AD Validation Guide Vol1 2015 En

8/21/2019 AD Validation Guide Vol1 2015 En

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Advance Design

Validation Guide Part 1

Version: 2015

Tests passed on: 16 April 2014

Number of tests: 519





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 automatictests, 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 ofquality management by extending the scope and automating the testing process.

While in previous versions, the tests were always about the calculation results which

were compared to a reference set, starting with version 2012, tests have been

extended to user interface behavior, import/export procedures, etc.

The next major improvement is the capacity to pass the tests automatically. These

current tests have obviously been passed on the “operational version”, but they are

actually passed on a daily basis during the development process, which helps improve

the daily quality by solving potential issues, immediately after they have been

introduced in the code.

In the field of structural analysis and design, software users must keep in mind that theresults 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 toEurocode 5, seismic analysis according to Eurocode 8, report generation, import /

export procedures and user interface behavior.

We hope that this guide will highly contribute to the knowledge and the confidence you

are placing in Advance Design.

Manuel LIEDOT

Chief Product Office





ADVANCE VALIDATION GUIDE

7

Table of Contents

1 FINITE ELEMENT METHOD ................................................................................................. 17

1.1 Cantilever rectangular plate (01-0001SSLSB_FEM) ....................................................................................18

1.2 System of two bars with three hinges (01-0002SSLLB_FEM) ......................................................................21

1.3 Thin lozenge-shaped plate fixed on one side (alpha = 15 °) (01-0008SDLSB_FEM) ...................................24

1.4 Thin circular ring fixed in two points (01-0006SDLLB_FEM) ........................................................................27

1.5 Thin lozenge-shaped plate fixed on one side (alpha = 30 °) (01-0009SDLSB_FEM) ...................................31

1.6 Thin lozenge-shaped plate fixed on one side (alpha = 0 °) (01-0007SDLSB_FEM) .....................................34

1.7 Vibration mode of a thin piping elbow in plane (case 2) (01-0012SDLLB_FEM) ..........................................37

1.8 Double fixed beam (01-0016SDLLB_FEM) .................................................................................................. 40

1.9 Vibration mode of a thin piping elbow in plane (case 3) (01-0013SDLLB_FEM) ..........................................44

1.10 Short beam on simple supports (on the neutral axis) (01-0017SDLLB_FEM).............................................47

1.11 Rectangular thin plate simply supported on its perimeter (01-0020SDLSB_FEM)......................................51

1.12 Cantilever beam in Eulerian buckling (01-0021SFLLB_FEM).....................................................................55

1.13 Vibration mode of a thin piping elbow in plane (case 1) (01-0011SDLLB_FEM).........................................57

1.14 Double fixed beam with a spring at mid span (01-0015SSLLB_FEM) ........................................................60

1.15 Thin square plate fixed on one side (01-0019SDLSB_FEM).......................................................................63

1.16 Thin lozenge-shaped plate fixed on one side (alpha = 45 °) (01-0010SDLSB_FEM) .................................67

1.17 Thin circular ring hanged on an elastic element (01-0014SDLLB_FEM) ....................................................70

1.18 Short beam on simple supports (eccentric) (01-0018SDLLB_FEM) ...........................................................74

1.19 Slender beam on two fixed supports (01-0024SSLLB_FEM)......................................................................78

1.20 Slender beam on three supports (01-0025SSLLB_FEM)............................................................................83

1.21 Fixed thin arc in out of plane bending (01-0028SSLLB_FEM) ....................................................................87

1.22 Double hinged thin arc in planar bending (01-0029SSLLB_FEM)...............................................................89

1.23 Bending effects of a symmetrical portal frame (01-0023SDLLB_FEM).......................................................92

1.24 Fixed thin arc in planar bending (01-0027SSLLB_FEM).............................................................................95

1.25 Beam on elastic soil, free ends (01-0032SSLLB_FEM)..............................................................................98

1.26 EDF Pylon (01-0033SFLLA_FEM)............................................................................................................ 101

1.27 Truss with hinged bars under a punctual load (01-0031SSLLB_FEM) .....................................................105

1.28 Annular thin plate fixed on a hub (repetitive circular structure) (01-0022SDLSB_FEM)............................108

1.29 Bimetallic: Fixed beams connected to a stiff element (01-0026SSLLB_FEM) ..........................................110

1.30 Portal frame with lateral connections (01-0030SSLLB_FEM) ...................................................................113

1.31 Caisson beam in torsion (01-0037SSLSB_FEM) ...................................................................................... 116

1.32 Thin cylinder under a uniform radial pressure (01-0038SSLSB_FEM) .....................................................119

1.33 Beam on two supports considering the shear force (01-0041SSLLB_FEM) .............................................121

1.34 Thin cylinder under a uniform axial load (01-0042SSLSB_FEM)..............................................................124




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1.35 Simply supported square plate (01-0036SSLSB_FEM)............................................................................ 127

1.36 Stiffen membrane (01-0040SSLSB_FEM)................................................................................................ 130

1.37 Torus with uniform internal pressure (01-0045SSLSB_FEM)................................................................... 133

1.38 Spherical shell under internal pressure (01-0046SSLSB_FEM)............................................................... 136

1.39 Thin cylinder under its self weight (01-0044SSLSB_MEF)....................................................................... 139

1.40 Beam on elastic soil, hinged ends (01-0034SSLLB_FEM)....................................................................... 141

1.41 Square plate under planar stresses (01-0039SSLSB_FEM) .................................................................... 145

1.42 Thin cylinder under a hydrostatic pressure (01-0043SSLSB_FEM) ......................................................... 148

1.43 Spherical dome under a uniform external pressure (01-0050SSLSB_FEM) ............................................ 151

1.44 Simply supported square plate under a uniform load (01-0051SSLSB_FEM).......................................... 154

1.45 Simply supported rectangular plate loaded with punctual force and moments (01-0054SSLSB_FEM)....156

1.46 Shear plate perpendicular to the medium surface (01-0055SSLSB_FEM)............................................... 158

1.47 Spherical shell with holes (01-0049SSLSB_FEM).................................................................................... 160 1.48 Simply supported rectangular plate under a uniform load (01-0053SSLSB_FEM)................................... 163

1.49 A plate (0.01333 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0058SSLSB_FEM) .......... 165

1.50 A plate (0.02 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0059SSLSB_FEM) .... 167

1.51 A plate (0.01 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0057SSLSB_FEM) .... 169

1.52 Pinch cylindrical shell (01-0048SSLSB_FEM).......................................................................................... 171

1.53 Simply supported rectangular plate under a uniform load (01-0052SSLSB_FEM)................................... 173

1.54 Triangulated system with hinged bars (01-0056SSLLB_FEM) ................................................................. 175

1.55 A plate (0.01 m thick), fixed on its perimeter, loaded with a punctual force (01-0062SSLSB_FEM)......... 178 1.56 A plate (0.01333 m thick), fixed on its perimeter, loaded with a punctual force (01-0063SSLSB_FEM)...181

1.57 A plate (0.1 m thick), fixed on its perimeter, loaded with a punctual force (01-0066SSLSB_FEM)........... 183

1.58 Vibration mode of a thin piping elbow in space (case 1) (01-0067SDLLB_FEM) ..................................... 185

1.59 A plate (0.1 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0061SSLSB_FEM) ...... 188

1.60 A plate (0.05 m thick), fixed on its perimeter, loaded with a punctual force (01-0065SSLSB_FEM)......... 190

1.61 Reactions on supports and bending moments on a 2D portal frame (Rafters) (01-0077SSLPB_FEM)....192

1.62 Reactions on supports and bending moments on a 2D portal frame (Columns) (01-0078SSLPB_FEM)............ 194

1.63 Vibration mode of a thin piping elbow in space (case 3) (01-0069SDLLB_FEM) ..................................... 196 1.64 A plate (0.05 m thick), fixed on its perimeter, loaded with a uniform pressure (01-0060SSLSB_FEM) .... 199

1.65 A plate (0.02 m thick), fixed on its perimeter, loaded with a punctual force (01-0064SSLSB_FEM)......... 201

1.66 Vibration mode of a thin piping elbow in space (case 2) (01-0068SDLLB_FEM) ..................................... 203

1.67 Slender beam of variable rectangular section (fixed-fixed) (01-0086SDLLB_FEM).................................. 206

1.68 Plane portal frame with hinged supports (01-0089SSLLB_FEM) ............................................................. 209

1.69 A 3D bar structure with elastic support (01-0094SSLLB_FEM)................................................................ 211

1.70 Fixed/free slender beam with centered mass (01-0095SDLLB_FEM)...................................................... 218

1.71 Slender beam of variable rectangular section with fixed-free ends (ß=5) (01-0085SDLLB_FEM)............ 223

1.72 Cantilever beam in Eulerian buckling with thermal load (01-0092HFLLB_FEM) ...................................... 228




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1.73 Simple supported beam in free vibration (01-0098SDLLB_FEM) .............................................................230

1.74 Membrane with hot point (01-0099HSLSB_FEM)..................................................................................... 233

1.75 Double cross with hinged ends (01-0097SDLLB_FEM)............................................................................236

1.76 Short beam on two hinged supports (01-0084SSLLB_FEM) ....................................................................240

1.77 Double fixed beam in Eulerian buckling with a thermal load (01-0091HFLLB_FEM)................................242

1.78 Fixed/free slender beam with eccentric mass or inertia (01-0096SDLLB_FEM).......................................244

1.79 Beam on 3 supports with T/C (k = -10000 N/m) (01-0102SSNLB_FEM) ..................................................248

1.80 Linear system of truss beams (01-0103SSLLB_FEM)..............................................................................251

1.81 Linear element in combined bending/tension - without compressed reinforcements - Partially tensionedsection (02-0158SSLLB_B91) .............................................................................................................................. 254

1.82 Linear element in simple bending - without compressed reinforcement (02-0162SSLLB_B91)................260

1.83 Beam on 3 supports with T/C (k -> infinite) (01-0101SSNLB_FEM) .........................................................264

1.84 Study of a mast subjected to an earthquake (02-0112SMLLB_P92) ........................................................267

1.85 Design of a concrete floor with an opening (03-0208SSLLG_BAEL91)....................................................272

1.86 Design of a 2D portal frame (03-0207SSLLG_CM66)............................................................................... 280

1.87 Beam on 3 supports with T/C (k = 0) (01-0100SSNLB_FEM)...................................................................288

1.88 Non linear system of truss beams (01-0104SSNLB_FEM).......................................................................291

1.89 Design of a Steel Structure according to CM66 (03-0206SSLLG_CM66).................................................295

1.90 Slender beam with variable section (fixed-free) (01-0004SDLLB_FEM)...................................................304

1.91 Tied (sub-tensioned) beam (01-0005SSLLB_FEM).................................................................................. 307

1.92 Circular plate under uniform load (01-0003SSLSB_FEM) ........................................................................312

1.93 Verifying the displacement results on linear elements for vertical seism (TTAD #11756) .........................315

1.94 Verifying constraints for triangular mesh on planar elements (TTAD #11447)..........................................315

1.95 Verifying forces results on concrete linear elements (TTAD #11647) .......................................................315

1.96 Verifying diagrams after changing the view from standard (top, left,...) to user view (TTAD #11854).......315

1.97 Verifying forces for triangular meshing on planar element (TTAD #11723) ..............................................315

1.98 Verifying stresses in beam with "extend into wall" property (TTAD #11680) .............................................317

1.99 Generating planar efforts before and after selecting a saved view (TTAD #11849)..................................317

1.100 Verifying results on punctual supports (TTAD #11489)........................................................................... 317

1.101 Verifying the level mass center (TTAD #11573, TTAD #12315) .............................................................317

1.102 Verifying diagrams for Mf Torsors on divided walls (TTAD #11557) .......................................................317

1.103 Verifying Sxx results on beams (TTAD #11599) ..................................................................................... 318

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

1.105 Verifying nonlinear analysis results for frames with semi-rigid joints and rigid joints (TTAD #11495) .....318

1.106 Generating a report with torsors per level (TTAD #11421) .....................................................................318

1.107 Verifying tension/compression supports on nonlinear analysis (TTAD #11518) .....................................318

1.108 Verifying tension/compression supports on nonlinear analysis (TTAD #11518) .....................................319

1.109 Verifying the display of the forces results on planar supports (TTAD #11728)........................................319

1.110 Verifying results of a steel beam subjected to dynamic temporal loadings (TTAD #14586)....................320




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1.111 Verifying the main axes results on a planar element (TTAD #11725)..................................................... 324

1.112 Verifying torsors on a single story coupled walls subjected to horizontal forces..................................... 324

1.113 Calculating torsors using different mesh sizes for a concrete wall subjected to a horizontal force (TTAD #13175).324

1.114 Verifying the internal forces results for a simple supported steel beam.................................................. 324

1.115 Verifying forces on a linear elastic support which is defined in a user workplane (TTAD #11929) ......... 325

2 CAD, RENDERING AND VISUALIZATION ........................................................................ 327

2.1 Verifying hide/show elements command (TTAD #11753) .......................................................................... 328

2.2 Verifying the dimensions and position of annotations on selection when new analysis is made.(TTAD #12807)....... 328

2.3 Verifying the saved view of elements with annotations. (TTAD #13033)................................................... 328

2.4 Verifying the visualisation of supports with rotational or moving DoFs.(TTAD #13891) ............................. 328

2.5 Verifying the annotations of a wind generated load. (TTAD #13190) ......................................................... 328

2.6 System stability during section cut results verification (TTAD #11752) ...................................................... 329

2.7 Generating combinations (TTAD #11721).................................................................................................. 329

2.8 Verifying the grid text position (TTAD #11704)........................................................................................... 329

2.9 Verifying descriptive actors after creating analysis (TTAD #11589) ........................................................... 329

2.10 Verifying the coordinates system symbol (TTAD #11611)........................................................................ 329

2.11 Creating a circle (TTAD #11525) .............................................................................................................. 330

2.12 Creating a camera (TTAD #11526) .......................................................................................................... 330

2.13 Verifying the representation of elements with HEA cross section (TTAD #11328) ................................... 330

2.14 Verifying the snap points behavior during modeling (TTAD #11458)........................................................ 330

2.15 Verifying the local axes of a section cut (TTAD #11681).......................................................................... 330

2.16 Verifying the descriptive model display after post processing results in analysis mode (TTAD #11475).. 331

2.17 Modeling using the tracking snap mode (TTAD #10979).......................................................................... 331

2.18 Verifying holes in horizontal planar elements after changing the level height (TTAD #11490) ................. 331

2.19 Verifying the display of elements with compound cross sections (TTAD #11486).................................... 331

2.20 Moving a linear element along with the support (TTAD #12110) .............................................................. 331

2.21 Turning on/off the "ghost" rendering mode (TTAD #11999)...................................................................... 332

2.22 Verifying the "ghost" display after changing the display colors (TTAD #12064) ....................................... 332

2.23 Verifying the grid text position (TTAD #11657)......................................................................................... 332

2.24 Verifying the "ghost display on selection" function for saved views (TTAD #12054) ................................ 332

2.25 Verifying the steel connections modeling (TTAD #11698)........................................................................ 332

2.26 Verifying the fixed load scale function (TTAD #12183)..... ........................................................................ 332

2.27 Verifying the saved view of elements by cross-section. (TTAD #13197) ................................................ 333

2.28 Verifying the annotations dimensions when new analysis is made.(TTAD #14825) ................................. 333

2.29 Verifying the default view.(TTAD #13248) ................................................................................................ 333

2.30 Verifying the dividing of planar elements which contain openings (TTAD #12229) .................................. 333

2.31 Verifying the program behavior when trying to create lintel (TTAD #12062) ............................................ 333 2.32 Verifying the program behavior when launching the analysis on a model with overlapped loads (TTAD #11837).... 333




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2.33 Verifying the display of punctual loads after changing the load case number (TTAD #11958) .................334

2.34 Verifying the display of a beam with haunches (TTAD #12299)...............................................................334

2.35 Creating base plate connections for non-vertical columns (TTAD #12170) ..............................................334

2.36 Verifying drawing of joints in y-z plan (TTAD #12453) ..............................................................................334

2.37 Verifying rotation for steel beam with joint (TTAD #12592).......................................................................334

2.38 Verifying annotation on selection (TTAD #12700)..................................................................................... 334

3 CLIMATIC GENERATOR .................................................................................................... 335

3.1 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12808)........................................336

3.2 EC1: wind load generation on a high building with horizontal roof using UK annex (DEV2013#4.1) (TTAD #12608) .336

3.3 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) ..........................................336

3.4 EC1: Generating snow loads on a single slope with lateral parapets (TTAD #12606)................................336

3.5 EC1: Generating snow loads on a 4 slopes shed with gutters (TTAD #12528) ..........................................336

3.6 EC1: generating wind loads on a square based lattice structure with compound profiles and automaticcalculation of "n" (NF EN 1991-1-4/NA) (TTAD #12744) ...................................................................................... 337

3.7 EC1: Generating snow loads on a 4 slopes with gutters building. (TTAD #12719).....................................337

3.8 EC1: Generating snow loads on two side by side buildings with gutters (TTAD #12806)...........................337

3.9 EC1: Generating snow loads on a 4 slopes with gutters building (TTAD #12716)......................................337

3.10 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12841)......................................337

3.11 EC1: Generating wind loads on a square based structure according to UK standards (BS EN 1991-1-4:2005) (TTAD #12608)........................................................................................................................................ 338

3.12 EC1: Generating snow loads on a 2 slope building with gutters and parapets. (TTAD #12878)...............338

3.13 EC1: Generating snow loads on 2 closed building with gutters. (TTAD #12835)......................................338

3.14 EC1: generating snow loads on a 3 slopes 3D portal frame with parapets (NF EN 1991-1-3/NA) (TTAD #11111)...338

3.15 EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.3 - Wind - Example C)....338

3.16 EC1: generating wind loads on a 2 slopes 3D portal frame (NF EN 1991-1-4/NA) (VT : 3.1 - Wind - Example A) ....339

3.17 EC1: wind loads on a triangular based lattice structure with compound profiles and user defined "n" (NFEN 1991-1-4/NA) (TTAD #12276) ........................................................................................................................ 339

3.18 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)................................................................................................................................................ 339

3.19 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) ...................................................................................... 339 3.20 EC1: generating snow loads on a 2 slopes 3D portal frame (NF EN 1991-1-3/NA) (VT : 3.4 - Snow - Example A) ..339

3.21 EC1: Verifying the wind loads generated on a building with protruding roof (TTAD #12071, #12278)......340

3.22 EC1: Verifying the geometry of wind loads on an irregular shed. (TTAD #12233) ....................................340

3.23 EC1: Generating snow loads on a 4 slopes shed with parapets. (TTAD #14578).....................................340

3.24 EC1: Generating 2D snow loads on a 2 slope portal with one lateral parapet. (TTAD #14530)................340

3.25 EC1: Generating wind loads on a 2 almost horizontal slope building. (TTAD #13663) .............................340

3.26 EC1: Generating 2D wind loads on a 2 slope portal. (TTAD #14531).......................................................341

3.27 EC1: Generating wind loads on a 4 slopes shed with parapets. (TTAD #14179)......................................341 3.28 EC1: Generating snow loads on 2 side by side single roof compounds with different height (TTAD 13158).............. 341




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3.29 EC1: Generating snow loads on 2 side by side single roof compounds with different height (TTAD 13159)...............341

3.30 EC1: snow load generation on double compound with gutters and parapets on all sides.(TTAD #13717) .................341

3.31 EC1: snow load generation on building with 2 slopes > 60 degrees according to Czech national annex. (TTAD#14235) ................................................................................................................................................................. 342

3.32 EC1: Generating snow loads on 2 side by side single roof compounds (TTAD #13286).......................... 342

3.33 EC1: Generating wind loads on a 2 slope building with parapets. (TTAD #13669) .................................. 342

3.34 EC1: Generating wind loads on a 2 slope building with increased height. (TTAD #13759) ...................... 342

3.35 EC1: Generating snow loads on a 2 slope building with custom pressure values. (TTAD #14004).......... 342

3.36 EC1: wind load generation on portal with CsCd set to auto according to Romanian national annex. (TTAD #13930w)343

3.37 EC1: snow load generation on a 3 compound building according to Romanian national annex. (TTAD #13930s)...343

3.38 EC1: Generating snow loads on a 2 slope building with gutters and lateral parapets. (TTAD #14005).... 343

3.39 EC1: Generating snow loads on a 2 slope building with parapets. (TTAD #13671) ................................. 343

3.40 EC1: snow load generation on compound with a double-roof volume close to a single-roof volume (TTAD #13559)343

3.41 EC1: wind load generation on multibay canopies (TTAD #11668) ........................................................... 344

3.42 EC1: wind load generation on portal with CsCd set to auto (TTAD #12823) ............................................ 344

3.43 EC1: generating wind loads on a 35m high structure according to Eurocodes 1 - French standard withCsCd min set to 0.7 and Delta to 0.15. (TTAD #11196) ....................................................................................... 344

3.44 EC1: generating wind loads on a canopy according to Eurocodes 1 - French standard. (TTAD #13855) 344

3.45 EC1: Generating wind loads on a single-roof volume compound with parapets. (TTAD #13672) ............ 344

3.46 EC1: Generating snow loads on a shed with parapets. (TTAD #12494) .................................................. 345

3.47 EC1: Generating snow loads on a shed with gutters building. (TTAD #13856) ........................................ 345

3.48 EC1: generating snow loads on a 3 slopes 3D portal frame.(TTAD #13169) ........................................... 345 3.49 EC1: Generating snow loads on 2 side by side single roof compounds with parapets (TTAD #13992) ...345

3.50 EC1: generating Cf and Cp,net wind loads on an isolated roof with double slope (DEV2013#4.3) .......... 345

3.51 EC1: wind load generation on a high building with double slope roof using different parameters defined perdirections (DEV2013#4.2) .................................................................................................................................... 346

3.52 EC1: generating Cf and Cp,net wind loads on an multibay canopy roof (DEV2013#4.3) ......................... 346

3.53 EC1: generating Cf and Cp,net wind loads on an isolated roof with one slope (DEV2013#4.3) ............... 346

3.54 EC1: wind load generation on a high building with a horizontal roof using different CsCd values for eachdirection (DEV2013#4.4) ...................................................................................................................................... 346

3.55 EC1: generating wind loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11687).... 347 3.56 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11569)... 347

3.57 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11531) ............................................ 347

3.58 EC1: generating snow loads on a 2 slopes 3D portal frame using the Romanian national annex (TTAD #11570)... 347

3.59 EC1: generating wind loads on a 2 slopes 3D portal frame (TTAD #11699) ........................................... 348

3.60 Generating the description of climatic loads report according to EC1 Romanian standards (TTAD #11688)348

3.61 EC1: generating snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapetheight (TTAD #11943).......................................................................................................................................... 348

3.62 EC1: verifying the snow loads generated on a monopitch frame (TTAD #11302) .................................... 348

3.63 EC1: generating wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls (TTAD #11937) ................349




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3.64 EC1: generating snow loads on two close roofs with different heights according to Czech standards (CSNEN 1991-1-3) (DEV2012 #3.18) ........................................................................................................................... 349

3.65 EC1: generating wind loads on double slope 3D portal frame according to Czech standards (CSN EN1991-1-4) (DEV2012 #3.18) ................................................................................................................................. 349

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

(TTAD #11735)..................................................................................................................................................... 349 3.67 EC1: generating snow loads on a 2 slopes 3D portal frame with gutter (TTAD #11113) ..........................350

3.68 EC1: generating snow loads on a 3D portal frame with horizontal roof and gutter (TTAD #11113) ..........350

3.69 EC1: generating snow loads on duopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)........................................................................................................................................ 350

3.70 EC1: generating wind loads on a 55m high structure according to German standards (DIN EN 1991-1-4/NA) (DEV2012 #3.12)........................................................................................................................................ 350

3.71 EC1: generating snow loads on two side by side roofs with different heights, according to Germanstandards (DIN EN 1991-1-3/NA) (DEV2012 #3.13) ............................................................................................ 351

3.72 EC1: snow on a 3D portal frame with horizontal roof and parapet with height reduction (TTAD #11191).351

3.73 EC1: generating snow loads on monopitch multispan roofs according to German standards (DIN EN 1991-1-3/NA) (DEV2012 #3.13)..................................................................................................................................... 351

3.74 EC1: generating wind loads on an isolated roof with two slopes (TTAD #11695) .....................................351

3.75 EC1: generating wind loads on double slope 3D portal frame with a fully opened face (DEV2012 #1.6) .352

3.76 EC1: generating wind loads on duopitch multispan roofs with pitch < 5 degrees (TTAD #11852) ............352

3.77 EC1 NF: generating wind loads on a 3D portal frame with 2 slopes roof (TTAD #11932) ........................352

3.78 EC1: wind load generation on simple 3D portal frame with 4 slopes roof (TTAD #11604)........................352

3.79 EC1: Generating 2D wind and snow loads on a 2 opposite slopes portal with Z down axis. (TTAD #15094)............ 353

3.80 EC1: Generating wind loads on a 3 compound building. (TTAD #13190).................................................353 3.81 EC1: Generating 2D wind loads on a double slope roof with an opening. (TTAD #15328) .......................353

3.82 EC1: Generating wind loads on a double slope with 5 degrees. (TTAD #15307) .....................................353

3.83 EC1: Generating wind loads on a 2 horizontal slopes building one higher that the other. (TTAD #13320).... ...........353

3.84 EC1: Generating snow loads on a custom multiple slope building. (TTAD #14285) .................................354

3.85 EC1: Generating 2D wind loads on a 2 slope isolated roof. (TTAD #14985) ............................................354

3.86 EC1: Generating 2D wind and snow loads on a 4 slope shed next to a higher one slope compound. (TTAD #15047).354

3.87 EC1: Generating 2D snow loads on a one horizontal slope portal. (TTAD #14975) .................................354

3.88

EC1: Generating 2D wind loads on a multiple roof portal. (TTAD #15140)...............................................354

3.89 EC1: wind load generation on a signboard ............................................................................................... 355

3.90 EC1: wind load generation on a building with multispan roofs ..................................................................355

3.91 EC1: wind load generation on a high building with horizontal roof ............................................................355

3.92 EC1: wind load generation on a simple 3D portal frame with 2 slopes roof (TTAD #11602).....................355

3.93 EC1: wind load generation on a simple 3D structure with horizontal roof .................................................355

4 COMBINATIONS ................................................................................................................. 357

4.1 Generating combinations (TTAD #11673) .................................................................................................. 358

4.2 Generating load combinations with unfavorable and favorable/unfavorable predominant action (TTAD #11357) ..........358 4.3 Defining concomitance rules for two case families (TTAD #11355)............................................................358




14

4.4 Generating combinations for NEWEC8.cbn (TTAD #11431)...................................................................... 358

4.5 Generating the concomitance matrix after adding a new dead load case (TTAD #11361)......................... 358

4.6 Generating load combinations after changing the load case number (TTAD #11359) ............................... 359

4.7 Performing the combinations concomitance standard test no.7 (DEV2012 #1.7) ...................................... 359



4.10 Performing the combinations concomitance standard test no.1 (DEV2010#1.7)...................................... 360

4.11 Performing the combinations concomitance standard test no. 5 (DEV2012 #1.7).................................... 360

4.12 Performing the combinations concomitance standard test no.6 (DEV2012 #1.7)..................................... 361

4.13 Performing the combinations concomitance standard test no.4 (DEV2012 #1.7).................................... 361

4.14 Performing the combinations concomitance standard test no.9 (DEV2012 #1.7)..................................... 361

4.15 Performing the combinations concomitance standard test no.10 (DEV2012 #1.7)................................... 362

4.16 Generating a set of combinations with different Q "Base" types (TTAD #11806) ..................................... 362 4.17 Performing the combinations concomitance standard test no.3 (DEV2012 #1.7)..................................... 363

4.18 Generating a set of combinations with Q group of loads (TTAD #11960)................................................. 363

4.19 Generating the concomitance matrix after switching back the effect for live load (TTAD #11806) ........... 363

4.20 Generating a set of combinations with seismic group of loads (TTAD #11889)........................................ 363

4.21 Verifying combinations for CZ localization (TTAD #12542) ...................................................................... 363

5 CONCRETE DESIGN.......................................................................................................... 365

5.1 EC2: Verifying the transverse reinforcement area for a beam subjected to linear loads ............................ 366

5.2 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - bilinear stress-strain diagram.......... 366

5.3 Modifying the "Design experts" properties for concrete linear elements (TTAD #12498)........................... 366

5.4 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - horizontal level behavior law.... .......... 366

5.5 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - inclined stress strain behavior law... 366

5.6 EC2: Verifying the longitudinal reinforcement area for a beam subjected to point loads............................ 367

5.7 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load ..................................... 367

5.8 EC2: Verifying the minimum reinforcement area for a simply supported beam.......................................... 367

5.9 EC2 Test 2: Verifying a rectangular concrete beam subjected to a uniformly distributed load, withoutcompressed reinforcement - Bilinear stress-strain diagram ................................................................................. 368

5.10 EC2 Test 4 I: Verifying a rectangular concrete beam subjected to Pivot A efforts – Inclined stress-strain diagram.... .... 376

5.11 EC2 Test 6: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram........... 383

5.12 EC2 Test 5: Verifying a T concrete section, without compressed reinforcement - Bilinear stress-strain diagram.......... 387

5.13 EC2 Test 8: Verifying a rectangular concrete beam without compressed reinforcement – Inclined stress-strain diagram392

5.14 EC2 Test 9: Verifying a rectangular concrete beam with compressed reinforcement – Inclined stress-strain diagram.. 400

5.15 EC2 Test 3: Verifying a rectangular concrete beam subjected to uniformly distributed load, withcompressed reinforcement- Bilinear stress-strain diagram .................................................................................. 411

5.16 EC2 Test 7: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram........... 422

5.17 EC2 Test 12: Verifying a rectangular concrete beam subjected to uniformly distributed load, withoutcompressed reinforcement- Bilinear stress-strain diagram (Class XD3) .............................................................. 426




15

5.18 EC2 Test 13: Verifying a rectangular concrete beam subjected to a uniformly distributed load, withoutcompressed reinforcement - Bilinear stress-strain diagram (Class XD1)..............................................................432

5.19 EC2 Test 16: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram.... ......438

5.20 EC2 Test 17: Verifying a rectangular concrete beam subjected to a uniformly distributed load, withoutcompressed reinforcement - Inclined stress-strain diagram (Class XD1) .............................................................444

5.21 EC2 Test 20: Verifying the crack openings for a rectangular concrete beam subjected to a uniformlydistributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) .......................450

5.22 EC2 Test 23: Verifying the shear resistance for a rectangular concrete - Bilinear stress-strain diagram (Class XC1) ....457

5.23 EC2 Test 10: Verifying a T concrete section, without compressed reinforcement - Inclined stress-strain diagram........463

5.24 EC2 Test 15: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram.... ......468

5.25 EC2 Test 19: Verifying the crack openings for a rectangular concrete beam subjected to a uniformlydistributed load, without compressed reinforcement - Bilinear stress-strain diagram (Class XD1) .......................475

5.26 EC2 Test 11: Verifying a rectangular concrete beam subjected to a uniformly distributed load, withoutcompressed reinforcement- Bilinear stress-strain diagram (Class XD1)...............................................................482

5.27 EC2 Test 14: Verifying a rectangular concrete beam subjected to a uniformly distributed load, withcompressed reinforcement- Bilinear stress-strain diagram (Class XD1)...............................................................489

5.28 EC2 Test 18: Verifying a rectangular concrete beam subjected to a uniformly distributed load, withcompressed reinforcement - Bilinear stress-strain diagram (Class XD1)..............................................................495

5.29 EC2 Test 26: Verifying the shear resistance for a rectangular concrete beam with vertical transversalreinforcement - Bilinear stress-strain diagram (Class XC1)..................................................................................500


5.31 EC2 Test29: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement -Inclined stress-strain diagram (Class XC1)........................................................................................................... 509

5.32 EC2 Test30: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement -

Bilinear stress-strain diagram (Class XC1) ........................................................................................................... 513

5.33 EC2 Test 25: Verifying the shear resistance for a rectangular concrete beam with inclined transversalreinforcement - Bilinear stress-strain diagram (Class XC1)..................................................................................517

5.34 EC2 Test 46 I: Verifying a square concrete beam subjected to a normal force of traction - Inclined stress-strain diagram (Class X0) ..................................................................................................................................... 521

5.35 EC2 Test 1: Verifying a rectangular cross section beam made from concrete C25/30 to resist simplebending - Bilinear stress-strain diagram ............................................................................................................... 524

5.36 EC2 Test33: Verifying a square concrete column subjected to compression by nominal rigidity method-Bilinear stress-strain diagram (Class XC1) ........................................................................................................... 529

5.37 EC2 Test34: Verifying a rectangular concrete column subjected to compression on the top – Method

based on nominal stiffness - Bilinear stress-strain diagram (Class XC1)..............................................................536 5.38 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).............................................544


5.40 EC2 Test28: Verifying the shear resistance for a T concrete beam with inclined transversal reinforcement -Bilinear stress-strain diagram (Class X0).............................................................................................................. 557

5.41 EC2 Test31: Verifying a square concrete column subjected to compression and rotation moment to the top- Bilinear stress-strain diagram (Class XC1)......................................................................................................... 561

5.42 EC2 Test 37: Verifying a square concrete column using the simplified method – Professional rules -Bilinear stress-strain diagram (Class XC1) ........................................................................................................... 575

5.43 EC2 Test 38: Verifying a rectangular concrete column using the simplified method – Professional rules -Bilinear stress-strain diagram (Class XC1) ........................................................................................................... 579




16

5.44 EC2 Test 41: Verifying a square concrete column subjected to a significant compression force and smallrotation moment to the top - Bilinear stress-strain diagram (Class XC1).............................................................. 582

5.45 EC2 Test 42: Verifying a square concrete column subjected to a significant rotation moment and smallcompression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1) ....... 595

5.46 EC2 Test36: Verifying a rectangular concrete column using the method based on nominal curvature-

Bilinear stress-strain diagram (Class XC1)........................................................................................................... 605 5.47 EC2 Test 40: Verifying a square concrete column subjected to a small compression force and significantrotation moment to the top - Bilinear stress-strain diagram (Class XC1).............................................................. 612

5.48 EC2 Test 45: Verifying a rectangular concrete beam supporting a balcony - Bilinear stress-strain diagram(Class XC1).......................................................................................................................................................... 620

5.49 EC2 Test 46 II: Verifying a square concrete beam subjected to a normal force of traction - Bilinear stress-strain diagram (Class X0)..................................................................................................................................... 628

5.50 EC2 Test 44: Verifying a rectangular concrete beam subjected to eccentric loading - Bilinear stress-straindiagram (Class X0)............................................................................................................................................... 631

5.51 EC2 Test35: Verifying a rectangular concrete column subjected to compression to top – Based on nominalrigidity method - Bilinear stress-strain diagram (Class XC1) ................................................................................ 637

5.52 EC2 Test 39: Verifying a circular concrete column using the simplified method – Professional rules -Bilinear stress-strain diagram (Class XC1)........................................................................................................... 648

5.53 EC2 Test 43: Verifying a square concrete column subjected to a small rotation moment and significantcompression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1) ....... 652

5.54 Verifying the capacity design results according to Eurocode EC2 and EC8 French standards. (DEV2013 #8.3)......... 662

5.55 EC2 Test 47: Verifying a rectangular concrete beam subjected to tension load - Bilinear stress-straindiagram (Class XD2)............................................................................................................................................ 663

5.56 EC2 Test 4 II: Verifying a rectangular concrete beam subjected to Pivot B efforts – Inclined stress-strain diagram....... 670

5.57 Testing the punching verification and punching reinforcement results on loaded analysis model (TTAD #14332).... .... 675

5.58 Verifying the peak smoothing influence over mesh, the punching verification and punching reinforcementresults when Z down axis is selected. (TTAD #14963)......................................................................................... 675

5.59 EC2: column design with “Nominal Stiffness method” square section (TTAD #11625) ............................ 675

5.60 Verifying the longitudinal reinforcement for a horizontal concrete bar with rectangular cross section...... 675

5.61 Verifying the minimum transverse reinforcement area results for articulated beams (TTAD #11342) ...... 676

5.62 Verifying the minimum transverse reinforcement area results for an articulated beam (TTAD #11342)... 676

5.63 EC2 : calculation of a square column in traction (TTAD #11892) ............................................................. 676

5.64 Verifying Aty and Atz for a fixed concrete beam (TTAD #11812) ............................................................. 677

5.65 Verifying the reinforced concrete results on a structure with 375 load cases combinations (TTAD #11683).................... 677

5.66 Verifying the longitudinal reinforcement for linear elements (TTAD #11636)............................................ 677

5.67 Verifying the longitudinal reinforcement bars for a filled circular column (TTAD #11678)......................... 677

5.68 Verifying concrete results for planar elements (TTAD #11583) ................................................................ 678

5.69 Verifying the reinforced concrete results on a fixed beam (TTAD #11836) .............................................. 678

5.70 Verifying the longitudinal reinforcement for a fixed linear element (TTAD #11700).................................. 678

5.71 Verifying concrete results for linear elements (TTAD #11556) ................................................................. 678

5.72 Verifying the reinforcement of concrete columns (TTAD #11635) ............................................................ 679

5.73 EC2 Test 47 I: Verifying a rectangular concrete beam subjected to a tension distributed load - Bilinear

stress-strain diagram (Class XD2)........................................................................................................................ 680



1 Finite element method




18

1.1 Cantilever rectangular plate (01-0001SSLSB_FEM)

Test ID: 2433

Test status: Passed

1.1.1 Description

Verifies the vertical displacement on the free extremity of a cantilever rectangular plate fixed on one side. The plate is1 m long, subjected to a uniform planar load.

1.1.2 Background

1.1.2.1 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: = 0.3.




19

Boundary conditions

■ Outer: Fixed at end x = 0,

■ Inner: None.

Loadings

■ External: Uniform load p = -1700 Pa on the upper surface,■ Internal: None.

1.1.2.2 Displacement of the model in the linear elastic range

Reference solution

The reference displacement is calculated for the unsupported end located at x = 1m.

u =bl

4p

8EIz =

0.1 x 14 x 1700

8 x 2.1 x 1011

x0.1 x 0.005

3

12

= -9.71 cm

Finite elements modeling

■ Planar element: plate, imposed mesh,

■ 1100 nodes,

■ 990 surface quadrangles.

Deformed shape

Deformed cantilever rectangular plate Scale =1/4

01-0001SSLSB_FEM




20

1.1.2.3 Theoretical results

Solver Result name Result description Reference value

CM2 DZ Vertical displacement on the free extremity [cm] -9.71

1.1.3 Calculated results

Result name Result description Value Error

DZ Vertical displacement on the free extremity [cm] -9.58696 cm 1.27%




21

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

Test ID: 2434

Test status: Passed

1.2.1 Description

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

1.2.2 Background


■ 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 =1/330002SSLLB_FEM

4 .5 0 0 m

3 0 ° 3

0 °

4. 5 0 0

m

A A BB

CC

FF

X

Y

Z X

Y

Z

Units

I. S.

Geometry

■ Bars angle relative to horizontal: = 30°,

■ Bars length: l = 4.5 m,

■ Bar section: A = 3 x 10-4

m2.



Pa.




22

Boundary conditions

■ Outer: Hinged in A and B,

■ Inner: Hinge on C

Loading

■ External: Punctual load in C: F = -21 x 103 N.

■ Internal: None.

1.2.2.2 Displacement of the model in C

Reference solution

uc = -3 x 10-3

m


■ Linear element: beam, imposed mesh,

■ 21 nodes,

■ 20 linear elements.

Displacement shape

System of two bars with three hinges Scale =1/33

Displacement in C 0002SSLLB_FEM




23

1.2.2.3 Bars stresses

Reference solutions

AC bar = 70 MPa

BC bar = 70 MPa



■ 21 nodes,


1.2.2.4 Shape of the stress diagram

System of two bars with three hinges Scale =1/34

Bars stresses 0002SSLLB_FEM



CM2 DZ Vertical displacement in point C [cm] -0.30

CM2 Sxx Tensile stress on AC bar [MPa] 70

CM2 Sxx Tensile stress on BC bar [MPa] 70



DZ Vertical displacement in point C [cm] -0.299954 cm 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%




24

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 ownweight only.

1.3.2 Background


■ Reference: Structure Calculation Software Validation Guide, test SDLS 02/89;

■ Analysis type: modal analysis;


Thin lozenge-shaped plate fixed on one side Scale =1/1001-0008SDLSB_FEM

Units

I. S.

Geometry

■ Thickness: t = 0.01 m,

■ Side: a = 1 m,

■ = 15°

■ Points coordinates:

► A ( 0 ; 0 ; 0 )

► B ( a ; 0 ; 0 )

► C ( 0.259a ; 0.966a ; 0 )► D ( 1.259a ; 0.966a ; 0 )




25



Pa,

■ Poisson's ratio: = 0.3,

■ Density: = 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 angle

Reference solution

M. V. Barton formula for a lozenge of side "a" leads to the frequencies:

f j = 2a2

1i

2

)1(12

Et2

2

where i = 1,2, or i

2 = g().

3.601

8.872 M. V. Barton noted the sensitivity of the result relative to the mode and the angle. He acknowledged that the i values were determined with a limited development of an insufficient order, which led to consider a reference valuethat is based on an experimental result, verified by an average of seven software that use the finite elementscalculation method.



■ 961 nodes,


Eigen mode shapes




26



CM2 Eigen mode Eigen mode 1 frequency [Hz] 8.999




Eigen mode 1 frequency [Hz] 8.95 Hz -0.54%





27

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


■ Reference: Structure Calculation Software Validation Guide, test SDLL 12/89;

■ Analysis type: modal analysis, plane problem;


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,

■ Point coordinates:

► O (0 ;0),

► A (-0.05 3 ; -0.05),

► B (0.05 3 ; -0.05).




28



Pa



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(i)

► v’i = sin (i)

► ’i =1-i

2

R sin (i)

■ Antisymmetrical mode:

► u’i = i sin(i)

► v’i = -cos (i)

► ’i =1-i

2

R cos (i)

From Green’s method results:

f j =2

1 j

2R

h

12

E

with a support angle of 120°.

i 1 2 3 4

Symmetrical mode 4.8497 14.7614 23.6157

Antisymmetrical mode 1.9832 9.3204 11.8490 21.5545


■ Linear element: beam, without meshing,

■ 32 nodes,





29

Eigen mode shapes




30

1.4.2.3 Theoretic results


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 6 frequency - 6 antisymmetric 4 [Hz] 2557




Eigen mode 1 frequency - 1 antisymmetric 1 [Hz] 236.32 Hz 0.43%

Eigen mode 2 frequency - 2 symmetric 1 [Hz] 578.52 Hz 0.56%



Eigen mode 5 frequency - 5 symmetric 2 [Hz] 1760 Hz 0.51%


Eigen mode 7 frequency - 7 symmetric 3 [Hz] 2777.43 Hz -0.86%




31


Test ID: 2441

Test status: Passed

1.5.1 Description


1.5.2 Background






Units

I. S.

Geometry


■ Side: a = 1 m,

■ = 30°


► A ( 0 ; 0 ; 0 )

► B ( a ; 0 ; 0 )

► C ( 0.5a ;32

a ; 0 )

► D ( 1.5a ;32

a ; 0 )




32



Pa,



Boundary conditions


■ Inner: None.

Loading

■ External: None,

■ Internal: None.

1.5.2.2 Eigen mode frequencies relative to the angle

Reference solution


f j = 2a2

1i

2

)1(12

Et2

2

where i = 1,2, or i

2 = g().

3.961

10.19

M. V. Barton noted the sensitivity of the result relative to the mode and the angle. He acknowledged that the i values were determined with a limited development of an insufficient order, which led to consider a reference valuethat is based on an experimental result, verified by an average of seven software that use the finite elementscalculation method.



■ 961 nodes,


Eigen mode shapes




33












34


Test ID: 2439

Test status: Passed

1.6.1 Description


1.6.2 Background






Units

I. S.

Geometry


■ Side: a = 1 m,

■ = 0°


► A ( 0 ; 0 ; 0 )

► B ( a ; 0 ; 0 )

► C ( 0 ; a ; 0 )

► D ( a ; a ; 0 )




35



Pa,



Boundary conditions


■ Inner: None.

Loading

■ External: None,

■ Internal: None.


Reference solution

M. V. Barton formula for a side "a" lozenge, leads to the frequencies:

f j = 2a2

1i

2

)1(12

Et2

2

where i = 1,2, and i

2 = g().

3.492

8.525

M.V. Barton noted the sensitivity of the result relative to the mode and the angle. He acknowledged that the i values were determined with a limited development of an insufficient order, which led to consider a reference valuethat is based on an experimental result, verified by an average of seven software that use the finite elementscalculation method.



■ 61 nodes,


Eigen mode shapes




36












37

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

Test ID: 2444

Test status: Passed

1.7.1 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.

1.7.2 Background



■ Analysis type: modal analysis (plane problem);


Vibration mode of a thin piping elbow Scale = 1/11Case 2 01-0012SDLLB_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,

■ Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10

-9

m

4

,■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10

-9 m

4,

■ Polar inertia: Ip = 9.274 x 10-9

m4.




38

■ Points coordinates (in m):

► O ( 0 ; 0 ; 0 )

► A ( 0 ; R ; 0 )

► B ( R ; 0 ; 0 )

► C ( -L ; R ; 0 )

► D ( R ; -L ; 0 )



Pa,



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.


Reference solution

The Rayleigh method applied to a thin curved beam is used to determine parameters such as:

■ in plane bending:

f j =2

2i

R2

A

EIz

where i = 1,2,


■ Linear element: beam,

■ 23 nodes,


Eigen mode shapes




39



CM2 Eigen mode Eigen mode frequency in plane 1 [Hz] 94




Eigen mode frequency in plane 1 [Hz] 94.62 Hz 0.66%





40

1.8 Double fixed beam (01-0016SDLLB_FEM)

Test ID: 2448

Test status: Passed

1.8.1 Description

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

1.8.2 Background


■ Reference: internal GRAITEC test (beams theory);

■ Analysis type: static linear, modal analysis;


Units

I. S.

Geometry


■ Axial section: S=0.06 m2

■ Inertia I = 0.0001 m4



N/m2,


■ Density: = 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.




41


Reference solution

The reference vertical displacement v5, is calculated at the middle of the beam at x = 2 m.

m05079.00001.0111.2192

1650000

192

33

5

E EI

Pl v



■ 9 nodes,

■ 8 elements.

Deformed shape

Double fixed beam

Deformed

1.8.2.3 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:

S

I E

Lf nn

.

.

..2 2

2

where for the first 4 eigen modes frequencies

Hz26.228=f 8.199

Hz15.871=f 9.120

Hz8.095=f 67.61

Hz2.937=f 37.22

424

3

2

3

222

121



■ 9 nodes,

■ 8 elements.




42

Modal deformations

Double fixed beam

Mode 1

Double fixed beam

Mode 2

Double fixed beam

Mode 3

Double fixed beam

Mode 4




43



CM2 Dz Vertical displacement on the middle of the beam [m] -0.05079







Dz Vertical displacement on the middle of the beam [m] -0.0507937 m -0.01%

Eigen mode 1 frequency [Hz] 2.94 Hz 0.10%







44


Test ID: 2445

Test status: Passed

1.9.1 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.

1.9.2 Background





Vibration mode of a thin piping elbow Scale = 1/12Case 3 01-0013SDLLB_FEM

Units

I. S.




45

Geometry



■ Outer diameter: de = 0.020 m,

■ Inner diameter: di = 0.016 m,

■ Section: A = 1.131 x 10-4 m2,

■ Flexure moment of inertia relative to the y-axis: Iy = 4.637 x 10-9

m4,

■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9

m4,


m4.


► 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 10

11 Pa,



Boundary conditions

■ Outer:

► Fixed at points C and Ds,



■ Inner: None.

Loading

■ External: None,

■ Internal: None.


Reference solution



f j =2

2

i

R2

A

EIz

where i = 1,2,



■ 41 nodes,





46

Eigen mode shapes



CM2 Eigen mode Eigen mode frequency in plane 1 [Hz] 25.300

CM2 Eigen mode Eigen mode frequency in plane 2 [Hz] 27.000



Eigen mode frequency in plane 1 [Hz] 24.96 Hz -1.34%

Eigen mode frequency in plane 2 [Hz] 26.71 Hz -1.07%




47

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

Test ID: 2449

Test status: Passed

1.10.1 Description

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

1.10.2 Background




Short beam on simple supports on the neutral axis Scale = 1/6

01-0017SDLLB_FEM

Units

I. S.

Geometry

■ Height: h = 0.2 m,


■ Width: b = 0.1 m,

■ Section: A = 2 x 10-2

m4,


m4.



Pa,

■ Poisson's ratio: = 0.3,■ Density: = 7800 kg/m

3.




48

Boundary conditions

■ Outer:

► Hinged at A (null horizontal and vertical displacements),

► Simple support in B.

■ Inner: None.

Loading

■ External: None.

■ Internal: None.

1.10.2.1 Eigen modes frequencies

Reference solution

The bending beams equation gives, when superimposing, the effects of simple bending, shear force deformationsand rotation inertia, Timoshenko formula.

The reference eigen modes frequencies are determined by a numerical simulation of this equation, independent ofany software.

The eigen frequencies in tension-compression are given by:

f i =

l2

i

E where i =

2

)1i2(


■ Linear element: S beam, imposed mesh,

■ 10 nodes,





49

Eigen mode shapes




50




















51

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

Test ID: 2452

Test status: Passed

1.11.1 Description

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

1.11.2 Background





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,


■ Points coordinates in m:

► A (0 ;0 ;0)

► B (0 ;1.5 ;0)

► C (1 ;1.5 ;0)

► D (1 ;0 ;0)




52



Pa,



Boundary conditions

■ Outer:

► Simple support on all sides,

► For the modeling: hinged at A, B and D.

■ Inner: None.

Loading

■ External: None.

■ Internal: None.


Reference solution

M. V. Barton formula for a rectangular plate with supports on all four sides, leads to:

f ij =2

[ (

a

i)2 + (

b

j)2]

)1(12

Et2

2

where:

i = number of half-length of wave along y ( dimension a)

j = number of half-length of wave along x ( dimension b)


■ Planar element: shell,■ 496 nodes,

■ 450 planar elements.




53

Eigen mode shapes




54



CM2 Eigen mode Eigen mode “i" - “j” frequency, for i = 1; j = 1. [Hz] 35.63








Eigen mode "i" - "j" frequency, for i = 1; j = 1 (Mode 1) [Hz] 35.58 Hz -0.14%


Eigen mode "i" - "j" frequency, for i = 1; j = 2 (Mode 3) [Hz] 109.98 Hz 0.33%







55

1.12 Cantilever beam in Eulerian buckling (01-0021SFLLB_FEM)

Test ID: 2453

Test status: Passed

1.12.1 Description

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

1.12.2 Background


■ Reference: internal GRAITEC test (Euler theory);

■ Analysis type: Eulerian buckling;


Units

I. S.

Geometry

■ L = 10 m

■ S=0.01 m2

■ I = 0.0002 m4



N/m2,


Boundary conditions


■ Inner: None.

Loading

■ External: Punctual load P = -100000 N at x = L,

■ Internal: None.




56

1.12.2.2 Critical load on node 5

Reference solution

The reference critical load established by Euler is:

98696.0100000

98696N98696

L4

EIP

2

2

critique


■ Planar element: beam, imposed mesh,

■ 5 nodes,

■ 4 elements.

Deformed shape



CM2 Fx Critical load on node 5. [N] -98696



Fx Critical load on node 5 (mode 1) [N] -100000 N -1.32%




57


Test ID: 2443

Test status: Passed

1.13.1 Description

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

1.13.2 Background





Vibration mode of a thin piping elbow in plane Scale = 1/7

Case 1 01-0011SDLLB_FEM

Units

I. S.

Geometry





■ Section: A = 1.131 x 10-4

m2,


m4,


m4,


m4.


► O ( 0 ; 0 ; 0 )

► A ( 0 ; R ; 0 )

► B ( R ; 0 ; 0 )




58



Pa,



Boundary conditions

■ Outer: Fixed at points A and B ,

■ Inner: None.

Loading

■ External: None,

■ Internal: None.


Reference solution



f j =2

2

i

R2

A

EIz

where i = 1,2,



■ 11 nodes,


Eigen mode shapes




59












60

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

Test ID: 2447

Test status: Passed

1.14.1 Description

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

1.14.2 Background


■ Reference: internal GRAITEC test;



Units

I. S.

Geometry

■ = 1 m

■ S = 0.01 m2

■ I = 0.0001 m4



Pa,


Boundary conditions

■ Outer:

■ Fixed at ends x = 0 and x = 4 m,

■ Elastic support with k = EI/ rigidity

■ Inner: None.

Loading

■ External: Punctual load P = -10000 N at x = 2m,

■ Internal: None.




61


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:

EI EI EI EI

EI EI EI EI

EI EI EI EI

EI EI EI EI

460

260

6120

6120

00l

ES00

ES

260

460

6120

6120

00ES

-00ES

K

22

2323

22

2323

e

Given the symmetry / X and load of the structure, it is unnecessary to consider the degrees of freedom associatedwith normal work (u2, u3, u4).

