 26230 * 2,133 *10−3 + 1 * 200000 * 7,06 *10−6 = 3.15MNm² Corrected stresses:

The total moment, including the second order effects is defined as a value and is added to the first order moment value:

M Ed

⎤ ⎡ ⎢ β ⎥ ⎥ = M 0 Ed * ⎢1 + NB ⎢ − 1⎥ ⎥⎦ ⎢⎣ N Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.3 (1)

M 0 Ed = 0.039MNm

(time of first order (ULS) taking into account the geometric imperfections, relative to

the center of gravity of concrete).

N Ed = 0.300MN (normal force acting at ULS). And:

β=

π² c0

with

c0 = 8

because the moment is constant (no horizontal force at the top of post).

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.3 (2)

β=

π² 8

= 1.234

NB = π ² *

EI 3.15 =π²* = 0.216 MN 2 l0 12²

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ADVANCE DESIGN VALIDATION GUIDE

Therefore the second order moment is:

M Ed

⎡ ⎤ ⎢ 1.234 ⎥ = −0.133MNm = 0.039 * ⎢1 + 0.216 ⎥ ⎢ − 1⎥ ⎣ 0.300 ⎦

There is a second order moment that is negative because it was critical that the normal force, NB, is less than the applied normal force => instability. A section corresponding to a ratio of 5 ‰ will be considered and the corresponding equivalent stiffness is recalculated.

EI = K c * Ecd * I c + K s * Es * I s With:

Ecd = 26230Mpa ; I c = 2,133 .10 −3 m 4 Es = 200000Mpa Is

: Inertia

ρ = 0,005 ⇒ As = 0,005 * Ac = 0,005 * 0,40² = 8cm² It sets up : 4HA16 =>

0.002 ≤ ρ =

ρ = 0,005 ⇒ As = 8,04cm²

As < 0.01 Ac 2

A ⎛h 8,04 *10−4 ⎛ 0.40 ⎞ ⎞ Is = 2 * s * ⎜ − c ⎟ = 2 * *⎜ − 0.05 ⎟ = 1,81 *10− 5 m 4 2 ⎝2 2 ⎠ ⎝ 2 ⎠

Ks = 1 and

Kc =

k1 * k 2 1.12 * 0.069 = = 0.031 1 + ϕ ef 1 + 1.49

Therefore:

EI = 0.031* 26230 * 2,133.10−3 + 1 * 200000 *1,81.10−5 = 5.35MNm² The second order effects must be recalculated:

M Ed

⎡ ⎤ ⎢ β ⎥ ⎥ = M 0 Ed * ⎢1 + NB ⎢ − 1⎥ ⎢⎣ N Ed ⎥⎦

β = 1.234 NB = π ² *

M Ed

464

EI 5.35 =π²* = 0.367 MN 2 l0 12²

⎡ ⎤ ⎢ 1.234 ⎥ = 0.254MNm = 0.039 * ⎢1 + 0.367 ⎥ ⎢ − 1⎥ ⎣ 0.300 ⎦

ADVANCE DESIGN VALIDATION GUIDE

There is thus a second order moment of 0.254MNm. This moment is expressed relative to the center of gravity. Reinforcement calculation for combined bending

The necessary frames for the second order stresses can now be determined. A column frames can be calculated from the diagram below:

The input parameters in the diagram are:

μ=

0.254 M Ed = = 0.238 2 b * h * f cd 0.40 * 0.40² *16.67

N Ed 0.300 = = 0.112 b * h * f cd 0.40 * 0.40 * 16.67

ω = 0 .485 is obtained, which gives:

∑A

ω * b * h * f cd f yd

0.485 * 0.40 ² * 16 .67 = 29 .75cm ² 434 .78

Therefore set up a section 14.87cm ² per side must be set, or 3HA32 per side (by excess) Buckling checking

The column in place will be verified without buckling. The new reinforcement area must be considered for the previous calculations: 6HA32 provides As=48.25cm2

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The normal rigidity evaluation:

It is estimated nominal rigidity of a post or frame member from the following formula:

EI = K c .Ecd . I c + K s .Es .I s With:

: Inertia

ρ=

As 48,25 *10−4 = = 0.03 Ac 0.40 * 0.40 2

A ⎛h 48,25 *10−4 ⎛ 0.40 ⎞ ⎞ Is = 2 * s * ⎜ − c ⎟ = 2 * *⎜ − 0.05 ⎟ = 1,086.10− 4 m 4 2 ⎝2 2 ⎠ ⎝ 2 ⎠

k1 = 1.12 k2 = 0.069 Kc =

K s = 1 and

k1 * k 2 1.12 * 0.069 = = 0.031 1 + ϕ ef 1 + 1.49

The following conditions are not implemented in Advance Design:

ρ= If:

Kc =

As ≥ 0.01 Ac

0,3 1 + 0,5 * ϕ ef

Then:

EI = 0.031* 26230 * 2,133 *10−3 + 200000 *1,086 *10−4 = 23.45MNm² Corrected stresses

The total moment, including second order effects, is defined as a value plus the moment of the first order:

M Ed

⎡ ⎤ ⎢ β ⎥ ⎥ = M 0 Ed * ⎢1 + NB ⎢ − 1⎥ ⎢⎣ N Ed ⎥⎦

M 0 Ed = 0.039MNm N Ed = 0.300MN

(normal force acting at ULS)