The same symmetry allows the deduction of:

■ v2 = v4

■ 2 = -4

■ 3 = 0

6

5

4

3

2

1

0

0

0

0

0

4626

612612

2680

26

6120

24612

2680

26

6120

124612

2680

26

6120

24612

2646

612612

5

5

1

1

5

5

4

4

3

3

2

2

1

1

22

22

22

22

22

22

22

22

22

22

M

R

P

M

R

v

v

v

v

v

EI

33

333

333

333

33




62

The elementary rigidity matrix of the spring in its local axis system, )(

)(

11

11

6

3

5U

U EI k

, must be expressed in

the global axis system by means of the rotation matrix (90° rotation):

6

6

6

3

3

3

5

000000

010010

000000

000000

010010

000000

v

u

v

u

EI K

344332 4

3 0

826v v

3443323320

24612v v v v

y)unnecessar (usually026826

244423222 vvvv

(3)

m1011905.03

612124612 03

2

3

34243332223

EI l

P v

EI

P v v v



■ 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 m

1.14.2.3 Theoretical resultsSolver Result name Result description Reference value

CM2 Dz Vertical displacement on the middle of the beam [mm] -0.11905



Dz Vertical displacement on the middle of the beam [mm] -0.119048 mm 0.00%




63

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

Test ID: 2451

Test status: Passed

1.15.1 Description

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

1.15.2 Background





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,


► A (0 ;0 ;0)

► B (1 ;0 ;0)

► C (1 ;1 ;0)

► D (0 ;1 ;0)




64



Pa,



Boundary conditions

■ Outer: Edge AD fixed.

■ Inner: None.

Loading

■ External: None.

■ Internal: None.


Reference solution

M. V. Barton formula for a square plate with side "a", leads to:

f j =2a2

1

i

2

)1(12

Et2

2

where i = 1,2, . . .

i 1 2 3 4 5 6

i 3.492 8.525 21.43 27.33 31.11 54.44


■ Planar element: shell,

■ 959 nodes,





65

Eigen mode shapes

Thin square plate fixed on one side

Mode 1


Mode 2


Mode 3




66




















67


Test ID: 2442

Test status: Passed

1.16.1 Description


1.16.2 Background






Units

I. S.




68

Geometry


■ Side: a = 1 m,

■ = 45°


► A ( 0 ; 0 ; 0 )

► B ( a ; 0 ; 0 )

► C (2

2a ;

2

2 a ; 0 )

► D (2

22 a ;

2

2a ; 0 )



Pa,



.

Boundary conditions


■ Inner: None.

Loading

■ External: None,

■ Internal: None.


Reference solution


f j = 2a2

1i

2

)1(12

Et2

2

where i = 1,2, or i

2 = g().

4.4502

10.56

M. V. Barton noted the sensitivity of the result relative to the mode and the angle. He acknowledged that the i values were determined with a limited development of an insufficient order, which led to consider a reference valuethat is based on an experimental result, verified by an average of seven software that use the finite elements

calculation method.



■ 961 nodes,





69

Eigen mode shapes



CM2 Eigen mode Eigen mode 1 frequency [Hz] 11.1212CM2 Eigen mode Eigen mode 2 frequency [Hz] 26.3897








70

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

Test ID: 2446

Test status: Passed

1.17.1 Description

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

1.17.2 Background



■ Analysis type: modal analysis, plane problem;


Thin circular ring hang from an elastic element Scale = 1/101-0014SDLLB_FEM

Units

I. S.




71

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 ).



Pa,



Boundary conditions

■ Outer: Fixed in A,■ Inner: None.

Loading

■ External: None,

■ Internal: None.


Reference solutions

The reference solution was established from experimental results of a mass manufactured aluminum ring.

Finite elements modeling■ Linear element: beam,

■ 43 nodes,





72

Eigen mode shapes




73



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 5 Symmetrical frequency [Hz] 682.00




Eigen mode 1 Asymmetrical frequency [Hz] 28.81 Hz 0.03%

Eigen mode 2 Symmetrical frequency [Hz] 189.69 Hz 0.21%



Eigen mode 5 Symmetrical frequency [Hz] 683.9 Hz 0.28%





74

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

Test ID: 2450

Test status: Passed

1.18.1 Description

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

1.18.2 Background



■ Analysis type: modal analysis, (plane problem);


Short beam on simple supports (eccentric) Scale = 1/501-0018SDLLB_FEM

Units

I. S.

Geometry

■ Height: h = 0.2m,


■ Width: b = 0.1 m,

■ Section: A = 2 x 10-2

m4,


m4.




75



Pa,



Boundary conditions

■ Outer:

► Hinged at A (null horizontal and vertical displacements),

► Simple support at B.

■ Inner: None.

Loading

■ External: None.

■ Internal: None.


Reference solution

The problem has no analytical solution, the solution is determined by averaging several software: Timoshenko modelwith shear force deformation effects and rotation inertia. The bending modes and the traction-compression arecoupled.


■ Linear element: S beam, imposed mesh,

■ 10 nodes,


Eigen modes shape

Short beam on simple supports (eccentric)

Mode 1




76


Mode 2


Mode 3


Mode 4




77


Mode 5


















78

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 shearforce, bending moment, vertical displacement and horizontal reaction are verified.

1.19.2 Background





Slender beam on two fixed supports Scale = 1/401-0024SSLLB_FEM

Units

I. S.

Geometry

■ Length: L = 1 m,

■ Beam inertia: I = 1.7 x 10-8

m4.



Pa.




79

Boundary conditions


■ 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 – 1.26L

C



■ 5 nodes,


Results shape

Slender beam on two fixed supports Scale = 1/5

Shear force




80

1.19.2.3 Bending moment in G

Reference solution


■ Bending moment at G: MG

MG =pL2

24 - 0.045LF – 0.3C



■ 5 nodes,


Results shape


Bending moment

1.19.2.4 Vertical displacement at G

Reference solution


■ Vertical displacement at G: vG

vG =pl

4

384EI +

0.003375FL3

EI +

0.015CL2

EI



■ 5 nodes,





81

Results shape


Deformed

1.19.2.5 Horizontal reaction at A

Reference solution


■ Horizontal reaction at A: H A

H A = -0.7F1 –0.3F2



■ 5 nodes,





82



CM2 Fz Shear force in point G. [N] -540

CM2 My Bending moment in point G. [Nm] -2800

CM2 Dz Vertical displacement in point G. [cm] -4.90CM2 Fx Horizontal reaction in point A. [N] 24000



Fz Shear force in point G [N] -540 N 0.00%

My Bending moment in point G [Nm] -2800 N*m 0.00%

DZ Vertical displacement in point G [cm] -4.90485 cm -0.10%

Fx Horizontal reaction in point A [N] 24000 N 0.00%




83

1.20 Slender beam on three supports (01-0025SSLLB_FEM)

Test ID: 2457

Test status: Passed

1.20.1 Description

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

1.20.2 Background



■ Analysis type: static (plane problem);


Slender beam on three supports Scale = 1/4901-0025SSLLB_FEM

Units

I. S.

Geometry


■ Beam inertia: I = 6.3 x 10-4

m4.




84


Longitudinal elastic modulus: E = 2.1 x 1011

Pa.

Boundary conditions

■ Outer:

► Hinged at A,► Elastic support at B (Ky = 2.1 x 10

6 N/m),

► Simple support at C.

■ Inner: None.

Loading

■ External: 2 punctual loads F = Fy = -42000N.

■ Internal: None.

1.20.2.2 Bending moment at B

Reference solution

The resolution of the hyperstatic system of the slender beam leads to:

k =Ky3L

EI6

■ Bending moment at B: MB

MB = ±2

L

)k8(

F)k26(



■ 5 nodes,


Results shape

Slender beam on three supports Scale = 1/49

Bending moment




85

1.20.2.3 Reaction in B

Reference solution

■ Compression force in the spring: VB

VB =-11F

8 + k



■ 5 nodes,


1.20.2.4 Vertical displacement at B

Reference solution

■ Deflection at the spring location: vB

vB =

11F

Ky(8 + k)



■ 5 nodes,


Results shape

Slender beam on three supports

Deformed




86



CM2 My Bending moment in point B. [Nm] -63000

CM2 DZ Vertical displacement in point B. [cm] -1.00

CM2 Fz Reaction in point B. [N] -21000



My Bending moment in point B [Nm] -63000 N*m 0.00%

DZ Vertical displacement in point B [cm] -1 cm 0.00%

Fz Reaction in point B [N] -21000 N 0.00%




87

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

Test ID: 2460

Test status: Passed

1.21.1 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 outof plane displacement, torsion moment and bending moment are verified.

1.21.2 Background



■ Analysis type: static linear;


Fixed thin arc in out of plane bending Scale = 1/6

01-0028SSLLB_FEM

Units

I. S.

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.


■ Longitudinal elastic modulus: E = 2 x 1011 Pa,





88

Boundary conditions

■ Outer: Fixed at A.

■ Inner: None.

Loading

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

1.21.2.2 Displacements at B

Reference solution

Displacement out of plane at point B:

uB =FR

3

EIx [

4

+EIx KT

(34

- 2)]

where KT is the torsional rigidity for a circular section (torsion constant is 2Ix).

KT = 2GIx =EIx

1 +

uB =FR

3

EIx [

4

+ (1 + ) (3

4

- 2)]



■ 46 nodes,


1.21.2.3 Moments at = 15°

Reference solution

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

■ Bending moment: Mz’ = Mf = -FRcos



■ 46 nodes,




CM2 D Displacement out of plane in point B [m] 0.13462

CM2 Mx Torsion moment in = 15° [Nm] 74.1180

CM2 Mz Bending moment in = 15° [Nm] -96.5925



D Displacement out of plane in point B [m] 0.135156 m 0.40%

Mx Torsion moment in Theta = 15° [Nm] 74.103 N*m -0.02%

Mz Bending moment in Theta = 15° [Nm] 96.5925 N*m 0.00%




89

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

Test ID: 2461

Test status: Passed

1.22.1 Description

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

1.22.2 Background



■ Analysis type: static linear (plane problem);


Double hinged thin arc in planar bending Scale = 1/8

01-0029SSLLB_FEM

Units

I. S.

Geometry



► de = 0.02 m,

► di = 0.016 m,

► A = 1.131 x 10-4

m2,

► Ix = 4.637 x 10-9

m4.




90


■ Longitudinal elastic modulus: E = 2 x 1011

Pa,


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.

1.22.2.2 Displacements at A, B and C

Reference solution

■ Rotation about z-axis

A = - B = (2 - 1)

FR22EI

■ Displacement;

Vertical at C: vC =8

FREA

+ (34

- 2)FR

3

2EI

Horizontal at B: uB =FR2EA

-FR

3

2EI



■ 37 nodes,■ 36 linear elements.




91

Displacements shape

Fixed thin arc in planar bending Scale = 1/11

Deformed



CM2 RY Rotation about Z-axis in point A [rad] 0.030774

CM2 RY Rotation about Z-axis in point B [rad] -0.030774

CM2 DZ Vertical displacement in point C [cm] -1.9206

CM2 DX Horizontal displacement in point B [cm] 5.3912



RY Rotation about Z-axis in point A [rad] 0.0307785 Rad 0.01%

RY Rotation about Z-axis in point B [rad] -0.0307785 Rad -0.01%

DZ Vertical displacement in point C [cm] -1.92019 cm 0.02%

DX Horizontal displacement in point B [cm] 5.386 cm -0.10%




92

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

Test ID: 2455

Test status: Passed

1.23.1 Description

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

1.23.2 Background





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,

■ Section: A = 1.392 x 10-4

m2,


m4,


A B C D E F

x -0.30 0.30 -0.30 0.30 -0.30 0.30y 0 0 0.36 0.36 0.81 0.81




93



Pa,



Boundary conditions


■ Inner: None.

Loading

■ External: None.

■ Internal: None.


Reference solution

Dynamic radius method (slender beams theory).



■ 60 nodes,


Deformed shape

Bending effects of a symmetrical portal frame Scale = 1/7

Mode 13




94



CM2 Eigen mode Eigen mode 1 antisymmetric frequency [Hz] 8.8


CM2 Eigen mode Eigen mode 3 symmetric frequency [Hz] 43.8CM2 Eigen mode Eigen mode 4 symmetric frequency [Hz] 56.3


CM2 Eigen mode Eigen mode 6 symmetric frequency [Hz] 102.6




CM2 Eigen mode Eigen mode 10 antisymmetric frequency [Hz] 206


CM2 Eigen mode Eigen mode 12 antisymmetric frequency [Hz] 320

CM2 Eigen mode Eigen mode 13 symmetric frequency [Hz] 335



Eigen mode 1 antisymmetric frequency [Hz] 8.78 Hz -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 6 symmetric frequency [Hz] 102.7 Hz 0.10%Eigen mode 7 antisymmetric frequency [Hz] 147.08 Hz -0.01%






Eigen mode 13 symmetric frequency [Hz] 334.96 Hz -0.01%




95

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

Test ID: 2459

Test status: Passed

1.24.1 Description

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

1.24.2 Background






01-0027SSLLB_FEM

Units

I. S.

Geometry



► de = 0.02 m,

► di = 0.016 m,

► A = 1.131 x 10-4

m2,

► Ix = 4.637 x 10-9

m4.




96


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.

1.24.2.2 Displacements at B

Reference solution

At point B:

■ displacement parallel to Ox: u =R

2

4EI [F1R + 2F2R + 4Mz]

■ displacement parallel to Oy: v =R

2

4EI [2F1R + (3 - 8)F2R + 2( - 2)Mz]

■ rotation around Oz: =R

4EI [4F1R + 2( - 2)F2R + 2Mz]







97

Results shape


Deformed


CM2 DX Horizontal displacement in point B [m] 0.3791

CM2 DZ Vertical displacement in point B [m] 0.2417

CM2 RY Rotation about Z-axis in point B [rad] -0.1654



DX Horizontal displacement in point B [m] 0.378914 m -0.05%

DZ Vertical displacement in point B [m] 0.241738 m 0.02%

RY Rotation about Z-axis in point B [rad] -0.165362 Rad 0.02%




98

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

Test ID: 2464

Test status: Passed

1.25.1 Description

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

1.25.2 Background





Beam on elastic soil, free ends Scale = 1/21

01-0032SSLLB_FEM

Units

I. S.

Geometry

■ L = ( 10 )/2,

■ I = 10-4

m4.



Pa.




99

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.

1.25.2.2 Bending moment and displacement at C

Reference solution

=4

K/(4EI)

= L/2

= sh (2) + sin (2)

■ Bending moment:

MC = (F/(4))(ch(2) - cos (2) – 8sh()sin())/

■ Vertical displacement:

vC = - (F/(2K))( ch(2) + cos (2) + 8ch()cos() + 2)/



■ 72 nodes,


Bending moment diagram

Beam on elastic soil, free ends Scale = 1/20

Bending moment




100

1.25.2.3 Displacements at A

Reference solution


v A = (2F/K)( ch()cos() + ch(2) + cos(2))/

■ Rotation about z-axis A = (-2F2

/K)( sh()cos() - sin()ch() + sh(2) - sin(2))/



■ 72 nodes,

■ 71 linear elements



CM2 My Bending moment in point C [Nm] 5759

CM2 Dz Vertical displacement in point C [m] -0.006844CM2 Dz Vertical displacement in point A [m] -0.007854

CM2 RY Rotation about z-axis in point A [rad] -0.000706



My Bending moment in point C [Nm] 5779.54 N*m 0.36%

Dz Vertical displacement in point C [m] -0.00684369 m 0.00%

Dz Vertical displacement in point A [m] -0.00786073 m -0.09%

RY Rotation Theta about z-axis in point A [rad] -0.000707427 Rad -0.20%




101

1.26 EDF Pylon (01-0033SFLLA_FEM)

Test ID: 2465

Test status: Passed

1.26.1 Description

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

1.26.2 Background


■ Reference: Internal GRAITEC test;

■ Analysis type: static linear, Eulerian buckling;

■ Element type: linear

Units

I. S.

Geometry



Pa,


Boundary conditions

■ Outer:

► Hinged support,

► For the modeling, a fixed restraint and 4 beams were added at the pylon supports level.

■ Inner: None.




102

Loading

■ External:

Punctual loads corresponding to a wind load.

► FX = 165550 N, FY = - 1240 N, FZ = - 58720 N on the mainarms,

► 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.


Reference solution




103

Software ANSYS 5.3 NE/NASTRAN 7.0

Max deflection (m) 0.714 0.714

dominating mode 2.77 2.77



■ 402 nodes,

■ 1034 elements.

Deformed shape




104

Buckling modal deformation (dominating mode)



CM2 D Displacement at the top of the pylon [m] 0.714

CM2 Dominating buckling - critical , mode 4 [Hz] 2.77



D 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%




105

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

Test ID: 2463

Test status: Passed

1.27.1 Description

Verifies the horizontal and the vertical displacement in several points of a truss with hinged bars, subjected to apunctual load.

1.27.2 Background





Truss with hinged bars under a punctual load Scale = 1/1001-0031SSLLB_FEM

Units

I. S.

Geometry

Elements Length (m) Area (m2)

AC 0.5 2 2 x 10-4

CB 0.5 2 2 x 10-4

CD 2.5 1 x 10-4

BD 2 1 x 10-4




106



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.

1.27.2.2 Displacements at C and D

Reference solution

Displacement method.


■ Linear element: beam,■ 4 nodes,


Displacements shape

Truss with hinged bars under a punctual load Scale = 1/9Deformed




107



CM2 DX Horizontal displacement in point C [mm] 0.26517

CM2 DX Horizontal displacement in point D [mm] 3.47902

CM2 DZ Vertical displacement in point C [mm] 0.08839

CM2 DZ Vertical displacement in point D [mm] -5.60084



DX Horizontal displacement in point C [mm] 0.264693 mm -0.18%

DX Horizontal displacement in point D [mm] 3.47531 mm -0.11%

DZ Vertical displacement in point C [mm] 0.0881705 mm -0.25%

DZ Vertical displacement in point D [mm] -5.595 mm 0.10%




108

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

Test ID: 2454

Test status: Passed

1.28.1 Description

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

1.28.2 Background




■ 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


Pa,



Boundary conditions

■ Outer: Fixed on a hub at any point r = Ri.

■ Inner: None.




109

Loading

■ External: None.

■ Internal: None.


Reference solution

The solution of determining the frequency based on Bessel functions leads to the following formula:

f ij =1

2Re2 ij

2

Et2

12(1-2)

where:

i = the number of nodal diameters

j = the number of nodal circles

and ij2 such as:

j \ i 0 1 2 3

0 13.0 13.3 14.7 18.51 85.1 86.7 91.7 100


■ Planar element: plate,

■ 360 nodes,













Result name Result description Value ErrorEigen mode "i" - "j" frequency, for i = 0; j = 0 (Mode 1) [Hz] 79.05 Hz -0.26%





Eigen mode “i" - “j” frequency, for i = 2; j = 1 (Mode 22) [Hz] 552.43 Hz -1.19%






110

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

Test ID: 2458

Test status: Passed

1.29.1 Description

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

1.29.2 Background





Fixed beams connected to a stiff element Scale = 1/10

01-0026SSLLB_FEM

Units

I. S.

Geometry

■ Lengths:

► L = 2 m,

► l = 0.2 m,

■ Beams inertia moment: I = (4/3) x 10-8

m4,

■ The beam sections are squared, of side: 2 x 10-2

m.



Pa.




111

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.

1.29.2.2 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:

vB = vD =FL

3

24EI



■ 4 nodes,


Results shape

Fixed beams connected to a stiff element Scale = 1/10

Deformed

1.29.2.3 Vertical reaction at A and C

Reference solution

Analytical solution.



■ 4 nodes,





112

1.29.2.4 Bending moment at A and C

Reference solution




■ 4 nodes,




CM2 D Deflection in point B [m] 0.125

CM2 D Deflection in point D [m] 0.125

CM2 Fz Vertical reaction in point A [N] -500

CM2 Fz Vertical reaction in point C [N] -500

CM2 My Bending moment in point A [Nm] 500CM2 My Bending moment in point C [Nm] 500



D Deflection in point B [m] 0.125376 m 0.30%

D Deflection in point D [m] 0.125376 m 0.30%

Fz Vertical reaction in point A [N] -500 N 0.00%

Fz Vertical reaction in point C [N] -500 N 0.00%

My Bending moment in point A [Nm] 500.083 N*m 0.02%

My Bending moment in point C [Nm] 500.083 N*m 0.02%




113

1.30 Portal frame with lateral connections (01-0030SSLLB_FEM)

Test ID: 2462

Test status: Passed

1.30.1 Description

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

1.30.2 Background





Portal frame with lateral connections Scale = 1/21

01-0030SSLLB_FEM

Units

I. S.




114

Geometry

Beam Length Moment of inertia

AB l AB = 4 mI AB =

643

x 10-8

m4

AC l AC = 1 mI AC =

112

x 10-8

m4

AD l AD = 1 mI AD =

112

x 10-8

m4

AE l AE = 2 mI AE =

43 x 10

-8 m

4

■ G is in the middle of DA.

■ The beams have square sections:

► A AB = 16 x 10-4

m

► A AD = 1 x 10-4

m

► A AC = 1 x 10-4

m

► A AE = 4 x 10-4

m


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 = - 10

3 N/m.

■ Internal: None.

1.30.2.2 Displacements at A

Reference solution

Rotation at A about z-axis:

We say: k An =EI An

l An where n = B, C, D or E

K = k AB + k AD + k AE +34 k AC

r An =k An

K

C1 =Fl AD

8 -

pl AB2

12

=C1

4K



■ 6 nodes,





115

Displacements shape

Portal frame with lateral connections

Deformed

1.30.2.3 Moments in A

Reference solution

■ M AB =pl AB

2

12 + r AB x C1

■ M AD = -Fl AD

8 + r AD x C1

■ M AE = r AE x C1

■ M AC = r AC x C1



■ 6 nodes,




CM2 RY Rotation about z-axis in point A [rad] -0.227118

CM2 My Bending moment in point A (M AB) [Nm] 11023.72

CM2 My Bending moment in point A (M AC) [Nm] 113.559

CM2 My Bending moment in point A (M AD) [Nm] 12348.588

CM2 My Bending moment in point A (M AE) [Nm] 1211.2994



RY Rotation Theta about z-axis in point A [rad] -0.227401 Rad -0.12%

My Bending moment in point A (Moment AB) [Nm] 11021 N*m -0.02%

My Bending moment in point A (Moment AC) [Nm] 113.704 N*m 0.13%

My Bending moment in point A (Moment AD) [Nm] 12347.5 N*m -0.01%

My Bending moment in point A (Moment AE) [Nm] 1212.77 N*m 0.12%




116

1.31 Caisson beam in torsion (01-0037SSLSB_FEM)

Test ID: 2468

Test status: Passed

1.31.1 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.

1.31.2 Background


■ Reference: Structure Calculation Software Validation Guide, test SSLS 05/89;



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.



Pa,





117

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.

1.31.2.2 Displacement and stress at two points

Reference solution

The reference solution is determined by averaging the results of several calculation software with implemented finiteelements 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).



■ 90 nodes,


Deformed shape

Caisson beam in torsion Scale = 1/4

Deformed




118



CM2 D Displacement in point A [m] -0.617 x 10-6

CM2 D Displacement in point B [m] -0.987 x 10-6

CM2 RY Rotation about Z-axis in point A [rad] 0.123 x 10-4

CM2 RY Rotation about Z-axis in point B [rad] 0.197 x 10-4

CM2 sxy_mid xy stress in point A [MPa] -0.11

CM2 sxy_mid xy stress in point B [MPa] -0.11



D Displacement in point A [µm] 0.615909 µm -0.18%

D Displacement in point B [µm] 0.986806 µm -0.02%

RY Rotation about Z-axis in point A [rad] -1.23211e-005 Rad -0.17%RY 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%




119

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

Test ID: 2469

Test status: Passed

1.32.1 Description

Verifies the stress, the radial deformation and the longitudinal deformation of a cylinder loaded with a uniform internalpressure.

1.32.2 Background



■ Analysis type: static elastic;


Thin cylinder under a uniform radial pressure Scale = 1/18

01-0038SSLSB_FEM

Units

I. S.

Geometry


■ Radius: R = 1 m,

■ Thickness: h = 0.02 m.



Pa,





120

Boundary conditions

■ Outer:

► Free conditions

► For the modeling, only ¼ of the cylinder is considered andthe symmetry conditions are applied. On the other side, we restrained the displacements at a fewnodes in order to make the model stable.

■ Inner: None.

Loading

■ External: Uniform internal pressure: p = 10000 Pa,

■ Internal: None.

1.32.2.2 Stresses in all points

Reference solution

Stresses in the planar elements coordinate system (x axis is parallel with the length of the cylinder):

■ xx = 0

■ yy = pRh



■ 209 nodes,


1.32.2.3 Cylinder deformation in all points

■ Radial deformation:

R =pR2Eh

■ Longitudinal deformation:

L =-pRL

Eh



CM2 syy_mid yy stress in all points [Pa] 500000.000000

CM2 Dz L radial deformation of the cylinder in all points [µm] 2.380000

CM2 DY L longitudinal deformation of the cylinder in all points [µm] -2.860000


Resultname

Result description Value Error

syy_mid Sigma yy stress in all points [Pa] 499521 Pa -0.10%

Dz Delta R radial deformation of the cylinder in all points [µm] 2.39213 µm 0.51%

DY Delta L longitudinal deformation of the cylinder in all points [µm] -2.85445 µm 0.19%




121

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

Test ID: 2472

Test status: Passed

1.33.1 Description

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

1.33.2 Background





Units

I. S.

Geometry

l = 300 cm

h = 20.04 cmb= 20.04 cm

tw = 1.46 cm

tf = 0.96 cm

Sx= 74.95 cm2

Iz = 5462 cm4

Sy = 16.43 cm2


■ Longitudinal elastic modulus: E = 2285938 daN/cm2,

■ Transverse elastic modulus G = 879207 daN/cm2





122

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.

1.33.2.2 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.

cm017.1105.0912.043.16

3.012

22859384

30020246

5462228593848

30020246

448

33

6

x x

x

x x

x

GS

Pl

EI

Pl v

shear

y

flexion

z


■ Planar element: S beam, imposed mesh,

■ 11 nodes,


Deformed shape




123



CM2 DZ Vertical displacement at node 6 [cm] -1.017



DZ Vertical displacement at node 6 [cm] -1.01722 cm -0.02%




124

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

Test ID: 2473

Test status: Passed

1.34.1 Description

Verifies the stress, the longitudinal deformation and the radial deformation of a cylinder under a uniform axial load.

1.34.2 Background



■ Analysis type: static elastic;


Thin cylinder under a uniform axial load Scale = 1/19

01-0042SSLSB_FEM

Units

I. S.

Geometry



■ Radius: R = 1 m.



Pa,





125

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.

1.34.2.2 Stress in all points

Reference solution

x axis of the local coordinate system of planar elements is parallel to the cylinders axis.

xx =qh

yy = 0


■ Planar element: shell, imposed mesh,

■ 697 nodes,


1.34.2.3 Cylinder deformation at the free end

Reference solution

■ L longitudinal deformation of the cylinder:

L = qLEh

■ R radial deformation of the cylinder:

R =-qREh



■ 697 nodes,





126

Deformation shape

Thin cylinder under a uniform axial load Scale = 1/22

Deformation shape



CM2 sxx_mid xx stress at all points [Pa] 5 x 105

CM2 syy_mid yy stress at all points [Pa] 0

CM2 DY L longitudinal deformation at the free end [m] 9.52 x 10-6

CM2 Dz R radial deformation at the free end [m] -7.14 x 10

-7



sxx_mid 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%

DY Delta L longitudinal deformation at the free end [mm] -0.00952381 mm -0.04%

Dz Delta R radial deformation at the free end [mm] 0.000710887 mm -0.44%




127

1.35 Simply supported square plate (01-0036SSLSB_FEM)

Test ID: 2467

Test status: Passed

1.35.1 Description

Verifies the vertical displacement in the center of a simply supported square plate.

1.35.2 Background





Simply supported square plate Scale = 1/9

01-0036SSLSB_FEM

Units

I. S.

Geometry

■ Side = 1 m,

■ Thickness h = 0.01m.



Pa,






128

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.

1.35.2.2 Vertical displacement at O

Reference solution

According to Love- Kirchhoff hypothesis, the displacement w at a point (x,y):

w(x,y) = wmnsinmxsinny

where wmn =

192g(1 - 2)

mn(m2 + n2)6Eh2



■ 441 nodes,


Deformed shape

Simply supported square plate Scale = 1/6

Deformed




129



CM2 Dz Vertical displacement in point O [m] -0.158



Dz Vertical displacement in point O [µm] -0.164901 µm -4.37%




130

1.36 Stiffen membrane (01-0040SSLSB_FEM)

Test ID: 2471

Test status: Passed

1.36.1 Description

Verifies the horizontal displacement and the stress on a plate (8 x 12 cm) fixed in the middle on 3 supports with apunctual load at its free node.

1.36.2 Background


■ Reference: Klaus-Jürgen Bathe - Finite Element Procedures in Engineering Analysis, Example 5.13;


■ Element type: planar (membrane).

1;1,

Units

I. S.

Geometry

■ Thickness: e = 0.1 cm,

■ Length: l = 8 cm,

■ Width: B = 12 cm.


■ Longitudinal elastic modulus: E = 30 x 106 N/cm

2,


Boundary conditions

■ Outer: Fixed on 3 sides,

■ Inner: None.

Loading

■ External: Uniform load Fx = F = 6000 N at A,

■ Internal: None.




131

1.36.2.2 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.

MPa96.17 N/cm1796;1

MPa98.8 N/cm898;0

0;1

for181

MPa55.11 N/cm1155;1

MPa77.5 N/cm577;0

0;1

for

MPa49.38 N/cm3849;1

MPa24.19 N/cm1924;0

0;1

for121

3410.97510.367410.2

6000

21

1

1

2

3

2

xy1

2

xy1

xy1

1

2

yy1

2

yy1

yy1

11

2

xx1

2

xx1

xx1

21

4

66

222

b

u E

a

u E

cm

a

ES

ba

eabE

F

K

F u

A xy

xx yy

A xx

A


■ Planar element: membrane, imposed mesh,

■ 6 nodes,

■ 2 quadrangle planar elements and 1 bar.

Deformed shape




132



CM2 DX Horizontal displacement Element 1 in A [cm] 9.340000

CM2 sxx_mid xx stress Element 1 in y = 0 cm [MPa] 38.490000

CM2 sxx_mid xx stress Element 1 in y = 6 cm [MPa] 0

CM2 syy_mid yy stress Element 1 in y = 0 cm [MPa] 11.550000

CM2 syy_mid yy stress Element 1 in y = 6 cm [MPa] 0

CM2 sxy_mid xy stress Element 1 in x = 0 cm [MPa] 0

CM2 sxy_mid xy stress Element 1 in x = 4 cm [MPa] -8.980000

CM2 sxy_mid xy stress Element 1 in x = 8 cm [MPa] -17.960000



DX 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%




133

1.37 Torus with uniform internal pressure (01-0045SSLSB_FEM)

Test ID: 2476

Test status: Passed

1.37.1 Description

Verifies the stress and the radial deformation of a torus with uniform internal pressure.

1.37.2 Background



■ Analysis type: static, linear elastic;


Torus with uniform internal pressure

01-0045SSLSB_FEM

Units

I. S.

Geometry


■ Transverse section radius: b = 1 m,

■ Average radius of curvature: a = 2 m.


■ Longitudinal elastic modulus: E = 2.1 x 1011 Pa,





134

Boundary conditions

■ Outer: For the modeling, only1/8 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.

1.37.2.2 Stresses

Reference solution

(See stresses description on the first scheme of the overview)

If a – b r a + b

11 =pb2h

r + a

r

22 =pb2h



■ 361 nodes,


1.37.2.3 Cylinder deformation

Reference solution

■ R radial deformation of the torus:

R = pb2Eh

(r - (r + a))



■ 361 nodes,





135



CM2 syy_mid 11 stresses for r = a - b [Pa] 7.5 x 105

CM2 syy_mid 11 stresses for r = a + b [Pa] 4.17 x 105

CM2 sxx_mid 22 stress for all r [Pa] 2.50 x 105

CM2 Dz L radial deformations of the torus for r = a - b [m] 1.19 x 10

-7

CM2 Dz L radial deformations of the torus for r = a + b [m] 1.79 x 10-6



syy_mid Sigma 11 stresses for r = a - b [Pa] 742770 Pa -0.96%

syy_mid 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%

Dz Delta L radial deformations of the torus for r = a - b [mm] -0.000117352 mm 1.38%

Dz Delta L radial deformations of the torus for r = a + b [mm] 0.00180274 mm 0.71%




136

1.38 Spherical shell under internal pressure (01-0046SSLSB_FEM)

Test ID: 2477

Test status: Passed

1.38.1 Description

A spherical shell is subjected to a uniform internal pressure. The stress and the radial deformation are verified.

1.38.2 Background





Spherical shell under internal pressure

01-0046SSLSB_FEM

Units

I. S.

Geometry


■ Radius: R2 = 1 m,

■ = 90° (hemisphere).




137



Pa,


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 restrainsplaced in the vertical plane xy in translation along z and in rotation along x and y). In addition, the node at thetop 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.

1.38.2.2 Stresses

Reference solution

(See stresses description on the first scheme of the overview)

If 0° 90°

11 = 22 =pR2

2

2h



■ 343 nodes,



Reference solution

■ R radial deformation of the calotte:

R =pR22 (1 - ) sin

2Eh



■ 343 nodes,





138

Deformed shape

Spherical shell under internal pressure Scale = 1/11

Deformed



CM2 sxx_mid 11 stress for all [Pa] 2.50 x 105

CM2 syy_mid

22 stress for all [Pa]2.50 x 10

5

CM2 Dz R radial deformations for = 90° [m] 8.33 x 10-7



sxx_mid Sigma 11 stress for all Theta [Pa] 250202 Pa 0.08%

syy_mid Sigma 22 stress for all Theta [Pa] 249907 Pa -0.04%

Dz Delta R radial deformations for Theta = 90° [mm] 0.000832794 mm -0.02%




139

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

Test ID: 2475

Test status: Passed

1.39.1 Description

Verifies the stress, the longitudinal deformation and the radial deformation of a thin cylinder subjected to its selfweight only.

1.39.2 Background





A cylinder of R radius and L length subject of self weight only.

Thin cylinder under its self weight Scale = 1/24

01-0044SSLSB_FEM

Units

I. S.

Geometry






Pa,


■ Density: = 7.85 x 104

N/m3.




140

Boundary conditions

■ Outer:

► Null axial displacement at z = 0,

► For the modeling, we consider only a quarter of the cylinder, so we impose the symmetryconditions on the nodes that are parallel with the cylinder’s axis.

■ Inner: None.

Loading

■ External: Cylinder self weight,

■ Internal: None.

1.39.2.2 Stresses

Reference solution


xx = z

yy = 0



■ 697 nodes,



Reference solution


L =z

2

2E

■ R radial deformation of the cylinder:

R =-Rz

E



CM2 sxx_mid xx stress for z = L [Pa] -314000.000000

CM2 DY L longitudinal deformation for z = L [mm] 0.002990

CM2 Dz R 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.



sxx_mid Sigma xx stress for z = L [Pa] -309143 Pa 1.55%

DY Delta L longitudinal deformation for z = L [mm] 0.00298922 mm -0.03%

Dz Delta R radial deformation for z = L [mm] -0.000443587 mm -0.82%




141

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

Test ID: 2466

Test status: Passed

1.40.1 Description

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

1.40.2 Background





Beam on elastic soil, hinged ends Scale = 1/2701-0034SSLLB_FEM

Units

I. S.

Geometry

■ L = ( 10 )/2,

■ I = 10-4

m4.




142



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: f y = p = - 5000 N/m,

► Torque at A: Cz = -C = -15000 Nm,

► Torque at B: Cz = C = 15000 Nm.

■ Internal: None.

1.40.2.2 Displacement and support reaction at A

Reference solution

=4

K/(4EI)

= L/2

= ch(2) + cos(2)

■ Vertical support reaction:

V A = -p(sh(2) + sin(2)) - 2Fch()cos() + 22C(sh(2) - sin(2)) x

1

2

■ Rotation about z-axis:

A = p(sh(2) – sin(2)) + 2Fsh()sin() - 22C(sh(2) + sin(2)) x

1

(K/)



■ 50 nodes,





143

Deformed shape

Beam on elastic soil, hinged ends Scale = 1/20

Deformed

1.40.2.3 Displacement and bending moment at D

Reference solution


vD = 2p( - 2ch()cos()) + F(sh(2) – sin(2)) - 82Csh()sin() x

1

2K

■ Bending moment:

MD = 4psh()sin() + F(sh(2) + sin(2)) - 82Cch()cos() x

1

42



■ 50 nodes,





144

Bending moment diagram

Beam on elastic soil, hinged ends Scale = 1/20

Bending moment



CM2 RY Rotation around z-axis in point A [rad] 0.003045

CM2 Fz Vertical reaction in point A [N] -11674

CM2 Dz Vertical displacement in point D [cm] -0.423326

CM2 My Bending moment in point D [Nm] -33840



RY Rotation around z-axis in point A [rad] 0.00304333 Rad -0.05%

Fz Vertical reaction in point A [N] -11709 N -0.30%

Dz Vertical displacement in point D [cm] -0.423297 cm 0.01%

My Bending moment in point D [Nm] -33835.9 N*m 0.01%




145

1.41 Square plate under planar stresses (01-0039SSLSB_FEM)

Test ID: 2470

Test status: Passed

1.41.1 Description

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

1.41.2 Background




■ Element type: planar (membrane).

Square plate under planar stresses Scale = 1/19

Modeling

1;1,

Units

I. S.

Geometry

■ Thickness: e = 1 m,

■ 4 square elements of side h = 1 m.



Pa,





146

Boundary conditions

■ Outer: Fixed on 3 sides,

■ Inner: None.

Loading

■ External: Uniform load p = -1. 108 N/ml on the upper surface,

■ Internal: None.


Reference solution

The reference displacements are calculated on nodes 7 and 9.

v9 =-6ph(3 + )(1 - 2

)

E(8(3 - )2 - (3 + )

2) = -0.1809 x 10

-3 m,

v7 =4(3 - )

3 + v9 = -0.592 x 10-3

m,

For element 1.4:

(For the stresses calculated above, the abscissa point (x = 0; y = 0) corresponds to node 8.)

yy =E

1 - 2 (v9 - v7)

2h (1 + ) for

xx = yy for

xy =E

1 + (v9 + v7) + (v9 - v7)

4h (1 + ) for


■ Planar element: membrane, imposed mesh,

■ 9 nodes,


= -1 ; xx = 0

= 0 ; xx = -14.23 MPa

= 1 ; xx = -28.46 MPa

= -1 ; = 0 ; xy = -47.82 MPa

= 0 ; = 0 ; xy = -31.21 MPa

= 1 ; = 0 ; xy = -14.61 MPa

= -1 ; yy = 0

= 0 ; yy = -47.44 MPa

= 1 ; yy = -94.88 MPa




147

Deformed shape



CM2 DZ Vertical displacement on node 7 [mm] -0.592

CM2 DZ Vertical displacement on node 5 [mm] -0.1809

CM2 sxx_mid xx stresses on Element 1.4 in x = 0 m [MPa] 0

CM2 sxx_mid xx stresses on Element 1.4 in x = 0.5 m [MPa] -14.23

CM2 sxx_mid xx stresses on Element 1.4 in x = 1 m [MPa] -28.46

CM2 syy_mid yy stresses on Element 1.4 in x = 0 m [MPa] 0

CM2 syy_mid yy stresses on Element 1.4 in x = 0.5 m [MPa] -47.44

CM2 syy_mid yy stresses on Element 1.4 in x = 1 m [MPa] -94.88

CM2 sxy_mid xy stresses on Element 1.4 in y = 0 m [MPa] -14.66

CM2 sxy_mid xy stresses on Element 1.4 in y = 1 m [MPa] -47.82



DZ Vertical displacement on node 7 [mm] -0.59203 mm -0.01%DZ Vertical displacement on node 5 [mm] -0.180898 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%




148

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

Test ID: 2474

Test status: Passed

1.42.1 Description

Verifies the stress, the longitudinal deformation and the radial deformation of a thin cylinder under a hydrostaticpressure.

1.42.2 Background





Thin cylinder under a hydrostatic pressure Scale = 1/25

01-0043SSLSB_FEM

Units

I. S.

Geometry






Pa,





149

Boundary conditions

■ Outer: For the modeling, we consider only a quarter of the cylinder, so we impose thesymmetry conditions on the nodes that are parallel with the cylinder’s axis.

■ Inner: None.

Loading

■ External: Radial internal pressure varies linearly with the "p" height, p = p0 zL ,

■ Internal: None.

1.42.2.2 Stresses

Reference solution


xx = 0

yy =p0Rz

Lh



■ 209 nodes,



Reference solution


L =

-p0Rz2

2ELh

■ L radial deformation of the cylinder:

R =p0R2zELh



■ 209 nodes,





150

Deformation shape

Thin cylinder under a hydrostatic pressure

Deformed



CM2 syy_mid yy stress in z = L/2 [Pa] 500000.000000

CM2 DY L longitudinal deformation of the cylinder at the inferiorextremity [mm]

-0.002860

CM2 Dz L radial deformation of the cylinder in z = L/2 [mm] 0.002380



syy_mid Sigma yy stress in z = L/2 [Pa] 504489 Pa 0.90%

DY Delta L longitudinal deformation of the cylinder at the inferiorextremity [mm]

-0.00285442 mm 0.20%

Dz Delta L radial deformation of the cylinder in z = L/2 [mm] 0.00238372 mm 0.16%




151

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 theexternal meridian stresses are verified.

1.43.2 Background





Spherical dome under a uniform external pressure

01-0050SSLSB_FEM

Units

I. S.




152

Geometry

■ Radius: a = 2.54 m,


■ Angle: = 75°.



Pa,


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 finiteelements method. 2% uncertainty about the reference solution.



■ 401 nodes,


Deformed shape




153



CM2 DX Horizontal displacements in = 15° 1.73 x 10-3

CM2 DX Horizontal displacements in = 45° -1.02 x 10-3

CM2 syy_mid yy external meridian stresses in = 15° -74

CM2 sxx_mid XX external meridian stresses in = 45° -68



DX Horizontal displacements in Psi = 15° [mm] 1.73064 mm 0.04%

DX 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%




154

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

Test ID: 2481

Test status: Passed

1.44.1 Description

A square plate simply supported is subjected to a uniform load. The vertical displacement and the bending momentsat the plate center are verified.

1.44.2 Background





Simply supported square plate under a uniform load Scale = 1/9

01-0051SSLSB_FEM

Units

I. S.

Geometry

■ Side: a =b = 1 m,








155

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.

1.44.2.2 Vertical displacement and bending moment at the center of the plate

Reference solution

Love-Kirchhoff thin plates theory.



■ 361 nodes,


1.44.2.3 Theoretical result


CM2 DZ 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



DZ 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%




156

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

Test ID: 2484

Test status: Passed

1.45.1 Description

A rectangular plate simply supported is subjected to a punctual force and moments. The vertical displacement isverified.

1.45.2 Background





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,


■ Longitudinal elastic modulus: E =1000 Pa,


Boundary conditions

■ Outer: Punctual support at A, B and D (null displacement along z-axis),

■ Inner: None.




157

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.

1.45.2.2 Vertical displacement at C

Reference solution




■ 867 nodes,


Deformed shape

Simply supported rectangular plate loaded with punctual force and moments

Deformed



CM2 Dz Vertical displacement in point C [m] -12.480



Dz Vertical displacement in point C [m] -12.6677 m -1.50%




158

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

Test ID: 2485

Test status: Passed

1.46.1 Description

Verifies the vertical displacement of a rectangular shear plate fixed at one end, loaded with two forces.

1.46.2 Background



■ Analysis type: static;


Shear plate Scale = 1/50

01-0055SSLSB_FEM

Units

I. S.

Geometry


■ Width: l = 1 m,





Boundary conditions

■ Outer: Fixed AD edge,

■ Inner: None.




159

Loading

■ External:

► At B: Fz = -1.0 N,

► At C: FZ = 1.0 N,

■ Internal: None.


Reference solution




■ 497 nodes,


Deformed shape

Shear plate Scale = 1/35Deformed



CM2 Dz Vertical displacement in point C [m] 35.37 x 10-3



Dz Vertical displacement in point C [m] 35.6655 mm 0.84%




160

1.47 Spherical shell with holes (01-0049SSLSB_FEM)

Test ID: 2479

Test status: Passed

1.47.1 Description

A spherical shell with holes is subjected to 4 forces, opposite 2 by 2. The horizontal displacement is verified.

1.47.2 Background





Spherical shell with holes01-0049SSLSB_FEM

Units

I. S.

Geometry

■ Radius: R = 10 m


■ Opening angle of the hole: 0 = 18°.




161




Boundary conditions

■ Outer: For modeling, we consider only a quarter of the shell, so we impose symmetryconditions (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.

1.47.2.2 Horizontal displacement at point A

Reference solution

The reference solution is determined by averaging the results of several calculation software with implemented finiteelements method. 2% uncertainty about the reference solution.



■ 99 nodes,


Deformed shape

Spherical shell with holes Scale = 1/79

Deformed




162

1.47.2.3 Theoretical background


CM2 DX Horizontal displacement at point A(R,0,0) [mm] 94.0



DX Horizontal displacement at point A(R,0,0) [mm] 92.6205 mm -1.47%




163

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

Test ID: 2483

Test status: Passed

1.48.1 Description

A rectangular plate simply supported is subjected to a uniform load. The vertical displacement and the bendingmoments at the plate center are verified.

1.48.2 Background





Simply supported rectangular plate under a uniform load Scale = 1/25

01-0053SSLSB_FEM

Units

I. S.

Geometry

■ Width: a = 1 m,

■ Length: b = 5 m,





Boundary conditions


■ Inner: None.




164

Loading

■ External: Normal pressure of plate p = pZ = -1.0 Pa,

■ Internal: None.


Reference solution




■ 793 nodes,




CM2 DZ 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



DZ 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%




165

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

Test ID: 2488

Test status: Passed

1.49.1 Description

Verifies the vertical displacement for a square plate (0.01333 m thick), of side "a", fixed on its perimeter, loaded witha uniform pressure.

1.49.2 Background


■ Reference: Structure Calculation Software Validation Guide, test SSLV 09/89;



Square plate of side "a", for the modeling, only a quarter of the plate is considered.

Units

I. S.

Geometry

■ Side: a = 1 m,


■ Slenderness: =ah

= 75.


■ Reinforcement,


Pa,


Boundary conditions

■ Outer:

Fixed sides: AB and BD,

For the modeling, we impose symmetry conditions at the CB side (restrained displacement along x andrestrained rotation around y and z) and CD side (restrained displacement along y and restrained rotationaround x and z),

■ Inner: None.




166

Loading

■ External: 1 MPa uniform pressure,

■ Internal: None.


Reference solution

This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained withSerendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between thesevalues at ± 5%.



■ 289 nodes,




CM2 Dz Vertical displacement in point C [m] -2.8053 x 10-2



Dz Vertical displacement in point C [cm] -2.79502 cm 0.37%




167


Test ID: 2489

Test status: Passed

1.50.1 Description

Verifies the vertical displacement for a square plate (0.02 m thick), of side "a", fixed on its perimeter, loaded with auniform pressure.

1.50.2 Background





Units

I. S.

Geometry

■ Side: a = 1 m,



= 50.


■ Reinforcement,



Boundary conditions

■ Outer:

Fixed edges: AB and BD,


■ Inner: None.




168

Loading


■ Internal: None.


Reference solution




■ 289 nodes,










169


Test ID: 2487

Test status: Passed

1.51.1 Description


1.51.2 Background





Units

I. S.

Geometry

■ Side: a = 1 m,



= 100.


■ Reinforcement,



Boundary conditions

■ Outer:

Fixed sides: AB and BD,


■ Inner: None.




170

Loading


■ Internal: None.


Reference solution

This problem has a precise analytical solution only for thin plates. Therefore we propose the solutions obtained withSerendip elements with 20 nodes or thick plate elements of 4 nodes. The expected result should be between these

values at 5%.



■ 289 nodes,











171

1.52 Pinch cylindrical shell (01-0048SSLSB_FEM)

Test ID: 2478

Test status: Passed

1.52.1 Description

A cylinder of length L is pinched by 2 diametrically opposite forces (F). The vertical displacement is verified.

1.52.2 Background





A cylinder of length L is pinched by 2 diametrically opposite forces (F).

Pinch cylindrical shell

01-0048SSLSB_FEM

Units

I. S.

Geometry

■ Length: L = 10.35 m (total length),■ Radius: R = 4.953 m,

■ Thickness: h = 0.094 m.




172




Boundary conditions

■ Outer: For the modeling, we consider only half of the cylinder, so we impose symmetryconditions (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.

1.52.2.2 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.



■ 777 nodes,


1.52.2.3 Theoretical result


CM2 DZ Vertical displacement in point A [m] -113.9 x 10-3



DZ Vertical displacement in point A [mm] -113.3 mm 0.53%




173

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

Test ID: 2482

Test status: Passed

1.53.1 Description

A rectangular plate simply supported is subjected to a uniform load. The vertical displacement and the bendingmoments at the plate center are verified.

1.53.2 Background





Simply supported rectangular plate under a uniform load Scale = 1/11

01-0052SSLSB_FEM

Units

I. S.

Geometry

■ Width: a = 1 m,

■ Length: b = 2 m,








174

Boundary conditions


■ Inner: None.

Loading

■ External: Normal pressure of plate p = pZ = -1.0 Pa,■ Internal: None.


Reference solution




■ 435 nodes,


1.53.2.3 Theoretical background


CM2 DZ 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



DZ 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%




175

1.54 Triangulated system with hinged bars (01-0056SSLLB_FEM)

Test ID: 2486

Test status: Passed

1.54.1 Description

A truss with hinged bars is placed on three punctual supports (subjected to imposed displacements) and is loadedwith two punctual forces. A thermal load is applied to all the bars. The traction force and the vertical displacement areverified.

1.54.2 Background



■ Analysis type: static (plane problem);


Units

I. S.

Geometry

■ = 30°,■ Section A1 = 1.41 x 10

-3 m

2,

■ Section A2 = 2.82 x 10-3

m2.


■ Longitudinal elastic modulus: E =2.1 x 1011

Pa,

■ Coefficient of linear expansion: = 10-5

°C-1

.

Boundary conditions

■ Outer:

► Hinge at A (u A = v A = 0),

► Roller supports at B and C ( uB = v’C = 0),

■ Inner: None.




176

Loading

■ External:

► Support displacement: v A = -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 assemblytemperature (specified geometry),

■ Internal: None.

1.54.2.2 Tension force in BD bar

Reference solution

Determining the hyperstatic unknown with the section cut method.


■ Linear element: S beam, automatic mesh,

■ 11 nodes,

■ 17 S beams + 1 rigid S beam for the modeling of the simple support at C.

1.54.2.3 Vertical displacement at D

Reference solution

vD displacement was determined by several software with implemented finite elements method.



■ 11 nodes,

■ 17 S beams + 1 rigid S beam for the modeling of simple support at C.

Deformed shape

Triangulated system with hinged bars01-0056SSLLB_FEM




177



CM2 Fx FX traction force on BD bar [N] 43633

CM2 DZ Vertical displacement on point D [m] -0.01618



Fx Fx traction force on BD bar [N] 42870.9 N -1.75%

DZ Vertical displacement on point D [m] -0.0162358 m -0.34%




178

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

Test ID: 2492

Test status: Passed

1.55.1 Description

Verifies the vertical displacement for a square plate (0.01 m thick), of side "a", fixed on its perimeter, loaded with apunctual force in the center.

1.55.2 Background





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,


■ Slenderness: =ah = 100.




179


■ Reinforcement,


Pa,


Boundary conditions

■ Outer: Fixed edges,

■ Inner: None.

Loading

■ External: Punctual force applied on the center of the plate: FZ = -106 N,

■ Internal: None.

1.55.2.2 Vertical displacement at point C (center of the plate)

Reference solution




■ 961 nodes,





180



CM2 DZ Vertical displacement in point C [m] -0.29579



DZ Vertical displacement in point C [m] -0.292146 m 1.23%




181


Test ID: 2493

Test status: Passed

1.56.1 Description

Verifies the vertical displacement for a square plate (0.01333 m thick), of side "a", fixed on its perimeter, loaded witha punctual force in the center.

1.56.2 Background






01-0063SSLSB_FEM

Units

I. S.

Geometry

■ Side: a = 1 m,



= 75.


■ Reinforcement,






182

Boundary conditions

■ Outer: Fixed sides,

■ Inner: None.

Loading

■ External: Punctual force applied on the center of the plate: FZ = -10

6

N,■ Internal: None.

1.56.2.2 Vertical displacement at point C (the center of the plate)

Reference solution




■ 961 nodes,■ 900 surface quadrangles.










183


Test ID: 2496

Test status: Passed

1.57.1 Description


1.57.2 Background





0.1 m thick plate fixed on its perimeter Scale = 1/501-0066SSLSB_FEM

Units

I. S.

Geometry

■ Side: a = 1 m,


■ Slenderness: = 10.


■ Reinforcement,


Pa,





184

Boundary conditions


■ Inner: None.