β = 1.234 NB = π ² *

466

EI 23.45 =π²* = 1.607 MN 2 l0 12²

ADVANCE DESIGN VALIDATION GUIDE

It was therefore a moment of second order which is:

M Ed

⎤ ⎡ ⎢ 1.234 ⎥ = 0.05MNm = 0.039 * ⎢1 + 1.607 ⎥ ⎢ − 1⎥ ⎣ 0.300 ⎦

Reinforcement calculation:

M Ed = 0.05 MNm => μ = N Ed = 0.300 MN => v =

M Ed 0.05 = = 0.047 2 b * h * f cd 0.40 * 0.40² * 16.67

N Ed 0.300 = = 0.112 b * h * f cd 0.40 * 0.40 * 16.67

The minimum reinforcement percentage conditions are not satisfied therefore there will be one more iteration. Additional iteration:

The additional iteration will be made for a section corresponding to 1%; As=0.009*0.40=14.4cm2, which is 7.2cm2 per side. A 3HA16 reinforcement will be chosen by either side (6HA16 for the entire column), which will give As=12.03cm2.

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ADVANCE DESIGN VALIDATION GUIDE

Nominal rigidity evaluation:

EI = K c .Ecd . I c + K s .Es .I s . Is

: Inertia

ρ=

As 12,06 *10−4 = = 0.00754 Ac 0.40 * 0.40

0.002 ≤ ρ = if

As < 0.01 Ac 2

As ⎛ h 12,06 * 10−4 ⎛ 0.40 ⎞ ⎞ Is = 2 * * ⎜ − c ⎟ = 2 * *⎜ − 0.05 ⎟ = 2,71.10− 5 m 4 2 ⎝2 2 ⎠ ⎝ 2 ⎠

K s = 1 and

Kc =

k1 * k 2 1.12 * 0.069 = = 0.031 1 + ϕ ef 1 + 1.49

Therefore:

EI = 0.031 * 26230 * 2,133 *10−3 + 1 * 200000 * 2,71 * 10−5 = 7.15MNm² Second order loads calculation:

M Ed

⎡ ⎤ ⎢ β ⎥ ⎥ = M 0 Ed * ⎢1 + NB ⎢ − 1⎥ ⎢⎣ N Ed ⎥⎦

M 0 Ed = 0.039MNm N Ed = 0.300MN

(normal force acting at ULS).

β = 1.234 NB = π ² *

M Ed

468

EI 7.15 =π²* = 0.49 MN 2 l0 12²

⎡ ⎤ ⎢ 1.234 ⎥ = 0.039 * ⎢1 + ⎥ = 0.115 MNm 0.49 ⎢ − 1⎥ ⎣ 0.300 ⎦

ADVANCE DESIGN VALIDATION GUIDE

Reinforcement calculation:

Frames are calculated again from the interaction diagram:

μ=

0.115 M Ed = = 0.108 2 b * h * f cd 0.40 * 0.40² *16.67

N Ed 0.300 = = 0.112 b * h * f cd 0.40 * 0.40 * 16.67

ω = 0 .18 is obtained, which gives:

∑A

ω * b * h * f cd f yd

0.18 * 0.40² *16.67 = 11.04cm ² 434 .78

Therefore set up a section 5.52cm ² per side must be set; this will be the final column reinforcement Finite elements modeling

■ ■ ■

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

Theoretical reinforcement area(cm2)

(reference value: 11.04cm2)

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ADVANCE DESIGN VALIDATION GUIDE

Theoretical value (cm2)

(reference value: 11.16 cm2)

5.44.2.3 Reference results Result name

Az R

Result description

Reference value 2

5.58 cm2

Reinforcement area [cm ] 2

11.16 cm2

Theoretical reinforcement area [cm ]

5.44.3 Calculated results

Result name Az

470

Result description Az

Value -5.58 cmÂ˛

Error -0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.45 EC2 Test35: Verifying a rectangular concrete column subjected to compression to top – Based on nominal rigidity method - Bilinear stress-strain diagram (Class XC1) Test ID: 5123 Test status: Passed

5.45.1 Description Verifies the adequacy of a rectangular concrete column made of concrete C30/37 subjected to compression to top – Based on nominal rigidity method- Bilinear stress-strain diagram (Class XC1). Based on nominal rigidity method The purpose of this test is to determine the second order effects by applying the nominal rigidity method, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by a fixed connection and free to the top part.

5.45.2 Background Based on nominal rigidity method Verify the adequacy of a rectangular cross section made from concrete C30/37. The purpose of this test is to determine the second order effects by applying the nominal rigidity method, and then calculate the frames by considering a section symmetrically reinforced.

5.45.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: 0.45 MN axial force ► 0.10 MNm rotation moment applied to the column top ► The self-weight is neglected Exploitation loadings: ► 0.50 MN axial force ► 0.06 MNm rotation moment applied to the column top ►

■

ψ 2 = 0,3

■ ■ ■ ■ ■ ■

The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q Characteristic combination of actions: CCQ = 1.0 x G + 0.3 x Q Concrete cover 5cm Concrete C30/37 Steel reinforcement S500B The column is considered isolated and braced

471

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.45 m, Width: b = 0.60 m, Length: L = 4.50 m, Concrete cover: c = 5cm along the long section edge and 3cm along the short section edge

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*0.45+1.5*0.50=1.3575MN MED=1.35*0.10+0.30*0.06=0.225MNm

■

e0 =

0.225 = 0.166 m 1.3575

5.45.2.2 Reference results in calculating the concrete column Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

l0 = 2 * l = 2 * 4.50 = 9m According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:

λ=

472

2 3 * l0 2 3 * 9 = = 69.28 a 0.45

ADVANCE DESIGN VALIDATION GUIDE

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

ϕ (∞,t0 )

creep coefficient

M EQP

serviceability firs order moment under quasi-permanent load combination

M Ed

ULS first order moment (including the geometric imperfections)

M 0 Eqp = 0.10 + 0.3 * 0.06 = 0.118 MNm

M 0 Ed = 0.225MNm The moment report becomes:

M 0 Eqp 0.118 = = 0.524 M 0 Ed 0.225 The creep coefficient

ϕ (∞,t0 )

is defined as follows:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )

β ( f cm ) = β (t0 ) =

ϕ RH

16.8 16.8 = = 2.73MPa f cm 30 + 8

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

(for t0= 28 days concrete age).

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

RH =relative humidity; RH=50%

α = α 2 = 1 if f ≤ 35MPa Where 1 cm h0 =

⎛ 35 ⎞ if not α1 = ⎜ ⎜ f ⎟⎟ ⎝ cm ⎠

0.7

and

⎛ 35 ⎞ ⎟⎟ α 2 = ⎜⎜ ⎝ f cm ⎠

0.2

2 * Ac 2 * 450 * 700 = = 274 mm u 2 * (450 + 700 )

f cm = 38MPa > 35MPa , Therefore:

⎛ 35 ⎞ ⎟⎟ α1 = ⎜⎜ ⎝ f cm ⎠

0.7

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.7

⎛ 35 ⎞ ⎟⎟ = 0.944 and α 2 = ⎜⎜ ⎝ f cm ⎠

0.2

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.2

= 0.984

473

ADVANCE DESIGN VALIDATION GUIDE

ϕ RH

50 RH ⎛ ⎞ ⎛ ⎞ 1− 1− ⎜ ⎟ ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 100 * 0.944 ⎟ * 0.984 = 1.70 = ⎜1 + 1 2 ⎜ 0.1 * 3 274 ⎟ ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.70 * 2.73 * 0.488 = 2.26 The effective creep coefficient calculation:

ϕef = ϕ (∞, t0 ) *

M EQP

= 2.26 * 0.524 = 1.18

M Ed

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

1 + ϕef = 1 + 1.18 = 2.18 The necessity of buckling calculation (second order effect):

For an isolated column, the slenderness limit check is done using the next formula:

λlim =

20 * A * B * C n According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

Where:

1.3575 N Ed = = 0.215 Ac * f cd 0.45 * 0.70 * 20

1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 *1.18 = 0.81 because the reinforcement ratio in not yet known because the ratio of the first order moment is not known

λlim =

20 * 0.85 * 1.1 * 0.7 = 26 .90 0.215

λ = 69.28 > λlim = 26.90 Therefore, the second order effects most be considered

5.45.2.3 The eccentricity calculation and the corrected loads on ULS: Initial eccentricity:

e0 =

0.225 = 0.166m 1.3575

Additional eccentricity:

ei =

474

l0 9 = = 2.25cm 400 400

ADVANCE DESIGN VALIDATION GUIDE

First order eccentricity- stresses correction:

N Ed = 1.3575MN e1 = e0 + ei = 0.1885m = 18.85cm M Ed = 0.256MNm Reinforcement calculation in the first order situation:

h 0.45 ⎞ ⎛ M ua = M G 0 + N * ( d − ) = 0.256 + 1.3575 * ⎜ 0.40 − ⎟ = 0.494 MNm 2 ⎠ 2 ⎝ Verification if the section is partially compressed:

μ BC = 0,8 *

μ cu =

h h 0,45 0,45 * (1 − 0,4 * ) = 0,8 * * (1 − 0,4 * ) = 0,495 d d 0,40 0,40

0.495 M ua = = 0.220 bw * d ² * f cd 0.70 * 0.40² * 20

μcu = 0.220 < 0.495 = μBC

therefore the section is partially compressed.

Calculations of steel reinforcement in pure bending:

μcu = 0.220

[

]

αu = 1,25 * 1 − (1 − 2 * 0,220) = 0,315 zc = d * (1 − 0,4 * α u ) = 0,40 * (1 − 0,4 * 0,315) = 0,350m A =

M ua 0,495 = = 32.46cm ² zc * f yd 0,350 * 434,78

Calculations of steel reinforcement in combined bending:

For the combined bending:

A = A'−

N 1.3575 = 32.46 *10− 4 − = −1.24cm2 f yd 434.78 475

ADVANCE DESIGN VALIDATION GUIDE

The minimum reinforcement percentage:

As , min =

0,10 * N Ed 0.10 * 1.3575 = = 3.12cm ² ≥ 0.002 * Ac = 6.3cm 2 f yd 434 .78

The reinforcement will be 8HA10 representing a 6.28cm2 section The second order effects calculation:

The second order effect will be determined by applying the method of nominal rigidity: Calculation of nominal rigidity:

It is estimated nominal rigidity of a post or frame member from the following formula:

EI = K c * Ecd * I c + K s * Es * I s With:

Ecd =

Ecm 1.2

f cm = f ck + 8MPa = 38MPa Ecm

⎛f ⎞ = 22000 * ⎜ cm ⎟ ⎝ 10 ⎠

⎛ 38 ⎞ = 22000 * ⎜ ⎟ ⎝ 10 ⎠

0.3

= 32837 MPa

Ecm 32837 = = 27364MPa 1.2 1.2

Ecd = Ic =

0.3

b * h 3 0.70 * 0.45 3 = = 5.316 * 10 − 3 m 4 12 12

(concrete only inertia)

Es = 200000MPa Is

: Inertia

ρ=

As 6,28 *10−4 = = 0.002 Ac 0.70 * 0.40

0.002 ≤ ρ =

k1 = n=

f ck = 20

As < 0.01 Ac

30 = 1.22 Mpa 20

N Ed 1.3575 = = 0.215 Ac . f cd 0.45 * 0.70 * 20

k2 = n *

λ 170

= 0.215 *

69.28 = 0.088 ≤ 0.20 170

A ⎛h 6,28.10−4 ⎛ 0.45 ⎞ ⎞ *⎜ Is = 2 * s * ⎜ − c ⎟ = 2 * − 0.05 ⎟ = 1,92.10− 5 m 4 2 2 ⎝2 ⎠ ⎝ 2 ⎠

Ks = 1

476

and

Kc =

k1 * k 2 1.22 * 0.088 = = 0.049 1 + ϕ ef 1 + 1.18

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

EI = 0.049 * 27364 * 5.316 *10−3 + 1 * 200000 *1.92 *10−5 = 10.97 MNm² Stresses correction:

The total moment, including second order effects, is defined as a value plus the time of the first order:

M Ed

⎡ ⎤ ⎢ β ⎥ ⎥ = M 0 Ed * ⎢1 + NB ⎢ − 1⎥ ⎢⎣ N Ed ⎥⎦

M 0 Ed = 0.0256MNm (moment of first order (ULS) taking into account geometric imperfections N Ed = 1.3575MN (normal force acting at ULS).

β=

π² c0

π² 8

and

c0 = 8

the moment is constant (no horizontal force at the top of post).

= 1.234

NB = π ² *

EI 10.82 =π²* = 1.32 MN 2 l0 9²

The second order moment is:

M Ed 2

⎡ ⎤ ⎢ 1.234 ⎥ = 0.256 * ⎢1 + ⎥ = −11.18MNm 1.32 ⎢ − 1⎥ ⎣ 1.3575 ⎦

An additional iteration must be made by increasing the ratio of reinforcement.

477

ADVANCE DESIGN VALIDATION GUIDE

Additional iteration:

The iteration is made considering 8HA12 or As=9.05cm2

ρ= Therefore

9,05 * 10 −4 = 0.0029 0.70 * 0.45

Is obtained:

Therefore:

N Ed 2 = 1.3575MN M Ed 2 = 2.48MNm Given the mentioned reports, there must be another reiteration: The reinforcement section must be increase to 8HA20 or As=25.13cm2 and ρ Is obtained:

478

= 0.008 :

ADVANCE DESIGN VALIDATION GUIDE

Therefore:

N Ed 2 = 1.3575MN M Ed 2 = 0.563MNm Using the EC2 calculation tools, combined bending analytical calculation, a 28.80cm2 value is found. There must be another iteration: The reinforcement section must be increase to As=30cm2 and ρ

= 0.0095 :

Therefore:

N Ed 2 = 1.3575MN M Ed 2 = 0.499MNm Using the EC2 calculation tools, combined bending analytical calculation, a 22.22cm2 value is found. The theoretical reinforcement section of 30cm2 will be adopted. Finite elements modeling

■ ■ ■

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

479

ADVANCE DESIGN VALIDATION GUIDE

Theoretical reinforcement area(cm2)

(reference value: 16cm2 )

Theoretical value (cm2)

(reference value: 31.99cm2)

5.45.2.4 Reference results Result name

Az R

Result description

Reference value 2

16 cm2

Reinforcement area [cm ] 2

31.99 cm2

Theoretical reinforcement area [cm ]

5.45.3 Calculated results

Result name Az

480

Result description Az

Value -15.995 cmÂ˛

Error 0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.46 EC2 Test36: Verifying a rectangular concrete column using the method based on nominal curvature- Bilinear stress-strain diagram (Class XC1) Test ID: 5125 Test status: Passed

5.46.1 Description Verifying a rectangular concrete column using the method based on nominal curvature - Bilinear stress-strain diagram (Class XC1) Verifies the adequacy of a rectangular cross section column made from concrete C30/37. Method based on nominal curvature The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by an articulated connection (all the translations are blocked and all the rotations are permitted) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked.

This example is provided by the â&#x20AC;&#x153;Calcul des Structures en betonâ&#x20AC;? book, by Jean-Marie Paille, edition Eyrolles.

481

ADVANCE DESIGN VALIDATION GUIDE

5.47 EC2 Test 37: Verifying a square concrete column using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1) Test ID: 5126 Test status: Passed

5.47.1 Description Verifies the adequacy of a square concrete column made of concrete C25/30, using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1). Simplified Method The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked.