Loading

■ External: punctual force applied in the center of the plate: FZ = -10

6


1.57.2.2 Vertical displacement at point C (center of the plate)

Reference solution







CM2 DZ Vertical displacement in point C [m] -0.42995 x 10-3



DZ Vertical displacement in point C [mm] -0.412094 mm 4.15%




185

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

Test ID: 2497

Test status: Passed

1.58.1 Description

Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, fixed on its ends andsubjected to its self weight only.

1.58.2 Background



■ Analysis type: modal analysis (space problem);


Vibration mode of a thin piping elbow Scale = 1/701-0067SDLLB_FEM

Units

I. S.

Geometry





■ Section: A = 1.131 x 10-4

m2,


m4,

■ Flexure moment of inertia relative to z-axis: Iz = 4.637 x 10-9 m4,


m4.




186


► O ( 0 ; 0 ; 0 )

► A ( 0 ; R ; 0 )

► B ( R ; 0 ; 0 )



Pa,



Boundary conditions

■ Outer: Fixed at points A and B,

■ Inner: None.

Loading

■ External: None,

■ Internal: None.


Reference solution


■ transverse bending:

f j =i

2

2 R2

GIp A where i = 1,2.




Eigen mode shapes




187



CM2 Eigen mode Eigen mode Transverse 1 frequency [Hz] 44.23

CM2 Eigen mode Eigen mode Transverse 2 frequency [Hz] 125



Eigen mode Transverse 1 frequency [Hz] 44.12 Hz -0.25%





188


Test ID: 2491

Test status: Passed

1.59.1 Description


1.59.2 Background





Units

I. S.

Geometry

■ Side: a = 1 m,


■ Slenderness: =ah = 10.


■ Reinforcement,



Boundary conditions

■ Outer:



■ Inner: None.




189

Loading


■ Internal: None.


Reference solution




■ 289 nodes,







Dz Vertical displacement in point C [mm] -0.0781846 mm 0.61%




190


Test ID: 2495

Test status: Passed

1.60.1 Description


1.60.2 Background






01-0065SSLSB_FEM

Units

I. S.

Geometry

■ Side: a = 1 m,




■ Reinforcement,






191

Boundary conditions

■ Outer: Fixed sides,

■ Inner: None.

Loading

■ External: Punctual force applied at the center of the plate: FZ = -10

6


1.60.2.2 Vertical displacement at point C center of the plate)

Reference solution






Solver Result name Result description Reference valueCM2 DZ Vertical displacement in point C [m] -0.2595 x 10

-2







192

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

Test ID: 2500

Test status: Passed

1.61.1 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.

1.61.2 Background


■ Reference: Design and calculation of metal structures.






193

1.61.2.2 Moments and actions on supports M.R. calculation on a 2D portal frame.

RDM results, for the linear load perpendicular on the rafters, are:

2

qLVV E A

H

f h3f 3k²h

f 5h8

32

²qLHH E A

HhMM DB f hH8

²qLMC


Comparison between theoretical results and the results obtained by Advance Design for a linear load perpendicular on the chords


CM2 Fz Vertical reaction V in A [DaN] -1000

CM2 Fz Vertical reaction V in E [DaN] -1000

CM2 Fx Horizontal reaction H in A [DaN] -332.9

CM2 Fx Horizontal reaction H in E [DaN] -332.9CM2 My Moment in node B [DaNm] 2496.8

CM2 My Moment in node D [DaNm] -2496.8

CM2 My Moment in node C [DaNm] -1671



Fz Vertical reaction V on node A [daN] -1000 daN 0.00%

Fz Vertical reaction V on node E [daN] -1000 daN 0.00%

Fx Horizontal reaction H on node A [daN] -332.665 daN 0.07%

Fx Horizontal reaction H on node E [daN] -332.665 daN 0.07%

My Moment in node B [daNm] 2494.99 daN*m -0.07%

My Moment in node D [daNm] -2494.99 daN*m 0.07%

My Moment in node C [daNm] -1673.35 daN*m -0.14%




194

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

Test ID: 2501

Test status: Passed

1.62.1 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.

1.62.2 Background


■ Reference: Design and calculation of metal structures.






195

1.62.2.2 Moments and reactions on supports M.R. calculation on a 2D portal frame.

RDM results, for the linear load perpendicular on the column, are:

L2

²qhVV E A

f h3f 3k²h

f h26kh5

16

²qhHE

qhHH E A

h H qh

M E B 2

² f hH

4

²qhM EC hHM ED



CM2 Fz Vertical reaction V in A [DaN] 140.6

CM2 Fz Vertical reaction V in E [DaN] -140.6

CM2 Fx Horizontal reaction H in A [DaN] 579.1

CM2 Fx Horizontal reaction H in E [DaN] 170.9

CM2 My Moment in B [DaNm] -1530.8

CM2 My Moment in D [DaNm] -1281.7

CM2 My Moment in C [DaNm] 302.7



Fz Vertical reaction V on node A [daN] 140.625 daN 0.02%

Fz Vertical reaction V on node E [daN] -140.625 daN -0.02%

Fx Horizontal reaction H on node A [daN] 579.169 daN 0.01%

Fx Horizontal reaction H on node E [daN] 170.831 daN -0.04%

My Moment in node B [daNm] -1531.27 daN*m -0.03%

My Moment in node D [daNm] -1281.23 daN*m 0.04%

My Moment in node C [daNm] 302.063 daN*m -0.21%




196


Test ID: 2499

Test status: Passed

1.63.1 Description

Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, extended with twostraight elements (2 m long) and subjected to its self weight only.



■ Analysis type: modal analysis (space problem);


Vibration mode of a thin piping elbow Scale = 1/12

01-0069SDLLB_FEM

Units

I. S.




197

Geometry


■ L = 2 m,




■ Section: A = 1.131 x 10-4

m2,


m4,


m4,


m4.


► O ( 0 ; 0 ; 0 )

► A ( 0 ; R ; 0 )

► B ( R ; 0 ; 0 )

► C ( -L ; R ; 0 )

► D ( R ; -L ; 0 )



Pa,



Boundary conditions

■ Outer:




■ Inner: None.

Loading

■ External: None,

■ Internal: None.




198


Reference solution



f j =i

2

2 R2

GIp A where i = 1,2 with i = 1,2:



■ 41 nodes,


Eigen mode shapes


Reference











199


Test ID: 2490

Test status: Passed

1.64.1 Description


1.64.2 Background





Units

I. S.

Geometry

■ Side: a = 1 m,



= 20.


■ Reinforcement,



Boundary conditions

■ Outer:



■ Inner: None.




200

Loading


■ Internal: None.


Reference solution




■ 289 nodes,











201


Test ID: 2494

Test status: Passed

1.65.1 Description


1.65.2 Background






01-0064SSLSB_FEM

Units

I. S.

Geometry

■ Side: a = 1 m,




■ Reinforcement,






202

Boundary conditions


■ Inner: None.

Loading

■ External: punctual force applied in the center of the plate: FZ = -10

6


1.65.2.2 Vertical displacement at point C (the center of the plate)

Reference solution














203


Test ID: 2498

Test status: Passed

1.66.1 Description

Verifies the eigen mode transverse frequencies for a thin piping elbow with a radius of 1 m, extended with twostraight elements (0.6 m long) and subjected to its self weight only.

1.66.2 Background



■ Analysis type: modal analysis (in space);


Vibration mode of a thin piping elbow Scale = 1/1101-0068SDLLB_FEM

Units

I. S.

Geometry


■ L = 0.6 m,




■ Section: A = 1.131 x 10-4

m2,


m4,


m4

,■ Polar inertia: Ip = 9.274 x 10

-9 m

4.




204


► O ( 0 ; 0 ; 0 )

► A ( 0 ; R ; 0 )

► B ( R ; 0 ; 0 )

► C ( -L ; R ; 0 )

► D ( R ; -L ; 0 )



Pa,



Boundary conditions

■ Outer:


► In A: translation restraint along y and z,

► In B: translation restraint along x and z,

■ Inner: None.

Loading

■ External: None,

■ Internal: None.


Reference solution



f j = i2

2 R2

GIp A where i = 1,2.



■ 23 nodes,


Eigen mode shapes




205




CM2 Eigen mode Eigen mode Transverse 2 frequency [Hz] 100








206

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





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 ( = 1)

► b = b0e-2x

► A = A0e-2x


■ E = 2 x 1011

Pa

■ = 0.3

■ = 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.




207

1.67.2.2 Reference results

Calculation method used to obtain the reference solution

i pulsation is given by the roots of the equation:

0rlsinslshrs2

²r ²sslchrlcos1

with

0²²s;²r ;EI

A 2i

si2i

2i

zo

2i04

i

Therefore, the translation components of i(x) mode, are:

))sx(rsh)rxsin(s()rlsin(s)sl(rsh

)sl(ch)rlcos(sxchrxcosex x

i

Uncertainty about the reference: analytical solution:

Reference values

Eigen mode i(x)*Eigen modeorder

Frequency (Hz)

x = 0 0.1 0.2 0.3 0.4 0.5 0.6

1 143.303 0 0.237 0.703 1 0.859 0.354 0

2 396.821 0 -0.504 -0.818 0 0.943 0.752 0

3 779.425 0 0.670 0.210 -0.831 0.257 1 0

4 1289.577 0 -0.670 0.486 0 -0.594 1 0

* i(x) eigen modes* standardized to 1 at the point of maximum amplitude.

Eigen modes




208



CM2 Eigen mode Frequency of eigen mode 1 [Hz] 143.303






Frequency of eigen mode 1 [Hz] 145.88 Hz 1.80%







209

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

Test ID: 2505

Test status: Passed

1.68.1 Description

Calculation of support reactions of a 2D portal frame with hinged supports.

1.68.2 Background





Units

I. S.

Geometry


■ I1 = 5.0 x 10-4

m4

■ a = 4 m

■ h = 8 m

■ b = 10.77 m

■ I2 = 2.5 x 10-4

m4


■ Isotropic linear elastic material.

■ E = 2.1 x 1011

Pa




210

Boundary conditions

Hinged base plates A and B (u A = v A = 0 ; uB = vB = 0).

Loading

■ p = -3 000 N/m

■ F1 = -20 000 N

■ F2 = -10 000 N

■ M = -100 000 Nm

1.68.2.2 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

■ V A = 3pl/8 + F1/2 – M/l + F2h/l

■ H A = pl²(3 + 5m)/(32Nh) + (F1l/(4h))(C/N) + F2(1-(B + C)/(2N)) + (3M/h)((1 + m)/(2N))

1.68.2.3 Reference values

Point Magnitudes and units Value

A V, vertical reaction (N) 31 500.0

A H, horizontal reaction (N) 20 239.4

C vc (m) -0.03072



CM2 Fz Vertical reaction V in point A [N] -31500

CM2 Fx Horizontal reaction H in point A [N] -20239.4

CM2 DZ vc displacement in point C [m] -0.03072



Fz Vertical reaction V in point A [N] -31500 N 0.00%

Fx Horizontal reaction H in point A [N] -20239.3 N 0.00%

DZ Displacement in point C [m] -0.0307191 m 0.00%




211

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

Test ID: 2508

Test status: Passed

1.69.1 Description

A 3D bar structure with elastic support is subjected to a vertical load of -100 kN. The V2 magnitude on node 5, thenormal force magnitude, the reaction magnitude on supports and the action magnitude are verified.


■ Reference: Internal GRAITEC;



Units

I. S.

Geometry

For all bars:

■ H = 3 m

■ B = 3 m■ S = 0.02 m

2

Element Node i Node j

1 (bar) 1 5

2 (bar) 2 5

3 (bar) 3 5

4 (bar) 4 5

5 (spring) 5 6


■ Isotropic linear elastic materials

■ Longitudinal elastic modulus: E = 2.1 E8 N/m2,




212

Boundary conditions

■ Outer: At node 5: K = 50000 kN/m ;

■ Inner: None.

Loading

■ External: Vertical load at node: P = -100 kN,■ Internal: None.


System solution

2

22 B

H L . Also, U1 = V1 = U5 = U6 = V6 = 0

■ Stiffness matrix of bar 1

1L

2x= where

.12

1.1

2

1)(.).1()(

ji ji uuuu L

xu

L

x xu




213

in the local coordinate system:

)(

)(

)(

)(

0000

0101

0000

0101

)(

)(

11

11

4

1

4

14

1

4

12

=

2

2

1

2

1

2

12

1

1

1

1

101

j

j

i

i

j

i

L T

e

T

v

v

u

v

u

L

ES

u

u

L

ES d

L

ES

d L

ES dx B B ES dV B H Bk

e

where

)v(

)u(

)v(

)u(

0000

0101

0000

0101

L

ESk

5

5

1

1

1

The elementary matrix ek expressed in the global coordinate system XY is the following: ( angle allowing

the transition from the global base to the local base):

22

22

22

22

e

eee

T

ee

sinsincossinsincossincoscossincoscos

sinsincossinsincos

sincoscossincoscos

L

ESK

cossin00

sincos00

00cossin

00sincos

RavecRkRK

Knowing thatL

Hsinand

2cos

L

B, then:

2

2

22

2

22

L2

HBcossin

L

Hsin

L2

Bcos

)(

)(

)(

)(

22

2222

22

2222

:)D

H(arctan=5,1nodes1elementfor

5

5

1

1

22

22

22

22

31

V

U

V

U

H HB

H HB

HB B HB B

H HB

H HB

HB B HB B

L

ES K




214

■ Stiffness matrix of spring support 5

)(

)(

)(

)(

0000

01010000

0101

)(

)(

1111 :systemcoordinatelocalin the

4

K K sayWe

5

j

j

i

i

j

i

v

u

v

u

K u

u K k

)(

)(

)(

)(

1010

0000

1010

0000

':90=6,5nodes5elementfor

6

6

5

5

5

V

U

V

U

K K

■ System FQK

6Y

6X

5X

1Y

1X

6

6

5

5

1

1

2

33

2

33

3

2

33

2

3

2

33

2

33

3

2

33

2

3

R

R

P

R

R

R

V

U

V

U

V

U

K0K000

000000

K0KHL

ES

2

HB

L

ESH

L

ES

2

HB

L

ES

002

HB

L

ES

2

B

L

ES

2

HB

L

ES

2

B

L

ES

00HL

ES

2

HB

L

ESH

L

ES

2

HB

L

ES

002

HB

L

ES

2

B

L

ES

2

HB

L

ES

2

B

L

ES

If U1 = V1 = U5 = U6 = V6 = 0, then:

m001885.0

4

KH

L

ES4

P

KHL

ES4

P

V2

3

2

3

5

And

N23563V4

KRN1436VH

L

ESR

0RN1015V2

HB

L

ESRN1015V

2

HB

L

ESR

56Y52

31Y

6X535X531X

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,

■ x1 value is similar to the one found by Advance Design by dividing this by 2




215

Effort in bar 1:

1759

1759

11

11

and

200

200

00

2

002

5

1

5

1

5

5

1

1

5

5

1

1

N

N

u

u

L

ES

V

U

V

U

L

B

L

H L

H

L

B L

B

L

H L

H

L

B

v

u

v

u

5

1

5

1

5

5

1

1

5

5

1

1

11

11 and

cossin00

sincos00

00cossin

00sincos

N

N

u

u

L

ES

V

U

V

U

v

u

v

u

Reference values

Point Magnitude Units Value

5 V2 m -1.885 10-3

All bars Normal force N -1759

Supports 1, 3, 4 and 5 Fz action N -1436

Supports 1, 3, 4 and 5 Action Fx=Fy N 7182/1015

Support 6 Fz action N 23563 x 4=94253


■ Linear element: beam, automatic mesh,

■ 5 nodes,





216

Deformed shape

Normal forces diagram




217

1.69.1.3 Reference values


CM2 D V2 magnitude on node 5 [m] -1.885 10-3

CM2 Fx Normal force magnitude on bar 1 [N] -1759




CM2 Fz Fz reaction magnitude on support 1 [N] 1436




CM2 Fx Action Fx magnitude on support 1 [N] -718

CM2 Fx Action Fx magnitude on support 3 [N] 718

CM2 Fx Action Fx magnitude on support 4 [N] 718

CM2 Fx Action Fx magnitude on support 5 [N] -718

CM2 Fy Action Fy magnitude on support 1 [N] -718CM2 Fy Action Fy magnitude on support 3 [N] -718

CM2 Fy Action Fy magnitude on support 4 [N] 718

CM2 Fy Action Fy magnitude on support 5 [N] 718

CM2 Fz Fz reaction magnitude on support 6 [N] 23563 x 4=94253



D Displacement on node 5 [mm] 1.88508 mm 0.00%

Fx Normal force magnitude on bar 1 [N] -1759.4 N -0.02%




Fy Fz reaction magnitude on support 1 [N] 1436.55 N 0.04%




Fx Action Fx magnitude on support 1 [N] -718.274 N -0.04%

Fx Action Fx magnitude on support 2 [N] 718.274 N 0.04%

Fx Action Fx magnitude on support 3 [N] 718.274 N 0.04%

Fx Action Fx magnitude on support 4 [N] -718.274 N -0.04%

Fz Action Fy magnitude on support 1 [N] -718.274 N -0.04%

Fz Action Fy magnitude on support 2 [N] -718.274 N -0.04%

Fz Action Fy magnitude on support 3 [N] 718.274 N 0.04%

Fz Action Fy magnitude on support 4 [N] 718.274 N 0.04%

Fy Action Fy magnitude on support 6 [N] 94253.8 N 0.00%




218

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

Test ID: 2509

Test status: Passed

1.70.1 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.

1.70.2 Background




■ Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion,

plane bending, transverse bending, punctual mass.

1.70.2.1 Test data

Units

I. S.

Geometry

■ Outer diameter de = 0.35 m,


■ Beam length: l = 10 m,

■ Area: A = 1.57865 x 10-2 m2

■ Polar inertia: IP = 4.43798 x 10-4

m4

■ Inertia: Iy = Iz = 2.21899 x 10-4

m4

■ Punctual mass: mc = 1000 kg■ Beam self-weight: M




219



Pa,


■ Poisson's ratio: =0.3 (this coefficient was not specified in the AFNOR test , the value 0.3seems 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


Reference frequency

For the first mode, the Rayleigh method gives the approximation formula

)M24.0m(I

EI3x2/1f

c3

z1

Mode Shape Units Reference

1 Flexion Hz 1.65

2 Flexion Hz 1.65

3 Flexion Hz 16.07

4 Flexion Hz 16.07

5 Flexion Hz 50.02

6 Flexion Hz 50.02

7 Traction Hz 76.47

8 Torsion Hz 80.47

9 Flexion Hz 103.210 Flexion Hz 103.2

Comment: The mass matrix associated with the beam torsion on two nodes, is expressed as:

12/1

2/11

3

Il P

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:3

Il p

Uncertainty about the reference frequencies

■ Analytical solution mode 1

■ Other modes: 1%


■ Linear element AB: Beam

■ Beam meshing: 20 elements.




220

Modal deformations




221

Observation: the deformed shape of mode No. 8 that does not really correspond to a torsion deformation, isactually the display result of the translations and not of the rotations. This is confirmed by the rotation values of thecorresponding mode.

Eigen modes vector 8

Node DX DY DZ RX RY RZ

1 -3.336e-033 6.479e-031 -6.316e-031 1.055e-022 5.770e-028 5.980e-0282 -5.030e-013 1.575e-008 -1.520e-008 1.472e-002 6.022e-008 6.243e-0083 -1.005e-012 6.185e-008 -5.966e-008 2.944e-002 1.171e-007 1.214e-0074 -1.505e-012 1.365e-007 -1.317e-007 4.416e-002 1.705e-007 1.769e-0075 -2.002e-012 2.381e-007 -2.296e-007 5.887e-002 2.206e-007 2.289e-0076 -2.495e-012 3.648e-007 -3.517e-007 7.359e-002 2.673e-007 2.774e-0077 -2.983e-012 5.149e-007 -4.963e-007 8.831e-002 3.106e-007 3.225e-0078 -3.464e-012 6.867e-007 -6.618e-007 1.030e-001 3.506e-007 3.641e-0079 -3.939e-012 8.785e-007 -8.464e-007 1.177e-001 3.873e-007 4.023e-00710 -4.406e-012 1.088e-006 -1.049e-006 1.325e-001 4.207e-007 4.371e-00711 -4.863e-012 1.315e-006 -1.267e-006 1.472e-001 4.508e-007 4.684e-00712 -5.310e-012 1.556e-006 -1.499e-006 1.619e-001 4.777e-007 4.964e-00713 -5.746e-012 1.811e-006 -1.744e-006 1.766e-001 5.015e-007 5.210e-00714 -6.169e-012 2.077e-006 -2.000e-006 1.913e-001 5.221e-007 5.423e-00715 -6.580e-012 2.353e-006 -2.265e-006 2.061e-001 5.396e-007 5.605e-00716 -6.976e-012 2.637e-006 -2.539e-006 2.208e-001 5.541e-007 5.755e-00717 -7.357e-012 2.928e-006 -2.819e-006 2.355e-001 5.658e-007 5.874e-00718 -7.723e-012 3.224e-006 -3.104e-006 2.502e-001 5.746e-007 5.965e-00719 -8.072e-012 3.524e-006 -3.393e-006 2.649e-001 5.808e-007 6.028e-00720 -8.403e-012 3.826e-006 -3.685e-006 2.797e-001 5.844e-007 6.065e-00721 -8.717e-012 4.130e-006 -3.977e-006 2.944e-001 5.856e-007 6.077e-007

With NE/NASTRAN, the results associated with mode No. 8, are:




222



CM2 Eigen mode 1 frequency [Hz] 1.65









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 thecorresponding description sheet).







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%





223

1.71 Slender beam of variable rectangular section with fixed-free ends (ß=5) (01-0085SDLLB_FEM)

Test ID: 2503

Test status: Passed

1.71.1 Description

Verifies the eigen modes (bending) for a slender beam with variable rectangular section (fixed-free).

1.71.2 Background





Units

I. S.

Geometry


■ 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


■ E = 2 x 1011

Pa

■ = 7800 kg/m3

Boundary conditions

■ Outer:

► Fixed at end x = 0,

► Free at end x = 1

■ Inner: None.

Loading

■ External: None,

■ Internal: None.




224



Precise calculation by numerical integration of the differential equation of beams bending (Euler-Bernoulli theories):

²t

² A

²x

²EIz

x

2

2

where Iz and A vary with the abscissa.

The result is:

12

E²l

1h,i

2

1fi with

51b

b

41h

h

1 2 3 4 5

= 5 24.308 75.56 167.21 301.9 480.4


Reference values

Eigen mode type Frequency (Hz)

1 56.55

2 175.79

3 389.01

4 702.36

Flexion

5 1117.63

MODE 1 Scale = 1/4




225

MODE 2 Scale = 1/4

MODE 3 Scale = 1/4




226

MODE 4 Scale = 1/4

MODE 5 Scale = 1/4




227


TheoreticalFrequency

Result name Result description Reference value

Eigen mode Frequency of eigen mode 1 [Hz] 56.55









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%




228

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

Test ID: 2507

Test status: Passed

1.72.1 Description

Verifies the vertical displacement and the normal force on a cantilever beam in Eulerian buckling with thermal load.

1.72.2 Background


■ Reference: Euler theory;



Units

I. S.

Geometry

■ L = 10 m

■ S=0.01 m2

■ I = 0.0002 m4



N/m2,


■ Coefficient of thermal expansion: = 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)

( 5000001.0T0005.001.010.2

100000

ES

N100

)




229


Reference solution


98696.0100000

98696

986964P 2

2

critical

N L

EI

Observation: in this case, the thermal load has no effect over the critical coefficient



■ 5 nodes,

■ 4 elements.

Deformed shape



CM2 DZ Vertical displacement v5 on Node 5 - Case 101 [cm] -1.0

CM2 Fx Normal Force on Node A - Case 101 [N] -100000

CM2 Fx Normal Force on Node A - Case 102 [N] -98696



DZ Vertical displacement on Node 5 [cm] -1 cm 0.00%

Fx Normal Force - Case 101 [N] -100000 N 0.00%

Fx Normal Force - Case 102 [N] -98699.3 N 0.00%




230

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

Test ID: 2512

Test status: Passed

1.73.1 Description

Simple supported beam in free vibration.

Tested functions: Shear force, eigen frequencies.

1.73.2 Background

■ Reference: NAFEMS, FV5


■ Tested functions: Shear force, eigen frequencies.

1.73.2.1 Problem data

Units

I. S.

Geometry

Full square section:

■ Dimensions: a x b = 2m x 2 m

■ Area: A = 4 m

2

■ Inertia: IP = 2.25 m4

Iy = Iz = 1.333 m4




231



Pa,



Boundary conditions

■ Outer:

► x = y = z = Rx = 0 at A ;

► y = z =0 at B ;

■ Inner: None.

Loading


1.73.2.2 Reference frequencies

Mode Shape Units Reference

1 Flexion Hz 42.6492 Flexion Hz 42.649

3 Torsion Hz 77.542

4 Traction Hz 125.00

5 Flexion Hz 148.31

6 Flexion Hz 148.31

7 Torsion Hz 233.10

8 Flexion Hz 284.55

9 Flexion Hz 284.55

Comment: Due to the condensed (lumped) nature of the mass matrix of Advance Design, the frequencies values of 3and 7 modes cannot be found by this software. The same modeling done with NE/NASTRAN gave respectively formode 3 and 7: 77.2 and 224.1 Hz.


■ Straight elements: linear element

■ Imposed mesh: 5 meshes




232

Modal deformations



CM2 Frequency of eigen mode 1 [Hz] 42.649CM2 Frequency of eigen mode 2 [Hz] 42.649

CM2 Frequency of eigen mode 3 [Hz] 77.542





Comment: The torsion modes No. 3 and 7 that are calculated with NASTRAN cannot be calculated with Advance DesignCM2 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.













233

1.74 Membrane with hot point (01-0099HSLSB_FEM)

Test ID: 2513

Test status: Passed

1.74.1 Description

Membrane with hot point.

Tested functions: Stresses.

1.74.2 Background

■ Reference: NAFEMS, Test T1

■ Analysis type: static, thermo-elastic;

■ Tested functions: Stresses.


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.




234

Units

I. S.

Geometry / meshing

A quarter of the structure is modeled by incorporating the terms of symmetries.

Thickness: 1 m



Pa,


■ Elongation coefficient = 0.00001.

Boundary conditions

■ Outer:

► For all nodes in y = 0, uy =0;

► For all nodes in x = 0, ux =0;

■ Inner: None.

Loading

■ External: None,

■ Internal: Hot point, thermal load T = 100°C;




235

1.74.3 yy stress at point A:

Reference solution:

Reference value: yy = 50 MPa in A


■ Planar elements: membranes,

■ 28 planar elements,

■ 39 nodes.



CM2 syy_mid yy in A [MPa] 50

Note: This value (50.87) is obtained with a vertical cross section through point A. The value represents yy at the leftend of the diagram.

With CM2, it is essential to display the results with the “Smooth results on planar elements” option deactivated.



syy_mid Sigma yy in A [MPa] 50.8666 MPa 1.73%




236

1.75 Double cross with hinged ends (01-0097SDLLB_FEM)

Test ID: 2511

Test status: Passed

1.75.1 Description

Double cross with hinged ends.

Tested functions: Eigen frequencies, crossed beams, in plane bending.

1.75.2 Background

■ Reference: NAFEMS, FV2 test


■ Tested functions: Eigen frequencies, Crossed beams, In plane bending.


Units

I. S.

Geometry

Full square section:

■ Arm length: L = 5 m

■ Dimensions: a x b = 0.125 x 0.125

■ Area: A = 1.563 10-2

m2

■ Inertia: IP = 3.433 x 10-5

m4

Iy = Iz = 2.035 x 10-5

m4




237



Pa,


Boundary conditions

■ Outer: A, B, C, D, E, F, G, H points restraint along x and y;

■ Inner: None.

Loading



Mode Units Reference

1 Hz 11.336

2,3 Hz 17.709

4 to 8 Hz 17.709

9 Hz 45.345

10,11 Hz 57.39012 to 16 Hz 57.390


■ Linear elements type: Beam

■ Imposed mesh: 4 Elements / Arms




238

Modal deformations




239



CM2 Frequency Frequency of Eigen Mode 1 [Hz] 11.336













CM2 Frequency Frequency of Eigen Mode 14 [Hz] 57.390CM2 Frequency Frequency of Eigen Mode 15 [Hz] 57.390




Frequency of Eigen Mode 1 [Hz] 11.33 Hz -0.05%



















240

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

Test ID: 2502

Test status: Passed

1.76.1 Description

Verifies the deflection magnitude on a non-slender beam with two hinged supports.

1.76.2 Background


■ Reference: Structure Calculation Software Validation Guide, test SSLL 02/89



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


■ E = 2 x 1011

Pa

■ = 0.3

Boundary conditions

■ Hinge at end x = 0,

■ Hinge at end x = 1.44 m.

LoadingUniformly distributed force of p = -1. X 10

5 N/m on beam AB.




241



The deflection on the middle of a non-slender beam considering the shear force deformations given by theTimoshenko function:

G A8

pl

EI

pl

384

5

vr

24

where

12

EG and

zr a

A A

where "Ar " is the reduced area and "az" the shear coefficient calculated on the transverse section.


Reference values

Point Magnitudes and units Value

C V, deflection (m) -1.25926 x 10-3



CM2 Dz Deflection magnitude in point C [m] -0.00125926



Dz Deflection magnitude in node C [m] -0.00125926 m 0.00%




242

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

Test ID: 2506

Test status: Passed

1.77.1 Description

Verifies the normal force on the nodes of a double fixed beam in Eulerian buckling with a thermal load.

1.77.2 Background


■ Reference: Euler theory;



Units

I. S.

Geometry

L = 10 m

Cross Section Sx m² Sy m² Sz m² Ix m4 Iy m4 Iz m4Vx m3 V1y m3 V1z m3 V2y m3 V2z m3

IPE200 0.002850 0.001400 0.001799 0.0000000646 0.0000014200 0.0000194300 0.00000000 0.00002850 0.00019400 0.00002850 0.00019400



N/m2,


■ Coefficient of thermal expansion: = 0.00001

Boundary conditions


■ Inner: None.




243

Loading

■ External: Punctual load FZ = 1 N at = L/2 (load that initializes the deformed shape),

■ Internal: T = 5°C corresponding to a compression force of:

kN925.29500001.000285.011E1.2TESN


Reference solution


93.3724.117

925.29k 724.117

2

P2

2

critical

N L

EI

Observation: in this case, the thermal load has no effect over the critical coefficient



■ 11 nodes,

■ 10 elements.

Deformed shape of mode 1



CM2 Fx Normal Force on Node 6 - Case 101 [kN] -29.925

CM2 Fx Normal Force on Node 6 - Case 102 [kN] -117.724



Fx Normal Force Fx on Node 6 - Case 101 [kN] -29.904 kN 0.07%

Fx Normal Force Fx on Node 6 - Case 102 [kN] -118.081 kN -0.30%




244

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

Test ID: 2510

Test status: Passed

1.78.1 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.

1.78.2 Background




■ Element type: linear.■ Tested functions: Eigen mode frequencies, straight slender beam, combined bending-torsion,

plane bending, transverse bending, punctual mass..


Units

I. S.

Geometry

■ Outer diameter: de= 0.35 m,



■ Distance BC: lBC = 1 m

■ Area: A =1.57865 x 10-2 m2

■ Inertia: Iy = Iz = 2.21899 x 10-4

m4

■ Polar inertia: Ip = 4.43798 x 10-4m4

■ Punctual mass: mc = 1000 kg




245


■ Longitudinal elasticity modulus of AB element: E = 2.1 x 1011

Pa,

■ Density of the linear element AB: = 7800 kg/m3

■ Poisson's ratio =0.3(this coefficient was not specified in the AFNOR test , the value 0.3 seemsto 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 = 10

21

Pa■ Density of the linear element BC: = 0 kg/m3

Boundary conditions

Fixed at point A, x = 0,

Loading



Reference solutions

The different eigen frequencies are determined using a finite elements model of Euler beam (slender beam).

f z + t0 = flexion x,z + torsion

f y + tr = flexion x,y + traction

Mode Units Reference

1 (f z + t0) Hz 1.636

2 (f y + tr ) Hz 1.642

3 (f y + tr ) Hz 13.460

4 (f z + t0) Hz 13.590

5 (f z + t0) Hz 28.900

6 (f y + tr ) Hz 31.960

7 (f z + t0) Hz 61.610

1 (f z + t0) Hz 63.930

Uncertainty about the reference solutions

The uncertainty about the reference solutions: 1%


■ Linear element AB: Beam

■ Imposed mesh: 50 elements.

■ Linear element BC: Beam

■ Without meshing




246

Modal deformations




247



CM2 Frequency Eigen mode 1 frequency (f z + t0) [Hz] 1.636

CM2 Frequency Eigen mode 2 frequency (f y + tr ) [Hz] 1.642







Note:

f z + t0 = flexion x,z + torsion

f y + tr = flexion x,y + traction

Observation: because the mass matrix of Advance Design is condensed and not consistent, the torsion modesobtained are not taking into account the self rotation mass inertia of the beam.














248

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

Test ID: 2516

Test status: Passed

1.79.1 Description

Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length andidentical characteristics with 3 T/C supports (k = -10000 N/m).

1.79.2 Background



■ Analysis type: static non linear;

■ Element type: linear, T/C.

Units

I. S.

Geometry

■ L = 10 m

■ Section: IPE 200, Iz = 0.00001943 m4



N/m2,


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

■ External: Vertical punctual load P = -100 N at x = 5 m,

■ Internal: None.




249

1.79.2.2 References solutions

Displacements

rad000034.0Lk2EI3EI32

LkEI6PL

m00058.0Lk2EI316

PL3v


LkEI3PL


LkEI2PL3

3

yzz

3yz

2

3

3yz

3

3

3yzz

3yz

2

2

3

yzz

3yz

2

1

M z Moments

N.m9.2202

MM

4

PL)m5x(M

N.m15.58Lk2EI316

PLk3M

0M

1z2zz

3yz

4y

2z

1z



■ 3 nodes,

■ 2 linear elements + 1 T/C.

Deformed shape




250

Moment diagram



CM2 RY Rotation Ry in node 1 [rad] -0.000129

CM2 RY Rotation Ry in node 2 [rad] 0.000106

CM2 DZ Displacement - node 3 [m] 0.00058


CM2 My M moment - node 1 [Nm] 0

CM2 My M moment - node 2 [Nm] -58.15

CM2 My M moment - middle span 1 [Nm] -220.9



RY Rotation Ry in node 1 [rad] 0.000129488 Rad 0.38%

RY Rotation Ry in node 2 [rad] -0.000105646 Rad 0.33%

DZ Displacement - node 3 [m] 0.000581169 m 0.20%

RY Rotation Ry in node 3 [rad] -3.44295e-005 Rad -1.26%

My M moment - node 1 [Nm] 1.77636e-015 N*m 0.00%My M moment - node 2 [Nm] 58.1169 N*m -0.06%

My M moment - middle span 1 [Nm] -220.942 N*m -0.02%




251

1.80 Linear system of truss beams (01-0103SSLLB_FEM)

Test ID: 2517

Test status: Passed

1.80.1 Description

Verifies the displacement and the normal force for a bar system containing 4 elements of the same length and 2diagonals.

1.80.2 Background




■ Element type: linear, bar.

Units

I. S.

Geometry■ L = 5 m

■ Section S = 0.005 m2



N/m2.

Boundary conditions

■ Outer:

► Support at node 1 restrained along x and y,


■ Inner: None.




252

Loading

■ External: Horizontal punctual load P = 50000 N at node 3,

■ Internal: None.


Displacements

m000108.0ES11

PL5v

m000541.0ES11

PL25u

m000129.0ES11

PL6v

m000649.0ES11

PL30u

4

4

3

3

N normal forces

N32141P11

25N N22727P

11

5N

N38569P11

26N N27272P

11

6N

N22727P11

5N 0N

4243

1323

1412


■ Linear element: bar, without meshing,

■ 4 nodes,


Deformed shape




253

Normal forces



CM2 DX u3 displacement on Node 3 [m] 0.000649

CM2 DZ v3 displacement on Node 3 [m] -0.000129


CM2 DZ v4 displacement on Node 4 [m] 0.000108

CM2 Fx N12 normal force on Element 1 [N] 0

CM2 Fx N23 normal force on Element 2 [N] -27272



CM2 Fx N13 normal effort on Element 5 [N] 38569




DX u3 displacement on Node 3 [m] 0.000649287 m 0.04%

DZ v3 displacement on Node 3 [m] -0.000129871 m -0.68%

DX u4 displacement on Node 4 [m] 0.000541063 m 0.01%

DZ v4 displacement on Node 4 [m] 0.000108224 m 0.21%

Fx N12 normal force on Element 1 [N] 0 N 0.00%

Fx N23 normal force on Element 2 [N] -27272.9 N 0.00%

Fx N43 normal force on Element 3 [N] 22727.1 N 0.00%

Fx N14 normal force on Element 4 [N] 22727.1 N 0.00%

Fx N13 normal effort on Element 5 [N] 38569.8 N 0.00%

Fx N42 normal force on Element 6 [N] -32140.9 N 0.00%




254

1.81 Linear element in combined bending/tension - without compressed reinforcements - Partiallytensioned section (02-0158SSLLB_B91)

Test ID: 2520

Test status: Passed

1.81.1 Description

Verifies the reinforcement results for a concrete beam with 8 isostatic spans subjects to uniform loads andcompression normal forces.


■ Reference: J. Perchat (CHEC) reinforced concrete course



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,


■ Longitudinal elastic modulus: E = 20000 MPa,

■ Poisson's ratio: = 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 = -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




255

► 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 = -8 kN at x = 36m (case 8 span 7


► Case 9 (ACC): uniform linear load a = -25 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 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 Concrete Reinforcement Application Concrete Cracking

1 B20 HA fe500 D>24h No Nonprejudicial

2 B35 Adx fe235 1h<D<24h No Nonprejudicial

3 B50 HA fe 400 D<1h Yes Non

prejudicial4 B25 HA fe500 D>24h Yes Prejudicial

5 B25 HA fe500 D>24h No Veryprejudicial

6 B30 Adx fe235 D>24h Yes Prejudicial

7 B40 HA fe500 D>24h Yes 160 MPa

8 B45 HA fe500 D<1h Yes Nonprejudicial




256

1.81.1.2 Reinforcement calculation

Reference solution

Span 1 Span 2 Span 3 Span 4 Span 5 Span 6 Span 7 Span 8

fc28 20 35 50 25 25 30 40 45ft28 1.8 2.7 3.6 2.1 2.1 2.4 3 3.3

fe 500 235 400 500 500 235 500 500

teta 1 0.9 0.85 1 1 1 1 0.85

gamb 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.15

gams 1.15 1.15 1.15 1.15 1.15 1.15 1.15 1

h 1.6 1 1.6 1.6 1.6 1 1.6 1.6

fbu 11.33 22.04 33.33 14.17 14.17 17.00 22.67 39.13

fed 434.78 204.35 347.83 434.78 434.78 204.35 434.78 500.00

sigpreju 250.00 156.67 264.00 250.00 250.00 156.67 160.00 252.76

sigtpreju 200.00 125.33 211.20 200.00 200.00 125.33 160.00 202.21

g 5.00 5.00 5.00 5.00 5.00 5.00 5.00 5.00

q 9.00 15.00 9.00 9.00 9.00 9.00 9.00 25.00

pu 20.25 29.25 20.25 20.25 20.25 20.25 20.25 30.00

pser 14.00 20.00 14.00 14.00 14.00 14.00 14.00

G -30.00 -100.00 -50.00 -100.00 -150.00 -10.00 -10.00 -10.00

Q -30.00 -50.00 -40.00 -100.00 -100.00 -8.00 -8.00 -8.00

l 6.00 6.00 6.00 6.00 6.00 6.00 6.00 6.00

Mu 91.13 131.63 91.13 91.13 91.13 91.13 91.13 135.00

Nu -85.50 -210.00 -127.50 -285.00 -352.50 -25.50 -25.50 -18.00

Mser 63.00 90.00 63.00 63.00 63.00 63.00 63.00

Nser -60.00 -150.00 -90.00 -200.00 -250.00 -18.00 -18.00

Vu 60.75 87.75 60.75 60.75 60.75 60.75 60.75 90.00

Main reinforcement calculation according to ULS

Mu/A 74.03 89.63 65.63 34.13 20.63 86.03 86.03 131.40

ubu 0.161 0.100 0.049 0.059 0.036 0.125 0.094 0.083

a 0.221 0.133 0.062 0.077 0.046 0.167 0.123 0.108

z 0.410 0.426 0.439 0.436 0.442 0.420 0.428 0.430

Au 6.12 20.57 7.97 8.35 9.18 11.27 5.21 6.46

Main reinforcement calculation with prejudicial cracking according to SLS

Mser/A 51.000 60.000 45.000 23.000 13.000 59.400 59.400 0.000

a 0.4186 0.6678 0.6303 0.4737 0.4737 0.6328 0.6923 0.6157

Mrb 87.53 220.78 302.44 121.16 121.16 182.01 258.82 267.55

A 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

B -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000

C -0.4533 -0.8511 -0.3788 -0.2044 -0.1156 -0.8426 -0.8250 0.0000

D 0.4533 0.8511 0.3788 0.2044 0.1156 0.8426 0.8250 0.0000

alpha1 0.238 0.432 0.428

z 0.414 0.385 0.386

Aserp 10.22 10.99 10.75




257


Main reinforcement calculation with very prejudicial cracking according to SLS

Mser/A 51.00 60.00 45.00 23.00 13.00 59.40 59.40 0.00

a 0.47 0.72 0.68 0.53 0.53 0.68 0.69 0.67

Mrb 96.93 231.67 319.66 132.43 132.43 192.27 258.82 283.60

A 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000B -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000 -3.0000

C -0.5667 -1.0638 -0.4735 -0.2556 -0.1444 -1.0532 -0.8250 0.0000

D 0.5667 1.0638 0.4735 0.2556 0.1444 1.0532 0.8250 0.0000

alpha1 0.203

z 0.420

Asertp 14.049

Transverse reinforcement calculation

tu 0.68 0.98 0.68 0.68 0.68 0.68 0.68 1.00

k 0.57 0.40 0.00 -0.14 -0.41 0.00 0.00 0.00

At/st 1.87 7.08 4.31 3.90 4.77 7.34 3.45 4.44

Recapitulation

Aflex 6.12 20.57 7.97 10.22 14.05 11.27 10.75 6.46

e0 -0.95 -1.67 -1.43 -3.17 -3.97 -0.29 -0.29 -0.13

Aminfsimp 0.75 2.38 1.86 0.87 0.87 2.11 1.24 1.37

Aminfcomp 0.83 2.54 2.01 0.90 0.90 2.80 1.65 0.30

At 1.87 7.08 4.31 3.90 4.77 7.34 3.45 4.44

Atmin 1.60 3.40 2.00 1.60 1.60 3.40 1.60 1.60


■ Linear elements: beams with imposed mesh

■ 29 nodes,





258



CM2 Az Inf. main reinf. T1 [cm2] 6.12

CM2 Amin Min. main reinf. T1 [cm2] 0.75

CM2 Atz Trans. reinf. T1 [cm2

] 1.87CM2 Az Inf. main reinf. T2 [cm

2] 20.57


CM2 Atz Trans. reinf. T2 [cm2] 7.08














CM2 Amin Min. main. reinf. T7 [cm2] 1.24





The "Mu limit" method must be applied in order to achieve the same results.




259



Az 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%


Amin Min. main reinf. T2 [cm²] 2.3783 cm² -0.07%

Atz Trans. reinf. T2 [cm²] 7.07943 cm² -0.01%


Amin Min. main reinf. T3 [cm²] 1.863 cm² 0.16%

Atz Trans. reinf. T3 [cm²] 4.3125 cm² 0.06%

Az Inf. main reinf. T4 [cm²] -10.2301 cm² -0.10%


Atz Trans. reinf. T4 [cm²] 3.9008 cm² 0.02% Az Inf. main reinf. T5 [cm²] -14.0512 cm² -0.01%







Amin Min. main. reinf. T7 [cm²] 1.242 cm² 0.16%


Az Inf. main reinf. T8 [cm²] -6.47718 cm² -0.27% Amin Min. main reinf. T8 [cm²] 1.3662 cm² -0.28%





260

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

Test ID: 2521

Test status: Passed

1.82.1 Description

Verifies the reinforcement results for a concrete beam with 8 isostatic spans subjected to uniform loads.


■ Reference: J. Perchat (CHEC) reinforced concrete course



Units

■ Forces: kN

■ Moment: kN.m■ Stresses: MPa

■ Reinforcement density: cm2

Geometry

■ Beam dimensions: 0.2 x 0.5 ht

■ Length: l = 42 m in 7 spans of 6m,


■ Longitudinal elastic modulus: E = 20000 MPa,

■ Poisson's ratio: = 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)

■ Internal: None.




261

Reinforced concrete calculation hypothesis:

■ All concrete covers are set to 5 cm

■ BAEL 91 calculation with the revised version 99

Span Concrete Reinforcement Application Concrete Cracking

1 B20 HA fe500 D>24h No Nonprejudicial

2 B35 Adx fe235 1h<D<24h No Nonprejudicial

3 B50 HA fe 400 D<1h Yes Nonprejudicial

4 B25 HA fe500 D>24h Yes Prejudicial

5 B60 HA fe500 D>24h No Veryprejudicial

6 B30 Adx fe235 D>24h Yes Prejudicial

7 B40 HA fe500 D>24h Yes 160 MPa

8 B45 HA fe500 D<1h Yes Nonprejudicial

1.82.1.2 Reinforcement calculation

Reference solution


fc28 20 35 50 25 60 30 40 45ft28 1.8 2.7 3.6 2.1 4.2 2.4 3 3.3fe 500 235 400 500 500 235 500 500

teta 1 0.9 0.85 1 1 1 1 0.85gamb 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.15gams 1.15 1.15 1.15 1.15 1.15 1.15 1.15 1

h 1.6 1 1.6 1.6 1.6 1 1.6 1.6

fbu 11.33 22.04 33.33 14.17 34.00 17.00 22.67 39.13fed 434.78 204.35 347.83 434.78 434.78 204.35 434.78 500.00

sigpreju 250.00 156.67 264.00 250.00 285.15 156.67 160.00 252.76

sigtpreju 200.00 125.33 211.20 200.00 228.12 125.33 160.00 202.21

g 5.00 5.00 5.00 5.00 5.00 5.00 5.00 5.00q 9.00 15.00 9.00 9.00 9.00 9.00 9.00 25.00pu 20.25 29.25 20.25 20.25 20.25 20.25 20.25 30.00

pser 14.00 20.00 14.00 14.00 14.00 14.00 14.00l 6.00 6.00 6.00 6.00 6.00 6.00 6.00 6.00

Mu 91.13 131.63 91.13 91.13 91.13 91.13 91.13 135.00Mser 63.00 90.00 63.00 63.00 63.00 63.00 63.00Vu 60.75 87.75 60.75 60.75 60.75 60.75 60.75 90.00

Longitudinal reinforcement calculation according to ELU

ubu 0.199 0.147 0.068 0.159 0.066 0.132 0.099 0.085a 0.279 0.200 0.087 0.217 0.086 0.178 0.131 0.111z 0.400 0.414 0.434 0.411 0.435 0.418 0.426 0.430

Au 5.24 15.56 6.03 5.10 4.82 10.67 4.91 6.28Main reinforcement calculation with prejudicial cracking according to SLS

A 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000B -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 -3.000 -3.000C -0.56000 -0.89362 -0.87500D 0.56000 0.89362 0.87500

alpha1 0.367 0.442 0.438z 0.395 0.384 0.384

Aserp 6.38 10.48 10.25

Main reinforcement calculation with very prejudicial cracking according to SLS

A 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00B -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 -3.00 -3.00C -0.70 -1.60 -0.66 -0.70 -0.61371 -1.12 -0.88 0.00D 0.70 1.60 0.66 0.70 0.61371 1.12 0.88 0.00

alpha1 0.381z 0.393

Asertp 7.030




262


Transversal reinforcement calculation

tu 0.68 0.98 0.68 0.68 0.68 0.68 0.68 1.00k 1.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00

At/st 0.69 1.79 4.31 3.45 3.45 7.34 3.45 4.44

Recapitulation

Aflex 5.24 15.56 6.03 6.38 7.03 10.67 10.25 6.28

Aminflex 0.75 2.38 1.86 0.87 1.74 2.11 1.24 1.37 At 0.69 1.79 4.31 3.45 3.45 7.34 3.45 4.44

Atmin 1.60 3.40 2.00 1.60 1.60 3.40 1.60 1.60


■ Linear elements: beams with imposed mesh

■ 29 nodes,



Reference










CM2 Atz Trans. reinf. T3 [cm

2

] 4.31CM2 Az Inf. main reinf. T4 [cm

2] 6.38










CM2 Amin Min. main. reinf. T7 [cm2] 1.24





The "Mu limit" method must be applied to attain the same results.




263



Az Inf. main reinf. T1 [cm²] -5.24348 cm² -0.07% Amin Min. main reinf. T1 [cm²] 0.7452 cm² -0.64%









Amin Min. main reinf. T4 [cm²] 0.8694 cm² -0.07% Atz Trans. reinf. T4 [cm²] 3.45 cm² 0.00%








Amin Min. main. reinf. T7 [cm²] 1.242 cm² 0.16%

Atz Trans. reinf. T7 [cm²] 3.45 cm² 0.00% Az Inf. main reinf. T8 [cm²] -6.29338 cm² -0.21%






264

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

Test ID: 2515

Test status: Passed

1.83.1 Description

Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length andidentical characteristics with 3 T/C supports (k -> infinite).

1.83.2 Background





Units

I. S.

Geometry

■ L = 10 m

■ Section: IPE 200, Iz = 0.00001943 m4



N/m2,


Boundary conditions

■ Outer:



► T/C stiffness ky (1.1030

N/m),

■ Inner: None.

Loading


■ Internal: None.




265


ky being infinite, the non linear model behaves the same way as a beam on 3 supports.

Displacements


LkEI6PL

0Lk2EI316

PL3v


LkEI3PL


LkEI2PL3

3yzz

3yz

2

3

3yz

3

3

3yzz

3

yz

2

2

3yzz

3yz

2

1

M z Moments

N.m13.2032

MM

4

PL)m5x(M

N.m75.93Lk2EI316

PLk3M

0M

1z2zz

3

yz

4y

2z

1z



■ 3 nodes,


Deformed shape




266

Moment diagram



CM2 RY Rotation Ry - node 1 [rad] 0.000115

CM2 RY Rotation Ry - node 2 [rad] -0.000077

CM2 DZ Displacement - node 3 [m] 0

CM2 RY Rotation Ry - node 3 [rad] 0.000038

CM2 My Moment M - node 1 [Nm] 0

CM2 My Moment M - node 2 [Nm] 93.75

CM2 My Moment M - middle span 1 [Nm] -203.13




RY Rotation Ry in node 2 [rad] -7.65875e-005 Rad 0.54%

DZ Displacement - node 3 [m] 9.36486e-030 m 0.00%

RY Rotation Ry in node 3 [rad] 3.81695e-005 Rad 0.45%

My Moment M - node 1 [Nm] 1.3145e-013 N*m 0.00%

My Moment M - node 2 [Nm] 93.6486 N*m -0.11%

My Moment M - middle span 1 [Nm] -203.176 N*m -0.02%




267

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

Test ID: 2519

Test status: Passed

1.84.1 Description

A structure consisting of 2 beams and 2 punctual masses, subjected to a lateral earthquake along X. The frequencymodes, the eigen vectors, the participation factors, the displacement at the top of the mast and the forces at the topof the mast are verified.



■ Analysis type: modal and spectral analyses;

■ Element type: linear, mass.

1.84.1.2 Material strength model

Units

I. S.

Geometry


■ Outer radius: Rext = 3.00 m

■ Inner radius: Rint = 2.80 m

■ Axial section: S= 3.644 m2

■ Polar inertia: Ip = 30.68 m4

■ Bending inertias: Ix =15.34 m4

Iy = 15.34 m4

Masses

■ M1 =203873.6 kg

■ M2 =101936.8 kg



N/m2,


■ Density: = 25 kN/m3




268

Boundary conditions

■ Outer: Fixed in X = 0, Y = 0 m,

Loading

■ External: Seismic excitation on X direction


Linear element: beam, automatic mesh,

1.84.1.3 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).

1.84.1.4 Modal analysis

Eigen periods reference solution

Substract the value of structure’s specific horizontal periods by solving the following equation:

0MKdet 2

2

1

3

M0

0MM

25

516

L7

EI48K

Eigen modes Units Reference

1 Hz 2.0852 Hz 10.742

Modal vectors

For 1:

055.3

1

U

U0

U

U

M0

0M

25

516

L7

EI48

2

1

1

2

1

12

2

2

11

3

For 2:

655.0

1

U

U

2

1

2

Normalizing relative to the mass

3

4

110842.2

10305.9 ;

3

3

210316.1

1001.2




269

Modal deformations

1.84.1.5 Spectral study

Design spectrum

Nominal acceleration:

2n1 sm5411.5a 2.085Hzf

2n2 sm25.6aHz742.01f

Observation: the gap between pulses is greater than 10%, so the modal responses can be regarded as independent.




270

Reference participation factors

Mii

séismedudirectionladedirecteur Vecteur :

Eigen modes Reference

1 479.4272 275.609

Pseudo-acceleration

iiii a in (m/s2)

4.0

%5

: Damping correction factor.

: Structure damping.

8.2556

2.70261

2.4783-

3.78521

Reference modal displacement

024.814E

021.576E1

045.446E-

048.318E2

Equivalent static forces

058.415E

055.510EF1

052.526E-

057.717EF2

Displacement at the top of the mast

22

1 04E446.502E81.4U

Units Reference

m 4.814 E-02

Shear force at the top of the mast

3

05E526.205E415.8T

22

1

3: Being the behavior coefficient of forces

Units Reference

N 2.929

E+05

Moment at the base

Units Reference

N.m 1.578 E+07




271


Reference


CM2 Frequency Frequency Mode 1 [Hz] 2.085

CM2 Frequency Frequency Mode 2 [Hz] 10.742CM2 D Displacement at the top of the mast [cm] 4.814

CM2 Fz Forces at the top of the mast [N] 2.929E+05



Frequency Mode 1 [Hz] 2.08 Hz -0.24%

Frequency Mode 2 [Hz] 10.74 Hz -0.02%

D Displacement at the top of the mast [cm] 4.81159 cm -0.05%

Fz Forces at the top of the mast [N] 292677 N -0.08%




272

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

Test ID: 2524

Test status: Passed

1.85.1 Description

Verifies the displacements, bending moments and reinforcement results for a 2D concrete slab with supports andpunctual loads.

1.85.2 Background


■ 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

Slab geometry




273

Support positions

Positions of punctual loads




274

Global loading overview

Load Combinations

Code Numbers Type Title

BAGMAX 1 Static Permanent loads + self weightBAQ 2 Static Usage overloads

BAELS 101 Comb_Lin Gmax+Q

BAELU 102 Comb_Lin 1.35Gmax+1.5Q




275

1.85.2.2 Effel Structure Results

SLS max displacements (load combination 101)

Mx bending moment for ULS load combination




276

My bending moment for ULS load combination

Mxy bending moment for ULS load combination

1.85.2.3 Effel RC Expert Results

Main hypothesis

■ 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.




277

Axi reinforcements

Ayi reinforcements




278

Axs reinforcements

Ays reinforcements




279


Solver Resultname

Result description Referencevalue

CM2 D 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.



D 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%




280

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

Test ID: 2523

Test status: Passed

1.86.1 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.

1.86.2 Background


■ 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 Numbers Type Title

CMP 1 Static Permanent load + self weight

CMS 2 Static Usage overloads

CMCFN 101 Comb_Lin 1.333P

CMCFN 102 Comb_Lin 1.333P+1.5S

CMCFN 103 Comb_Lin P+1.5S

CMCD 104 Comb_Lin P+S




281

1.86.2.2 Effel Structure Results

Ridge displacements (combination 104)

Diagram of normal force envelope




282

Envelope of bending moments diagram

1.86.2.3 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



■ Buckling: XY 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

■ Work ratio optimization between 90 and 100%

■ Labels optimization (on Advance Design templates)




283

Deflection verification

Ratio

CM Stress diagrams

Work ratio

Stresses




284

Buckling lengths

Lfy

Lfz

Lateral-torsional buckling lengths

Ldi




285

Lds

Optimization




286



CM2 D Displacement at the ridge [cm] 9.36

CM2 Fx Envelope normal forces on Columns (min) [T] -15.77

CM2 Fx Envelope normal forces on Rafters (max) [T] -1.02

CM2 My Envelope bending moments on Columns (min) [T.m] -42.41

CM2 My 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] 6

CM2 Lfy Buckling lengths on Rafters - Lfy [m] 7.08

CM2 Lfz Buckling lengths on Rafters - Lfz [m] 10.11

Warning, the local axes in Effel Structure have different orientation in Advance Design.


CM2 Ldi Lateral-torsional buckling lengths on Columns Ldi [m] 6

CM2 Lds Lateral-torsional buckling lengths on Columns Lds [m] 2

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 Rate (%) Final section

CM2 Work ratio IPE500 columns - section optimization 98 IPE500

CM2 Work ratio IPE400 rafters - section optimization 195 IPE550




287



D Displacement at the ridge [cm] 9.36473 cm 0.05%

Fx Envelope normal forces on Columns (min) [T] -15.7798 T -0.06%Fx Envelope normal forces on Rafters (max) [T] -1.0161 T 0.38%

My Envelope bending moments on Columns (min) [Tm] 42.4226 T*m 0.03%

My 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] 6 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] 6 m 0.00%

Lds Lateral-torsional buckling lengths on Columns Lds [m] 2 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%




288

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

Test ID: 2514

Test status: Passed

1.87.1 Description

Verifies the rotation, the displacement and the moment on a beam consisting of two elements of the same length andidentical characteristics with 3 T/C supports (k = 0).

1.87.2 Background





UnitsI. S.

Geometry

■ L = 10 m

■ Section: IPE 200, Iz = 0.00001943 m4



N/m2,


Boundary conditions

■ Outer:



► T/C stiffness ky = 0,

■ Inner: None.

Loading


■ Internal: None.




289


ky being null, the non linear model behaves the same way as the structure without support 3.

Displacements


LkEI6PL

m00153.0Lk2EI316

PL3v


LkEI3PL


LkEI2PL3

3yzz

3yz

2

3

3yz

3

3

3yzz

3yz

2

2

3yzz

3yz

2

1

M z Moments

N.m2502

MM

4

PL)m5x(M

0Lk2EI316

PLk3M

0M

1z2zz

3

yz

4y

2z

1z



■ 3 nodes,


Deformed shape




290

Moment diagrams




CM2 RY Rotation Ry in node 2 [rad] -0.000153

CM2 DZ Displacement V in node 3 [m] 0.00153


CM2 My Moment M in node 1 [Nm] 0

CM2 My Moment M - middle span 1 [Nm] -250




RY Rotation Ry in node 2 [rad] -0.000153175 Rad -0.11%

DZ Displacement V in node 3 [m] 0.00153175 m 0.11%

RY Rotation Ry in node 3 [rad] -0.000153175 Rad -0.11%

My Moment M in node 1 [Nm] 5.05771e-014 N*m 0.00%

My Moment M - middle span 1 [Nm] -250 N*m 0.00%




291

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

Test ID: 2518

Test status: Passed

1.88.1 Description

Verifies the displacement and the normal force for a bar system containing 4 elements of the same length and 2diagonals.

1.88.2 Background




■ Element type: linear, bar, tie.

Units

I. S.

Geometry■ L = 5 m

■ Section S = 0.005 m2



N/m2.

Boundary conditions

■ Outer:



■ Inner: None.




292

Loading

■ External: Horizontal punctual load P = 50000 N at node 3,

■ Internal: None.


In non linear analysis without large displacement, the introduction of ties for the diagonal bars removes bar 5 (testNo. 0103SSLLB_FEM allows finding an compression force in this bar at the linear calculation).

Displacements

0v

m000238.0ES

PLv

m001195.0ES11

PL5uu

4

3

43

N normal forces

0N 0N

N70711P2N N50000PN

0N 0N

4243

1323

1412


■ Linear element: bar, without meshing,

■ 4 nodes,


Deformed shape




293

Normal forces




294




CM2 DZ v3 displacement on Node 3 [m] -0.000238


CM2 DZ v4 displacement on Node 4 [m] 0





CM2 Fx N13 normal effort on Element 5 [N] 70711




DX u3 displacement on Node 3 [m] 0.00119035 m -0.39%

DZ v3 displacement on Node 3 [m] -0.000238095 m -0.04%

DX u4 displacement on Node 4 [m] 0.00119035 m -0.39%

DZ v4 displacement on Node 4 [m] 9.94686e-316 m 0.00%


Fx N23 normal force on Element 2 [N] -50000 N 0.00%


Fx N14 normal force on Element 4 [N] 2.08884e-307 N 0.00%

Fx N13 normal effort on Element 5 [N] 70710.7 N 0.00%





295

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

Test ID: 2522

Test status: Passed

1.89.1 Description

Verifies the steel calculation results (maximum displacement, normal force, bending moment, deflections, bucklinglengths, lateral-torsional buckling and cross section optimization) for a simple metallic framework with a concretefloor, according to CM66.

1.89.2 Background


■ 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




296

Structure’s load case

Code No. Type Title

CMP 1 Static SW + Dead loadsCMS 2 Static Overloads for usageCMV 3 Static Wind overloads along +X in overpressureCMV 4 Static Wind overloads along +X in depression

CMV 5 Static Wind overloads along -X in overpressureCMV 6 Static Wind overloads along -X in depressionCMV 7 Static Wind overloads along +Z in overpressureCMV 8 Static Wind overloads along +Z in depressionCMV 9 Static Wind overloads along -Z in overpressureCMV 10 Static Wind overloads along -Z in depressionCMN 11 Static Normal snow overloads

1.89.2.2 Effel Structure results

Displacement Envelope (“CMCD" load combinations)

Envelope of linear element forces

D DX DY DZ Env. Case No. Max.location (cm) (cm) (cm) (cm)

Max(D) 213 148 CENTER 12.115 0.037 12.035 -1.393

Min(D) 188 1.1 START 0.000 0.000 0.000 0.000

Max(DX) 204 72.1 START 3.138 3.099 0.434 0.244

Min(DX) 204 313 END 2.872 -1.872 -0.129 -2.174

Max(DY) 213 148 CENTER 12.115 0.037 12.035 -1.393

Min(DY) 213 61.5 END 9.986 -0.118 -9.985 0.046

Max(DZ) 201 371 CENTER 4.149 -0.006 -0.188 4.145

Min(DZ) 203 370 CENTER 4.124 -0.006 -0.240 -4.118

Envelope of forces on linear elements (“CMCFN” load combinations)

Envelope of linear element forces

Fx Fy Fz Mx My Mz Env. Case No. MaxSite (T) (T) (T) (T*m) (T*m) (T*m)

Max (Fx) 120 4.1 START 19.423 -4.108 -1.384 -0.003 1.505 7.551

Min (Fx) 138 98 START -41.618 -0.962 -0.192 0.000 0.000 0.000

Max(Fy) 120 57 END -13.473 16.349 -0.016 -0.003 0.002 55.744

Min(Fy) 120 60 START -15.994 -16.112 -0.006 -3E-004 6E-006 53.096

Max(Fz) 177 371 START -3.486 -0.118 2.655 0.000 0.000 0.000

Min(Fz) 187 370 START -3.666 -0.147 -2.658 0.000 0.000 0.000

Max(Mx) 120 111 END 3.933 4.840 0.278 0.028 -4E-005 11.531

Min(Mx) 120 21 END -22.324 13.785 -0.191 -0.028 -0.004 42.562

Max(My) 177 371 CENTER -3.099 -0.118 -0.323 0.000 4.403 -0.500

Min(My) 179 370 CENTER -3.283 -0.155 0.321 0.000 -4.373 -0.660Max (Mz) 120 57 END -13.473 16.349 -0.016 -0.003 0.002 55.744

Min (Mz) 120 59.2 END -19.455 -8.969 -0.702 -0.003 -0.001 -57.105

Envelope of linear element stresses (“CMCFN” load combinations)

Envelope of linear element stresses

sxxMax sxyMax sxzMax sFxx sMxxMax Env. Case No. MaxSite

(MPa) (MPa) (MPa) (MPa) (MPa)

Max(sxxMax) 120 59.2 END 273.860 -14.696 -1.024 -16.453 290.312

Min(sxxMax) 120 292 START -150.743 0.000 0.000 -150.743 0.000

Max(sxyMax) 120 57 START 262.954 37.139 -0.030 -15.609 278.562

Min(sxyMax) 120 60 END 241.643 -36.595 -0.011 -18.536 260.179

Max(sxzMax) 185 371 START -2.949 -0.183 3.876 -2.949 0.000

Min(sxzMax) 179 370 START -3.104 -0.255 -3.882 -3.104 0.000

Max(sFxx) 120 293 END 161.095 9E-005 -0.002 161.095 0.000

Min(sFxx) 120 292 START -150.743 0.000 0.000 -150.743 0.000

Max(sMxxMax) 120 59.2 END 273.860 -14.696 -1.024 -16.453 290.312

Min(sMxxMax) 1 1.1 START -4.511 3.155 -0.646 -4.511 0.000




297

1.89.2.3 CM66 Effel Expertise results

Hypotheses

For columns



■ Buckling XY plane: Automatic calculation of the structure on displaceable nodes


■ Lateral-torsional buckling: Ldi automatic calculation: hinged restraint

Lds automatic calculation: hinged restraint

For rafters



■ Buckling: XY plane: Automatic calculation of the structure on displaceable nodes


■ Lateral-torsional buckling: Ldi automatic calculation: no restraint

Lds automatic calculation: hinged restraint

For columns



■ 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 restraintLds 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.




298

Deflection verification

Ratio

Max values on the element

■ Columns: L / 168

■ Rafter: L / 96

■ Column: L / 924

CM Stress diagrams




299

Work ratio

Stresses

Max values on the element■ Columns: 375.16 MPa

■ Rafter: 339.79 MPa

■ Column: 180.98 MPa




300

Buckling lengths

Lfy

Lfz




301

Lateral-torsional buckling lengths

Ldi

Lds




302

Optimization



CM2 D Maximum displacement (CMCD) [cm] 12.115

CM2 Fx Envelope normal force (CMCFN) Min (Fx) [T] -41.618

CM2 Fx Envelope normal force (CMCFN) Max (Fx) [T] 19.423

CM2 My Envelope bending moment (CMCFN) Min (Mz) [Tm] -57.105CM2 My 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.


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 Rafters [MPa] 339.74


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




303

Warning, the local axes in Effel Structure are opposite to those in Advance Design.


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] 2

CM2 Lds Lateral-torsional buckling lengths on Columns Lds [m] 2

Solver Result name Result description Rate (%) Final section

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



D Maximum displacement (CMCD) [cm] 12.1379 cm 0.19%

Fx Envelope normal force (CMCFN) Min (Fx) [T] -41.6547 T -0.09%

Fx Envelope normal force (CMCFN) Max (Fx) [T] 19.4819 T 0.30%

My Envelope bending moment (CMCFN) Min (Mz)[Tm] -57.1196 T*m -0.03%

My Envelope bending moment (CMCFN) Max (Mz) [Tm] 55.7697 T*m 0.05%

Deflection CM deflections on Columns [adm] 167.707 Adim. -0.17%

Deflection CM deflections on Rafters [adm] 96.1075 Adim. 0.11%

Deflection CM deflections on Columns [adm] 924.933 Adim. 0.10%

Stress CM stresses on Columns [MPa] 374.61 MPa -0.02%

Stress CM stresses on Rafters [MPa] 352.36 MPa 3.71%

Stress CM stresses on Columns [MPa] 180.693 MPa -0.16%

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] 2 m 0.00%

Lds Lateral-torsional buckling lengths on Columns Lds [m] 2 m 0.00%

Work ratio IPE500 columns - section optimization [adm] 1.59408 Adim. 0.26%

Work ratio IPE400 rafters - section optimization [adm] 1.4994 Adim. 3.41%

Work ratio IPE400 columns - section optimization [adm] 0.768905 Adim. -0.14%




304

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

Test ID: 2436

Test status: Passed

1.90.1 Description

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

1.90.2 Background





Slender beam with variable section (fixed-free) Scale =1/4

01-0004SDLLB_FEM

Units

I. S.

Geometry


■ 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,




305

■ 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.



Pa,

■ Density: 7800 kg/m3.

Boundary conditions

■ Outer: Fixed in A,

■ Inner: None.

Loading

■ External: None,

■ Internal: None.


Reference solutions

Precise calculation by numerical integration of the differential equation of beams bending (Euler-Bernoulli theories):

2

x2 (EIz

2v

x2 ) = - A

2v

x2 where Iz and A vary with the abscissa.

The result is: f i =1

2 i

h2

l2

E

12

1 2 3 4 5

23.289 73.9 165.23 299.7 478.1


■ Linear element: variable beam, imposed mesh,

■ 31 nodes,


Eigen mode shapes




306






CM2 Eigen mode Eigen mode 4 frequency [Hz] 697.24CM2 Eigen mode Eigen mode 5 frequency [Hz] 1112.28











307

1.91 Tied (sub-tensioned) beam (01-0005SSLLB_FEM)

Test ID: 2437

Test status: Passed

1.91.1 Description

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

1.91.2 Background



■ Analysis type: static, thermoelastic (plane problem);


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,




308

■ 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.


■ Isotropic linear elastic material,


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 = 6.52 x 10-3

m (dilatation coefficient: CE = 1 x 10-5

/°C

and temperature variation T = -163°C).

1.91.2.2 Compression force in CE bar

Reference solution

The solution is established by considering the deformation effects due to the shear force and normal force:

= 1 -43 x

aL

k = A Ar

= 2.5

t =I

A

= (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)

= k x [(2Et2) / (GaL)]

= + +

0 = 1 – (a/L)2

x (2 – a/L)

0 = 6k x (E/G) x (t/L)2 x (1 + b/L)

0 = 0 + 0

NCE = - (1/12) x (pL2/c) x (0 /) + (EI/(Lc

2)) x (/) = 584584 N


Linear element: without meshing,

■ AD, DF, FB: S beam (considering the shear force deformations),

■ AC, CD, EF, EB: bar,

■ CE: beam,

■ 6 nodes.




309

Force diagrams


Compression force in CE bar

1.91.2.3 Bending moment at point H

Reference solution

MH = - (1/8) x pL2 x [1- (2/3) x (0/)] – (EI/(Lc)) x (/p) = 49249.5 N


Linear element: without meshing,

■ AD, DF, FB: S beam (considering the shear force deformations),

■ AC, CD, EF, EB: bar,

■ CE: beam,

■ 6 nodes.




310

Shape of the bending moment diagram


Mz bending moment

1.91.2.4 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


■ Linear element: without meshing,

► AD, DF, FB: S beam (considering the shear force deformations),

► AC, CD, EF, EB: bar,

► CE: beam,

■ 6 nodes.




311

Deformed shape


Deformed


Solver Result name Result description Reference valueCM2 FX Tension force on CE bar [N] 584584



Fx Tension force on CE bar [N] 584580 N 0.00%




312

1.92 Circular plate under uniform load (01-0003SSLSB_FEM)

Test ID: 2435

Test status: Passed

1.92.1 Description

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

1.92.2 Background


Reference: Structure Calculation Software Validation Guide, test SSLS 03/89;



Circular plate under uniform load Scale =1/1001-0003SSLSB_FEM

Units

I. S.

Geometry

■ Circular plate radius: r = 1m,

■ Circular plate thickness: h = 0.005 m.



Pa,





313

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 rotationrestrained nodes along y and z: translation restrained nodes along x and rotation restrained nodes along y andz:

■ Inner: None.

Loading

■ External: Uniform loads perpendicular on the plate: pZ = -1000 Pa,

■ Internal: None.

1.92.2.2 Vertical displacement of the model at the center of the plate

Reference solution

Circular plates form:

u =pr

4

64D

=-1000 x 1

4

64 x 2404

= - 6.50 x 10-3

m

with the plate radius coefficient: D =Eh

3

12(1-2) =

2.1 x 1011

x 0.0053

12(1-0.32)

D = 2404



■ 70 nodes,


Circular plate under uniform load Scale =1.5

Meshing 01-0003SSLSB_FEM




314

Deformed shape

Circular plate under uniform load Scale =1.5

Deformed 01-0003SSLSB_FEM



CM2 DZ Vertical displacement on the plate center [mm] -6.50



DZ Vertical displacement on the plate center [mm] -6.47032 mm 0.46%




315

1.93 Verifying the displacement results on linear elements for vertical seism (TTAD #11756)

Test ID: 3442

Test status: Passed

1.93.1 Description

Verifies the displacements results on an inclined steel bar for vertical seism according to Eurocodes 8 localizationand 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 Zdirection (vertical).

1.94 Verifying constraints for triangular mesh on planar elements (TTAD #11447)

Test ID: 3460

Test status: Passed

1.94.1 Description

Performs the finite elements calculation, verifies the stresses for triangular mesh on a planar element and generatesa 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 onFX direction.

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

Test ID: 3551

Test status: Passed

1.95.1 Description

Verifies forces results on concrete beams consisting of a linear element and on beams consisting of two linearelements. Generates the linear elements forces by load case report.

1.96 Verifying diagrams after changing the view from standard (top, left,...) to user view (TTAD#11854)

Test ID: 3539

Test status: Passed

1.96.1 Description

Verifies the results diagrams display after changing the view from standard (top, left,...) to user view.

1.97 Verifying forces for triangular meshing on planar element (TTAD #11723)

Test ID: 3463

Test status: Passed

1.97.1 Description

Performs the finite elements calculation, verifies the forces for triangular meshing on a planar element and generatesa 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. Alinear load of -10.00 kN is applied on FZ direction.




316




317

1.98 Verifying stresses in beam with "extend into wall" property (TTAD #11680)

Test ID: 3491

Test status: Passed

1.98.1 Description

Verifies the results on two concrete beams which have the "Extend into the wall" option enabled. One of the beams isconnected to 2 walls on both sides and one with a wall and a pole. Generates the linear elements forces by elementsreport.

1.99 Generating planar efforts before and after selecting a saved view (TTAD #11849)

Test ID: 3454

Test status: Passed

1.99.1 Description

Generates efforts for all planar elements before and after selecting the third saved view.

1.100 Verifying results on punctual supports (TTAD #11489)

Test ID: 3693

Test status: Passed

1.100.1 Description

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 actionsby 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.

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

Test ID: 3609

Test status: Passed

1.101.1 Description

Performs the finite elements calculation on a model with two planar concrete elements with a linear support. Verifiesthe 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: selfweight, a planar live load of -1 kN and seism loads according to French standards of Eurocodes 8.

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

Test ID: 3610

Test status: Passed

1.102.1 Description

Performs the finite elements calculation and verifies the results diagrams for Mf torsors on a high wall divided in 6walls (by height).

The loads applied on the model: self weight, two live load cases and seism loads according to Eurocodes 8.




318

1.103 Verifying Sxx results on beams (TTAD #11599)

Test ID: 3595

Test status: Passed

1.103.1 Description

Performs the finite elements calculation on a complex model with concrete, steel and timber elements. Verifies theSxx 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 onthe structure: dead loads, live loads, snow loads, wind loads and temperature loads (according to Eurocodes).

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

Test ID: 3594

Test status: Passed

1.104.1 Description

Performs the finite elements calculation on a complex concrete structure with four levels. Generates results forTorsors NZ/Group. Verifies the legend results.

The structure has 88 linear elements, 30 planar elements, 48 windwalls, etc.

1.105 Verifying nonlinear analysis results for frames with semi-rigid joints and rigid joints (TTAD#11495)

Test ID: 3795

Test status: Passed

1.105.1 Description

Verifies the nonlinear analysis results for two frames with one level. One of the frames has semi-rigid joints and theother has rigid joints.

1.106 Generating a report with torsors per level (TTAD #11421)

Test ID: 3774

Test status: Passed

1.106.1 Description

Generates a report with the torsors per level results.

1.107 Verifying tension/compression supports on nonlinear analysis (TTAD #11518)

Test ID: 4198

Test status: Passed

1.107.1 Description

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 atthe base and a T/C punctual support at the top. A value of 15000.00 kN/m is defined for the KTX and KTZ stiffenersof the T/C support. Two loads of 500.00 kN are applied.




319

1.108 Verifying tension/compression supports on nonlinear analysis (TTAD #11518)

Test ID: 4197

Test status: Passed

1.108.1 Description

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 atthe base and a T/C punctual support at the top. A large value of the KTX stiffener of the T/C support is defined. Twoloads of 500.00 kN are applied.

1.109 Verifying the display of the forces results on planar supports (TTAD #11728)

Test ID: 4375

Test status: Passed

1.109.1 Description

Performs the finite elements calculation and verifies the display of the forces results on a planar support. The modelconsists of a concrete vertical element with a planar support.




320

1.110 Verifying results of a steel beam subjected to dynamic temporal loadings (TTAD #14586)

Test ID: 5853

Test status: Passed

1.110.1 Description

Verifies a double-end fixed steel beam subjected to harmonic concentrated loadings.

2 Hz and 3 Hz excitation frequencies are studied.

1.110.2 Background

The harmonic response of a steel beam fixed at both ends is studied. The beam contains 8 elements having thesame length and identical characteristics. Harmonic concentrated loadings (a vertical load and a bending moment)are applied in the middle of the beam. Two excitation frequencies are studied: 2.0 and 3.0 Hz.


■ Reference: NE/Nastran V8;■ Analysis type: modal analysis;


Units

I. S.

Geometry

Below are described the beam cross section characteristics:

■ Beam length: L = 16 m,

■ Square shaped cross section: b = 0.05 m,■ Section area: A = 0.06 m2,

■ Flexion inertia moment about the y (or z) axis: I = 0.0001 m4.


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

■ Poisson coefficient: = 0.3,


Boundary conditions

■ Outer:

► Fixed support at start point (x = 0),




321

► Fixed s

upport at end point (x= 16.00).

■ Inner: None.

Loading

■ External:

► Point load at x=8: P = Fz = -50 000 sin (2π f t) N

► Bending moment at x=8: M = My = 10000 sin (2π f t) Nm

■ Internal: None.

1.110.2.2 Harmonic response from NE/Nastran V8

Reference solution

Considering a natural frequency (modal) analysis for a double-end fixed beam, the first four natural frequencies canbe determined using the following formula:

A

I E

L f n

n

2

2

2

The modal response is determined considering 14 modes.

The first four mode shapes and their frequencies are:

12 = 22.37 f1 = 2.937 Hz

22 = 61.67 f2 = 8.095 Hz

32 = 120.9 f3 = 15.871 Hz

42 = 199.8 f4 = 26.228 Hz

The vertical reference displacement is calculated in the middle of the beam at x = 8 m.

Software NE/NASTRAN 8.0

Vertical maximum displacement (f = 2 Hz) in the middleof the beam

0.155 m

Vertical maximum displacement (f = 3 Hz) in the middleof the beam

2.266 m




322

Response in the middle of the beam with f = 2 Hz

Response in the middle of the beam with f = 3 Hz


■ Linear element: beam, imposed mesh

■ 9 nodes,





323



Deformed – D Vertical maximum displacement in the middle of the beam2 Hz m

0.155 m

Deformed – D Vertical maximum displacement in the middle of the beam(3 Hz) [m]

2.266 m



D Vertical maximum displacement in the middle of the beam (2 Hz) 15.3783 cm -0.7853 %

D Vertical maximum displacement in the middle of the beam (3 Hz) 211.41 cm -6.7036 %



1.111 Verifying the main axes results on a planar element (TTAD #11725)

Test ID: 4310

Test status: Passed

1.111.1 Description

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 forcesresults on the planar element main axes.

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

Test ID: 4804

Test status: Passed

1.112.1 DescriptionVerifies torsors on a single story coupled walls subjected to horizontal forces

1.113 Calculating torsors using different mesh sizes for a concrete wall subjected to a horizontalforce (TTAD #13175)

Test ID: 5088

Test status: Passed

1.113.1 Description

Calculates torsors using different mesh sizes for a concrete wall subjected to a horizontal force.

1.114 Verifying the internal forces results for a simple supported steel beam

Test ID: 4533

Test status: Passed

1.114.1 Description

Performs the finite elements calculation for a horizontal element (S235 material and IPE180 cross section) with twohinge rigid supports at each end. One of the supports has translation restraints on X, Y and Z, the other support hasrestraints 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.



325

1.115 Verifying forces on a linear elastic support which is defined in a user workplane (TTAD#11929)

Test ID: 4553

Test status: Passed

1.115.1 Description

Verifies forces on a linear elastic support, which is defined in a user workplane, and generates a report with forces forlinear support in global and local workplane.





2 CAD, rendering and visualization




328

2.1 Verifying hide/show elements command (TTAD #11753)

Test ID: 3443

Test status: Passed

2.1.1 Description

Verifies the hide/show elements command for the whole structure using the right-click option.

2.2 Verifying the dimensions and position of annotations on selection when new analysis ismade.(TTAD #12807)

Test ID: 6201

Test status: Passed

2.2.1 Description

Verifies the dimensions and position of annotations on selection when new analysis is made. It takes a printscreen ofthe loaded saved view.

2.3 Verifying the saved view of elements with annotations. (TTAD #13033)

Test ID: 6209

Test status: Passed

2.3.1 Description

Verifies the saved view of elements with annotations. It makes saved views and takes printscreens of them afterswitching between them.

2.4 Verifying the visualisation of supports with rotational or moving DoFs.(TTAD #13891)

Test ID: 6213

Test status: Passed

2.4.1 Description

Verifies the visualisation of supports with rotational or moving DoFs by taking a printscreen.

2.5 Verifying the annotations of a wind generated load. (TTAD #13190)

Test ID: 6210

Test status: Passed

2.5.1 Description

Verifies annotations of a wind generated load. It generates wind on the current 2 slope building( the remarks arefilled) then hides all the elements, makes visible only the load with identifier 1 and takes a printscreen.




329

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

Test ID: 3457

Test status: Passed

2.6.1 Description

Performs the finite elements calculation and verifies the section cut results on a concrete planar element with anopening.

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 andbase effect); generates the combinations description report.

2.8 Verifying the grid text position (TTAD #11704)

Test ID: 3464

Test status: Passed

2.8.1 Description

Verifies the grid text position 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 descriptiveactors 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 Verifying the coordinates system symbol (TTAD #11611)Test ID: 3550

Test status: Passed

2.10.1 Description

Verifies the coordinates system symbol display from different views.




330

2.11 Creating a circle (TTAD #11525)

Test ID: 3607

Test status: Passed

2.11.1 Description

Creates a circle.

2.12 Creating a camera (TTAD #11526)

Test ID: 3608

Test status: Passed

2.12.1 Description

Verifies the camera creation and visibility.

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

Test ID: 3701

Test status: Passed

2.13.1 Description

Verifies the representation of elements with HEA340 cross section.

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

Test ID: 3644

Test status: Passed

2.14.1 Description

Verifies the snap points behavior when the "Allowed deformation" function is enabled (stretch points) and when it isdisabled (grip points).

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

Test ID: 3637

Test status: Passed

2.15.1 Description

Changes the local axes of a section cut in the descriptive model and verifies if the local axes are kept in analysismodel.




331

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

Test ID: 3733

Test status: Passed

2.16.1 Description

Performs the finite elements calculation and displays the forces results on linear elements. Returns to the modelmode to verify the descriptive model display.

2.17 Modeling using the tracking snap mode (TTAD #10979)

Test ID: 3745

Test status: Passed

2.17.1 DescriptionEnables the "tracking" snap mode to model structure elements.

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

Test ID: 3740

Test status: Passed

2.18.1 Description

Verifies holes in horizontal planar elements after changing the level height.

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

Test ID: 3742

Test status: Passed

2.19.1 Description

Creates an element with compound cross section (CS1 IPE400 IPE240) and verifies the cross section display.

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.




332

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

Test ID: 4304

Test status: Passed

2.21.1 Description

Verifies the on/off function for the "ghost" rendering mode when the workplane display is disabled.

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

Test ID: 4349

Test status: Passed

2.22.1 Description

Verifies the "ghost" display on selected elements after changing the element display color.

2.23 Verifying the grid text position (TTAD #11657)

Test ID: 3465

Test status: Passed

2.23.1 Description

Verifies the grid text position from different views.

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

Test ID: 4347

Test status: Passed

2.24.1 Description

Verifies the display of saved views which contain elements with the "ghost on selection" function enabled.

2.25 Verifying the steel connections modeling (TTAD #11698)

Test ID: 4440

Test status: Passed

2.25.1 Description

Verifies the modeling of steel connections.

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

Test ID: 4429

Test status: Passed

2.26.1 Description

Verifies the "fixed load scale" function.




333

2.27 Verifying the saved view of elements by cross-section. (TTAD #13197)

Test ID: 6180

Test status: Passed

2.27.1 Description

Verifies the saved view of elements by cross-section. It takes a printscreen of the loaded saved view.

2.28 Verifying the annotations dimensions when new analysis is made.(TTAD #14825)

Test ID: 6199

Test status: Passed

2.28.1 Description

Verifies the annotations dimensions when new analysis is made. It takes a printscreen of the loaded saved view.

2.29 Verifying the default view.(TTAD #13248)

Test ID: 6198

Test status: Passed

2.29.1 Description

Verifies the default view to be the top view. It takes a printscreen of the loaded saved view.

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

Test ID: 4483

Test status: Passed

2.30.1 Description

Verifies the dividing of planar elements which contain openings.

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

Test ID: 4507

Test status: Passed

2.31.1 Description

Verifies the program behavior when trying to create lintel on a planar element with an inappropriate opening.

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

Test ID: 4511

Test status: Passed

2.32.1 Description

Verifies the program behavior when launching the analysis on a model that had overlapped loads.




334

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

Test ID: 4508

Test status: Passed

2.33.1 Description

Creates a punctual load and verifies the display of the load after placing it in another load case using the load casenumber from the properties window.

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

Test ID: 4513

Test status: Passed

2.34.1 Description

Verifies the display of a beam with haunches, in the "Linear contour" rendering mode.

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

Test ID: 4534

Test status: Passed

2.35.1 Description

Creates a base plate connection on a non-vertical column.

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

Test ID: 4551

Test status: Passed

2.36.1 Description

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

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

Test ID: 4560

Test status: Passed

2.37.1 Description

Verifying rotation for steel beam with joint at one end (TTAD #12592)

2.38 Verifying annotation on selection (TTAD #12700)

Test ID: 4575

Test status: Passed

2.38.1 Description

Verifying annotation on selection (TTAD #12700)



3 Climatic generator




336

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

Test ID: 4849

Test status: Passed

3.1.1 Description

Generates snow loads on 2 closed building with gutters, according to the Eurocodes 1 - French standard (NF EN1991-1-3/NA).

3.2 EC1: wind load generation on a high building with horizontal roof using UK annex(DEV2013#4.1) (TTAD #12608)

Test ID: 5560

Test status: Passed

3.2.1 Description

Generates wind loads on the windwalls of a concrete structure, according to the BS EN 1991-1-4:2005 standard.

The structure is 63m high, has 4 columns and 4 beams (R20*30 cross section and C20/25 material), rigid supportsand horizontal roof.

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

Test ID: 4569

Test status: Passed

3.3.1 Description

Generates snow loads on a 4 slopes shed with gutters on each slope and middle parapets, according to theEurocodes 1 - French standard (NF EN 1991-1-3/NA).

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

Test ID: 4570

Test status: Passed

3.4.1 Description

Generates snow loads on a single slope with lateral parapets, according to the Eurocodes 1 - French standard (NFEN 1991-1-3/NA).

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

Test ID: 4568

Test status: Passed

3.5.1 Description

Generates snow loads on a 4 slopes shed with gutters on each slope and lateral parapets, according to theEurocodes 1 - French standard (NF EN 1991-1-3/NA).




337

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

Test ID: 4580

Test status: Passed

3.6.1 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 EN1991-1-4/NA).

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

Test ID: 4847

Test status: Passed

3.7.1 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). It verifies snow fall from higher to lower close building.

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

Test ID: 4848

Test status: Passed

3.8.1 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).

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

Test ID: 4846

Test status: Passed

3.9.1 Description

Generates snow loads on a model from CTCIM which contains 4 slopes with gutters and lateral parapets, accordingto the Eurocodes 1 - French standard (NF EN 1991-1-3/NA). It also verifies snow fall from higher to lower closebuilding.


Test ID: 4851

Test status: Passed

3.10.1 Description

Generates snow loads on 2 closed building with gutters. The lower building is longer and has a 4 slope shed and thehigher building has a 2 slope roof. The snow loads are generated according to the Eurocodes 1 - French standard(NF EN 1991-1-3/NA).




338

3.11 EC1: Generating wind loads on a square based structure according to UK standards (BS EN1991-1-4:2005) (TTAD #12608)

Test ID: 4845

Test status: Passed

3.11.1 Description

Generates the wind loads on a square based structure. The wind loads are generated according to Eurocodes 1 - UKstandards (BS EN 1991-1-4:2005).

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

Test ID: 4852

Test status: Passed

3.12.1 DescriptionGenerates snow loads on a 2 slope building with gutters and lateral parapets on all sides, according to the Eurocodes1 - French standard (NF EN 1991-1-3/NA).


Test ID: 4850

Test status: Passed

3.13.1 Description

Generates snow loads on 2 closed building with gutters. The lower building is longer. The wind loads are generatedaccording to Eurocodes 1 - French standards (NF EN 1991-1-3/NA).

3.14 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

3.14.1 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). The third slope is an extension of the roof with a different angle.

3.15 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

3.15.1 Description

Generates wind loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA).




339

3.16 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

3.16.1 Description

Generates wind loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA).

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

Test ID: 4526

Test status: Passed

3.17.1 Description

Generates wind loads on a triangular based lattice structure with compound profiles using user defined "n" - eigenmode frequency - (NF EN 1991-1-4/NA) (TTAD #12276).

3.18 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

3.18.1 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).

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

Test ID: 4525

Test status: Passed

3.19.1 Description

Generates the wind loads on a triangular based lattice structure with compound profiles, using automatic calculationof "n" - eigen mode frequency (NF EN 1991-1-4/NA). The wind loads are generated according to Eurocodes 1 -

French standards.

3.20 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

3.20.1 Description

Generates snow loads on a 2 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA)




340

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

Test ID: 4510

Test status: Passed

3.21.1 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.

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

Test ID: 4478

Test status: Passed

3.22.1 Description

Verifies the geometry of wind loads on an irregular shed. The wind loads are generated according to Eurocodes 1 -French standard.

3.23 EC1: Generating snow loads on a 4 slopes shed with parapets. (TTAD #14578)

Test ID: 6193

Test status: Passed

3.23.1 Description

Generates snow loads on a 4 slopes shed with parapets on Y+/- sides, according to the Eurocodes 1 - Frenchstandard (NF EN 1991-1-3/NA).

3.24 EC1: Generating 2D snow loads on a 2 slope portal with one lateral parapet. (TTAD #14530)

Test ID: 6195

Test status: Passed

3.24.1 Description

Generates 2D snow loads on a 2 slope portal with one lateral parapet, according to the Eurocodes 1 - Frenchstandard (NF EN 1991-1-3/NA).

3.25 EC1: Generating wind loads on a 2 almost horizontal slope building. (TTAD #13663)

Test ID: 6196

Test status: Passed

3.25.1 Description

Generates wind loads on a 2 almost horizontal slope building, according to the Eurocodes 1 - French standard (NFEN 1991-1-4/NA).




341

3.26 EC1: Generating 2D wind loads on a 2 slope portal. (TTAD #14531)

Test ID: 6194

Test status: Passed

3.26.1 Description

Generates 2D wind loads on a 2 slope portal, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA).

3.27 EC1: Generating wind loads on a 4 slopes shed with parapets. (TTAD #14179)

Test ID: 6184

Test status: Passed

3.27.1 Description

Generates wind loads on a 4 slopes shed with parapets on all sides, according to the Eurocodes 1 - French standard(NF EN 1991-1-4/NA).

3.28 EC1: Generating snow loads on 2 side by side single roof compounds with different height(TTAD 13158)

Test ID: 6188

Test status: Passed

3.28.1 Description

Generates snow loads on on 2 side by side single roof compounds, according to the Eurocodes 1 - French standard(NF EN 1991-1-3/NA). One compound is higher than the other and the slopes have opposite sign. The model isreversed in comparison with the one from 6187 and has only the exceptional snow fall is checked.

3.29 EC1: Generating snow loads on 2 side by side single roof compounds with different height(TTAD 13159)

Test ID: 6187

Test status: Passed

3.29.1 Description

Generates snow loads on on 2 side by side single roof compounds, according to the Eurocodes 1 - French standard(NF EN 1991-1-3/NA). One compound is higher than the other and the slopes have opposite sign. Exceptional snowfalls and accumulations are checked.

3.30 EC1: snow load generation on double compound with gutters and parapets on all sides.(TTAD#13717)

Test ID: 6186

Test status: Passed

3.30.1 Description

Generates snow loads on a metal based double compound with gutters and parapets on all sides, according to theEurocodes 1 France. One compound is a double-roof volume and the second is a single-roof volume with the sameslope as the one it is next to.




342

3.31 EC1: snow load generation on building with 2 slopes > 60 degrees according to Czech nationalannex. (TTAD #14235)

Test ID: 6185

Test status: Passed

3.31.1 Description

Generates snow load on building with 2 slopes > 60 degrees, according to the Eurocodes 1 - Czech standard.

3.32 EC1: Generating snow loads on 2 side by side single roof compounds (TTAD #13286)

Test ID: 6177

Test status: Passed

3.32.1 Description

Generates snow loads on on 2 side by side single roof compounds, according to the Eurocodes 1 - French standard(NF EN 1991-1-3/NA). It verifies the normal and accidental snow loads on Y+/- wind directions.

3.33 EC1: Generating wind loads on a 2 slope building with parapets. (TTAD #13669)

Test ID: 6168

Test status: Passed

3.33.1 Description

Generates wind loads on a 2 slope building with lateral parapets on all sides, according to the Eurocodes 1 - Frenchstandard (NF EN 1991-1-4/NA).

3.34 EC1: Generating wind loads on a 2 slope building with increased height. (TTAD #13759)

Test ID: 6172

Test status: Passed

3.34.1 Description

Generates wind loads on a 2 slope building with increased height to 26m, according to the Eurocodes 1 - Frenchstandard (NF EN 1991-1-4/NA).

3.35 EC1: Generating snow loads on a 2 slope building with custom pressure values. (TTAD

#14004)

Test ID: 6176

Test status: Passed

3.35.1 Description

Generates snow loads on a 2 slope building, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).It verifies the accidental accumulation from exceptional drifted snow when other region is selected.




343

3.36 EC1: wind load generation on portal with CsCd set to auto according to Romanian nationalannex. (TTAD #13930w)

Test ID: 6182

Test status: Passed

3.36.1 Description

Generates wind loads on a 3 compound building according to the Eurocodes 1 Romanian standard (CR 1-1-4/2012)using auto CsCd values and CsCd min to 0.7. 2 compounds are double sloped and one is single sloped.

3.37 EC1: snow load generation on a 3 compound building according to Romanian national annex.(TTAD #13930s)

Test ID: 6183

Test status: Passed

3.37.1 Description

Generates snow load generation on a 3 compound building according to the Eurocodes 1 Romanian standard (CR 1-1-3/2012). 2 compounds are double sloped and one is single sloped. It also verifies parapet and valleyaccumulations.

3.38 EC1: Generating snow loads on a 2 slope building with gutters and lateral parapets. (TTAD#14005)

Test ID: 6181

Test status: Passed

3.38.1 Description

Generates snow loads on a 2 slope building with gutters and lateral parapets, according to the Eurocodes 1 - Frenchstandard (NF EN 1991-1-3/NA).

3.39 EC1: Generating snow loads on a 2 slope building with parapets. (TTAD #13671)

Test ID: 6167

Test status: Passed

3.39.1 Description

Generates snow loads on a 2 slope building with lateral parapets on all sides, according to the Eurocodes 1 - French

standard (NF EN 1991-1-3/NA).

3.40 EC1: snow load generation on compound with a double-roof volume close to a single-roofvolume (TTAD #13559)

Test ID: 6166

Test status: Passed

3.40.1 Description

Generates snow loads on a metal based compound with a double-roof volume close to a single-roof volumeaccording to the Eurocodes 1 France.




344

3.41 EC1: wind load generation on multibay canopies (TTAD #11668)

Test ID: 6164

Test status: Passed

3.41.1 Description

Generates wind loads on multibay canopies, according to the Eurocodes 1 France.

3.42 EC1: wind load generation on portal with CsCd set to auto (TTAD #12823)

Test ID: 6165

Test status: Passed

3.42.1 Description

Generates wind loads on a 3 slope building according to the Eurocodes 1 France standard using auto CsCd values

and CsCd min to 0.7.

3.43 EC1: generating wind loads on a 35m high structure according to Eurocodes 1 - Frenchstandard with CsCd min set to 0.7 and Delta to 0.15. (TTAD #11196)

Test ID: 6170

Test status: Passed

3.43.1 Description

Generates wind loads on the roof a 35m high structure, according to Eurocodes 1 - French standard (NF EN 1991-1-4/NA) with CsCd min set to 0.7 and Delta to 0.15.

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

3.44 EC1: generating wind loads on a canopy according to Eurocodes 1 - French standard. (TTAD#13855)

Test ID: 6171

Test status: Passed

3.44.1 Description

Generates wind loads on a canopy, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA) only forselected wind directions.

3.45 EC1: Generating wind loads on a single-roof volume compound with parapets. (TTAD #13672)

Test ID: 6169

Test status: Passed

3.45.1 Description

Generates wind loads on a single-roof volume compound with lateral parapets on all sides, according to theEurocodes 1 - French standard (NF EN 1991-1-4/NA).




345

3.46 EC1: Generating snow loads on a shed with parapets. (TTAD #12494)

Test ID: 6175

Test status: Passed

3.46.1 Description

Generates snow loads on a model which contains 4 slopes with lateral parapets on X+/- direction, according to theEurocodes 1 - French standard (NF EN 1991-1-3/NA). It verifies also the valley accumulation.

3.47 EC1: Generating snow loads on a shed with gutters building. (TTAD #13856)

Test ID: 6174

Test status: Passed

3.47.1 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). It verifies also the valley accumulation.

3.48 EC1: generating snow loads on a 3 slopes 3D portal frame.(TTAD #13169)

Test ID: 6178

Test status: Passed

3.48.1 Description

Generates snow loads on a 3 slopes 3D portal frame, according to the Eurocodes 1 - French standard (NF EN 1991-1-3/NA).

3.49 EC1: Generating snow loads on 2 side by side single roof compounds with parapets (TTAD#13992)

Test ID: 6179

Test status: Passed

3.49.1 Description

Generates snow loads on on 2 side by side single roof compounds with parapets, according to the Eurocodes 1 -French standard (NF EN 1991-1-3/NA). One compound is much higher than the other and has the slope < 15degrees causing the drifted snow to dissipate.

3.50 EC1: generating Cf and Cp,net wind loads on an isolated roof with double slope (DEV2013#4.3)

Test ID: 5594

Test status: Passed

3.50.1 Description

Generates wind loads on a concrete structure, according to the Eurocodes 1 French standard. The obstruction isdifferent for each direction: X+ 1; X- 0.9; Y+ 0.8 and Y- 0.

The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and an isolated roofwith double slope.




346

3.51 EC1: wind load generation on a high building with double slope roof using differentparameters defined per directions (DEV2013#4.2)

Test ID: 5561

Test status: Passed

3.51.1 Description

Generates wind loads on the windwalls of a concrete structure, according to the Eurocodes 1 FR standard usingdifferent parameters defined per directions.

The structure is 22m high, has 4 columns and 7 beams (R20*30 cross section and C20/25 material), rigid supportsand a double slope roof.

3.52 EC1: generating Cf and Cp,net wind loads on an multibay canopy roof (DEV2013#4.3)

Test ID: 5595

Test status: Passed

3.52.1 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 a multibay canopyroof.

3.53 EC1: generating Cf and Cp,net wind loads on an isolated roof with one slope (DEV2013#4.3)

Test ID: 5593

Test status: Passed

3.53.1 Description

Generates wind loads on a concrete structure, according to the Eurocodes 1 French standard. The obstruction isdifferent for each direction: X+ 1; X- 0; Y+ 0.8 and Y- 1.

The structure has concrete columns and beams (R20*30 cross section and C20/25 material) and an isolated roofwith one slope.

3.54 EC1: wind load generation on a high building with a horizontal roof using different CsCdvalues for each direction (DEV2013#4.4)

Test ID: 5596

Test status: Passed

3.54.1 Description

Generates wind loads on the windwalls of a concrete structure, according to the Eurocodes 1 France standard usingdifferent CsCd values for each direction : X+ auto; X- imposed to 0.9; Y+ auto and Y- no.

The structure is 35m high, has 4 columns and 4 beams (R20*30 cross section and C20/25 material), rigid supportsand a horizontal roof.




347

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

Test ID: 4055

Test status: Passed

3.55.1 Description

Generates wind loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanianstandards.

The structure consists of concrete (C20/25) beams and columns with rigid fixed supports.

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

Test ID: 4087

Test status: Passed

3.56.1 Description

Generates snow loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanianstandards.

The structure consists of concrete (C20/25) beams and columns with rigid fixed supports, with rectangular crosssection (R20*30).

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

Test ID: 4090

Test status: Passed

3.57.1 Description

Generates wind loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 French standards -Martinique wind speed.


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

Test ID: 4086

Test status: Passed

3.58.1 Description

Generates snow loads on the windwalls of a 2 slopes 3D portal frame, according to Eurocodes 1 Romanianstandards.





348

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

Test ID: 4085

Test status: Passed

3.59.1 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.

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

Test ID: 4104

Test status: Passed

3.60.1 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.

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

Test ID: 3706

Test status: Passed

3.61.1 Description

Generates snow loads on a 2 slopes 3D portal frame with roof thickness greater than the parapet height, according toEurocodes 1 French standard.


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

Test ID: 3713

Test status: Passed

3.62.1 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.




349

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

Test ID: 3705

Test status: Passed

3.63.1 Description

Generates wind loads on a 2 slopes 3D portal frame with 2 fully opened windwalls, according to the Eurocodes 1French standards.

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

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

Test ID: 3623

Test status: Passed

3.64.1 Description

Generates snow loads on two close roofs with different heights, according to Eurocodes 1 Czech standards.


3.65 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

3.65.1 Description

Generates wind loads on the windwalls of a double slope 3D portal frame, according to the Eurocodes 1 Czechstandard (CSN EN 1991-1-4).

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

3.66 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

3.66.1 Description

Generates snow loads on the windwalls of a 3D portal frame with a roof which has a small span (< 5m) and aparapet, according to Eurocodes 1.





350

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

Test ID: 3603

Test status: Passed

3.67.1 Description

Generates snow loads on the windwalls of a 2 slopes 3D portal frame with gutter, according to Eurocodes 1. Thestructure consists of concrete (C20/25) beams and columns with rigid fixed supports.

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

Test ID: 3604

Test status: Passed

3.68.1 DescriptionGenerates snow loads on the windwalls of a 3D portal frame with horizontal roof and gutter, according to Eurocodes1.


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

Test ID: 3613

Test status: Passed

3.69.1 DescriptionGenerates snow loads on the windwalls of duopitch multispan roofs structure, according to Eurocodes 1 Germanstandards.


3.70 EC1: generating wind loads on a 55m high structure according to German standards (DIN EN1991-1-4/NA) (DEV2012 #3.12)

Test ID: 3618

Test status: Passed

3.70.1 DescriptionGenerates wind loads on the windwalls of a 55m high structure, according to Eurocodes 1 German standards.





351

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

Test ID: 3615

Test status: Passed

3.71.1 Description

Generates snow loads on two side by side roofs with different heights, according to Eurocodes 1 German standards.


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

Test ID: 3605

Test status: Passed

3.72.1 Description

Generates snow loads on the windwalls of a 3D portal frame with horizontal roof and 2 parapets, according toEurocodes 1. The height of one parapet is reduced.


3.73 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

3.73.1 Description

Generates snow loads on the windwalls of a monopitch multispan roofs structure, according to Eurocodes 1 Germanstandards.


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

Test ID: 3529

Test status: Passed

3.74.1 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 withtwo slopes.




352

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

Test ID: 3535

Test status: Passed

3.75.1 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 anda fully opened face.

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

Test ID: 3530

Test status: Passed

3.76.1 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 multispanroofs with pitch < 5 degrees.

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

Test ID: 3004

Test status: Passed

3.77.1 DescriptionGenerates 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.

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

Test ID: 3104

Test status: Passed

3.78.1 Description

Generates wind loads on the windwalls of a concrete structure with 4 slopes roof, according to the Eurocodes 1standard.

The structure has 6 concrete columns (R20*30 cross section and C20/25 material) with rigid supports and C20/25concrete walls.




353

3.79 EC1: Generating 2D wind and snow loads on a 2 opposite slopes portal with Z down axis.(TTAD #15094)

Test ID: 6206

Test status: Passed

3.79.1 Description

Generates 2D wind and snow loads on a 2 opposite slopes portal, according to the Eurocodes 1 - French standard(NF EN 1991-1-4/NA / 1991-1-3/NA), with Z down axis.

3.80 EC1: Generating wind loads on a 3 compound building. (TTAD #13190)

Test ID: 6211

Test status: Passed

3.80.1 DescriptionGenerates wind loads on a 3 compound building, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA).

3.81 EC1: Generating 2D wind loads on a double slope roof with an opening. (TTAD #15328)

Test ID: 6216

Test status: Passed

3.81.1 Description

Generates 2D wind loads on a double slope roof with an opening, according to the Eurocodes 1 - French standard(NF EN 1991-1-4/NA).

3.82 EC1: Generating wind loads on a double slope with 5 degrees. (TTAD #15307)

Test ID: 6217

Test status: Passed

3.82.1 Description

Generates wind loads on a double slope with 5 degrees, according to the Eurocodes 1 - French standard (NF EN1991-1-4/NA).

3.83 EC1: Generating wind loads on a 2 horizontal slopes building one higher that the other. (TTAD#13320)

Test ID: 6212

Test status: Passed

3.83.1 Description

Generates wind loads on a 2 horizontal slopes building one higher that the other, according to the Eurocodes 1 -French standard (NF EN 1991-1-4/NA).