5.47.2 Background Simplified Method Verifies the adequacy of a square cross section made from concrete C25/30.

5.47.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: 800kN axial force ► The self-weight is neglected Exploitation loadings: ► 800kN axial force Concrete cover 5cm Concrete C25/30 Steel reinforcement S500B Relative humidity RH=50% Buckling length L0=0.70*4=2.80m ►

■ ■ ■ ■ ■ ■

482

ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.40 m, Width: b = 0.400 m, Length: L = 4.00 m, Concrete cover: c=5cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection (all the translations and rotations are blocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked. Loading

The beam is subjected to the following load combinations: ■

Load combinations:

N ED = 1.35 * 800 + 1.5 * 800 = 2280kN 5.47.2.2 Reference results in calculating the concrete column Scope of the method:

λ=

L L L L * 12 2.8 * 12 L0 L = 0 = 0 = 0 = 0 = 0 = = 24.25 a a 0.4 i I a4 a2 12 A 12 12 2 a According to Eurocode 2 EN 1992-1-1 (2004) Chapter 5.8.3.2(1)

The method of professional rules can be applied as:

λ < 120 20 < f ck < 50 MPa h > 0 .15 m Reinforcement calculation:

λ = 24.25 < 60 , therefore: α=

0.86 ⎛λ ⎞ 1+ ⎜ ⎟ ⎝ 62 ⎠

0.86 ⎛ 24.25 ⎞ 1+ ⎜ ⎟ ⎝ 62 ⎠

= 0.746

According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283 Not knowing the values for ρ and δ , we can considered k h = 0.93

483

ADVANCE DESIGN VALIDATION GUIDE

k s = 1 .6 − 0 .6 *

As =

1 f yd

f yk 500

= 1 .6 − 0 .6 *

500 =1 500

⎛ N ed ⎞ 1 2.280 ⎛ ⎞ * ⎜⎜ − b * h * f cd ⎟⎟ = *⎜ − 0.4 * 0.4 *16 .67 ⎟ = 14,24cm ² ⎠ ⎝ kh * ks *α ⎠ 434 .78 ⎝ 0.93 *1 * 0.746 According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283

Finite elements modeling

■ ■ ■

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

Theoretical reinforcement area(cm2)

(reference value: 14.24cm2=4*3.56cm2)

5.47.2.3 Reference results Result name

Result description

Reference value 2

3.57cm2

Reinforcement area [cm ] 2

14.26 cm2

Theoretical reinforcement area [cm ]

5.47.3 Calculated results

Result name Ay

484

Result description

Value

-3.56562 cm²

Error

0.0001 %

ADVANCE DESIGN VALIDATION GUIDE

5.48 EC2 Test 26: Verifying the shear resistance for a rectangular concrete beam with vertical transversal reinforcement - Bilinear stress-strain diagram (Class XC1) Test ID: 5072 Test status: Passed

5.48.1 Description Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the cross-sectional area of the shear reinforcement, Asw. For the calculation, the reduced shear force values will be used.

5.48.2 Background Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear force diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be calculated, along with the cross-sectional area of the shear reinforcement, Asw. For the calculation, the reduced shear force values will be used.

5.48.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Concrete C25/30 ■ Reinforcement steel: S500B ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: The dead load will be represented by a linear load of 40kN/m

■

Exploitation loadings: The live load will be considered from one linear load of 25kN

■ ■ ■

Structural class: S1 Reinforcement steel ductility: Class B The reinforcement will be displayed like in the picture below:

485

ADVANCE DESIGN VALIDATION GUIDE

The objective is to verify: ■ ■

The shear stresses results The cross-sectional area of the shear reinforcement, Asw

Units

Metric System Geometry

Beam cross section characteristics: ■ ■ ■ ■ ■

Height: h = 0.70 m, Width: b = 0.35 m, Length: L = 5.75 m, Concrete cover: c=3.5cm Effective height: d=h-(0.06*h+ebz)=0.623m; d’=ehz=0.035m

■ ■

Stirrup slope: α= 90° Strut slope: θ=45˚

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 end point (x = 5.30) restrained in translation along Y, Z and rotation along X Inner: None. ►

■

Loading:

For a beam subjected to a point load, Pu, the shear stresses are defined by the formula below:

VEd =

Pu * l 2 According to EC2 the Pu point load is defined by the next formula: Pu = 1,35*Pg + 1,50*Pq=1.35*40kN+1.50*25kN=91.5kN

In this case :

VEd =

486

5.75 * 91.5 = 263 KN 2

ADVANCE DESIGN VALIDATION GUIDE

5.48.2.2 Reference results in calculating the lever arm zc: The lever arm will be calculated from the design formula for pure bending:

Pu = 91.5kN / ml M Ed =

91.5 * 5.75² = 378.15kNm 8

M Ed 0.378 = = 0.167 bw * d ² * f cd 0.35 * 0.623 ² * 16.67

μ cu =

(

)

(

)

αu = 1.25 * 1 − 1 − 2 * μcu = 1.25 * 1 − 1 − 2 * 0.167 = 0.230 zc = d * (1 − 0.4 * α u ) = 0.623 * (1 − 0.4 * 0.230) = 0.566m Calculation of reduced shear force

When the shear force curve has no discontinuities, the EC2 allows to consider, as a reduction, the shear force to a horizontal axis In case of a member subjected to a distributed load, the equation of the shear force is:

V ( x) = Pu * x −

Pu * l 2

Therefore:

VRd, max = α cw * ν1 * fcd * z u * b w *

(cotα + cotθ) 1 + cot 2θ According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 6.2.3.(3)

Where:

α cw = 1

coefficient taking account of the state of the stress in the compression chord and

f ⎤ ⎡ v1 = 0,6 * ⎢1 − ck ⎥ ⎣ 250 ⎦ When the transverse frames are vertical, the above formula simplifies to: VRd,max =

v 1 * f cd * z u * b w tgθ + cot θ

In this case:

θ = 45°

and

α = 90°

v1 strength reduction factor for concrete cracked in shear 25 ⎤ f ⎤ ⎡ ⎡ = 0.54 v1 = 0,6 * ⎢1 − ck ⎥ = 0.6 * ⎢1 − ⎣ 250 ⎦ ⎣ 250 ⎥⎦

487

ADVANCE DESIGN VALIDATION GUIDE

zu = 0.9 * d = 0.9 * 0.623 = 0.56m VRd , max =

0.54 *16.67 * 0.566 * 0.35 = 0.891MN = 891kN 2

VEd = 236 kN < VRd , max = 891kN Calculation of transversal reinforcement:

Given the vertical transversal reinforcement (α = 90°), the following formula is used:

Asw VEd . * tgθ 0,211 * tg 45 = = = 8,67cm² / ml 500 s zu * f ywd 0,566 * 1,15 Finite elements modeling

■ Linear element: S beam, ■ 3 nodes, ■ 1 linear element. Advance Design gives the following results for Atz (cm2/ml)

5.48.2.3 Reference results Result name

Result description

Reference value

Fz corresponding to the 101 combination (ULS) [kNm]

263 kN

At,z

Theoretical reinforcement area [cm2/ml]

8.67 cm2/ml

5.48.3 Calculated results

Result name Fz

Atz

488

Result description

Value

Error

-263.062 kN

-0.0002 %

Atz

8.68636 cm²

0.0000 %

ADVANCE DESIGN VALIDATION GUIDE

5.49 EC2 Test30: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement - Bilinear stress-strain diagram (Class XC1) Test ID: 5098 Test status: Passed

5.49.1 Description Verifies a T cross section beam made from concrete C30/37 to resist simple bending. For this test, the shear force diagram and the moment diagram will be generated. The design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts, VRd,max, will be determined, along with the cross-sectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value.

5.50 EC2 Test34: Verifying a rectangular concrete column subjected to compression on the top – Method based on nominal curvature - Bilinear stress-strain diagram (Class XC1) Test ID: 5114 Test status: Passed

5.50.1 Description Verifies the adequacy of a rectangular cross section column made of concrete C30/37. The verification of the axial stresses applied on top, at ultimate limit state is performed. Method based on nominal curvature - The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced. The column is considered connected to the ground by a fixed connection and free at the top part.

5.50.2 Background Method based on nominal curvature Verifies the adequacy of a rectangular cross section made from concrete C30/37. The purpose of this test is to determine the second order effects by applying the method of nominal curvature, and then calculate the frames by considering a section symmetrically reinforced.

5.50.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: 300kN axial force ► The self-weight is neglected Exploitation loadings: ► 500kN axial force ►

■ ■

ψ 2 = 0,3

■ ■ ■ ■ ■ ■

489

ADVANCE DESIGN VALIDATION GUIDE

■

The column is considered isolated and braced

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.40 m, Width: b = 0.60 m, Length: L = 4.00 m, Concrete cover: c = 5 cm along the long section edge and 3cm along the short section edge

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a fixed connection and free to the top part. Loading

The beam is subjected to the following load combinations: ■

Load combinations: The ultimate limit state (ULS) combination is: NEd =1.35*0.30+1.5*0.50=1.155MN NQP=1.35*0.30+0.30*0.50=0.450MN

■

490

ADVANCE DESIGN VALIDATION GUIDE

5.50.2.2 Reference results in calculating the concrete column Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

l0 = 2 * l = 2 * 4 = 8m According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b) Calculating the slenderness of the column:

λ=

2 3 * l0 2 3 * 8 = = 69.28 a 0.40

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

ϕ ef = ϕ (∞, t0 ).

M EQP M Ed According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

ϕ (∞,t0 )

creep coefficient

M EQP

serviceability firs order moment under quasi-permanent load combination

M Ed

ULS first order moment (including the geometric imperfections)

The moment report becomes:

M 0 Eqp N eqp * e1 N eqp 0.450 = = = = 0.39 M 0 Ed N ed *e1 N ed 1.155 The creep coefficient

ϕ (∞,t0 )

is defined as:

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 )

β ( f cm ) = β (t0 ) =

ϕ RH

16.8 16.8 = = 2.73MPa f cm 30 + 8

1 1 = = 0.488 0.20 0.1 + 28 0.20 0 .1 + t 0

(for t0= 28 days concrete age).

RH ⎛ ⎞ 1− ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 1 2 ⎜ 0.1 * 3 h0 ⎟ ⎜ ⎟ ⎝ ⎠

RH = relative humidity; RH=50% Where α1

= α 2 = 1 if f cm

⎛ 35 ⎞ ⎟⎟ ≤ 35MPa if not α1 = ⎜⎜ f ⎝ cm ⎠

0.7

and

⎛ 35 ⎞ ⎟⎟ α 2 = ⎜⎜ f cm ⎝ ⎠

0.2

491

ADVANCE DESIGN VALIDATION GUIDE

h0 =

2 * Ac 2 * 400 * 600 = = 240 mm u 2 * (400 + 600 )

f cm = 38MPa > 35MPa , Therefore:

⎛ 35 ⎞ ⎟⎟ α1 = ⎜⎜ ⎝ f cm ⎠

ϕ RH

0.7

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.7

⎛ 35 ⎞ ⎟⎟ = 0.944 and α 2 = ⎜⎜ ⎝ f cm ⎠

0.2

⎛ 35 ⎞ =⎜ ⎟ ⎝ 38 ⎠

0.2

= 0.984

50 RH ⎛ ⎞ ⎛ ⎞ 1− 1− ⎜ ⎟ ⎜ ⎟ 100 * α ⎟ * α = ⎜1 + 100 * 0.944 ⎟ * 0.984 = 1.73 = ⎜1 + 1 2 ⎜ 0.1 * 3 h0 ⎟ ⎜ 0.1 * 3 240 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

ϕ (∞, t0 ) = ϕ RH * β ( f cm ) * β (t0 ) = 1.73 * 2.73 * 0.488 = 2.30 The effective creep coefficient calculation:

ϕef = ϕ (∞, t0 ) *

M EQP M Ed

= 2.30 * 0.39 = 0.90 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

The necessity of buckling calculation (second order effect):

For an isolated column, the slenderness limit check is done using the next formula:

λlim =

20 * A * B * C n

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1) Where:

N Ed 1.155 = = 0.241 Ac * f cd 0.40 * 0.60 * 20

1 1 = (1 + 0,2 * ϕef ) 1 + 0.2 * 0.90 = 0.85

B = 1 + 2 * ω = 1.1 C = 1,7 − rm = 0.70

λlim =

because the reinforcement ratio in not yet known

because the ratio of the first order moment is not known

20 * 0.85 * 1.1 * 0.7 = 26 .66 0.112

λ = 69.28 > λlim = 26.66 Therefore, the second order effects most be considered

492

ADVANCE DESIGN VALIDATION GUIDE

5.50.2.3 The eccentricity calculation and the corrected loads on ULS: Initial eccentricity:

No initial eccentricity because the post is only applied in simple compression. Additional eccentricity:

ei =

8 l0 = = 0.02m 400 400

First order eccentricity- stresses correction:

⎧⎪ 20mm ⎧⎪20mm ⎧ 20mm = 20mm e0 = max ⎨ h = max ⎨ 400mm = max ⎨ 13 . 3 mm ⎩ ⎪⎩ 30 ⎪⎩ 30 According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4) Solicitations corrected to take into account when calculating flexural combined, are: NEd= 1.155MN MEd= 1.155*0.02=0.0231MNm Reinforcement calculation in the first order situation:

h 0.40 ⎞ ⎛ M ua = M G 0 + N * ( d − ) = 0.0231 + 1.155 * ⎜ 0.35 − ⎟ = 0.196 MNm 2 2 ⎠ ⎝ Verification if the section is partially compressed:

μ BC = 0,8 *

μ cu =

h h 0,40 0,40 * (1 − 0,4 * ) = 0,8 * * (1 − 0,4 * ) = 0,496 d d 0,35 0,35

M ua 0.196 = = 0.133 bw * d ² * f cd 0.60 * 0.35² * 20

μcu = 0.133 < 0.496 = μBC

therefore the section is partially compressed.

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ADVANCE DESIGN VALIDATION GUIDE

Calculations of steel reinforcement in pure bending:

μcu = 0.133

[

]

α u = 1,25 * 1 − (1 − 2 * 0,133) = 0,179 zc = d * (1 − 0,4 * α u ) = 0,35 * (1 − 0,4 * 0,179) = 0,325m A =

M ua 0,196 = = 13.87cm ² zc * f yd 0,325 * 434,78

Calculations of steel reinforcement in combined bending:

For the combined bending:

A = A'−

N 1.155 = 13.87 *10− 4 − = −12.69cm2 f yd 434.78

The minimum reinforcement percentage:

As , min =

0,10 * N Ed 0.10 * 1.155 = = 2.66 cm ² ≥ 0.002 * Ac = 4.80cm 2 f yd 434 .78

The reinforcement will be 8HA10 representing a 6.28cm2 section The second order effects calculation:

The second order effect will be determined by applying the method of nominal rigidity: Calculation of nominal rigidity:

It is estimated nominal rigidity of a post or frame member from the following formula:

EI = K c * Ecd * I c + K s * Es * I s With:

Ecd =

Ecm 1.2

f cm = f ck + 8MPa = 38MPa Ecm

⎛f ⎞ = 22000 * ⎜ cm ⎟ ⎝ 10 ⎠

Ecd = Ic =

0.3

⎛ 38 ⎞ = 22000 * ⎜ ⎟ ⎝ 10 ⎠

b * h 3 0.60 * 0.40 3 = = 3,2.10 − 3 m 4 12 12

: Inertia

ρ=

494

= 32837 MPa

Ecm 32837 = = 27364MPa 1.2 1.2

Es = 200000MPa Is

0.3

As 6,28.10 −4 = = 0.0026 Ac 0.60 * 0.40

(concrete only inertia)

ADVANCE DESIGN VALIDATION GUIDE

0.002 ≤ ρ =

f ck = 20

k1 = n=

As < 0.01 Ac

30 = 1.22 Mpa 20

N Ed 1.155 = = 0.241 Ac . f cd 0.40 * 0.60 * 20

k2 = n *

λ 170

= 0.241*

69.28 = 0.098 ≤ 0.20 170

6,28.10 −4 ⎛ 0.40 A ⎛h ⎞ ⎞ *⎜ Is = 2 * s *⎜ − c ⎟ = 2 * − 0.05 ⎟ = 1,41.10 −5 m 4 2 ⎝2 2 2 ⎠ ⎝ ⎠