354

3.84 EC1: Generating snow loads on a custom multiple slope building. (TTAD #14285)

Test ID: 6197

Test status: Passed

3.84.1 Description

Generates wind loads on a 2 almost horizontal slope building, according to the Eurocodes 1 - French standard (NFEN 1991-1-4/NA).

3.85 EC1: Generating 2D wind loads on a 2 slope isolated roof. (TTAD #14985)

Test ID: 6204

Test status: Passed

3.85.1 Description

Generates 2D wind loads on a 2 slope isolated roof, according to the Eurocodes 1 - French standard (NF EN 1991-1-4/NA). The depth for one slope is set to 20m and for the other to 40m.

3.86 EC1: Generating 2D wind and snow loads on a 4 slope shed next to a higher one slopecompound. (TTAD #15047)

Test ID: 6205

Test status: Passed

3.86.1 Description

Generates 2D wind and snow loads on a 4 slope shed next to a higher one slope compound, according to theEurocodes 1 - French standard (NF EN 1991-1-4/NA / 1991-1-3/NA).

3.87 EC1: Generating 2D snow loads on a one horizontal slope portal. (TTAD #14975)

Test ID: 6203

Test status: Passed

3.87.1 Description

Verifies the generation of 2D snow loads on a one horizontal slope portal.

3.88 EC1: Generating 2D wind loads on a multiple roof portal. (TTAD #15140)

Test ID: 6208

Test status: Passed

3.88.1 Description

Generates 2D wind loads on a one slope compound next to a higher double slope compound, according to theEurocodes 1 - French standard (NF EN 1991-1-4/NA).



3.89 EC1: wind load generation on a signboard

Test ID: 3107

Test status: Passed

3.89.1 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.

3.90 EC1: wind load generation on a building with multispan roofs

Test ID: 3106

Test status: Passed

3.90.1 Description

Generates wind loads on the windwalls of a concrete structure with multispan roofs, according to the Eurocodes 1standard.

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

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

Test ID: 3101

Test status: Passed

3.91.1 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 supportsand horizontal roof.

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

Test ID: 3103

Test status: Passed

3.92.1 Description

Generates wind loads on the windwalls of a concrete structure with 2 slopes roof, according to the Eurocodes 1standard.

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

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

Test ID: 3099

Test status: Passed

3.93.1 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.





4 Combinations




358

4.1 Generating combinations (TTAD #11673)

Test ID: 3471

Test status: Passed

4.1.1 Description

Generates concomitance between three types of loads applied on a structure (live loads, dead loads and seismicloads - EN 1998-1), using the quadratic combination function. Generates the combinations description report and thepoint support actions by element report.

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

Test ID: 3751

Test status: Passed

4.2.1 Description

Generates load combinations with unfavorable and favorable/unfavorable predominant action. Predominant action isa case family with 2 static load cases.

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 casefamilies and generates the concomitance matrix.

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

Test ID: 3746

Test status: Passed

4.4.1 Description

Generates combinations for NEWEC8.cbn.

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

Test ID: 3766

Test status: Passed

4.5.1 Description

Creates a new dead load case, after two case families were created, and generates the concomitance matrix. Areport with the combinations description is generated.




359

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 thecombinations description is generated.

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

Test ID: 4405

Test status: Passed

4.7.1 Description

Creates loads:

1 - G (Favorable or Unfavorable)

2 - Q (Base or Acco)


4 - Q (Acco)

5 - Acc (Base)

Generates combinations. Generates the combinations report.


Test ID: 4407Test status: Passed

4.8.1 Description

Generates loads:


2 - Q Group(Base or Acco)

2 - Q

3 - Q

4 - Q


6 - Snow (Base or Acco)


Un-group 2-Q Group to independent 2-Q, 3-Q, 4-Q loads.





360


Test ID: 4384

Test status: Passed

4.9.1 Description

Generates loads:




3 - Q (Base)


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

Test ID: 4382

Test status: Passed

4.10.1 Description

Generates loads:



3 - Q (Acco)


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

Test ID: 4394

Test status: Passed

4.11.1 Description

Generates loads:



3 - E (Base) - only one seismic load !


Set value "0" (exclusive) between seismic and all Q loads (seism only combination).





361


Test ID: 4397

Test status: Passed

4.12.1 Description

Generates loads:



3 - E (Base) - only one seismic load



Test ID: 4391

Test status: Passed

4.13.1 Description

Generates loads:


2 - Q (Base)


4 - Q (Base)



Test ID: 4408

Test status: Passed

4.14.1 Description

Generates loads:



3 - E Group (Base)

3 - EX

4 - EY

5 - EZ

Un-group 3 - E Group to independent loads : 3 - EX 4 - EY 5 - EZ





362


Test ID: 4409

Test status: Passed

4.15.1 Description

Generates loads:



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.

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

Test ID: 4357

Test status: Passed

4.16.1 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 -BaseGenerates the second set of combinations and the second combinations report.



363


Test ID: 4386

Test status: Passed

4.17.1 Description

Generates loads:






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

Test ID: 4353

Test status: Passed

4.18.1 Description

Generates a set of combinations with Q group of loads.

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

Test ID: 4219

Test status: Passed

4.19.1 Description

Generates the concomitance matrix and the combinations description reports after switching back the effect for liveload.

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

Test ID: 4350

Test status: Passed

4.20.1 Description

Generates a set of combinations with seismic group of loads.

4.21 Verifying combinations for CZ localization (TTAD #12542)

Test ID: 4550

Test status: Passed

4.21.1 Description

Verifies simplified combinations for CZ localization.





5 Concrete Design




366

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

Test ID: 4555

Test status: Passed

5.1.1 Description

Verifies the transverse reinforcement area for a beam subjected to linear loads. The verification is made according toEC2 norm with French Annex.

5.2 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - bilinearstress-strain diagram

Test ID: 4541

Test status: Passed

5.2.1 Description

Verifies the longitudinal reinforcement area of a simply supported beam under a linear load - bilinear stress-straindiagram.

Verification is done according to Eurocodes 2 norm with French Annex.

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

Test ID: 4542

Test status: Passed

5.3.1 Description

Defines the "Design experts" properties for a concrete (EC2) linear element in analysis model and verifies propertiesfrom descriptive model.

5.4 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - horizontallevel behavior law

Test ID: 4557

Test status: Passed

5.4.1 Description

Verifies the longitudinal reinforcement area of a beam under self-weight and linear loads - horizontal level behaviorlaw. The verification is made according to EC2 norm with French Annex.

5.5 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - inclinedstress strain behavior law

Test ID: 4522

Test status: Passed

5.5.1 Description

Verifies the longitudinal reinforcement area of a beam under a linear load - inclined stress strain behavior law.

Verification is done according to Eurocodes 2 norm with French Annex.




367

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

Test ID: 4527

Test status: Passed

5.6.1 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.

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

Test ID: 4519

Test status: Passed

5.7.1 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.

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

Test ID: 4517

Test status: Passed

5.8.1 Description

Verifies the minimum reinforcement area for a simply supported concrete beam subjected to self weight. Theverification is made with Eurocodes 2 - French annex.




368

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

Test ID: 4970

Test status: Passed

5.9.1 Description

Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, the determination of stresses is made along with the determination of the longitudinal reinforcement and theverification of the minimum reinforcement percentage.

5.9.2 Background

Simple Bending Design for Ultimate Limit State

Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, the calculation of stresses will be made along with the calculation of the longitudinal reinforcement and theverification of the minimum reinforcement percentage.


■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002;



The following load cases and load combination are used:

■ Loadings from the structure: G = 15 kN/m + dead load,

■ Exploitation loadings (category A): Q = 20kN/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

Simply supported beam

Units

Metric System




369

Geometry

Beam cross section characteristics:

■ Height: h = 0.60 m,

■ Width: b = 0.25 m,

■ Length: L = 5.80 m,

■ Section area: A = 0.15 m2 ,

■ Concrete cover: c=4.5cm

■ Effective height: d=h-(0.6*h+ebz)=0.519m; d’=ebz=0.045m


Rectangular solid concrete C25/30 and S500A reinforcement steel is used. The following characteristics are used inrelation to this material:

■ Exposure class XD1

■ Concrete density: 25kN/m3

■ Stress-strain law for reinforcement: Bilinear stress-strain diagram

■ Cracking calculation required

■ Concrete C25/30: MPa,,

f f

c

ckcd 6716

51

25

γ

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

■ MPa.*.f *.f //ckctm 56225300300 3232

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1

■ Steel S500 : MPa,,

f f

s

ykyd 78434

151

500

γ

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

■ MPa*f

*E

..

ckcm 3147610

8252200010

822000

3030


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 rotationalong X.

■ Inner: None.




370

Loading

The beam is subjected to the following load combinations:

■ Permanent loads:

G’=0.25*0.6*2.5=3.75kN/ml

■ Load combinations:The ultimate limit state (ULS) combination is:

Cmax = 1.35 x G + 1.5 x Q=1.35*(15+3.75)+1.5*20=55.31kN/ml

Characteristic combination of actions:

CCQ = 1.0 x G + 1.0 x Q=15+3.75+20=38.75kN/ml

Quasi-permanent combination of actions:

CQP = 1.0 x G + 0.3 x Q=15+3.75+0.3*20=24.75kN/ml

■ Load calculations:

mkN M Ed .59,232

8

²80,5*31,55

mkN M Ecq .94,1628

²80,5*75,38

mkN M Eqp .07,1048

²80,5*75,24




371

5.9.2.2 Reference results in calculating the equivalent coefficient

To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at28 days and 50% humidity:

)t(*)f (*)t,( cmRH 00 ββ

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.2

9252825

816816β .

.

f

.)f (

cm

cm


48802810

1

10

1β

2002000

0 ..t.

)t(..

at t0=28 days


The value of the φRH coefficient depends of the concrete quality:

213

0

αα10

100

1

1 **h*.

RH

RH

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.3a

1αα 21 if MPaf CM 35

If not:

70

135

α

.

cmf

and

20

235

α

.

cmf

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.8c

In this case: 1αα338 21 MpaMpaf f ckcm

8914717610

100

501

1471766002502

60025022

30 ..*.

mm.)(*

**

u

Ac*h RH


70248809252891ββ 00 ..*.*.)t(*)f (*)t,( cmRH

The coefficient of equivalence is determined by the following formula:

Ecar

Eqp

cm

se

M

M*)t,(

E

E

01

α

Defined earlier:

m.kN,²,*,

MEcar 941628

8057538

m.kN,²,*,

MEqp 071048

8057524

This gives:

32.17

94.16207.104*70.21

31476

200000

e




372

5.9.2.3 Reference results in calculating the concrete beam reduced moment limit

Due to exposure class XD1, we will verify the section having non-compressed steel reinforcement, determining thereduced 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 MPa f ck 50 valid for a constitutive law to horizontal plateau:

)*..(f *)*..(f *)(K

ck

ckeluc

γ207690159γ701694αμ

24αα10α eee *c*ba*K

The values of the coefficients "a", "b" and "c" are defined in the following table:

Diagram for inclined tier Diagram for horizontal plateau

a 8,189*3,75 ck f

108*2,71 ck f

b 5,874*6,5 ck f

4,847*2,5 ck f

c 13*04,0 ck f

5,12*03,0 ck f

This gives us:

■ 188810825*2,71 a

■ 4.7174,84725*2,5 b

■ 75.115,1225*03,0 c

■ 079.1²27.17*75.1127.17*4.717188810 4 e K

Then:

■ 43,1

94,162

58,232

■ 2504.0)*20.7690.159(*)*70.169.4(

).(

ck

ck eluc

f

f K

Reference reinforcement calculation at SLU:

The calculation of the reinforcement is detailed below:

■ Effective height: d=0.9*h=0.53m

■ Calculation of reduced moment:

207,067,16*²519,0*25,0

233,0

*²*

cd w

Ed

cu f d b

M

■ Calculation of the lever arm zc:

md z uc 458,0)294,0*4,01(*519,0)*4,01(*

■ Calculation of the reinforcement area:




373

Reference reinforcement calculation at SLS:

We will also conduct a SLS design to ensure that we do not get an upper section of longitudinal reinforcement.

We must determine the position of the neutral axis by calculating (position corresponding to the state of maximum

stress on the concrete and reinforcement).

■ 3940400153217

153217σσα

σαα1 ,*.

*.*

*sce

ce

rbcq ser M kNm M 163, there in no compressed steel reinforcement. This confirms what we previously

found by determining the critical moment limit depending on the coefficient of equivalence and .

Then we calculate the reinforcement resulted from the SLS efforts (assuming the maximum constrain is reached onsteel and concrete):

■ Neutral axes position :394,01

■ Lever arm : md z c 451.0

3

394.01*519.0

3

1* 1

■ Reinforcement section : ²03.9400*451.0

163.0

*,1 cm

z

M A

sc

Ecq

ser s

This shows that the tensile reinforcement obtained in SLS are not dimensioning compared to SLU.

Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is:

■

d*b*.

d*b*f

f *.

Max A

w

wyk

eff ,ct

min,s

00130

260

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2

Where:

MPa.f f ctmeff ,ct 562 from cracking conditions

Therefore:

►

²cm.

²m*..*.*.

²m*..*.*.

*.max A min,s 731

10701519025000130

107315190250500

562260

4

4


■ Linear element: S beam,

■ 7 nodes,

■ 1 linear element.




374

ULS and SLS load combinations(kNm)

Simply supported beam subjected to bending

ULS (reference value: 232.59kNm)

SLS (reference value: 162.94kNm)

Theoretical reinforcement area(cm2 )

(reference value: 11.68cm2)

Minimum reinforcement area(cm2 )




My,ULS My corresponding to the 101 combination (ULS) [kNm] 232.59 kNm

My,SLS My corresponding to the 102 combination (SLS) [kNm] 162.94 kNm

Az Theoretical reinforcement area [cm2] 11.68 cm

2

Amin Minimum reinforcement area [cm2] 1.73 cm

2




375



My My USL -232.578 kN*m -0.0001 %

My My SLS -162.936 kN*m 0.0001 %

Az Az -11.6779 cm² 0.0004 %

Amin Amin -1.73058 cm² -0.0001 %




376

5.10 EC2 Test 4 I: Verifying a rectangular concrete beam subjected to Pivot A efforts – Inclinedstress-strain diagram

Test ID: 4977

Test status: Passed

5.10.1 Description


The purpose of this test is to verify the software results for Pivot A efforts. For these tests, the constitutive law forreinforcement steel, on the inclined stress-strain diagram is applied.

The objective is to verify:

- The stresses results

- The longitudinal reinforcement corresponding to Class A reinforcement steel ductility

- The minimum reinforcement percentage

5.10.2 Background

This test performs the verification of the value (hence the position of the neutral axis) to determine the Pivot efforts(A or B) to be considered for the calculations.

The distinction between the Pivot A and Pivot B efforts is from the following diagram:

d d d

x

cuud

cuu

cuud

cuu ..x 2

2

2

2

The limit for depends of the ductility class:

■ For a Class A steel: 13460α5022ε .. uud

■ For a Class B steel: 0720α45ε .uud

■ For a Class C steel: 0490α5067ε .. uud

The purpose of this test is to verify the software results for Pivot A efforts. For these tests, it will be used theconstitutive law for reinforcement steel, on the inclined stress-strain diagram.

MPa AS su su 454.38,95271,432500

MPa BS su su 466.27,72771,432500

MPaC S su su 493.52,89571,432500




377

The Pivot efforts types are described below:

■ Pivot A: Simple traction and simple bending or combined

■ Pivot B: Simple or combined bending

■ Pivot C: Combined bending with compression and simple compression






■ Loadings from the structure: G = 25 kN/m+ dead load,




■ There will be considered a Class A reinforcement steel

■ The calculation will be made considering inclined stress-strain diagram


■ The stresses results

■ The longitudinal reinforcement corresponding to Class A reinforcement steel ductility

■ The minimum reinforcement percentage




378


Units

Metric System

Geometry


■ Height: h = 0.90 m,

■ Width: b = 0.50 m,

■ Length: L = 5.80 m,


■ Concrete cover: c = 4.00 cm

■ Effective height: d = h - (0.6 * h + ebz)=0.806 m; d’ = ebz = 0.040 m


Rectangular solid concrete C25/30 and S500B reinforcement steel is used. The following characteristics are used inrelation to this material:

■ Exposure class XC1


■ There will be considered a Class A reinforcement steel ductility




f f

c

ckcd 6716

51

25

γ


■ MPa.*.f *.f //ckctm

56225300300 3232



f f

s

ykyd 78434

151

500

γ


Boundary conditions


■ Outer:

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

► Support at end point (x = 5.8) restrained in translation along Y and Z, and restrained rotationalong X.

■ Inner: None.




379

Loading


Dead load:

G’=0.9*0.5*2.5=11.25 kN/ml

Load combinations:

■ The ultimate limit state (ULS) combination is:

Cmax = 1.35 x G + 1.5 x Q=1.35*(25+11.25)+1.5*50=123.94 kN/ml

■ Characteristic combination of actions:

CCQ = 1.0 x G + 1.0 x Q=25+11.25+50=86.25kN/ml


kNm M Ed 16.5218

²80.5*94.123

kNm M Ecq 68.3628

²80.5*25.86


For a S500B reinforcement steel, we have 372.0lu (since we consider no limit on the compression concrete to

SLS).

Reference reinforcement calculation:


■ Effective height: d=h-(0.6*h+ebz)=0.806 m


096,067,16*²806.0*50,0

10*52116.0

*²* 2

3

MPamm

Nm

f d b

M

cd w

Ed

cu

■ The α value:

■ The condition: 13460αα .ucu is satisfied, therefore the Pivot A effort conditions are true

There will be a design in simple bending, by considering an extension on the reinforced steel tension equal to

5022ε .ud

, which gives the following available stress:

■ MPa.,, AS su 454022503895271432σ500




380

The calculation of the concrete shortening:

27.3

127.01

127.0*50.22

1*

cu

cu sucu

It is therefore quite correct to consider a design stress of concrete equal to cd f .

md z cuc 765.0)127.0*4.01(*806.0)*4,01(*

²00.15454*765.0

52116.0

*cm

f z

M A

yd c

Ed u



d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


Where:

MPa f f ctmeff ct 56.2, from cracking conditions

Therefore:

²38.5

²10*24.5806.0*50.0*0013.0

²10*38.5806.0*50.0*500

56.2*26.0

max4

4

min, cm

m

m A s



■ 7 nodes,





381

ULS and SLS load combinations (kNm)




Theoretical reinforcement area (cm2 )

For Class A reinforcement steel ductility (reference value: A=15.00cm2)






382





Az Theoretical reinforcement area (Class A) [cm2] 15.00 cm

2

Amin Minimum reinforcement area [cm2] 5.38 cm2



My My USL -521.125 kN*m 0.0000 %

My My SLS -362.657 kN*m -0.0001 %

Az Az Class A -14.9973 cm² -0.0001 %

Az Az Class B -14.9973 cm² 0.7491 % Amin Amin -5.37514 cm² 0.0000 %




383

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

Test ID: 4979

Test status: Passed

5.11.1 Description

The purpose of this test is to verify the software results for the My resulted stresses for the USL load combination andfor the results of the theoretical reinforcement area Az.

5.11.2 Background

This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to thedefined loads. The test confirms the absence of the compressed reinforcement for this model.






■ Loadings from the structure: G = 45 kN/m

■ Exploitation loadings (category A): Q = 37.4kN/m,



■

■ Reinforcement steel ductility: Class B

■ The calculation is made considering bilinear stress-strain diagram


■ The theoretical reinforcement area results





384

Units

Metric System

Geometry


■ Beam length: 8m■ Concrete cover: c=3.50 cm

■ Effective height: d=h-(0.6*h+ebz)=0.435 m; d’=ebz=0.035m







■ Concrete C25/30: MPa f f c

ck cd 67.10

5,125


■ MPa f f ck ctm 56.225*30.0*30.0 3/23/2


■ Steel S500B : MPa f

f s

yk

yd 78.43415,1

500


Boundary conditions


■ Outer:


► Support at end point (x = 8) restrained in translation along Y and Z, and restrained rotationalong X.

■ Inner: None.




385

Loading


■ Load combinations:

The ultimate limit state (ULS) combination is:

Cmax = 1.35 x G + 1.5 x Q=1.35*45+1.5*37.4=116.85kN/ml


CCQ = 1.0 x G + 1.0 x Q=45+34.7=79.7kN/ml


kNm M Ed 8.9348

²8*85.116

5.11.2.2 Reference results in calculating the concrete beam moment

At first it will be determined the moment resistance of the concrete section only:

kNm f hd hb M cd f

f eff btu 967.16*215.0435.0*15.0*00.1*

2**

Comparing Mbtu with MEd:

Therefore, the concrete section is not entirely compressed;

Therefore, the calculations considering the T section are required.

5.11.2.3 Reference reinforcement calculation:

Theoretical section 2:The moment corresponding to this section is:

kNm

hd f hbb M

f

cd f weff Ed

720

2

15.0435.0*67.16*15.0*)20.01(

2***)(2

According to this value, the steel section is:

²00.46

8.347*

2

15.0435.0

720.0

*

2

22 cm

f h

d

M A

yd

f

Ed

Theoretical section 1:

The theoretical section 1 corresponds to a calculation for a rectangular shape beam section.

kNm M M M Ed Ed Ed 21572093521

341.067.16*²435.0*20.0

215.0

*²*

1 cd w

Ed cu

F d b

M

kNm M kNm M Ed btu 8.934900




386

For a S500B reinforcement and for a XC1 exposure class, there will be a: 3720μμ .lucu , therefore there will be no

compressed reinforcement (the compression concrete limit is not exceeded because of the exposure class)

There will be a calculation without considering compressed reinforcement:

544.0341.0*211*25.1)*21(1*25.1 cuu

md z uc 340.0)544.0*40.01(*435.0)*4.01(*1

²52.1478.347*340.0

215.0

*1

11 cm

f z

M A

yd c

Ed


In conclusion, the entire reinforcement steel area is A=A1+A2=46+14.52=60.52cm2



■ 9 nodes,


ULS load combinations(kNm)



For Class B reinforcement steel ductility (reference value: A=60.52cm2)



Az (Class B) Theoretical reinforcement area [cm2] 60.52 cm

2



Az Az -60.5167 cm² 0.0001 %




387

5.12 EC2 Test 5: Verifying a T concrete section, without compressed reinforcement - Bilinearstress-strain diagram

Test ID: 4978

Test status: Passed

5.12.1 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 thetheoretical reinforcement area Az.



- The theoretical reinforcement area results

5.12.2 Background







■ Loadings from the structure: G = 52.3 kN/m








■ The theoretical reinforcement area results





388

Units

Metric System

Geometry


■ Beam length: 7m

■ Concrete cover: c=3.50 cm






■ Reinforcement steel ductility: Class A


■ Concrete C16/20: MPa f

f c

ck cd 67.10

5,1

16


■ MPa.*.f *.f //ckctm

90116300300 3232



f s

yk

yd 8.34715,1

400


Boundary conditions


■ Outer:



■ Inner: None.




389

Loading




Cmax = 1.35 x G + 1.5 x Q=1.35*352.3+1.5*13=90.105kN/ml


CCQ = 1.0 x G + 1.0 x Q=52.3+13=65.3kN/ml

Load calculations:

kNm M Ed 89.5518

²7*105.70

kNm M Ecq 96.3998

²7*3.65

5.12.2.2 Reference results in calculating the concrete beam moment At first, it will be determined the moment resistance of the concrete section only:

kNm,*,

,*,*,f *h

d*h*bM cdf

f eff btu 52367102

100595010090

2


Therefore the concrete section is not entirely compressed;

Therefore, calculations considering the T section are required.

5.12.2.3 Reference reinforcement calculation:


■ The moment corresponding to this section is:

MNm

hd f hbb M

f

cd f weff Ed

418.0

2

1.00595*67.10*1.0*)18.09.0(

2***)(2

■ According to this value, the steel section is:

²08.22

8.347*21.0595.0

418.0

*2

22 cm

f hd

M A

yd f

Ed

kNm M kNm M Ed btu 89.551523




390


The theoretical section 1 corresponds to a calculation for a rectangular shape beam section

kNm M M M Ed Ed Ed 133418.0552.021

197.067.10*²595.0*18.0

133.0

*²*

1

cd w

Ed

cu F d b

M

For a S400B reinforcement and for a XC1 exposure class, there will be a: 3720μμ .lucu , therefore there will be no

compressed reinforcement.

There will be a calculation without compressed reinforcement:

276.0197.0*211*25.1)*21(1*25.1 cuu

md z uc 529.0)276.0*40.01(*595.0)*4.01(*1

²25.78.347*529.0

133.0

*1

11 cm

f z

M A

yd c

Ed


In conclusion the entire reinforcement steel area is A=A1+A2=7.25+22.08=29.33cm2



■ 8 nodes,




ULS (reference value: 552kNm)






391



My,ULS My corresponding to the 101 combination (ULS) [kNm] 552 kNm


2



My My USL -551.893 kN*m -0.0000 %

Az Az -29.3334 cm² 0.0002 %




392

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

Test ID: 4981

Test status: Passed

5.13.1 Description


The purpose of the test is to verify the results using a constitutive law for reinforcement steel, on the inclined stress-strain diagram.



- The longitudinal reinforcement corresponding to Class A reinforcement steel ductility

- The minimum reinforcement percentage

5.13.2 Background

Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During thistest, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and theverification of the minimum reinforcement percentage.

The purpose of this test is to verify the software results for using a constitutive law for reinforcement steel, on theinclined stress-strain diagram.



■ Analysis type: static linear (plane problem);■ Element type: linear.


■ Loadings from the structure: G = 25 kN/m (including the dead load)


■




■ Reinforcement steel: Class A

■ The calculation is performed considering inclined stress-strain diagram




■ The longitudinal reinforcement corresponding to doth Class A reinforcement steel ductility





393


Units

Metric System

Geometry


■ Height: h = 0.65 m,

■ Width: b = 0.28 m,

■ Length: L = 6.40 m,





Rectangular solid concrete C30/37 and S500A reinforcement steel is used. The following characteristics are used in

relation to this material:






■ Concrete C25/30: MPa f

f c

ck cd 20

5,1

30


■ MPa f f ck ctm 90.230*30.0*30.0 3/23/2


■ MPa*f

*E

..ck

cm 3283710

83022000

10

822000

3030


f f

s

ykyd 78434

151

500

γ





394

Boundary conditions


■ Outer:



■ Inner: None.

Loading




Cmax = 1.35 x G + 1.5 x Q=1.35*(25)+1.5*30=78.75kN/ml


CCQ = 1.0 x G + 1.0 x Q=25+30=55kN/ml


CQP = 1.0 x G + 0.6 x Q=25+0.6*30=43kNm


kNm M Ed 20.4038

²40.6*75.78

kNm M Ecq 60.2818

²40.6*55

kNm M Eqp 16.2208

²40.6*43




395



)(*)(*),( 00 t f t cm RH



at t0=28 days



213

0

αα10

1001

1 **h*.

RH

RH


1αα 21 if MPaf CM 35

If not:

70

135

α

.

cmf

and

20

235

α

.

cmf


In this case, MpaMpaf f ckcm 388

944038

3535α

7070

1 .f

..

cm

984038

3535α

2020

2 .f

..

cm

7819840719510

100

501

171956502802

65028022

30 ..*.*.

mm.)(*

**

u

Ac*h RH


3724880732781ββ 00 ..*.*.)t(*)f (*)t,( cmRH

The coefficient of equivalence is determined using the following formula:

4017

60281

162203721

32837

200000

1

α

0

.

.

.*.

M

M*)t,(

E

E

Ecar

Eqp

cm

se




396



This value can be determined by the next formula if MPa f ck 50 and valid for a constitutive law considering

inclined stress-strain diagram:

)*62.7969.165(*)*66.162.4(*)(

ck

ck eluc

f

f K

24 ***10 eee cba K

The values of the coefficients "a", "b" and "c" are defined in the following table:

Diagram for inclined stress-strain diagram Diagram for bilinear stress-strain diagram

a 8,189*3,75 ck f

108*2,71 ck f

b 5,874*6,5 ck f

4,847*2,5 ck f

c 13*04,0 ck f

5,12*03,0 ck f

This gives us:

22069818925375 ..*.a

570658743025 ..*,b

8111330040 .*,c

087.1²60.17*80.1160.17*5.7062.206910 4 e K

Then:

425,12.307

76.437

272.0)*62.7969.165(*)*66.162.4(

).(

ck

ck eluc

f

f K

Calculation of reduced moment:

225,020*²566.0*50,0

10*403.0

*²* 2

3

MPamm

Nm

f d b

M

cd w

Ed cu

272.0225.0 luccu therefore, there is no compressed reinforcement



■ Effective height: d = 0.566m


225.0cu


md z uc 493,0)323,0*4,01(*566,0)*4,01(*




397


²68,11²10*68,1178,434*458,0

233,0

*

4 cmm f z

M A

yd c

Ed u

■ Calculation of the stresses from the tensioned reinforcement:

Reinforcement steel elongation:

35.75.3*323,0

323,01*

12

cu

u

u su

‰

Tensioned reinforcement efforts:

MPa MPa su su 45471.43900735.0*38,95271,432*38,95271,432

■ Reinforcement section calculation:

²60.18²10*60.1871.439*493.0

403.0

*

4 cmm f z

M A

yd c

Ed u



d*b*.

d*b*f

f *.

Max A

w

wyk

eff ,ct

min,s

00130

260


Where:


Therefore:

²39.2

²10*06.2566.0*28.0*0013.0

²10*39.2566.0*28.0*500

90.2*26.0

max4

4

min, cm

m

m A s



■ 7 nodes,





398






For Class A reinforcement steel ductility (reference value: A=18.60cm2)






399





Az (Class A) Theoretical reinforcement area [cm2] 18.60 cm2


2



My My USL -403.2 kN*m 0.0000 %

My My SLS -281.6 kN*m 0.0000 %

Az Az -18.6006 cm² -0.0001 %

Amin Amin -2.38697 cm² 0.0001 %




400

5.14 EC2 Test 9: Verifying a rectangular concrete beam with compressed reinforcement – Inclinedstress-strain diagram

Test ID: 4982

Test status: Passed

5.14.1 Description


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 tothe defined loads. The test confirms the presence of the compressed reinforcement for this model.

5.14.2 Background

Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, the calculation of stresses, the calculation of the longitudinal reinforcement and the verification of the minimumreinforcement 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 tothe defined loads. The test confirms the presence of the compressed reinforcement for this model.






■ Linear loadings :

Loadings from the structure: G = 55 kN/m+ dead load,

Live loads: Q=60kN/m

■ Point loads:

Loadings from the structure: G = 35kN;

Live loads: Q=25kN

■ 8,02



■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.6 x Q■ Reinforcement steel ductility: Class A




■ The longitudinal reinforcement





401


Units

Metric System

Geometry


■ Height: h = 0.80 m,

■ Width: b = 0.40 m,

■ Length: L = 6.30 m,











■ Concrete C25/30:

MPa,,

f f

c

ckcd 6716

51

25

γ



MPa f

E ck cm 31476

10

82522000

10

822000

3.03.0

■ Steel S500 :





402

Boundary conditions


■ Outer:


► Support at end point (x = 6.3) restrained in translation along Y and Z, and restrained rotation

along X.■ Inner: None.

Loading


■ Dead load:

G’=0.4*0.8*2.5=8.00 kN/ml

■ Linear load combinations:


Cmax = 1.35 x G + 1.5 x Q=1.35*(55+8)+1.5*60=175.05 kN/ml


CCQ = 1.0 x G + 1.0 x Q=55+8+60=123 kN/ml


CQP = 1.0 x G + 0.6 x Q=55+8+0.8*60=111 kN/m

■ Point load combinations:


Cmax = 1.35 x G + 1.5 x Q=1.35*35+1.5*25=84.75 kN


CCQ = 1.0 x G + 1.0 x Q=35+25=60 kN


CQP = 1.0 x G + 0.6 x Q=35+0.8*25=55 kN


kNm M Ed 10024

30.6*75.184

8

²30.6*05.175

kNm M Ecq 7054

30.6*60

8

²30.6*123

kNm M Eqp 6374

30.6*55

8

²30.6*111




403



)(*)(*),( 00 t f t cm RH


925.2825

8.168.16)(

cm

cm f

f


at t0 = 28 days



213

0

***1.0

1001

1

h

RH

RH


121 if MPa f CM 35

If not:

7.0

1

35

cm f and

2.0

2

35

cm f


In this case,

Mpa Mpa f f ck cm 338

121

78.167.266*1.0

100

501

167.266)800400(*2

800*400*2*230

RH mm

u

Ach


54.2488.0*92.2*78.1)(*)(*),( 00 t f t cm RH


94.20

705

637*54..21

31476

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E




404



This value can be determined by the next formula if MPa f ck 50 valid for a constitutive law to horizontal plateau:

)*62.7969.165(*)*66.162.4(*)(

ck

ck eluc

f

f K

24 ***10 eee cba K

Where:

7.16928,18925*3,75 a

5.7345,87425*6,5 b

121325*04,0 c

181.1²94.20*1294.20*5.7347.1692*10 4 e K

Then:

422,1705

1002

271.0)*62.7969.165(*)*66.162.4(

*)(

ck

ck eluc

f

f K

Calculation of reduced moment:

301.067.16*²707.0*40.0

002.1

*²* cd w

Ed cu f d b

M

271.0301.0 luccu 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 lu :

Nm f d b M cd wlu Ed

6

1 10*902.067.16*²707.0*40.0*271.0*²**

■ The α value:

404.0)271.0*21(1*25.1)*21(1*25.1 lulu


md z lulu 593.0)404.0*4.01(*707.0)*4.01(*

■ Tensioned reinforcement elongation calculation:

17.55.3*404.0

404.01*

12

cu

u

u su

‰




405

■ Tensioned reinforcement efforts calculation(S500A):

MPa su su 454*38,95271,432

Mpa MPa su 45463.43700517.0*38,95271,432


²76.3463.437*593.0

10*902.0

.

6

11 cm

MPam

Nm

f z

M A

yd lu

Ed

Compressed steel reinforcement reduction (Section As2 ):

Reduction coefficient:

9572045070704040707040401000

53α

α1000

53ε ...*.

.*.*

.)'dd*(

d**

,lu

lusc ‰

MPa sc yd sc 52.43500295.0*38.95271.43200217.000295.0

Compressed reinforcement calculation:

²47.352.435*)045.0707.0(

902.0002.1

)'(

12 cm

d d

M M A

sc

Ed Ed s

The steel reinforcement condition:

²48.364.437

52.435*47.3.22 cm

f A A

yd

sc s

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)


The concrete beam design at SLS will be made considering a limitation of the characteristic compressive cylinderstrength of concrete at 28 days, at 0.6*f ck

The assumptions are:

■ The SLS moment: kNmMEcq 705

■ The equivalence coefficient: 9420α .e

■ The stress on the concrete will be limited at 0.6*fck=15Mpa and the stress on steel at 0.8*fyk, or 400MPa




406

Calculation of the resistance moment M Rd for detecting the presence of the compressed reinforcement:

md x sce

ce 311.0707.0*40015*94.20

15*94.20*

*

*1

N xb F cwc

6

1 10*933.015311.0*40.0*2

1

***2

1

m x

d z c 603.03

311.0707.0

3

1

Nm z F M ccrb

6610*563.0603.0*10*933.0*

Therefore the compressed reinforced established earlier was correct.

Theoretical section 1 (tensioned reinforcement only)

Nm M M rb

6

1 10*563.0

440.0707.0

311.011

d

x

m x

d z c 603.03

311.0707.0

3

1

²34.23400603.0

563.011 cm

z

M A

sc

Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)

Nm M M M rbcq ser

66

,2 10*142.010*)591.0705.0(

Compressed reinforcement stresses:

d

d d ce sc

*

'***

1

1

MPa sc 66.268707.0*440.0

045.0707.0*440.0*15*94.20

Compressed reinforcement area:

²98.766.268*045.0707.0

142.0

*)'('

2

cmd d

M

A sc

Complementary tensioned reinforcement area:

²36.5400

66.268*98.7*' cm A A

s

sc s




407

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 oftensioned reinforcement (after ULS: Au=38.24cm

2)

Reference reinforcement third calculation at SLS:

For this third iteration, the calculation will begin considering the section of the tensile reinforcement found whencalculating for ULS: Au=38.24cm

2

From this value,it will be calculated the stress obtained in the tensioned reinforcement:

MPa

A

A s

ELU

ELS s 21.300400*

24.38

70.28*

Calculating the moment resistance Mrb for detecting the presence of compressed steel reinforcement:

md x sce

ce 361.0707.0*21.30015*94.20

15*94.20*

*

*1

N xb F cwc

6

1 10*083.115361.0*40.0*2

1***

2

1

m x

d z c 587.03

361.0707.0

3

1

Nm z F M ccrb

66 10*636.0587.0*10*083.1*



Nm M M rb

6

1 10*636.0

511.0707.0

361.011

d

x

m

x

d z c 587.03

361.0

707.03

1

²09.36400*578.0

36.0

*

11 cm

z

M A

sc




408


Nm M M M rbcq ser

66

,2 10*069.010*)636.0705.0(


MPad d d ce sc 275

707.0*511.0045.0707.0*511.0*15*94.20

*'***

1

1


²79.3

275*045.0707.0

069.0

*)'(' 2 cm

d d

M A

sc


²47.321.300

275*79.3*' cm A A

s

sc s

Section area:





d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


Where:


Therefore:

²77.3

²10*68.3707.0*40.0*0013.0

²10*77.3707.0*40.0*500

56.2*26.0

max4

4

min, cm

m

m A s



■ 9 nodes,





409




SLS (reference value: 705kNm)

SLS –Quasi-permanent (reference value: 637kNm)


For Class A reinforcement steel ductility (reference value: A=39.56cm2 and A’=3.79cm

2)






410




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 [cm2] 39.56 cm2


2



My My USL -1001.92 kN*m -0.0001 %

My My SLS cq -704.714 kN*m 0.0000 %

My My SLS qp -637.304 kN*m 0.0001 % Az Az -39.5904 cm² -0.0001 %

Amin Amin 3.77193 cm² 0.0001 %




411

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

Test ID: 4976

Test status: Passed

5.15.1 Description




- The longitudinal reinforcement

- The verification of the minimum reinforcement percentage

5.15.2 Background







The following load cases and load combinations are used:

■ Loadings from the structure: G = 70 kN/m,


■






■ The longitudinal reinforcement

■ The verification of the minimum reinforcement percentage




412


Units

Metric System

Geometry

Beam cross section characteristics:■ Height: h = 1.25 m,

■ Width: b = 0.65 m,




■ Effective height: d = h-(0.6*h+ebz) = 1.130 m; d’ = ebz = 0.045m



■ Exposure class XD1■ Concrete density: 25kN/m

3




f f

c

ckcd 6716

51

25

γ


■ MPa.*.f *.f //ckctm 56225300300 3232



f

f s

yk

yd 78434151

500

γ


■ MPa*f

*E

..ck

cm 3147610

82522000

10

822000

3030





413

Boundary conditions


■ Outer:


► Support at end point (x = 14) restrained in translation along Z.

■ Inner: None.

Loading




Cmax = 1.35 x G + 1.5 x Q=1.35*70+1.5*80=214.5*103 N/ml


CCQ = 1.0 x G + 1.0 x Q=70+80=150*103 N/ml


CQP = 1.0 x G + 0.3 x Q=70+0.3*80=94*103 N/ml


Nm*.²*.

MEd6102555

8

145214

Nm*.²*

MEcq6106753

8

14150150

Nm*.²*

MEqp6103032

8

1494




414



■ )t(*)f (*)t,( cmRH 00 ββ


■ 9252825

816816β .

.

f

.)f (

cm

cm


■ 48802810

1

10

1β

2002000

0 ..t.

)t(..

at t0=28 days



■ 2130

αα10

1001

1 **h*.

RH

RH


■ 121 if MPaf cm 35

■ If not:

70

135

α

.

cmf

and

20

235

α

.

cmf


■ In our case we have: 1338 21 Mpa Mpa f f ck cm

■ 6616342710

100501

16342712506502

125065022

30 ..*.

mm.)(*

**

u

Ac*h RH


■ 375248809252661ββ 00 ..*.*.)t(*)f (*)t,( cmRH


■

Ecar

Eqp

cm

se

M

M*)t,(

E

E

01

α

Defined earlier:

Nm M Ecar 3

10*3675

Nm M Eqp3

10*2303

This gives:

8215

103975

1023037021

31476

200000α

3

3

.

*

**.

e




415


For the calculation of steel ULS, we consider the moment reduced limit of 372.0lu for a steel grade 500Mpa.

Therefore, make sure to enable the option "limit )500/372.0( S " in Advance Design.

Reference reinforcement calculation at SLU:The calculation of the reinforcement is detailed below:

■ Effective height: d=0.9*h=1.125 m


383,067,16*²125.1*65,0

10*25.5255

*²* 2

3

MPamm

Nm

f d b

M

cd w

Ed cu

372.0383.0 lucu therefore the compressed reinforced must be resized and then the concrete section

must be adjusted.

Calculation of the tension steel section (Section A1):

The calculation of tensioned steel section must be conducted with the corresponding moment of :

Nm f d b M cd wlu Ed

6

1 10*10.567.16*²125.1*65.0*372.0*²**

■ The α value: 61803720211251μ211251α .).*(*.)*(*. lulu

■ Calculation of the lever arm zc: m.).*.(*.)*.(*dz lulu 851061804011301α401


²35.13778.434*851.0

10*10.5

.

6

11 cm

MPam

Nm

f z

M A

yd lu

Ed




416

Compressed steel reinforcement reduction (Section As2):

Reduction coefficient:

00327.0045.0130.1*618.0130.1*618.0*1000

5.3)'*(

**1000

5,3 d d

d lu

lu

sc

MPa f yd sc yd sc 78.43400217.000327.0

Compressed reinforcement calculation:

²32.278.434)045.0130.1(

15.5255.5

)'(

12 cm

d d

M M A

sc

Ed Ed s

The steel reinforcement condition:

²32.2.22 cm f

A A yd

sc s

Total area to be implemented:

In the lower part: As1=A1+A2=142.42 cm2

In the top part: As2=2.32 cm2


The concrete beam design at SLS will be made considering a limitation of the characteristic compressive cylinderstrength of concrete at 28 days, at 0.6*f ck

The assumptions are:

■ The SLS moment: Nm*.²*

MEcq6106753

8

14150150

■ The equivalence coefficient: 8215α .e

■ 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:

m..*.

*.d.

*

*x

sce

ce 41901251400158215

158215

σσα

σα1

N*..*.**x*b*F cwc6

1 100421541906502

1σ

2

1

m..

.x

dzc 98503

41901251

3

1

Nm*..**.z*FMccrb

66 10012985010042





417


Nm M M rb

6

1 10*01.2

372.0125.1

419.011

d

x

m x

d z c 985.03

419.0125.1

31

²01.51400985.0

01.211 cm

z

M A

sc


Nm M M M rbcq ser

66

,2 10*665.110*)01.2675.3(


d

d d ce sc

*

'***

1

1

MPa sc 78.211125.1*372.0

045.0125.1*372.0*15*82.15


²92.71

78.211*045.0125.1

645.1

*)'(' 2 cm

d d

M A

sc


²08.38400

78.211*92.71*' cm A A

s

sc s

Section area:



Considering an envelope calculation of ULS and SLS, it will be obtained:

Tensioned reinforcement ULS: A=141.42cm2

Compressed reinforcement SLS: A=71.92cm2

To optimize the reinforcement area, it is preferable a third iteration by recalculating with SLS as a baseline amount oftensioned reinforcement (after ULS: Au=141.44cm

2)




418

Reference reinforcement additional iteration for calculation at SLS:

For this iteration the calculation will be started from the section of the tensioned reinforcement found when calculatingthe SLS: Au=141.44cm

2.

For this particular value, it will be calculated the resistance obtained for tensioned reinforcement:

MPa A A s

ELU

ELS s 252400

42.14118.89

This is a SLS calculation, considering this limitation:

Calculating the moment resistance MRb for detecting the presence of compressed steel reinforcement:

md x sce

ce 55.0130.1*25215*82.15

15*82.15*

*

*1

N xb F cwc

6

1 10*67.215*55.0*65.0*2

1***

2

1

m xd z c 94.0355.0125.1

31

Nm x

d xb z F M cwccrb

611 10*52.2

3

55.0125.1*15*55.0*65.0*

2

1

3****

2

1*

According to the calculation above Mser,cq is grater then Mrb the compressed steel reinforcement is set.


Nm M M rb

6

1 10*52.2

488.0125.155.01

1 d x

m x

d z c 94.03

548.0125.1

31

²38.10625294.0

52.211 cm

z

M A

sc


Nm M M M rbcq ser

66

,2 10*155.110*)52.2675.3(


d

d d ce sc

*

'***

1

1

MPa sc 73.217125.1*485.0

045.0125.1*485.0*15*82.15




419


²12.49

73.217*045.0125.1

155.1

*)'(' 2 cm

d d

M A

sc


²44.42252

73.217*12.49*'cm A A

s

sc s

Section area:





d b

d b f

f

Max A

w

w yk

eff ct

s

**0013.0

***26.0

,

min,


Where:


Therefore:

²73.9

²10*51.9125.1*65.0*0013.0

²10*73.9125.1*65.0*500

56.2*26.0

max

4

4

min, cm

m

m A s



■ 15 nodes,





420




SLS –Characteristic (reference value: 3675kNm)

SLS –Quasi-permanent (reference value: 2303kNm)


(reference value: A’=148.82cm2 A=49.12cm

2)


(reference value: 9.73cm2

)




421




My,SLS,c My corresponding to the 102 combination (SLS) [kNm] 3675 kNm

My,SLS,q My corresponding to the 103 combination (SLS) [kNm] 2303 kNm

Az (A’) Tensioned theoretical reinforcement area [cm2] 148.82 cm2


2



My My USL -5255.25 kN*m 0.0000 %

My My SLS cq -3675 kN*m 0.0000 %

My My SLS pq -2303 kN*m 0.0000 % Az Az -147.617 cm² -0.0003 %

Amin Amin 9.79662 cm² 0.0000 %




422

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

Test ID: 4980

Test status: Passed

5.16.1 Description

The purpose of this test is to verify the My resulted stresses for the USL load combination and the results of thetheoretical reinforcement area Az.

This test performs verification of the theoretical reinforcement area for the T concrete beam subjected to the definedloads. The test confirms the absence of the compressed reinforcement for this model.

5.16.2 Background

Verifies the theoretical reinforcement area for a T concrete beam subjected to the defined loads. The test confirmsthe absence of the compressed reinforcement for this model.






■ Loadings from the structure: G = 0 kN/m (the dead load is not taken into account)

■ Exploitation loadings (category A): Q = 18.2kN/m,



■



The objective is to test:

■ The theoretical reinforcement area





423

Units

Metric System

Geometry


■ Beam length: 8m

■ Concrete cover: c=3.50 cm■ Effective height: d=h-(0.6*h+ebz)=0.482 m; d’=ebz=0.035m







■ Concrete C25/30: MPa

f

f c

ck

cd 67.105,1

25


■ MPa f f ck ctm 56.225*30.0*30.0 3/23/2



f s

yk

yd 83.34715,1

400


Boundary conditions


■ Outer:



■ Inner: None.




424

Loading




Cmax = 1.35 x G + 1.5 x Q=1.35*0+1.5*18.2=27.3kN/ml


CCQ = 1.0 x G + 1.0 x Q=0+18.2=18.2kN/ml


kNm M Ed 40.2188

²8*3.27

kNm M Ecq 60.1458

²8*20.18



kNm f h

d hb M cd

f

f eff btu 92867.16*2

12.0482.0*12.0*10.1*

2**


Therefore, the concrete section is not entirely compressed; This requires a calculation considering a rectangularsection of b=110cm and d=48.2cm.

Reference longitudinal reinforcement calculation:

051.067.16*²482.0*10.1

218.0

*²*

1 cd w

Ed cu

F d b

M

066.0051.0*211*25.1)*21(1*25.1 cuu

md z uc 469.0)066.0*40.01(*482.0)*4.01(*

²38.1383.347*469.0

218.0

*cm

f z

M A

yd c

Ed



■ 9 nodes,


kNm M kNm M btu Ed 92840.218




425








2



Az Az -13.3793 cm² 0.0002 %




426

5.17 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

5.17.1 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. Theverification 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 reinforcementand the verification of the minimum reinforcement percentage.

5.17.2 Background


Verify the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During thistest, the calculation of stresses, the calculation of the longitudinal reinforcement and the verification of the minimumreinforcement percentage are performed.






■ Loadings from the structure: G = 25 kN/m (including dead load),


■




Units

Metric System




427

Geometry


■ Height: h = 0.65 m,

■ Width: b = 0.28 m,

■ Length: L = 6.40 m,



■ Effective height: d = h-(0.6*h+ebz) = 0.57m; d’ = ebz = 0.045 m




■ Concrete density: 25 kN/m3


■ The concrete age t0 = 28 days

■ Humidity 50%

Boundary conditions


■ Outer:



■ Inner: None.

Loading




CCQ = 1.0 x G + 1.0 x Q = 25 + 15 = 40 kN/ml


CQP = 1.0 x G + 0.3 x Q = 25 + 0.3*15 = 29.5 kN/ml


kNm².*)(

MEcq 2058

4061525

kNm

².*)*.(

MEqp 1518

406153025




428

5.17.2.2 Reference results in calculating the concrete final value of creep coefficient

)()(),( 00 t f t cm RH

Where:

MPa f f cm

cm 725.2830

8.168.16

)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t

t0 : concrete age t0=28days

21

0

3 **

*1.0

1001

1

h

RH

RH

121 if Mpa f cm 35

If not

70

135

α

.

cmf

and

20

235

α

.

cmf

In this case therefore

944.038

3535 7.07.0

1

cm f

984.0383535

2.02.0

2

cm f

In this case:

Humidity RH = 50 %

mm

u

Ach 70.195

650280*2

650*280*220

37.2488.0*73.2*78.1)(*)(*),(78.1984.0*

70.195*1.0

100

501

1 003

t f t cm RH RH




429

Calculating the equivalence coefficient:


Ecar

Eqp

cm

se

M M t

E

E

*),(1 0

Where:

MPa f

E ck cm 32837

10

830*22

10

8*22

3.03.0

MPa E s 200000

75.2

205

151*37.21*),(1 0

Ecar

Eqp

M

M t

76.16

205

151*37.21

32837

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E

Material characteristics:

The maximum compression on the concrete is: Mpa f ck bc 1830*6,0*6,0

For the maximum stress on the steel taut, we consider the constraint limit Mpa f yk s

400*8,0

Neutral axis position calculation:

The position of the neutral axis must be determined by calculating 1 (position corresponding to the state of

maximum stress on the concrete and reinforcement):

430.040018*76.16

15*76.16

*

*1

sce

ce

Moment resistance calculation:

Knowing the 1α value, it can be determined the moment resistance of the concrete section, using the following

formulas:

243057004300α11 ..*.d*x m

MNm..

.**.*.*.x

d**x*b*M cwrb 29703

2430570018243028050

3σ

2

1 11

Where:

Utile height : d = h – (0.06h + ebz) = 0.57 m

The moment resistance Mrb = 297 KNm

Because kNmMkNmM rbEcq 297205 the supposition of having no compressed reinforcement is correct.




430

Calculation of reinforcement area with max constraint on steel and concrete

The reinforcement area is calculated using the SLS load combination

Neutral axis position: 4300α1 ,

Lever arm: m..

*.*dzc 48503

430015703

α1

1

Reinforcement section: ²cm.*.

.

*z

M A

sc

ser ser ,s 5610

4004850

2050

σ1



d*b*.

d*b*f

f *.

Max A

w

wyk

eff ,ct

min,s

00130

260

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2Where:


Therefore:

²cm.

²m*..*.*.

²m*..*.*.

.max A min,s 402

1007257028000130

10402570280500

8962260

4

4



■ 7 nodes,■ 1 linear element.




431






Minimum reinforcement area (cm2 )




My,SLS My corresponding to the 102 combination (SLS) [kNm] 205 kNm

Az Theoretical reinforcement area [cm2] 10.50 cm2


2



My My SLS -204.8 kN*m 0.0000 %

Az Az -10.3489 cm² -0.0001 %

Amin Amin -2.38697 cm² 0.0001 %




432

5.18 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

5.18.1 Description

Simple Bending Design for Service State Limit - Verifies the adequacy of a rectangular cross section made fromconcrete C25/30 to resist simple bending.


The verification of the bending stresses at service limit state is performed.

5.18.2 Background


Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, the calculation of stresses will be made along with the calculation of the longitudinal reinforcement and theverification of the minimum reinforcement percentage.








■




Units

Metric System




433

Geometry


■ Height: h = 0.60 m,

■ Width: b = 0.25 m,

■ Length: L = 5.80 m,









■ The concrete age t0=28 days

■ Humidity RH=50%

Boundary conditions


■ Outer:



■ Inner: None.

Loading




CCQ = 1.0 x G + 1.0 x Q = 25 + 30 = 55 kN/ml


CQP = 1.0 x G + 0.3 x Q = 25 + 0.3*30 = 34 kN/ml


kNm M Ecq 28.2318

²80.5*)3025(

kNm M Eqp 97.1428

²80.5*)30*3.025(




434


)()(),( 00 t f t cm RH

Where:

MPa f f cm

cm 92.2825

8.168.16

)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t

t0 : concrete age t0 = 28 days

21

0

3 **

*1.0

1001

1

h

RH

RH

121 if Mpa f cm 35

If not

7.0

1

35

cm f

and

2.0

2

35

cm f

In this case, therefore:

121

In this case:

Humidity RH = 50 %

mm

u

Ach 176

600250*2

600*250*220

70.2488.0*92.2*89.1)(*)(*),(89.1176*1.0

100

501

1 003

t f t cm RH RH



Ecar

Eqp

cm

se

M

M t

E E

*),(1 0

Where:

MPa f

E ck cm 31476

10

825*22

10

8*22

3.03.0

MPa E s 200000




435

67.2231

143*69.21*),(1 0

Ecar

Eqp

M

M t

97.16

231

143*69.21

31476

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E



For the maximum stress on the steel taut, we consider the constraint limit Mpa f yk s 400*8,0


The position of the neutral axis must be determined by calculating 1 (position corresponding to the state ofmaximum stress on the concrete and reinforcement):

389.040018*97.16

15*97.16

*

*1

sce

ce



formulas:

202051903890α11 ..*.d*x m

MNm x

d xb M cwrb 171.03

202.0519.0*15*202.0*25.0*5.0)

3(****

2

1 11

Where:

Utile height : d = h – (0.06h + ebz) = 0.519m

The moment resistance Mrb = 171KNm

Because kNm M kNm M rb Ecq 171231 , the supposition of having no compressed reinforcement is

incorrect.

The calculation of the tension reinforcement theoretical section A1

m x

d z c 452.03

202.0519.0

31

²46.9400*452.0

171.0

*1 cm

z

M A

sc

rb




436

Stress calculation for steel reinforcement σ sc :

087.0519.0

045.0''

d

d

MPace sc 75.197389.0

087.0389.0*15*97.16

'**

1

1

Calculation of the steel compressed reinforcement A’:

²44.675.197*)045.0519.0(

171.0231.0

*)'(' cm

d d

M M A

sc

rb ser

Calculation of the steel tensioned reinforcement A2 :

²18.3400

75.197*44.6'*2 cm A A

s

sc

Calculation of the steel reinforcement:

²64.1218.346.921 cm A A A

²44.6' cm A




Where:


Therefore:

²73.1

²10*69.1519.0*25.0*0013.0

²10*73.1519.0*25.0*500

56.2*26.0

max4

4

min, cm

m

m A s



■ 6 nodes,





437





(reference value: As=12.64cm2; A’=6.44cm

2)





My,SLS My corresponding to the 102 combination (SLS) [kNm] 231.28 kNm Az Theoretical reinforcement area [cm

2] 12.64 cm

2


2



My My SLS -231.275 kN*m -0.0000 %

Az Az -12.6425 cm² -0.0002 %

Amin Amin -1.73058 cm² -0.0001 %




438

5.19 EC2 Test 16: Verifying a T concrete section, without compressed reinforcement- Bilinearstress-strain diagram

Test ID: 4999

Test status: Passed

5.19.1 Description


The purpose of this test is to verify the My resulted stresses for the SLS load combination and the results of thetheoretical reinforcement area Az and of the minimum reinforcement percentage. This test performs verification forthe theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms theabsence of the compressed reinforcement for this model.

5.19.2 Background












■ The calculation is performed considering bilinear stress-strain diagram









439

Units

Metric System

Geometry


■ Beam length: 8m

■ Beam height: h=0.67m





■ Concrete: f ck = 20MPa




■ Humidity RH=50%

Boundary conditions


■ Outer:



■ Inner: None.

Loading




CCQ = 1.0 x G + 1.0 x Q=40+10=50kN/ml


CQP = 1.0 x G + 0.3 x Q=40+0.3*10=43kN/ml


kNm

²*M cq,ser 400

8

81040

kNm

²**.M qp,ser 344

8

8103040




440


)()(),( 00 t f t cm RH

Where:

MPa f f cm

cm 17.3820

8.168.16

)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t


21

0

3 **

*1.0

1001

1

h

RH

RH

if

If not

7.0

1

35

cm f and

2.0

2

35

cm f

In this case,

therefore:

121

In this case:

Humidity RH=50 %

mmu

Ach 123

3140

192600*220

11.3488.0*17.3*2)(*)(*),(2123*1.0

100

501

1 003

t f t cm RH RH



Ecar

Eqp

cm

se

M

M*)t,(

E

E

01

α

Where:

MPa*f

*E

..ck

cm 2996210

82022

10

822

3030

MPaEs 200000

683400

344

11311 0 .*.M

M

*)t,( Ecar

Eqp




441

54.24

400

344*11.31

29962

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E


The maximum compression on the concrete is:

For the maximum stress on the steel taut, we consider the constraint limit

Mpa f yk s 400*8,0

Neutral axis position calculation; Calculation of M tser :

MNmhbhd

hd

M f eff

f

f

e

stser 083,010,0*90.10*

10,0585,0

3

10,0585,0

*54.24*2

400**3*

*2

22

kNm M kNm M tser ser 83400 the neutral axes is on the beam body

Concrete compressive stresses

MPah

d hb

M

f

f eff

ser m 31.8

)2

10,0585,0(*10,0*90.0

400,0

)2

(**

MPahd

d e

s

f

e

sm

c

61.1054.24

400

210,0585,0

54.24

40031.8

*585,0

2

*

MPa MPa cc 1261.10 => there is no compressed reinforcement

The calculation of the tension reinforcement theoretical section As1

and




442

Calculation of the steel compressed reinforcement As2 :

mh x x f 1306,010,02306,012

MPa

x

xcc 01.6

2306,0

1306,0*61.1*

1

22

MN xbb N cweff c 282,0

2

01.6*1306,0*)18,090.0(

2*)( 2

22

Nm z N M cc 125,0441,0*282,0* 222

with m z c 441,03

1306,010,0585,02

²06.7400*441,0

125,0

*2

22 cm

z

M A

sc

s

Notions of serviceability moment M 0 :

Theoretical steel reinforcement section :

²83.18435,0

400,0*46.20*

0

0 cm M

M A A ser s

s



d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


Where:

MPa f f ctmeff ct 21.2,

from cracking conditions

Therefore:

²37.1

²10*37.1585.0*18.0*0013.0

²10*21.1585.0*18.0*500

21.2*26.0

max4

4

min, cm

m

m A s



■ 9 nodes,





443

SLS load combinations(kNm)




(reference value: As=18.83cm2)







2


2



My My SLS -400 kN*m 0.0000 %

Az Az -18.8948 cm² 0.0001 %

Amin Amin -1.36843 cm² -0.0001 %




444

5.20 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

5.20.1 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 thistest, the determination of stresses is made along with the determination of compressing stresses in concrete sectionand compressing stresses in the steel reinforcement section.

5.20.2 Background


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 andcompressing stresses in the steel reinforcement section σs.






■ 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




■ The compressing stresses in concrete section σc

■ The compressing stresses in the steel reinforcement section σs.


Units

Metric System




445

Geometry


■ Height: h = 0.50 m,

■ Width: b = 0.20 m,

■ Length: L = 8.00 m,



■ Effective height: d=44cm;


Rectangular solid concrete C20/25 is used. The following characteristics are used in relation to this material:


■ Stress-strain law for reinforcement: Inclined stress-strain diagram


■ Humidity RH=50%

■ Characteristic compressive cylinder strength of concrete at 28 days: Mpa f ck 20

■ Characteristic yield strength of reinforcement: Mpa f yk 400

■ ²42,9 cm A st for 3 HA20

Boundary conditions


■ Outer:


► Support at end point (x = 8) restrained in translation along Y and Z

■ Inner: None.

Loading



CCQ = 1.0 x G =9.375kN/ml


Mfq = Mcar = Mqp = 75 kNm




446


)()(),( 00 t f t cm RH

Where:

MPa f f cm

cm 17.3820

8.168.16

)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t


21

0

3 **

*1.0

1001

1

h

RH

RH

121 if Mpa f cm 35

If not,

7.0

1

35

cm f and

2.0

2

35

cm f

In this case,

Therefore121

In this case:

Humidity RH=50 %

mm

u

Ach 86.142

500200*2

500*200*220

03.3488.0*17.3*96.1)(*)(*),(96.186.142*1.0

100

501

1 003

t f t cm RH RH


Ecar

Eqp

cm

se

M M t

E

E

*),(1 0




447

Where:

MPa f

E ck cm 29962

10

820*22

10

8*22

3.03.0

MPa E s 200000

03.41*03.31*),(1 0 Ecar

Eqp

M

M t

90.26

1*03.31

29962

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E


The maximum compression on the concrete is: Mpa*,f , ckbc 15256060σ

For the maximum stress on the steel taut, we consider the constraint limit Mpaf *, yks 32080σ

Checking inertia cracked or not:

Before computing the constraints, check whether the section is cracked or not. For this, we determine the crackingmoment which corresponds to a tensile stress on the concrete equal to

ctm f :

v

I f M ctm

cr

*

Where:

433

00208,012

50,0*20,0

12

*m

hb I

mh

v 25,02

The average stress in concrete is:

Mpa f f ck ctm 21,220*30.0*30.0 3

2

3

2

The critical moment of cracking is therefore:

MNmv

I f M ctmcr 018,0

25,000208,0*21,2*

The servility limit state moment is 0.075MNm therefore the cracking inertia is present.




448


Neutral axis equation: 0)'(**)(**²**2

1111 d x A xd A xb e sce st w

w

st scew st sce st sce

b

Ad Ad b A A A A

x

)*'*(***2)²(*)(* 2

1

By simplifying the previous equation, by considering 0 sc A , it will be obtained:

w

st ew st e st e

b

Ad b A A x

)****2²*)(* 2

1

cm x 04,2320

)42,9/44(*90,26*20*2²42,9*²90.2642,9*90.261

Calculating the second moment:

443

1

3

1 001929,0)²230,044,0(*90.26*10*42,93

230,0*20,0)²(**

3

*m xd A

xb I e st

w

Stresses calculation:

Mpa Mpa x I

M c

ser c 1294,8230,0*

001929,0

075,0* 1

Mpa Mpa x

xd sce st 32057.219

230,0

230,044,0*94,8*90.26**

1

1



■ 9 nodes,





449

SLS load combinations (kNm) and stresses (MPa)


SLS (reference value: Mscq=75kNm)

Compressing stresses in concrete section σ c

(reference value: σc =8.94MPa )

Compressing stresses in the steel reinforcement section σ s

(reference value: σs=218.57MPa)




σc Compressing stresses in concrete section σc (MPa) 8.94 MPa

σs Compressing stresses in the steel reinforcement section σs (MPa) 219.67 MPa



My My SLS -75 kN*m -0.0000 %

Sc CQ Sc CQ 8.99322 MPa 0.0359 %

Ss CQ Ss CQ -219.639 MPa 0.0006 %




450

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


5.21.1 Description


Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, the determination of stresses is made along with the determination of compressing stresses in concrete sectionand compressing stresses in the steel reinforcement section; the maximum spacing of cracks and the crack openingsare verified.

5.21.2 Background


Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, 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.







■ Exploitation loadings (category A): Q = 37.5 kN/m,







■ The maximum spacing of cracks

■ The crack opening





451

Units

Metric System

Geometry


■ Height: h = 0.80 m,

■ Width: b = 0.40 m,

■ Length: L = 8.00 m,







■ 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: Mpaf ck 25

■ Characteristic yield strength of reinforcement: Mpaf yk 500

■ ²cm. Ast 1630 for 3 beds of 5HA16

Boundary conditions


■ Outer:



■ Inner: None.

Loading



► M0Ed = 774 kNm

► Mcar = 540 kNm

► Mfq = 390 kNm

► Mqp = 330 kNm




452


)()(),( 00 t f t cm RH

Where:

MPa f f cm

cm 92.2825

8.168.16

)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t


21

0

3 **

*1.0

1001

1

h

RH

RH

If Mpa f cm 35 ,

121

If not,

7.0

1

35

cm f

and

2.0

2

35

cm f

In this case,

, therefore 121

In this case:

Humidity RH=50 %

mm

u

Ach 67.266

800400*2

800*400*220

55.2488.0*92.2*78.1)(*)(*),(78.167.266*1.0

100

501

1 003

t f t cm RH RH


Under quasi-permanent combinations:

),(1 0t

E

E

cm

se

Where:

MPa f

E ck cm 31476

10

825*22

10

8*22

3.03.0

MPa E s 200000

55.2),( 0 t




453

56.22

55.21

31476

200000

),(1 0

t

E

E

cm

se


The maximum compression on the concrete is: Mpa f ck bc 1525*6,06,0



Neutral axis equation: 0αα2

1111 )'dx(** A)xd(** A²x*b* escestw

w

stscewstscestsce

b

) A*d A'*d(**b*)² A A(*) A A(*x

α2αα 2

1


cm.**.**².*²..*.

b

) A*d**b*² A* A*x

stewsteste

3540

16307156224021630562216305622

α2αα 2

1


443

11

31

01450350071056221016303

350040

αα3

m.)²..(*.**,.*.

)²'dx(** A)²xd(** Ax*b

I escestw


Mpa x I

M ser c 96.7350,0*

0145,0

330,0* 1

Mpa Mpa x

xd sce st 4007.184

350,0

350,071,0*96.7*56.22**

1

1




454

Maximum spacing of cracks:

Bottom reinforcement 3HA20+3HA16=15.46cm2

2, 06.015.0*4.0

4.02

8.0

15.03

)350.08.0(

255.0)71.08.0(*5.2

min*40.0

2

3)(

)(*5.2

min* m

h

xh

d h

b A eff c

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(2); Figure 7.1

0500.006.0

10*16.30 4

,

,

eff c

seff p

A

A

mmnn

nneq 16

**

**

2211

2

22

2

11

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(3)

eff p

r

k k ck s

,

213max,

***425.0*

Where:

c=0.051m

113.251

25*4.3

25*4.3

3/23/2

3

c

k

Therefore:

mmk k

ck seff p

r 162050.0

)10*16(*5.0*8.0*425.0051.0*113.2

***425.0*

3

,

213max,

Calculation of average strain:

41.631187

200000

cm

se

E

E

44

,

,

,

10*54.5*6,010*88.7200000

)050.0*41.61(*050.0

56.2*4.071.184).1(**

s

s

s

eff pe

eff p

eff ct

t s

cm sm E E

f k


Calculation of crack widths:

mm sw cm smr k 128.0)1088.7(*162.0)(* 4

max,


For an exposure class XD1, using the French national annex, we retained an opening crack of 0.20mm max.

This criterion is satisfied.




455



■ 11 nodes,




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 σc-qp

(reference value: σc-qp =7.96MPa )

Compressing stresses in the steel reinforcement section σs-qp

(reference value: σs-qp=185MPa)




456

Maximum cracking space Sr,max

(reference value: Sr,max=162mm)

Maximum crack opening Wk

(reference value: Wk=0.128mm)



MEd My corresponding to 102 combination (ULS) [kNm] 774 kNmMser-cq My corresponding to104 combination (SLS) [kNm] 540 kNm

Mser-qp My corresponding to 108 combination (SLS) [kNm] 330 kNm

σc Compressing stresses in concrete section σc [MPa] 7.96 MPa

σs Compressing stresses in the steel reinforcement section σs [MPa] 185 Mpa

Sr,max Maximum cracking space Sr,max [cm] 16.2 cm

Wk Maximum crack opening Wk [cm] 0.0128 cm



My My ULS -774 kN*m 0.0000 %

My My SLS cq -540 kN*m 0.0000 %

My My SLS qp -330 kN*m 0.0000 %

Sc QP Sc QP 7.76924 MPa 0.0000 %

Ss QP Ss QP -181.698 MPa 0.0001 %

Sr,max Sr,max 16.1577 cm -0.0003 %

wk Wk -0.0125137 cm 0.0000 %




457

5.22 EC2 Test 23: Verifying the shear resistance for a rectangular concrete - Bilinear stress-straindiagram (Class XC1)

Test ID: 5053

Test status: Passed

5.22.1 Description

Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, theshear force diagram will be generated. The design value of the maximum shear force which can be sustained by themember, limited by crushing of the compression struts (VRd,max) will be determined, along with the cross-sectionalarea of the shear reinforcement (Asw) calculation.

5.22.2 Background




■ Concrete C25/30

■ Reinforcement steel: S500B




■ Loadings from the structure:

The dead load family is be considered from three loads: two point loads of 55kN and 65kN and one linear loadof 25kN/m, placed along the beam as described in the picture:




458

■ Exploitation loadings:

The live load will consist of three loads: two point loads of 40kN and 35kN and one linear load of 20kN/m,placed along the beam as described in the picture:



■ The reinforcement will be displayed like in the picture below:


■ The shear stresses results

■ The cross-sectional area of the shear reinforcement, Asw

■ The theoretical reinforcement value

Units

Metric System

Geometry


■ Height: h = 0.70 m,

■ Width: b = 0.30 m,

■ Length: L = 5.30 m,


■ Effective height: d=h-(0.06*h+ebz)=0.623m; d’=ehz=0.0335m

■ Stirrup slope: = 90°■ Stirrup slope: 45˚




459

Boundary conditions


■ Outer:


► Support at end point (x = 5.30) restrained in translation along Y, Z and rotation along X

■ Inner: None.

Loading

The maximum shear stresses from the concrete beam:

Using the software structure calculation the following values were obtained:

x = 0m => KNVEd 406

x = 0.50m (for the first point load) => KN.VEd 76373 and KN.VEd 51239

x = 0.95m (for the second point load) => KN.VEd 82210 and KN.VEd 5770

x = 2.06m the shear force is null => KNVEd 0

Note: In Advance Design, the shear reduction is not taken into account. These are the values corresponding to an unreducedshear that will be used for further calculations.




460

5.22.2.2 Reference results in calculating the maximum design shear resistance

21max,cot1

cotcot*****

wucd cw Rd b z f V


Where:

1cw coefficient taking account of the state of the stress in the compression chord and

250

1*6,01ck f

v

When the transverse frames are vertical, the above formula simplifies to:

cot

***1max,

tg

b z f vV wucd Rd

In this case:

45 and 90

1v strength reduction factor for concrete cracked in shear

54.0250

251*6.0

2501*6,01

ck f

v

md z u 561.0623.0*9.0*9.0

MN V Rd 757.02

30.0*561.0*67.16*54.0max,

MN V MN V Rd Ed 757.0406.0 max,

Calculation of transversal reinforcement:

Is determined by considering the transverse reinforcement steels vertical ( = 90°) and connecting rods inclinedat 45 °, at different points of the beam.

Before the first point load:

ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²65,16

15,1

500*561,0

45*406,0

*

*.

Between the first and the second point load:

ml cmtg s A sw /²82,9

15,1

500*561,0

45*23951,0

It also calculates the required reinforcement area to the right side of the beam:

ml cmtg

s

A sw /²48,8

15,1

500*561,0

45*20674,0




461



■ 7 nodes,


Advance Design gives the following results for Atz (cm2

/ml)

Note: after the second point load, the minimum transverse reinforcement is set (noted with A tmin in ADVANCE Design)(in cm²/ml):

The reinforcement theoretical value is calculated using the formula:

ml cmb f

f b

s

Aw

yk

ck

ww sw /²4.290sin*30.0*

500

25*08,0sin**

*08,0sin**min,




462



Fz Fz corresponding to 101 combination (ULS) x=0m [kN] 405.63 kN

Fz Fz corresponding to 101 combination (ULS) x=0.5m [kN] 373.76 kN




Atz Transversal reinforcement area x=0m [cm2/ml] 16,65 cm

2/ml

Atz Transversal reinforcement area x=0.501m [cm2/ml] 9.82 cm

2/ml

Atz Transversal reinforcement area x=5.3m [cm2/ml] 8.84 cm

2/ml

At,min,z Theoretical reinforcement area [cm2/ml] 2.40 cm

2/ml



Fz Fz,0 -405.633 kN -0.0001 %

Fz Fz,1 -373.758 kN -0.0001 %

Fz Fz,1' -239.445 kN 0.0002 %

Fz Fz,2 -210.821 kN 0.0001 %

Fz Fz,2' -70.5644 kN -0.0001 %

Atz Atz,0 16.6391 cm² 0.0002 %

Atz Atz,1 9.82205 cm² 0.0000 %

Atz Atz,r 8.48058 cm² -0.0000 %

Atminz At,min,z 2.4 cm² 0.0000 %




463

5.23 EC2 Test 10: Verifying a T concrete section, without compressed reinforcement - Inclinedstress-strain diagram

Test ID: 4984

Test status: Passed

5.23.1 Description

Simple Bending Design for Ultimate Limit State - The purpose of this test is to verify the My resulted stresses for theULS load combination, the results of the theoretical reinforcement area, "Az" and the minimum reinforcementpercentage, "Amin".


5.23.2 Background

Verifies the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirmsthe absence of the compressed reinforcement for this model.











■ 30ψ2 ,






■ The reinforcement minimum percentage area





464

Units

Metric System

Geometry


■ Beam length: 6m






■ The calculation is made considering the inclined stress-strain diagram


MPa

,

f f

c

ckcd 20

51

30

γ


MPa.*.f *.f //ckctm 90230300300 3232


■ Steel S400B :

MPa.,

f f

s

ykyd 78434

151

500

γ


Boundary conditions


■ Outer:



■ Inner: None.




465

Loading


Load combinations:


Cmax = 1.35 x G + 1.5 x Q=1.35*500+1.5*300=1125 kN/m


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:

kNm.²*

MEd 550628

61125

kNm²*

MEcq 36008

6800

kNm²*

MEqp 26558

6590



kNm*.*,

,*,*.f *h

d*h*bM cdf

f eff btu3

104804202

20090200401

2


Therefore the concrete section is not entirely compressed;

There are required calculations considering the T section.


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


The moment corresponding to this section is:

kNm*.

..**.*)..(

hd*f *h*)bb(M f

cdf weff Ed

3

2

1023

2

20090020200400401

2

Stress from the compressed steel reinforcement considering a steel grade S500B:

MPa*,, susu 466ε2772771432σ

In order to determine the suε , the neutral axis position must be determined:

kNm*.*)..(MMM EdEdEd33

21 1086311020030635

28702090400

8631

μ 1

.*².*.

.

F*²d*b

M

cdw

Edcu

kNm*.MkNm*.M Edbtu33 1006255104804




466

43502870211251μ211251α ..**.)*(*. cuu

Depending of uα , the neutral axis position can be established:

5545034350

43501ε

α

α1ε 2 ..*

.

.* cu

u

usu

‰

Then we calculate the stress in the tensioned steel reinforcement:

MPaMPa,.,,su 466024360045502772771432σ

According to those above, the theoretical reinforcement can be calculated:

²cm.

.*.

.

.

F*h

d

M A

ydf

Ed 7491

024362

20090

2003

2

22


The theoretical section 1 corresponds to a calculation for a rectangular shape beam section

kNm*.*)..(MMM EdEdEd 3321 1086311020030635

3720μ28702090400

8631μ 1 ..

*².*.

.

F*²d*b

Mlu

cdw

Edcu

For a S500B reinforcement and for a XC2 exposure class, there will be a: 3720μ2870μ .. lucu , therefore there will be

no compressed reinforcement.

There will be a calculation without compressed reinforcement:

43502870211251μ211251α ..**.)*(*. cuu

m.).*.(*.)*.(*dz uc 74304350400190α4011

²cm..*.

.

f *z

M A

ydcEd 4657024367430

8631

111


In conclusion the entire reinforcement steel area is A=A1+A2=91.74+57.46=149.20cm2



d*b*.

d*b*f

f *.

Max A

w

wyk

eff ,ct

min,s

00130

260


Where:


Therefore:

²cm.

²m*..*.*.

²m*..*.*.

*.max A min,s 425

1068490040000130

10425900400500

902260

4

4



■ 7 nodes,





467












2


2



My My USL -5062.5 kN*m 0.0000 %

Az Az -149.031 cm² 0.0001 %

Amin Amin -5.42219 cm² 0.0000 %




468

5.24 EC2 Test 15: Verifying a T concrete section, without compressed reinforcement- Bilinearstress-strain diagram

Test ID: 4998

Test status: Passed

5.24.1 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 thetheoretical reinforcement area Az and of the minimum reinforcement percentage.

5.24.2 Background

This test performs the verification of the theoretical reinforcement area for a T concrete beam subjected to thedefined loads. The test confirms the absence of the compressed reinforcement for this model.












The objective is to verify:■ The stresses results







469

Units

Metric System

Geometry


■ Beam length: 8m

■ Beam height: h=0.76m■ Concrete cover: c=4.50 cm









■ Humidity RH=50%

■ Concrete: f ck = 20MPa

Boundary conditions


■ Outer:



■ Inner: None.

Loading




CCQ = 1.0 x G + 1.0 x Q=20+10=30kN/ml


CQP = 1.0 x G + 0.3 x Q=20+0.3*10=23kN/ml





470

kNm M cq ser 240

8

²8*1020,

kNm M qp ser 184

8

²8*10*3.020,

5.24.2.2 Reference results in calculating the concrete final value of the creep coefficient

)()(),( 00 t f t cm RH

Where:

MPa f

f cm

cm 17.3820

8.168.16)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t


21

0

3 **

*1.0

1001

1

h

RH

RH

121 if Mpa f cm 35

If not

7.0

1

35

cm f and

2.0

2

35

cm f

In this case Mpa Mpa Mpa f f ck cm 35288 therefore

121

In this case:

Humidity RH=50 %

mmu

Ach 24.162

3920

318000*220

97.2488.0*17.3*92.1)(*)(*),(92.124.162*1.0

100

50

11 003 t f t cm RH RH



Ecar

Eqp

cm

se

M

M t

E

E

*),(1 0




471

Where:

MPa f

E ck cm 29962

10

820*22

10

8*22

3.03.0

MPa E s 200000

28.3240

184*97.21*),(1 0

Ecar

Eqp

M

M t

89.21

240

184*97.21

2962

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E




Neutral axis position calculation; Calculation of Mtser :

MNmhbhd

hd

M f eff

f

f

e

stser 122,010,0*20,1*

10,0669,0

3

10,0669,0

*77.43

400**3*

*2

22

kNm M kNm M tser ser 122240 the neutral axes is on the beam body

Concrete compressive stresses

MPah

d hb

M

f

f eff

ser m 23,3

)2

10,0669,0(*10,0*20,1

240,0

)2

(**

MPah

d

d e

s

f

e

sm

c 96,489.21

400

2

10,0669,0

89.21

40023,3

*669,0

2

*

MPa MPa cc 1296.4 => there is no compressed reinforcement




472

The calculation of the tension reinforcement theoretical section As1

cmd x sce

ce 30.14669,0*40096,4*89.21

96,4*89.21

*

*1

MN xb N ceff c 426,0

296,4*1430,0*20,1

2** 11

m.MN,,*,z*NM cc 265062204260111 and m,,

,zc 62203

1430066901

²65,10400*622,0

265,0

*1

11 cm

z

M A

sc

s

Calculation of the steel compressed reinforcement As2 :

m,,,hxx f 043001001430012

MPa,,

,*,x

x*cc 49114300

04300964σσ1

22

MN,,

*,*),,(x*)bb(N cweff c 0290

2

4910430030201

2

σ 222

MNm,,*,z*NM cc 016055500290222

with m,,

,,zc 55503

0430010066902

²cm,*,

,

*z

M A

sc

s 7204005550

0160

σ2

22

Notions of serviceability moment M 0 :

²cm... A A A sss 9397206510210

m.MN,,,MMM 249001602650210

Theoretical steel reinforcement section :

²cm,,

,*,

M

M* A A ser s

s 5892490

2400939

0

0



d*b*.

d*b*f

f *.

Max A

w

wyk

eff ,ct

min,s

00130

260


Where:


Therefore:

²cm.

²m*..*.*.

²m*..*.*.

*.max A min,s 612

10612669030000130

103126690300500

212260

4

4




473



■ 9 nodes,


SLS load combinations(kNm)




(reference value: As=9.58cm2)

Minimum reinforcement area (cm2

)





474





2




My My SLS -240 kN*m 0.0000 %

Az Az -9.64217 cm² -0.6490 %

Amin Amin -2.574 cm² -0.0000 %




475

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


5.25.1 Description


Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, the determination of stresses is made along with the determination of compressing stresses in concrete sectionand compressing stresses in the steel reinforcement section; the maximum spacing of cracks and the crack openingsare verified.

5.25.2 Background


Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, 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.







■ Exploitation loadings (category A): Q = 20 kN/m,







■ The maximum spacing of cracks

■ The crack opening





476

Units

Metric System

Geometry


■ Height: h = 0.60 m,

■ Width: b = 0.20 m,

■ Length: L = 5.80 m,








■ Humidity RH=50%

■ Characteristic compressive cylinder strength of concrete at 28 days: Mpaf ck 25

■ Characteristic yield strength of reinforcement: Mpaf yk 500

■ ²cm. Ast 4615 for 3 HA20+3 HA16

Boundary conditions


■ Outer:



■ Inner: None.

Loading



► M0Ed = 228 kNm

► Mcar = 160 kNm

► Mfq = 118 kNm

► Mqp = 101 kNm




477


)()(),( 00 t f t cm RH

Where:

MPa f f cm

cm 92.2825

8.168.16

)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t


21

0

3 **

*1.0

1001

1

h

RH

RH

121 if Mpa f cm 35

If not

7.0

1

35

cm f

and

2.0

2

35

cm f

In this case,

, therefore:

In this case:

Humidity RH=50 %

mm

u

Ach 150

600200*2

600*200*220

77.2488.0*92.2*94.1)(*)(*),(94.1150*1.0

100

501

1 003

t f t cm RH RH


Ecar

Eqp

cm

se

M

M t

E

E

*),(1 0




478

Where:

MPa f

E ck cm 31476

10

825*22

10

8*22

3.03.0

MPaEs 200000

75.2160

101*77.21*),(1 0

Ecar

Eqp

M

M t

75.2

160

101*77.21

31476

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E







w

stscewstscestsce

b

) A*d A'*d(**b*)² A A(*) A A(*x

α2αα 2

1


cm..**.**².*²..*.

b

) A*d**b*² A* A*x

stewsteste

672620

46155347172024615471746154717

α2αα 2

1


443

11

31

003140267053047171046153

267020

αα3

m.)²..(*.**,

.*.

)²'dx(** A)²xd(** Ax*b

I escestw


Mpa x I

M ser c 59.8267,0*

00314,0

101,0* 1

Mpa Mpa x

xd sce st 40082.147

267,0

267,053,0*59.8*47.17**

1

1




479

Maximum spacing of cracks:

Bottom reinforcement 3HA20+3HA16=15.46cm2

2

, 0222.0111.0*2.0

3.02

6.0

111.03

)267.06.0(175.0)53.06.0(*5.2

min*20.0

2

3

)()(*5.2

min* m

h

xhd h

b A eff c

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 7.3.4.(2); Figure 7.1

070.00222.0

10*46.15 4

,

,

eff c

seff p

A

A

mmnn

nneq 22.18

**

**

2211

2

22

2

11


eff p

r

k k ck s

,

213max,

***425.0*

Where:

c=0.051m

114.251

25*4.3

25*4.3

3/23/2

3

c

k

Therefore:

mmk k

ck seff p

r 15207.0

)10*22.18(*5.0*8.0*425.0051.0*114.2

***425.0*

3

,

213max,

Calculation of average strain:

35.631476

200000

cm

se

E

E

44

,

,

,

10*43.4*6,010*33.6

200000

)07.0*35.61(*07.0

56.2*4.082.147).1(**

s

s

s

eff pe

eff p

eff ct

t s

cm sm

E E

f k



mm sw cm smr k 096.0)1033.6(*152)(* 4

max,


For an exposure class XD1, using the French national annex, we retained an opening crack of 0.20mm max.This criterion is satisfied.




480



■ 11 nodes,





Compressing stresses in concrete section σc

(reference value: σc =8.59MPa )

Compressing stresses in the steel reinforcement section σs

(reference value: σs=148MPa)

Maximum cracking space Sr,max

(reference value: Sr,max=152mm)

Maximum crack opening Wk

(reference value: Wk=0.096mm)




481



Mser-cq My corresponding to the 104 combination (SLS) [kNm] 160 kNm

σc Compressing stresses in concrete section σc [MPa] 8.50 MPa

σs Compressing stresses in the steel reinforcement section σs [MPa] 148 MPa

Sr,max Maximum cracking space Sr,max [cm] 15.2 cm

Wk Maximum crack opening Wk [cm] 0.0096 cm



My My SLS cq -159.546 kN*m -0.0001 %

Sc QP Sc QP 8.50328 MPa -0.0000 %

Ss QP Ss QP -147.055 MPa 0.0003 %

Sr,max Sr,max 15.3633 cm 0.9139 %

wk Wk -0.00966483 cm -0.9138 %




482

5.26 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

5.26.1 Description



5.26.2 Background



test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and theverification of the minimum reinforcement percentage.








■




Units

Metric System




483

Geometry


■ Height: h = 0.60 m,

■ Width: b = 0.25 m,

■ Length: L = 5.80 m,









■ The concrete age t0 =28 days



MPa,,

f f

c

ckcd 6716

51

25

γ


MPa.*.f *.f //ckctm 56225300300 3232


■ Steel S500 :

MPa,,

f f

s

yk

yd 78434151

500

γ


MPa f

E ck cm 31476

10

825*22000

10

8*22000

3.03.0


Boundary conditions


■ 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.




484

Loading



G’=0.25*0.6*2.5=3.75kN/ml

■ Load combinations:Characteristic combination of actions:

CCQ = 1.0 x G + 1.0 x Q=15+3.75+20=38.75kN/ml


CQP = 1.0 x G + 0.3 x Q=15+3.75+0.3*20=24.75kN/ml


kNm M Ecq 94,1628

²80,5*75,38

kNm M Eqp 7.104

8

²80,5*75.24




485


)()(),( 00 t f t cm RH

Where:

MPa..

f

.)f (

cmcm

9252825

816816β

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t

t0 : concrete age t0 = 28days

21

0

3 **

*1.0

1001

1

h

RH

RH

1αα 21 if Mpaf cm 35

If not,

70

135

α

.

cmf

and

20

235

α

.

cmf

In this case,

, therefore 1αα 21

In this case,

Humidity RH=50 %

mm.

*

**

u

Ach 47176

6002502

600250220

70248809252891ββ8914717610

100

501

1 003..*.*.)t(*)f (*)t,(.

.*.cmRHRH

Therefore:

73.2163

104*70.21*),(1 0

Ecar

Eqp

M

M t


32.17

163

104*70.21

31476

200000

*),(1 0

Ecar

Eqp

cm

s

e

M

M t

E

E







486


The position of the neutral axis must be determined by calculating 1 (position corresponding to the state of

maximum stress on the concrete and reinforcement):

394.040015*32.17

15*32.17

*

*1 sce

ce


Knowing the 1 value, it can be determined the moment resistance of the concrete section, using the following

formulas:

204051903940α11 ..*.d*x m

MNm..

.**.*.*.)x

d(**x*b*M cwrb 17303

2040519015204025050

3σ

2

1 11

Where:



Because kNm M kNm M rb Ecq 17394.162 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: 394,01

Lever arm: m..

*.*dzc 45103

394015190

3

α1 1

Reinforcement section: ²03.9400*451.0

163.0

*,1 cm

z

M A

sc

ser ser s



d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


Where:


Therefore:

²73.1

²10*69.1519.0*25.0*0013.0

²10*73.1519.0*25.0*500

56.226.0

max4

4

min, cm

m

m A s




487



■ 11 nodes,












488





2




My My USL -162.936 kN*m 0.0001 %

Az Az -9.02822 cm² -0.0000 %

Amin Amin -1.73058 cm² -0.0001 %




489

5.27 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

5.27.1 Description

Simple Bending Design for Service State Limit - Verifies the adequacy of a rectangular cross section made fromconcrete C25/30 to resist simple bending.


The verification of the bending stresses at service limit state is performed.

5.27.2 Background








■ 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

3,02

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





490

Units

Metric System

Geometry


■ Height: h = 0.80 m,

■ Width: b = 0.40 m,

■ Length: L = 6.30 m,










■ Humidity RH=50%

Boundary conditions


■ Outer:



■ Inner: None.

Loading


■ Dead load:

0.40*0.80*25 = 8kN/ml



CCQ = 1.0 x G + 1.0 x Q=8+50+60=118kN/ml


CQP = 1.0 x G + 0.3 x Q=8+50+0.3*60=76kN/ml


mkN M cq ser .05.672

8

²3.6*60508

4

3.6*2530,

mkN M qp ser .11.436

8

²3.660*3.0508

4

3.6*25*3.030,




491


)()(),( 00 t f t cm RH

Where:

MPa f f cm

cm 92.2825

8.168.16

)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t


21

0

3 **

*1.0

1001

1

h

RH

RH

121 if Mpa f cm 35

If not

7.0

1

35

cm f and

2.0

2

35

cm f

In this case therefore

121

In this case:

Humidity RH=50 %

mm

u

Ach 267

800400*2

800*400*220

54.2488.0*92.2*78.1)(*)(*),(78.1267*1.0

100

501

1 003

t f t cm RH RH



Ecar

Eqp

cm

se

M

M t

E

E

*),(1 0

Where:

MPa f

E ck cm 31476

10

825*22

10

8*22

3.03.0

MPa E s 200000




492

65.2672

435*54.21*),(1 0

Ecar

Eqp

M

M t

82.16

672

435*54.21

31476

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E





The position of the neutral axis must be determined by calculating (position corresponding to the state ofmaximum stress on the concrete and reinforcement):

387.040018*82.16

15*82.16

*

*1

sce

ce



formulas:

273.0707.0*387.0*11 d x m

MNm x

d xb M cwrb 505.03

273.0707.0*15*273.0*4.0*5.0)

3(****

2

1 11

Where:



Because kNm M kNm M rb Ecq 505672 , the supposition of having no compressed reinforcement is

incorrect.

The calculation of the tension reinforcement theoretical section A1

m x

d z c 616.03

273.0707.0

31

²51.20400*616.0

505.0

*1 cm

z

M A

sc

rb




493

Stress calculation for steel reinforcement σ sc :

064.0707.0

045.0''

d

d

MPace sc 78.210387.0

067.0387.0*15*82.16

'**

1

1

Calculation of the steel compressed reinforcement A’:

²96.1178.210*)045.0707.0(

505.0672.0

*)'(' cm

d d

M M A

sc

rb ser

Calculation of the steel tensioned reinforcement A2 :

²30.6400

78.210*96.11'*2 cm A A

s

sc

Calculation of the steel reinforcement :

²81.2630.651.2021 cm A A A



d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,


Where:


Therefore:

²76.3

²10*68.3707.0*40.0*0013.0

²10*76.3707.0*40.0*500

56.2*26.0

max4

4

min, cm

m

m A s



■ 7 nodes,





494




Theoretical reinforcement area (cm2)

(reference value: As=26.81cm2; A’=11.96cm

2)

Minimum reinforcement area (cm2 )






2




My My SLS -672.032 kN*m -0.0001 %

Az Az -26.8086 cm² -0.0001 %

Amin Amin -3.77193 cm² -0.0001 %




495

5.28 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

5.28.1 Description

Simple Bending Design for Serviceability State Limit - Verifies the adequacy of a rectangular cross section made fromconcrete C25/30 to resist simple bending.

During this test, the determination of stresses is made along with the determination of compressing stresses inconcrete section and compressing stresses in the steel reinforcement section.

5.28.2 Background



test, the calculation of stresses is made along with the calculation of compressing stresses in concrete section σc andcompressing stresses in the steel reinforcement section σs.






■ Loadings from the structure: G = 37.5 kN/m (including the dead load),

■ Exploitation loadings (category A): Q = 37.5 kN/m,



■ For the stress calculation the French annexes was used






Units

Metric System




496

Geometry


■ Height: h = 0.80 m,

■ Width: b = 0.35 m,

■ Length: L = 8.00 m,


■ Concrete cover: c=4 cm







■ Humidity RH=50%

■ Characteristic compressive cylinder strength of concrete at 28 days: Mpaf ck 25 ■ Characteristic yield strength of reinforcement: Mpaf yk 500

■ ²cm. Ast 7037 for 3 HA20

■ ²cm. Asc 286 for 2 HA10

Boundary conditions


■ Outer:



■ Inner: None.

Loading



► Characteristic combination of actions:

CCQ = 1.0*G+1.0*Q =75 kN/ml


► M0Ed = 855 kNm

► Mcar = 600 kNm

► Mqp = 390 kNm




497


)()(),( 00 t f t cm RH

Where:

MPa f f cm

cm 92.2825

8.168.16

)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t


21

0

3 **

*1.0

1001

1

h

RH

RH

if

If not,

7.0

1

35

cm f and

2.0

2

35

cm f

In this case,

, therefore121

In this case,

Humidity RH=50 %

mm

u

Ach 48.243

800350*2

800*350*220

56.2488.0*92.2*80.1)(*)(*),(80.148.243*1.0

100

501

1 003

t f t cm RH RH


Ecar

Eqp

cm

se

M

M t

E

E

*),(1 0




498

Where:

MPa f

E ck cm 31476

10

825*22

10

8*22

3.03.0

MPa E s 200000

664.2600

390*56.21*),(1 0

Ecar

Eqp

M

M t

90.16

664.2

31476

200000

*),(1 0

Ecar

Eqp

cm

se

M

M t

E

E







cm,).*.*(*.**)²..(*².)..(*.

b

) A'*d A*d(**b*)² A A(*²) A A(*x

scstewscstescste

413435

286470377290163522867037901628670379016

α2αα1


443

11

31

0147201472097445349016286453472901670373

413435

αα3

m.cm)².(*.*,)²,(.*,,*

)²'dx(** A)²xd(** Ax*b

I escestw


Mpa Mpa x I

M

c

ser

c 12143441,0*01472,0

600,0

* 1

Mpa Mpa x

xd sce st 400259

3441,0

3441,072,0*14*90.16**

1

1



■ 9 nodes,





499




Compressing stresses in concrete section σc

(reference value: σc =14MPa )

Compressing stresses in the steel reinforcement section σs

(reference value: σs=260.15MPa)




σc Compressing stresses in concrete section σc (MPa) 14 MPa

σs Compressing stresses in the steel reinforcement section σs (MPa) 260.15 MPa



My My SLS -600 kN*m 0.0000 %

Sc CQ Sc CQ 14.122 MPa 0.0142 %

Ss CQ Ss CQ -260.116 MPa 0.0014 %




500

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

Test ID: 5072

Test status: Passed

5.29.1 Description

Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, theshear force diagram will be generated. The design value of the maximum shear force which can be sustained by themember, limited by crushing of the compression struts, VRd,max, will be calculated, along with the cross-sectionalarea of the shear reinforcement, Asw. For the calculation, the reduced shear force values will be used.

5.29.2 Background

Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, theshear force diagram will be generated. The design value of the maximum shear force which can be sustained by themember, limited by crushing of the compression struts, VRd,max, will be calculated, along with the cross-sectional areaof the shear reinforcement, Asw. For the calculation, the reduced shear force values will be used.



■ Concrete C25/30






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










501

Units

Metric System

Geometry


■ Height: h = 0.70 m,■ Width: b = 0.35 m,

■ Length: L = 5.75 m,



■ Stirrup slope: = 90°

■ Strut slope: θ=45˚

Boundary conditions


■ 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:

2

* l P V u Ed

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 :

KN V Ed 2632

5.91*75.5




502

5.29.2.2 Reference results in calculating the lever arm zc:

The lever arm will be calculated from the design formula for pure bending:

ml kN P u /5.91

kNm M Ed 15.3788

²75.5*5.91

167.067.16*²623.0*35.0

378.0

*²*

cd w

Ed cu

f d b

M

230.0167.0*211*25.1*211*25.1 cuu

md z uc 566.0230.0*4.01*623.0*4.01*

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:

2

**)( l P x P xV u

u

Therefore:

m z x 566.045cot*566.0cot*

kN V red Ed 211263566.0*5.91,

Warning: Advance Design does not apply the reduction of shear force corresponding to the distributed loads (cutting

edge at x = d).

Calculation of maximum design shear resistance:

θ1

θα να 1 2wucdcwmax,Rd

cot

cotcot*b*z*f **V


Where:


250

1*6,01

ck f v


θθ

1

cottg

b*z*f *vV wucd

max,Rd

In this case:

45 and 90


54.0250

25

1*6.02501*6,01

ck f

v




503

md z u 56.0623.0*9.0*9.0

kN MN V Rd 891891.02

35.0*566.0*67.16*54.0max,

kN V kN V Rd Ed

891236max,


Given the vertical transversal reinforcement ( = 90°), the following formula is used:

ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²67,8

15,1

500*566,0

45*211,0

*

*.



■ 3 nodes,■ 1 linear element.

Advance Design gives the following results for Atz (cm2/ml)



Fz Fz corresponding to the 101 combination (ULS) [kNm] 263 kN

At,z Theoretical reinforcement area [cm2/ml] 8.67 cm

2/ml



Fz Fz -263.062 kN -0.0002 % Atz Atz 8.68636 cm² 0.0000 %




504


Test ID: 5076

Test status: Passed

5.30.1 Description

Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For this test, theshear force diagram is generated. The design value of the maximum shear force which can be sustained by themember, limited by crushing of the compression struts, VRd,max., will be calculated, along with the cross-sectionalarea of the shear reinforcement, Asw, and the theoretical reinforcement. For the calculation, the reduced shear forcevalues will be used.

5.30.2 Background

Description: Verifies a rectangular cross section beam made from concrete C25/30 to resist simple bending. For thistest, the shear force diagram is generated. The design value of the maximum shear force which can be sustained bythe member, limited by crushing of the compression struts, VRd,max. will be calculated, along with the cross-sectionalarea of the shear reinforcement, Asw, and the theoretical reinforcement.



■ Concrete C25/30






The dead load will be represented by a linear load of 40kN/m


The live load will be considered from one linear load of 25kN






■ The theoretical reinforcement value




505

Units

Metric System

Geometry



■ Length: L = 5.75 m,





Boundary conditions


■ 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:


2

* l P V u Ed


Pu = 1,35*Pg + 1,50*Pq=1.35*40kN+1.50*25kN=91.5kN

In this case :

KN V Ed 2632

5.91*75.5




506

5.30.2.2 Reference results in calculating the lever arm zc:

The lever arm is calculated using the design formula for pure bending:

ml kN P u /5.91

kNm M Ed 15.3788

²75.5*5.91

167.067.16*²623.0*35.0

378.0

*²*

cd w

Ed cu

f d b

M

230.0167.0*211*25.1*211*25.1 cuu

md z uc 566.0230.0*4.01*623.0*4.01*

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 cot. z x .

In the case of a member subjected to a distributed load, the equation of the shear force is:

2

**)( l P x P xV u

u

Therefore:

m z x 566.045cot*566.0cot*

kN V red Ed 211263566.0*5.91,

Warning: Advance Design does not apply the reduction of shear force corresponding to the distributed loads (cutting

edge at x = d).

Calculation of maximum design shear resistance:

θ1


cot

cotcot*b*z*f **V


Where:


250

1*6,01ck f

v


cot

***1max,

tg

b z f vV wucd Rd

In this case:

45 and 90


54.0

250

251*6.0

250

1*6,01

ck f v




507

md z u 56.0623.0*9.0*9.0

kN MN V Rd 891891.02

35.0*566.0*67.16*54.0max,

kN V kN V Rd Ed

891236max,


Given the vertical transversal reinforcement ( = 90°), the transverse reinforcement is calculated using the followingformula:

ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²67,8

15,1

500*566,0

45*211,0

*

*.

Calculation of theoretical reinforcement value:

The French national annex indicates the formula:

sin**min, ww sw b s

A

With:

4

min, 1015.7500

20*08,0*08,0 yk

ck

w f

f

ml cmb s

Aww

sw /²43.190sin*2.0*10*15.7sin** 4

min,



■ 3 nodes,


Advance Design gives the following results for At,min,z (cm2/ml)




508



Fz Fz corresponding to the 101 combination (ULS) [kNm] 263 kN

At,min,z Theoretical reinforcement area [cm2/ml] 1.43 cm

2/ml



Fz Fz -168.937 kN -0.0003 %

Atminz Atmin,z 1.43108 cm² 0.0002 %




509

5.31 EC2 Test29: Verifying the shear resistance for a T concrete beam with inclined transversalreinforcement - Inclined stress-strain diagram (Class XC1)

Test ID: 5092

Test status: Passed

5.31.1 Description

Verifies a T cross section beam made from concrete C35/40 to resist simple bending. For this test, the shear forcediagram and the moment diagram will be generated. The design value of the maximum shear force which can besustained by the member, limited by crushing of the compression struts, VRd,max., will be determined, along with thecross-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.

5.31.2 Background

Verifies a T cross section beam made from concrete C35/40 to resist simple bending. For this test, the shear forcediagram 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, V Rd,max, will be determined, along with thecross-sectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value.




■ Concrete: C35/40






The dead load will be represented by two linear loads of 22.63kN/m and 47.38kN/m


The live load will be considered from one linear load of 80.00kN/m



■ Exposure class: XC1

■ Characteristic compressive cylinder strength of concrete at 28 days:

■ Partial factor for concrete:




510

■ 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: 80.0


■ 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


■ Length: L = 10.00 m,


■ Strut slope: θ=29.74˚

Boundary conditions


■ Outer:



■ Inner: None.




511

Loading:


5.31.2.2 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:

5.31.2.3 Reference results in calculating the transversal reinforcement

ywd u

Ed sw

f z

tg V

s

A

*

*.

Where:

md z u 0125.1125.1*9.0*9.0

74.2975.1cot

90

ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²50.19

15,1

500*0125.1

74.29*502.1

*

*.




512

The minimum reinforcement percentage calculation:

ml cmb f

f b

s

Aw

yk

ck

ww sw /²21.590sin*55.0*

500

35*08,0sin**

*08,0sin**min,



■ 15 nodes,





My My corresponding to the 101 combination (ULS) [kNm] 5255.58 kNm

Fz Fz corresponding to the 101 combination (ULS) [kNm] 1501.59 kN

Az(5cm cover) Az longitudinal reinforcement corresp. to a 5cm cover [cm2/ml] 108.30 cm

2/ml

Az(2.7cm cover) Az longitudinal reinforcement corresp. to a 2.7cm cover [cm2/ml] 105.69 cm

2/ml

At,z,1 At,z transversal reinforcement for the beam end [cm2/ml] 19.10 cm

2/ml

At,z,2 At,z transversal reinforcement for the middle of the beam [cm2/ml] 5.21 cm

2/ml



My My -5255.58 kN*m -0.0000 %

Fz Fz -1501.59 kN -0.0003 %

Az Az 5cm -105.694 cm² 2.4071 %

Az Az 2.7cm -105.694 cm² -0.0000 %

Atz Atz end 19.0973 cm² 0.0000 %

Atz Atz middle 5.20615 cm² 0.0000 %




513

5.32 EC2 Test30: Verifying the shear resistance for a T concrete beam with inclined transversalreinforcement - Bilinear stress-strain diagram (Class XC1)

Test ID: 5098

Test status: Passed

5.32.1 Description

Verifies a T cross section beam made from concrete C30/37 to resist simple bending. For this test, the shear forcediagram and the moment diagram will be generated. The design value of the maximum shear force which can besustained by the member, limited by crushing of the compression struts, VRd,max, will be determined, along with thecross-sectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value.


5.32.2 Background

Verifies a T cross section beam made from concrete C30/37 to resist simple bending. For this test, the shear forcediagram 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, V Rd,max, will be determined, along with thecross-sectional area of the shear reinforcement, Asw. The test will not use the reduced shear force value.










The dead load will be represented by two linear loads of 23.50kN/m and 43.50kN/m






■ Characteristic compressive cylinder strength of concrete at 28 days:

■ Partial factor for concrete:




514

■ Relative humidity: RH=50%

■ Concrete age: t0=28days

■ Design value of concrete compressive strength: fcd=20 MPa

■ Secant modulus of elasticity of concrete: Ecm=33000 MPa

■ Concrete density:

■ Mean value of axial tensile strength of concrete: fctm=2.9 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:



■ The longitudinal reinforcement corresponding to a 6.4cm concrete cover

■ The transverse reinforcement corresponding to a 4.3cm concrete cover

Units

Metric System

Geometry


■ Length: L = 8.10 m,


■ Strut slope: θ=22.00˚

Boundary conditions


■ Outer:



■ Inner: None.




515

Loading:


5.32.2.2 Reference results in calculating the longitudinal reinforcement

For a 6.4 cm concrete cover and dinit=0.50m, the reference value will be 57.1 cm2:

For a 4.3 cm concrete cover and dinit=0.521, the reference value will be 54.3 cm2:

5.32.2.3 Reference results in calculating the transversal reinforcement

ywd u

Ed sw

f z

tg V

s

A

*

*.

Where:

md z u 469.0521.0*9.0*9.0

225.2cot

90

ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²67.11

15,1500*469.0

22*589.0

*

*.