K s = 1 and K c =

k1 * k 2 1.22 * 0.098 = = 0.063 1 + ϕ ef 1 + 0.90

Therefore:

EI = 0.063 * 27364 * 3,2 *10−3 + 1 * 200000 *1,41 *10−5 = 8.34MNm² Stresses correction:

The total moment, including second order effects, is defined as a value plus the time of the first order:

M Ed

⎡ ⎤ ⎢ β ⎥ ⎥ = M 0 Ed * ⎢1 + NB ⎢ − 1⎥ ⎢⎣ ⎥⎦ N Ed

M 0 Ed = 0.0231MNm (moment of first order (ULS) taking into account geometric imperfections (normal force acting at ULS).

β=

π² c0

π² 8

and

c0 = 8

the moment is constant (no horizontal force at the top of post).

= 1.234

NB = π ² *

EI 8.25 = π ²* = 1.27 MN 2 l0 8²

The second order efforts are:

N Ed 2 = 1.155MN M Ed 2 = 0.31MN .m The reinforcement calculations in the second order effect:

The reinforcement calculations for the combined flexural, under the second order effect, are done using the Graitec EC2 tools. The obtained section is null, therefore the minimum reinforcement defined above is sufficient to absorb all the forces.

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ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling

■ ■ ■

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

Theoretical reinforcement area (cm2) and minimum reinforcement area (cm2)

(reference value: 2.4cm2 and 4.8cm2)

Theoretical value (cm2)

(reference value: 4.80 cm2)

5.50.2.4 Reference results Result name

496

Result description

Reference value 2

Reinforcement area [cm ]

2.40 cm2

Amin

Minimum reinforcement area

4.80 cm2

Theoretical reinforcement area [cm2]

4.80 cm2

ADVANCE DESIGN VALIDATION GUIDE

5.50.3 Calculated results

Result name Az

Amin

Result description

Value

Error

-2.4 cm²

0.0000 %

Amin

4.8 cm²

0.0000 %

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ADVANCE DESIGN VALIDATION GUIDE

5.51 EC2 Test 38: Verifying a rectangular concrete column using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1) Test ID: 5127 Test status: Passed

5.51.1 Description Verifies a rectangular cross section concrete C25/30 column using the simplified method –Professional rules Bilinear stress-strain diagram (Class XC1). The column is considered connected to the ground by an articulated connection (all the translations are blocked). At the top part, the translations along X and Y axis are blocked and the rotation along the Z axis is also blocked.

5.51.2 Background Simplified Method Verifies the adequacy of a rectangular cross section made from concrete C25/30.

5.51.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used: ■

Loadings from the structure: 800kN axial force ► The self-weight is neglected Concrete cover 5cm Concrete C25/30 Steel reinforcement S500B Relative humidity RH=50% Buckling length L0=6.50m ►

■ ■ ■ ■ ■

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ADVANCE DESIGN VALIDATION GUIDE

Units

Metric System Geometry

Below are described the beam cross section characteristics: ■ ■ ■ ■

Height: h = 0.30 m, Width: b = 0.50 m, Length: L = 6.50 m, Concrete cover: c=5cm

Boundary conditions

The boundary conditions are described below: The column is considered connected to the ground by a articulated connection (all the translations are blocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked. Loading

The beam is subjected to the following load combinations: ■

Load combinations:

N ED = 1.35 * 800 = 1080kN 5.51.2.2 Reference results in calculating the concrete column Scope of the method:

L L L L * 12 6.5 * 12 L0 L = 0 = 0 = 0 = 0 = 0 = = 75.06 4 2 a a 0.3 i I a a 12 A 12 12 2 a

λ=

According to Eurocode 2 EN 1992-1-1 (2004) Chapter 5.8.3.2(1) The method of professional rules can be applied as:

λ < 120 20 < f ck < 50 MPa h > 0 .15 m Reinforcement calculation:

60 < λ = 75.06 ≤ 120 1.3

⎛ 32 ⎞ ⎟ ⎝λ ⎠

α =⎜

1.3

⎛ 32 ⎞ =⎜ ⎟ ⎝ 75.06 ⎠

= 0.33

According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283

k h = (0.75 + 0.5 * h) * (1 − 6 * ρ * δ ) = 0.93 k s = 1 .6 − 0 . 6 *

As =

1 f yd

f yk 500

= 1 .6 − 0 .6 *

500 =1 500

⎛ N ed ⎞ 1 1.08 ⎛ ⎞ * ⎜⎜ − b * h * f cd ⎟⎟ = *⎜ − 0.3 * 0.5 *16.67 ⎟ = 23 .43cm ² ⎠ ⎝ kh * ks *α ⎠ 434 .78 ⎝ 0.93 *1 * 0.33

499

ADVANCE DESIGN VALIDATION GUIDE

Finite elements modeling

■ ■ ■

Linear element: S beam, 8 nodes, 1 linear element.

Theoretical reinforcement area(cm2)

(reference value: 23.43cm2=4*5.86cm2)

5.51.2.3 Reference results Result name

Result description

Reference value 2

5.85cm2

Reinforcement area [cm ] 2

Theoretical reinforcement area [cm ]

23.40 cm2

5.51.3 Calculated results

Result name Ay

500

Result description

Value

-5.85106 cm²

Error

0.0001 %