516

The minimum reinforcement percentage calculation:

ml cmb f

f b

s

Aw

yk

ck

ww sw /²01.790sin*80.0*

500

30*08,0sin**

*08,0sin**min,



■ 9 nodes,





My My corresponding to the 101 combination (ULS) [kNm] 1193.28 kNm



2/ml


2/ml

At,z,1 At,z transversal reinforcement for the beam end [cm2/ml] 11.68 cm

2/ml

At,z,2 At,z transversal reinforcement for the middle of the beam [cm2/ml] 7.01 cm

2/ml



My My -1193.28 kN*m -0.0002 %

Fz Fz -589.275 kN -0.0000 %

Az Az 4.3cm cover -54.2801 cm² 0.0000 %

Atz Atz beam end 11.6782 cm² -0.0002 %

Atz Atz middle 7.01085 cm² -0.0000 %




517

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

Test ID: 5065

Test status: Passed

5.33.1 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 besustained by the member, limited by crushing of the compression struts (VRd,max) is calculated, along with thecross-sectional area of the shear reinforcement (Asw).

5.33.2 Background

Verifies a rectangular cross section beam made from concrete C20/25 to resist simple bending. For this test, theshear 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 ofthe shear reinforcement (Asw).



■ Concrete C20/25






The dead load will be represented by a punctual load of 105kN


The live load will be considered from one point load of 95kN







518




Units

Metric System

Geometry


■ Height: h = 0.50 m,

■ Width: b = 0.20 m,

■ Length: L = 3.00 m,





Boundary conditions


■ Outer:



■ Inner: None.

Loading:


l

a

P V u Ed 1


Pu = 1,35*Pg + 1,50*Pq=1.35*105kN=1.50*95kN=284.25kN

In this case (a=1m; l=3m):

KN V Ed 5,1893

11*25,284





519



Where:


2501*6,01

ck f v

In this case,

30 and 45


55.0250

201*6.0

2501*6,01

ck f

v

md z u 392.0435.0*9.0*9.0

3940

301

3420039203313540 .

cot

0cot5cot*.*.*.*.V

2max,Rd


Calculation of transverse reinforcement:

The transverse reinforcement is calculated using the following formula:

ml cm f z

V

s

A

ywd u

Ed sw /²76.545sin*)45cot30(cot*78.434*392.0

18950.0

sin*)cot(cot**

Beyond the point load, the shear force is constant and equal to Rb, therefore,

it also calculates the required reinforcement area to the right side of the beam:

ml cm s

A sw /²88.245sin*)45cot30(cot*78.434*392.0

09475.0



■ 3 nodes,






520



Fz,1 Fz corresponding to the 101 combination (ULS) x=1.0m [kN] 189.5 kN

Fz,2 Fz corresponding to the 101 combination (ULS) x=1.01m [kN] 94.75 kN

At,z,1Theoretical reinforcement area at x=1.0m [cm

2/ml]

5.76 cm2/ml At,z,2 Theoretical reinforcement area at x=1.01m [cm

2/ml] 2.88 cm2/ml



Fz Fz,1 -189.5 kN -0.0000 %

Fz Fz,2 94.75 kN 0.0000 %

Atz Atz,1 5.76277 cm² 0.0001 %

Atz Atz,2 2.88139 cm² -0.0001 %




521

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

Test ID: 5232

Test status: Passed

5.34.1 Description

Verifies a square cross section beam made of concrete C20/25 subjected to a normal force of traction - Inclinedstress-strain diagram (Class X0).

Tie sizing

Inclined stress-strain diagram

Determines the armature of a pulling reinforced concrete, subjected to a normal force of traction.

The load combinations will produce the following efforts:

NEd=1.35*233.3+1.5+56.67=400kN


- Support at start point (x=0) fixed connection

- Support at end point (x = 5.00) translation along the Z axis is blocked

5.34.2 Background

Tie sizing

Bilinear stress-strain diagram / Inclined stress-strain diagram

Verifies the armature of a pulling reinforced concrete, C20/25, subjected to a normal force of traction.


■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002;■ Analysis type: static linear (plane workspace);



■ Loadings from the structure: Fx,G = 233.33 kN

The dead load is neglected

■ Exploitation loadings (category A): Fx,Q = 56.67kN


Units

Metric System




522

Geometry


■ Height: h = 0.15 m,

■ Width: b = 0.15 m,

■ Length: L = 5.00 m,


■ Concrete cover: c=3cm

■ Reinforcement S400, Class: B, ss=400MPa

■ Fck=20MPa

■ The load combinations will produce the following efforts:

NEd=1.35*233.3+1.5+56.67=400kN

Boundary conditions


■ Outer:

► Support at start point (x=0) fixed connection,

► Support at end point (x = 5.00) translation along the Z axis is blocked

■ Inner: None.

5.34.2.2 Reference results in calculating the concrete beam

There will be two successive calculations, considering a bilinear stress-strain diagram constitutive law and then ainclined stress-strain diagram constitutive law.

Calculations according a bilinear stress-strain diagram

²50,11²10*50,11

15.1

400

400.0 4

,

, cmm N

AU s

Ed U s

It will be used a 4HA20=A=12.57cm2

Calculations according a inclined stress-strain diagram

MPaClassBS U s 373400 ,

²70,10²10*72,10373

400.0 4

,

, cmm N

AU s

Ed U s

It can be seen that the gain is not negligible (about 7%).

Checking the condition of non-fragility:

yk

ctmc s

f

f A A *

MPa f f ck ctm 21,220*30.0*30.0 3/23/2

²0225.015.0*15.0 m Ac

²24.1400

21.2*0225.0* cm

f

f A A

yk

ctmc s




523



■ 6 nodes,


ULS load combinations (kNm)

In case of using the bilinear stress-strain diagram, the reinforcement will result: (Az=11.50cm2=2*5.75

cm

2)

In case of using the inclined stress-strain diagram, the reinforcement will result: (Az=10.70cm2=2*5.35 cm2)



Az,1 Longitudinal reinforcement obtained using the bilinear stress-straindiagram [cm

2]

5.75 cm2

Az,2 Longitudinal reinforcement obtained using the inclined stress-straindiagram [cm

2]

5.37 cm2


Result name Result description Value Error Az Az-ii -5.36526 cm² 0.0001 %




524

5.35 EC2 Test 1: Verifying a rectangular cross section beam made from concrete C25/30 to resistsimple bending - Bilinear stress-strain diagram

Test ID: 4969

Test status: Passed

5.35.1 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 longitudinalreinforcement and the verification of the minimum reinforcement percentage.

5.35.2 Background


Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the

verification of the minimum reinforcement percentage.








■





Units

Metric System

Geometry


■ Height: h = 0.60 m,

■ Width: b = 0.25 m,

■ Length: L = 5.80 m,







525





■ Stress-strain law for reinforcement: Bilinear stress-strain diagram■ Cracking calculation required


f f

c

ckcd 6716

51

25

γ


■ MPa.*.f *.f //ckctm 56225300300 3232



f f

s

ykyd 78434

151

500

γ


■ MPa*f

*E

..ck

cm 3147610

82522000

10

822000

3030


Boundary conditions


■ Outer:



■ Inner: None.

Loading



G’=0.25*0.6*2.5=3.75kN/ml



Cmax = 1.35 x G + 1.5 x Q=1.35*(15+3.75)+1.5*20=55.31kN/ml


CCQ = 1.0 x G + 1.0 x Q=15+3.75+20=38.75kN/ml


CQP = 1.0 x G + 0.3 x Q=15+3.75+0.3*20=24.75kN/ml


mkN M Ed .59,2328

²80,5*31,55

mkN M Ecq .94,1628

²80,5*75,38




526


Due to exposure class XC1, and 500Mpa steel resistance, we will consider a moment limit:

Calculating the reduced moment we will consider ULS moment:

■

■ because there is no compressed reinforcement



■ Effective height: d=h-(0.6*h+ebz)=0.524m


203,067,16*²524,0*25,0

233,0

*²*

cd w

Ed cu

f d b

M

285.0203.0 lucu


md z uc 464,0)287,0*4,01(*524,0)4,01(*


²55,11²10*55,1178,434*464,0

233,0

*

4cmm

f z

M A

yd c

Ed u



d*b*.

d*b*f

f *.

Max A

w

wyk

eff ,ct

min,s

00130

260


Where:


Therefore:

²cm.

²m...*.*.

²m...*.*.

.Max A min,s 761

10701524025000130

107615240250500

562260

4

4



■ 7 nodes,





527












528





Az Theoretical reinforcement area [cm2] 11.55 cm2


2



My My USL -232.578 kN*m -0.0001 %

My My SLS cq -162.936 kN*m 0.0001 %

Az Az -11.5329 cm² -0.0003 %

Amin Amin -1.74725 cm² -0.0002 %




529

5.36 EC2 Test33: Verifying a square concrete column subjected to compression by nominal rigiditymethod- Bilinear stress-strain diagram (Class XC1)

Test ID: 5109

Test status: Passed

5.36.1 Description

Verifies a square concrete column subjected to compression by nominal rigidity method- Bilinear stress-straindiagram (Class XC1).

The column is made of concrete C30/37. The verification of the axial force, applied on top, at ultimate limit state isperformed.

Nominal rigidity method.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and thencalculate 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).

5.36.2 Background


Verify the adequacy of a square cross section made from concrete C30/37.








► 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




530

Units

Metric System

Geometry


■ Height: h = 0.30 m,

■ Width: b = 0.30 m,

■ Length: L = 4.74 m,


Boundary conditions


The column is considered connected to the ground by a fixed connection (all the translations and rotations areblocked) and to the top part, the translations along X and Y axis are also blocked.

Loading




NEd =1260kN

■ Proposed reinforcement:

2*6.28cm2

=12.56cm2




531

5.36.2.2 Reference results in calculating the concrete column

Load calculation:

NEd= 1.35*NG = 1700 KN

Initial eccentricity: cm M

M e

u

u 07.1

0

0

Additional eccentricity due to geometric imperfections:

cmcmcmcm L

cmei 2805.0;2max400

322;2max

400;2max 0

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.2(7)

The first order eccentricity : meee i 02.001

The necessity of buckling calculation (second order effect):

For an isolated column, the slenderness limit check is done using the next formula:

n

C B A ***20lim

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

Where:

944.020*²30.0

7.1

*

cd c

Ed

f A

N n

7.0 A if ef is not known, if it is,

ef

A *2,01

1

1.1 B if ω (reinforcement ratio) is not known, if it is, cd c

yd s

f A

f A B

*

**21*21

In this case 27.120*²3.0

78.434*10*56.12*21

4

B

70.0C if r m is not known, if it is, mr C 7,1

In this case:

81.12944.0

7.0*27.1*7.0*20***20lim

n

C B A

34.383.0

12*32.312*0 h

L

81.1234.38 lim

Therefore, the second order effects most be considered




532

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

Ed

EQP

ef M

M t ., 0

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

0,t creep coefficient

EQP M serviceability firs order moment under quasi-permanent load combination

Ed M ULS first order moment (including the geometric imperfections)

The ratio of the moment is in this case:

74.0

7.1

26.1

*

*

1

1

0

0 ed

eqp

ed

eqp

Ed

Eqp

N

N

e N

e N

M

M

),( 0t is the final value of the creep:

)(*)(*),( 00 t f t cm RH

With:

t0 is the concrete age

213

0

***1.0

1001

1

h

RH

RH

121 if

MPa f cm 35 if not

7.0

1

35

cm f

and

2.0

2

35

cm f

RH relative humidity: RH=50%

h0= mean radius of the element in mm

mm

u

Ach 150

300300*2

300*300*2*20

u=column section perimeter

MPa MPa f cm 3538944.0

38

35 7.0

1

and

984.038

35 2.0

2




533

86.1984.0*944.0*150*1.0

100

501

1***1.0

1001

1321

30

h

RH

RH

48.2488.0*73.2*86.1)(*)(*),( 00 t f t cm RH

835.174.0*48.2*),(0

0

0 Ed

Eqp

ef M

M t

835.2835.111 ef

Calculation of nominal rigidity:

The nominal rigidity of a post or frame member, it is estimated from the following formula:

s s sccd c I E K I E K EI ****

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.7.2 (1)

e

c

k k K

1

* 21

835.21 ef

22.120

30

201 ck f

k

170

*2

nk

944.020*²3.0

7.1

*

cd c

Ed

f A

N n

34.38

20.0213.0170

34.38*944.0

170

*22 k

nk

086.0835.2

20.022.1

1

21

e

c

k k K

CE

cmcd

E E

MPa f

E ck cm 32837

10

825*22

10

8*22

3.03.0

MPa E

E CE

cmcd 27364

2.1

32837

433

000675.012

3.0*3.0

12

*m

hb I c

Ks=1




534

Es = 200 Gpa

45

242

10*256.12

05.025.0*

2

10*56.12*2

2

'*

2*2 m

d d A I theo s

²1.410*256.1*200000*1000675.0*27364*086.0 5 MNm EI

MN L

EI N

f

B 67.3²32.3

10.4*²

²

*²

234.18

²²

0

c

MNm

N

N M M

Ed

B Ed Ed 070.0

17.1

67.3

234.11*034.0

1

1*0

The calculation made with flexural:

MN N Ed 7.1and

MNm M Ed 070.0,

Therefore, a 2*6.64cm2 reinforcement area is obtained.

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)



■ 7 nodes,





535



Theoretical value (cm2 )

(reference value: 13.08 cm2)



Az Reinforcement area [cm2] 6.54 cm

2



Az Az -6.53808 cm² 0.0001 %




536

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

Test ID: 5114

Test status: Passed

5.37.1 Description

Verifies the adequacy of a rectangular cross section column made of concrete C30/37. The verification of the axialstresses applied on top, at ultimate limit state is performed.

Method based on nominal stiffness - The purpose of this test is to determine the second order effects by applying themethod of nominal stiffness, 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.

5.37.2 Background

Method based on nominal stiffness

Verifies the adequacy of a rectangular cross section made of concrete C30/37.

The purpose of this test is to determine the second order effects by applying the method of nominal stiffness, andthen calculate the frames by considering a section symmetrically reinforced.











■ 3,02




■ Transversal reinforcement spacing a=30cm

■ Concrete C30/37


■ The column is considered isolated and braced




537

Units

Metric System

Geometry


■ 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 column is considered connected to the ground by a fixed connection and free to the top part.

Loading




NEd =1.35*0.30+1.5*0.50=1.155MN

NQP=1.35*0.30+0.30*0.50=0.450MN

■




538


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 isconsidered to be:

ml l 84*2*20

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:

28.6940.0

8*32*32 0 a

l



Ed

EQP

ef

M

M t ., 0


Where:


EQP M serviceability first order moment under quasi-permanent load combination


The ratio between moments becomes:

39.0155.1450.0

**

1

1

0

0 ed

eqp

ed

eqp

Ed

Eqp

N N

e N e N

M M

The creep coefficient 0,t is defined as:

)(*)(*),( 00 t f t cm RH

MPa f

f cm

cm 73.2830

8.168.16)(

488.0

281.0

1

1.0

1)(

20.020.0

0

0

t

t (for t0= 28 days concrete age).

213

0

***1.0

1001

1

h

RH

RH

RH = relative humidity; RH=50%

Where 121 if MPa f cm 35 if not

7.0

1

35

cm f and

2.0

2

35

cm f




539

mm

u

Ach 240

600400*2

600*400*2*20

MPa MPa f cm 3538 ,

Therefore:

944.038

3535 7.07.0

1

cm f and 984.0

38

3535 2.02.0

2

cm f

73.1984.0*944.0*240*1.0

100

501

1***1.0

1001

1321

30

h

RH

RH

30.2488.0*73.2*73.1)(*)(*),( 00 t f t cm RH

The effective creep coefficient calculation:

90.039.0*30.2*, 0 Ed

EQP

ef M

M t




n

C B A ***20lim


Where:

241.020*60.0*40.0

155.1

*

cd c

Ed

f A

N n

85.0

90.0*2.01

1

*2,01

1

ef

A

1.1*21 B because the reinforcement ratio in not yet known

70.07,1 mr C because the ratio of the first order moment is not known

66.26112.0

7.0*1.1*85.0*20lim

66.2628.69 lim

Therefore, the second order effects must be considered.




540

5.37.2.3 The eccentricity calculation and the corrected loads on ULS:

Initial eccentricity:

No initial eccentricity because the post is subjected only in simple compression.

Additional eccentricity:

ml

ei 02.0400

8

400

0

First order eccentricity- stresses correction:

mmmm

mmmmmm

hmm

e 203.13

20max

30

40020

max

30

20max0

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4)

The corrected solicitations which are taken into account when calculating the column under combined bending effortand compression are:

NEd= 1.155MN

MEd= 1.155*0.02=0.0231MNm

Reinforcement calculation in the first order situation:

When using the nominal stiffness method, a starting section frame is needed. For this it will be used a concretesection 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 abovewere obtained in the center of gravity for the concrete section alone. The stresses must be reduced in the centroid ofthe tensioned steel.

First order moment in the centroid of the tensioned reinforcement is:

MNmh

d N M M Gua 196.02

40.035.0*155.10231.0)

2(*0

Verification if the section is partially compressed:

496,0)35,0

40,0*4,01(*

35,0

40,0*8,0)*4,01(**8,0

d

h

d

h BC

133.020*²35.0*60.0

196.0

*²*

cd w

uacu

f d b

M

BC cu 496.0133.0 therefore the section is partially compressed.




541

Calculations of steel reinforcement in pure bending:

133.0cu

179,0)133,0*21(1*25,1 u

md z uc 325,0)179,0*4,01(*35,0)*4,01(*

²87.1378,434*325,0

196,0

*cm

f z

M A

yd c

ua

Calculations of steel reinforcement in combined bending:

For the combined bending:

24 69.1278.434

155.110*87.13' cm

f

N A A

yd

The minimum reinforcement percentage:

2

min, 80.4*002.0²66.278.434

155.1*10.0*10,0cm Acm

f

N A c

yd

Ed s

The reinforcement will be 4HA12 + 2HA10 representing 6.09cm2.

The second order effects calculation:

The second order effect will be determined by applying the method of nominal stiffness:

Calculation of nominal stiffness:

It is estimated nominal stiffness of a post or frame member from the following formula:


With:

2.1

cmcd

E E

MPa MPa f f ck cm 388

MPa f

E cmcm 32837

10

38*22000

10*22000

3.03.0

MPa E E cmcd 27364

2.132837

2.1

410.2,312

40.0*60.0

12

* 333

mhb

I c

(concrete only inertia)

MPa E s 200000

s I : Inertia of the steel rebars

00254.0

40.0*60.0

10.09,6 4

c

s

A

A




542

01.0002.0 c

s

A

A

Mpa f

k ck 22.120

30

201

241.020*60.0*40.0

155.1

.

cd c

Ed

f A

N n

20.0098.0170

28.69*241.0

170*2

nk

410.37,105.02

40.0*

2

10.09,6*2

2*

2*2 5

242

mch A

I s s

1 s K and 063.0

90.01

098.0*22.1

1

* 21

ef

c

k k K

Therefore:

²2566.810*37,1*200000*110*2,3*27364*063.0 53

MNm EI

Stresses correction:

The total moment, including second order effects, is defined as a value plus the time of the first order:

1

1*0

Ed

B Ed Ed

N

N M M

MNm M Ed 0231.00 (moment of first order (ULS) taking into account geometric imperfections)

(normal force acting at ULS).

0

²

c

and 80 c the moment is constant (no horizontal force at the top of post).

234.18

²

MN l

EI N B 27327.1

²8

2566.8*²*²

2

0

The second order efforts are:

MN N Ed 155.12

m MN M Ed .3015.02

The reinforcement calculations considering the second order effect:

The reinforcement calculations for the combined flexural, under the second order effect, are done using the GraitecEC2 tools. The obtained section is null, therefore the minimum reinforcement area defined above is sufficient for the

defined efforts.




543



■ 5 nodes,


Theoretical reinforcement area (cm2 ) and minimum reinforcement area (cm

2 )

(reference value: 2.66cm2and 4.8cm

2)





Ay Reinforcement area [cm2] 2.66 cm

2


2



Ay Ay -2.66 cm² -0.0000 %

Amin Amin 4.8 cm² 0.0000 %




544

5.38 EC2 Test32: Verifying a square concrete column subjected to compression and rotationmoment to the top – Method based on nominal curvature- Bilinear stress-strain diagram (ClassXC1)


5.38.1 Description

Verifies a square concrete column subjected to compression and rotation moment to the top – Method based onnominal 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, atultimate limit state is performed.

The purpose of this test is to determine the second order effects by applying the method of nominal curvature, andthen calculate the frames by considering a section symmetrically reinforced.

5.38.2 Background

Method based on nominal curvature

Verifies the adequacy of a square cross section made from concrete C25/30.







► 15t axial force

► 1.5tm rotation moment applied to the column top► The self-weight is neglected


► 6.5t axial force

► 0.7tm rotation moment applied to the column top

► 3,02

► 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




545

Units

Metric System

Geometry


■ Height: h = 0.40 m,

■ Width: b = 0.40 m,

■ Length: L = 6.00 m,

■ Concrete cover: c = 5 cm

Boundary conditions


The column is considered connected to the ground by a fixed connection and free at the top.

Loading




NEd =1.35*15+1.5*6.5=30t=0.300MN

MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm

■ m..

.

N

M

eEd

Ed

1003000

03100




546




ml l 12*20



10440.0

12*32*32 0 a

l



Ed

EQP

ef M

M

t ., 0


Where:




First order eccentricity evaluation:

ieee 01

mei 03.0

The first order moment provided by the quasi-permanent loads:

me N

M eee i

Eqp

Eqp

i 13.030.065.0*30.015

70.0*30.050.1

0

0

01

t N Eqp 95.1670.0*30.050.11

MNmtme N M Eqp Eqp 022.020.213.0*95.16* 111

The first order ULS moment is defined latter in this example:

The creep coefficient 0,t is defined as follows:

)(*)(*),( 00 t f t cm RH

92.2825

8.168.16)(

cm

cm f

f

488.0281.0

11.0

1)(20.020.0

0

0

t





547

30*1.0

1001

1h

RH

RH

85.1

200*1.0

100

50

11200400400*2400*400*2*2

30

RH mmu Ach

64.2488.0*92.2*85.1)(*)(*),( 00 t f t cm RH


49.1039.0

022.0*64.2*, 0

Ed

EQP

ef M

M t



For an isolated column, the slenderness limit verification is done using the next formula:

n

C B A ***20lim


Where:

112.067.16*²40.0

300.0

*

cd c

Ed

f A

N n

77.049.1*2.01

1

*2,01

1

ef A


70.07,1 mr C because the ratio of the first order moment is unknown

43.35112.0

7.0*1.1*77.0*20lim

43.35104 lim

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




548


m N

M e

Ed

Ed 10.0300.0

031.00


ml ei 03.0400

12

400

0

The first order eccentricity: stresses correction:

The forces correction, used for the combined flexural calculations:

MN N Ed 300.0

meee i 13.001

MNm N e M Ed Ed 039.0300.0*)03.010.0(*1

0*e N M Ed

mmmm

mmmmmm

hmm

e 203.13

20max

30

40020

max

30

20max0



To play the nominal rigidity method a starting section frame is needed. For this it will be used a concrete sectionconsidering only the first order effects.



MNmh

d N M M Gua 084.02

40.035.0*300.0039.0)

2(*0


496,0)35,0

40,0*4,01(*

35,0

40,0*8,0)*4,01(**8,0

d

h

d

h BC

103.067.16*²35.0*40.0

084.0

*²*

cd w

uacu

f d b

M





549




24 06.178.434

300.010*84.5' cm

f

N A A

yd


2

min, 02.3*002.0²69.078.434

300.0*10.0*10,0cm Acm

f

N A c

yd

Ed s

The reinforcement will be 4HA10 representing a 3.14cm2 section


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 followingformula:

0

1**

1

r K K

r r


With:

1

0

0138.035.0*45.0

200000

78.434

*45,0*45,0

1 md

E

f

d r s

yd

yd

r K : is a correction factor depending on axial load,=> 1

bal u

u

r nn

nn

K

112.067.16*²40.0

300.0

*

cd c

Ed

f A

N n

0512.067.16*²40.0

78.434*10.14,3

*

* 4

cd c

yd s

f A

f A

0512.10512.011 un

4,0bal n




550

1144.140.00512.1

112.00512.1

r r K K

K : is a factor for taking account of creep=> 1*1 ef K

218.0150

104

200

2535.015020035,0

ck

f

11675.049.1*218.01*1 K K ef

Therefore the curvature becomes:

1

0

0138.01

**1 m

r K K

r r

Calculation moment:

20 M M M Ed Ed

Where:

Ed M 0 : first order moment including the geometrical imperfections.

2 M : second order moment

The second order moment must be calculated from the curvature:

22 *e N M Ed

mc

l

r e 199.0

10

²12*0138.0*

1 2

02

MNme N M Ed 0597.0199.0*300.0* 22

MNm M M M Ed Ed 099.00597.0039.020

The frame must be sized corresponding the demands of the second order effects:

MN N Ed 300.0

MNm M Ed 099.0

Reinforcement calculation corresponding to the second order:

The input parameters in the diagram are:

093.067.16*²40.0*40.0

099.0** 2

cd

Ed

f hb M

112.067.16*40.0*40.0

300.0

**

cd

Ed

f hb

N v




551

150ω . ; => ²cm..

.*².*.

f

f *h*b* A

yd

cds 209

78434

6716400150ω; which means 4.60cm2 per side.

Buckling checking

The verification will be made considering the reinforcement found previously (9.20cm2)

The curvature evaluation:

The frames have an influence on the r K parameter only:

1

0

0138.01 m

r

1

bal u

ur

nn

nn K

112.0n

15.067.16*²40.0

78.434*10*20,9

*

* 4

cd c

yd s

f A

f A

15.115.011 un

4,0bal n

11384.140.015.1

112.015.1

r r K K

1 K : coefficient that takes account of the creep => 1*1 ef K

The curvature becomes:

1

0

0138.01

**1 m

r K K

r r

It was obtained the same curvature and therefore the same second order moment, which validates the sectionreinforcement found.




552



■ 7 nodes,


Theoretical reinforcement area (cm2

)

(reference value: 9.20cm2 = 2 x 4.64 cm

2)


(reference value: 9.28 cm

2

)




2



Az Az -4.64 cm² -0.0000 %




553


Test ID: 5058

Test status: Passed

5.39.1 Description


5.39.2 Background




■ Concrete C20/25






The dead load will be represented by a punctual load of 105kN

■ Exploitation loadings:The live load will be considered from one point load of 95kN










554

Units

Metric System

Geometry



■ Length: L = 3.00 m,





Boundary conditions


■ 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:


l

a P V u Ed 1


Pu = 1,35*Pg + 1,50*Pq=1.35*105kN=1.50*95kN=284.25kN

In this case, (a = 1 m; l = 3 m):

KN V Ed 5,1893

11*25,284





555


θ1


cot

cotcot*b*z*f **V


Where:

1α cw coefficient taking account of the state of the stress in the compression chord and

2501601

ckf *,v


θθ

1

cottg

b*z*f *vV wucd

max,Rd

In this case,

45 and 90


55.0250

201*6.0

2501*6,01

ck f

v

md z u 392.0435.0*9.0*9.0

MN V Rd 288.02

20.0*392.0*33.13*55.0max,


Calculation of transversal reinforcement:■ is determined by considering the transverse reinforcement steel vertical (=90°) and connecting rods inclined

at 45 °, at different points of the beam

■ before the first point load: ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²13.11

15,1

500*392,0

45*18950,0

*

*.

■ beyond the point load, the shear force is constant and equal to Rb, therefore:

KN l

a P V uu 75,94

3

1*25,284*

■ it also calculates the required reinforcement area to the right side of the beam:



■ 3 nodes,





556

Advance Design gives the following results for Atz (cm2 /ml)



Fz,1 Fz corresponding to the 101 combination (ULS) x=1m [kNm] 189.5 kN

Fz,2 Fz corresponding to the 101 combination (ULS) x=1.01m [kNm] 94.75 kN At,z,1 Theoretical reinforcement area x=1m [cm

2/ml] 11.13 cm

2/ml

At,z,2 Theoretical reinforcement area x=1.01m [cm2/ml] 5.57 cm

2/ml



Fz Fz,1 -189.5 kN -0.0000 %

Fz Fz,2 94.75 kN 0.0000 % Atz Atz,1 11.1328 cm² 0.0002 %

Atz Atz,2 5.56641 cm² 0.0000 %




557

5.40 EC2 Test28: Verifying the shear resistance for a T concrete beam with inclined transversalreinforcement - Bilinear stress-strain diagram (Class X0)

Test ID: 5083

Test status: Passed

5.40.1 Description

Verifies a T cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear forcediagram 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 theshear reinforcement, Asw, calculation.

The test will not use the reduced shear force value.

5.40.2 Background

Verifies a T cross section beam made from concrete C25/30 to resist simple bending. For this test, the shear forcediagram 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 theshear reinforcement, Asw, calculation. The test will not use the reduced shear force value.









The dead load will be represented by a linear load of 13.83kN/m





■ Exposure class: X0






558

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


■ Length: L = 10.00 m,





Boundary conditions


■ Outer:



■ Inner: None.




559

Loading:

For a beam under a uniformly distributed load Pu, the shear force is defined by the following equation:

2

**)( l P x P xV u

u

For the beam end, (x=0) the shear force will be:

kN l P

V u Ed 9,292

2

10*57,58

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.


cot

***1max,

tg

b z f vV wucd Rd

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 6.2.3.(3)In this case:

30 and 90


54.0250

251*6.0

2501*6,01

ck f

v

md z u 688.0764.0*9.0*9.0

MN tg V Rd 590.030cot30

22.0*688.0*67.16*54.0max,



The transverse reinforcement is calculated using the following formula:

ml cmtg

f z

tg V

s

A

ywd u

Ed sw /²66.5

15,1

500*688.0

30*293.0

*

*.




560



■ 11 nodes,


Advance Design gives the following results for Atz (cm2

/ml)




At,z Theoretical reinforcement area [cm2/ml] 5.66 cm

2/ml



Fz Fz -292.852 kN -0.0002 %

Atz Atz 5.65562 cm² 0.0000 %




561

5.41 EC2 Test31: Verifying a square concrete column subjected to compression and rotationmoment to the top - Bilinear stress-strain diagram (Class XC1)

Test ID: 5101

Test status: Passed

5.41.1 Description


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 thencalculate 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.


5.41.2 Background


Verifies the adequacy of a rectangular cross section made from concrete C25/30.








► 15t axial force




► 6.5t axial force


■ 3,02




■ Transversal reinforcement spacing a=30cm■ Concrete C25/30






562

Units

Metric System

Geometry


■ Height: h = 0.40 m,

■ Width: b = 0.40 m,

■ Length: L = 6.00 m,


Boundary conditions



Loading




NEd =1.35*15+150*6.5=30t=0.300MN

MEd=1.35*1.50+1.5*0.7=3.075t=0.31MNm

■ m..

.

N

Me

Ed

Ed 1003000

03100




563




ml l 12*20



10440.0

12*32*32 0 a

l



Ed

EQP

ef

M

M t ., 0


Where:


serviceability first order moment under quasi-permanent load combination

ULS first order moment (including the geometric imperfections)


ieee 01

mei 03.0


me N

M eee i

Eqp

Eqp

i 13.030.065.0*30.015

70.0*30.050.1

0

0

01

t N Eqp 95.1670.0*30.050.11

MNmtme N M Eqp Eqp 022.020.213.0*95.16* 111


)(*)(*),( 00 t f t cm RH

92.2825

8.168.16)(

cm

cm f

f

488.0281.0

1

1.0

1)(

20.020.0

0

0

t

t

(for t0= 28 days concrete age).




564

30*1.0

1001

1h

RH

RH

85.1

200*1.0

100

50

11200400400*2400*400*2*2

30

RH mmu Ach

64.2488.0*92.2*85.1)(*)(*),( 00 t f t cm RH


49.1039.0

022.0*64.2*, 0

Ed

EQP

ef M

M t




n

C B A ***20lim


Where:

112.067.16*²40.0

300.0

*

cd c

Ed

f A

N n

77.0

49.1*2.01

1

*2,01

1

ef

A



43.35112.0

7.0*1.1*77.0*20lim

43.35104 lim

Therefore, the second order effects most be considered.



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:






565


m N

M e

Ed

Ed 10.0300.0

031.00


ml ei 03.0400

12

400

0



MN N Ed 300.0

meee i 13.001

MNm N e M Ed Ed 039.0300.0*)03.010.0(*1

0*e N M Ed






MNmh

d N M M Gua 084.02

40.035.0*300.0039.0)

2(*0


496,0)35,0

40,0*4,01(*

35,0

40,0*8,0)*4,01(**8,0

d

h

d

h BC

103.067.16*²35.0*40.0

084.0

*²*

cd w

uacu

f d b

M





566


103.0cu

136,0)103,0*21(1*25,1 u

md z uc 331,0)136,0*4,01(*35,0)*4,01(*



24 06.178.434

300.010*84.5' cm

f

N A A

yd


2

min, 02.3*002.0²69.078.434

300.0*10.0*10,0cm Acm

f

N A c

yd

Ed s



The second order effect will be determined by applying the method of nominal rigidity:


The nominal rigidity of a post or frame member, it is estimated from the following formula:



With:

2.1

cmcd

E E


Mpa f

E cmcm 31476

10

33*22000

10*22000

3.03.0

Mpa E

E cmcd 26230

2.1

31476

2.1

410.133,212

40.0

12

* 343

mhb

I c

(cross section inertia)

Mpa E s 200000

s I : Inertia

002.040.040.0

10.14,3 4

c

s

A

A




567

01.0002.0 c

s

A

A

Mpa f

k ck 12.120

25

201

112.067.16*²40.0

300.0

*

cd c

Ed

f A

N n

20.0069.0170

104*112.0

170*2

nk

410*06,705.02

40.0*

2

10*14,3*2

2*

2*2 6

242

mch A

I s s

1 s K and

031.0

49.11

069.0*12.1

1

* 21

ef

c

k k K

Therefore:

²15.310*06,7*200000*110*133,2*26230*031.0 63 MNm EI

Corrected stresses:

The total moment, including the second order effects is defined as a value and is added to the first order momentvalue:

11*0

Ed

B Ed Ed

N N M M


MNm M Ed 039.00 (time of first order (ULS) taking into account the geometric imperfections, relative to the

center of gravity of concrete).

MN N Ed 300.0 (normal force acting at ULS).

And:

0

²

c

with 8

0

c because the moment is constant (no horizontal force at the top of post).


234.18

²

MN l

EI N B 216.0

²12

15.3*²*²

2

0




568

Therefore the second order moment is:

MNm M Ed 133.0

1300.0

216.0

234.11*039.0

There is a second order moment that is negative because it was critical that the normal force, N B, is less than theapplied normal force => instability.

A section corresponding to a ratio of 5 ‰ will be considered and the corresponding equivalent stiffness isrecalculated.


With:

Mpa E cd 26230 ;

410.133,2 3m I c

Mpa E s 200000

s I : Inertia

²8²40,0*005,0*005,0005,0 cm A A c s

It sets up : 4HA16 =>²04,8005,0 cm A s

01.0002.0 c

s

A

A

45

242

10*81,105.0

2

40.0*

2

10*04,8*2

2

*

2

*2 mch A

I s s

1 s K and

031.049.11

069.0*12.1

1

* 21

ef

c

k k K

Therefore:

²35.510.81,1*200000*110.133,2*26230*031.0 53

MNm EI

The second order effects must be recalculated:

11*0

Ed

B Ed Ed

N

N M M

234.1

MN l

EI N B 367.0

²12

35.5*²*²

2

0

MNm M Ed 254.0

1300.0

367.0

234.11*039.0




569

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:


238.067.16*²40.0*40.0

254.0

** 2

cd

Ed

f hb

M

112.067.16*40.0*40.0

300.0

**

cd

Ed

f hb

N v

485.0 is obtained, which gives: ²75.2978.434

67.16*²40.0*485.0***cm

f

f hb A

yd

cd s

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 theprevious calculations: 6HA32 provides As=48.25cm

2

The normal rigidity evaluation:

It is estimated nominal rigidity of a post or frame member from the following formula:

s s sccd c I E K I E K EI ....

With:

s I : Inertia

03.040.0*40.0

10*25,48 4

c

s

A A




570

44

242

10.086,105.02

40.0*

2

10*25,48*2

2*

2*2 mc

h A I s s

12.11 k

069.02 k

1 s K and

031.049.11

069.0*12.1

1

* 21

ef

c

k k K

The following conditions are not implemented in Advance Design:

If:

01.0c

s

A

A

ef

c K *5,01

3,0

Then:

²45.2310*086,1*20000010*133,2*26230*031.0 43

MNm EI

Corrected stresses

The total moment, including second order effects, is defined as a value plus the moment of the first order:

1

1*0

Ed

B Ed Ed

N

N M M

MNm M Ed 039.00

MN N Ed 300.0 (normal force acting at ULS)

234.1

MN l

EI N B 607.1

²12

45.23*²*²

2

0

It was therefore a moment of second order which is:

MNm M Ed 05.0

1300.0

607.1

234.11*039.0

Reinforcement calculation:

047.067.16*²40.0*40.0

05.0

**05.0

2

cd

Ed Ed

f hb

M MNm M

112.0

67.16*40.0*40.0

300.0

**

300.0 cd

Ed Ed

f hb

N v MN N




571

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%; A s=0.009*0.40=14.4cm2, which is 7.2cm

2 per

side. A 3HA16 reinforcement will be chosen by either side (6HA16 for the entire column), which will give As=12.03cm

2.

Nominal rigidity evaluation:

s s sccd c I E K I E K EI .... .

s I : Inertia

00754.040.0*40.0

10*06,12 4

c

s

A

A

if

01.0002.0 c

s

A

A

45

242

10.71,205.0240.0*

210*06,12*2

2*

2*2 mch A I s

s

1 s K and

031.049.11

069.0*12.1

1

* 21

ef

c

k k K

Therefore:

²15.710*71,2*200000*110*133,2*26230*031.0 53

MNm EI




572

Second order loads calculation:

1

1*0

Ed

B Ed Ed

N

N M M

MNm M Ed 039.00


234.1

MN l

EI N B 49.0

²12

15.7*²*²

2

0

MNm M Ed 115.0

1300.0

49.0234.11*039.0


Frames are calculated again from the interaction diagram:

108.067.16*²40.0*40.0

115.0

** 2

cd

Ed

f hb

M

112.0

67.16*40.0*40.0

300.0

**

cd

Ed

f hb

N v

18.0 is obtained, which gives: ²04.1178.434

67.16*²40.0*18.0***cm

f

f hb A

yd

cd s

Therefore set up a section 5.52cm ² per side must be set; this will be the final column reinforcement



■ 7 nodes,





573



Theoretical value (cm2

)

(reference value: 11.16 cm2 = 2 x 5.58 cm

2)




574




2



Az Az -5.58 cm² -0.0000 %




575

5.42 EC2 Test 37: Verifying a square concrete column using the simplified method – Professionalrules - Bilinear stress-strain diagram (Class XC1)

Test ID: 5126

Test status: Passed

5.42.1 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 areblocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is alsoblocked.

5.42.2 Background

Simplified MethodVerifies the adequacy of a square cross section made from concrete C25/30.









■ Exploitation loadings:► 800kN axial force


■ Concrete C25/30


■ Relative humidity RH=50%

■ Buckling length L0=0.70*4=2.80m




576

Units

Metric System

Geometry


■ Height: h = 0.40 m,

■ Width: b = 0.400 m,

■ Length: L = 4.00 m,


Boundary conditions


The column is considered connected to the ground by a fixed connection (all the translations and rotations areblocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is alsoblocked.

Loading



kN N ED 2280800*5.1800*35.1




577


Scope of the method:

25.244.0

12*8.212*

121212

00

2

0

2

4

000 a

L

a

L

a

L

a

a

L

A

I

L

i

L

According to Eurocode 2 EN 1992-1-1 (2004) Chapter 5.8.3.2(1)

The method of professional rules can be applied as:

120

MPa f ck 5020

mh 15.0


6025.24 , therefore:

746.0

62

25.241

86.0

621

86.022

According to “Conception Et Calcul Des Structures De Batiment”, by Henry Thonier, page 283

Not knowing the values for and , we can considered 93.0hk

1500

500*6.06.1

500*6.06.1 yk

s

f k

²24,1467.16*4.0*4.0746.0*1*93.0

280.2*

78.434

1**

***

1cm f hb

k k

N

f A cd

sh

ed

yd

s




■ 5 nodes,





578

Theoretical reinforcement area(cm2)

(reference value: 14.24cm2=4*3.56cm

2)



Ay Reinforcement area [cm2] 3.57cm

2



Ay Az -3.56562 cm² 0.0001 %




579

5.43 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

5.43.1 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). Atthe top part, the translations along X and Y axis are blocked and the rotation along the Z axis is also blocked.

5.43.2 Background

Simplified Method











■ Concrete C25/30

■ Steel reinforcement S500B■ Relative humidity RH=50%

■ Buckling length L0=6.50m

Units

Metric System




580

Geometry


■ Height: h = 0.30 m,

■ Width: b = 0.50 m,

■ Length: L = 6.50 m,


Boundary conditions


The column is considered connected to the ground by a articulated connection (all the translations are blocked) andto the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked.

Loading



kN N ED 1080800*35.1



06.753.0

12*5.612*

121212

00

2

0

2

4

000 a

L

a

L

a

L

a

a

L

A

I

L

i

L

According to Eurocode 2 EN 1992-1-1 (2004) Chapter 5.8.3.2(1)


120

MPa f ck 5020

mh 15.0


12006.7560

33.0

06.75

3232 3.13.1


93.0)**61(*)*5.075.0( hk h

1500

500*6.06.1

500*6.06.1 yk

s

f k

²43.2367.16*5.0*3.033.0*1*93.0

08.1*

78.434

1**

***

1cm f hb

k k

N

f A cd

sh

ed

yd

s




581



■ 8 nodes,


Theoretical reinforcement area(cm2

)


2)



Ay Reinforcement area [cm2] 5.85cm2



Ay Ay -5.85106 cm² 0.0001 %




582

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

Test ID: 5195

Test status: Passed

5.44.1 Description

Verifies a square cross section concrete column made of concrete C30/37 subjected to a significant compressionforce and small rotation moment to the top - Bilinear stress-strain diagram (Class XC1)


The purpose of this test is to determine the second order effects by applying the method of nominal rigidity and thencalculate the frames by considering a symmetrically reinforced section.


5.44.2 Background











► 15kMm rotation moment applied to the column top




► 7kNm rotation moment applied to the column top

■ 3,02


■ Concrete cover 3cm and 5cm


■ Concrete C30/37■ Steel reinforcement S500B





583

Units

Metric System

Geometry


■ Height: h = 0.50 m,

■ Width: b = 0.50 m,

■ Length: L = 5.80 m,

■ Concrete cover: c = 5cm

Boundary conditions



Loading




NEd = 1.35*150+1.50*100 = 352.50kN = 0.035MN

MEd = 1.35*15+1.50*7 = 30.75kNm = 0.03075MNm

■ m..

.

N

Me

Ed

Ed 08703530

0370500




584




ml l 60.11*20



37.8050.0

60.11*32*32 0 a

l



Ed

EQP

ef

M

M t ., 0


Where:





ieee 01

mei 03.0


me N

M eee i

Eqp

Eqp

i 125.030.0100*30.0150

7*30.015

0

0

01

kN N Eqp 180100*30.01501

MNmkNme N M Eqp Eqp 0225.050.22125.0*180* 111


MNm M Ed 041.01


)(*)(*),( 00 t f t cm RH

72.2830

8.168.16)(

cm

cm f

f

488.0281.0

1

1.0

1

)( 20.020.00

0 t t (for t0= 28 days concrete age).




585

213

0

***1.0

1001

1

h

RH

RH

MPa f cm 35 therefore: 944.038

3535 7.07.0

1

cm f and 984.0

38

3535 2.02.0

2

cm f

72.1984.0*944.0*

250*1.0

100

501

1250500500*2

500*500*2*230

RH mm

u

Ach

28.2488.0*72.2*72.1)(*)(*),( 00 t f t cm RH


25.1041.0

0225.0*28.2*, 0

Ed

EQP

ef M

M t



For an isolated column, the slenderness limit verification is done using the next formula:

n

C B A ***20lim


Where:

071.020*²50.0

353.0

*

cd c

Ed

f A

N n

80.0

25.1*2.01

1

*2,01

1

ef

A



24.46071.0

7.0*1.1*80.0*20lim

24.4637.80 lim

Therefore, the second order effects must be taken into account.

Calculation of the eccentricities and solicitations corrected for ULS:


■ NEd= 1.35*150 + 1.50*100= 352.5kN = 0,3525 MN

■ MEd= 1.35*15 + 1.50*7= 352.5kN= 0.03075MNm

Therefore, we must calculate:






586


m N

M e

Ed

Ed 087.0353.0

03075.00


ml ei 03.0400

6.11

4000



MN N Ed 353.0

meee i 117.001

MNm N e M Ed Ed 041.0353.0*117.0*1

0*e N M Ed

mmmm

mmmmmm

hmm

e 207.16

20max

30

50020

max

30

20max0



To apply the nominal rigidity method, we need an initial reinforcement area to start from. For this, the concretesection will be sized considering only the first order effect.

The Advance Design calculation is different: it is iterating as many time as necessary starting from the minimum

percentage area.

The reinforcement will be determined using a compound bending with compressive stress. The determinedsolicitations were calculated from the center of gravity of the concrete section alone. Those stresses must be reducedto the centroid of tensioned steel:

MNmh

d N M M Gua 112.02

50.045.0*353.0041.0)

2(*0

Verification about the partially compressed section:

494,0)45,0

50,0*4,01(*

45,0

50,0*8,0)*4,01(**8,0

d

h

d

h BC

055.020*²45.0*50.0

112.0

*²*

cd w

uacu

f d b

M

494.0055.0 BC cu therefore the section is partially compressed




587

The calculation for the tensioned steel in pure bending:

055.0cu

071,0)055,0*21(1*25,1 u

md z uc 437,0)071,0*4,01(*45,0)*4,01(*

²89,578,434*437,0

112,0

*cm

f z

M A

yd c

ua

The calculation for the compressed steel in bending:

For the compound bending:

2423.2

78.434

353.010*89.5 cm

F

N A A

yd

The minimum column percentage reinforcement must be considered:

²81.078.434

353.0*10.0*10,0min, cm

F

N A

yd

Ed s

Therefore, a 5cm2reinforcement area will be considered.




588

5.44.2.3 Calculation of the second order effects:

Estimation of the nominal rigidity:

It is estimated the nominal rigidity of a post or frame member from the following formula:


Where:

2.1

cmcd

E E


Mpa f

E cmcm 57.32836

10

38*22000

10*22000

3.03.0

Mpa E

E cmcd 27364

2.1

32837

2.1

4343

10.208,512

50.0

12

*m

hb I c

inertia of the concrete section only

Mpa E s 200000

s I : Inertia

002.050.0*50.0

10.5 4

c

s

A

A

01.0002.0 c

s

A

A

22.120

30

201 ck f k

071.020*²50.0

353.0

*

cd c

Ed

f A

N n

20.00336.0170

37.80*071.0

170*2

nk

45

242

10.205.02

50.0*

2

10*5*2

2*

2*2 mc

h A I s s

1 s K and 018.0

25.11

0336.0*22.1

1

* 21

ef

c

k k K

Therefore:

²56.610*2*200000*110*208,5*27364*018.0 53 MNm EI




589

Stress correction:


11*0

Ed

B Ed Ed

N N M M



In addition:

0

²

c

and 80 c because the moment is constant (no horizontal force at the top of post).

234.18

²

MN l

EI N B 48.0

²60.11

56.6*²*²

2

0

It was therefore a moment of 2nd order which is:

MNm M Ed 182.0

1

353.0

48.0

234.11*041.0

There is thus a second order moment of 0.182MNm




590

Calculation of the flexural combined reinforcement

The theoretical reinforcement will be determined by the following diagram:

MNm M Ed 182.0

MN N Ed 353.0

The input parameters of the diagram are:

073.020*50.0*50.0

182.0

** 22

cd

Ed

f hb

M

071.020*5.0*5.0

353.0

**

cd

Ed

f hb

N

Therefore:

11.0

The reinforcement area will be:

22

65.1278.434

20*50.0*11.0***cm

f

f hb A

yd

cd s

This means a total of 12.65cm2

The initial calculations must be repeated by increasing the section; a 6cm2 reinforcement section will be considered.


One more iteration by considering an initial section of 6.5cm ²




591

Estimation of the nominal rigidity:


Where:

2.1cmcd

E E


Mpa f

E cmcm 57.32836

10

38*22000

10*22000

3.03.0

Mpa E

E cmcd 27364

2.1

32837

2.1

4343

10.208,512

50.0

12

*

m

hb

I c

considering only the concrete section only

Mpa E s 200000

s I : Inertia

0026.050.0*50.0

10*5.6 4

c

s

A

A

01.0002.0 c

s

A

A

22.120

30

201 ck f k

071.020*²50.0

353.0

*

cd c

Ed

f A

N n

20.00336.0170

37.80*071.0

170*2

nk

45

242

106.205.0

2

50.0

2

10*5.6*2

2

*

2

*2 mch A

I s s

1 s K and 018.0

25.11

0336.0*22.1

1

* 21

ef

c

k k K

Therefore:

²78.710*6.2*200000*110*208,5*27364*018.0 53 MNm EI




592

Stress correction:


11*0

Ed

B Ed Ed

N N M M



In addition:

0

²

c

and 80 c because the moment is constant (no horizontal force at the top of post).

234.18

²

MN l

EI N B 57.0

²60.11

78.7*²*²

2

0

It was therefore a moment of 2nd order which is:

MNm M Ed 123.0

1

353.0

57.0

234.11*041.0

There is thus a second order moment of 0.123MNm

Calculation of the flexural compound reinforcement

The theoretical reinforcement will be determined, from the following diagram:

MNm M Ed 123.0

MN N Ed 353.0




593


049.020*50.0*50.0

123.0

** 22

cd

Ed

f hb

M

071.020*5.0*5.0

353.0

** cd

Ed

f hb

N

Therefore:

05.0


22

75.578.434

20*50.0*05.0***cm

f

f hb A

yd

cd s

This means a total of 5.75cm2



■ 7 nodes,





594



2)





Az Reinforcement area [cm2

] 2.88 cm2



Az Az -3.01 cm² -0.0000 %




595

5.45 EC2 Test 42: Verifying a square concrete column subjected to a significant rotation momentand small compression force to the top with Nominal Curvature Method - Bilinear stress-straindiagram (Class XC1)


5.45.1 Description

Verifies a square cross section column made of concrete C30/37 subjected to a significant rotation moment and smallcompression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1).

The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed.

Nominal curvature method.



5.45.2 Background


Verifies the adequacy of a square cross section made from concrete C30/37.












► 7kN axial force


■ 3,02


■ Concrete cover 3cm and 5cm■ Transversal reinforcement spacing a=40cm

■ Concrete C30/37






596

Units

Metric System

Geometry


■ Height: h = 0.50 m,

■ Width: b = 0.50 m,

■ Length: L = 5.80 m,


Boundary conditions



Loading




► NEd =1.35*15+1.50*7=30.75kN=0.03075MN

► MEd=1.35*150+1.50*100=352.50kNm=0.352MNm

■ m..

.

N

Me

Ed

Ed 4511030750

35200




597




ml l 60.11*20



37.8050.0

60.11*32*32 0 a

l



Ed

EQP

ef

M

M t ., 0


Where:





ieee 01

mei 03.0


me N

M eee i

Eqp

Eqp

i 56.1030.07*30.015

100*30.0150

0

0

01

kN N Eqp 10.177*30.0151

MNmkNme N M Eqp Eqp 181.058.18056.10*10.17* 111


MNm M Ed 3523.01


)(*)(*),( 00 t f t cm RH

72.2830

8.168.16)(

cm

cm f

f

488.0281.0

1

1.0

1)( 20.020.0

0





598

213

0

***1.0

1001

1

h

RH

RH


3535 7.07.0

1

cm f and 984.0

38

3535 2.02.0

2

cm f

72.1984.0*944.0*

250*1.0

100

501

1250500500*2

500*500*2*230

RH mm

u

Ach

28.2488.0*72.2*72.1)(*)(*),( 00 t f t cm RH


17.1352.0

181.0*28.2*, 0

Ed

EQP

ef M

M t




n

C B A ***20lim


Where:

0062.020*²50.0

031.0

*

cd c

Ed

f A

N n

81.0

17.1*2.01

1

*2,01

1

ef

A



42.1580062.0

7.0*1.1*81.0*20lim

42.15837.80 lim

Therefore, the second order effects can be neglected.




599



■ NEd= 1.35*15 + 1.50*7= 30.75kN = 0,03075 MN

■ MEd= 1.35*150 + 1.50*100= 352.5kN= 0.3525 MNm





m N

M e

Ed

Ed 46.1103075.0

3525.00


ml

ei 03.0400

6.11

4000

The first order eccentricity: stresses correction:The forces correction, used for the combined flexural calculations:

MN N Ed 03075.0

meee i 49.1101

MNm N e M Ed Ed 353.003075.0*49.11*1

0*e N M Ed

mm)mm.;mmmax(mm

;mmmaxh

;mmmaxe 207162030

50020

30200





600


The theoretical reinforcement will be determined by the following diagram:


141.020*50.0*50.0

353.0

** 22

cd

Ed

f hb

M

00615.020*5.0*5.0

03075.0**

cd

Ed

f hb N

Therefore:

35.0


22

25.4078.434

20*50.0*cm A s

which means 20.13cm2 per face.

The total area will be 40.25cm2.




601

The nominal curvature methods (second order effect):

The curvature calculation:

Considering a reinforcement of 40.25cm² (considered symmetric), one can determine the curvature from the followingformula:

0

1

**

1

r K K r r

According to Eurocode 2 – EN 1992-1-1(2004): Chapter 5.8.8.3(1)

1

0

0107.045.0*45,0

1000

78.434

)*45,0()*45,0(

1 md

E

f

d r

s

yd

yd

Kr is the correction coefficient depending of the normal force:

1

bal u

ur

nn

nn K


00615.020*²50.0

03075.0

*

cd c

Ed

f A

N n

350.020*²50.0

78.434*10*25.40

*

* 4

cd c

yd s

f A

f A

350.1350.011 un

4,0bal n

41.140.0350.1

00615.0350.1

r K

Condition: 1r K , therefore it will be considered: 1r K

K creep coefficient:

1*1 ef K


therefore

Calculation moment:

The moment of calculation is estimated from the formula:

20 M M M Ed Ed

Where:

Ed M 0 is moment of the first order including geometric imperfections.




602

2 M is nominal moment of second order.

The 2nd order moment is calculated from the curvature:

22 *e N M Ed

mcl

r e 144.0

10²60.11*0107.0*1

2

02

MNme N M Ed 00443.0144.0*03075.0* 22

MNm M M M Ed Ed 357.000443.0353.020

The reinforcement must be sized considering the demands of the second degree effects, as follows:

MN N Ed 03075.0

MNm M Ed 357.0

Reinforcement calculation according the second order effects:

The interaction diagram will be used.


35.0 will be obtained, which gives: ²25.40078.434

20*²50.0*35.0***cm

f

f hb A

yd

cd s

Meaning a 20.13 cm2 per side (which confirms the initial section)




603



■ 7 nodes,


Theoretical reinforcement area(cm2

)


2)


(reference value: 38.64 cm2 = 2 x 19.32cm

2)




604




2



Az Az -19.32 cm² 0.0000 %




605

5.46 EC2 Test36: Verifying a rectangular concrete column using the method based on nominalcurvature- 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-straindiagram (Class XC1)

Verifies the adequacy of a rectangular cross section column made from concrete C30/37.



The column is considered connected to the ground by an articulated connection (all the translations are blocked andall 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 “Calcul des Structures en beton” book, by Jean-Marie Paille, edition Eyrolles.

5.46.2 Background














► The quasi permanent coefficient 3,02

■ Wind loads:

► 250kN axial force► The wind loads are applied horizontally on the middle of the column to the wider face (against the

positive sense of the Y local axis)

► The quasi permanent coefficients 10 and02


■ Concrete C30/37



■ Concrete age t0=30days

■ The reinforcement is set to 8HA16 (16.08cm2)




606

Units

Metric System

Geometry


■ Height: h = 0.35 m,

■ Width: b = 0.60 m,

■ Length: L = 5.00 m,


Boundary conditions


The column is considered connected to the ground by a articulated connection (all the translations are blocked andall the rotations are permitted) and to the top part the translations along X and Y axis are blocked and the rotationalong Z axis is also blocked.

Loading



MN N ED 99.03.0*5.14.0*35.1

cmN

M

eu

u

00

cm;cmmaxl

;cmmaxei 2400

5002

4002 0

First order eccentricity: m.cmeee i 020201




607




ml l 50



5.4935.0

5*32*32 0 a

l


The creep coefficient is determined by the next formula:

Ed

EQP

ef

M

M t ., 0


Where:



The EQP M value is calculated using the

W QG M EQP *03.0

MNme N M qp Eqp 0098.002.0*)3.0*3.04.0(* 10


The ED M 0 value is calculated using the W QG M oED *5.1*5.135.1

MNm L

H e N M W ed Eqp 48855.04

5*25.0*5.102.0*99.0

4**5.1* 10

The moment report becomes:

02.048855.0

0098.0

0

0 Ed

Eqp

M

M

The creep coefficient 0,t is:

)(*)(*),( 00 t f t cm RH

MPa f

f cm

cm 73.2830

8.168.16)(

482.0301.0

1

1.0

1)(

20.020.0

0

0

t





608

213

0

***1.0

1001

1

h

RH

RH

RH = relative humidity; RH = 50%

Where 121 if MPa f cm 35 if not

7.0

1

35

cm f and

2.0

2

35

cm f

mm

u

Ach 221

600350*2

600*350*2*20

MPa MPa f cm 3538 therefore 944.038

3535 7.07.0

1

cm f and

984.0383535

2.02.0

2

cm f

752.1984.0*944.0*221*1.0

100

501

1***1.0

1001

1321

30

h

RH

RH

30.2482.0*73.2*752.1)(*)(*),( 00 t f t cm RH


046.002.0*30.2*, 0 Ed

EQP

ef M

M t


046.1046.011 ef


The M2 moment is calculated from the curvature formula:22 * e N M Ed

c

l

r e

2

02 *

1

l0 is the buckling length: l0=l=5m

“c” is a factor depending on the curvature distribution. Because the cross section is constant, c = 10.

0

1**

1

r K K

r r


).45,0(

1

0 d r

yd

and1000

1739.2

200000

78.434

s

yd

yd E

f




609

1

0

0161.0)30.0*45,0(

1000

1739.2

)*45,0(

1 md r

yd


1

bal u

ur

nnnn K


236.020*35.0*6.0

99.0

*

cd c

Ed

f A

N n

1un

166.020*35.0*6.0

78.434*10*08.16 4

cd c

yd s

f A

f A

166.1166.0.11 un

4,0bal n

214.14.0166.1

236.0166.1

bal u

ur

nn

nn K

Condition: 1r K , therefore it will be considered: 1r K

K creep coefficient: 1*1 ef K


15020035,0

ck f

5.4935.0

125120 h

l

17.0150

5.49

200

3035.0

15020035,0

ck f

008.1046.0*17.01*1 ef K

Therefore:

0162.00161.0*008.1*11

**1

0

r

K K r

r

mc

l

r e 04057.0

10

²5*0162.0*

1 2

02

MNme N M Ed 04017.004057.0*990.0* 22

MNm M M M Ed Ed 5287.004017.048855.020

The frames must be sized considering the demands of the second degree effects, as follows:

MN N Ed 990.0




610

MNm M Ed 5287.0


The calculations are made using the following values:

MN N Ed 990.0

MNm M Ed 5287.0

The calculations resulted in 38.02cm2 tensioned reinforcement and a 20cm

2 compressed reinforcement, meaning a

total of 58.02cm2 reinforcement area.

Buckling checking:

The calculation is performed considering the reinforcement found previously (58.02cm2):

Curvature calculation:

The reinforcement area has an influence only over the Kr parameter:

1

00161.0

1

mr

1

bal u

ur

nn

nn K

112.0n

60.020*35.0*60.0

78.434*10*02,58

*

* 4

cd c

yd s

f A

f A

60.160.011 un

4,0bal n

1124.140.060.1

112.060.1

r r K K

1 K the creep coefficient

1.1 ef K

Therefore the curvature becomes:

1

0

0161.01

**1 m

r K K

r r

Considering this result, the same curvature is obtained, which means that the second order moment is the same. Thereinforcement section is correctly chosen.



■ 5 nodes,





611








2



Ay Ay -38.01 cm² -0.0000 %




612

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

Test ID: 5153

Test status: Passed

5.47.1 Description

Verifies a square cross section column made of concrete C30/37 subjected to a small compression force andsignificant rotation moment to the top - Bilinear stress-strain diagram (Class XC1)

Nominal rigidity method - The purpose of this test is to determine the second order effects by applying the method ofnominal rigidity, and then calculate the frames by considering a symmetrically reinforced section.


5.47.2 Background














► 7kN axial force


■




■ Transversal reinforcement spacing a = 40cm






613

Units

Metric System

Geometry


■ Height: h = 0.50 m,

■ Width: b = 0.50 m,

■ Length: L = 5.80 m,

■ Concrete cover: c = 5cm

Boundary conditions



Loading




NEd =1.35*15+150*7=30.75kN=0.03075MN

MEd=1.35*150+1.50*100=352.50kNm=0.352MNm

■ m..

.

N

Me

Ed

Ed 4511030750

35200




614




ml l 60.11*20



37.8050.0

60.11*32*32 0 a

l



Ed

EQP

ef

M

M t ., 0


Where:





ieee 01

mei 03.0


kN N Eqp 10.177*30.0151

MNmkNme N M Eqp Eqp 181.058.18056.10*10.17* 111



)(*)(*),( 00 t f t cm RH

72.2830

8.168.16)(

cm

cm f

f

488.0281.0

1

1.0

1

)( 20.020.0

0





615

213

0

***1.0

1001

1

h

RH

RH

MPa f cm 35 therefore:

944.038

3535 7.07.0

1

cm f

and

984.038

3535 2.02.0

2

cm f

72.1984.0*944.0*

250*1.0

100

501

1250500500*2

500*500*2*230

RH mm

u

Ach

28.2488.0*72.2*72.1)(*)(*),( 00 t f t cm RH


17.1352.0

181.0*28.2*, 0

Ed

EQP

ef M

M t




n

C B A ***20lim


Where:

0062.020*²50.0

031.0

*

cd c

Ed

f A

N n

81.0

17.1*2.01

1

*2,01

1

ef

A



42.1580062.0

7.0*1.1*81.0*20lim

42.15837.80 lim

Therefore, the second order effects can be neglected.



NEd= 1.35*15 + 1.50*7= 30.75kN = 0,03075 MN

MEd= 1.35*150 + 1.50*100= 352.5kN= 0.3525MNm




616

Therefore, we must calculate:




m N

M

e Ed

Ed

46.1103075.0

3525.00


ml

ei 03.0400

6.11

400

0



MN N Ed 03075.0

meee i 49.1101

MNm N e M Ed Ed 353.003075.0*49.11*1

0*e N M Ed

mmmm

mmmmmm

hmm

e 207.16

20max

30

50020

max

30

20max0





617


The theoretical reinforcement will be determined by the following diagram


141.020*50.0*50.0

353.0

** 22

cd

Ed

f hb

M

00615.020*5.0*5.0

03075.0

**

cd

Ed

f hb

N

Therefore:

35.0


22

25.4078.434

20*50.0*cm A s

which means 20.13cm2 per face.

The total area will be 40.25cm2.


■ Linear element: S beam,■ 7 nodes,





618




(reference value: 38.64 cm2

= 2 x 19.32cm2

)




619




2



Az Az -19.32 cm² 0.0000 %




620

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

Test ID: 5225

Test status: Passed

5.48.1 Description

Verifies the adequacy of a rectangular cross section beam made of concrete C25/30 supporting a balcony - Bilinearstress-strain diagram (Class XC1).


Verifies the column resistance to rotation moment along its length. During this test, the determination of stresses ismade along with the determination of the longitudinal and transversal reinforcement.

5.48.2 Background


Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist rotation moment along itslength. During this test, the calculation of stresses is performed, along with the calculation of the longitudinal andtransversal reinforcement.





■ The geometric dimension of the beam are:


■ Concrete type: C25/30

■ Reinforcement type: S500B


■ Balcony load: 1kN/m2

■ The weight of the beam will be considered in calculation


■ The beam is considered fixed at both ends

■ Concrete cover: 40mm

■ Beam length: 4.00m




621

Units

Metric System

Geometry



■ Length: L = 4.00 m,



Boundary conditions


■ Outer:


► Support at end point (x = 4.00) fixed connection

■ Inner: None.


The ULS load calculation:

The first step of the calculation is to determine charges transmitted to the beam:

■ Vertical loads applied to the beam (kN/ml) from the load distribution over the balcony

■ Rotation moment applied to the beam (kN/ml) from the load distribution over the balcony

Each action (self-weight and distributed load) is determined by summing the resulting vertical loads along the eavesand the torsional moment by multiplying the resultant by the corresponding lever arm.

CAUTION, different lever arm must be considered from the center of the beam (by adding therefore the half-width).

The results are displayed in the table below:




622

Load calculation:

From previously calculated results, the following stresses can be determined:

■ Shear: Ed V

■ Bending moment: Ed M

■ Torque: Ed T

Shear and bending moment:

One can determine the load at ULS taken over by the beam:

m KN P u /14.212*5,144.13*35,1

For a beam fixed on both ends, the following values will be obtained:

Maximum shear (ULS):

MN KN l P V u Ed 042,03,42

24*14,21

2*

Bending moment at the supports:

MNm KNml P

M u Ed 028,02,28

12

²4*14,21

12

²*

Maximum Moment at middle of span:

MNm KNml P

M u Ed 014,01,14

24

²4*14,21

24

²*

Torsion moment:

For a beam subjected to a torque constant:

m MNmm KNmmtu /015,0/97,1425,2*5,159,8*35,1

MNml

mT tu Ed 03,02

4*015,0

2*




623

5.48.2.3 Bending rebar

Span reinforcement (at bottom fiber)

007,067.16*²70.0*25,0

014,0cu

0088,0007,0*211*25,1 u

m z c 697,00088,0*4,01*70.0

²46.0²10*62.478,434*697,0

014,0 4 cmm Au

Minimum reinforcement percentage verification:

d b

d b f

f

Max A

w

w

yk

eff ct

s

**0013.0

***26.0 ,

min,

Cracking matrix required (calculation hypothesis): Mpa f f ctmeff ct 56.2,

²27.2

²27.270.0*25.0*0013.0**0013.0

²27.270.0*25.0*500

56.2*26.0***26.0

,

min, cm

cmd b

cmd b f

f

Max A

w

w

yk

eff ct

s

Therefore, it retains 2.27 cm ².

Reinforcement on supports (at top fiber)

014,0

67.16*²70.0*25,0

028,0

cu

018,0014,0*211*25,1 u

m z c 695,0018,0*4,01*70.0

²93.0²10*27.978,434*695,0

028,0 4 cmm Au

It also retains 2.27 cm ² (minimum percentage).

Shear reinforcement

MN V Ed 042,0

The transmission to the support is not direct; it is considered a connecting rod inclined by 45˚, therefore 1cot

Concrete rod verification:

²cot1

cotcot****max,

cd w Rd f v z bV

According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 6.2.3(4)

m z z c 695.0 (from the design in simple bending of support)




624

Vertical frames:

54.0250

251*6,0

v

MN tg

f v z b f v z bV cd cwcd cw Rd 78.0

2

67.16*54.0*695.0*25.0

cot

***

²cot1

cotcot****max,

cot

***1max,

tg

b z f vV wucd Rd




ml cm f z

tg V

s

A

yd u

Ed sw /²39.178.434*695.0

042.0

*

*.

(over shear)

From the minimum reinforcement percentage:

sin..min, ww sw b s

A


With:

0008.0500

25*08,0*08,0min, yk

ck w

f f

ml cm s

A sw /²225.0*0008.0

Therefore:

ml cm s

A sw /²2

Torsion calculation:

Torsion moment was calculated before: MNmT Ed 03,0




625

Torsional shear stress:

k ief

Ed it

At

T

**2 ,

,


cmcmu

Acmc

t ief 375.9375.9)7525(2

)75*25(8*2

max,

²1025.0²1025)375.975(*)375.925( mcm Ak

Mpa..*.*

.

A*t*

T

ki,ef

Edi,t 561

102500937502

030

2τ

Concrete verification:

Calculate the maximum allowable stress in the rods:

cos*sin******2 ,max, ief k cd cw Rd t A f vT


54.0250

1*6,0

ck f v

MN T Rd 085.070.0*70.0*09375.0*1025.0*67.16*54.0*2max,

Because of the combined share/moment effect, we must calculate:

0,1max,max,

Rd

Ed

Rd

Ed

V V

T T


0,1399.078.0

042.0

085.0

03.0

Torsion longitudinal reinforcement

cot***2

*

yd k

k Ed l

f A

uT A


mcmt ht bu ef ef k 625.15.162)]375.975()375.925[(*2)]()[(*2

²47.578.434*1025.0*2

625.1*03.0cm Al




626

Torsion transversal reinforcement

ml cm f A

T

s

A

yd k

Ed

T

swT /²36.378.434*1025.0*2

03.0

cot***2

Therefore ml cm s

A

T

swT /²36,3 for each face.



■ 11 nodes,



Torsional moment (Ted=29.96kNm)

Longitudinal reinforcement (5.46cm2)

Transversal reinforcement (3.36cm2/ml)




627



Mx Torsional moment [kNm] 29.96 cm2

Al Longitudinal reinforcement [cm2] 5.46 cm

2

Ator,y Transversal reinforcement [cm

2

/ml] 3.36 cm

2

/ml



Mx Mx 29.9565 kN*m 0.0000 %

Al Al 5.4595 cm² 0.0000 %

Ator,y / face Ator,y/face 3.35969 cm² 0.0001 %




628

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

Test ID: 5231

Test status: Passed

5.49.1 Description

Verifies a square cross section beam made of concrete C20/25 subjected to a normal force of traction - Bilinearstress-strain diagram (Class X0).

Tie sizing

Bilinear stress-strain diagram

Determines the armature of a pulling reinforced concrete, subjected to a normal force of traction.

The load combinations will produce the following rotation efforts:

NEd=1.35*233.3+1.5+56.67=400kNm


- Support at start point (x=0) fixed connection

- Support at end point (x = 5.00) translation along the Z axis is blocked

5.49.2 Background

Tie sizing

Bilinear stress-strain diagram / Inclined stress-strain diagram

Determine the armature of a pulling reinforced concrete, C20/25, subjected to a normal force of traction.


■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002;■ Analysis type: static linear (plane problem);



■ Loadings from the structure: Fx,G = 233.33 kN


■ Exploitation loadings (category A): Fx,Q = 56.67kN


Units

Metric System




629

Geometry


■ Height: h = 0.15 m,

■ Width: b = 0.15 m,

■ Length: L = 5.00 m,



■ Reinforcement S400, Class: B

■ Fck=20MPa

■ The load combinations will produce the following rotation efforts:

NEd=1.35*233.3+1.5+56.67=400kNm

Boundary conditions


■ Outer:

► Support at start point (x = 0) fixed connection,

► Support at end point (x = 5.00) translation along the Z axis is blocked

■ Inner: None.


There will be two successive calculations, considering a bilinear stress-strain diagram constitutive law and then ainclined stress-strain diagram constitutive law.

Calculations according a bilinear stress-strain diagram

²50,11²10*50,11

15.1

400

400.0 4

,

, cmm N

AU s

Ed U s

It will be used a 4HA20=A=12.57cm2

Calculations according a inclined stress-strain diagram

MPaClassBS U s 373400 ,

²70,10²10*72,10373

400.0 4

,

, cmm N

AU s

Ed U s

It can be seen that the gain is not negligible (about 7%).


yk

ctmc s

f

f A A *

MPa f f ck ctm 21,220*30.0*30.0 3/23/2

²0225.015.0*15.0 m Ac

²24.1400

21.2*0225.0* cm

f

f A A

yk

ctmc s




630



■ 6 nodes,


ULS load combinations (kNm)

In case of using the bilinear stress-strain diagram, the reinforcement will result: (Az=11.50cm2=2*5.75

cm

2)

In case of using the inclined stress-strain diagram, the reinforcement will result: (Az=10.70cm2=2*5.35

cm

2)



Az,1 Longitudinal reinforcement obtained using the bilinear stress-straindiagram [cm

2]

5.75 cm2

Az,2 Longitudinal reinforcement obtained using the inclined stress-straindiagram [cm

2]

5.37 cm2



Az Az-i -5.75001 cm² 0.0000 %




631

5.50 EC2 Test 44: Verifying a rectangular concrete beam subjected to eccentric loading - Bilinearstress-strain diagram (Class X0)

Test ID: 5213

Test status: Passed

5.50.1 Description

Verifies a rectangular cross section beam made of concrete C30/37 subjected to eccentric loading - Bilinear stress-strain diagram (Class X0).

The verification of the bending stresses at ultimate limit state is performed.


During this test, the determination of stresses is made along with the determination of the longitudinal and transversalreinforcement.

- Support at start point (x = 0) fixed connection

- Support at end point (x = 5.00) fixed connection

5.50.2 Background


Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist eccentric loading. Duringthis test, the calculation of stresses is made along with the calculation of the longitudinal and transversalreinforcement.










Units

Metric System




632

Geometry


■ Height: h = 0.30 m,

■ Width: b = 0.18 m,

■ Length: L = 5.00 m,



■ Effective height: d=h-(0.6*h+ebz)=0.25m; ebz=0.035m

■ The load eccentricity will be considered of 0.50m

■ The load eccentricity will produce the following rotation moments:

Mx,G=6.00kNm

Mx,Q=1.50kNm

Boundary conditions


■ Outer:


► Support at end point (x = 5.00) fixed connection

■ Inner: None.




633

Reference results in calculating the concrete beam

The moment is defined by the following formulas:

P P ba

be P T A

10

3

5

3**5,0**

P P ba

ae P T B51

52**5,0**

Note: the diagram of bending moment is the same as the shear force multiplied by the eccentricity.

The ULS load calculation:

kN P u 70.203*5.112*35.1

Before the point of application of torque, the torque is:

KNm P T u Ed 21.67,20*10

3*

10

3

After the point of application of torque, the torque is:

KNm P

T u Ed 14.4

5

7.20

5

Result ADVANCE Design 2012 - Moment: (in kNm)

Reference value: 6.21 kNm and -4.14kNm




634

5.50.2.2 Calculation in pure bending

Bending moment at ULS:

KN P u 7,203*5,112*35,1

MNm KNm P

M u Ed 0248,084,24

5

7,20*6

5

*6

Simple bending design:

110,020*²25.0*18,0

0248,0cu

146,0110,0*211*25,1 u

m z c 235,0146,0*4,01*25.0

²43,2²10*43,278,434*235,0

0248,0 4 cmm Au

5.50.2.3 Torsion reinforcement

Torsion longitudinal reinforcement - Before application of the moment

cot***2

*

yd k

k Ed l

f A

uT A


))()((*2 ef ef k t ht bu

)(*)( ef ef k t ht b A

with:

m

u

Act ef 07.005625.0;07.0max

3.018.0*2

30.0*18.0;035.0*2max;*2max

mcmuk 68.068))730()718((*2

²0253.0²253)730(*)718( mcm Ak

²92.178.434*0253.0*2

68.0*0062.0cm Al

Torsion longitudinal reinforcement - After application of the moment

²28.178.434*0253.0*2

68.0*00414.0cm Al

Torsion transversal reinforcement - Before application of the moment

ml cm f A

T

s

A

yd k

Ed

T

swT /²82.278.434*0253.0*2

0062.0

cot***2




635

Torsion transversal reinforcement - After application of the moment

ml cm f A

T

s

A

yd k

Ed

T

swT /²88.178.434*0253.0*2

00414.0

cot***2

Concrete verification:

Calculation of the maximum allowable stress under torsional moment:

cos*sin******2 ,max, ief k cd cw Rd t A f vT


528.0250

1*6,0

ck f

v

MN T Rd 018.070.0*70.0*07.0*0253.0*20*528.0*2max,

Calculation of the maximum allowable stress under shear:

²cot1

cotcot****max,

cd w Rd f v z bV


m z z c 235.0 (from the design in simple bending)

Vertical reinforcement: 1cot

528.0250

301*6,0

v

MN f v z bV cd cw Rd 223.02

1*20*528.0*235.0*18.0

²cot1

cotcot****max,

Because of the combined shear/moment effect, it must be calculated:

0,1max,max,

Rd

Ed

Rd

Ed

V

V

T

T


0,1400.0

223.0

0124.0

018.0

0062.0



■ 6 nodes,





636


Longitudinal reinforcement (Al=1.92cm2and Al=1.28cm

2)

Transversal reinforcement (2.82cm2/ml

and 1.88cm

2/ml)



Al,1 Longitudinal reinforcement for the first part [cm2] 1.92 cm

2

Al,2 Longitudinal reinforcement for the second part [cm2] 1.28 cm

2

Ator,y,1 Transversal reinforcement for the first part [cm2/ml] 2.82 cm

2/ml

Ator,y,2 Transversal reinforcement for the second part [cm2/ml] 1.88 cm

2/ml



Al Al,1 1.91945 cm² 0.0002 %

Al Al,2 1.27964 cm² -0.0003 %

Ator,y / face Ator,y,1 2.82273 cm² -0.0001 %

Ator,y / face Ator,y,2 1.88182 cm² -0.0001 %




637

5.51 EC2 Test35: Verifying a rectangular concrete column subjected to compression to top – Basedon nominal rigidity method - Bilinear stress-strain diagram (Class XC1)

Test ID: 5123

Test status: Passed

5.51.1 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 thencalculate the frames by considering a section symmetrically reinforced.


5.51.2 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 thencalculate the frames by considering a section symmetrically reinforced.







► 0.45 MN axial force

► 0.10 MNm rotation moment applied to the column top



► 0.50 MN axial force

► 0.06 MNm rotation moment applied to the column top

■ 3,02









638

Units

Metric System

Geometry


■ 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



Loading




NEd =1.35*0.45+1.5*0.50=1.3575MN

MED=1.35*0.10+0.30*0.06=0.225MNm

■ me 166.0

3575.1

225.00




639




ml l 950.4*2*20



28.6945.0

9*32*32 0 a

l



Ed

EQP

ef

M

M t ., 0


Where:




MNm M Eqp 118.006.0*3.010.00

MNm M Ed 225.00

The moment report becomes:

524.0225.0

118.0

0

0 Ed

Eqp

M

M

The creep coefficient 0,t

is defined as follows:

)(*)(*),( 00 t f t cm RH

MPa f f cm

cm 73.2830

8.168.16)(

488.0281.0

1

1.0

1)(

20.020.0

0

0

t


213

0

***1.0

1001

1

h

RH

RH

RH =relative humidity; RH=50%




640

Where 121

if MPa f cm 35 if not

7.0

1

35

cm f and

2.0

2

35

cm f

mm

u

Ach 274

700450*2

700*450*2*20

MPa MPa f cm 3538 ,

Therefore:

944.038

3535 7.07.0

1

cm f and 984.0

38

3535 2.02.0

2

cm f

70.1984.0*944.0*274*1.0

100

501

1***1.0

1001

1321

30

h

RH

RH

26.2488.0*73.2*70.1)(*)(*),( 00 t f t cm RH


18.1524.0*26.2*, 0 Ed

EQP

ef M

M t


18.218.111 ef



n

C B A ***20lim


Where:

215.020*70.0*45.0

3575.1

*

cd c

Ed

f A

N n

81.0

18.1*2.01

1

*2,01

1

ef

A

because the reinforcement ratio in not yet known

because the ratio of the first order moment is not known

90.26215.0

7.0*1.1*85.0*20lim

90.2628.69 lim

Therefore, the second order effects most be considered




641

5.51.2.3 The eccentricity calculation and the corrected loads on ULS:


me 166.03575.1

225.00


First order eccentricity- stresses correction:

MN N Ed 3575.1

cmmeee i 85.181885.001

MNm M Ed 256.0





MNmh

d N M M Gua 494.02

45.040.0*3575.1256.0)

2(*0


495,0)40,0

45,0*4,01(*

40,0

45,0*8,0)*4,01(**8,0

d

h

d

h BC

220.020*²40.0*70.0

495.0

*²*

cd w

uacu

f d b

M





642


220.0cu

315,0)220,0*21(1*25,1 u

md z uc 350,0)315,0*4,01(*40,0)*4,01(*

²46.3278,434*350,0

495,0

*cm

f z

M A

yd c

ua



24 24.178.434

3575.110*46.32' cm

f

N A A

yd


2

min, 3.6*002.0²12.378.434

3575.1*10.0*10,0cm Acm

f

N A c

yd

Ed s



The second order effect will be determined by applying the method of nominal rigidity:


It is estimated nominal rigidity of a post or frame member from the following formula:


With:

2.1

cmcd

E E

MPa MPa f f ck cm 388

MPa f

E cmcm 32837

10

38*22000

10*22000

3.03.0

MPa E

E cmcd 273642.1

32837

2.1

410*316.512

45.0*70.0

12

* 333

mhb

I c (concrete only inertia)

MPa E s 200000

s I : Inertia

002.040.0*70.0

10*28,6 4

c

s

A

A




643

01.0002.0 c

s

A

A

Mpa f

k ck 22.120

30

201

215.020*70.0*45.0

3575.1

.

cd c

Ed

f A

N n

20.0088.0170

28.69*215.0

170*2

nk

45

242

10.92,105.02

45.0*

2

10.28,6*2

2*

2*2 mc

h A I s s

1 s K and 049.0

18.11

088.0*22.1

1

* 21

ef

c

k k K

Therefore:

²97.1010*92.1*200000*110*316.5*27364*049.0 53

MNm EI

Stresses correction:

The total moment, including second order effects, is defined as a value plus the time of the first order:

1

1*0

Ed

B Ed Ed

N

N M M

MNm M Ed 0256.00 (moment of first order (ULS) taking into account geometric imperfections


0

²

c

and 80 c the moment is constant (no horizontal force at the top of post).

234.18

²

MN l EI N B 32.1

²982.10*²*²

2

0

The second order moment is:

MNm M Ed 18.11

13575.1

32.1

234.11*256.02

An additional iteration must be made by increasing the ratio of reinforcement.




644


The iteration is made considering 8HA12 or As=9.05cm2

Therefore

0029.045.0*70.0

10*05,9 4

Is obtained:

Therefore:

MN N Ed 3575.12

MNm M Ed 48.22

Given the mentioned reports, there must be another reiteration:

The reinforcement section must be increase to 8HA20 or As=25.13cm2

and 008.0 :




645

Is obtained:

Therefore:

MN N Ed 3575.12

MNm M Ed 563.02

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=30cm2and 0095.0 :




646

Therefore:

MN N Ed 3575.12

MNm M Ed 499.02

Using the EC2 calculation tools, combined bending analytical calculation, a 22.22cm2 value is found.

The theoretical reinforcement section of 30cm2 will be adopted.



■ 6 nodes,



(reference value: 16cm2)


(reference value: 31.99cm2= 2 x 15.99cm

2)




647



Az Reinforcement area [cm2] 16 cm

2



Az Az -15.995 cm² 0.0000 %




648

5.52 EC2 Test 39: Verifying a circular concrete column using the simplified method – Professionalrules - Bilinear stress-strain diagram (Class XC1)

Test ID: 5146

Test status: Passed

5.52.1 Description

Verifies the adequacy of a concrete (C25/30) column with circular cross section using the simplified method –Professional rules - Bilinear stress-strain diagram (Class XC1).

Simplified Method

The column is considered connected to the ground by an articulated connection (all the translations are blocked). Tothe top part the translations along X and Y axis are blocked and the rotation along the Z axis is also blocked.

5.52.2 Background

Simplified Method

Verifies the adequacy of a circular cross section made from concrete C25/30.










■ Concrete C25/30



■ Buckling length L0=5.00m




649

Units

Metric System

Geometry


■ Radius: r = 0.40 m,

■ Length: L = 5.00 m,


Boundary conditions


The column is considered connected to the ground by a articulated connection (all the translations are blocked) andto the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked.

Loading



kN N ED 67505000*35.1




650



According to Eurocode 2 EN 1992-1-1 (2004): Chapter 5.8.3.2(1)

L0 :buckling length L0=L=5m

According to Eurocode 2 EN 1992-1-1 (2004): Chapter 5.8.3.2(2)

i :the radius of giration of uncraked concrete section

I : second moment of inertia for circular cross sections

A : cross section area

, therefore:


■

■

■


6025 , therefore:

722.0

62

251

84.0

621

84.022

)**81(*)*5.070.0( Dk h




651

If the ρ and δ values are unknown, the term will be considered:

95.0)**81( , therefore:

045.195.0*)8.0*5.070.0()**61(*)*5.070.0( Dk h

95.0500

500*65.06.1500*65.06.1 yk s

f k

MPa f

f ck cd 67.16

5.1

25

5.1

MPa f

f yk

yd 33.3335.1

500

5.1

²14.3167.16*4.0*722.0*95.0*045.1

750.6*

33.333

1**

***

1 22 cm f r k k

N

f A cd

sh

ed

yd

s



■ 6 nodes,







2



Az Az -17.4283 cm² 0.0002 %




652

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


5.53.1 Description

Verifies the adequacy of a square cross section column made of concrete C30/37 subjected to a small rotationmoment and significant compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram(Class XC1).

The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed.




5.53.2 Background


Verify the adequacy of a rectangular cross section made from concrete C30/37.








► 150 kN axial force

► 15 kMm rotation moment applied to the column top

► the self-weight is neglected


► 100 kN axial force

► 7 kNm rotation moment applied to the column top

■

■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q■ Concrete cover 3cm and 5cm


■ Concrete C30/37






653

Units

Metric System

Geometry


■ Height: h = 0.50 m,

■ Width: b = 0.50 m,

■ Length: L = 5.80 m,


Boundary conditions



Loading




NEd =1.35*150+1.50*100=352.50kN=0.353MN

MEd=1.35*15+1.50*7=30.75kNm=0.03075MNm

■




654




ml l 60.11*20



37.8050.0

60.11*32*32 0 a

l



Ed

EQP

ef

M

M t ., 0


Where:





ieee 01

mei 03.0


me N

M eee i

Eqp

Eqp

i 125.030.0100*30.0150

7*30.015

0

0

01

kN N Eqp 180100*30.01501

MNmkNme N M Eqp Eqp 0225.050.22125.0*180* 111


MNm M Ed 041.01

The creep coefficient 0,t

is defined as follows:

)(*)(*),( 00 t f t cm RH

72.2830

8.168.16)(

cm

cm f

f

488.0281.0

1

1.0

1)( 20.020.0

0





655

213

0

***1.0

1001

1

h

RH

RH


3535 7.07.0

1

cm f and 984.0

38

3535 2.02.0

2

cm f

72.1984.0*944.0*

250*1.0

100

501

1250500500*2

500*500*2*230

RH mm

u

Ach

28.2488.0*72.2*72.1)(*)(*),( 00 t f t cm RH


25.1041.0

0225.0*28.2*, 0

Ed

EQP

ef M

M t



n

C B A ***20lim


Where:

071.020*²50.0

353.0

*

cd c

Ed

f A

N n

80.0

25.1*2.01

1

*2,01

1

ef

A



24.46

071.0

7.0*1.1*80.0*20lim

24.4637.80 lim

Therefore, the second order effects cannot be neglected.

Calculation of the eccentricities and solicitations corrected for ULS :


NEd= 1.35*150 + 1.50*100= 352.5kN = 0,3525 MN

MEd= 1.35*15 + 1.50*7= 352.5kN= 0.03075MNm







656


m N

M e

Ed

Ed 087.0353.0

03075.00


ml ei 03.0400

6.11

400

0



MN N Ed 353.0

meee i 117.001

MNm N e M Ed Ed 041.0353.0*117.0*1

0*e N M Ed

mmmm

mmmmmm

hmm

e 207.16

20max

30

50020

max

30

20max0



To apply the nominal rigidity method, we need an initial reinforcement area to start from. For this, the concretesection will be sized considering only the first order effect.

The Advance Design calculation is different: it is iterating as many time as necessary starting from the minimum

percentage area.

The reinforcement is determined using a compound bending with compressive stress. The determined solicitationswere calculated from the center of gravity of the concrete section alone. Those stresses must be reduced to thecentroid of tensioned steel:

MNmh

d N M M Gua 112.02

50.045.0*353.0041.0)

2(*0




657

Verification about the partially compressed section:

494,0)45,0

50,0*4,01(*

45,0

50,0*8,0)*4,01(**8,0

d

h

d

h BC

055.020*²45.0*50.0

112.0

*²*

cd w

uacu

f d b

M

494.0055.0 BC cu therefore the section is partially compressed

The calculation for the tensioned steel in pure bending:

055.0cu

071,0)055,0*21(1*25,1 u

md z uc 437,0)071,0*4,01(*45,0)*4,01(*

²89,578,434*437,0

112,0*

cm f z

M A yd c

ua

The calculation for the compressed steel in bending:

For the compound bending:

The minimum column percentage reinforcement must be considered:

Therefore a 5cm2 reinforcement area will be considered




658

5.53.2.3 The nominal curvature method (second order effect):

The curvature calculation:

Considering a reinforcement of 5cm ² (considered symmetric), one can determine the curvature from the followingformula:

0

1**1r

K K r

r


1

0

0107.045.0*45,0

200000

78.434

)*45,0()*45,0(

1 md

E

f

d r s

yd

yd


1

bal u

ur

nn

nn K


0706.020*²50.0

353.0

*

cd c

Ed

f A

N n

0435.020*²50.0

78.434*10*5

*

* 4

cd c

yd s

f A

f A

0435.10435.011 un

4,0bal n

51.140.00435.1

0706.00435.1

r K

Condition: 1r K , therefore we consider: 1r K

K creep coefficient:

1*1 ef K


,

Therefore:

1

0

0107.00107.0*1*11

**1 m

r K K

r r

Calculation moment:

The moment of calculation is defined by the formula:

20 M M M Ed Ed




659

Where:

Ed M 0 is moment of the first order including geometric imperfections.

2 M is nominal moment of second order.

The 2nd order moment is calculated from the curvature:

22 *e N M Ed

mc

l

r e 144.0

10

²60.11*0107.0*

1 2

02

MNme N M Ed 051.0144.0*353.0* 22

MNm M M M Ed Ed 09.0051.0041.020

The reinforcement must be sized considering the demands of the second degree effects, as follows:


The interaction diagram will be used.





660

According to the diagram, it will be obtained a minimal percentage reinforcement:

2

min, 5*002.0²81.078.434

353.0*10.0*10,0cm Acm

f

N A c

yd

Ed s



■ 7 nodes,





661


(reference value: Au=5cm2)


(reference value: 5 cm2)



Amin Reinforcement area [cm2] 5 cm

2



Amin Amin 5 cm² 0.0000 %




662

5.54 Verifying the capacity design results according to Eurocode EC2 and EC8 French standards.(DEV2013 #8.3)

Test ID: 5602

Test status: Passed

5.54.1 Description

Verifies the capacity design results according to Eurocode EC2 and EC8 French standards.




663

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

Test ID: 5964

Test status: Passed

5.55.1 Description

Verifies a rectangular cross section beam made from concrete C25/30 to resist simple tension. Verification of thebending stresses at ultimate limit state and serviceability limit state is performed.

Verification is done according to EN 1992-1-1 French annex.

5.55.2 Background

Simple Bending Design for Ultimate and Service State Limit

Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple tension. During thistest, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the

verification of the minimum reinforcement percentage.








■



Units

Metric System

Geometry


■ Height: h = 0.20 m,

■ Width: b = 0.20 m,

■ Length: L = 5.00 m,






664


Rectangular solid concrete C25/30 and S500, class A reinforcement steel is used. The following characteristics areused in relation to this material:




■ Humidity RH=50%

Boundary conditions


■ Outer:

► Support at start point (x=0) , fixed connection,

► Support at end point (x = 5.80) restrained in translation along Z.

■ Inner: None.

Loading


Load combinations:

■ Ultimate Limit State:

kN*.*.Q*.G*.NEd 4071505113535151351


KNQGN cq,ser 285150135

■ Quasi-permanent combination of actions:

KN*.Q*.GN qp,ser 1801503013530




665

5.55.2.2 Reference results in calculating the longitudinal reinforcement and the crack width

Calculating for ultimate limit state:

²37.9²10*37.9

15.1

500

407.0 4

,

, cmm N

AU s

Ed U s

Calculating for serviceability limit state:

S

ser ser s

N A

,

Mpa f yk s 400500*8,0*8,0

²13.7²10*13.7400

285.0 4

, cmm A ser s

Final reinforcement

Is retained the steel section between the maximum theoretical calculation at ULS and calculating the SLS or

Ath = 9.37 cm ².

Therefore, 4HA20 = A = 12.57 cm ².

Constraint checking to the ELS:

MPa MPa A

N s s

ser s

ser s 40080.226

001257.0

285.0

,

Crack opening verification:

It will be checked the openings of cracks by considering 4HA20.

Calculating the maximum spacing between the cracks

The maximum spacing of cracks is given by the formula:

eff p

r

k k ck s

,

213max,

***425.0*


The effective depth “d” must be estimated by considering the real reinforcement of the beam:

cmd 4.13

2

26.0520

020.0100.0

2

20.0165.0)134.020.0(*5.2

min*20.0

2

)(*5.2min*,

h

d hb A eff c

According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 7.3.4(2); Figure: 7.1

²040.0, m A eff c this value was multiplied by two because the section is fully stretched

031.0040.0

10*57.12 4

,

,

eff c

seff p

A

A





666

mmnn

nneq 20

**

**

2211

2

22

2

11


eff pr

k k

ck s,

21

3max,

***425.0

*

Concrete cover of 5cm and HA6 reinforcement, therefore c=5+0.6=5.6cm

801 .k HA bars

12 k pure tension sizing

99156

2543

2543

3232

3 .*.c

*.k

//

mmk k

ck s

eff p

r 331

031.0

20*1*8.0*425.056*99.1

***425.0*

,

213max,

Calculation of average deformations

35.631476

200000

cm

se

E

E


Steel stresses in SLSqp:

Mpa A

N

ser s

qp ser

s 2.143001257.0

180.0

,

44 10304σ

6010175

200000

031035610310

56240143ρα1

ρσ

εε

*.E

*,*.

).*.(*.

.*.

E

)*(*f

*k

s

s

s

eff ,peeff ,p

eff ,ctts

cmsm



mm sw cm smr k 171.0)10*17.5(*331)(* 4

max,


For an exposure class XD2, applying the French national annex, it retained a maximum opening of 0.20mm crack.

This criterion is satisfied.




667


In case one control for cracking is required:

yk

ctmc s f

f A A .

Because h<0.3m

Mpa f f ck ctm 56,225*30.0*30.0 3/23/2

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2);Table 3.1

²04.020.0*20.0 m Ac

²05.2500

56.2*04.0* cm

f

f A A

yk

ctmc s



■ 6 nodes,





668

Longitudinal reinforcement for ULS and SLS load combinations (kNm)

ULS (reference value: Az=9.36 cm2=2*4.68 cm

2)

SLS (reference value: Az=7.12 cm2=2*3.56 cm

2)

Stresses in the steel reinforcement

(reference value: s= 226.80Mpa)

Maximum spacing of cracks S r, max (m)

(reference value: Sr,max= 0.331m)

Crack width w k (mm)

(reference value: wk= 0.171mm)




669



Az,ULS Theoretical reinforcement area for the ULS load combination [cm2] 4.68 cm

2

s Stresses in steel reinforcement [MPa] 226.77 MPa

Sr, max Maximum spacing of cracks [m] 33.1 cm

wk Crack width [mm] 0.017 cm



Az Az,USL -4.68337 cm² -0.0001 %

Ss CQ SigmaS -226.766 MPa 0.0016 %

Sr,max Sr,max 32.7639 cm -0.7155 %

wk wk -0.0170451 cm -0.2650 %




670

5.56 EC2 Test 4 II: Verifying a rectangular concrete beam subjected to Pivot B efforts – Inclinedstress-strain diagram

Test ID: 5893

Test status: Passed

5.56.1 Description

Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During thistest, the determination of stresses is made along with the determination of the longitudinal reinforcement.

The purpose of this test is to verify the software results for Pivot B efforts. For these tests, the constitutive law forreinforcement steel, on the inclined stress-strain diagram is applied.

The objective is to verify the longitudinal reinforcement corresponding to Class B reinforcement steel ductility.

5.56.2 Background

This test performs the verification of the value (hence the position of the neutral axis) to determine the Pivot efforts

(A or B) to be considered for the calculations.The distinction between the Pivot A and Pivot B efforts is from the following diagram:

d d d

x

cuud

cu

ucuud

cuu ..x 2

2

2

2

The limit for depends of the ductility class:

■ For a Class A steel: 13460α5022ε .. uud

■ For a Class B steel: 0720α45ε .uud

■ For a Class C steel: 0490α5067ε .. uud

The purpose of this test is to verify the software results for Pivot A efforts. For these tests, it will be used theconstitutive law for reinforcement steel, on the inclined stress-strain diagram.

MPa AS su su 454.38,95271,432500

MPa BS su su 466.27,72771,432500

MPaC S su su 493.52,89571,432500

The Pivot efforts types are described below:

■ Pivot A: Simple traction and simple bending or combined

■ Pivot B: Simple or combined bending

■ Pivot C: Combined bending with compression and simple compression




671






■ Loadings from the structure: G = 25 kN/m+ dead load,




■ There will be considered a Class B reinforcement steel




■ The longitudinal reinforcement corresponding to Class B reinforcement steel ductility





672


Units

Metric System

Geometry


■ Height: h = 0.90 m,

■ Width: b = 0.50 m,

■ Length: L = 5.80 m,



■ Effective height: d = h - (0.6 * h + ebz)=0.806 m; d’ = ebz = 0.040 m





■ There will be considered a Class B reinforcement steel ductility




f f

c

ckcd 6716

51

25

γ


■ MPa.*.f *.f //ckctm 56225300300 3232



f f

s

ykyd 78434

151

500

γ


Boundary conditions


■ Outer:



■ Inner: None.




673

Loading


Dead load:

G’=0.9*0.5*2.5=11.25 kN/ml

Load combinations:

■ The ultimate limit state (ULS) combination is:

Cmax = 1.35 x G + 1.5 x Q=1.35*(25+11.25)+1.5*50=123.94 kN/ml


CCQ = 1.0 x G + 1.0 x Q=25+11.25+50=86.25kN/ml


kNm M Ed 16.5218

²80.5*94.123

kNm M Ecq 68.3628

²80.5*25.86


For S500B reinforcement steel, we have 372.0lu (since we consider no limit on the compression concrete to

SLS).

Reference solution for reinforcement from Class B steel ductility

If the Class B reinforcement steel is chosen, the calculations must be done considering the Pivot B efforts.


■ Effective height: d=h-(0.6*h+ebz)=0.806 m


096,067,16*²806.0*50,0

10*52116.0*²*

2

3

MPamm Nm

f d b M

cd w

Ed cu

■ The α value:

A design in simple bending is performed and it will be considered a design stress of concrete equal to cd f .

md z cuc 765.0)127.0*4.01(*806.0)*4,01(*

■ Tensioned reinforcement steel elongation :

1.245.3*

127,0

127,01*

12cu

u

u su

■ Reinforcement steel stresses :

MPa su su 466*27,72771,432

MPa Mpa su 46625.4500241,0*27,72771,432

²13.1525.450*765.0

52116.0

*cm

f z

M A

yd c

Ed u



■ 7 nodes,





674









My,ULS My corresponding to 101 combination (ULS) [kNm] 521.16 kNm

My,SLS My corresponding to 102 combination (SLS) [kNm] 362.68 kNm

Az Theoretical reinforcement area (Class B) [cm2] 15.13 cm

2



My My corresponding to 101 combination (ULS) -521.125 kN*m 0.0000 %

My My corresponding to 102 combination (SLS) -362.657 kN*m -0.0001 %

Az Theoretical reinforcement area (Class B) -15.1105 cm² -0.0002 %




675

5.57 Testing the punching verification and punching reinforcement results on loaded analysismodel (TTAD #14332)

Test ID: 6189

Test status: Passed

5.57.1 Description

Tests the punching verification and punching reinforcement results on loaded analysis mode and generates thecorresponding report.

5.58 Verifying the peak smoothing influence over mesh, the punching verification and punchingreinforcement results when Z down axis is selected. (TTAD #14963)

Test ID: 6200

Test status: Passed

5.58.1 Description

Verifies the peak smoothing influence over mesh, the punching verification and punching reinforcement results whenZ down axis is selected.

The model consists in a c25/30 concrete planar element and four concrete columns (with R20/30, IPE400, D40,L60*20 cross-sections) and is subjected to self weight and 1 live load of -100 KN.

5.59 EC2: column design with “Nominal Stiffness method” square section (TTAD #11625)

Test ID: 3001

Test status: Passed

5.59.1 Description

Verifies and generates the corresponding report for the longitudinal reinforcement bars of a column. The column isdesigned with "Nominal stiffness method", with a square cross section (C40).

5.60 Verifying the longitudinal reinforcement for a horizontal concrete bar with rectangular crosssection

Test ID: 4179

Test status: Passed

5.60.1 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: "Longitudinalreinforcement linear elements".

The model consists of a concrete linear element with rectangular cross section (R18*60) with rigid hinge supports atboth ends and two linear vertical loads: -15.40 kN and -9.00 kN.




676

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

Test ID: 3639

Test status: Passed

5.61.1 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 "Transversereinforcement 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.

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

Test ID: 3638

Test status: Passed

5.62.1 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 "Transversereinforcement linear elements" report.

The beam has a rectangular cross section (R20*50), B25 material and two hinge rigid supports at both ends.

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

Test ID: 3509

Test status: Passed

5.63.1 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 longitudinalreinforcement report.




677

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

Test ID: 3528

Test status: Passed

5.64.1 Description

Performs the finite elements calculation and the reinforced concrete calculation of a model with a horizontal concretebeam.

The beam has a R20*50 cross section and two hinge rigid supports.

Verifies Aty and Atz for the fixed concrete beam.

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

Test ID: 3475

Test status: Passed

5.65.1 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) andaccidental loads. A number of 375 combinations are obtained.

Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforcement areasplanar elements report.

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

Test ID: 3545

Test status: Passed

5.66.1 Description

Verifies the longitudinal reinforcement for a vertical concrete bar.

Performs the finite elements calculation and the reinforced concrete calculation. Verifies the longitudinalreinforcement 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 translationrestraints on X and Y. A vertical punctual load of -1260.00 kN is applied.

5.67 Verifying the longitudinal reinforcement bars for a filled circular column (TTAD #11678)Test ID: 3543

Test status: Passed

5.67.1 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.




678

5.68 Verifying concrete results for planar elements (TTAD #11583)

Test ID: 3548

Test status: Passed

5.68.1 Description

Verifies the reinforcement results on planar elements.

Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforced concreteanalysis report: data and results.

The model consists of two planar elements (C20/25 material) with rigid fixed linear supports. On each element, apunctual load of 50.00 kN on FX is applied.

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

Test ID: 3542

Test status: Passed

5.69.1 Description

Verifies the concrete results on a fixed horizontal beam.

Performs the finite elements calculation and the reinforced concrete calculation. Generates the reinforced concretecalculation 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 supportwith translation restraints on Y and Z at the other end. A linear dead load (-28.75 kN) and a live load (-50.00 kN) areapplied.

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

Test ID: 3547

Test status: Passed

5.70.1 Description

Verifies the longitudinal reinforcement for a horizontal concrete bar.

Performs the finite elements calculation and the reinforced concrete calculation. Verifies the longitudinalreinforcement 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 withtranslation restraints on Y and Z. Four loads are applied: a linear dead load of -50.00 kN on FZ, a punctual dead loadof -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.

5.71 Verifying concrete results for linear elements (TTAD #11556)

Test ID: 3549

Test status: Passed

5.71.1 Description

Verifies the reinforcement results for a horizontal concrete bar.

Performs the finite elements calculation and the reinforced concrete calculation. Verifies the reinforcement andgenerates 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.




679

5.72 Verifying the reinforcement of concrete columns (TTAD #11635)

Test ID: 3564

Test status: Passed

5.72.1 Description

Verifies the reinforcement of a concrete column.



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

Test ID: 5887

Test status: Passed

5.73.1 Description

Verifies a rectangular cross section beam made of concrete C25/30 subjected to a tension distributed load - Bilinearstress-strain diagram (Class XD2).

During this test, the determination of stresses is made along with the determination of the longitudinal reinforcementarea and the verification of the minimum reinforcement area.

5.73.2 Background

Simple Bending Design for Serviceability Limit State

Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple tension. During this

test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and theverification of the minimum reinforcement percentage.










Units

Metric System

Geometry


■ Height: h = 0.20 m,

■ Width: b = 0.20 m,

■ Length: L = 5.00 m,






681


Rectangular solid concrete C25/30 and S500, class A reinforcement steel is used. The following characteristics areused in relation to this material:




■ Humidity RH=50%

Boundary conditions


■ Outer:

► Support at start point (x=0) , fixed connection,

► Support at end point (x = 5.80) restrained in translation along Z.

■ Inner: None.

Loading


Load combinations:


KNQGN cq,ser 285150135

■ Quasi-permanent combination of actions:

KN*.Q*.GN qp,ser 1801503013530

5.73.2.2 Reference results in calculating the longitudinal and the minimum reinforcement

Calculating the longitudinal reinforcement for serviceability limit state:

S

ser ser s

N A

,

Mpa f yk s 400500*8,0*8,0

²13.7²10*13.7400

285.0 4

, cmm A ser s

Calculating the minimum reinforcement:

²04.020.0*20.0 m Ac

²05.2500

56.2*04.0* cm

f

f A A

yk

ctmc s



■ 6 nodes,





Longitudinal reinforcement for SLS load combinations

SLS (reference value: Az=7.12 cm2=2*3.56 cm

2)

Minimum longitudinal reinforcement

(reference value: Amin= 2.05 cm2)



Az,SLS Theoretical reinforcement area for the SLS load combination [cm2] 3.56 cm

2

Amin Minimum longitudinal reinforcement [cm2] 2.05 cm

2



Az Theoretical reinforcement area for the SLS load combination -3.5625 cm² 0.0000 %

Amin Minimum longitudinal reinforcement 2.05197 cm² 0.0001 %

Documents

AD Validation Guide Vol1 2015 En